Market Finance
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ESCP EUROPE Business School
Finance
University Year: 2011-2012
Introduction:
- Target Audience
- Objectives of the Course
- Prerequisite
Recommended Textbooks (in French):
- Piermay M. et A. Lazimi, (1989), Mathématiques
FINANCIAL MATHS REVIEW COURSE
Financières, 2ème édition, Economica, 258 pages.
- Schlacther
D.,
(2007),
Comprendre
les
Mathématiques Financières, Hachette, 158
Emmanuel F. Jurczenko
Pages.
- Hull J., (2007), Options, futures et autres actifs
Slides
dérivés, 6ième édition, Pearson Edition, 815 pages.
(Preliminary version - do not diffuse)
Recommended Textbooks (English):
- Jorion P., (2007), Financial Risk Manager
Handbook, 4th edition, Wiley Finance, 736 pages. Associate Professor of Finance at ESCP EUROPE and Associate Researcher at HEC-Lausanne and at the University of Paris-1 (CES/CNRS).
E-mails: ejurczenko@escpeurope.eu with cc to emmanuel.jurczenko@gmail.com
- Hull J., (2008), Options, Futures, and Other
Derivatives, 7th edition, Prentice Hall, 816 pages.
- Neftci S., (2009), Principles of Financial
Engineering, Academic Press, 2nd Edition, 696 pages. ©
Jurczenko - Copyright 11. Preliminary Version (September 2011) - Usual disclaimers apply.
Do not quote or diffuse without permission.
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Three parts:
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1. Basic Tools in Financial Mathematics
- Basic Tools in Financial Mathematics:
• Compounding and future value (discrete and continuous compounded interest rates/rate of
payments and receipts, which are spread over time.
However, one Euro obtained (invested) today is not
returns)
• Discounting and present value (bond, stock and derivatives pricing)
• Applications (bond pricing, stock pricing and derivatives) • Function of one variable, derivatives and Taylor series expansion (dollar duration)
• Function of more than one variable, partial derivatives and total differentials (“The Greeks”)
valued):
• Preference for present
• Inflation and purchase power erosion
Two identical sums of money perceived at different moments are not equal (Time Value of Money)
- Compounding and discounting permit us to make
• Integral and integration by parts
comparison between cash-flows perceived or paid
- Tools in probability variables, equivalent to one Euro obtained (paid) tomorrow (less
• Credit risk
- Standard calculus
• Random
- Financial decisions are based on a stream of
at different times by fixing conventionally one density functions
and
statistical moments (portfolio diversification)
valuation period - the future or the present – to obtain equivalent sums at a common date:
• Binomial, Normal and Lognormal Distributions
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• Compounding gives the future (terminal) value
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• Discounting gives the present value V0 that
VT that can be obtained from an initial
must be invested (or borrowed) today to
investment (loan) V0 :
obtain (repay) a given future value VT :
− V0
{
+ VT ?
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2
Present
Future
− V0 ?
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2
+ VT
{
Future
Present
(Lending)
(Lending)
+ V0
{
− VT ?
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2
Present
Future
+ V0 ?
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2
− VT
{
Future
Present
(Borrowing)
(Borrowing)
where 0 and T are respectively the current date and the maturity date of the financial operation; + and - represent respectively a cash
where 0 and T are respectively the current date and the maturity
inflow (receipt) and a cash outflow (payment).
date of the financial operation;+ and - represent respectively a cash inflow (receipt) and a cash outflow (payment).
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1.1. Discrete and continuous compounding
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Substituting leads to:
V2 = V0× (1 + i ) × (1 + i )
= V0× (1 + i )
2
- Compounding a single cash-flow:
We speak of compounding since interests obtained periodically
• Principle: Compounding gives the future value
are compounded, that is:
of a cash-flow (or of a stream of cash flows)
V2 = V0 + 2iV0 +
which has been invested for many periods
(years)
without
any
intermediary
Compoundin g
At the end of T years (maturity), the future value is:
VT = V0× (1 + i )T
supplementary contribution or withdrawal.
• Proposition: The future value VT of an initial
i 2V0
{
Remark:
amount V0 compounded annually for T years
If the maturity of the operation is less than one year, the future
at a constant yearly interest rate i (p.a.) is:
value will be given by:
VT = V0 × (1 + i )T
days
VT = V0× 1 + i ×
360
• Proof : The future value of the initial
where i corresponds to the simple annual interest rate (prorata temporis). investment V0 at the end of the first year is:
See price of depos, Eurocurrency deposits,…
V1 = V0 + i × V0 = V0 × (1 + i )
At the end of the second year, the principal (initial amount invested plus interest of the first period) is also compounded, that is:
Example: Consider an investment of 1000 EUR in a bank deposit for six years compounded annually at an interest rate of 10%
p.a.. What is the future value of your investment?
V2 = V1 × (1 + i )
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V6 = 1000× (1 + 0.1)
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
• Using the future value formula with discrete
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= 1771.56 EUR
compounding:
♦ Compute the sum of the interest receipts after T years of
Table 1. Future Value and compounded interest
investment:
[
]
Year
Initial Value
Interest
Future Value
1
1000
100
1100
2
1100
110
1210
♦ Compute the annual discretely compounded interest rate
3
1210
121
1331
knowing the initial investment, the terminal value and the
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1331
133.1
1464.1
5
1464
146.41
1610.51
6
1610.51
161.051
1771.561
IT = V0 × (1 + i ) − 1
T
investment horizon:
VT = V0 × (1 + i )
T
⇒ (1 + i )T =
Example: The Dutch Peter Minuit have bought in 1626
1/ T
V
⇔i= T
V
0
Manhattan island to the Manhattes Indians for 24 dollars worth trinkets. If Indians have required instead a cash payment and invested it at a discrete compounded interest rate of 6% per annum (see B. Malkiel, A Random Walk Down Wall Street, p. 116,
V385 = 24 × (1.06)
385
today 15000 EUR. What is the annual rate of return on this
1 / 10
= 132.73 Billions USD
−1
Example: A bank offers you 30000 EUR in 10 years if you invest
30000 investment? i =
15000
Norton Company), they will have obtained today:
VT
V0
− 1 = 0.0718 = 7.18%
Example : Your father is asking you if it
More than 132 billions of dollars!! !
is better to buy
government bonds priced at 0,86 EUR (zero-coupon OAT) that will be redeemed at 1 EUR in 2 years or to invest in a money banking account at 5%/year? The yearly return on the bond investment is equal to:
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♦ Investment maturity determination, knowing the initial
1 1 / 2
− 1 = 0.0783 = 7.83%!! i =
0.86
amount of money invested, the terminal value of the investment
Example : From J. Siegel (Stocks for the Long run, Mc Graw-Hill,
and the interest rate:
VT = V0 × (1 + i )
1998), 1 dollar invested in 1802 on the US stock market have
T
yield in 1997 (by including reinvested dividends) 7,47 millions of
⇒ (1 + i )T =
dollars. The same investment on the US bond has yield in 1997
(by including reinvested coupons) 10744 dollars, 3679 dollars on
VT
V0
V
⇔ T ln (1 + i ) = ln T
V0
the money market and 1337 dollars with gold! Compute the
(nominal) annual rate of returns on these different asset classes.
⇒T =
US Stocks:
ln(VT ) − ln (V0 ) ln (1 + i )
Example: Consider an individual who invests today 60000 EUR
7470000 1 / 195
− 1 = 0.084 = 8.4% i =
1
on a time-deposit at an annual interest rate of 6% p.a.. What must be the maturity of its investment so that the terminal value
US Bonds:
of its investment equals 80000 EUR ? Same question to double
10744 1/ 195
− 1 = 0.048 = 4.8% i =
1
the initial amount of money invested?
From the future value formula, we have:
US Money Market:
VT = V0 × (1 + i )
T
3679 1 / 195
− 1 = 0.043 = 4.3% i =
1
⇔ 80000 = 60000 (1.06 )T
Gold:
Taking the logarithm on both sides of previous expression, leads
1337 1 / 195
− 1 = 0.037 = 3.7% i =
1
to: ln(80000 ) = ln (60000 ) + T × ln(1.06)
That is:
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8 ln ln (80000 ) − ln (60000 )
6
T=
= = 4.93 ≈ 5 years ln (1.06) ln (1.06 )
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10 ln ln (10000000 ) − ln (800000 )
8
T=
= = 7.54 ≈ 8 years ln (1.03) ln (1.03)
That is five years. To double the initial investment, we find:
T = 11.896
Stock crashes discussion
That is 12 years approximatively.
Remark: A practical rule to compute the investment horizon which is necessary to double the initial amount of money invested is given by:
72 i × 100
Check:
72
= 12 ≈ 11.896
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Example: You want to buy a house which costs 1 MEUR. You possess actually 800000 EUR. If we assume that the interest rates stand at 12% p.a., how many years would you have to wait to be able to buy this house?
Solution:
10 ln ln (10000000 ) − ln(800000 )
8
T=
= = 1.96 ≈ 2 years ln(1.12) ln (1.12 )
Same question with a yearly rate of return of 3% on investments.
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Remarque: Si les intérêts composés constituent la référence pour
Exemple : Soit un placement de 10000 EUR sur un compte à un
toutes les opérations financières d’une durée supérieure ou égale
taux d’intérêt simple annuel de 5 %, à intérêts simples pendant 3
à un an, pour toutes les opérations de durée inférieure à un an
mois. Au terme des 3 mois, les intérêts perçus sont de :
fois, ne portant pas eux-mêmes intérêts et étant proportionnels à
90
V3mths = 1000 × 1 + 0.05 ×
= 1125
360
la durée totale du placement (Opérations sur le marché
Exemple : Un trésorier d’entreprise place une somme de 500000
monétaire, paiement de coupon couru sur le marché obligataire,
EUR sur un compte rapportant 6% / an, pendant 4 ans et 7
calcul
mois. De combien disposera t-il à l’issu de son placement ?
on utilise les intérêts simples, les intérêts étant payés une seule
d’escompte,
d’agios
bancaires….).
La
valeur
future/acquise sera donnée dans ce cas par la formule suivante
A l’issu de la 4ème année la somme placée est devenue :
V4 = 500000 × (1.06)4 = 631238.48 EUR
(capitalisation à intérêt simple post comptés) :
jours
VT = V0× 1 + i ×
360
Pendant les 7/12 d’année restant à courir cette somme procure des intérêts simples calculés prorata temporis.
où i correspond au taux d’intérêt annuel simple (prorata temporis
Le montant des intérêts acquis sur les 7 mois est de :
Dans le cas où la durée ne correspond pas à un nombre entier d’années, les intérêts sont alors calculés sur un nombre entier
I = 631238.48 × 0.06 ×
7
= 22093.347 EUR
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d’années selon la technique des intérêts composés et la fraction d’année restante donne naissance à un calcul d’intérêts selon la technique des intérêts simples.
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- Compounding at different frequencies (discrete and continuous compounding): While annual
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i
VT2 = V0 × 1 +
2
2×T
compounding is the norm in finance, other
• Periodic Compounding at the proportional annual
conventions exist however. Yearly interest rate can
interest rate: If the yearly interest rate i is
be
(semi-annual
compounded p times per year, the terminal
compounding, see US and GB bonds), twelve times
value VT of the initial investment V0 is for T
a year (monthly compounding), or either an infinite
years:
compounded
number
of
twice
times
a
in
a
year
year
(continuous
compounding). It is then necessary to define an equivalent basis to compare interest rates computed or compounded on different frequencies. This is done by working with an equivalent yearly return.
p
1 + i
VT = V0
p
i{ p
p ×T
where p is the number of period considered in the year.
• Annual Compounding at effective annual interest rate:
The terminal value VT of an initial amount V0 invested for T years at a fixed yearly interest rate i compounded annually (once a year) is:
VT = V0 × (1 + i )
T
• Semi-annual Compounding at the proportional annual
interest rate: If the yearly interest rate i is now compounded for T years semi-annually (twice a year), the future value VT of the initial investment V0 then becomes:
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Remarque: On appelle le taux annuel i composé p fois l’an le
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• Continuous
compounding
-
daily
compounding
« taux annuel proportionnel ». Ce dernier étant similaire à
(Derivatives, Repo,…): As the frequency of
intérêts simple il ne permet pas de connaître directement le
compounding increases ( p tends to infinity),
montant des intérêts reçu à maturité puisqu’il n’inclut pas l’effet
the yearly interest rate i becomes continuously
de capitalisation des intérêts (les intérêts sur les intérêts). Pour ce
compounded (paid every seconds) and it can
faire on le doit convertir en un taux période effectif i p définit
be shown that the initial investment V0 accrues
comme:
to: ip =
i p VTc
sur lequel on applique alors la formule de capitalisation à intérêts composés. :
VTp
i = V0 1 + p
p×T
i = V0 × eiT
= lim V0 × 1 + p p →∞
where e = 2.71828 represent the exponential function (see infra).
p×T
Remark: For most practical use (Derivatives Pricing, see Black and Scholes, 1973), continuous compounding is equivalent to daily compounding
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• Equivalent interest rates (conversion between annual
Effect of compounding frequency on the value of 1000 EUR at the end of one year when the yearly interest rate is 10%
Compounding frequency
and periodic compounded rate of returns): Two
Value of 1000 EUR at end of one year
interest
(i=10% p.a.)
rates
compounded
at
different
Annually (p=1)
1100
frequencies are equivalent if they lead to the
Semi-annually (p=2)
1102.5
same future value after one year (reference
Quarterly (p=4)
1103.8
period). That is:
Monthly (p=12)
1104.7
Weekly (p=52)
1105.1
i (1 + ia ) = 1 + = (1 + i p ) p p
Daily (p=365)
1105.16
Continuously
1105.17
p
where ia is the effective annual interest rate and i p is the
Where the future value in one year with continuous
effective periodic interest rate.
compounding is obtained as:
♦ Annual equivalent interest rate to a periodic rate (discrete
V1c = V0 × ei = 1000 × e0.1 = 1105.2 EUR
compounding): The effective annual interest rate (compounded
Remark: Don't confuse the base at which an interest rates are measured (annual basis) and the frequency at which it is paid
once a year) i e which is equivalent to the effective periodic a interest rate i p =
i compounded (compounded p times a year) p can then be defined such as: p (annual, semi-annual, continuous,…)
e ia 21
p
ip
ip
= 1 + − 1 = 1 + − 1
p p
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e where ia is the annual equivalent interest rate compounded once
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Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Example: A Treasurer can invest 20 MEUR for one year. He
a year, i is the proportional annual interest rate compounded
considers the two following different commercial propositions.
p times a year and i p is effective periodic interest rate.
A time-deposit with bank 1 at an interest rate of 6.10% p.a. compounded each quarter and a time-deposit with bank 2 at an interest rate of 6.12% compounded semi-annually. What is the
That is : e ia
most interested investment?
= (1 + im ) − 1 if im is a monthly rate;
e ia 12
Investment 1:
= (1 + iq ) − 1 if iq is quarterly rate...
4
To answer the question, we have to determine the annual equivalent interest rate, that is:
Example: What is the annual equivalent rate of a proportional annual rate of 2% compounded quarterly?
0.02 e
1 + ia = 1 +
4
e ia 4
4
i
0.061
= 1 + 3 − 1 = 1 +
− 1 = 6.241%
4
4
4
=TAUX.EFFECTIF(0,061;4)
e
⇒ ia = (1.005) − 1 = 2.015%
Investment 2:
4
Any agent will be indifferent between an investment with an annual rate of 2.015% compounded once a year and an
2
0.0612 e ia = (1 + i6 )2 − 1 = 1 +
− 1 = 6.214%
2
investment with an annual rate of 2% quarterly compounded.
=TAUX.EFFECTIF(0,0612;2)
We choose the first solution!!!
see fonction finances Excel Taux.Effectif.
The equivalent annual interest rate is higher than the
=TAUX.EFFECTIF(0,02;4)
proportional annual interest rate.
The difference increase with the compounding frequency
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♦ Periodic equivalent rate to an effective annual rate (discrete
♦ Continuous equivalent rate to an effective annual interest rate:
compounding): The periodic equivalent interest rate i e p e
The continuous annual interest rate ic which is equivalent to the
compounded p times a year to an effective annual rate
effective annual interest rate i is defined such as: e ic = ln(1 + i )
compounded once a year ia is given by: ip = e i
e
p
= (1 + ia )
These equations are used to convert interest rates
1/ p
−1
compounded
where ia is the annual interest rate compounded once a year and i e is the periodic equivalent interest rate compounded p times a p year.
at
different
frequencies
per
annum
(annually).
Example: Consider an effective annual interest quoted at 8% per annum. What is its equivalent continuously compounded yearly
That is:
[
i = p × (1 + ia ) e rate?
]
−1
1/ p
e ic = ln(1 + i ) = ln(1.08) = 0.0769
That is 7.69% p.a. .
♦ Annual equivalent rate to a continuous rate: The effective
Example:
e annual interest rate ia which is equivalent to the proportional
Suppose that a bank quotes the interest rate on 10-year maturity
annual interest rate continuously compounded (every second) ic
loans at 6% with continuous compounding. What is the equivalent annual rate?
can then be defined such as:
(1 + i ) = e i = (e
(
e a ic
) (
)
e
⇒ ia = eic − 1 = e0.06 − 1 = 0.0618
ic
e a )
That is 6.18% p.a. .
−1
Remark: For fixed present and future values, increasing the frequency of the compounding decrease the annual equivalent yearly interest rate
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- Compounding a sequence of cash-flows (variable and constant):
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♦ Proof: Let’s consider at date 0 an integer number T of annual
periods and assumes that an agent invest at each date t=1,..,T, a
• Definition: We refer to a sequence of annuities,
any sequence of regular payments at . If the
cash-flow at .
Diagram of cash-flows:
cash-flows are paid at the end of each period,
a1 (1 + i )T −1
ordinary
a2 (1 + i )T − 2
the
annuities
are
labelled
as
annuities (bank loan repayment). If the cash-flows are paid at the beginning of each
a1
a2
1
2
aT
period, they are labelled as immediate annuities (life-insurance or saving plans). I
0
• Future value of a sequence of ordinary annuities :
T
♦ Proposition : The future value in t = T of a sequence of
That is:
different monetary cash-flows, denoted by at , invested at the
VT = ∑ at (1 + i )
T
t =1
T −t
=
constant interest rate i (flat term structure of interest rates), is
(T-1 ) − th payment at interest rate i for 1 year
flows at i, that is:
T
aT −1×(1+i )1
a2 ×(1+i )T − 2
+
first payment compounded at interest rate i for (T-1) years
+L+
given by compounding each element of the sequence of cashVT = ∑ at × (1 + i )
a1×(1+i )T −1
second payment compounded at interest rate i for (T- 2 ) years
+
aT last payment
Remark: If the annuities can be free (general case), more often they are
T −t
constant or increasing at a constant rate following an arithmetical or
t =1
Remark: We are assuming here that the cash-flows at are paid at
geometrical progression.
the end of each period and compounded at the unique fixed interest rate i for the remaining periods.
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• Future value of a sequence of immediate annuities:
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
• Simplification (Compounding a sequence of identical
♦ Immediate annuities : The future value in t = T , of a sequence of
cash-flows) : see insurance contracts
different monetary cash-flows, denoted by at , invested at the constant interest rate i (flat term structure of interest rates), is
♦ Future value of a sequence of constant ordinary
given by compounding each element of the sequence of cash-
annuities:
flows at i, that is:
T
V 'T = VT × (1 + i ) = ∑ at (1 + i )
Proposition: The future value in t = T of a sequence of constant
T +1−t
annuities a (with at = a ), compounded at a constant interest rate
t =1
We are assuming here that the cash-flows at are paid at the beginning of each period and compounded at the unique fixed interest rate i for the remaining periods.
♦ Proof:
I (flat term structure of interest rates), is given by :
(1 + i )T − 1
VT = a ×
i
Proof: We are now assuming that the future payments a are the
Same line of reasoning…
same at the end of each period and compounded at the same interest rate i for the remaining number of periods.
The future value of this compounded sequence of constant cashflows VT after T periods is given by:
VT = a × (1 + i )T −1+ a × (1 + i )
T −2
+ K+ a
This expression corresponds to the sum of the T first terms of a geometric progression.
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(1.045)30 − 1
V30 = 100 ×
0.045
= 6100,71 EUR
Math Review: The sum of the n first terms of a geometric progression with first term u1 and common ratio q, denoted Sn , is given by :
See fonction finances Excel VC.=VC(0,045 ;30 ;-100 ; 0)
S n = u1 + u2 + K + un
(
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
)
1 − qn
= u1 × 1 + q + K + q n −1 = u1 ×
1− q
♦ Future value of a sequence of constant immediate annuities: If the cash-flow a is invested at the beginning of
Future value of a sequence of ordinary annuities:
VT = a (1 + i )T −1+ a (1 + i )
T −2
each period, the future value after T years is given by:
+K+ a
(1 + i )T − 1
VT' = (1 + i ) × VT = a (1 + i ) ×
i
Sum of the first T terms of a geometric progression with common ratio (1 + i) and first term a. That is:
Cf. supra.
1 − (1 + i )T
(1 + i )T − 1
VT = a ×
= a×
−i i
4243
1
• Future value of a sequence of constant
ordinary annuities, k periods after the last
annuity factor
t = T of a
Example: You can deposit each year 100 EUR in an interest
investment: The future value in
bearing bank account without withdrawals for 30 years at an
sequence of constant annuities a (with at = a ),
annual interest rate of 4.5%, starting in one year (ordinary
compounded at a constant interest rate i (flat
annuities). How much will you received at the end of your
term structure of interest rates), k-periods after the
investment?
last T-th investment is given by:
We have to compute the future value of a sequence of 30 ordinary annuities of 100 Euros compounded at an annual
(1 + i)T − 1
VTk = a (1 + i) k ×
i
interest rate of 4.5%, that is:
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Exemple : Si l’épargnant considéré précédemment cesse ses
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1.2. Discounting, Present Value and Asset valuation
versements à l’issue de la cinquième année, la valeur acquise du capital au bout de 30 ans est égale à :
V525
- Discounting of a single cash-flow:
(1,045)5 − 1
= 100 (1,045) ×
0.045
= 1644,19 EUR
• Use: Discounting determines what initial
25
amount must be invested today in order to obtain a given future value (the present value)
Cf. Excel.
or if we adopt the point of view of the
=(1+0.45)^25*VC(0,045 ;30 ;-100 ; 0)
borrower, the maximum amount of money that can be borrowed initially if we want to be
Exemple : Vous avez besoin de 50.000 EUR dans 7 ans. Vous
able to meet a given future repayment. This
prévoyez de faire 7 versements identiques chaque (fin) année, sur
approach permits to compare stream of cash
un compte qui rapporte du 11% par an capitalisé annuellement.
flows perceived at different dates.
Quel doit être le montant de chaque versement annuel ?
• Proposition : The present value V0 of a future
1) Diagramme de flux
fixed cash-flow in T years VT is the amount of
2) Réponse :
money that must be invested today at a fixed
(1.11) − 1
V7 = 50000 = a ×
0.11
7
yearly interest rate i in order to obtain VT in
T years, that is:
0.11
⇒ a = 50000 ×
= 5110.76 EUR
7
(1.11) − 1
V0 =
VT
(1 + i )
T
= VT × (1 + i )−T
• Proof : The future value of an initial investment
Cf. fonction finances Excel VC.
=VPM(0,11 ;7 ;0 ; 50000)
V0 , compounded annually at the yearly interest rate i for T years is:
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VT = V 0× (1 + i )T
The initial investment of VT / (1 + i )T
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⇒ V0 =
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
V1
1000
=
= 961.54 EUR < 1000 EUR
(1 + i ) 1.04
at the annually
compounded yearly interest rate i over T years gives:
Example: As a Treasurer, you know that your company will
VT
T
× (1 + i ) = VT
T
(1 + i )
receive 10 MEUR in 5 years. What is the maximum amount of
Which is precisely the future value VT that we want to obtain.
year borrowing interest rate is actually 4% (lending conditions)?
QED.
The maximum amount of money that the company can borrow
Remark: If the maturity of the operation is less than one year, the
V0 is the present value of 10 MEUR repaid in 5 years (principal
funds that you can be borrow from the company's bank if the 5-
present value will be given by:
V0 =
plus interests), that is:
VT days
1 + i ×
360
V1 = 10 MEUR = V0 × (1 + i )5
⇒ V0 =
where i corresponds to the simple annual interest rate (prorata
V1
(1 + i )5
=
10 MEUR
(1.04)5
≈ 8.22 MEUR
temporis).
Remark: If the yearly interest rate i is continuous (continuous compounding Vs discrete compounding, see supra), the present
See price of T-Bills, CDs,…
value V0 of a future cash-flow VT becomes:
Example: You need 1000 EUR in one year. What is the amount of money that you must invest today if interest rate for one year is 4%? The amount of money to invest V0 is the present value of
V0 =
VT e iT
= VT × e−iT
1000 EUR in one year, that is:
V1 = 1000 = V0 × (1 + i )
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• Proof :
Example: Same as before but with a 5-year borrowing interest rate continuously compounded at 4%.
Diagram of cash flows:
V0 = 10 MEUR × e − 0.04×5 ≈ 8.18 MEUR aT aT
- Discounting a stream of future (constant) cash-
(1 + i)T
• Proposition : The present value at t = 0 of a
sequence of different cash-flows at (with t = [1,..., T ]) discounted at the constant annual
interest rate I, is given by discounting each
(1 + i)t
T at = ∑ at
1 + i t t =1
(
)
a3
M a1 (1 + i)
cash-flow of the sequence at this rate, that is :
T
V0 = ∑ t =1
at
at
flows:
a2 a1 1
× (1 + i )−t
2
3
t
…
T
Since any future cash flow at has by definition a present value
where we make here the assumption that the interest applicable
equal to: at to 1, 2, 3, ..T year horizons is constant and equal to i (flat and
(1 + i )t
constant term structure of interest rates).
This present value is exactly the sum that must be invested
(obtained) today if one wants to obtain (pay) the stream of cash flows at when the constant annual interest rate is i .
The present value of the stream of cash-flows at , with t = [1,..., T ], is:
Vo =
=
37
a1 aT +L+
(1 + i )
(1 + i )T
T
∑ t =1
at
(1 + i )t
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• Simplifications:
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
1 T
1 −
1
1 + i
V0 = a ×
×
(1 + i ) 1 − 1
1 + i
♦Sequence of constant cash-flows (bullet coupon bond, see
infra): If the future cash-flows are, that is if at = a ,
∀t = [1,K, T ], the present value becomes:
1 − (1 + i )−T
= a×
1+ i −1
at t t =1 (1 + i ) a a a =
+
+K+
2
(1 + i ) (1 + i )
(1 + i )T
T
V0 = ∑
So that:
1 − (1 + i )−T
V0 = a ×
i
1
1
1
= a×
+
+K+
2
(1 + i )T
(1 + i ) (1 + i )
Example: An individual want to finance his housing by a 15-year
1
1
1
= a×
× 1 +
+K+
T −1
(1 + i ) (1 + i )
(1 + i )
loan at an annual interest rate of 4%. The loan is repaid by a sequence of constant monthly interest payments. If we assume
where the terms in the bracket correspond to the sum of the T first terms of a geometric progression of common ratio 1 (1 + i) and first term 1.
that the agent can repay 1800 Eur/month, what will be the maximum amount that he can borrowed today from the bank?
First, compute the monthly interest rate, that is:
Math review: The sum of the n first terms of a geometric
im 0.04
=
= 0.0033
12 12
progression of first term u1 and common ratio q is given by:
From the present value equality we have:
1 − qn
S n = u1 ×
1− q
V0 =
So that:
1800
1800
1800
+
+K+
180
0.04 0.04 2
0.04
1 +
1 +
1+
12
12
12
where V0 corresponds to the maximum amount of borrowing.
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From the formula of the present value of a sequence of constant
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That is in thousand of euros:
1 − (1.08)− 6
1 − (1.08)− 2
1
+ 300 ×
×
V0 = 500 + 200 ×
(1,08)6 0.08
0.08
≈ 1.762 M EUR
cash-flows a , we get:
0.04 −180
1 − 1 +
12
V0 = 1800 × 12 ×
0.04
= 243 345.87 EUR
Example: An agent must repay in four years 200000 EUR. He wants to postpone the repayment of its debt in 10 years. What
See fonction finances Excel VA.
will be the new amount of money X at he will have to repay in
10 years (6% interest rate)?
=VA(0,04/12 ;-1800)
Two sequences of cash-flows are equivalent if they possess exactly the same present value, that is:
Example : An investor purchase a plot of land by paying 500000
200000
X
=
4
(1.06) (1.06)10
EUR cash, 200000 EUR each year for the six following years and
300000 EUR year 7 and 8. The discount rate for this project is equal to 8% (opportunity cost if the investment is financed by the investor’s capital or borrowing cost if the investment is
So That :
X = 200000 × (1.06 ) = 283703.82 EUR
6
financed by borrowing). What is the present value of the land?
