CHAPTER 23



Chapter 27 Solutions It?o’s lemma is a useful result because it allows the computation of stochastic differentials of arbitrary functions having as an argument a stochastic process that itself is assumed to possess a stochastic differential. In this respect, It?o’s formula is as useful as the chain rule of ordinary calculus. Given an It?o stochastic process S(t,w) with respect to a given Wiener process Z(t,w), and letting Y (t,w) = u[t, S(t,w)] be a new process, then based on It?o’s lemma and Taylor’s theorem, we can obtain dt × dt = 0, dZ× dZ = dt, dt × dZ = 0dY = utdt + uSdS + 12 uSS (dS)2A stochastic process is an It?o process if the random variable dS(t,w) can be represented bydS(t,w) = μ[t, S(t,w)]dt + σ[t, S(t,w)]dZ(t,w)where The first term, μ[t, S(t,w)]dt, is the expected change in S(t,w) at time t. The second term, σ[t, S(t,w)]dZ(t,w), reflects the uncertain term. dZ(t,w), is called white noise followed standard normal distribution; it denotes an infinitesimal change in the Wiener process, models financial uncertainty in continuous time.The stock price has a constant expected return μ and the volatility of its return is constant σ times the Wiener process dZ(t,w) which follows normal distribution with zero mean and variance, t. Therefore, by equation (27.12) in section 27.3, the stock price follows a lognormal distribution with mean (μ-12 σ2)t and variance σ2t.Assume the stock price follows the stochastic process as followsdS(t,w) = μS(t,w)dt + σS(t,w)dZ(t,w)The expected value of a call option is C=E[e-rT(ST-k)+] where k is the strike price in call option contract.By equation (27.12), we can arrange the call option formula asC=E[e-rTST-k)+=E[e-rT(S0eμ-12σ2T+σZ-k)+] (1)Where z is the normal distribution N(0, T).To derive the Black–Scholes call option model, first we should prove μ equal to risk-free rater, r, in risk-neutral framework.That is, EQe-rTST=S0 under the risk-neutral measure Q.Proof:EQe-rTST=EQe-rTS0eμ-12σ2T+σZ=S0e(μ-r)T=S0 Therefore, μ=r and replace r into equation (1), we can obtain equation (2) as follows:S0er-12σ2T+σZ-k≥0 (2)Let standard normal distribution Y=ZT, then we can rearrange equation (2) asY< d2=1σT[lnkS0+r-12σ2T] (3) Therefore, equation (2) holds if and only if equation (3) holds. Then equation (1) can be written asC=E[ST-k)+=-∞d212πe-rTS0er-12σ2T+σTy-e-rTke-12y2dy =-∞d212πS0e-12y2-σTy-12σ2Tdy-e-rTk-∞d212πe-12y2dy =S0-∞d212πe-12y+σT2dy-e-rTkNd2 =S0-∞d2+σT12πe-12y'2dy-e-rTkNd2 where y'=y+σT =S0Nd1-e-rTkNd2 where d1=d2+σTlet Y=Sn(t,w) and use It?o’s lemma to find dY dSnt,w= dY=?Y?tdt+?Y?SdS+12?2Y(?S)2(dS)2 dY=0dt+nSn-1t,wdS+12(n)(n-1)Sn-2t,w(dS)2 =nSn-1t,wμSt,wdt+σSt,wdZt,w +12nn-1Sn-2t,wσ2S2t,wdt=nμ+12nn-1σ2Snt,wdt+nσSnt,wdZt,w The stock price ST=S0eμ-12σ2t+σZ follows lognormal distribution where Z follows normal distribution with zero mean and variance, T. Given the information μ=0.15, σ=0.2 and S0=$50 in the first two years, then the stock price at the end of two years is S2y=50e0.15-120.22(2)+0.2Z=50e0.26+(0.2)ZWhere Z follows normal distribution with mean zero and variance, 2. Therefore, the stock price follows lognormal distribution with mean 4.172 (ln50+0.26 = 4.172) and variance 0.8 (0.22(2) = 0.8).At the end of three years, given the information μ=0.28, σ=0.35 in the last year, the stock price can be written in terms of the stock price at the end of two years asS3y=S2ye0.28-120.352+0.35Z'=S2ye0.4375+0.35Z' Where Z’ is the normal distribution with mean zero and variance, 1. From the part (a), we can