# Financial Theory and Corporate Policy (4th Edition) Solution Chapter 7

Topics: Option, Put option, Stock Pages: 11 (3383 words) Published: September 10, 2011
Chapter 7
Pricing Contingent Claims: Option Pricing Theory and Evidence 1. We can use the Black-Scholes formula (equation 7.36 pricing European calls. C = SN(d1 ) − Xe− rf T N(d 2 ) where d1 = ln(S/ X) + rf T + (1/ 2 )σ T σ T

d2 = d1 – σ T Substituting the correct values into d1, we have d1 = d1 = ln(28/40) + .06(.5) + .5( .5)( .5) .5 .5 −.356675 + .03 + .25 = –.40335 .5

d2 = –.40335 – ( .5)( .5) = –.90335 Using the Table for Normal Areas, we have N(d1) = .5 – .1566 = .3434 N(d2) = .5 – .3171 = .1829 Substituting these values back into the Black-Scholes formula, we have C = 28(.3434) – 40e = \$2.52 2. We know from put-call parity that the value of a European put can be determined using the value of a European call with the same parameters, according to equation 7.8 C0 – P0 = (1 + rf )S0 − X (1 + rf )

–.06(.5)

(.1829)

C = 9.6152 – 40(.970446)(.1829)

Solving for P0, the present value of a put, and converting the formula to continuous rather than discrete compounding, we have P0 = C0 − S0 + X − rf T e

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Copeland/Shastri/Weston • Financial Theory and Corporate Policy, Fourth Edition

First, calculate the value of the corresponding call option, according to the Black-Scholes formula (equation 7.36) C = SN(d1) – e − rf T XN(d2) where: d1 = Solving for d1, we have d1 = ln(20 / 20) + (.08)(.5) (.6)( .5) + (1/ 2)(.6)( .5)

ln(S/X) + rf T + (1/ 2)σ T σ T

= .3064 and d2 is equal to d2 = d1 – .6 .5 = .3064 – .4243 = –.1179 From the Table of Normal Areas, we have N(d1) = .5 + .12172 = .62172 N(d2) = .5 – .04776 = .45224 Substituting these values into the Black-Scholes formula yields C = 20(.62172) – e–(.08) (.5)(20)(.45224) = 12.434 – 8.690 = 3.744 Solving for the value of a put, P = 3.744 – 20 + (20)e–(.08)(.5) = –16.256 + 19.216 = \$2.96

Chapter 7

Pricing Contingent Claims: Option Pricing Theory and Evidence

79

3. (a) Figure S7.1a shows the payoffs from selling one call (–C), selling one put (–P), and from the combination (–C–P).

Figure S7.1a Payoffs from selling one call and one put The portfolio (–C–P) is the opposite of a straddle. It earns a positive rate of return if the stock price does not change much from its original value. If the instantaneous variance of the stock ∂C > 0. Suppose the value of the call increases by increases, the value of the call increases, since ∂σ some amount a > 0, C1 = C0 + a Then by put-call parity, the value of the put also increases by a. P1 = C1 – S + Xe = P0 + a If you sold one call and one put for prices of C0 and P0, and the options’ true values were C0 + a and P0 + a, this represents an opportunity loss to you of –2a. Given the inside information, the portfolio strategy would be to buy both the put and the call (at P0 and C0) for a gain of 2a. (b) Figure S7.1b shows the payoffs from buying one call (+C), selling one put (–P) and from the combination (C – P). –rt

= C0 + a – S + Xe–rt

Figure S7.1b Payoffs from buying one call and selling one put

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Copeland/Shastri/Weston • Financial Theory and Corporate Policy, Fourth Edition

The return from this portfolio (C – P) remains unchanged by an increase in the instantaneous variance, since, by put-call parity, the increase in the value of the long position in the call option is exactly offset by an increase in the value of the shorted put option. In the algebra of part (a) above, C0 + a – (P0 + a) = C0 + P0. Therefore, the strategy of this portfolio is neither advantageous nor disadvantageous, given inside information of an increase in the instantaneous variance. 4. The value of the call option can be calculated directly by the Black-Scholes formula for a European call, equation 7.32. Using the correct values from the problem, C = 44.375N(d1) – 45e First calculate d1 and d2: d1 = =

– (.07) (156/365)

N(d2).

ln(44.375 / 45) + (.07)(156 / 365) + (1/ 2)( .0961)( 156 / 365) ( .0961)( 156 / 365) −.01398 + (.07)(.4274) + (1/ 2)(.31)(.6538) (.31)(.6538)

= .18 d2 = .18 – ( .0961) (.6538) =...

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