Using the present value formula, we have:
8
6
1
1
V0 = 500000 + 200000 × ∑
+ 300000 × ∑ t t
t = 7 (1.08)
t =1 (1.08)
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♦Infinite sequence of cash flows:
Proof: Si g est le taux de croissance constant à l'infini des
a) Discounting an infinite sequence of constant cash-flows:
versements et a est le premier versement effectué à la fin de la
If the sequence of constant cash flows is infinite, the present
première année, la valeur actuelle est alors donnée par : a a × (1 + g ) a × (1 + g )2 a × (1 + g )T −1
V0 =
+K
+K+
+
+
(1 + i ) (1 + i )2
(1 + i )3
(1 + i )T
value can be simplified as:
1 − (1 + i )−T a
V0 = lim a ×
= i T →∞
i
b) Discounting an infinite sequence of cash-flows progressing at a constant growth rate: If the sequence of cash flows is infinite, with each individual cash-flow progressing at a constant growth rate g (perpetuity), the present value can be
(1 + g ) (1 + g )2
(1 + i )T −1 + K a +
=
× 1+
+K+
2
(1 + i ) 144i44(1 + i42444+ i4−44
(1 4 )T 1 3
(1 + ) 44) somme infinie de termes d' une suite
géométrique
Somme infinie de termes issus d'une suite géométrique de premier terme 1 et de raison (1 + g ) / (1 + i ), i.e. :
1 + g T
1 − a 1+ i
× lim
V0 =
(1 + i ) T →∞ 1 − 1 + g
1+ i
simplified as:
V0 =
a
(i − g )
With i the discount rate and g the growth rate.
si i > g , la série versements est convergente et on obtient finalement :
V0 =
a
(i − g )
Cf. Gordon-Shapiro.
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Exemple : Vous souhaitez déterminer la valeur d’un bien
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
- Financial pricing applications (Bonds, Stocks and
Linear Derivatives):
immobilier locatif d’une surface de 100m2. On suppose que le loyer annuel courant (charges déduites) est de 20.000 EUR/an.
• Bond pricing: A bond is a long-term debt
Le taux d’actualisation est de 6% et le taux de croissance des
instrument where the issuer (borrower) has the
loyers est de 2% par an. A quel bien accepteriez-vous d’acheter
contractual
ce bien ?
bondholder (lender) regular fixed interest
obligation
to
make
to
the
payments (coupon payments) and to repay the
La valeur de ce bien correspond à la valeur actuelle de ses cash-
bond’s principal (face value) at the maturity
flows futurs, soit en supposant une durée de vie infinie :
date. The price of a bond must always (on the
V0 =
20000
= 500 000
(6% − 2% )
primary or the secondary market) be equal to the discounted present value of its stream of
Soit 5000 EUR le m2 !!!
cash flows (i.e. coupon payments and principal repayment). But what kind of discount
rate(s) do we have to use?
♦ The economic valuation and the term structure of spot rates
(exact valuation): Any bond can be decomposed as a portfolio of unique cash-flows, so that the equilibrium value of a bond must be equal to the sum of the value of each cash-flow discounted by its appropriate spot interest rate (zero coupon bond annual interest rate), that is:
P0 =
C
C
C+F
+
+K+
2
1+ 0 i1 (1+ 0 i2 )
(1+ 0 iT )T
with:
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C = c× F
Remark: It is also possible to use the term structure of the
where c is the coupon rate, C is the coupon payment and 0 it
implied forward rates to value the stream of coupons (interest
the t-period annual spot interest rate for a maturity of
rates implied by today’s spot interest rates, that are applicable in
t = [1,....,T ] years (rate of interest that is realized on an
one point of time in the future to another point of time in the
investment that starts today and lasts for t-periods with no
future), so that:
intermediate payments.
P0 =
C
C
C+F
+
+K+
1+ 0 i1 (1+ 0 i1 )(1+ 0 f1,1 )
(1+ 0 i1 )(1+ 0 f1,1 )K 1+ 0 fT −1,1
(
)
Diagram of cash flows for a typical coupon bond (plain vanilla or straight
where 0 f s , n − s is the actual annual forward interest rate that starts
bond):
at a future date s and ends at n with a maturity of (n − s ).
F
+C
+ C ……………………
+C
− P0
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♦ The financial valuation and the yield to maturity
The yield to maturity can be interpreted as the average
(approximate): In practice since the determination of the array of
annual return obtained on a bond bought at the initial
all spot /zero coupon bond interest rates is a very difficult and
market price P0 and held until maturity (realized yield
long task, traders prefer to use only one constant interest rate –
assuming that all intermediate coupon payments will be
i.e. the yield to maturity of the most newly issued bonds on the
reinvested at that yield, that is a flat and constant term
primary market ("on the run" issues) - to evaluate the price of
structure of spot interest rates).
old issued bonds which the same characteristics traded on the secondary market ("off the run" issues).
The YTM acts in fact as a “proxy” for the whole term
1) Definition yield to maturity (YTM): The yield to maturity on a
structure of the discount rates (complex average of the spot
bond is the single/constant discount rate, denoted y , which
discount rates).
makes the present value of the bond’s payments equal to its market price (internal rate of return or IRR of the series of
One should be emphasized that an YTM is not a very
the bond’s cash-flows), that is:
meaningful number. This is because there is no reason one should discount cash-flows occurring at different dates with a unique discount rate. The YTM is just a convenient
Ct
N
+ t t =1 (1 + y )
(1 + y )T
T
y / P0 = P0 ( y ) = ∑
way for practitioners to express bond prices (much like the implied volatility for options).
T
Ct
N
+
=0
⇔ NPV0 ( y ) = P0 − ∑ i i = t (1 + y )
(144T
+ y)
144 24 3
4
P0 ( y )
with y the yield to maturity of the bond
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2) The most newly issued bond yield to maturity and secondary bond pricing
(relative bond pricing): The price of a bond on the secondary market can then be obtained by taking the yield-to-maturity on
Primary / benchmark market:
the most newly/active issued bond (government or corporate
P0new
benchmark) which possesses the same characteristics (maturity,
yInew
Secondary Market:
coupon rate, default risk, tax rate,…), that is:
Ct
N
+
t t =1 (1 + y )
(1 + y )T
T
P0 = P0 ( y ) = ∑
P0old
yInew
where T represents this time the residual life of the old-issue
See:
bond and y is the yield to maturity of the new issued bond (the
http://www.aft.gouv.fr/aft_fr_23/dette_etat_24/adjudicati
market interest rate).
ons_98/dernieres_adjudications_101/index.html
Remark: Do we use the discount rate(s) / yield to maturity to
http://www.aft.gouv.fr/article_308.html?id_article=308&id
obtain bond prices or the bond prices to infer the implied
_rubrique=100
discount rate(s) / yield to maturity. The answer of this “chickenand-egg” question depends on the situation. Roughly speaking,
http://www.treasurydirect.gov/RT/RTGateway?page=ins
one would use the competitive (auctioned) market prices on the
titHome
benchmark securities as given and infer the implied discount rate(s)/yield to maturity; and then use that market information
(the government yield or swap rate curve, see below) to price any other fixed-income security .
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Example: Consider a 30-year coupon bond with a principal value
Example: Consider a 10-year residual maturity, 4% coupon bond
of 2000 EUR, 6% annual coupon rate and a residual maturity of
with a par value of 2000 EUR. Suppose that the required yield
10 years. Assume that the interest rate on the most newly issued
on the most newly issued bond for a maturity of 10 years and a
bond for a maturity of 10 years and a similar class of risk stands
similar class of risk stands at 3%.
actually at 3%. At what price are you ready to buy (sell) the first
a) At what price are you ready to buy (sell) the first bond on the
bond on the secondary market?
secondary market?
First, we have to compute the coupon payments on the first
b) What will be the impact of semi-annual payment instead of
bond, that is:
annual one?
C = c × F = 2000 × 0.06 = 120 EUR
a) The annual or periodic coupon payment is:
C = 2000 × 0.04 = 80 EUR
Interest rate 3%:
The price of the bond is equal to the sum of the present values
C
F
+ t (1 + y ) (1 + y )T
120
120
120
2000
=
+
+K
+
2
10
(1,03) (1.03)
(1.03) (1.03)10
P0 =
T
∑
t =1
of its cash flows, that is:
C
F
+
t
(1 + y ) (1 + y )T
80
80
80
2000
=
+
+K
+
2
10
(1,03) (1.03)
(1.03) (1.03)10
P0 =
1 − (1.03)
2000
= 120 ×
+
0.03 (1.03)10
2000
= 120 × 8.53 +
(1.03)10
= 2511.81 EUR
−10
T
∑
t =1
1 − (1.03)−10
2000
= 80 ×
+
0.03 (1.03)10
2000
= 80 × 8.53 +
(1.03)10
= 2170.6 EUR
see fonction finances Excel VA.
=VA(0,03;10 ;-120 ;-2000)
See fonction finances Excel VA.
=VA(0,03;10 ;-80 ;-2000)
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♦ The inverse relationship between the market bond prices
b) The semi-annual or periodic coupon payment is:
C = 2000 ×
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0.04
= 40 EUR
2
(quoted on the secondary market) and market yields (on new or most newly issued bonds on the primary market): A fundamental
The price of the bond is equal to the sum of the present values of its cash flows that is:
property that characterizes the bond market, is that there exists an inverse relation between the quoted market price of any
1 − (1.015)
2000
P0 = 40 ×
+
20
0.015
(1.015)
2000
= 40 × 17.57 +
(1.015)20
= 2171.69 EUR
− 20
traded bond (secondary market) and the required market yield on the most newly issued bonds (primary market). That is the bond market prices change in the opposite direction from the change in the required yields:
See fonction finances Excel VA.
1) Present value explanation: The price of a bond is the present
=VA(0,03;10 ;-80 ;-2000)
value of its cash flows. As the required market yield increases,
We will retain y as a synthetic indicator of the interest rates level the present value of the future cash flows decreases, hence the quoted market price decreases. The opposite is true when the required yield increases. If we name CFt , the cash flow of a bond (coupons and redemption price), the price of a bond traded on the secondary market becomes:
T
CFt
−t
= ∑ CFt × (1 + y ) t t =1 t =1 (1 + y )
T
P0 = ∑
y ↑ P0 ↓
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2) Economic explanation: When at a given point in time the interest
resulting lack of demand and the excess offer would cause the
rate on the most newly issued comparable bond(s) is increasing
price of the old issued bonds to fall and thus their yield to
(decreasing), the price on the old issued bond must decrease
maturity to rise.
(increase) to the point when the old issued bond gives to its bondholder the same yield to maturity than the most newly
When the required yield on the most newly issued
issued one(s). Indeed, as the required yield on the newly issued
comparable bond(s) is below the coupon rate of an old issued
bond(s) changes in the market place, the only variable that can
bond, the price of the old bond must sell above its face value.
change to compensate an investor holding old bonds is their
This is because if investors have the opportunity to purchase the
market prices (the coupon rate is fixed by definition).
old issued bond(s) at par value, they would get a coupon rate in excess of what the market actually required. As the result,
When the coupon rate on an old issued bond is equal to the
investors will bid up the prices of the old issued bonds because
required market yield on the most newly issued comparable
their yields are so attractive, up to the point where the old bonds
bond(s) (same maturity, risk,..), the price of the old issued bond
offer the market required rate.
will be equal to its face value.
⇒ old bond price on the run required yield ⇒ old bond price >face value
Coupon rate old bonds < on the run required yield
When the required yield on the most newly issued comparable bond(s) (same maturity, risk,…) rises above the coupon rate of an old issued bond, the price of the old bond must adjusts so that the investor contemplating its purchase can realize some capital appreciation by holding the bond until maturity in order to compensate for a coupon rate which is lower than required. If not, nobody will want to buy the old issued bond(s) since they offer a below market yield. The
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Example: Take the previous example, but assume now that the
The 1% interest rate increase (100 basis points or bps) is
market interest rate is increasing from 4% to 5%, that is:
associated with a price decrease of:
Po =
∆P = 1845.57 − 2000 = −154.43 EUR
80
80
2080
+
+K+
2
(1,05) (1.05)
(1.05)10
1 − (1.05)−10
2080
= 80 ×
+
10
0.05 (1.05)
2080
= 80 × (7.72) +
(1.05)10
= 1845.57 EUR
Price
(EUR)
Bond price Vs interest rate
see fonction finances Excel VA.
=VA(0,05;10 ;-80 ;-2000)
Yield(%)
The value of a bond is a non-linear (convex) decreasing function of the yield to maturity.
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Remark: There exist different approaches to get y in the valuation
Example: Consider a 4-year residual maturity, 10% coupon bond
formula, such as:
with a par value of 1000 EUR, selling on the market at 968.98
1. Trial and error (iterative method): Try different interest rates
EUR. What is the YTM of this bond?
until the present value of the cash flows equals its market
To compute the yield to maturity, different discount rates must
price.
be tried until the present value of the cash flows is equal to
2. Linear interpolation (Galbrun): Let y = f ( x) and two
903.10 (90.31%).
points (x1 , f (x1 )) and (x2 , f (x2 )) which surround the value
Trying an annual interest rate of 10% gives:
y = f ( x) , with x1 < x2 . Assuming the equality of the slopes between x1 and x2 and y and x1 we get a linear approximation of y :
P(10% ) =
100
100
100
100
1000
+
+
+
+
2
3
4
1.10 (1.10 ) (1.10 ) (1.10 ) (1.10 )4
1 − (1.10 )− 4 1000
= 100
+
4
0.10 (1.10 )
= 1000 > 968.98 = P0
y − f (x1 ) f (x2 ) − f (x1 ) y − f (x1 )
⇒ x = x1 + (x2 − x1 ) ×
=
x2 − x1 x − x1
f (x2 ) − f (x1 )
Because the present value computed using a 10% discount rate is
Which in the case of the determination of the yield to maturity is
higher than the market price, a higher interest rate must be
equivalent to:
considered (inverse relation between price and interest rates).
0 − NPV ( y1) y = y1 + ( y2 − y1) ×
NPV ( y2 ) − NPV ( y1)
Trying an annual interest rate of 12%, we obtained:
1 − (1.12 )− 4 1000
P(12% ) = 100 ×
+
0.12 (1.12 )4
= 939.25 < 968.98 = P0
3. Newton-Raphson algorithm (iterative method): yn +1 = yn −
NPV ( yn )
NPV ' ( yn )
Since at 12% the present value of the cash flow is less than the
Two trials are generally sufficient.
market quoted price, a lower interest is needed.
Trying an annual interest rate of 11%, we obtained:
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1 − (1.11)− 4 1000
P(11% ) = 100
+
0.11 (1.11)4
= 968.98 = P0
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For y1 = 6%, we get:
NPV (6% ) = f (0.06 ) = 3.99
For y2 = 7%, we have:
NPV (7% ) = f (0.07 ) = −4.05
Therefore the YTM on this bond is 11%
see fonction finances Excel TRI
Using linear interpolation we obtain:
=TRI(B5:B9)
y ≈ 0.06 + (0.07 − 0.06 )
Example: Consider a 12-year maturity, 6 % coupon bond with a par value of 1000 EUR, issued at discount price 970 EUR and
0 − 3.99
= 6.49%
− 4.05 − 3.99
Iterating the process, we obtain for YTM:
6.48%
repaid at premium at 1020 EUR. What is the YTM of an investor who purchases this bond and holds it until maturity?
The yield received by the investor is higher than the coupon
Since the definition of the YTM, we are searching y such as:
yield since the bond has been issued at discount and redeemed at
970 =
premium.
60
60
1020
+
+K+
2
(1 + y ) (1 + y )
(1 + y )12
1 − (1 + y )−12
1020
− 970 + 60
=0
+
12
y
(1 + y )
1 − (1 + y )
− 97 + 6 y
−12
see fonction finances Excel TRI
=TRI(B12:B24)
102
=0
+
(1 + y )12
Let's write:
1 − (1 + y ) f ( y ) = −97 + 6 y
−12
102
+
12
(1 + y )
We are searching the value of y such as:
f (y) = 0
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• Stock pricing (Gordon Shapiro, 1956): A stock
Démonstration : Le point de départ de cette approche porte sur la
gives to its holder a stream of dividend
définition du taux de rentabilité attendu d'un investissement en
payments (distributed profits). Its market price
action au cours d'une période :
is also equal to the discounted value of its stream of future cash flows (i.e. dividend
La rentabilité espérée d'un placement en action sur une période à
payments), that is:
deux composantes : Le dividende et une plus ou moins-value en capital, i.e. :
[1 + E (R )] = E0 (D1 ) + E0 (P1 )
E0 (Dt ) E0 ( PT ) ∞ E0 ( Dt )
+
=∑ t t =1(1 + k )
(1 + k )T t =1(1 + k )t
T
P0 = ∑
P0
Les agents n'accepteront d'investir dans une action que si le taux
with:
lim
t →∞
Pt
(1 + k )t
de rentabilité attendu est égal au taux de rentabilité exigé par le
=0
marché compte tenu du risque que fait subir ce titre à son détenteur, i.e. :
where P0 is the initial stock price, E0 (Dt ) is the expected resell price of the stock at terminal date T , E0 (P ) is the expected
T
dividend of the stock received in period t and k is the required
[1 + E (R )] = E0 (D1 ) + E0 (P1 ) = (1 + k )
P0
avec k taux de rentabilité exigé par les actionnaires.
expected rate of return by the investors (CAPM or APT).
De cette égalité on tire la valeur du prix courant de l'action :
P0 =
E0 (D1 ) + E0 (P )
1
(1)
(1 + k )
En appliquant le même raisonnement en t = 1 on obtient l'expression de E0 ( P ), soit :
1
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E0 ( P ) =
1
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E0 (D2 ) + E0 (P2 )
( 2)
(1 + k )
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If we assume moreover that the dividends progress at a constant growth rate, denoted g , and that the first expected dividend payment is E0 (D1 ), the price of the stock is then equal to (see
Soit en remplaçant dans l'équation (1) :
E0 ( D2 ) + E0 ( P2 )
E ( D ) + E0 (P )
(1 + k )
1
P0 = 0 1
=
(1 + k )
(1 + k )
E ( D ) E (D ) + E0 (P2 )
= 0 1 + 0 2
(3)
(1 + k )
(1 + k )2
E0 ( D1 ) +
perpetuity):
P0 =
E0 ( D1 ) E0 (D2 ) E0 (D3 )
E (D )
K + 0 nn + K
+
+
2
3
(1 + k ) (1 + k ) (1 + k )
(1 + k )
=
E0 (D1 ) E0 (D1 ) × (1 + g ) E0 (D1 ) × (1 + g )2
E (D ) × (1 + g )n −1
+K
+K+ 0 1
+
+
(1 + k )
(1 + k )2
(1 + k )3
(1 + k )n
=
(1 + g )n −1
E0 (D1 ) (1 + g ) (1 + g )2
+ K
+K+
× 1 +
+
(1 + k ) (1 + k ) (1 + k )2
(1 + k )n −1
En généralisant, on obtient alors :
P0 =
∞ E (D )
E0 ( D1 ) E0 (D2 )
E (D )
+
+ K + 0 ∞∞ + K = ∑ 0 t t
2
t =1 (1 + k )
(1 + k ) (1 + k )
(1 + k )
avec : lim
T →∞
E0 ( PT )
(1 + k )
T
(6)
where the sequence of terms in the bracket corresponds to an infinite sum of the n first terms of a geometric progression with
=0
a common ratio (1 + g ) / (1 + k ) and first term equal to 1.
That is (see consol):
1 + g n
1 −
1 + g n
E0 ( D1 )
1 + k E0 (D1 )
× lim
P0 =
=
× lim 1 −
(1 + k ) n → ∞ 1 − 1 + g (k − g ) n →∞ 1 + k
1+ k
Provided that the dividend growth rate is lower than the required rate of return, k > g , the sum converges and the stock price can be expressed as:
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E0 (D1 )
(k − g )
D × (1 + g )
= 0
(k − g )
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Exemple : Un analyste suivant la société CCC effectue les
P0 =
prévisions suivantes : pour l'année prochaine il s’attend à un dividende de 1.2 EUR, ainsi qu’un taux de croissance des dividendes constant de 5%. Si le taux de rentabilité exigé par les
with:
actionnaires est de 7.7%%, a quel prix l'analyste évaluera le prix
E0 (D1 ) = D0 × (1 + g )
de l'action de cette société?
P0 =
(Gordon-Shapiro DDM formula)
1.2
= 44.4 EUR
(0.077 − 0.05)
Remarque : Le dividende correspondant à la part du bénéfice
Remark: From the DDM pricing formula one can infer the
distribué aux actionnaires, soit ;
expected rate of return by the investors on stocks as:
Dt =
E (D ) k= 0 1 +g
P0
bénéficest
×
nbre 44 444 d' actions émises
14
2
3
dt
{
taux de distrib
BPAt
On en déduit que pour augmenter les dividendes, trois stratégies sont possibles : soit augmenter les bénéfices, soit accroître le
See Market Timing applications.
taux de distribution des dividendes ou encore réduire le
nombre d’actions en circulation (rachat d’actions).
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• Derivatives
pricing:
Linear
and
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Terminal payoff on a long forward contract position
non-linear
derivatives instruments:
♦ Forward (linear derivatives): A forward or a futures contract is
ΠT
K > ST
K = ST
K < ST
an instrument which gives to the holder the obligation to buy
(long position) or to sell (short position) at a specific time in the future T (maturity date) an underlying asset for a certain delivery
Slope = +1
price (the forward delivery or strike price), denoted K , which is
0
fixed today.
Payoff = ST - K
a)Terminal pay-off associated with a long forward/futures
Slope = +1
K
Exercise
price
Payoff = ST - K
ST
position: The terminal payoff associated with a long position in a forward/futures position is given by:
ΠT = (ST − K )
Remark: It is useful to designate the terminal payoffs on long or where ST is the terminal price of the underlying asset and K is the strike (forward delivery) price for a maturity of T years.
short derivatives positions in terms of directions vectors (see also below complex trading strategies with options) where + 1, 0 and -1, indicate respectively a positive correlation between the terminal payoff of the derivatives and the terminal price of the underlying asset (a rise/fall in the underlying asset price results in a rise/fall in the derivatives payoff); a zero correlation and a negative correlation (a rise/fall in the underlying asset price is associated with a fall/rise in derivative payoff).
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b)Terminal pay-off associated with a short forward/futures
c) Forward strike price: The forward delivery price K of a
position: The terminal payoff associated with a long position in
forward contract is generally equal to the fair value of the
a forward/futures position is given by:
forward price (e.g. the cost of carrying the underlying asset until
Π T = ( K − ST )
maturity of the forward contract), so that the value of the
where ST is the terminal price of the underlying asset and K is
forward contract at inception is equal to zero, that is for a non-
the strike (forward delivery) price for a maturity of T years.
income paying underlying asset:
K = F0,T = S 0 × e r T
Terminal payoff on a short forward contract position
K > ST
where S0 is the actual spot price of the underlying asset, T is the
K < ST
K = ST
maturity of forward contract, and r is the continuous interest
ΠT
rate for T years (constant interest rate hypothesis).
Indeed suppose that:
Slope = -1
K > S0 × e r T
0
Payoff = K- ST
Slope = -1
K
Exercise
price
Payoff = K-ST
ST
An investor can borrow S0 Euros for T periods at the risk-free rate r to buy one unit of the underlying asset and take simultaneously a short position in the forward contract (sell the forward). At maturity, the asset is sold under the terms of the forward contract for K and S0 × e rT is used to repay the loan leading at time T to an arbitrage profit of:
(K − S
73
0
)
er T > 0
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(
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[
)
]
f 0 S 0 × e r ×T , T = e − rT EQ [(ST − K )] = e − rτ S 0 × e r ×T − S 0 × e r ×T = 0
Using the same line of reasoning it is possible to show that it is not possible to have (short the underlying and be long the
Remark: The value of a forward contract at inception with a
forward): K < S0 × e r T
d) Forward contract value: To avoid arbitrage opportunities,
strike price of K = S0 and a maturity date of T years is not equal
the value at time t of a forward contract with a strike price K
to zero since:
must be equal to the discounted value of its expected terminal
[
]
(
f 0 (S 0 , T ) = e − rT EQ [(ST − K )] = e − rτ S 0 × e r ×T − S 0 = S 0 × 1 − e − r T
)
payoff under the risk-neutral probability distribution, that is:
Remark: when the underlying asset is paying an additional income
f t ( K , T ) = e − r τ EQ [(ST − K )] = e − rτ × (Ft ,T − K )
(dividend/coupon and stock/bond loans), the fair forward delivery price must be modified such as:
with:
K = S 0 × e (r − g − b ) T
Ft ,T = EtQ [ST ] = St × e r τ
τ = (T − t )
where S0 is the actual spot price of the underlying asset, T is the
where Q is the so-called risk-neutral probability measure,
maturity of forward contract, and r , g and b are respectively
f t ( K , T ) is the value at time t of a long forward contract with a
the continuous interest rate, the dividend yield and the repo rate
forward delivery price of K and a residual maturity of τ , r is
for T years.
the continuous annualized continuous interest rate for τ years
(constant interest rate hypothesis).
Remark: The value of a forward contract at inception with a strike price of K = S0 × e rT (fair cost-of-carry forward price) and a maturity date of T years is equal to zero since:
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♦ Options (Non-linear Derivatives): An option is a contract
The actual market prices of a European call and a put option are
which confers on its holder the right, but not the obligation, to
then given by the discounted values of their terminal payoffs,
purchase (call option) or to sell (a put option) an underlying asset
that is:
for a prescribed amount, known as the exercise or strike price, denoted K , at the expiration date T (European option), or
c0 = e − rT EQ [max(ST − K , 0 )]
− rT
p0 = e EQ [max( K − ST ,0 )]
within a specified period of time T (American option). By convention, if the current underlying asset price exceeds the exercise price of the option, the call is said to be in-the-money and the put is out-of-the-money. If the current underlying asset price is below the exercise price of the option, the call is out-themoney and the put is in-the-money. When the current underlying asset price is approximately equal to the exercise price of the option, both the call and the put are at the money.
The value at maturity of a European call and a put option are given by (terminal payoffs):
0
cT = max[ST − K , 0] = (
ST − K )
p = max[K − S ,0] = ( K − ST )
T
T
0
if
ST ≤ K
if
ST > K
if
ST < K
if
S≥K
where ST is the terminal price of the underlying asset and K is the exercise price.
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Terminal payoff on a long European call option position
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ΠTTerminal
payoff on a short European call option position
K > ST
In-the-money
K = ST
At-the-money
K < ST
Out-the-money
ΠT
K > ST
Out-the-money
K < ST
In-the-money
K = ST
At-the-money
Slope = 0
Slope = +1
0
ST
0
P&L = 0
K
Exercise price
ST
P&L = (ST - K)
K
Exercise price
Slope = 0
P&L = 0
Slope = -1
P&L = -( ST -K)
Terminal payoff on a long European put option position
Terminal payoff on a short European put option position
ΠT
K > ST
In-the-money
K = ST
At-the-money
K < ST
Out-the-money
ΠT
K > ST
In-the-money
K = ST
At-the-money
K < ST
Out-the-money
K
0
Exercise price
ST
K
Exercise price
Slope = -1
0
ST
Slope = 0
Slope = 0
Slope = -+1
P&L = (K - ST)
P&L = 0
79
P&L = -(K - ST )
P&L = 0
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Remark: All these graphs represent the terminal payoffs and not
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2. Standard Calculus
the terminal P&L. For an option buyer, the terminal P&L is computed as the difference between the terminal payoff and the
-
Function of one variable, Derivatives and
(compounded) initial option’s premium, whereas it is equal to
Taylor series expansion (continuous interest
the opposite for an option seller. That is for a long position in a
rate, bond duration and convexity)
European call or put option:
-
Function of more than one variable, partial derivatives and total differential (option’s
c
(ST − K ) −c 0
Π T = Max[0, (ST − K )] − c0 =
− c 0
−p
p
Π T = Max[0, (K − ST )] − p0 = 0
(K − ST ) − p0
if
ST > K
if
ST ≤ K
if
ST < K
-
where ST is the terminal price of the underlying asset, K is the exercise price, and c0 and p0 are respectively the initial premium/price of the call and put option.
Integral, integration by parts and ordinary differential equations (ODE)
ST ≥ K
if
delta and the “Greeks”)
2.1. Function, Derivatives and Taylor Series Expansion
-
Function of one variable:
• Mathematical definition: Formally a function of one variable is a mathematical relation between two sets that associates to each element of the first set a unique element of the second one, that is:
A → B f :
x → f ( x) = y
The set of permissible values of x is called the domain of the
function and the set of values of y which results when the
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function is applied to values of x is called the range of the
♦ The Logarithmic function: The logarithmic function is
function.
defined as the inverse of the exponential function.
• Important functions:
Given:
♦ The Exponential function: The exponential function is obtained by raising the mathematical constant e ( ≈ 2.71828 ), to
y = ex the natural logarithm of y is given by:
ln( y ) = x
a power x , that is:
*
IR → IR+
f
x → f ( x) = e x
with y > 0 .
It gives the power at which the exponential number must be raised in order to obtain the value of x .
Some useful properties:
Some useful properties:
ex × ez = ex+ z
e 0 = 1 ⇔ ln (1) = 0
e =1
0
e1 = e ⇔ ln (e ) = 1
e1 = e
*
*
and ∀( x, y ) ∈ IR+ × IR+ ,
ln( x × y ) = ln( x ) + ln( y )
ln( x y ) = ln( x ) − ln( y )
( )
ln 1 x = −ln ( x )
ln x n = n × ln( x )
Representation of the exponential function
40
( )
ex
30
20
10
0
-4
-3
-2
-1
0
1
2
3
4
83
x
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The (first) derivative: The derivative of a function
• Geometric interpretation: The derivative of a
f ( x ) gives the change of the function f for
function f (.) at a point x corresponds to the
small (infinitesimal) change in the value of the
slope of the line which is just tangent to the
variable x . It is the slope of the function f ( x ) at
function f (.) at the point x (the derivative/slope
a particular point x .
-
is generally variable).