obtainS3y=S2ye0.4375+0.35Z'=50e0.26+0.2Ze0.4375+0.35Z' =50e0.6975+0.2Z+0.35Z'=eln50+0.6975+0.2Z+0.35Z'=e4.6095+(0.2)Z+0.35Z' Since Z and Z’ are two independent normal distributions, therefore the sum of these two independent normal distribution Z”= 0.2Z+0.35Z' is following a normal distribution with mean zero and variance 0.2025 (0.22(2)+0.352 = 0.2025).Then the stock price at the end of three years follows lognormal distribution with mean 4.6095 and variance 0.2025 (σ=0.45).The stock price change following the stochastic process in question 2 is more appropriate than the stochastic process in question 6 because the parameters μ and σ are meaningful for the return of stock price and can be estimated by the expected return and standard deviation of the stock’s returns, respectively. In addition, investors prefer to know how much percent of return for their investment rather than the change of the investment. When the stock price follows the stochastic process as follows:dS(t,w) = μdt + σdZ(t,w), μ=1.5, σ=2 and S0 = $110Then St follows a normal distribution N(S0+ μt, σ2t). Therefore, for next year, S1~ N(110+1.5, 22) = N(111.5, 4). It implies that the stock price in the next year follows a normal distribution with mean 111.5 and standard deviation 4.95% confidence limits for the stock price in the next year are between the mean±standard deviation, that is, [ 111.5- 4, 111.5+ 4] =[107.5, 115.5]When the stock price follows the stochastic process as follows:dS(t,w) = μ S(t,w) dt + σ S(t,w) dZ(t,w), μ=0.7, σ2=25% and S0 = $100Then, the stock price ST=S0eμ-12σ2t+σZ follows lognormal distribution where Z follows normal distribution with zero mean and variance, TAt the end of six months, S0.5y=100e0.7-120.25(0.5)+0.5Z1Where Z1 follows normal distribution with zero mean and variance 0.5.Therefore, the stock price follows lognormal distribution with mean 4.893 (ln100+0.7-120.25(0.5) = 4.893) and variance 0.125 (0.52(0.5) = 0.125)The expected stock price at the end of six months is e 4.893=133.353The standard deviation of the stock price at the end of six months is e 0.125=1.133In next year, S1y=100e0.7-120.25+0.5Z2Where Z2 follows normal distribution with zero mean and variance 1. Therefore, the expected stock price is 100e0.7-120.25=177.713 Assume the stock price follows the stochastic process as follows:dS(t,w) = μ S(t,w) dt + σ S(t,w) dZ(t,w)Y=3 S(t,w), then dY=3 dS(t,w). Therefore, Y follows the stochastic process as dY=3dS(t,w) = 3μ S(t,w) dt + 3σ S(t,w) dZ(t,w)Y= Sn(t,w), then dY= nSn-1(t,w) dS(t,w)+12 n(n-1)Sn-2(t,w) [dS(t,w)]2 = nSn-1(t,w) [μ S(t,w) dt + σ S(t,w) dZ(t,w)]+ 12 n(n-1)Sn-2(t,w) (σ2 ) S2(t,w)dt= (nμ+12 n(n-1) σ2 )Sn(t,w) dt + (n σ) Sn(t,w) dZ(t,w)Y=e r(T-t) Sn(t,w), then dY= -e r(T-t) Sn(t,w)dt+ e r(T-t) {nSn-1(t,w) dS(t,w)+12 n(n-1)Sn-2(t,w) [dS(t,w)]2}=(nμ+12 n(n-1) σ2 -1) e r(T-t) Sn(t,w) dt + e r(T-t) (n σ) Sn(t,w) dZ(t,w) The stock price ST=S0eμ-12σ2t+σZ follows lognormal distribution where Z follows normal distribution with zero mean and variance, T. The stock price in two years in equation 5 is S2y=50e0.15-120.22(2)+0.2Z=50e0.26+(0.2)ZTherefore, the stock price follows lognormal distribution with mean 4.172 (ln50+0.26 = 4.172) and variance 0.8 (0.22(2) = 0.8).The probability that stock price larger than $130 in two year can be calculated asP(S2y>$130)=P(50e0.26+(0.2)Z>130)= PZ>ln13050-0.260.2=P(X>ln13050-0.260.20.2)=P(X>7.7761)=0Where X is the standard normal distribution and Z is the normal distribution with zero mean and variance, 2. ................
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