Investment Science Chapter 3 Sols

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Investment Science Chapter 3
Dr. James A. Tzitzouris

3.1 Use A= 1− rP
1 (1+r)n

with r = 7/12 = 0.58%, P = $25, 000, and n = 7 × 12 = 84, to obtain A = $377.32. 3.2 Observe that since the net present value of X is P , the cash flow stream arrived at by cycling X is equivalent to one obtained by receiving payment of P every n + 1 periods (since k = 0, . . . , n). Let d = 1/(1 + r). Then ∞

P∞ = P
k=0

(dn+1 )k .

Solving explicitly for the geometric series, we have that P∞ = Denoting the annual worth by A, we must have A= rP , 1 − dn P . 1 − dn+1

so that solving for P as a function of P∞ and substituting the result into the equation for A, we arrive at 1 − dn+1 A=r P∞ . 1 − dn 1

Investment Science Chapter 4 Solutions to Suggested Problems Dr. James A. Tzitzouris

4.1 (One forward rate) f1,2 = (1 + s2 )2 1.0692 −1= − 1 = 7.5% (1 + s1 ) 1.063

4.2 (Spot Update) Use f1,k . Hence, for example, f1,k . All values are f1,2 f1,3 f1,4 f1,5 f1,6 5.60 5.90 6.07 6.25 6.32 (1.061)6 = 1.05 1/5

(1 + sk )k = 1 + s1

1/(k−1)

−1

− 1 = 6.32%

1

4.3 (Construction of a zero) Use a combination of the two bonds: let x be the number of 9% bonds, and y teh number of 7% bonds. Select x and y to satisfy 9x + 7y = 0, x + y = 1. The first equation makes the net coupon zero. The second makes the face value equal to 100. These equations give x = −3.5, and y = 4.5, respectively. The price is P = −3.5 × 101.00 + 4.5 × 93.20 = 65.90. 4.5 (Instantaneous rates) (a) es(t2 )t2 = es(t1 )t1 eft1 ,t2 (t2 −t1 ) =⇒ ft1 ,t2 = (b) r(t) = limt→t1 (c) We have d(ln x(t)) = r(t)dt, = s(t)dt + s (t)dt, = d[s(t)t]. Hence, ln x(t) = ln x(0) + s(t)t, and finally that x(t) = x(0)es(t)t . This is in agreement with the invariance property of expectation dynamics. Investing continuously give the same result as investing in a bond that matures at time t. 4.6 (Discount conversion) s(t)t−s(t1 )t1 t−t1

s(t2 )t2 −s(t1 )t1 t2 −t1

=

d[s(t)t] dt

= s(t) + s (t)

2

The discount factors are found by successive multiplication. For example, d0,2 = d0,1 d1,2 = 0.950 × 0.940 = 0.893. The complete set is 0.950, 0.893, 0.770, 0.707, 0.646. 4.7 (Bond taxes) Let t be the tax rate, xi be the number of bond i purchased, ci be the coupon of bond i, pi be the price of bond i. To create a zero coupon bond, we require, first, that the after tax coupons match. Hence x1 (1 − t)c1 + x2 (1 − t)c2 = 0, which reduces to x1 c1 + x2 c2 = 0. Next, we require that the after tax final cash flows match. Hence p0 = x1 p1 + x2 p2 . Using this last relation in the equationfor final cash flow, we find x1 + x2 = 1. Combining these equations, we find that p0 = c2 p1 − c1 p2 . c2 − c1

After plugging in the given values, we find that p0 = 37.64.

4.8 (Real zeros) We assume that with coupon bonds there is a capital gains tax at maturity. We replicate the zerocoupon bond’s after-tax cash flows using bonds 1 and 2. Let xi be the amount of bond i required 3

4.11 (Running PV examples)

(a)

d0,1 d0,2 d0,3 d0,4 d0,5 d0,6 0.9524 0.9018 0.8492 0.7981 0.7472 0.7010 0 −40 0.9524 9.497 1 10 0.9469 51.970 2 10 0.9416 44.324 34 10 0.9399 36.453

=⇒ N P V = 9.497 6 10 10 9381 19.381 10.000

Year Cash Flow (b) Discount PV(n)

5 10 0.9362 28.144

4.12 (Pure duration)
n n −k

P (λ) = dP (λ) dλ
k=0 n

xk (1 + sk /m) xk
k=0 n

−k

=
k=0

xk (1 + s0 /m)eλ/m k
−k−1

,

= =

−k m −k m

(1 + s0 /m)eλ/m k (1 + sk /m)−k ,

(1 + s0 /m)eλ/m , k

xk



1 dP (λ) P (λ) dλ

=

k=0 n k −k k=0 xk m (1 + sk /m) n −k k=0 xk (1 + sk /m)

≡ D.

This D exactly corresponds to the original definition of duration as a cash flow weighted average of the times of cash payments. No modification factor is needed even though we are working in discrete time. 4.14 (Mortgage division) (a) (1 + r)k−1 − 1 r

P (k) = B − rM (k − 1) = B − r (1 + r)k−1 M (0) − = (1 + r)k−1 (B − rM ), 6

B ,

Solution of HW3
Problem 6.1
X0 = outlay (in this case it is equal to the...
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