• Mathematical definition: The derivative of f ( x) with
Remark: The first derivative can then be used in approximations.
respect to x is given by:
Using the definition of derivative and if ∆x is small enough, we
f ' (x ) = f x =
df ( x ) f ( x + ∆x ) − f ( x ) Change of f ( x )
= lim
=
dx
Small Change in x
∆x
∆x → 0
can write approximately (at a first order):
f ( x + ∆x ) ≈ f ( x ) + f ' ( x ) × ∆x
where ∆x is a small increment (infinitesimal change) in x .
Example: Suppose x represents time, and f ( x ) is the value of an asset, say a stock, at time t , denoted by St . The first derivative then gives the change of the price of the asset for a small interval of time, that is :
The value of f (.) at point x + ∆x can then be linearly approximated by the value of f (.) at point x plus the derivative
f ' (.) evaluated at x multiplied by ∆x . Note that when one
does not know the exact value of f ( x + ∆x ), the knowledge of f ( x ) , f ' ( x ) and ∆x is sufficient to obtain a linear
∆St S (t + ∆t ) − S (t )
=
∆t
∆t
approximation.
for a small interval of time ∆t (a second!!).
Example: Crude prediction of the future stock price knowing the price today, that is:
S (t + ∆t ) ≈ S (t ) + S ' (t ) × ∆S (t )
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• Basic rules:
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Example: The change of the price of a European call option c written on a stock S (t ) with respect to the change in time. That
♦The power function rule:
is, with c( S (t )) :
f ( x) = c ⇒ f ' ( x ) = 0
dc[S (t )] dc[S (t )] dS (t )
=
× dt dS (t ) dt f ( x ) = x n ⇒ f ' ( x ) = n × x n −1 f ( x ) = 1 / x n = x − n ⇒ f ' ( x ) = − n × x − n −1
Higher-order
-
♦The sum-difference, product and quotient rules:
derivatives
and
Taylor
series
expansion: The derivative of a function f ( x )
h( x ) = f ( x ) ± g ( x ) ⇒ h ' ( x ) = f ' ( x) ± g ' ( x )
is in general also a function of x . This
h( x ) = f ( x ) × g ( x ) ⇒ h ' ( x ) = f ' ( x) × g ( x ) + f ( x ) × g ' ( x )
function can then be differentiated again to
f ' ( x) × g ( x) − f ( x) × g ' ( x ) h( x ) = f ( x ) / g ( x ) ⇒ h ( x ) =
[g ( x )]2
'
obtain the derivative of the derivative - the second derivative. Just as the first derivative is the expression of the slope of the original
♦Chain rule: The chain rule is applied to expressions which are
function, the second derivative gives the slope of
function of a function.
the first derivative (the slope of the slope).
Definition: For:
With a Taylor series expansion, the second
y = f [g (x)]
derivative enables us to approximate more
the derivative of the function f (.) at a point x corresponds to:
efficiently a non-linear relationship.
df df dg f ' (x ) =
=
× dx dg dx
• Mathematical definition of the second derivative: The second derivative of f ( x) with respect to x is
It is a useful tool in approximating the response of one variable
given by:
to changes in other variables.
f '' ( x ) = f xx
87
df ( x ) d '
'
d f (x) dx
= lim f ( x + ∆x ) − f ( x )
=
= dx ∆x → 0
∆x
dx 2
2
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∆x = ( x − x0 ) ≈ small < 1
where ∆x is a small increment (infinitesimal change) in x .
Then we are sure that:
Remark: this process can be continued to find higher derivatives when they exist (third, fourth order derivatives,…).
f ( x ) is an infinitely
differentiable analytic function of x , the Taylor series expansion of f ( x) around any arbitrary
terms that are negligible (smaller than) in the Taylor series expansion. The second-order Taylor series expansion of f ( x) around x0 is given by:
f ( x) ≈ f ( x0 ) + f ' ( x0 ) × ( x − x0 ) +
value x0 can be defined as:
f ( x) = f ( x0 ) + f ' ( x0 ) × ( x − x0 ) +
+
∞
= ∑
k =0
3
Under these conditions it is possible to eliminate some of the
• Mathematical definition of the Taylor series expansion (B.
Taylor, 1685-1731): If
2
x − x0 > x − x0 > x − x0 > ...
1 ''
2
f ( x0 ) × ( x − x0 )
2!
1 '' '
3
f ( x0 ) × ( x − x0 ) + ...
3!
1 '' f ( x0 ) × ( x − x0 )2
2!
which becomes equality as x → x0 .
The value of a non-linear function f (.) at x + ∆x can then be approximated by the value of f (.) at x plus the first derivative
1 k k f ( x0 ) × ( x − x0 ) k! f ' (.) evaluated at x multiplied by ∆x (linear term) and the
where f k ( x0 ) is the k-th order derivative of f (.) with respect to
second derivative f ' ' (.) evaluated at x multiplied by 0.5 × (∆x )2
(quadratic term). Note that when one does not know the exact
x evaluated at x0 , with k ∈ N .
Remark: While the Taylor series expansion is an infinite series, it
value of f ( x + ∆x ). The knowledge of f ( x ) , f ' ( x ) , f ' ' ( x )
is usually truncated at a finite order (second) to obtain useful
and
approximation of a non-linear function f ( x) . Assuming that x is
approximation of it (more accurate than a linear one).
∆x is
then
sufficient
to
obtain
a
non-linear
close to x0 , that is:
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When the increment is very small, even the quadratic term is negligible and we have a first-order Taylor series expansion
(linear term) that is:
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• Financial applications:
♦Demonstration of the continuous compounding formula:
The Taylor series expansion can be used to demonstrate the
f ( x) ≈ f ( x0 ) + f ' ( x0 ) × ( x − x0 )
formula of the future value of an investment with continuous
which becomes equality as x → x0 .
compounding, that is:
V1 = V0 × ei
Remark: The second order Taylor series expansion is one of the most fundamental tool in finance. It is a powerful shortcut for practitioners to assess the impact of small changes in key variables on asset/portfolio values such as changes in yield for bonds (non-linear relation), changes in the underlying asset price for options (non-linear derivatives), changes the weights for portfolios (see incremental Value At Risk) …, with no need to recompute the new value of the underlying position given the new value of the risk factor. It allows us to
where V1 is the terminal value of an initial investment V0 invested for one year at a yearly continuous interest rate i .
Proof: If the initial investment V0 is compounded p times per year at the yearly interest rate i , the terminal value of the investment is:
i V1 = V0 × 1 + p
p
p
approximate the future value of an individual asset (portfolio)
If we take a first-order Taylor series expansion of the function
given the change of one of its constituent (risk-management).
ln(1 + x) around 1, we can write: f ( z ) ≈ f ( z 0 ) + f ' ( z0 ) × ( z − z0 )
1
ln(1 + x ) ≅ ln(1) + × (1 + x − 1) = x
1
with f ( z ) = ln(1 + x ) , z = 1 + x , z0 = 1, n = 1 and f ' ( z ) = 1 z .
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♦Bond price sensivity to a yield change (Dollar Duration
For x « small » (near of zero), we then have:
lim [ln(1 + x )] = x
and Dollar Convexity): The second-order Taylor series
x →0
expansion can be used to approximate the bond price reaction
That is, for an interest rate i fixed, if we increase the frequency
for an interest rate change.
of compounding ( p tends to infinity) we obtain:
The price of a T -year coupon-bond is given by:
C
C
C
F
+
+K
+
T
2
(1 + y ) (1 + y )
(1 + y ) (1 + y )T
T
C
F
+
=∑
t t =1 (1 + y )
(1 + y )T
P0 =
i i lim ln1 + = p →+∞
p p
So that:
p × lim ln1 + p → +∞
i
= lim p × ln 1 + p p →+∞
where P0 is the actual market price and y refers to the yield to
i
=i p
maturity of the considered bond.
Deriving twice this expression it is then possible to approximate
Using the logarithmic function properties leads to:
the impact of a change of interest rate on the bond price.
1) The first derivative of the bond price with respect to the yield
p
i i
lim p × ln 1 + = lim ln 1 + = i p p → +∞ p p → +∞
is (dollar duration):
Taking exponential on both sides leads to the desired result, that
P' ( y ) =
is:
(− )T × p p
i i i
lim expln 1 + = lim 1 + = e p → +∞ p p → +∞ p
which
dP0
C
(− )2 × C 3 K(− )T × C T +1
= (− )1 × dy (1 + y )2
(1 + y )
(1 + y )
corresponds
to
the
formula
of
F
(1 + y )T +1
T
C
F
= − ∑t
+T
t +1
(1 + y )T +1
t =1 (1 + y )
P ( y + ∆y ) − P ( y )
= − DD ≈
∆y
continuous
compounding. with: 93
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D$ =
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∆P = P( y + ∆y ) − P( y )
T
F
C
1
× ∑t ×
+T × t (1 + y ) t =1 (1 + y )
(1 + y )T
1
2
≈ P ' ( y ) × ∆y + × P '' ( y ) × (∆y )
2
1
2
≈ − D$ × ∆y + × C$ × (∆y )
2
The negative of the first derivative is called the dollar duration of the bond (DD). It is related to the absolute change in the bond price associated with a small absolute
The dollar duration measures the first-order (linear)
change in yield.
absolute effect of changes in yield. It is the negative of the slope of tangent to the price-yield curve at the current
2) The second derivative of the bond price with respect to the
interest rate. For small movements in the yield, this linear
yield is equal to:
approximation provides a reasonable fit to the exact price.
P '' ( y ) =
F
d 2 P0 T
C
=C
= ∑ t (t + 1)
+ T (T + 1) t +2
T +2
2
dy
(1 + y )
(1 + y ) $
t =1
But the price value function which relates bond price to yield is not a linear function. It is a rather convex one. The dollar duration, which attempts to estimate a convex relationship
The second derivative is called the dollar convexity of the bond
with a straight line, will then systematically understate the actual
(DC). It takes into account in fact the curvature of the price-
price: when interest drop, the slope of the value function or
yield function of the bond (non-linear relation, see supra). It is
equivalently minus the dollar duration will underestimate the
always positive for straight bonds.
price rise and when, interest rise it will overestimate the price drop. The convexity measures the second-order (quadratic term) effect of changes in yield. It captures the curvature of price function.
For large movements in the interest rates, the quadratic
Putting together all these equations in a second-order Taylor
approximation improves the fit since the price-yield relation
series expansion leads to an approximation for the absolute price
becomes more curved.
change of a bond, that is:
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Remark: Note that the convexity characteristic of the bond price-
Remark: Beside the dollar duration and convexity, there exist
yield relation is always good feature. Ceteris paribus, the higher the
other definitions of bond risk such as the Macaulay Duration
convexity of a bond, the higher will be its price reaction to an
and the Modified Duration
interest rate decrease and the lower will be its price reaction to an interest rate increase (greater convexity is both beneficial for
The Macaulay duration (1938) gives the absolute value of
both falling and rising market yields).
the percentage change of a bond price for a small
(infinitesimal) percentage change in (one plus) the market
Price
(EUR)
Price Approximation
required yield (minus price elasticity measure). That is:
Real price DMac = −η P / 1+ y
P( y + ∆y)
P ( y + ∆y )
with:
Value increases more than duration model
C
C+F
+ t t =1 (1 + y )
(1 + y )T
T −1
P( y)
P0 = ∑
Duration + convexity estimate
Duration
estimate
y + ∆y
Ct
CT + F
T −1
t∑1 t (1 + y )t + T (1 + y )T dP / P0
=
=−
=
P0 d (1 + y ) / (1 + y )
That is:
Value drops less than duration model
T
DMac = ∑ t × wt
y
t =1
Yield(%)
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with:
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wt = PVt / P0
Ct
(1 + y )t if t < T
PVt = C + N
T
if t = T
(1 + y )T
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Due to this feature, zero-coupon bonds (ZCBs) tend to provide the most price movement for a given change in interest rate,
(most attractive bonds in declining interest rate environments).
The Macaulay duration of perpetual bonds (infinite number of identical coupon payments) is given instead by:
and:
D=
Ct
N
+ t t =1 (1 + y )
(1 + y )T
T
P0 = P0 ( y ) = ∑
(1 + y ) y which is much less than its infinite maturity (indeed the cash-
(Expressed in period units, in years with annual
flows paid early dominate the duration of the perpetual bond).
compounding, in semesters with semi-annual,…)
It corresponds to the bond’s weighted average life of cash-flows, where each weight reflects the relative importance of the cashflow payment with respect to the value of the bond. It can also
be viewed as the holding investment period where capital
The Modified duration (1938)
gives the
absolute value of the percentage change of a bond price for a small (infinitesimal) absolute change in the market required yield. That is:
and coupon reinvestment risk are offsetting g each other.
Remark: The Macaulay duration of a zero coupon bond is given by its residual maturity since there are no regular intermediate
DMod
dP
D
1
P
DMac = $
=− 0 =−
(1 + y )
P0
dy
(Expressed in % units)
coupon payments. That is:
T×
Dmac =
F
F
T
(1 + y ) = T × (1 + y )T = T
F
P0
(1 + y )T
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To sum up:
[
w pi =
]
dP C
C
( 1)
= 2 × 1 − (1 + y )−T + T × F − × (1 + y )− T +
D$ = − dy y y
2CT
d 2 P 2C
C
C$ = 2 = 3 × 1 − (1 + y )−T − 2
+ T (T + 1) F − × (1 + y
T +1 y dy y (1 + y )
y
dP 1 D$
DMod = − dy × P = P
0
0
D$
(1 + y ) dP
DMac = − d (1 + y ) × P = (1 + y ) × DMod = (1 + y ) × P
0
0
where we have used here the closed formula of the bond
[
]
ni × Pi
∑ ni × Pi
i =1
N
where w pi , ni and Pi represent respectively the weight, the number and market price of bond i in portfolio p.
Market Beta analogy
(annuity simplification) to compute the dollar duration and convexity). Example: Consider a 10-Year 4.5% coupon bonds selling on the
Remark: The Macaulay duration (modified duration) of a portfolio of bonds is equal to the weighted average of the
Macaulay (modified) durations of the bonds in the portfolio
(if the yield-to-maturity is the same for all bonds), the duration of each bond being weighted by its market value divided by the total market value of the portfolio, that is:
market at par value (100 EUR). What will be its dollar and percentage price change if the interest rate level increases by 1%?
Remark: A basis point (often denoted as bp) is equal to 0.01%
(1/100th of a percent) or 0.0001 in decimal form. So, a bond whose yield increases from 3% to 3.5% is said to increase by 50 basis points; or interest rates that have risen 1% are said to have
N
DMac , p = ∑ wip × Dmac ,i
increased by 100 basis points.
i =1
with:
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The dollar duration and convexity of the bond are given respectively by:
[
]
[
]
C
C
−T
− (T +1)
= 791.27
D$ = 2 × 1 − (1 + y ) + T × F − × (1 + y ) y
y
2CT
C
− (T + 2 )
C = 2C × 1 − (1 + y )−T −
+ T (T + 1) F − × (1 + y )
T +1
2
$ y3 y y (1 + y )
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Check:
P(4.5% ) =
4.5
4. 5
104.5
+
+K+
2
(1.045) (1.045)
(1.045)10
1 − (1.045)−10
100
= 4. 5 ×
+
10
0.045
(1.045)
= 100 EUR
If the required yield level increase by one percent (infinitesimal change), the bond price will be equal to:
See fonction finances Excel VA.
=100*DUREE.MODIFIEE(DATE(2011;1;1);DATE(2021;1;
1);0,045;0,045;1)
That means that the impact of a 1% (100 b.p.) change in interest rates on the 10-year bond price can be approximated to a first
P(5.5% ) =
4.5
4. 5
104.5
+
+K+
2
(1.055) (1.055)
(1.055)10
1 − (1.055)−10
100
= 4. 5 ×
+
10
0.055
(1.055)
= 92.46 EUR
That is an absolute decrease in the bond value of:
92.46 − 100 = −7.54 EUR
order as:
∆P = P(5.5% ) − P(4.5% ) ≈ − D$ × ∆y
≈ −791.27 × 0.01 = −7.91EUR
Which compares with -7.52 (see before).
And to a second order by:
1
2
∆P = P(5.5% ) − P(4.5% ) ≈ − D$ × ∆y + × C$ × (∆y )
2
1
2
≈ −791.27 × 0.01 + × 7781.02 × (0.01) = −7.52 EUR
2
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Example: An investor is considering the two following five-year
Modified duration of bond B:
coupon bonds. A 9% bond A selling on the market at par value
(100 EUR) and a 6% bond B selling to yield 9%.
B
DMod
Which bond the investor will choose if she is expecting a
[
[
]
(
)
with:
P0B
1 − (1.09 )− 5
100
= 6×
= 88.33EUR
+
5
0.09 (1.09 )
See fonction finances Excel VA.
Modified duration of bond A:
A
DMod
]
6
6
−6
× 1 − (1.09 )−5 + 5 × 100 −
× 1.09
2
0.09
(0.09)
= 4.06%
=
88.33
decrease in the general level of interest rates. Same question if he is expecting now an increase in the required market yields.
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
(
)
9
9
−6
× 1 − (1.09 )−5 + 5 × 100 −
× 1.09
2
0.09
(0.09)
= 3.89
=
100
with:
=DUREE.MODIFIEE(DATE(2011;1;1);DATE(2016;1;1);0,
09;0,06;1)
If the required yield increase (decrease) by 100 basis points, the price of bond A will decrease (increase) approximatively by
P0A = 100
3.89%; while the price of bond B will decrease approximatively
(increase) by 4.06%, since:
See fonction finances Excel VA.
=DUREE.MODIFIEE(DATE(2011;1;1);DATE(2016;1;1);0,
Any rational investor will then invest in bond A if she is
09;0,09;1)
expecting an increase of interest rate (less sensitive bond) and in bond B if she is expecting a decrease of interest rates (more sensitive).
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Example: Consider a 10-year 3.5% coupon bonds selling on the market at par value (100 EUR). What is the Macaulay duration?
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CFt
10
−t
∑ t × CFt (1 + y ) t 860.77
(1 + y ) = t =1
=
≈ 8.61
10
P0 ( y )
100
CFt (1 + y )− t
∑
∑ t×
DMac =
t =1
What will be its percentage and dollar price change if the interest
(in years)
t =1
rate level rises by 35 basis points (absolute change of 0.35%)? To compute the Macaulay duration makes the following table:
So that:
∆y
∆P / P0 ≈ − D × (1 + y ) = −2.91%
∆P ≈ − P0 × D × ∆y = −2.91 EUR
(1 + y )
Calculation of Macaulay Duration
Cash flows
PV of 1 EUR 3.5%
PV of CFt
t x PVt
CFt
dt = (1+y ) - t
CFt (1+y ) - t
t * CFt (1+y ) - t
1
3.50
0.9662
3.38
4.30622
2
3.50
0.9335
3.27
8.24156
That means that if the interest rates (required yield) increase (or
3
3.50
0.9019
3.16
11.83000
decrease) by 35 bps, the 10-year bond will experienced a -2.91%
4
3.50
0.8714
3.05
15.09410
5
3.50
0.8420
2.95
18.05514
6
3.50
0.8135
2.85
20.73318
EUR loss (gain) per bond.
7
3.50
0.7860
2.75
23.14709
Check:
8
3.50
0.7594
2.66
25.31466
9
3.50
0.7337
2.57
27.25262
10
103.50
0.7089
73.37
672.90442
100
860.77
Year
decrease (increase) in its value which can be translated a 2.91
P(3.5% ) =
3.5
3.5
103.5
+
+K+
2
(1.035) (1.035)
(1.035)10
1 − (1.035)−10
100
= 3. 5 ×
+
10
0.035
(1.035)
= 100 EUR
From it we then compute the Macaulay duration of this bond
If the required yield level increase by 35 bps, the bond price will
such as:
be equal to:
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P(3.85% ) =
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♦Option price sensivity to an underlying asset price
3. 5
3. 5
103.5
+
+K+
2
(1.0385) (1.0385)
(1.0385)10
1 − (1.0385)
= 3. 5 ×
0.0385
= 97.14 EUR
−10
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
movement:
100
+
10
(1.0385)
∆c = c ' (St )∆S +
1 '' c (St ) × (∆S )2
2!
where St is the price of the underlying asset, c ' (St ) is the first
That is a relative decrease in the bond value of:
derivative of the call price with respect to the price of the
97.14
− 1 = −0.0286 = −2.86%
100
underlying (i.e. the delta of the option) and c" (St ) is the second derivative of the call price with respect to the price of the
which translates into an absolute euro loss of:
underlying (i.e. the gamma of the option).
97.14 − 100 = −2.86 EUR
Remark: Using now the compact formula of the Macaulay duration, we find:
[
DMac
]
3.5
3.5
1 − (1.035)−10 + 10 × 100 −
(1.035)
2
0.035
(0.035)
= (1.035) ×
100
−11
= 8.61
which is precisely what we obtained previously!!!
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• Application (Greek derivatives): The selling or buying
2.2. Function of More than One Variable, Partial
Derivatives and Total Differential
of an option creates a position with various sources of market risk, some of which may be
Function of more than one variable:
-
unwanted. In order to manage those risk factors,
• Mathematical definition: Formally a function of
a trader will need to compute the sensitivities of
several variables is a mathematical relation
the derivative’s price to the various parameters
between two sets of elements that associates a
that impact its values (the “Greeks”).
unique element to the second set with each n-uple of elements of the first set, that is:
Consider the price of call option. It is determined by the
A → B f :
(x1, x2 , x3, K, xn ) → f (x1, x2 , x3, K, xn ) = y
underlying asset price, the exercise price, the time to maturity,
Partial derivative: A partial derivative gives the
asset price and the time to maturity are interconnected since in
change of the function when only one variable of
practice one needs some time to pass before the underlying asset
interest change.
can change).
-
• Mathematical
definition:
The
derivative
of
f ( x1, x2 ,K, xn ) with respect to x1 is given by:
f x1 =
∂f (x1, x2 , x3, K, xn )
= lim
∆x1→ 0
the asset's log-return volatility and the risk-free interest rate (the
The price of the call is a function of five different variables:
ct = c(St , K , τ , σ , r
)
where St is the current price of the underlying asset, K is the exercise price of the call option, τ = (T − t ) the maturity of the
∂x1
f (x1 + ∆x1, x2 , x3, K, xn ) − f (x1, x2 , x3, K, xn )
option contract, σ the volatility of the log-return of the
∆x1
underlying asset and r the risk-free rate.
where ∆x1 is a small change in the x1 direction.
If we fix all the parameters except one and differentiate the option price with respect the later one, we can obtain the following partial derivatives:
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♦Delta (linearity): It measures the first-order sensitivity of the option to a movement in the underlying’ asset price, i.e. its
Call Price as a function of the underlying asset
sensitivity to small changes in the price of the underlying asset.
Mathematically it is the first partial derivative of the option price with respect to the underlying, so it gives the change in option price for a one-unit increase in the underlying price. The deltas for call and put options are given by (Black and Scholes model):
c ∂ct
−g T
Ν (d1 ) > 0
∆ t = ∂S = e t
∂p
∆p = t = −e − g T Ν (− d1 ) = e − g T [Ν (d1 ) − 1] < 0 t
∂St
with:
d1 =
(
)
ln(St / K ) + r − g + 0.5 σ 2 τ
σ τ
The figure depicts the case with K = 100 , σ = 0.56 and r = 0.05 .
and τ = (T − t )
♦Gamma (option’s convexity): It measures the second-order
where ∆ct and ∆p are respectively the delta for the call and put t sensitivity of the option to a movement in the underlying’ asset
option, N (.) is the standard normal cumulative probability
price, i.e. sensitivity of the delta to small changes in the price of
distribution function, g is the continuous dividend yield and τ
the underlying asset or of the option price to the realized volatility. Mathematically it is the first partial derivative of the
is the maturity of the option.
delta or the second partial derivative of the option price with
Note that:
∆ct, OTM → 0 as St → 0
c, ATM
≈ 0.5
∆ t
c, ITM
→ +1 as St → ∞
∆ t
and
respect to the underlying price, so it gives an indication of how
∆p , ITM → −1 as St → 0
t
p , ATM
≈ −0.5
∆ t
p ,OTM
→ 0 as St → ∞
∆ t
convex the option is in the underlying asset price or how quickly the delta of the option moves as the underlying asset price moves. The Gamma is highest for ATM (spot or forward)
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options and lowest for deep ITM or OTM options (ceteris
Gamma of call options with different maturities (1 week solid line;
paribus), since for deep ITM and deep OTM options, delta will
1 month dashed line and 1 quarter dotted line)
not change much for small changes in the underlying price. Note however that a higher volatility lowers the Gamma of ATM
European call/put options but raises it for ITM and OTM options, since high volatility means that ITM and OTM have more time value and so their Gamma sensitivities should not be too different from the Gamma sensitivity near the strike price.
The Gammas for both call and put options are given by (Black and Scholes model):
♦Vega (implied volatility): It measures the first-order
Γtc, p =
∂ 2 ct ∂ 2 pt n(d1 ) × e − gτ
=
>0
=
Stσ τ
∂St2 ∂St2
sensitivity of the option to a movement in the implied volatility of the underlying asset, i.e. its sensitivity to small changes in the
where Γtc , p is the Black-Scholes Gamma for both the call and
implied volatility of the underlying asset. Mathematically it is the
put option, n(.) is the standard normal density function, g is the
first partial derivative of the option price with respect to the
continuous dividend yield and τ is the maturity of the option.
underlying implied volatility, so it gives the change in option price for a one-unit increase in the implied volatility (1%).
Similar to the Gamma, the Gamma tends to be highest for ATM
Note that:
(spot or forward) options and lowest for deep ITM or OTM
Γtc , p ≈ 0 for St > K
options (ceteris paribus). But contrary to the Gamma, the Vega is largest when the maturity is longer (greater effect of the volatility on the call price). The Vega for both call and put options is given by (Black and Scholes model):
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Vega of call options with different maturities
υtc , p =
with:
∂ct ∂pt
=
= St × e − gτ n(d1 ) τ = Ke − rτ n(d1 ) τ > 0
∂σ ∂σ
(
(1 week solid line; 1 month dashed line and 1 quarter dotted line) )
ln(St / K ) + r − g + 0.5 σ 2 τ d1 =
σ τ
d 2 = d1 − σ τ
τ = (T − t )
where υtc , p is the Black-Scholes Vega for both the call and put option, n(.) is the standard normal density function, g is the
♦Theta (time decay): It measures the sensitivity of the option
continuous dividend yield and τ is the maturity of the option.
to a small change in time to maturity, i.e. its sensitivity to the passage of time. Mathematically, as time to maturity decreases, it
Note that:
is minus the first partial derivative of the option price with
υtc , p ≈ 0 for St > K
respect to the time maturity, so it gives the change in option price for a one-unit decrease (1 day). The thetas for vanilla call and put options are given by (Black and Scholes model):
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c ∂ct − ∂ct
S e − gτ n(d1 ) σ
=− t
− g St e − gτ N (d1 ) − r Ke − rτ N (d 2 ) < 0
=
Θt =
∂τ
∂t
2 τ
− gτ
Θtp = ∂pt = − ∂pt = − St e n(d1 ) σ + g St e − gτ N (− d1 ) + r Ke − rτ N (− d 2 ) 0
∂τ
2 τ
∂t
with:
(
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Theta sensivities of call options with different maturities
(1 week solid line; 1 month dashed line and 1 quarter dotted line)
)
ln(St / K ) + r − g + 0.5 σ 2 τ
d1 = σ τ
τ = (T − t )
where Θtc and Θtp are respectively the theta for the call and put option, N (.) is the standard normal cumulative probability distribution function, g is the continuous dividend yield and τ
♦Rho (interest rate sensitivity): It measures the sensitivity of
is the maturity of the option.
the option to a small change in the risk-free interest rate.
Mathematically, it is the first partial derivative of the option price with respect to the interest rate, so it gives the change in option price for a one-unit change (1%) of interest rate. The Rho for vanilla call and put options are given by (Black and Scholes model): c ∂ct
− rτ
ρt = ∂r = τ Ke N (d 2 ) > 0
∂p
ρtp = t = −τ Ke − rτ N (− d 2 ) < 0
∂r
with:
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(
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)
ln(St / K ) + r − g + 0.5 σ 2 τ d1 =
σ τ
d 2 = d1 − σ τ
τ = (T − t )
where
ρtc
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Total differential: The total differentiation gives
-
the total change of the function f ( x1, x2 ,K, xn ) due to the changes of its independent variables
xi , that is:
and ρt are respectively the Rho for the call and put p option, N (.) is the standard normal cumulative probability distribution function, g is the continuous dividend yield and τ
df ( x1, x2 ,K, xn ) = f x1dx1 + f x2 dx2 + K + f xn dxn
Example: The total change of a call option price for an infinitesimal interval of time is given by:
is the maturity of the option.
∂ct
∂c
∂c
∂c
dSt + t dτ + t dσ + t dr
∂r
∂St
∂τ
∂σ
dct =
Since the call option value depends on in St a non-linear
See also Vomma, Vanna, Cross Gamma, DdeltaDtime
(Charm), DgammaDtime (Color or Gamma decay),
DgammaDvol
(Zomma),
DgammaDspot
fashion, it is better to use a second-order Taylor series expansion, that is:
(speed),
DvommaDvol (ultima, and Omega or Lambda (delta elasticity),.... dct =
∂c
∂c
∂c
∂ 2ct
∂ct
2 dSt + 0.5 × 2 dSt + t dτ + t dσ + t dr
∂r
∂σ
∂τ
∂St
∂St
{
{
{
{
{
∆t
Γt
Θt
υt
ρt
Remark: The Taylor series expansion version of a function with several variables is:
∂f
∂f
1 ∂2 f dx + K + × 2 (dx1)2 dx + df =
2 ∂x1
∂x1 1 ∂x2 2
+
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2
∂2 f f (dx1dx2 ) + 1 × ∂ 2 (dx2 ) + K
2 ∂x2
∂x1∂x2
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2.3.
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Integral,
Integration
and
Ordinary
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Differential
-
Equations
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Integration: The integration is the process by which an original function is recovered from its derivative (ante derivative):
-
• The constant function rule:
Integral: It is the mathematical tool used for calculating the sum of uncountable infinite
∫c
objects
dx = cx + k
• The power rule:
• Definition (Riemann Integral): We are given a
n
dx =
x n +1
+k
n +1
deterministic function f (t) of time t ∈ [ 0, T ] ,
∫x
where the interval is partitioned into n disjoint
• Integration by parts:
subintervals:
T
∫0
t0 = 0 < t1 < t2 < K < tn = T
If:
T f ( x )g ' ( x )dx = [ f (T )g (T ) − f (0 )g (0 )] − ∫0 f ' ( x )g ( x )dx
lim max ti − t(i −1) = 0
n→∞
Ordinary differential equations: An equation which contains one or more derivatives of a
i
function is called a differential equation. If the the (Riemann) integral will be defined as: n ∑ n → ∞ i =1
lim
function is of a single variable, the equation is an
n
ti + t(i −1) f × (ti − t(i −1) ) = lim ∑ f (ti −1 ) × (ti − t(i −1) )
2 n → ∞ i =1
n ∑ n →∞ i =1
= lim
ordinary differential equation (ODE), in contrast to the partial equation which arises from functions of several variables (PDE).
T f (ti ) × (ti − ti −1 ) = ∫0 f (t ) dt
Example:Consider the expression:
Area under the curve defined by f (t ) over a given part of the domain.
123
dBt = −rt Bt dt with known B0 , rt > 0 .
⇒
dBt
= − rt dt
Bt
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3. Tools in Probability and Statistics
Its solution is given by:
− ∫ t ru du
Bt = e 0
3.1. Random Variables and Probability Densities
If rt = r , then:
Bt = e − rt
-
Random variables (R.V.):
• Mathematical definition: Formally, a random variable
X is a function that associates to each element of the sample space
(set of all possible
outcomes) a unique element of the (sub)set of real numbers (events) IR, that is:
Ω → IR
X :
ω → X (ω ) = x where the sample space lists all possible outcomes of a random experiment. Example: Consider an experiment in which a coin is tossed three times. In this experiment, the sample space is:
Ω = {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT }
where H represents head and T tail. The random variable could then be defined as the number of heads obtained on the three tosses. For each possible sequence (outcome) ω that consist of three tosses, this random variable would assign a number X (ω)
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equal to the number of heads in it. Thus if ω is the sequence
• Discrete probability density and distribution functions:
(outcome){ HHH } , then:
For any discrete random variable, it is possible to
X = X (ω ) = 3
-
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
compute simultaneously the probability that the random variable takes a specific value and that its
Univariate distribution function (discrete and
• The distribution function: The mathematical model
realisations belong to a specific interval:
~
♦ Discrete random variable: A random variable X is called
for the probabilities that are associated to a
discrete if it can take only a finite number of different values
random variable X is given by the distribution
(discrete values).
continuous cases):
function
F ( x)
(also called the cumulative
distribution function or the c.d.f), defined such
~
♦ Discrete probability function (frequency function): If X
~
has a discrete distribution, the discrete probability function of X is defined as the function p(.) such that for every real number x :
~
p( x) = P( X = x)
as:
F ( x) = P( X ≤ x) with: with:
F (+ ∞) = 1
F (− ∞) = 0
p ( xi ) ≥ 0 ∀xi
∞
i∑ p( xi ) = 1
=1
where − ∞ < x < +∞ .
This is the probability that the realisation of the random variable
~
X ends up less or equal to the given number x .
where p ( xi ) is the probability of observing xi .
~
♦ Probability of an event: If X has a discrete probability distribution, the probability of each subset of the real line (event) can be determined from the relation:
~
P (a ≤ X ≤ b ) = P( X ∈ A) = ∑ p( x ) = ∑ p( xi ) = [F (a ) − F (b )]
A
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a ≤ xi ≤ b
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• Continuous density and distribution functions: For any
In the case of a continuous random variable, the
continuous random variable, it is only possible to
probability function is represented by the area under the
compute the probability that the random variable
curve f ( x) on the interval [a, b] .
belongs to a specific interval:
♦ Continuous random variable: A random variable is called
~
As a consequence the probability that X takes a specific
continuous if it can take an uncountable infinite number of
value x is zero (not impossible but negligible), that is:
different values.
x
P( X = x ) = ∫ f ( x)dx = 0 x Examples: Daily rainfall, asset return (-100% to + infinity),…
Nevertheless, the probability that the continuous random
~
variable X belongs to a small interval around x can be
~
♦ Continuous probability density function (p.d.f): If X has
~
a continuous distribution, the probability density function of X
approximated by:
is defined as the nonnegative valued function f (.) such that, for
~
every interval [a, b] , the probability that X takes value in it is
x+
dx
2 dx ~ dx
P x − ≤ X ≤ x + = ∫ f ( x)dx
2
2 x − dx
2
x+
given by the integral of f (.) over this interval, that is:
dx
2
≅ f ( x) ∫
dx
dx x− 2
b
P[ a ≤ X ≤ b] = ∫ f ( x) dx
dx dx
= f ( x) × x + − x +
2
2
= f ( x) × dx
a
with:
f ( x) ≥ 0 ∀x ∈ IR
+∞
+∞
∫− ∞ f ( x ) dx = ∫− ∞ dF ( x ) = 1
Or equivalently:
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x + dx
~
P(x ≤ X ≤ xdx ) = ∫ f ( x)dx
-
x x + dx
≅ f ( x) ∫
dx
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Multivariate distribution function (bivariate case):
~
~
In the case of two random variables X and Y , it is possible to compute the joint distribution
= f ( x) × [x + dx − x ]
function F (x, y) , the marginal probability
= f ( x) × dx
functions
x
since, while f ( x) is a non-linear curve, for small ∆x (that is for
dx
dx ), f ( x) will not change very much as x varies from x −
2
dx
/ x to x − / x + dx and it can be approximated
2
of f X ( x)
and
fY ( y )
and
their
conditional distributions.
♦ Joint cumulative distribution: In the discrete case, the joint
cumulative bivariate probability function is written as:
F (a, b ) = P ( X ≤ a, Y ≤ b ) = ∑ ∑ p( X = xi , Y = yi ) xi ≤ a yi ≤ b
reasonably well by f ( x) during this interval.
where p( x, y ) is the joint discrete probability (frequency)
The rectangle obtained would then be very close to the
function.
probability that x will fall within a small neighbourhood of x
In the continuous case, the joint cumulative bivariate probability
represented by the quantity dx .
function is written as: a b
F (a, b ) = P( X ≤ a, Y ≤ b ) = ∫ ∫ f ( x, y )dxdy
So that:
−∞ −∞
P[ a ≤ X ≤ b] = lim ∑ f ( xi ) × ∆xi = ∫ f ( x) dx
n →∞ n i =1
b
where f ( x, y ) is the joint continuous probability (frequency)
a
where xi ∈ [a, b], and the interval is partitioned into n disjoint
function.
subintervals:
Example: Probability of observing simultaneously tomorrow a return of -20% on the SP500 Index and a return of - 30% on the
a = x0 < x1 < x2 < K < xn = b
CAC40 French Index, that is:
P ( RSP500 ≤ −0.2, RCAC40 ≤ −0.3)
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♦ Marginal distribution function: In the discrete case, we can
~
~ recover the marginal probability frequency function of X and Y by: Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
♦Conditional probability distributions: In the discrete case, the conditional probability density function of the random
~
~ variable Y given the random variable X is given by:
p ( x, y )
~
~ pY = y X =x = pX (x)
p X ( x ) = ∑ p( X = x, Y = yi ) yi
pY ( y ) = ∑ p( X = xi , Y = y ) xi
(
)
with p X ( x) > 0 .
In the continuous case, we can recover the marginal probability
~
~ density function of X and Y from:
∞
( x) = ∫ f ( x, y ) dy
fX
−∞
∞
( ) ∫ ( fY y = f x, y ) dx
−∞
~
This is the probability of Y = y conditional on the fact that
~
X = x.
~
~
If X and Y are independent, then: p( x, y ) = p X ( x ) × pY ( y )
p ( x ) × pY ( y )
~
~
⇒ pY = y X =x = X
= pY ( y ) pX (x )
~
That is we learn nothing about Y from observing a
~
particular realisation of X .
(
Remark: Any multivariate distribution can be factored into two components: on one hand, the marginal distributions which represent the univariate characteristics of the multivariate
)
distribution, and on the other hand, the copula function which summarizes the joint features / dependence structure of the multivariate distribution. That is in the continuous case: a b
F (a, b ) = P( X ≤ a, Y ≤ b ) = ∫ ∫ f ( x ) f ( y ) c[F ( x ), F ( y )]dx dy
In the continuous case, the conditional probability density
~
~ function of the random variable Y given the random variable X is given by:
f (y x) =
−∞ −∞
where:
f ( x, y ) f X (x)
with f X ( x) > 0 .
f ( x, y ) = c[F ( x ), F ( y )] × f ( x ) × f ( y )
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This is the density of Y = y conditional on the fact that X = x .
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
3.2. Moments of Random Variable
If X and Y are independent, then: f (x, y) = f X ( x) × fY ( y )
⇒ f ( y x) =
-
f X ( x) × fY ( y )
= fY ( y ) f X ( x)
While any random variable is characterised by its distribution function, in many instances it is not necessary to draw the entire probability density
That is we learn nothing about Y from observing a particular
(frequency) function to have an idea of its shape.
realisation of X .
Many random variables have indeed probability distributions which can be uniquely determined
Example: Probability of observing tomorrow a return of -30 on
by their moments (K. Pearson, 1857-1936). Some
the CAC40 French Index conditional on a return today of - 20%
can be determined parametrically only by their
on the SP500 Index, that is:
firsts two moments. Others need higher-order
P ( RCAC40 = −0.3 RSP500 = −0.2, )
central moments for a full characterisation.
-
The first two moments:
• The mean or the expected value of a random variable (first
central moment): It is a measure of the central tendency or center of gravity of the distribution of the random variable of interest.
♦ Mathematical definition:
µ = E (X ) = ∑ xi p ( xi ) (discrete case) n ~
i =1
~ +∞ µ = E (X ) = ∫ x f ( x ) dx (continuous case)
−∞
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♦ Properties of the expectation operator (linear operator):
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
• The variance of a random variable (second central
moment): It gives information about the way the
~
~
E (a × X + b ) = a × E (X ) + b
~ ~
~
~
E (X ± Y ) = E (X ) ± E (Y )
~ ~
~ ~
~
E (X Y ) = E (X )E (Y ) if X and Y are independent
distribution of the random variable is spread out.
It is a measure of the average squared deviation of the future realizations around the mean.
♦ Empirical counterpart (sample estimates): In practice, the
♦ Mathematical definition:
σ 2 ( X ) = V ( X ) = E [( X − µ ) 2 ]
mean of a random variable is estimated by its sample counterpart. Let {x1,K, xT } be a random sample of T
~
realisations of X . Assuming that all observations are identically and independently distributed (I.I.D., i.e. they are independent
~
draws from the same random variable X whose distribution is
~
constant in time), the first moment of the random variable X ,
~
µ = E (X ), can then be estimated from a sample of T realisations
(general definition)
σ 2 ( X ) = Var( X ) = ∑ p ( xi ) × [ xi − E( X )] 2 n i =1
(discrete case)
+∞
σ 2 ( X ) = Var( X ) = ∫ [ x − E( X )] 2 f ( x) dx
−∞
by the sample mean, that is:
(continuous case)
~ 1 T
ˆ ˆ µ = E (X ) = ∑ xt
T t =1
Remark: The square root of the variance is the standard deviation, that is:
Std ( X ) = σ ( X ) = Var ( X )
It is a more convenient measure of dispersion to use as it has the same units as the original variable X .
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♦ Properties of the variance (quadratic operator) and
♦ Empirical counterpart (sample estimates):
1 T
2
ˆ
ˆ 2
∑ ( xt − µ X ) σ (X ) =
T − 1 t =1
T
σ XY = 1 ∑ ( xt − µ X )( yt − µY )
ˆ
ˆ
ˆ
T − 1 t =1
covariance (bilinear operator):
( )
Var ( X ) = E X 2 − [E ( X )]
(Huygens Formula's )
Var (a × X + b ) = a 2 × Var ( X ) σ (a + b × X ) = b × σ ( X )
2
[
Var ( X + Y ) = E ( X + Y − E ( X ) − E (Y ))
2
]
[
]
= E [( X − E ( X )) ] + E [(Y − E (Y )) ] + 2 E [( X − E ( X ))(Y − E (Y ))]
• Financial application (portfolio performance, expected
= E ( X − E ( X )) + (Y − E (Y )) + 2( X − E ( X ))(Y − E (Y ))
2
2
2
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
2
= Var ( X ) + Var (Y ) + 2 × Cov ( X , Y )
return and risk diversification, Markowitz, 1952):
♦Portfolio rate of return (one period): The one period rate of return over a selected horizon of a portfolio consisting of fixed
where Cov( X , Y ) is the covariance between X and Y (positive or negative comovements between the two variables) defined
positions in N assets is a linear combination of the returns on the underlying assets involved, where the weights on each asset are given by the relative amounts invested at the beginning of
such as:
the period, that is for two assets:
R p ,t +1 = w1 R1,t +1 + w2 R2,t +1
Cov( X , Y ) = σ XY = E{[ X − E ( X )][Y − E (Y )]}
= E ( XY ) − E ( X ) E (Y )
with:
with (bi-linear operator):
Ri ,t +1 =
Cov( X , Y ) = Cov(Y , X )
Cov( X , a + b Y ) = b × Cov( X , Y )
2
Cov( X , X ) = σ XX = σ X = Var ( X )
(Pi,t +1 − Pi,t ) + Di,t +1
Pi ,t
where Ri ,t +1 is the rate of return of the asset i with i = [1, 2] and
wi is its relative weight in the portfolio (determined in t ).
Or more generally for N assets:
R p ,t +1 =
139
N
∑ wi i =1
Ri ,t +1
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♦Portfolio expected rate of return (one period): The
( )
Var R p =
portfolio expected rate of return is a linear combination of the asset expected returns which are involved in it, that is for two assets: N
∑ wi i =1
[wi w j Cov(Ri , R j )]
[wi w j Cov(Ri , R j )]
(
Cov Ri , R p
)
variances of the individual securities but also on all the
[wi × E (Ri )]
covariances (risk reduction by diversification).
♦Portfolio variance: The variance of the portfolio return is a linear combination of the covariances between the return of
Remark:
σ ( Ri ) = Var ( Ri )
The standard deviation of the asset return is of the same order
each asset and those of the portfolio, that is for two assets:
[
N N
∑ ∑ i =1 j =1
N N
∑ ∑ i =1 j =1 j ≠i
That is, the variance of a portfolio depends not only on the
Or more generally for N assets:
( )
Var ( Ri ) +
=
= w1 E (R1 ) + w2 E (R2 )
E
N
∑ wi2 i =1
=
E (R p ) = E (w1R1 + w2 R2 )
N
Rp = ∑ i =1
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
]2
Var (R p ) = E R p − E (R p )
than the expected return (some % per annum).
= E [w1R1 + (1 − w1 )R2 − (w1E (R1 ) + (1 − w1 )E (R2 ))]
= E [w1 (R1 − E (R1 )) + (1 − w1 )(R1 − E (R1 ))]
2
2
= w1 E [R1 − E (R1 )] + (1 − w1 ) E [R2 − E (R2 )]
2
2
2
2
+ 2 w1 (1 − w1 ) E{[R1 − E ( R1 )] [R2 − E ( R2 )]}
2
= w1 Var (R1 ) + (1 − w1 ) Var (R2 ) + 2 w1 (1 − w1 ) Cov(R1 , R2 )
2
2
2
= w1 Var (R1 ) + w2 Var (R2 ) + 2 w1 w2 Cov(R1 , R2 )
with: w2 = (1 − w1 ) ⇔ (w1 + w2 ) = 1 (budget constraint)
Generally for N assets :
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Higher order moments: The third central moment of
positive outcomes and large number of small negative
a distribution is the skewness and is informative
-
outcomes)
of the presence of asymmetry in the distribution.
The fourth central moment of a probability distribution is the kurtosis and measures the width of its tails (extreme risks):
• Skewness (third-order moment):
♦Mathematical definition:
[
• Kurtosis (fourth-order moment):
♦Mathematical definition:
[
♦Interpretation:
]
+∞
3
3
3 s ( X ) = E ( X − E ( X )) = ∫ [ X − E ( X )] f ( x )dx
−∞
3
γ ( X ) = s ( X ) (standardized)
1 σ ( X )3
]
+∞
4 κ ( X ) = E ( X − E ( X ))4 = ∫ [ X − E ( X )]4 f ( x )dx
−∞
4
γ ( X ) = κ ( X )
2 σ ( X )4
(symmetric
distribution)
the
kurtosis
coefficient is a measure of the tail thickness / peakedness of the probability distribution.
1) A Kurtosis lower than 3 is called platykurtic. The distribution
♦Interpretation: The skewness is a measure of the imbalance
has lower tails and a flatter peak than the Normal distribution
of the tails of the probability distribution (the direction of
(larger fraction of outcomes are cluster around the mean).
2) A Kurtosis equal to 3 is said to be mesokurtic. The distribution
skewness is to the tails).
1) A negative skewness (left-skewed) indicates that the left-tail
has same tails and peak than the Normal distribution.
of the distribution is longer that the right (few extreme
3) A Kurtosis greater than 3 is called leptokurtic The distribution
negative outcomes and large number of small positive
has fatter tails and more pointed peak than the Normal
outcomes)
distribution (larger fraction of outcomes is at the extremes.
2) A zero skewness indicates that the same amount of data
Discussion: Dynamic strategies, Options, Hedge funds,….
tails on both sides of the mean
3) A positive skewness (right-skewed) indicates that the righttail of the distribution is longer that the right (few extreme
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The quantile: A probability distribution can also be described by its quantile which is the cut-off
0.6
point with an associated probability α %:
0.5
♦Mathematical definition: the α % quantile ( 0 ≤ α ≤ 1) of a
~
random variable X , corresponds to the cut-off point Qα such
Probability Density
0.4
Empirical Transaction-time Return
0.3
Normal Law
that there is a probability of α % that the random variable will
0.2
fall below it:
F ( x ) = P[ X ≤ Qα ] = α
0.1
-4
-3
-2
-1
0
0
Centered Reduced Return
1
2
3
4
See Value-At-Risk
Source: Maillet et Michel, (2001); densité associée aux rentabilités du titre Elf mesurées en intra-day sur les six premiers mois (+ de 180 000 points).
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Figure 1: Normal Return Density Functions
3.3. Some Important Probability Distribution:
for Two Asset Returns
-
The normal distribution (Abraham de Moivre, 1733 and
3,00
Karl Gauss, 1809): The normal distribution plays a
2,50
central role in statistics as well as in finance because of
its
convenient
properties.
2,00
In
particular, the normal distribution is stable under
1,50
addition (any linear combination of jointly
1,00
normally distributed random variables has a
0,50
normal
distribution),
provides
the
limiting
distribution of the average of independent
0,00
-1
-0,5
0
0,5
1
1,5
random variables (through central limit theorem or CLT) and can be entirely characterised by its first two moments only, the mean and the
♦ Symmetrical distribution: It is symmetrical around the mean, which is the same as its mode (most likely or highest point
variance:
• The Normal Distribution: A random variable X is said to follow a normal distribution (e.g. to be normally Three useful properties
distributed)
with
mean
µ
and
variance σ if it such as:
(
N (x µ , σ ) = f (x µ , σ ) = 2πσ
)
2 −1 / 2
1 x − µ 2 exp −
2 σ
where E ( X ) = µ , Var ( X ) = σ and − ∞ < x < +∞ .
2
147
of the distribution) and median (which has a 50% of occurrence). The skewness of a normal distribution is null which indicates its symmetry around the mean
♦ Lower tails: Its standardized kurtosis is equal to three
(distributions with fatter tails have a greater standardized coefficient). 148
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• The Standard Normal Distribution: The use of the
♦ Confidence intervals: If X is normally distributed with mean µ and standard deviation σ , then we know by
normal
construction that the following inequalities are necessary true
simplified by transforming a normal random
(confidence intervals):
variables X with mean µ and standard deviation
distribution
can
be
considerably
σ into a standard normal random variables Z
P[µ − σ ≤ X ≤ µ + σ ] = 0.683
P[µ − 2 × σ ≤ X ≤ µ + 2 × σ ] = 0.955
P[µ − 3 × σ ≤ X ≤ µ + 3 × σ ] = 0.997 i
P[µ − 4 × σ ≤ X ≤ µ + 4 × σ ] = 0.999
which has been standardized or normalized so that it possesses a zero mean and an unitary variance, that is:
Z=
X −µ
σ
where E (Z ) = 0 , Var (Z ) = 1.
The standard normally distributed random variable Z has a probability density function denoted f (.) and a probability
That is: 68%, 95.5%, 99.7% and 99.9% of the realizations are respectively contained
in
[µ − σ , µ + σ ], [µ − 2σ , µ + 2σ ], and [µ − 4σ , µ + 4σ ].
the
closed
sets:
distribution denoted F (.) such as:
1
N (z 0, 1) = f (z 0, 1) = (2π )−1 / 2 exp − z 2
2
[µ − 3σ , µ + 3σ ] and: F (z ) =
z
∫
−∞
(2π )−1/ 2 exp − 1 z 2 dz
2
with − ∞ < z < +∞ .
The standard normal distribution is characterized by the three following properties:
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♦ Symmetrical distribution: It is symmetrical around the
Remark: The distribution of any random variable X can be
mean which is zero.
recovered from that of the standard normal Z by defining:
X = µ + Z ×σ
f (− z ) = f ( z ) by symmetry
P(Z ≥ z ) = [1 − F ( z )]
F (− z ) = 1 − F ( z ) by symmetry
X has indeed the desired moments, that is:
E ( X ) = µ + E (Z ) × σ = µ
2
2
Var ( X ) = Var (Z ) × σ = σ
♦ Lower tails: Its kurtosis is equal to three (distributions with fatter tails have a greater coefficient).
♦ Confidence intervals: If Z is a standard normal, then we know that:
Figure: Standard Normal Return Density Function
0,40
X −µ
P[− 1 < Z < +1] = P − 1 ≤ σ ≤ +1 = 0.683
X −µ
< +2 = 0.955
P[− 2 < Z < +2] = P − 2 <
σ
P[− 3 < Z < +3] = P − 3 < X − µ < 3 = 0.997
σ
P[− 4 < Z < +4] = P − 4 < X − µ < 4 = 0.999
σ
0,30
68.3 %
95 %
0,20
0,10
99 %
That is 68%, 95% and 99% of the distribution are respectively
0,00
-4
contained between ± 1, ± 2 and ± 3 .
151
-3
-2
-1
0
1
2
3
4
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• Financial application (distribution of future discrete asset
returns): A traditional assumption made in finance is that the discrete one period asset returns are normally distributed, that is:
(
~
Ri ,t +1 ~ N (µi , σ i ) = 2πσ i2
)
−1 / 2
2
1 R i ,t +1 − µi
exp −
σi 2
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
1)The probability of the negative return can be computed by using the standardized variate:
R p ,t +1 − µ p 0 − µ p
≤
P (R p ,t +1 ≤ 0 ) = P
σp σp
− 0.1 0 − 0.1
R
= P H ,t +1
≤
0. 2
0. 2
= P(Z ≤ −0.5) = 0.3085
where:
That is 30.85% (read tabulated standard normal cumulative
− ∞ < Ri ,t +1 < +∞
distribution) !!!!
with a fixed mean µi and standard deviation σ i .
The 95.5% confidence interval on the future value of the fund
Example: Consider a mutual fund (or a stock index, a
can be computed as follows:
commodity,….) which is normally distributed with an expected return of 10% per annum and a standard deviation of 20% per annum. P(0.1 − 2 × 0.2 ≤ RH ≤ 0.1 + 2 × 0.2 ) = 0.955
⇔ P(− 0.3 ≤ RH ,t +1 ≤ 0.5) = 0.95
Since the future NAV is equal to:
NAVt +1 = NAVt × (1 + R p ,t +1 )
1) What is the probability that the fund (asset) realizes a loss next year? The 95.5% confidence interval for the NAV of the Hedge fund
2) If the NAV (net asset value) of the mutual fund is actually
1000, define a 95.5% confidence interval for its future value at the end of the year.
in one year is:
1000 × (1 − 0.3) ≤ NAV p ,t +1 ≤ 1000 × (1 + 0.5)
⇒ 700 ≤ NAV p ,t +1 ≤ 1500
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The Lognormal Distribution: A random variable
X is said to have a lognormal distribution if its logarithm Y = ln X is normally distributed. The
Lognormal Density Function
essence of lognormality is the idea that if a random variable Y is normally distributed, then the random variable
X = eY
is distributed
lognormally.
• Mathematical definition:
0,9
If the random variable
Y = ln( X ) is normally distributed with a mean
0,8
µ and a standard deviation σ , X is lognormally
0,7
distributed random variable with a lognormal
0,5
density function lN (.) :
(ln x − µ )2
−1
X ~ LN ( µ , σ ) = (σx 2π ) exp −
2σ 2
Y = ln ( X ) ~ N (µ , σ )
0,4
0,3
0,2
with mean µ X and variance σ X given by:
1
µ+ σ2
µ X = E( X ) = e 2
σ 2 = Var ( X ) = e 2× µ +σ 2
X
0,6
0,1
0
0
2
× eσ − 1
1
2
3
4
5
6
7
8
9
10
This distribution is skewed to the right with x > 0 . Its tail increases for greater values of σ . This explains why as the variance of ln X increases, the mean of X is pulled up.
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• Financial application (distribution of future discrete asset
Example: Consider previous mutual fund (or a stock index, a
returns): Another commonly used assumption
commodity,….) but supposed now that its future discrete return
made in finance is that the log-returns are
is lognormally distributed with an expected return of 10% per
normally distributed which implies that the
annum and a standard deviation of 20% per annum.
discrete future asset returns are lognormally
1) What is the probability that the fund (asset) realizes a loss next
distributed, that is:
year?
2) If the NAV (net asset value) of the mutual fund is actually
(ri ,t +1 − µ r )2
−1
−1
Ri ,t +1 ~ LN ( µ , σ ) = ( 2π σ r ) (1 + Ri ,t +1 ) Exp −
−1
2
2σ r
~
Pi ,t +1
(
)
ri ,t +1 = ln
P ~ N µr , σ r i ,t
1000, define a 95.5% confidence interval for its future value at the end of the year.
1)The probability of the negative return can be computed by using the gross discrete return, that is:
P (R p ,t +1 ≤ 0 ) ⇔ P (1 + R p ,t +1 ≤ 1)
with discrete (log-normal) return mean and variance given by:
(
rp ,t +1 = ln (1 + R p ,t +1 )
Since
2
µr + σ r −1
µ = exp
2
2
2
σ = exp 2 µ r + σ r exp σ r − 1
is
by
definition
normally
distributed, we have to determine its first two moments µ and
)[ ( ) ]
σ.
where µ r and σ r represent respectively the mean and the
Recall that:
2
µ + σ = 1.1
E 1 + Ri , t +1 = exp
2
2
2
2
Var 1 + R i ,t1 = exp 2 µ + σ exp σ − 1 = (0.2 ) = 0.04
(
standard deviation of the log-return ri ,t +1 and:
− 1 < Ri ,t +1 < +∞
)
(
157
)
(
)[ ( ) ]
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So that:
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Since the future NAV is equal to:
µ = ln E (1 + Ri ,t +1 ) − 0.5 × σ 2 = ln(1.1) − 0.5 × 0.0325 ≈ 0.08
Var (1 + Ri ,t +1 )
2
= 0.0325
σ = ln 1 +
E (1 + Ri ,t +1 )2
2
σ = σ = 0.18
The probability of the negative return then can be computed by using the standardised variable:
[
P (1 + R p ,t +1 ≤ 1) = P ln(1 + R p ,t +1 ) ≤ 0
NAVt +1 = NAVt e
(
ln 1+ R p , t +1
)
The 95.5% confidence interval for the NAV of the mutual fund in one year is:
1000 × exp − 0.28 ≤ NAVH ,t +1 ≤ 1000 × exp + 0.44
⇒ 755.78 ≤ NAVH ,t +1 ≤ 1552.71
]
ln(1 + R p ,t +1 ) − µ 0 − µ
= P
≤
σ σ
ln(1 + R p ,t +1 ) − 0.08 0 − 0.08
= P
≤
0.18
0.18
= P(Z ≤ −0.44 ) = 0.33
Thank you for your attention!!!!!
That is 33% (instead of 30.85% in the previous normally distributed discrete return hypothesis) !!!!
The 95.5% confidence interval on the future value of the fund can be computed as follows:
P (0.08 − 2 × 0.18 ≤ ln (1 + R p ,t +1 ) ≤ 0.08 + 2 × 0.18) = 0.955
⇔ P(− 0.28 ≤ ln (1 + RH ,t +1 ) ≤ 0.44 ) = 0.955
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ESCP EUROPE Business School
Finance
University Year: 2011-2012
Introduction:
- Target Audience
- Objectives of the Course
- Prerequisite
Recommended Textbooks (in French):
- Piermay M. et A. Lazimi, (1989), Mathématiques
FINANCIAL MATHS REVIEW COURSE
Financières, 2ème édition, Economica, 258 pages.
- Schlacther
D.,
(2007),
Comprendre
les
Mathématiques Financières, Hachette, 158
Emmanuel F. Jurczenko
Pages.
- Hull J., (2007), Options, futures et autres actifs
Slides
dérivés, 6ième édition, Pearson Edition, 815 pages.
(Preliminary version - do not diffuse)
Recommended Textbooks (English):
- Jorion P., (2007), Financial Risk Manager
Handbook, 4th edition, Wiley Finance, 736 pages. Associate Professor of Finance at ESCP EUROPE and Associate Researcher at HEC-Lausanne and at the University of Paris-1 (CES/CNRS).
E-mails: ejurczenko@escpeurope.eu with cc to emmanuel.jurczenko@gmail.com
- Hull J., (2008), Options, Futures, and Other
Derivatives, 7th edition, Prentice Hall, 816 pages.
- Neftci S., (2009), Principles of Financial
Engineering, Academic Press, 2nd Edition, 696 pages. ©
Jurczenko - Copyright 11. Preliminary Version (September 2011) - Usual disclaimers apply.
Do not quote or diffuse without permission.
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Three parts:
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1. Basic Tools in Financial Mathematics
- Basic Tools in Financial Mathematics:
• Compounding and future value (discrete and continuous compounded interest rates/rate of
payments and receipts, which are spread over time.
However, one Euro obtained (invested) today is not
returns)
• Discounting and present value (bond, stock and derivatives pricing)
• Applications (bond pricing, stock pricing and derivatives) • Function of one variable, derivatives and Taylor series expansion (dollar duration)
• Function of more than one variable, partial derivatives and total differentials (“The Greeks”)
valued):
• Preference for present
• Inflation and purchase power erosion
Two identical sums of money perceived at different moments are not equal (Time Value of Money)
- Compounding and discounting permit us to make
• Integral and integration by parts
comparison between cash-flows perceived or paid
- Tools in probability variables, equivalent to one Euro obtained (paid) tomorrow (less
• Credit risk
- Standard calculus
• Random
- Financial decisions are based on a stream of
at different times by fixing conventionally one density functions
and
statistical moments (portfolio diversification)
valuation period - the future or the present – to obtain equivalent sums at a common date:
• Binomial, Normal and Lognormal Distributions
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• Compounding gives the future (terminal) value
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• Discounting gives the present value V0 that
VT that can be obtained from an initial
must be invested (or borrowed) today to
investment (loan) V0 :
obtain (repay) a given future value VT :
− V0
{
+ VT ?
13
2
Present
Future
− V0 ?
13
2
+ VT
{
Future
Present
(Lending)
(Lending)
+ V0
{
− VT ?
13
2
Present
Future
+ V0 ?
13
2
− VT
{
Future
Present
(Borrowing)
(Borrowing)
where 0 and T are respectively the current date and the maturity date of the financial operation; + and - represent respectively a cash
where 0 and T are respectively the current date and the maturity
inflow (receipt) and a cash outflow (payment).
date of the financial operation;+ and - represent respectively a cash inflow (receipt) and a cash outflow (payment).
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1.1. Discrete and continuous compounding
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Substituting leads to:
V2 = V0× (1 + i ) × (1 + i )
= V0× (1 + i )
2
- Compounding a single cash-flow:
We speak of compounding since interests obtained periodically
• Principle: Compounding gives the future value
are compounded, that is:
of a cash-flow (or of a stream of cash flows)
V2 = V0 + 2iV0 +
which has been invested for many periods
(years)
without
any
intermediary
Compoundin g
At the end of T years (maturity), the future value is:
VT = V0× (1 + i )T
supplementary contribution or withdrawal.
• Proposition: The future value VT of an initial
i 2V0
{
Remark:
amount V0 compounded annually for T years
If the maturity of the operation is less than one year, the future
at a constant yearly interest rate i (p.a.) is:
value will be given by:
VT = V0 × (1 + i )T
days
VT = V0× 1 + i ×
360
• Proof : The future value of the initial
where i corresponds to the simple annual interest rate (prorata temporis). investment V0 at the end of the first year is:
See price of depos, Eurocurrency deposits,…
V1 = V0 + i × V0 = V0 × (1 + i )
At the end of the second year, the principal (initial amount invested plus interest of the first period) is also compounded, that is:
Example: Consider an investment of 1000 EUR in a bank deposit for six years compounded annually at an interest rate of 10%
p.a.. What is the future value of your investment?
V2 = V1 × (1 + i )
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V6 = 1000× (1 + 0.1)
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
• Using the future value formula with discrete
6
= 1771.56 EUR
compounding:
♦ Compute the sum of the interest receipts after T years of
Table 1. Future Value and compounded interest
investment:
[
]
Year
Initial Value
Interest
Future Value
1
1000
100
1100
2
1100
110
1210
♦ Compute the annual discretely compounded interest rate
3
1210
121
1331
knowing the initial investment, the terminal value and the
4
1331
133.1
1464.1
5
1464
146.41
1610.51
6
1610.51
161.051
1771.561
IT = V0 × (1 + i ) − 1
T
investment horizon:
VT = V0 × (1 + i )
T
⇒ (1 + i )T =
Example: The Dutch Peter Minuit have bought in 1626
1/ T
V
⇔i= T
V
0
Manhattan island to the Manhattes Indians for 24 dollars worth trinkets. If Indians have required instead a cash payment and invested it at a discrete compounded interest rate of 6% per annum (see B. Malkiel, A Random Walk Down Wall Street, p. 116,
V385 = 24 × (1.06)
385
today 15000 EUR. What is the annual rate of return on this
1 / 10
= 132.73 Billions USD
−1
Example: A bank offers you 30000 EUR in 10 years if you invest
30000 investment? i =
15000
Norton Company), they will have obtained today:
VT
V0
− 1 = 0.0718 = 7.18%
Example : Your father is asking you if it
More than 132 billions of dollars!! !
is better to buy
government bonds priced at 0,86 EUR (zero-coupon OAT) that will be redeemed at 1 EUR in 2 years or to invest in a money banking account at 5%/year? The yearly return on the bond investment is equal to:
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♦ Investment maturity determination, knowing the initial
1 1 / 2
− 1 = 0.0783 = 7.83%!! i =
0.86
amount of money invested, the terminal value of the investment
Example : From J. Siegel (Stocks for the Long run, Mc Graw-Hill,
and the interest rate:
VT = V0 × (1 + i )
1998), 1 dollar invested in 1802 on the US stock market have
T
yield in 1997 (by including reinvested dividends) 7,47 millions of
⇒ (1 + i )T =
dollars. The same investment on the US bond has yield in 1997
(by including reinvested coupons) 10744 dollars, 3679 dollars on
VT
V0
V
⇔ T ln (1 + i ) = ln T
V0
the money market and 1337 dollars with gold! Compute the
(nominal) annual rate of returns on these different asset classes.
⇒T =
US Stocks:
ln(VT ) − ln (V0 ) ln (1 + i )
Example: Consider an individual who invests today 60000 EUR
7470000 1 / 195
− 1 = 0.084 = 8.4% i =
1
on a time-deposit at an annual interest rate of 6% p.a.. What must be the maturity of its investment so that the terminal value
US Bonds:
of its investment equals 80000 EUR ? Same question to double
10744 1/ 195
− 1 = 0.048 = 4.8% i =
1
the initial amount of money invested?
From the future value formula, we have:
US Money Market:
VT = V0 × (1 + i )
T
3679 1 / 195
− 1 = 0.043 = 4.3% i =
1
⇔ 80000 = 60000 (1.06 )T
Gold:
Taking the logarithm on both sides of previous expression, leads
1337 1 / 195
− 1 = 0.037 = 3.7% i =
1
to: ln(80000 ) = ln (60000 ) + T × ln(1.06)
That is:
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8 ln ln (80000 ) − ln (60000 )
6
T=
= = 4.93 ≈ 5 years ln (1.06) ln (1.06 )
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10 ln ln (10000000 ) − ln (800000 )
8
T=
= = 7.54 ≈ 8 years ln (1.03) ln (1.03)
That is five years. To double the initial investment, we find:
T = 11.896
Stock crashes discussion
That is 12 years approximatively.
Remark: A practical rule to compute the investment horizon which is necessary to double the initial amount of money invested is given by:
72 i × 100
Check:
72
= 12 ≈ 11.896
6
Example: You want to buy a house which costs 1 MEUR. You possess actually 800000 EUR. If we assume that the interest rates stand at 12% p.a., how many years would you have to wait to be able to buy this house?
Solution:
10 ln ln (10000000 ) − ln(800000 )
8
T=
= = 1.96 ≈ 2 years ln(1.12) ln (1.12 )
Same question with a yearly rate of return of 3% on investments.
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Remarque: Si les intérêts composés constituent la référence pour
Exemple : Soit un placement de 10000 EUR sur un compte à un
toutes les opérations financières d’une durée supérieure ou égale
taux d’intérêt simple annuel de 5 %, à intérêts simples pendant 3
à un an, pour toutes les opérations de durée inférieure à un an
mois. Au terme des 3 mois, les intérêts perçus sont de :
fois, ne portant pas eux-mêmes intérêts et étant proportionnels à
90
V3mths = 1000 × 1 + 0.05 ×
= 1125
360
la durée totale du placement (Opérations sur le marché
Exemple : Un trésorier d’entreprise place une somme de 500000
monétaire, paiement de coupon couru sur le marché obligataire,
EUR sur un compte rapportant 6% / an, pendant 4 ans et 7
calcul
mois. De combien disposera t-il à l’issu de son placement ?
on utilise les intérêts simples, les intérêts étant payés une seule
d’escompte,
d’agios
bancaires….).
La
valeur
future/acquise sera donnée dans ce cas par la formule suivante
A l’issu de la 4ème année la somme placée est devenue :
V4 = 500000 × (1.06)4 = 631238.48 EUR
(capitalisation à intérêt simple post comptés) :
jours
VT = V0× 1 + i ×
360
Pendant les 7/12 d’année restant à courir cette somme procure des intérêts simples calculés prorata temporis.
où i correspond au taux d’intérêt annuel simple (prorata temporis
Le montant des intérêts acquis sur les 7 mois est de :
Dans le cas où la durée ne correspond pas à un nombre entier d’années, les intérêts sont alors calculés sur un nombre entier
I = 631238.48 × 0.06 ×
7
= 22093.347 EUR
12
d’années selon la technique des intérêts composés et la fraction d’année restante donne naissance à un calcul d’intérêts selon la technique des intérêts simples.
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- Compounding at different frequencies (discrete and continuous compounding): While annual
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i
VT2 = V0 × 1 +
2
2×T
compounding is the norm in finance, other
• Periodic Compounding at the proportional annual
conventions exist however. Yearly interest rate can
interest rate: If the yearly interest rate i is
be
(semi-annual
compounded p times per year, the terminal
compounding, see US and GB bonds), twelve times
value VT of the initial investment V0 is for T
a year (monthly compounding), or either an infinite
years:
compounded
number
of
twice
times
a
in
a
year
year
(continuous
compounding). It is then necessary to define an equivalent basis to compare interest rates computed or compounded on different frequencies. This is done by working with an equivalent yearly return.
p
1 + i
VT = V0
p
i{ p
p ×T
where p is the number of period considered in the year.
• Annual Compounding at effective annual interest rate:
The terminal value VT of an initial amount V0 invested for T years at a fixed yearly interest rate i compounded annually (once a year) is:
VT = V0 × (1 + i )
T
• Semi-annual Compounding at the proportional annual
interest rate: If the yearly interest rate i is now compounded for T years semi-annually (twice a year), the future value VT of the initial investment V0 then becomes:
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Remarque: On appelle le taux annuel i composé p fois l’an le
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• Continuous
compounding
-
daily
compounding
« taux annuel proportionnel ». Ce dernier étant similaire à
(Derivatives, Repo,…): As the frequency of
intérêts simple il ne permet pas de connaître directement le
compounding increases ( p tends to infinity),
montant des intérêts reçu à maturité puisqu’il n’inclut pas l’effet
the yearly interest rate i becomes continuously
de capitalisation des intérêts (les intérêts sur les intérêts). Pour ce
compounded (paid every seconds) and it can
faire on le doit convertir en un taux période effectif i p définit
be shown that the initial investment V0 accrues
comme:
to: ip =
i p VTc
sur lequel on applique alors la formule de capitalisation à intérêts composés. :
VTp
i = V0 1 + p
p×T
i = V0 × eiT
= lim V0 × 1 + p p →∞
where e = 2.71828 represent the exponential function (see infra).
p×T
Remark: For most practical use (Derivatives Pricing, see Black and Scholes, 1973), continuous compounding is equivalent to daily compounding
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• Equivalent interest rates (conversion between annual
Effect of compounding frequency on the value of 1000 EUR at the end of one year when the yearly interest rate is 10%
Compounding frequency
and periodic compounded rate of returns): Two
Value of 1000 EUR at end of one year
interest
(i=10% p.a.)
rates
compounded
at
different
Annually (p=1)
1100
frequencies are equivalent if they lead to the
Semi-annually (p=2)
1102.5
same future value after one year (reference
Quarterly (p=4)
1103.8
period). That is:
Monthly (p=12)
1104.7
Weekly (p=52)
1105.1
i (1 + ia ) = 1 + = (1 + i p ) p p
Daily (p=365)
1105.16
Continuously
1105.17
p
where ia is the effective annual interest rate and i p is the
Where the future value in one year with continuous
effective periodic interest rate.
compounding is obtained as:
♦ Annual equivalent interest rate to a periodic rate (discrete
V1c = V0 × ei = 1000 × e0.1 = 1105.2 EUR
compounding): The effective annual interest rate (compounded
Remark: Don't confuse the base at which an interest rates are measured (annual basis) and the frequency at which it is paid
once a year) i e which is equivalent to the effective periodic a interest rate i p =
i compounded (compounded p times a year) p can then be defined such as: p (annual, semi-annual, continuous,…)
e ia 21
p
ip
ip
= 1 + − 1 = 1 + − 1
p p
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24
e where ia is the annual equivalent interest rate compounded once
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Example: A Treasurer can invest 20 MEUR for one year. He
a year, i is the proportional annual interest rate compounded
considers the two following different commercial propositions.
p times a year and i p is effective periodic interest rate.
A time-deposit with bank 1 at an interest rate of 6.10% p.a. compounded each quarter and a time-deposit with bank 2 at an interest rate of 6.12% compounded semi-annually. What is the
That is : e ia
most interested investment?
= (1 + im ) − 1 if im is a monthly rate;
e ia 12
Investment 1:
= (1 + iq ) − 1 if iq is quarterly rate...
4
To answer the question, we have to determine the annual equivalent interest rate, that is:
Example: What is the annual equivalent rate of a proportional annual rate of 2% compounded quarterly?
0.02 e
1 + ia = 1 +
4
e ia 4
4
i
0.061
= 1 + 3 − 1 = 1 +
− 1 = 6.241%
4
4
4
=TAUX.EFFECTIF(0,061;4)
e
⇒ ia = (1.005) − 1 = 2.015%
Investment 2:
4
Any agent will be indifferent between an investment with an annual rate of 2.015% compounded once a year and an
2
0.0612 e ia = (1 + i6 )2 − 1 = 1 +
− 1 = 6.214%
2
investment with an annual rate of 2% quarterly compounded.
=TAUX.EFFECTIF(0,0612;2)
We choose the first solution!!!
see fonction finances Excel Taux.Effectif.
The equivalent annual interest rate is higher than the
=TAUX.EFFECTIF(0,02;4)
proportional annual interest rate.
The difference increase with the compounding frequency
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♦ Periodic equivalent rate to an effective annual rate (discrete
♦ Continuous equivalent rate to an effective annual interest rate:
compounding): The periodic equivalent interest rate i e p e
The continuous annual interest rate ic which is equivalent to the
compounded p times a year to an effective annual rate
effective annual interest rate i is defined such as: e ic = ln(1 + i )
compounded once a year ia is given by: ip = e i
e
p
= (1 + ia )
These equations are used to convert interest rates
1/ p
−1
compounded
where ia is the annual interest rate compounded once a year and i e is the periodic equivalent interest rate compounded p times a p year.
at
different
frequencies
per
annum
(annually).
Example: Consider an effective annual interest quoted at 8% per annum. What is its equivalent continuously compounded yearly
That is:
[
i = p × (1 + ia ) e rate?
]
−1
1/ p
e ic = ln(1 + i ) = ln(1.08) = 0.0769
That is 7.69% p.a. .
♦ Annual equivalent rate to a continuous rate: The effective
Example:
e annual interest rate ia which is equivalent to the proportional
Suppose that a bank quotes the interest rate on 10-year maturity
annual interest rate continuously compounded (every second) ic
loans at 6% with continuous compounding. What is the equivalent annual rate?
can then be defined such as:
(1 + i ) = e i = (e
(
e a ic
) (
)
e
⇒ ia = eic − 1 = e0.06 − 1 = 0.0618
ic
e a )
That is 6.18% p.a. .
−1
Remark: For fixed present and future values, increasing the frequency of the compounding decrease the annual equivalent yearly interest rate
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- Compounding a sequence of cash-flows (variable and constant):
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♦ Proof: Let’s consider at date 0 an integer number T of annual
periods and assumes that an agent invest at each date t=1,..,T, a
• Definition: We refer to a sequence of annuities,
any sequence of regular payments at . If the
cash-flow at .
Diagram of cash-flows:
cash-flows are paid at the end of each period,
a1 (1 + i )T −1
ordinary
a2 (1 + i )T − 2
the
annuities
are
labelled
as
annuities (bank loan repayment). If the cash-flows are paid at the beginning of each
a1
a2
1
2
aT
period, they are labelled as immediate annuities (life-insurance or saving plans). I
0
• Future value of a sequence of ordinary annuities :
T
♦ Proposition : The future value in t = T of a sequence of
That is:
different monetary cash-flows, denoted by at , invested at the
VT = ∑ at (1 + i )
T
t =1
T −t
=
constant interest rate i (flat term structure of interest rates), is
(T-1 ) − th payment at interest rate i for 1 year
flows at i, that is:
T
aT −1×(1+i )1
a2 ×(1+i )T − 2
+
first payment compounded at interest rate i for (T-1) years
+L+
given by compounding each element of the sequence of cashVT = ∑ at × (1 + i )
a1×(1+i )T −1
second payment compounded at interest rate i for (T- 2 ) years
+
aT last payment
Remark: If the annuities can be free (general case), more often they are
T −t
constant or increasing at a constant rate following an arithmetical or
t =1
Remark: We are assuming here that the cash-flows at are paid at
geometrical progression.
the end of each period and compounded at the unique fixed interest rate i for the remaining periods.
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• Future value of a sequence of immediate annuities:
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
• Simplification (Compounding a sequence of identical
♦ Immediate annuities : The future value in t = T , of a sequence of
cash-flows) : see insurance contracts
different monetary cash-flows, denoted by at , invested at the constant interest rate i (flat term structure of interest rates), is
♦ Future value of a sequence of constant ordinary
given by compounding each element of the sequence of cash-
annuities:
flows at i, that is:
T
V 'T = VT × (1 + i ) = ∑ at (1 + i )
Proposition: The future value in t = T of a sequence of constant
T +1−t
annuities a (with at = a ), compounded at a constant interest rate
t =1
We are assuming here that the cash-flows at are paid at the beginning of each period and compounded at the unique fixed interest rate i for the remaining periods.
♦ Proof:
I (flat term structure of interest rates), is given by :
(1 + i )T − 1
VT = a ×
i
Proof: We are now assuming that the future payments a are the
Same line of reasoning…
same at the end of each period and compounded at the same interest rate i for the remaining number of periods.
The future value of this compounded sequence of constant cashflows VT after T periods is given by:
VT = a × (1 + i )T −1+ a × (1 + i )
T −2
+ K+ a
This expression corresponds to the sum of the T first terms of a geometric progression.
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(1.045)30 − 1
V30 = 100 ×
0.045
= 6100,71 EUR
Math Review: The sum of the n first terms of a geometric progression with first term u1 and common ratio q, denoted Sn , is given by :
See fonction finances Excel VC.=VC(0,045 ;30 ;-100 ; 0)
S n = u1 + u2 + K + un
(
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
)
1 − qn
= u1 × 1 + q + K + q n −1 = u1 ×
1− q
♦ Future value of a sequence of constant immediate annuities: If the cash-flow a is invested at the beginning of
Future value of a sequence of ordinary annuities:
VT = a (1 + i )T −1+ a (1 + i )
T −2
each period, the future value after T years is given by:
+K+ a
(1 + i )T − 1
VT' = (1 + i ) × VT = a (1 + i ) ×
i
Sum of the first T terms of a geometric progression with common ratio (1 + i) and first term a. That is:
Cf. supra.
1 − (1 + i )T
(1 + i )T − 1
VT = a ×
= a×
−i i
4243
1
• Future value of a sequence of constant
ordinary annuities, k periods after the last
annuity factor
t = T of a
Example: You can deposit each year 100 EUR in an interest
investment: The future value in
bearing bank account without withdrawals for 30 years at an
sequence of constant annuities a (with at = a ),
annual interest rate of 4.5%, starting in one year (ordinary
compounded at a constant interest rate i (flat
annuities). How much will you received at the end of your
term structure of interest rates), k-periods after the
investment?
last T-th investment is given by:
We have to compute the future value of a sequence of 30 ordinary annuities of 100 Euros compounded at an annual
(1 + i)T − 1
VTk = a (1 + i) k ×
i
interest rate of 4.5%, that is:
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Exemple : Si l’épargnant considéré précédemment cesse ses
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1.2. Discounting, Present Value and Asset valuation
versements à l’issue de la cinquième année, la valeur acquise du capital au bout de 30 ans est égale à :
V525
- Discounting of a single cash-flow:
(1,045)5 − 1
= 100 (1,045) ×
0.045
= 1644,19 EUR
• Use: Discounting determines what initial
25
amount must be invested today in order to obtain a given future value (the present value)
Cf. Excel.
or if we adopt the point of view of the
=(1+0.45)^25*VC(0,045 ;30 ;-100 ; 0)
borrower, the maximum amount of money that can be borrowed initially if we want to be
Exemple : Vous avez besoin de 50.000 EUR dans 7 ans. Vous
able to meet a given future repayment. This
prévoyez de faire 7 versements identiques chaque (fin) année, sur
approach permits to compare stream of cash
un compte qui rapporte du 11% par an capitalisé annuellement.
flows perceived at different dates.
Quel doit être le montant de chaque versement annuel ?
• Proposition : The present value V0 of a future
1) Diagramme de flux
fixed cash-flow in T years VT is the amount of
2) Réponse :
money that must be invested today at a fixed
(1.11) − 1
V7 = 50000 = a ×
0.11
7
yearly interest rate i in order to obtain VT in
T years, that is:
0.11
⇒ a = 50000 ×
= 5110.76 EUR
7
(1.11) − 1
V0 =
VT
(1 + i )
T
= VT × (1 + i )−T
• Proof : The future value of an initial investment
Cf. fonction finances Excel VC.
=VPM(0,11 ;7 ;0 ; 50000)
V0 , compounded annually at the yearly interest rate i for T years is:
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VT = V 0× (1 + i )T
The initial investment of VT / (1 + i )T
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⇒ V0 =
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
V1
1000
=
= 961.54 EUR < 1000 EUR
(1 + i ) 1.04
at the annually
compounded yearly interest rate i over T years gives:
Example: As a Treasurer, you know that your company will
VT
T
× (1 + i ) = VT
T
(1 + i )
receive 10 MEUR in 5 years. What is the maximum amount of
Which is precisely the future value VT that we want to obtain.
year borrowing interest rate is actually 4% (lending conditions)?
QED.
The maximum amount of money that the company can borrow
Remark: If the maturity of the operation is less than one year, the
V0 is the present value of 10 MEUR repaid in 5 years (principal
funds that you can be borrow from the company's bank if the 5-
present value will be given by:
V0 =
plus interests), that is:
VT days
1 + i ×
360
V1 = 10 MEUR = V0 × (1 + i )5
⇒ V0 =
where i corresponds to the simple annual interest rate (prorata
V1
(1 + i )5
=
10 MEUR
(1.04)5
≈ 8.22 MEUR
temporis).
Remark: If the yearly interest rate i is continuous (continuous compounding Vs discrete compounding, see supra), the present
See price of T-Bills, CDs,…
value V0 of a future cash-flow VT becomes:
Example: You need 1000 EUR in one year. What is the amount of money that you must invest today if interest rate for one year is 4%? The amount of money to invest V0 is the present value of
V0 =
VT e iT
= VT × e−iT
1000 EUR in one year, that is:
V1 = 1000 = V0 × (1 + i )
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• Proof :
Example: Same as before but with a 5-year borrowing interest rate continuously compounded at 4%.
Diagram of cash flows:
V0 = 10 MEUR × e − 0.04×5 ≈ 8.18 MEUR aT aT
- Discounting a stream of future (constant) cash-
(1 + i)T
• Proposition : The present value at t = 0 of a
sequence of different cash-flows at (with t = [1,..., T ]) discounted at the constant annual
interest rate I, is given by discounting each
(1 + i)t
T at = ∑ at
1 + i t t =1
(
)
a3
M a1 (1 + i)
cash-flow of the sequence at this rate, that is :
T
V0 = ∑ t =1
at
at
flows:
a2 a1 1
× (1 + i )−t
2
3
t
…
T
Since any future cash flow at has by definition a present value
where we make here the assumption that the interest applicable
equal to: at to 1, 2, 3, ..T year horizons is constant and equal to i (flat and
(1 + i )t
constant term structure of interest rates).
This present value is exactly the sum that must be invested
(obtained) today if one wants to obtain (pay) the stream of cash flows at when the constant annual interest rate is i .
The present value of the stream of cash-flows at , with t = [1,..., T ], is:
Vo =
=
37
a1 aT +L+
(1 + i )
(1 + i )T
T
∑ t =1
at
(1 + i )t
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• Simplifications:
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
1 T
1 −
1
1 + i
V0 = a ×
×
(1 + i ) 1 − 1
1 + i
♦Sequence of constant cash-flows (bullet coupon bond, see
infra): If the future cash-flows are, that is if at = a ,
∀t = [1,K, T ], the present value becomes:
1 − (1 + i )−T
= a×
1+ i −1
at t t =1 (1 + i ) a a a =
+
+K+
2
(1 + i ) (1 + i )
(1 + i )T
T
V0 = ∑
So that:
1 − (1 + i )−T
V0 = a ×
i
1
1
1
= a×
+
+K+
2
(1 + i )T
(1 + i ) (1 + i )
Example: An individual want to finance his housing by a 15-year
1
1
1
= a×
× 1 +
+K+
T −1
(1 + i ) (1 + i )
(1 + i )
loan at an annual interest rate of 4%. The loan is repaid by a sequence of constant monthly interest payments. If we assume
where the terms in the bracket correspond to the sum of the T first terms of a geometric progression of common ratio 1 (1 + i) and first term 1.
that the agent can repay 1800 Eur/month, what will be the maximum amount that he can borrowed today from the bank?
First, compute the monthly interest rate, that is:
Math review: The sum of the n first terms of a geometric
im 0.04
=
= 0.0033
12 12
progression of first term u1 and common ratio q is given by:
From the present value equality we have:
1 − qn
S n = u1 ×
1− q
V0 =
So that:
1800
1800
1800
+
+K+
180
0.04 0.04 2
0.04
1 +
1 +
1+
12
12
12
where V0 corresponds to the maximum amount of borrowing.
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From the formula of the present value of a sequence of constant
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That is in thousand of euros:
1 − (1.08)− 6
1 − (1.08)− 2
1
+ 300 ×
×
V0 = 500 + 200 ×
(1,08)6 0.08
0.08
≈ 1.762 M EUR
cash-flows a , we get:
0.04 −180
1 − 1 +
12
V0 = 1800 × 12 ×
0.04
= 243 345.87 EUR
Example: An agent must repay in four years 200000 EUR. He wants to postpone the repayment of its debt in 10 years. What
See fonction finances Excel VA.
will be the new amount of money X at he will have to repay in
10 years (6% interest rate)?
=VA(0,04/12 ;-1800)
Two sequences of cash-flows are equivalent if they possess exactly the same present value, that is:
Example : An investor purchase a plot of land by paying 500000
200000
X
=
4
(1.06) (1.06)10
EUR cash, 200000 EUR each year for the six following years and
300000 EUR year 7 and 8. The discount rate for this project is equal to 8% (opportunity cost if the investment is financed by the investor’s capital or borrowing cost if the investment is
So That :
X = 200000 × (1.06 ) = 283703.82 EUR
6
financed by borrowing). What is the present value of the land?
Using the present value formula, we have:
8
6
1
1
V0 = 500000 + 200000 × ∑
+ 300000 × ∑ t t
t = 7 (1.08)
t =1 (1.08)
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♦Infinite sequence of cash flows:
Proof: Si g est le taux de croissance constant à l'infini des
a) Discounting an infinite sequence of constant cash-flows:
versements et a est le premier versement effectué à la fin de la
If the sequence of constant cash flows is infinite, the present
première année, la valeur actuelle est alors donnée par : a a × (1 + g ) a × (1 + g )2 a × (1 + g )T −1
V0 =
+K
+K+
+
+
(1 + i ) (1 + i )2
(1 + i )3
(1 + i )T
value can be simplified as:
1 − (1 + i )−T a
V0 = lim a ×
= i T →∞
i
b) Discounting an infinite sequence of cash-flows progressing at a constant growth rate: If the sequence of cash flows is infinite, with each individual cash-flow progressing at a constant growth rate g (perpetuity), the present value can be
(1 + g ) (1 + g )2
(1 + i )T −1 + K a +
=
× 1+
+K+
2
(1 + i ) 144i44(1 + i42444+ i4−44
(1 4 )T 1 3
(1 + ) 44) somme infinie de termes d' une suite
géométrique
Somme infinie de termes issus d'une suite géométrique de premier terme 1 et de raison (1 + g ) / (1 + i ), i.e. :
1 + g T
1 − a 1+ i
× lim
V0 =
(1 + i ) T →∞ 1 − 1 + g
1+ i
simplified as:
V0 =
a
(i − g )
With i the discount rate and g the growth rate.
si i > g , la série versements est convergente et on obtient finalement :
V0 =
a
(i − g )
Cf. Gordon-Shapiro.
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Exemple : Vous souhaitez déterminer la valeur d’un bien
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
- Financial pricing applications (Bonds, Stocks and
Linear Derivatives):
immobilier locatif d’une surface de 100m2. On suppose que le loyer annuel courant (charges déduites) est de 20.000 EUR/an.
• Bond pricing: A bond is a long-term debt
Le taux d’actualisation est de 6% et le taux de croissance des
instrument where the issuer (borrower) has the
loyers est de 2% par an. A quel bien accepteriez-vous d’acheter
contractual
ce bien ?
bondholder (lender) regular fixed interest
obligation
to
make
to
the
payments (coupon payments) and to repay the
La valeur de ce bien correspond à la valeur actuelle de ses cash-
bond’s principal (face value) at the maturity
flows futurs, soit en supposant une durée de vie infinie :
date. The price of a bond must always (on the
V0 =
20000
= 500 000
(6% − 2% )
primary or the secondary market) be equal to the discounted present value of its stream of
Soit 5000 EUR le m2 !!!
cash flows (i.e. coupon payments and principal repayment). But what kind of discount
rate(s) do we have to use?
♦ The economic valuation and the term structure of spot rates
(exact valuation): Any bond can be decomposed as a portfolio of unique cash-flows, so that the equilibrium value of a bond must be equal to the sum of the value of each cash-flow discounted by its appropriate spot interest rate (zero coupon bond annual interest rate), that is:
P0 =
C
C
C+F
+
+K+
2
1+ 0 i1 (1+ 0 i2 )
(1+ 0 iT )T
with:
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C = c× F
Remark: It is also possible to use the term structure of the
where c is the coupon rate, C is the coupon payment and 0 it
implied forward rates to value the stream of coupons (interest
the t-period annual spot interest rate for a maturity of
rates implied by today’s spot interest rates, that are applicable in
t = [1,....,T ] years (rate of interest that is realized on an
one point of time in the future to another point of time in the
investment that starts today and lasts for t-periods with no
future), so that:
intermediate payments.
P0 =
C
C
C+F
+
+K+
1+ 0 i1 (1+ 0 i1 )(1+ 0 f1,1 )
(1+ 0 i1 )(1+ 0 f1,1 )K 1+ 0 fT −1,1
(
)
Diagram of cash flows for a typical coupon bond (plain vanilla or straight
where 0 f s , n − s is the actual annual forward interest rate that starts
bond):
at a future date s and ends at n with a maturity of (n − s ).
F
+C
+ C ……………………
+C
− P0
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♦ The financial valuation and the yield to maturity
The yield to maturity can be interpreted as the average
(approximate): In practice since the determination of the array of
annual return obtained on a bond bought at the initial
all spot /zero coupon bond interest rates is a very difficult and
market price P0 and held until maturity (realized yield
long task, traders prefer to use only one constant interest rate –
assuming that all intermediate coupon payments will be
i.e. the yield to maturity of the most newly issued bonds on the
reinvested at that yield, that is a flat and constant term
primary market ("on the run" issues) - to evaluate the price of
structure of spot interest rates).
old issued bonds which the same characteristics traded on the secondary market ("off the run" issues).
The YTM acts in fact as a “proxy” for the whole term
1) Definition yield to maturity (YTM): The yield to maturity on a
structure of the discount rates (complex average of the spot
bond is the single/constant discount rate, denoted y , which
discount rates).
makes the present value of the bond’s payments equal to its market price (internal rate of return or IRR of the series of
One should be emphasized that an YTM is not a very
the bond’s cash-flows), that is:
meaningful number. This is because there is no reason one should discount cash-flows occurring at different dates with a unique discount rate. The YTM is just a convenient
Ct
N
+ t t =1 (1 + y )
(1 + y )T
T
y / P0 = P0 ( y ) = ∑
way for practitioners to express bond prices (much like the implied volatility for options).
T
Ct
N
+
=0
⇔ NPV0 ( y ) = P0 − ∑ i i = t (1 + y )
(144T
+ y)
144 24 3
4
P0 ( y )
with y the yield to maturity of the bond
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2) The most newly issued bond yield to maturity and secondary bond pricing
(relative bond pricing): The price of a bond on the secondary market can then be obtained by taking the yield-to-maturity on
Primary / benchmark market:
the most newly/active issued bond (government or corporate
P0new
benchmark) which possesses the same characteristics (maturity,
yInew
Secondary Market:
coupon rate, default risk, tax rate,…), that is:
Ct
N
+
t t =1 (1 + y )
(1 + y )T
T
P0 = P0 ( y ) = ∑
P0old
yInew
where T represents this time the residual life of the old-issue
See:
bond and y is the yield to maturity of the new issued bond (the
http://www.aft.gouv.fr/aft_fr_23/dette_etat_24/adjudicati
market interest rate).
ons_98/dernieres_adjudications_101/index.html
Remark: Do we use the discount rate(s) / yield to maturity to
http://www.aft.gouv.fr/article_308.html?id_article=308&id
obtain bond prices or the bond prices to infer the implied
_rubrique=100
discount rate(s) / yield to maturity. The answer of this “chickenand-egg” question depends on the situation. Roughly speaking,
http://www.treasurydirect.gov/RT/RTGateway?page=ins
one would use the competitive (auctioned) market prices on the
titHome
benchmark securities as given and infer the implied discount rate(s)/yield to maturity; and then use that market information
(the government yield or swap rate curve, see below) to price any other fixed-income security .
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Example: Consider a 30-year coupon bond with a principal value
Example: Consider a 10-year residual maturity, 4% coupon bond
of 2000 EUR, 6% annual coupon rate and a residual maturity of
with a par value of 2000 EUR. Suppose that the required yield
10 years. Assume that the interest rate on the most newly issued
on the most newly issued bond for a maturity of 10 years and a
bond for a maturity of 10 years and a similar class of risk stands
similar class of risk stands at 3%.
actually at 3%. At what price are you ready to buy (sell) the first
a) At what price are you ready to buy (sell) the first bond on the
bond on the secondary market?
secondary market?
First, we have to compute the coupon payments on the first
b) What will be the impact of semi-annual payment instead of
bond, that is:
annual one?
C = c × F = 2000 × 0.06 = 120 EUR
a) The annual or periodic coupon payment is:
C = 2000 × 0.04 = 80 EUR
Interest rate 3%:
The price of the bond is equal to the sum of the present values
C
F
+ t (1 + y ) (1 + y )T
120
120
120
2000
=
+
+K
+
2
10
(1,03) (1.03)
(1.03) (1.03)10
P0 =
T
∑
t =1
of its cash flows, that is:
C
F
+
t
(1 + y ) (1 + y )T
80
80
80
2000
=
+
+K
+
2
10
(1,03) (1.03)
(1.03) (1.03)10
P0 =
1 − (1.03)
2000
= 120 ×
+
0.03 (1.03)10
2000
= 120 × 8.53 +
(1.03)10
= 2511.81 EUR
−10
T
∑
t =1
1 − (1.03)−10
2000
= 80 ×
+
0.03 (1.03)10
2000
= 80 × 8.53 +
(1.03)10
= 2170.6 EUR
see fonction finances Excel VA.
=VA(0,03;10 ;-120 ;-2000)
See fonction finances Excel VA.
=VA(0,03;10 ;-80 ;-2000)
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♦ The inverse relationship between the market bond prices
b) The semi-annual or periodic coupon payment is:
C = 2000 ×
EFJ Copyright11
0.04
= 40 EUR
2
(quoted on the secondary market) and market yields (on new or most newly issued bonds on the primary market): A fundamental
The price of the bond is equal to the sum of the present values of its cash flows that is:
property that characterizes the bond market, is that there exists an inverse relation between the quoted market price of any
1 − (1.015)
2000
P0 = 40 ×
+
20
0.015
(1.015)
2000
= 40 × 17.57 +
(1.015)20
= 2171.69 EUR
− 20
traded bond (secondary market) and the required market yield on the most newly issued bonds (primary market). That is the bond market prices change in the opposite direction from the change in the required yields:
See fonction finances Excel VA.
1) Present value explanation: The price of a bond is the present
=VA(0,03;10 ;-80 ;-2000)
value of its cash flows. As the required market yield increases,
We will retain y as a synthetic indicator of the interest rates level the present value of the future cash flows decreases, hence the quoted market price decreases. The opposite is true when the required yield increases. If we name CFt , the cash flow of a bond (coupons and redemption price), the price of a bond traded on the secondary market becomes:
T
CFt
−t
= ∑ CFt × (1 + y ) t t =1 t =1 (1 + y )
T
P0 = ∑
y ↑ P0 ↓
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2) Economic explanation: When at a given point in time the interest
resulting lack of demand and the excess offer would cause the
rate on the most newly issued comparable bond(s) is increasing
price of the old issued bonds to fall and thus their yield to
(decreasing), the price on the old issued bond must decrease
maturity to rise.
(increase) to the point when the old issued bond gives to its bondholder the same yield to maturity than the most newly
When the required yield on the most newly issued
issued one(s). Indeed, as the required yield on the newly issued
comparable bond(s) is below the coupon rate of an old issued
bond(s) changes in the market place, the only variable that can
bond, the price of the old bond must sell above its face value.
change to compensate an investor holding old bonds is their
This is because if investors have the opportunity to purchase the
market prices (the coupon rate is fixed by definition).
old issued bond(s) at par value, they would get a coupon rate in excess of what the market actually required. As the result,
When the coupon rate on an old issued bond is equal to the
investors will bid up the prices of the old issued bonds because
required market yield on the most newly issued comparable
their yields are so attractive, up to the point where the old bonds
bond(s) (same maturity, risk,..), the price of the old issued bond
offer the market required rate.
will be equal to its face value.
⇒ old bond price on the run required yield ⇒ old bond price >face value
Coupon rate old bonds < on the run required yield
When the required yield on the most newly issued comparable bond(s) (same maturity, risk,…) rises above the coupon rate of an old issued bond, the price of the old bond must adjusts so that the investor contemplating its purchase can realize some capital appreciation by holding the bond until maturity in order to compensate for a coupon rate which is lower than required. If not, nobody will want to buy the old issued bond(s) since they offer a below market yield. The
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Example: Take the previous example, but assume now that the
The 1% interest rate increase (100 basis points or bps) is
market interest rate is increasing from 4% to 5%, that is:
associated with a price decrease of:
Po =
∆P = 1845.57 − 2000 = −154.43 EUR
80
80
2080
+
+K+
2
(1,05) (1.05)
(1.05)10
1 − (1.05)−10
2080
= 80 ×
+
10
0.05 (1.05)
2080
= 80 × (7.72) +
(1.05)10
= 1845.57 EUR
Price
(EUR)
Bond price Vs interest rate
see fonction finances Excel VA.
=VA(0,05;10 ;-80 ;-2000)
Yield(%)
The value of a bond is a non-linear (convex) decreasing function of the yield to maturity.
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Remark: There exist different approaches to get y in the valuation
Example: Consider a 4-year residual maturity, 10% coupon bond
formula, such as:
with a par value of 1000 EUR, selling on the market at 968.98
1. Trial and error (iterative method): Try different interest rates
EUR. What is the YTM of this bond?
until the present value of the cash flows equals its market
To compute the yield to maturity, different discount rates must
price.
be tried until the present value of the cash flows is equal to
2. Linear interpolation (Galbrun): Let y = f ( x) and two
903.10 (90.31%).
points (x1 , f (x1 )) and (x2 , f (x2 )) which surround the value
Trying an annual interest rate of 10% gives:
y = f ( x) , with x1 < x2 . Assuming the equality of the slopes between x1 and x2 and y and x1 we get a linear approximation of y :
P(10% ) =
100
100
100
100
1000
+
+
+
+
2
3
4
1.10 (1.10 ) (1.10 ) (1.10 ) (1.10 )4
1 − (1.10 )− 4 1000
= 100
+
4
0.10 (1.10 )
= 1000 > 968.98 = P0
y − f (x1 ) f (x2 ) − f (x1 ) y − f (x1 )
⇒ x = x1 + (x2 − x1 ) ×
=
x2 − x1 x − x1
f (x2 ) − f (x1 )
Because the present value computed using a 10% discount rate is
Which in the case of the determination of the yield to maturity is
higher than the market price, a higher interest rate must be
equivalent to:
considered (inverse relation between price and interest rates).
0 − NPV ( y1) y = y1 + ( y2 − y1) ×
NPV ( y2 ) − NPV ( y1)
Trying an annual interest rate of 12%, we obtained:
1 − (1.12 )− 4 1000
P(12% ) = 100 ×
+
0.12 (1.12 )4
= 939.25 < 968.98 = P0
3. Newton-Raphson algorithm (iterative method): yn +1 = yn −
NPV ( yn )
NPV ' ( yn )
Since at 12% the present value of the cash flow is less than the
Two trials are generally sufficient.
market quoted price, a lower interest is needed.
Trying an annual interest rate of 11%, we obtained:
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1 − (1.11)− 4 1000
P(11% ) = 100
+
0.11 (1.11)4
= 968.98 = P0
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For y1 = 6%, we get:
NPV (6% ) = f (0.06 ) = 3.99
For y2 = 7%, we have:
NPV (7% ) = f (0.07 ) = −4.05
Therefore the YTM on this bond is 11%
see fonction finances Excel TRI
Using linear interpolation we obtain:
=TRI(B5:B9)
y ≈ 0.06 + (0.07 − 0.06 )
Example: Consider a 12-year maturity, 6 % coupon bond with a par value of 1000 EUR, issued at discount price 970 EUR and
0 − 3.99
= 6.49%
− 4.05 − 3.99
Iterating the process, we obtain for YTM:
6.48%
repaid at premium at 1020 EUR. What is the YTM of an investor who purchases this bond and holds it until maturity?
The yield received by the investor is higher than the coupon
Since the definition of the YTM, we are searching y such as:
yield since the bond has been issued at discount and redeemed at
970 =
premium.
60
60
1020
+
+K+
2
(1 + y ) (1 + y )
(1 + y )12
1 − (1 + y )−12
1020
− 970 + 60
=0
+
12
y
(1 + y )
1 − (1 + y )
− 97 + 6 y
−12
see fonction finances Excel TRI
=TRI(B12:B24)
102
=0
+
(1 + y )12
Let's write:
1 − (1 + y ) f ( y ) = −97 + 6 y
−12
102
+
12
(1 + y )
We are searching the value of y such as:
f (y) = 0
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• Stock pricing (Gordon Shapiro, 1956): A stock
Démonstration : Le point de départ de cette approche porte sur la
gives to its holder a stream of dividend
définition du taux de rentabilité attendu d'un investissement en
payments (distributed profits). Its market price
action au cours d'une période :
is also equal to the discounted value of its stream of future cash flows (i.e. dividend
La rentabilité espérée d'un placement en action sur une période à
payments), that is:
deux composantes : Le dividende et une plus ou moins-value en capital, i.e. :
[1 + E (R )] = E0 (D1 ) + E0 (P1 )
E0 (Dt ) E0 ( PT ) ∞ E0 ( Dt )
+
=∑ t t =1(1 + k )
(1 + k )T t =1(1 + k )t
T
P0 = ∑
P0
Les agents n'accepteront d'investir dans une action que si le taux
with:
lim
t →∞
Pt
(1 + k )t
de rentabilité attendu est égal au taux de rentabilité exigé par le
=0
marché compte tenu du risque que fait subir ce titre à son détenteur, i.e. :
where P0 is the initial stock price, E0 (Dt ) is the expected resell price of the stock at terminal date T , E0 (P ) is the expected
T
dividend of the stock received in period t and k is the required
[1 + E (R )] = E0 (D1 ) + E0 (P1 ) = (1 + k )
P0
avec k taux de rentabilité exigé par les actionnaires.
expected rate of return by the investors (CAPM or APT).
De cette égalité on tire la valeur du prix courant de l'action :
P0 =
E0 (D1 ) + E0 (P )
1
(1)
(1 + k )
En appliquant le même raisonnement en t = 1 on obtient l'expression de E0 ( P ), soit :
1
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E0 ( P ) =
1
68
E0 (D2 ) + E0 (P2 )
( 2)
(1 + k )
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If we assume moreover that the dividends progress at a constant growth rate, denoted g , and that the first expected dividend payment is E0 (D1 ), the price of the stock is then equal to (see
Soit en remplaçant dans l'équation (1) :
E0 ( D2 ) + E0 ( P2 )
E ( D ) + E0 (P )
(1 + k )
1
P0 = 0 1
=
(1 + k )
(1 + k )
E ( D ) E (D ) + E0 (P2 )
= 0 1 + 0 2
(3)
(1 + k )
(1 + k )2
E0 ( D1 ) +
perpetuity):
P0 =
E0 ( D1 ) E0 (D2 ) E0 (D3 )
E (D )
K + 0 nn + K
+
+
2
3
(1 + k ) (1 + k ) (1 + k )
(1 + k )
=
E0 (D1 ) E0 (D1 ) × (1 + g ) E0 (D1 ) × (1 + g )2
E (D ) × (1 + g )n −1
+K
+K+ 0 1
+
+
(1 + k )
(1 + k )2
(1 + k )3
(1 + k )n
=
(1 + g )n −1
E0 (D1 ) (1 + g ) (1 + g )2
+ K
+K+
× 1 +
+
(1 + k ) (1 + k ) (1 + k )2
(1 + k )n −1
En généralisant, on obtient alors :
P0 =
∞ E (D )
E0 ( D1 ) E0 (D2 )
E (D )
+
+ K + 0 ∞∞ + K = ∑ 0 t t
2
t =1 (1 + k )
(1 + k ) (1 + k )
(1 + k )
avec : lim
T →∞
E0 ( PT )
(1 + k )
T
(6)
where the sequence of terms in the bracket corresponds to an infinite sum of the n first terms of a geometric progression with
=0
a common ratio (1 + g ) / (1 + k ) and first term equal to 1.
That is (see consol):
1 + g n
1 −
1 + g n
E0 ( D1 )
1 + k E0 (D1 )
× lim
P0 =
=
× lim 1 −
(1 + k ) n → ∞ 1 − 1 + g (k − g ) n →∞ 1 + k
1+ k
Provided that the dividend growth rate is lower than the required rate of return, k > g , the sum converges and the stock price can be expressed as:
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E0 (D1 )
(k − g )
D × (1 + g )
= 0
(k − g )
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Exemple : Un analyste suivant la société CCC effectue les
P0 =
prévisions suivantes : pour l'année prochaine il s’attend à un dividende de 1.2 EUR, ainsi qu’un taux de croissance des dividendes constant de 5%. Si le taux de rentabilité exigé par les
with:
actionnaires est de 7.7%%, a quel prix l'analyste évaluera le prix
E0 (D1 ) = D0 × (1 + g )
de l'action de cette société?
P0 =
(Gordon-Shapiro DDM formula)
1.2
= 44.4 EUR
(0.077 − 0.05)
Remarque : Le dividende correspondant à la part du bénéfice
Remark: From the DDM pricing formula one can infer the
distribué aux actionnaires, soit ;
expected rate of return by the investors on stocks as:
Dt =
E (D ) k= 0 1 +g
P0
bénéficest
×
nbre 44 444 d' actions émises
14
2
3
dt
{
taux de distrib
BPAt
On en déduit que pour augmenter les dividendes, trois stratégies sont possibles : soit augmenter les bénéfices, soit accroître le
See Market Timing applications.
taux de distribution des dividendes ou encore réduire le
nombre d’actions en circulation (rachat d’actions).
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• Derivatives
pricing:
Linear
and
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Terminal payoff on a long forward contract position
non-linear
derivatives instruments:
♦ Forward (linear derivatives): A forward or a futures contract is
ΠT
K > ST
K = ST
K < ST
an instrument which gives to the holder the obligation to buy
(long position) or to sell (short position) at a specific time in the future T (maturity date) an underlying asset for a certain delivery
Slope = +1
price (the forward delivery or strike price), denoted K , which is
0
fixed today.
Payoff = ST - K
a)Terminal pay-off associated with a long forward/futures
Slope = +1
K
Exercise
price
Payoff = ST - K
ST
position: The terminal payoff associated with a long position in a forward/futures position is given by:
ΠT = (ST − K )
Remark: It is useful to designate the terminal payoffs on long or where ST is the terminal price of the underlying asset and K is the strike (forward delivery) price for a maturity of T years.
short derivatives positions in terms of directions vectors (see also below complex trading strategies with options) where + 1, 0 and -1, indicate respectively a positive correlation between the terminal payoff of the derivatives and the terminal price of the underlying asset (a rise/fall in the underlying asset price results in a rise/fall in the derivatives payoff); a zero correlation and a negative correlation (a rise/fall in the underlying asset price is associated with a fall/rise in derivative payoff).
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b)Terminal pay-off associated with a short forward/futures
c) Forward strike price: The forward delivery price K of a
position: The terminal payoff associated with a long position in
forward contract is generally equal to the fair value of the
a forward/futures position is given by:
forward price (e.g. the cost of carrying the underlying asset until
Π T = ( K − ST )
maturity of the forward contract), so that the value of the
where ST is the terminal price of the underlying asset and K is
forward contract at inception is equal to zero, that is for a non-
the strike (forward delivery) price for a maturity of T years.
income paying underlying asset:
K = F0,T = S 0 × e r T
Terminal payoff on a short forward contract position
K > ST
where S0 is the actual spot price of the underlying asset, T is the
K < ST
K = ST
maturity of forward contract, and r is the continuous interest
ΠT
rate for T years (constant interest rate hypothesis).
Indeed suppose that:
Slope = -1
K > S0 × e r T
0
Payoff = K- ST
Slope = -1
K
Exercise
price
Payoff = K-ST
ST
An investor can borrow S0 Euros for T periods at the risk-free rate r to buy one unit of the underlying asset and take simultaneously a short position in the forward contract (sell the forward). At maturity, the asset is sold under the terms of the forward contract for K and S0 × e rT is used to repay the loan leading at time T to an arbitrage profit of:
(K − S
73
0
)
er T > 0
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[
)
]
f 0 S 0 × e r ×T , T = e − rT EQ [(ST − K )] = e − rτ S 0 × e r ×T − S 0 × e r ×T = 0
Using the same line of reasoning it is possible to show that it is not possible to have (short the underlying and be long the
Remark: The value of a forward contract at inception with a
forward): K < S0 × e r T
d) Forward contract value: To avoid arbitrage opportunities,
strike price of K = S0 and a maturity date of T years is not equal
the value at time t of a forward contract with a strike price K
to zero since:
must be equal to the discounted value of its expected terminal
[
]
(
f 0 (S 0 , T ) = e − rT EQ [(ST − K )] = e − rτ S 0 × e r ×T − S 0 = S 0 × 1 − e − r T
)
payoff under the risk-neutral probability distribution, that is:
Remark: when the underlying asset is paying an additional income
f t ( K , T ) = e − r τ EQ [(ST − K )] = e − rτ × (Ft ,T − K )
(dividend/coupon and stock/bond loans), the fair forward delivery price must be modified such as:
with:
K = S 0 × e (r − g − b ) T
Ft ,T = EtQ [ST ] = St × e r τ
τ = (T − t )
where S0 is the actual spot price of the underlying asset, T is the
where Q is the so-called risk-neutral probability measure,
maturity of forward contract, and r , g and b are respectively
f t ( K , T ) is the value at time t of a long forward contract with a
the continuous interest rate, the dividend yield and the repo rate
forward delivery price of K and a residual maturity of τ , r is
for T years.
the continuous annualized continuous interest rate for τ years
(constant interest rate hypothesis).
Remark: The value of a forward contract at inception with a strike price of K = S0 × e rT (fair cost-of-carry forward price) and a maturity date of T years is equal to zero since:
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♦ Options (Non-linear Derivatives): An option is a contract
The actual market prices of a European call and a put option are
which confers on its holder the right, but not the obligation, to
then given by the discounted values of their terminal payoffs,
purchase (call option) or to sell (a put option) an underlying asset
that is:
for a prescribed amount, known as the exercise or strike price, denoted K , at the expiration date T (European option), or
c0 = e − rT EQ [max(ST − K , 0 )]
− rT
p0 = e EQ [max( K − ST ,0 )]
within a specified period of time T (American option). By convention, if the current underlying asset price exceeds the exercise price of the option, the call is said to be in-the-money and the put is out-of-the-money. If the current underlying asset price is below the exercise price of the option, the call is out-themoney and the put is in-the-money. When the current underlying asset price is approximately equal to the exercise price of the option, both the call and the put are at the money.
The value at maturity of a European call and a put option are given by (terminal payoffs):
0
cT = max[ST − K , 0] = (
ST − K )
p = max[K − S ,0] = ( K − ST )
T
T
0
if
ST ≤ K
if
ST > K
if
ST < K
if
S≥K
where ST is the terminal price of the underlying asset and K is the exercise price.
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Terminal payoff on a long European call option position
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ΠTTerminal
payoff on a short European call option position
K > ST
In-the-money
K = ST
At-the-money
K < ST
Out-the-money
ΠT
K > ST
Out-the-money
K < ST
In-the-money
K = ST
At-the-money
Slope = 0
Slope = +1
0
ST
0
P&L = 0
K
Exercise price
ST
P&L = (ST - K)
K
Exercise price
Slope = 0
P&L = 0
Slope = -1
P&L = -( ST -K)
Terminal payoff on a long European put option position
Terminal payoff on a short European put option position
ΠT
K > ST
In-the-money
K = ST
At-the-money
K < ST
Out-the-money
ΠT
K > ST
In-the-money
K = ST
At-the-money
K < ST
Out-the-money
K
0
Exercise price
ST
K
Exercise price
Slope = -1
0
ST
Slope = 0
Slope = 0
Slope = -+1
P&L = (K - ST)
P&L = 0
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P&L = -(K - ST )
P&L = 0
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Remark: All these graphs represent the terminal payoffs and not
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2. Standard Calculus
the terminal P&L. For an option buyer, the terminal P&L is computed as the difference between the terminal payoff and the
-
Function of one variable, Derivatives and
(compounded) initial option’s premium, whereas it is equal to
Taylor series expansion (continuous interest
the opposite for an option seller. That is for a long position in a
rate, bond duration and convexity)
European call or put option:
-
Function of more than one variable, partial derivatives and total differential (option’s
c
(ST − K ) −c 0
Π T = Max[0, (ST − K )] − c0 =
− c 0
−p
p
Π T = Max[0, (K − ST )] − p0 = 0
(K − ST ) − p0
if
ST > K
if
ST ≤ K
if
ST < K
-
where ST is the terminal price of the underlying asset, K is the exercise price, and c0 and p0 are respectively the initial premium/price of the call and put option.
Integral, integration by parts and ordinary differential equations (ODE)
ST ≥ K
if
delta and the “Greeks”)
2.1. Function, Derivatives and Taylor Series Expansion
-
Function of one variable:
• Mathematical definition: Formally a function of one variable is a mathematical relation between two sets that associates to each element of the first set a unique element of the second one, that is:
A → B f :
x → f ( x) = y
The set of permissible values of x is called the domain of the
function and the set of values of y which results when the
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function is applied to values of x is called the range of the
♦ The Logarithmic function: The logarithmic function is
function.
defined as the inverse of the exponential function.
• Important functions:
Given:
♦ The Exponential function: The exponential function is obtained by raising the mathematical constant e ( ≈ 2.71828 ), to
y = ex the natural logarithm of y is given by:
ln( y ) = x
a power x , that is:
*
IR → IR+
f
x → f ( x) = e x
with y > 0 .
It gives the power at which the exponential number must be raised in order to obtain the value of x .
Some useful properties:
Some useful properties:
ex × ez = ex+ z
e 0 = 1 ⇔ ln (1) = 0
e =1
0
e1 = e ⇔ ln (e ) = 1
e1 = e
*
*
and ∀( x, y ) ∈ IR+ × IR+ ,
ln( x × y ) = ln( x ) + ln( y )
ln( x y ) = ln( x ) − ln( y )
( )
ln 1 x = −ln ( x )
ln x n = n × ln( x )
Representation of the exponential function
40
( )
ex
30
20
10
0
-4
-3
-2
-1
0
1
2
3
4
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x
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The (first) derivative: The derivative of a function
• Geometric interpretation: The derivative of a
f ( x ) gives the change of the function f for
function f (.) at a point x corresponds to the
small (infinitesimal) change in the value of the
slope of the line which is just tangent to the
variable x . It is the slope of the function f ( x ) at
function f (.) at the point x (the derivative/slope
a particular point x .
-
is generally variable).
• Mathematical definition: The derivative of f ( x) with
Remark: The first derivative can then be used in approximations.
respect to x is given by:
Using the definition of derivative and if ∆x is small enough, we
f ' (x ) = f x =
df ( x ) f ( x + ∆x ) − f ( x ) Change of f ( x )
= lim
=
dx
Small Change in x
∆x
∆x → 0
can write approximately (at a first order):
f ( x + ∆x ) ≈ f ( x ) + f ' ( x ) × ∆x
where ∆x is a small increment (infinitesimal change) in x .
Example: Suppose x represents time, and f ( x ) is the value of an asset, say a stock, at time t , denoted by St . The first derivative then gives the change of the price of the asset for a small interval of time, that is :
The value of f (.) at point x + ∆x can then be linearly approximated by the value of f (.) at point x plus the derivative
f ' (.) evaluated at x multiplied by ∆x . Note that when one
does not know the exact value of f ( x + ∆x ), the knowledge of f ( x ) , f ' ( x ) and ∆x is sufficient to obtain a linear
∆St S (t + ∆t ) − S (t )
=
∆t
∆t
approximation.
for a small interval of time ∆t (a second!!).
Example: Crude prediction of the future stock price knowing the price today, that is:
S (t + ∆t ) ≈ S (t ) + S ' (t ) × ∆S (t )
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• Basic rules:
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Example: The change of the price of a European call option c written on a stock S (t ) with respect to the change in time. That
♦The power function rule:
is, with c( S (t )) :
f ( x) = c ⇒ f ' ( x ) = 0
dc[S (t )] dc[S (t )] dS (t )
=
× dt dS (t ) dt f ( x ) = x n ⇒ f ' ( x ) = n × x n −1 f ( x ) = 1 / x n = x − n ⇒ f ' ( x ) = − n × x − n −1
Higher-order
-
♦The sum-difference, product and quotient rules:
derivatives
and
Taylor
series
expansion: The derivative of a function f ( x )
h( x ) = f ( x ) ± g ( x ) ⇒ h ' ( x ) = f ' ( x) ± g ' ( x )
is in general also a function of x . This
h( x ) = f ( x ) × g ( x ) ⇒ h ' ( x ) = f ' ( x) × g ( x ) + f ( x ) × g ' ( x )
function can then be differentiated again to
f ' ( x) × g ( x) − f ( x) × g ' ( x ) h( x ) = f ( x ) / g ( x ) ⇒ h ( x ) =
[g ( x )]2
'
obtain the derivative of the derivative - the second derivative. Just as the first derivative is the expression of the slope of the original
♦Chain rule: The chain rule is applied to expressions which are
function, the second derivative gives the slope of
function of a function.
the first derivative (the slope of the slope).
Definition: For:
With a Taylor series expansion, the second
y = f [g (x)]
derivative enables us to approximate more
the derivative of the function f (.) at a point x corresponds to:
efficiently a non-linear relationship.
df df dg f ' (x ) =
=
× dx dg dx
• Mathematical definition of the second derivative: The second derivative of f ( x) with respect to x is
It is a useful tool in approximating the response of one variable
given by:
to changes in other variables.
f '' ( x ) = f xx
87
df ( x ) d '
'
d f (x) dx
= lim f ( x + ∆x ) − f ( x )
=
= dx ∆x → 0
∆x
dx 2
2
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∆x = ( x − x0 ) ≈ small < 1
where ∆x is a small increment (infinitesimal change) in x .
Then we are sure that:
Remark: this process can be continued to find higher derivatives when they exist (third, fourth order derivatives,…).
f ( x ) is an infinitely
differentiable analytic function of x , the Taylor series expansion of f ( x) around any arbitrary
terms that are negligible (smaller than) in the Taylor series expansion. The second-order Taylor series expansion of f ( x) around x0 is given by:
f ( x) ≈ f ( x0 ) + f ' ( x0 ) × ( x − x0 ) +
value x0 can be defined as:
f ( x) = f ( x0 ) + f ' ( x0 ) × ( x − x0 ) +
+
∞
= ∑
k =0
3
Under these conditions it is possible to eliminate some of the
• Mathematical definition of the Taylor series expansion (B.
Taylor, 1685-1731): If
2
x − x0 > x − x0 > x − x0 > ...
1 ''
2
f ( x0 ) × ( x − x0 )
2!
1 '' '
3
f ( x0 ) × ( x − x0 ) + ...
3!
1 '' f ( x0 ) × ( x − x0 )2
2!
which becomes equality as x → x0 .
The value of a non-linear function f (.) at x + ∆x can then be approximated by the value of f (.) at x plus the first derivative
1 k k f ( x0 ) × ( x − x0 ) k! f ' (.) evaluated at x multiplied by ∆x (linear term) and the
where f k ( x0 ) is the k-th order derivative of f (.) with respect to
second derivative f ' ' (.) evaluated at x multiplied by 0.5 × (∆x )2
(quadratic term). Note that when one does not know the exact
x evaluated at x0 , with k ∈ N .
Remark: While the Taylor series expansion is an infinite series, it
value of f ( x + ∆x ). The knowledge of f ( x ) , f ' ( x ) , f ' ' ( x )
is usually truncated at a finite order (second) to obtain useful
and
approximation of a non-linear function f ( x) . Assuming that x is
approximation of it (more accurate than a linear one).
∆x is
then
sufficient
to
obtain
a
non-linear
close to x0 , that is:
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When the increment is very small, even the quadratic term is negligible and we have a first-order Taylor series expansion
(linear term) that is:
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• Financial applications:
♦Demonstration of the continuous compounding formula:
The Taylor series expansion can be used to demonstrate the
f ( x) ≈ f ( x0 ) + f ' ( x0 ) × ( x − x0 )
formula of the future value of an investment with continuous
which becomes equality as x → x0 .
compounding, that is:
V1 = V0 × ei
Remark: The second order Taylor series expansion is one of the most fundamental tool in finance. It is a powerful shortcut for practitioners to assess the impact of small changes in key variables on asset/portfolio values such as changes in yield for bonds (non-linear relation), changes in the underlying asset price for options (non-linear derivatives), changes the weights for portfolios (see incremental Value At Risk) …, with no need to recompute the new value of the underlying position given the new value of the risk factor. It allows us to
where V1 is the terminal value of an initial investment V0 invested for one year at a yearly continuous interest rate i .
Proof: If the initial investment V0 is compounded p times per year at the yearly interest rate i , the terminal value of the investment is:
i V1 = V0 × 1 + p
p
p
approximate the future value of an individual asset (portfolio)
If we take a first-order Taylor series expansion of the function
given the change of one of its constituent (risk-management).
ln(1 + x) around 1, we can write: f ( z ) ≈ f ( z 0 ) + f ' ( z0 ) × ( z − z0 )
1
ln(1 + x ) ≅ ln(1) + × (1 + x − 1) = x
1
with f ( z ) = ln(1 + x ) , z = 1 + x , z0 = 1, n = 1 and f ' ( z ) = 1 z .
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♦Bond price sensivity to a yield change (Dollar Duration
For x « small » (near of zero), we then have:
lim [ln(1 + x )] = x
and Dollar Convexity): The second-order Taylor series
x →0
expansion can be used to approximate the bond price reaction
That is, for an interest rate i fixed, if we increase the frequency
for an interest rate change.
of compounding ( p tends to infinity) we obtain:
The price of a T -year coupon-bond is given by:
C
C
C
F
+
+K
+
T
2
(1 + y ) (1 + y )
(1 + y ) (1 + y )T
T
C
F
+
=∑
t t =1 (1 + y )
(1 + y )T
P0 =
i i lim ln1 + = p →+∞
p p
So that:
p × lim ln1 + p → +∞
i
= lim p × ln 1 + p p →+∞
where P0 is the actual market price and y refers to the yield to
i
=i p
maturity of the considered bond.
Deriving twice this expression it is then possible to approximate
Using the logarithmic function properties leads to:
the impact of a change of interest rate on the bond price.
1) The first derivative of the bond price with respect to the yield
p
i i
lim p × ln 1 + = lim ln 1 + = i p p → +∞ p p → +∞
is (dollar duration):
Taking exponential on both sides leads to the desired result, that
P' ( y ) =
is:
(− )T × p p
i i i
lim expln 1 + = lim 1 + = e p → +∞ p p → +∞ p
which
dP0
C
(− )2 × C 3 K(− )T × C T +1
= (− )1 × dy (1 + y )2
(1 + y )
(1 + y )
corresponds
to
the
formula
of
F
(1 + y )T +1
T
C
F
= − ∑t
+T
t +1
(1 + y )T +1
t =1 (1 + y )
P ( y + ∆y ) − P ( y )
= − DD ≈
∆y
continuous
compounding. with: 93
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∆P = P( y + ∆y ) − P( y )
T
F
C
1
× ∑t ×
+T × t (1 + y ) t =1 (1 + y )
(1 + y )T
1
2
≈ P ' ( y ) × ∆y + × P '' ( y ) × (∆y )
2
1
2
≈ − D$ × ∆y + × C$ × (∆y )
2
The negative of the first derivative is called the dollar duration of the bond (DD). It is related to the absolute change in the bond price associated with a small absolute
The dollar duration measures the first-order (linear)
change in yield.
absolute effect of changes in yield. It is the negative of the slope of tangent to the price-yield curve at the current
2) The second derivative of the bond price with respect to the
interest rate. For small movements in the yield, this linear
yield is equal to:
approximation provides a reasonable fit to the exact price.
P '' ( y ) =
F
d 2 P0 T
C
=C
= ∑ t (t + 1)
+ T (T + 1) t +2
T +2
2
dy
(1 + y )
(1 + y ) $
t =1
But the price value function which relates bond price to yield is not a linear function. It is a rather convex one. The dollar duration, which attempts to estimate a convex relationship
The second derivative is called the dollar convexity of the bond
with a straight line, will then systematically understate the actual
(DC). It takes into account in fact the curvature of the price-
price: when interest drop, the slope of the value function or
yield function of the bond (non-linear relation, see supra). It is
equivalently minus the dollar duration will underestimate the
always positive for straight bonds.
price rise and when, interest rise it will overestimate the price drop. The convexity measures the second-order (quadratic term) effect of changes in yield. It captures the curvature of price function.
For large movements in the interest rates, the quadratic
Putting together all these equations in a second-order Taylor
approximation improves the fit since the price-yield relation
series expansion leads to an approximation for the absolute price
becomes more curved.
change of a bond, that is:
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Remark: Note that the convexity characteristic of the bond price-
Remark: Beside the dollar duration and convexity, there exist
yield relation is always good feature. Ceteris paribus, the higher the
other definitions of bond risk such as the Macaulay Duration
convexity of a bond, the higher will be its price reaction to an
and the Modified Duration
interest rate decrease and the lower will be its price reaction to an interest rate increase (greater convexity is both beneficial for
The Macaulay duration (1938) gives the absolute value of
both falling and rising market yields).
the percentage change of a bond price for a small
(infinitesimal) percentage change in (one plus) the market
Price
(EUR)
Price Approximation
required yield (minus price elasticity measure). That is:
Real price DMac = −η P / 1+ y
P( y + ∆y)
P ( y + ∆y )
with:
Value increases more than duration model
C
C+F
+ t t =1 (1 + y )
(1 + y )T
T −1
P( y)
P0 = ∑
Duration + convexity estimate
Duration
estimate
y + ∆y
Ct
CT + F
T −1
t∑1 t (1 + y )t + T (1 + y )T dP / P0
=
=−
=
P0 d (1 + y ) / (1 + y )
That is:
Value drops less than duration model
T
DMac = ∑ t × wt
y
t =1
Yield(%)
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wt = PVt / P0
Ct
(1 + y )t if t < T
PVt = C + N
T
if t = T
(1 + y )T
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Due to this feature, zero-coupon bonds (ZCBs) tend to provide the most price movement for a given change in interest rate,
(most attractive bonds in declining interest rate environments).
The Macaulay duration of perpetual bonds (infinite number of identical coupon payments) is given instead by:
and:
D=
Ct
N
+ t t =1 (1 + y )
(1 + y )T
T
P0 = P0 ( y ) = ∑
(1 + y ) y which is much less than its infinite maturity (indeed the cash-
(Expressed in period units, in years with annual
flows paid early dominate the duration of the perpetual bond).
compounding, in semesters with semi-annual,…)
It corresponds to the bond’s weighted average life of cash-flows, where each weight reflects the relative importance of the cashflow payment with respect to the value of the bond. It can also
be viewed as the holding investment period where capital
The Modified duration (1938)
gives the
absolute value of the percentage change of a bond price for a small (infinitesimal) absolute change in the market required yield. That is:
and coupon reinvestment risk are offsetting g each other.
Remark: The Macaulay duration of a zero coupon bond is given by its residual maturity since there are no regular intermediate
DMod
dP
D
1
P
DMac = $
=− 0 =−
(1 + y )
P0
dy
(Expressed in % units)
coupon payments. That is:
T×
Dmac =
F
F
T
(1 + y ) = T × (1 + y )T = T
F
P0
(1 + y )T
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To sum up:
[
w pi =
]
dP C
C
( 1)
= 2 × 1 − (1 + y )−T + T × F − × (1 + y )− T +
D$ = − dy y y
2CT
d 2 P 2C
C
C$ = 2 = 3 × 1 − (1 + y )−T − 2
+ T (T + 1) F − × (1 + y
T +1 y dy y (1 + y )
y
dP 1 D$
DMod = − dy × P = P
0
0
D$
(1 + y ) dP
DMac = − d (1 + y ) × P = (1 + y ) × DMod = (1 + y ) × P
0
0
where we have used here the closed formula of the bond
[
]
ni × Pi
∑ ni × Pi
i =1
N
where w pi , ni and Pi represent respectively the weight, the number and market price of bond i in portfolio p.
Market Beta analogy
(annuity simplification) to compute the dollar duration and convexity). Example: Consider a 10-Year 4.5% coupon bonds selling on the
Remark: The Macaulay duration (modified duration) of a portfolio of bonds is equal to the weighted average of the
Macaulay (modified) durations of the bonds in the portfolio
(if the yield-to-maturity is the same for all bonds), the duration of each bond being weighted by its market value divided by the total market value of the portfolio, that is:
market at par value (100 EUR). What will be its dollar and percentage price change if the interest rate level increases by 1%?
Remark: A basis point (often denoted as bp) is equal to 0.01%
(1/100th of a percent) or 0.0001 in decimal form. So, a bond whose yield increases from 3% to 3.5% is said to increase by 50 basis points; or interest rates that have risen 1% are said to have
N
DMac , p = ∑ wip × Dmac ,i
increased by 100 basis points.
i =1
with:
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The dollar duration and convexity of the bond are given respectively by:
[
]
[
]
C
C
−T
− (T +1)
= 791.27
D$ = 2 × 1 − (1 + y ) + T × F − × (1 + y ) y
y
2CT
C
− (T + 2 )
C = 2C × 1 − (1 + y )−T −
+ T (T + 1) F − × (1 + y )
T +1
2
$ y3 y y (1 + y )
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Check:
P(4.5% ) =
4.5
4. 5
104.5
+
+K+
2
(1.045) (1.045)
(1.045)10
1 − (1.045)−10
100
= 4. 5 ×
+
10
0.045
(1.045)
= 100 EUR
If the required yield level increase by one percent (infinitesimal change), the bond price will be equal to:
See fonction finances Excel VA.
=100*DUREE.MODIFIEE(DATE(2011;1;1);DATE(2021;1;
1);0,045;0,045;1)
That means that the impact of a 1% (100 b.p.) change in interest rates on the 10-year bond price can be approximated to a first
P(5.5% ) =
4.5
4. 5
104.5
+
+K+
2
(1.055) (1.055)
(1.055)10
1 − (1.055)−10
100
= 4. 5 ×
+
10
0.055
(1.055)
= 92.46 EUR
That is an absolute decrease in the bond value of:
92.46 − 100 = −7.54 EUR
order as:
∆P = P(5.5% ) − P(4.5% ) ≈ − D$ × ∆y
≈ −791.27 × 0.01 = −7.91EUR
Which compares with -7.52 (see before).
And to a second order by:
1
2
∆P = P(5.5% ) − P(4.5% ) ≈ − D$ × ∆y + × C$ × (∆y )
2
1
2
≈ −791.27 × 0.01 + × 7781.02 × (0.01) = −7.52 EUR
2
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Example: An investor is considering the two following five-year
Modified duration of bond B:
coupon bonds. A 9% bond A selling on the market at par value
(100 EUR) and a 6% bond B selling to yield 9%.
B
DMod
Which bond the investor will choose if she is expecting a
[
[
]
(
)
with:
P0B
1 − (1.09 )− 5
100
= 6×
= 88.33EUR
+
5
0.09 (1.09 )
See fonction finances Excel VA.
Modified duration of bond A:
A
DMod
]
6
6
−6
× 1 − (1.09 )−5 + 5 × 100 −
× 1.09
2
0.09
(0.09)
= 4.06%
=
88.33
decrease in the general level of interest rates. Same question if he is expecting now an increase in the required market yields.
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
(
)
9
9
−6
× 1 − (1.09 )−5 + 5 × 100 −
× 1.09
2
0.09
(0.09)
= 3.89
=
100
with:
=DUREE.MODIFIEE(DATE(2011;1;1);DATE(2016;1;1);0,
09;0,06;1)
If the required yield increase (decrease) by 100 basis points, the price of bond A will decrease (increase) approximatively by
P0A = 100
3.89%; while the price of bond B will decrease approximatively
(increase) by 4.06%, since:
See fonction finances Excel VA.
=DUREE.MODIFIEE(DATE(2011;1;1);DATE(2016;1;1);0,
Any rational investor will then invest in bond A if she is
09;0,09;1)
expecting an increase of interest rate (less sensitive bond) and in bond B if she is expecting a decrease of interest rates (more sensitive).
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Example: Consider a 10-year 3.5% coupon bonds selling on the market at par value (100 EUR). What is the Macaulay duration?
10
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
CFt
10
−t
∑ t × CFt (1 + y ) t 860.77
(1 + y ) = t =1
=
≈ 8.61
10
P0 ( y )
100
CFt (1 + y )− t
∑
∑ t×
DMac =
t =1
What will be its percentage and dollar price change if the interest
(in years)
t =1
rate level rises by 35 basis points (absolute change of 0.35%)? To compute the Macaulay duration makes the following table:
So that:
∆y
∆P / P0 ≈ − D × (1 + y ) = −2.91%
∆P ≈ − P0 × D × ∆y = −2.91 EUR
(1 + y )
Calculation of Macaulay Duration
Cash flows
PV of 1 EUR 3.5%
PV of CFt
t x PVt
CFt
dt = (1+y ) - t
CFt (1+y ) - t
t * CFt (1+y ) - t
1
3.50
0.9662
3.38
4.30622
2
3.50
0.9335
3.27
8.24156
That means that if the interest rates (required yield) increase (or
3
3.50
0.9019
3.16
11.83000
decrease) by 35 bps, the 10-year bond will experienced a -2.91%
4
3.50
0.8714
3.05
15.09410
5
3.50
0.8420
2.95
18.05514
6
3.50
0.8135
2.85
20.73318
EUR loss (gain) per bond.
7
3.50
0.7860
2.75
23.14709
Check:
8
3.50
0.7594
2.66
25.31466
9
3.50
0.7337
2.57
27.25262
10
103.50
0.7089
73.37
672.90442
100
860.77
Year
decrease (increase) in its value which can be translated a 2.91
P(3.5% ) =
3.5
3.5
103.5
+
+K+
2
(1.035) (1.035)
(1.035)10
1 − (1.035)−10
100
= 3. 5 ×
+
10
0.035
(1.035)
= 100 EUR
From it we then compute the Macaulay duration of this bond
If the required yield level increase by 35 bps, the bond price will
such as:
be equal to:
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P(3.85% ) =
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♦Option price sensivity to an underlying asset price
3. 5
3. 5
103.5
+
+K+
2
(1.0385) (1.0385)
(1.0385)10
1 − (1.0385)
= 3. 5 ×
0.0385
= 97.14 EUR
−10
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
movement:
100
+
10
(1.0385)
∆c = c ' (St )∆S +
1 '' c (St ) × (∆S )2
2!
where St is the price of the underlying asset, c ' (St ) is the first
That is a relative decrease in the bond value of:
derivative of the call price with respect to the price of the
97.14
− 1 = −0.0286 = −2.86%
100
underlying (i.e. the delta of the option) and c" (St ) is the second derivative of the call price with respect to the price of the
which translates into an absolute euro loss of:
underlying (i.e. the gamma of the option).
97.14 − 100 = −2.86 EUR
Remark: Using now the compact formula of the Macaulay duration, we find:
[
DMac
]
3.5
3.5
1 − (1.035)−10 + 10 × 100 −
(1.035)
2
0.035
(0.035)
= (1.035) ×
100
−11
= 8.61
which is precisely what we obtained previously!!!
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• Application (Greek derivatives): The selling or buying
2.2. Function of More than One Variable, Partial
Derivatives and Total Differential
of an option creates a position with various sources of market risk, some of which may be
Function of more than one variable:
-
unwanted. In order to manage those risk factors,
• Mathematical definition: Formally a function of
a trader will need to compute the sensitivities of
several variables is a mathematical relation
the derivative’s price to the various parameters
between two sets of elements that associates a
that impact its values (the “Greeks”).
unique element to the second set with each n-uple of elements of the first set, that is:
Consider the price of call option. It is determined by the
A → B f :
(x1, x2 , x3, K, xn ) → f (x1, x2 , x3, K, xn ) = y
underlying asset price, the exercise price, the time to maturity,
Partial derivative: A partial derivative gives the
asset price and the time to maturity are interconnected since in
change of the function when only one variable of
practice one needs some time to pass before the underlying asset
interest change.
can change).
-
• Mathematical
definition:
The
derivative
of
f ( x1, x2 ,K, xn ) with respect to x1 is given by:
f x1 =
∂f (x1, x2 , x3, K, xn )
= lim
∆x1→ 0
the asset's log-return volatility and the risk-free interest rate (the
The price of the call is a function of five different variables:
ct = c(St , K , τ , σ , r
)
where St is the current price of the underlying asset, K is the exercise price of the call option, τ = (T − t ) the maturity of the
∂x1
f (x1 + ∆x1, x2 , x3, K, xn ) − f (x1, x2 , x3, K, xn )
option contract, σ the volatility of the log-return of the
∆x1
underlying asset and r the risk-free rate.
where ∆x1 is a small change in the x1 direction.
If we fix all the parameters except one and differentiate the option price with respect the later one, we can obtain the following partial derivatives:
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♦Delta (linearity): It measures the first-order sensitivity of the option to a movement in the underlying’ asset price, i.e. its
Call Price as a function of the underlying asset
sensitivity to small changes in the price of the underlying asset.
Mathematically it is the first partial derivative of the option price with respect to the underlying, so it gives the change in option price for a one-unit increase in the underlying price. The deltas for call and put options are given by (Black and Scholes model):
c ∂ct
−g T
Ν (d1 ) > 0
∆ t = ∂S = e t
∂p
∆p = t = −e − g T Ν (− d1 ) = e − g T [Ν (d1 ) − 1] < 0 t
∂St
with:
d1 =
(
)
ln(St / K ) + r − g + 0.5 σ 2 τ
σ τ
The figure depicts the case with K = 100 , σ = 0.56 and r = 0.05 .
and τ = (T − t )
♦Gamma (option’s convexity): It measures the second-order
where ∆ct and ∆p are respectively the delta for the call and put t sensitivity of the option to a movement in the underlying’ asset
option, N (.) is the standard normal cumulative probability
price, i.e. sensitivity of the delta to small changes in the price of
distribution function, g is the continuous dividend yield and τ
the underlying asset or of the option price to the realized volatility. Mathematically it is the first partial derivative of the
is the maturity of the option.
delta or the second partial derivative of the option price with
Note that:
∆ct, OTM → 0 as St → 0
c, ATM
≈ 0.5
∆ t
c, ITM
→ +1 as St → ∞
∆ t
and
respect to the underlying price, so it gives an indication of how
∆p , ITM → −1 as St → 0
t
p , ATM
≈ −0.5
∆ t
p ,OTM
→ 0 as St → ∞
∆ t
convex the option is in the underlying asset price or how quickly the delta of the option moves as the underlying asset price moves. The Gamma is highest for ATM (spot or forward)
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options and lowest for deep ITM or OTM options (ceteris
Gamma of call options with different maturities (1 week solid line;
paribus), since for deep ITM and deep OTM options, delta will
1 month dashed line and 1 quarter dotted line)
not change much for small changes in the underlying price. Note however that a higher volatility lowers the Gamma of ATM
European call/put options but raises it for ITM and OTM options, since high volatility means that ITM and OTM have more time value and so their Gamma sensitivities should not be too different from the Gamma sensitivity near the strike price.
The Gammas for both call and put options are given by (Black and Scholes model):
♦Vega (implied volatility): It measures the first-order
Γtc, p =
∂ 2 ct ∂ 2 pt n(d1 ) × e − gτ
=
>0
=
Stσ τ
∂St2 ∂St2
sensitivity of the option to a movement in the implied volatility of the underlying asset, i.e. its sensitivity to small changes in the
where Γtc , p is the Black-Scholes Gamma for both the call and
implied volatility of the underlying asset. Mathematically it is the
put option, n(.) is the standard normal density function, g is the
first partial derivative of the option price with respect to the
continuous dividend yield and τ is the maturity of the option.
underlying implied volatility, so it gives the change in option price for a one-unit increase in the implied volatility (1%).
Similar to the Gamma, the Gamma tends to be highest for ATM
Note that:
(spot or forward) options and lowest for deep ITM or OTM
Γtc , p ≈ 0 for St > K
options (ceteris paribus). But contrary to the Gamma, the Vega is largest when the maturity is longer (greater effect of the volatility on the call price). The Vega for both call and put options is given by (Black and Scholes model):
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Vega of call options with different maturities
υtc , p =
with:
∂ct ∂pt
=
= St × e − gτ n(d1 ) τ = Ke − rτ n(d1 ) τ > 0
∂σ ∂σ
(
(1 week solid line; 1 month dashed line and 1 quarter dotted line) )
ln(St / K ) + r − g + 0.5 σ 2 τ d1 =
σ τ
d 2 = d1 − σ τ
τ = (T − t )
where υtc , p is the Black-Scholes Vega for both the call and put option, n(.) is the standard normal density function, g is the
♦Theta (time decay): It measures the sensitivity of the option
continuous dividend yield and τ is the maturity of the option.
to a small change in time to maturity, i.e. its sensitivity to the passage of time. Mathematically, as time to maturity decreases, it
Note that:
is minus the first partial derivative of the option price with
υtc , p ≈ 0 for St > K
respect to the time maturity, so it gives the change in option price for a one-unit decrease (1 day). The thetas for vanilla call and put options are given by (Black and Scholes model):
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c ∂ct − ∂ct
S e − gτ n(d1 ) σ
=− t
− g St e − gτ N (d1 ) − r Ke − rτ N (d 2 ) < 0
=
Θt =
∂τ
∂t
2 τ
− gτ
Θtp = ∂pt = − ∂pt = − St e n(d1 ) σ + g St e − gτ N (− d1 ) + r Ke − rτ N (− d 2 ) 0
∂τ
2 τ
∂t
with:
(
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Theta sensivities of call options with different maturities
(1 week solid line; 1 month dashed line and 1 quarter dotted line)
)
ln(St / K ) + r − g + 0.5 σ 2 τ
d1 = σ τ
τ = (T − t )
where Θtc and Θtp are respectively the theta for the call and put option, N (.) is the standard normal cumulative probability distribution function, g is the continuous dividend yield and τ
♦Rho (interest rate sensitivity): It measures the sensitivity of
is the maturity of the option.
the option to a small change in the risk-free interest rate.
Mathematically, it is the first partial derivative of the option price with respect to the interest rate, so it gives the change in option price for a one-unit change (1%) of interest rate. The Rho for vanilla call and put options are given by (Black and Scholes model): c ∂ct
− rτ
ρt = ∂r = τ Ke N (d 2 ) > 0
∂p
ρtp = t = −τ Ke − rτ N (− d 2 ) < 0
∂r
with:
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(
122 EFJ Copyright11
)
ln(St / K ) + r − g + 0.5 σ 2 τ d1 =
σ τ
d 2 = d1 − σ τ
τ = (T − t )
where
ρtc
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Total differential: The total differentiation gives
-
the total change of the function f ( x1, x2 ,K, xn ) due to the changes of its independent variables
xi , that is:
and ρt are respectively the Rho for the call and put p option, N (.) is the standard normal cumulative probability distribution function, g is the continuous dividend yield and τ
df ( x1, x2 ,K, xn ) = f x1dx1 + f x2 dx2 + K + f xn dxn
Example: The total change of a call option price for an infinitesimal interval of time is given by:
is the maturity of the option.
∂ct
∂c
∂c
∂c
dSt + t dτ + t dσ + t dr
∂r
∂St
∂τ
∂σ
dct =
Since the call option value depends on in St a non-linear
See also Vomma, Vanna, Cross Gamma, DdeltaDtime
(Charm), DgammaDtime (Color or Gamma decay),
DgammaDvol
(Zomma),
DgammaDspot
fashion, it is better to use a second-order Taylor series expansion, that is:
(speed),
DvommaDvol (ultima, and Omega or Lambda (delta elasticity),.... dct =
∂c
∂c
∂c
∂ 2ct
∂ct
2 dSt + 0.5 × 2 dSt + t dτ + t dσ + t dr
∂r
∂σ
∂τ
∂St
∂St
{
{
{
{
{
∆t
Γt
Θt
υt
ρt
Remark: The Taylor series expansion version of a function with several variables is:
∂f
∂f
1 ∂2 f dx + K + × 2 (dx1)2 dx + df =
2 ∂x1
∂x1 1 ∂x2 2
+
121
2
∂2 f f (dx1dx2 ) + 1 × ∂ 2 (dx2 ) + K
2 ∂x2
∂x1∂x2
122
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2.3.
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Integral,
Integration
and
Ordinary
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Differential
-
Equations
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Integration: The integration is the process by which an original function is recovered from its derivative (ante derivative):
-
• The constant function rule:
Integral: It is the mathematical tool used for calculating the sum of uncountable infinite
∫c
objects
dx = cx + k
• The power rule:
• Definition (Riemann Integral): We are given a
n
dx =
x n +1
+k
n +1
deterministic function f (t) of time t ∈ [ 0, T ] ,
∫x
where the interval is partitioned into n disjoint
• Integration by parts:
subintervals:
T
∫0
t0 = 0 < t1 < t2 < K < tn = T
If:
T f ( x )g ' ( x )dx = [ f (T )g (T ) − f (0 )g (0 )] − ∫0 f ' ( x )g ( x )dx
lim max ti − t(i −1) = 0
n→∞
Ordinary differential equations: An equation which contains one or more derivatives of a
i
function is called a differential equation. If the the (Riemann) integral will be defined as: n ∑ n → ∞ i =1
lim
function is of a single variable, the equation is an
n
ti + t(i −1) f × (ti − t(i −1) ) = lim ∑ f (ti −1 ) × (ti − t(i −1) )
2 n → ∞ i =1
n ∑ n →∞ i =1
= lim
ordinary differential equation (ODE), in contrast to the partial equation which arises from functions of several variables (PDE).
T f (ti ) × (ti − ti −1 ) = ∫0 f (t ) dt
Example:Consider the expression:
Area under the curve defined by f (t ) over a given part of the domain.
123
dBt = −rt Bt dt with known B0 , rt > 0 .
⇒
dBt
= − rt dt
Bt
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3. Tools in Probability and Statistics
Its solution is given by:
− ∫ t ru du
Bt = e 0
3.1. Random Variables and Probability Densities
If rt = r , then:
Bt = e − rt
-
Random variables (R.V.):
• Mathematical definition: Formally, a random variable
X is a function that associates to each element of the sample space
(set of all possible
outcomes) a unique element of the (sub)set of real numbers (events) IR, that is:
Ω → IR
X :
ω → X (ω ) = x where the sample space lists all possible outcomes of a random experiment. Example: Consider an experiment in which a coin is tossed three times. In this experiment, the sample space is:
Ω = {HHH , HHT , HTH , THH , HTT , THT , TTH , TTT }
where H represents head and T tail. The random variable could then be defined as the number of heads obtained on the three tosses. For each possible sequence (outcome) ω that consist of three tosses, this random variable would assign a number X (ω)
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equal to the number of heads in it. Thus if ω is the sequence
• Discrete probability density and distribution functions:
(outcome){ HHH } , then:
For any discrete random variable, it is possible to
X = X (ω ) = 3
-
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
compute simultaneously the probability that the random variable takes a specific value and that its
Univariate distribution function (discrete and
• The distribution function: The mathematical model
realisations belong to a specific interval:
~
♦ Discrete random variable: A random variable X is called
for the probabilities that are associated to a
discrete if it can take only a finite number of different values
random variable X is given by the distribution
(discrete values).
continuous cases):
function
F ( x)
(also called the cumulative
distribution function or the c.d.f), defined such
~
♦ Discrete probability function (frequency function): If X
~
has a discrete distribution, the discrete probability function of X is defined as the function p(.) such that for every real number x :
~
p( x) = P( X = x)
as:
F ( x) = P( X ≤ x) with: with:
F (+ ∞) = 1
F (− ∞) = 0
p ( xi ) ≥ 0 ∀xi
∞
i∑ p( xi ) = 1
=1
where − ∞ < x < +∞ .
This is the probability that the realisation of the random variable
~
X ends up less or equal to the given number x .
where p ( xi ) is the probability of observing xi .
~
♦ Probability of an event: If X has a discrete probability distribution, the probability of each subset of the real line (event) can be determined from the relation:
~
P (a ≤ X ≤ b ) = P( X ∈ A) = ∑ p( x ) = ∑ p( xi ) = [F (a ) − F (b )]
A
127
a ≤ xi ≤ b
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• Continuous density and distribution functions: For any
In the case of a continuous random variable, the
continuous random variable, it is only possible to
probability function is represented by the area under the
compute the probability that the random variable
curve f ( x) on the interval [a, b] .
belongs to a specific interval:
♦ Continuous random variable: A random variable is called
~
As a consequence the probability that X takes a specific
continuous if it can take an uncountable infinite number of
value x is zero (not impossible but negligible), that is:
different values.
x
P( X = x ) = ∫ f ( x)dx = 0 x Examples: Daily rainfall, asset return (-100% to + infinity),…
Nevertheless, the probability that the continuous random
~
variable X belongs to a small interval around x can be
~
♦ Continuous probability density function (p.d.f): If X has
~
a continuous distribution, the probability density function of X
approximated by:
is defined as the nonnegative valued function f (.) such that, for
~
every interval [a, b] , the probability that X takes value in it is
x+
dx
2 dx ~ dx
P x − ≤ X ≤ x + = ∫ f ( x)dx
2
2 x − dx
2
x+
given by the integral of f (.) over this interval, that is:
dx
2
≅ f ( x) ∫
dx
dx x− 2
b
P[ a ≤ X ≤ b] = ∫ f ( x) dx
dx dx
= f ( x) × x + − x +
2
2
= f ( x) × dx
a
with:
f ( x) ≥ 0 ∀x ∈ IR
+∞
+∞
∫− ∞ f ( x ) dx = ∫− ∞ dF ( x ) = 1
Or equivalently:
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x + dx
~
P(x ≤ X ≤ xdx ) = ∫ f ( x)dx
-
x x + dx
≅ f ( x) ∫
dx
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Multivariate distribution function (bivariate case):
~
~
In the case of two random variables X and Y , it is possible to compute the joint distribution
= f ( x) × [x + dx − x ]
function F (x, y) , the marginal probability
= f ( x) × dx
functions
x
since, while f ( x) is a non-linear curve, for small ∆x (that is for
dx
dx ), f ( x) will not change very much as x varies from x −
2
dx
/ x to x − / x + dx and it can be approximated
2
of f X ( x)
and
fY ( y )
and
their
conditional distributions.
♦ Joint cumulative distribution: In the discrete case, the joint
cumulative bivariate probability function is written as:
F (a, b ) = P ( X ≤ a, Y ≤ b ) = ∑ ∑ p( X = xi , Y = yi ) xi ≤ a yi ≤ b
reasonably well by f ( x) during this interval.
where p( x, y ) is the joint discrete probability (frequency)
The rectangle obtained would then be very close to the
function.
probability that x will fall within a small neighbourhood of x
In the continuous case, the joint cumulative bivariate probability
represented by the quantity dx .
function is written as: a b
F (a, b ) = P( X ≤ a, Y ≤ b ) = ∫ ∫ f ( x, y )dxdy
So that:
−∞ −∞
P[ a ≤ X ≤ b] = lim ∑ f ( xi ) × ∆xi = ∫ f ( x) dx
n →∞ n i =1
b
where f ( x, y ) is the joint continuous probability (frequency)
a
where xi ∈ [a, b], and the interval is partitioned into n disjoint
function.
subintervals:
Example: Probability of observing simultaneously tomorrow a return of -20% on the SP500 Index and a return of - 30% on the
a = x0 < x1 < x2 < K < xn = b
CAC40 French Index, that is:
P ( RSP500 ≤ −0.2, RCAC40 ≤ −0.3)
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♦ Marginal distribution function: In the discrete case, we can
~
~ recover the marginal probability frequency function of X and Y by: Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
♦Conditional probability distributions: In the discrete case, the conditional probability density function of the random
~
~ variable Y given the random variable X is given by:
p ( x, y )
~
~ pY = y X =x = pX (x)
p X ( x ) = ∑ p( X = x, Y = yi ) yi
pY ( y ) = ∑ p( X = xi , Y = y ) xi
(
)
with p X ( x) > 0 .
In the continuous case, we can recover the marginal probability
~
~ density function of X and Y from:
∞
( x) = ∫ f ( x, y ) dy
fX
−∞
∞
( ) ∫ ( fY y = f x, y ) dx
−∞
~
This is the probability of Y = y conditional on the fact that
~
X = x.
~
~
If X and Y are independent, then: p( x, y ) = p X ( x ) × pY ( y )
p ( x ) × pY ( y )
~
~
⇒ pY = y X =x = X
= pY ( y ) pX (x )
~
That is we learn nothing about Y from observing a
~
particular realisation of X .
(
Remark: Any multivariate distribution can be factored into two components: on one hand, the marginal distributions which represent the univariate characteristics of the multivariate
)
distribution, and on the other hand, the copula function which summarizes the joint features / dependence structure of the multivariate distribution. That is in the continuous case: a b
F (a, b ) = P( X ≤ a, Y ≤ b ) = ∫ ∫ f ( x ) f ( y ) c[F ( x ), F ( y )]dx dy
In the continuous case, the conditional probability density
~
~ function of the random variable Y given the random variable X is given by:
f (y x) =
−∞ −∞
where:
f ( x, y ) f X (x)
with f X ( x) > 0 .
f ( x, y ) = c[F ( x ), F ( y )] × f ( x ) × f ( y )
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This is the density of Y = y conditional on the fact that X = x .
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
3.2. Moments of Random Variable
If X and Y are independent, then: f (x, y) = f X ( x) × fY ( y )
⇒ f ( y x) =
-
f X ( x) × fY ( y )
= fY ( y ) f X ( x)
While any random variable is characterised by its distribution function, in many instances it is not necessary to draw the entire probability density
That is we learn nothing about Y from observing a particular
(frequency) function to have an idea of its shape.
realisation of X .
Many random variables have indeed probability distributions which can be uniquely determined
Example: Probability of observing tomorrow a return of -30 on
by their moments (K. Pearson, 1857-1936). Some
the CAC40 French Index conditional on a return today of - 20%
can be determined parametrically only by their
on the SP500 Index, that is:
firsts two moments. Others need higher-order
P ( RCAC40 = −0.3 RSP500 = −0.2, )
central moments for a full characterisation.
-
The first two moments:
• The mean or the expected value of a random variable (first
central moment): It is a measure of the central tendency or center of gravity of the distribution of the random variable of interest.
♦ Mathematical definition:
µ = E (X ) = ∑ xi p ( xi ) (discrete case) n ~
i =1
~ +∞ µ = E (X ) = ∫ x f ( x ) dx (continuous case)
−∞
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♦ Properties of the expectation operator (linear operator):
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
• The variance of a random variable (second central
moment): It gives information about the way the
~
~
E (a × X + b ) = a × E (X ) + b
~ ~
~
~
E (X ± Y ) = E (X ) ± E (Y )
~ ~
~ ~
~
E (X Y ) = E (X )E (Y ) if X and Y are independent
distribution of the random variable is spread out.
It is a measure of the average squared deviation of the future realizations around the mean.
♦ Empirical counterpart (sample estimates): In practice, the
♦ Mathematical definition:
σ 2 ( X ) = V ( X ) = E [( X − µ ) 2 ]
mean of a random variable is estimated by its sample counterpart. Let {x1,K, xT } be a random sample of T
~
realisations of X . Assuming that all observations are identically and independently distributed (I.I.D., i.e. they are independent
~
draws from the same random variable X whose distribution is
~
constant in time), the first moment of the random variable X ,
~
µ = E (X ), can then be estimated from a sample of T realisations
(general definition)
σ 2 ( X ) = Var( X ) = ∑ p ( xi ) × [ xi − E( X )] 2 n i =1
(discrete case)
+∞
σ 2 ( X ) = Var( X ) = ∫ [ x − E( X )] 2 f ( x) dx
−∞
by the sample mean, that is:
(continuous case)
~ 1 T
ˆ ˆ µ = E (X ) = ∑ xt
T t =1
Remark: The square root of the variance is the standard deviation, that is:
Std ( X ) = σ ( X ) = Var ( X )
It is a more convenient measure of dispersion to use as it has the same units as the original variable X .
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♦ Properties of the variance (quadratic operator) and
♦ Empirical counterpart (sample estimates):
1 T
2
ˆ
ˆ 2
∑ ( xt − µ X ) σ (X ) =
T − 1 t =1
T
σ XY = 1 ∑ ( xt − µ X )( yt − µY )
ˆ
ˆ
ˆ
T − 1 t =1
covariance (bilinear operator):
( )
Var ( X ) = E X 2 − [E ( X )]
(Huygens Formula's )
Var (a × X + b ) = a 2 × Var ( X ) σ (a + b × X ) = b × σ ( X )
2
[
Var ( X + Y ) = E ( X + Y − E ( X ) − E (Y ))
2
]
[
]
= E [( X − E ( X )) ] + E [(Y − E (Y )) ] + 2 E [( X − E ( X ))(Y − E (Y ))]
• Financial application (portfolio performance, expected
= E ( X − E ( X )) + (Y − E (Y )) + 2( X − E ( X ))(Y − E (Y ))
2
2
2
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
2
= Var ( X ) + Var (Y ) + 2 × Cov ( X , Y )
return and risk diversification, Markowitz, 1952):
♦Portfolio rate of return (one period): The one period rate of return over a selected horizon of a portfolio consisting of fixed
where Cov( X , Y ) is the covariance between X and Y (positive or negative comovements between the two variables) defined
positions in N assets is a linear combination of the returns on the underlying assets involved, where the weights on each asset are given by the relative amounts invested at the beginning of
such as:
the period, that is for two assets:
R p ,t +1 = w1 R1,t +1 + w2 R2,t +1
Cov( X , Y ) = σ XY = E{[ X − E ( X )][Y − E (Y )]}
= E ( XY ) − E ( X ) E (Y )
with:
with (bi-linear operator):
Ri ,t +1 =
Cov( X , Y ) = Cov(Y , X )
Cov( X , a + b Y ) = b × Cov( X , Y )
2
Cov( X , X ) = σ XX = σ X = Var ( X )
(Pi,t +1 − Pi,t ) + Di,t +1
Pi ,t
where Ri ,t +1 is the rate of return of the asset i with i = [1, 2] and
wi is its relative weight in the portfolio (determined in t ).
Or more generally for N assets:
R p ,t +1 =
139
N
∑ wi i =1
Ri ,t +1
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♦Portfolio expected rate of return (one period): The
( )
Var R p =
portfolio expected rate of return is a linear combination of the asset expected returns which are involved in it, that is for two assets: N
∑ wi i =1
[wi w j Cov(Ri , R j )]
[wi w j Cov(Ri , R j )]
(
Cov Ri , R p
)
variances of the individual securities but also on all the
[wi × E (Ri )]
covariances (risk reduction by diversification).
♦Portfolio variance: The variance of the portfolio return is a linear combination of the covariances between the return of
Remark:
σ ( Ri ) = Var ( Ri )
The standard deviation of the asset return is of the same order
each asset and those of the portfolio, that is for two assets:
[
N N
∑ ∑ i =1 j =1
N N
∑ ∑ i =1 j =1 j ≠i
That is, the variance of a portfolio depends not only on the
Or more generally for N assets:
( )
Var ( Ri ) +
=
= w1 E (R1 ) + w2 E (R2 )
E
N
∑ wi2 i =1
=
E (R p ) = E (w1R1 + w2 R2 )
N
Rp = ∑ i =1
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
]2
Var (R p ) = E R p − E (R p )
than the expected return (some % per annum).
= E [w1R1 + (1 − w1 )R2 − (w1E (R1 ) + (1 − w1 )E (R2 ))]
= E [w1 (R1 − E (R1 )) + (1 − w1 )(R1 − E (R1 ))]
2
2
= w1 E [R1 − E (R1 )] + (1 − w1 ) E [R2 − E (R2 )]
2
2
2
2
+ 2 w1 (1 − w1 ) E{[R1 − E ( R1 )] [R2 − E ( R2 )]}
2
= w1 Var (R1 ) + (1 − w1 ) Var (R2 ) + 2 w1 (1 − w1 ) Cov(R1 , R2 )
2
2
2
= w1 Var (R1 ) + w2 Var (R2 ) + 2 w1 w2 Cov(R1 , R2 )
with: w2 = (1 − w1 ) ⇔ (w1 + w2 ) = 1 (budget constraint)
Generally for N assets :
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Higher order moments: The third central moment of
positive outcomes and large number of small negative
a distribution is the skewness and is informative
-
outcomes)
of the presence of asymmetry in the distribution.
The fourth central moment of a probability distribution is the kurtosis and measures the width of its tails (extreme risks):
• Skewness (third-order moment):
♦Mathematical definition:
[
• Kurtosis (fourth-order moment):
♦Mathematical definition:
[
♦Interpretation:
]
+∞
3
3
3 s ( X ) = E ( X − E ( X )) = ∫ [ X − E ( X )] f ( x )dx
−∞
3
γ ( X ) = s ( X ) (standardized)
1 σ ( X )3
]
+∞
4 κ ( X ) = E ( X − E ( X ))4 = ∫ [ X − E ( X )]4 f ( x )dx
−∞
4
γ ( X ) = κ ( X )
2 σ ( X )4
(symmetric
distribution)
the
kurtosis
coefficient is a measure of the tail thickness / peakedness of the probability distribution.
1) A Kurtosis lower than 3 is called platykurtic. The distribution
♦Interpretation: The skewness is a measure of the imbalance
has lower tails and a flatter peak than the Normal distribution
of the tails of the probability distribution (the direction of
(larger fraction of outcomes are cluster around the mean).
2) A Kurtosis equal to 3 is said to be mesokurtic. The distribution
skewness is to the tails).
1) A negative skewness (left-skewed) indicates that the left-tail
has same tails and peak than the Normal distribution.
of the distribution is longer that the right (few extreme
3) A Kurtosis greater than 3 is called leptokurtic The distribution
negative outcomes and large number of small positive
has fatter tails and more pointed peak than the Normal
outcomes)
distribution (larger fraction of outcomes is at the extremes.
2) A zero skewness indicates that the same amount of data
Discussion: Dynamic strategies, Options, Hedge funds,….
tails on both sides of the mean
3) A positive skewness (right-skewed) indicates that the righttail of the distribution is longer that the right (few extreme
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-
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The quantile: A probability distribution can also be described by its quantile which is the cut-off
0.6
point with an associated probability α %:
0.5
♦Mathematical definition: the α % quantile ( 0 ≤ α ≤ 1) of a
~
random variable X , corresponds to the cut-off point Qα such
Probability Density
0.4
Empirical Transaction-time Return
0.3
Normal Law
that there is a probability of α % that the random variable will
0.2
fall below it:
F ( x ) = P[ X ≤ Qα ] = α
0.1
-4
-3
-2
-1
0
0
Centered Reduced Return
1
2
3
4
See Value-At-Risk
Source: Maillet et Michel, (2001); densité associée aux rentabilités du titre Elf mesurées en intra-day sur les six premiers mois (+ de 180 000 points).
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Figure 1: Normal Return Density Functions
3.3. Some Important Probability Distribution:
for Two Asset Returns
-
The normal distribution (Abraham de Moivre, 1733 and
3,00
Karl Gauss, 1809): The normal distribution plays a
2,50
central role in statistics as well as in finance because of
its
convenient
properties.
2,00
In
particular, the normal distribution is stable under
1,50
addition (any linear combination of jointly
1,00
normally distributed random variables has a
0,50
normal
distribution),
provides
the
limiting
distribution of the average of independent
0,00
-1
-0,5
0
0,5
1
1,5
random variables (through central limit theorem or CLT) and can be entirely characterised by its first two moments only, the mean and the
♦ Symmetrical distribution: It is symmetrical around the mean, which is the same as its mode (most likely or highest point
variance:
• The Normal Distribution: A random variable X is said to follow a normal distribution (e.g. to be normally Three useful properties
distributed)
with
mean
µ
and
variance σ if it such as:
(
N (x µ , σ ) = f (x µ , σ ) = 2πσ
)
2 −1 / 2
1 x − µ 2 exp −
2 σ
where E ( X ) = µ , Var ( X ) = σ and − ∞ < x < +∞ .
2
147
of the distribution) and median (which has a 50% of occurrence). The skewness of a normal distribution is null which indicates its symmetry around the mean
♦ Lower tails: Its standardized kurtosis is equal to three
(distributions with fatter tails have a greater standardized coefficient). 148
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• The Standard Normal Distribution: The use of the
♦ Confidence intervals: If X is normally distributed with mean µ and standard deviation σ , then we know by
normal
construction that the following inequalities are necessary true
simplified by transforming a normal random
(confidence intervals):
variables X with mean µ and standard deviation
distribution
can
be
considerably
σ into a standard normal random variables Z
P[µ − σ ≤ X ≤ µ + σ ] = 0.683
P[µ − 2 × σ ≤ X ≤ µ + 2 × σ ] = 0.955
P[µ − 3 × σ ≤ X ≤ µ + 3 × σ ] = 0.997 i
P[µ − 4 × σ ≤ X ≤ µ + 4 × σ ] = 0.999
which has been standardized or normalized so that it possesses a zero mean and an unitary variance, that is:
Z=
X −µ
σ
where E (Z ) = 0 , Var (Z ) = 1.
The standard normally distributed random variable Z has a probability density function denoted f (.) and a probability
That is: 68%, 95.5%, 99.7% and 99.9% of the realizations are respectively contained
in
[µ − σ , µ + σ ], [µ − 2σ , µ + 2σ ], and [µ − 4σ , µ + 4σ ].
the
closed
sets:
distribution denoted F (.) such as:
1
N (z 0, 1) = f (z 0, 1) = (2π )−1 / 2 exp − z 2
2
[µ − 3σ , µ + 3σ ] and: F (z ) =
z
∫
−∞
(2π )−1/ 2 exp − 1 z 2 dz
2
with − ∞ < z < +∞ .
The standard normal distribution is characterized by the three following properties:
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♦ Symmetrical distribution: It is symmetrical around the
Remark: The distribution of any random variable X can be
mean which is zero.
recovered from that of the standard normal Z by defining:
X = µ + Z ×σ
f (− z ) = f ( z ) by symmetry
P(Z ≥ z ) = [1 − F ( z )]
F (− z ) = 1 − F ( z ) by symmetry
X has indeed the desired moments, that is:
E ( X ) = µ + E (Z ) × σ = µ
2
2
Var ( X ) = Var (Z ) × σ = σ
♦ Lower tails: Its kurtosis is equal to three (distributions with fatter tails have a greater coefficient).
♦ Confidence intervals: If Z is a standard normal, then we know that:
Figure: Standard Normal Return Density Function
0,40
X −µ
P[− 1 < Z < +1] = P − 1 ≤ σ ≤ +1 = 0.683
X −µ
< +2 = 0.955
P[− 2 < Z < +2] = P − 2 <
σ
P[− 3 < Z < +3] = P − 3 < X − µ < 3 = 0.997
σ
P[− 4 < Z < +4] = P − 4 < X − µ < 4 = 0.999
σ
0,30
68.3 %
95 %
0,20
0,10
99 %
That is 68%, 95% and 99% of the distribution are respectively
0,00
-4
contained between ± 1, ± 2 and ± 3 .
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-3
-2
-1
0
1
2
3
4
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• Financial application (distribution of future discrete asset
returns): A traditional assumption made in finance is that the discrete one period asset returns are normally distributed, that is:
(
~
Ri ,t +1 ~ N (µi , σ i ) = 2πσ i2
)
−1 / 2
2
1 R i ,t +1 − µi
exp −
σi 2
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
1)The probability of the negative return can be computed by using the standardized variate:
R p ,t +1 − µ p 0 − µ p
≤
P (R p ,t +1 ≤ 0 ) = P
σp σp
− 0.1 0 − 0.1
R
= P H ,t +1
≤
0. 2
0. 2
= P(Z ≤ −0.5) = 0.3085
where:
That is 30.85% (read tabulated standard normal cumulative
− ∞ < Ri ,t +1 < +∞
distribution) !!!!
with a fixed mean µi and standard deviation σ i .
The 95.5% confidence interval on the future value of the fund
Example: Consider a mutual fund (or a stock index, a
can be computed as follows:
commodity,….) which is normally distributed with an expected return of 10% per annum and a standard deviation of 20% per annum. P(0.1 − 2 × 0.2 ≤ RH ≤ 0.1 + 2 × 0.2 ) = 0.955
⇔ P(− 0.3 ≤ RH ,t +1 ≤ 0.5) = 0.95
Since the future NAV is equal to:
NAVt +1 = NAVt × (1 + R p ,t +1 )
1) What is the probability that the fund (asset) realizes a loss next year? The 95.5% confidence interval for the NAV of the Hedge fund
2) If the NAV (net asset value) of the mutual fund is actually
1000, define a 95.5% confidence interval for its future value at the end of the year.
in one year is:
1000 × (1 − 0.3) ≤ NAV p ,t +1 ≤ 1000 × (1 + 0.5)
⇒ 700 ≤ NAV p ,t +1 ≤ 1500
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The Lognormal Distribution: A random variable
X is said to have a lognormal distribution if its logarithm Y = ln X is normally distributed. The
Lognormal Density Function
essence of lognormality is the idea that if a random variable Y is normally distributed, then the random variable
X = eY
is distributed
lognormally.
• Mathematical definition:
0,9
If the random variable
Y = ln( X ) is normally distributed with a mean
0,8
µ and a standard deviation σ , X is lognormally
0,7
distributed random variable with a lognormal
0,5
density function lN (.) :
(ln x − µ )2
−1
X ~ LN ( µ , σ ) = (σx 2π ) exp −
2σ 2
Y = ln ( X ) ~ N (µ , σ )
0,4
0,3
0,2
with mean µ X and variance σ X given by:
1
µ+ σ2
µ X = E( X ) = e 2
σ 2 = Var ( X ) = e 2× µ +σ 2
X
0,6
0,1
0
0
2
× eσ − 1
1
2
3
4
5
6
7
8
9
10
This distribution is skewed to the right with x > 0 . Its tail increases for greater values of σ . This explains why as the variance of ln X increases, the mean of X is pulled up.
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• Financial application (distribution of future discrete asset
Example: Consider previous mutual fund (or a stock index, a
returns): Another commonly used assumption
commodity,….) but supposed now that its future discrete return
made in finance is that the log-returns are
is lognormally distributed with an expected return of 10% per
normally distributed which implies that the
annum and a standard deviation of 20% per annum.
discrete future asset returns are lognormally
1) What is the probability that the fund (asset) realizes a loss next
distributed, that is:
year?
2) If the NAV (net asset value) of the mutual fund is actually
(ri ,t +1 − µ r )2
−1
−1
Ri ,t +1 ~ LN ( µ , σ ) = ( 2π σ r ) (1 + Ri ,t +1 ) Exp −
−1
2
2σ r
~
Pi ,t +1
(
)
ri ,t +1 = ln
P ~ N µr , σ r i ,t
1000, define a 95.5% confidence interval for its future value at the end of the year.
1)The probability of the negative return can be computed by using the gross discrete return, that is:
P (R p ,t +1 ≤ 0 ) ⇔ P (1 + R p ,t +1 ≤ 1)
with discrete (log-normal) return mean and variance given by:
(
rp ,t +1 = ln (1 + R p ,t +1 )
Since
2
µr + σ r −1
µ = exp
2
2
2
σ = exp 2 µ r + σ r exp σ r − 1
is
by
definition
normally
distributed, we have to determine its first two moments µ and
)[ ( ) ]
σ.
where µ r and σ r represent respectively the mean and the
Recall that:
2
µ + σ = 1.1
E 1 + Ri , t +1 = exp
2
2
2
2
Var 1 + R i ,t1 = exp 2 µ + σ exp σ − 1 = (0.2 ) = 0.04
(
standard deviation of the log-return ri ,t +1 and:
− 1 < Ri ,t +1 < +∞
)
(
157
)
(
)[ ( ) ]
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So that:
Preliminary Version. Financial Maths Review Course – MSGP ESCP Europe (2011-2012)
Since the future NAV is equal to:
µ = ln E (1 + Ri ,t +1 ) − 0.5 × σ 2 = ln(1.1) − 0.5 × 0.0325 ≈ 0.08
Var (1 + Ri ,t +1 )
2
= 0.0325
σ = ln 1 +
E (1 + Ri ,t +1 )2
2
σ = σ = 0.18
The probability of the negative return then can be computed by using the standardised variable:
[
P (1 + R p ,t +1 ≤ 1) = P ln(1 + R p ,t +1 ) ≤ 0
NAVt +1 = NAVt e
(
ln 1+ R p , t +1
)
The 95.5% confidence interval for the NAV of the mutual fund in one year is:
1000 × exp − 0.28 ≤ NAVH ,t +1 ≤ 1000 × exp + 0.44
⇒ 755.78 ≤ NAVH ,t +1 ≤ 1552.71
]
ln(1 + R p ,t +1 ) − µ 0 − µ
= P
≤
σ σ
ln(1 + R p ,t +1 ) − 0.08 0 − 0.08
= P
≤
0.18
0.18
= P(Z ≤ −0.44 ) = 0.33
Thank you for your attention!!!!!
That is 33% (instead of 30.85% in the previous normally distributed discrete return hypothesis) !!!!
The 95.5% confidence interval on the future value of the fund can be computed as follows:
P (0.08 − 2 × 0.18 ≤ ln (1 + R p ,t +1 ) ≤ 0.08 + 2 × 0.18) = 0.955
⇔ P(− 0.28 ≤ ln (1 + RH ,t +1 ) ≤ 0.44 ) = 0.955
159
160
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