Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 39 / 4

Use the usual Black-Scholes model with $\sigma >0$ and the filtration generated by $W$. SetFor $u>t$, independence and the Gaussian distribution exponential moment formula give ${\mathbb{E}}_P[Z_u/Z_t\mid {\mathcal{F}}_t]=1$. Thus $Z$ is a strictly positive martingale of mean one, defining an equivalent measure $Q$ on every finite horizon by $dQ/dP{|}_{{\mathcal{F}}_t}=Z_t$.

To verify the measure change directly, put ${\hat{W}}_t=W_t+\lambda t$. For any real $a$ and $h=u-t$,The conditional characteristic function is that of $N(0,h)$ and does not depend on past information. Therefore $\hat{W}$ has independent increments with Gaussian distribution and continuous paths, and is a Brownian motion under $Q$. Substituting $dW=d\hat{W}-\lambda dt$ gives $d{S}_t=S_t(rdt+\sigma d{\hat{W}}_t)$. Hence $S_t/B_t$ is a geometric Brownian motion that is a martingale and $Q$ is an equivalent martingale measure.

The form is also forced. By the Brownian martingale representation theorem, any density process that is a martingale has $d{Z}_t=a_{t}d{W}_t$. The change of measure identity makes $Z_t{S}_t/B_t$ a local martingale under $P$. Its drift, calculated by Itô formula, is $(S_t/B_t)[Z_t(\mu -r)+\sigma a_t]dt$, which must vanish. Thus $a_t=-\lambda Z_t$. Solving $d{Z}_t=-\lambda Z_{t}d{W}_t$ with $Z_0=1$ gives exactly the displayed stochastic exponential. Equivalence here is on each finite horizon, as usual for Black-Scholes pricing.

Let $\Phi$ denote the standard normal distribution cumulative distribution function, and $\varphi$ its density. Put $\tau =T-t$. For $K>0$ and $t<T$, defineUnder the risk-neutral measure, conditional on ${\mathcal{F}}_t$,Thus $Q(S_T<K\mid {\mathcal{F}}_t)=\Phi (-d_2)$. The identity $e^{az}\varphi (z)=e^{a^2/2}\varphi (z-a)$ for the standard Gaussian distribution probability density function givesSplit the payout according to whether the stock exceeds the floor and discount. The risk-neutral pricing process isIts discounted value is ${\mathbb{E}}_Q[{\xi }_T/B_T\mid {\mathcal{F}}_t]$, a true martingale. Thus adding this asset preserves the equivalent martingale measure and gives no arbitrage. This is the guaranteed terminal stock floor in the Black-Scholes model. For $K\le 0$, positivity of the stock makes the claim simply $S_T$, and the price is ${\xi }_t=S_t$.

Let $\varphi$ denote the standard Gaussian distribution probability density function. Directly from the definitions, $S\varphi (d_1)=K{e}^{-r\tau }\varphi (d_2)$. Differentiating the price in part (ii), the terms involving derivatives of $d_1,d_2$ cancel, yielding $V_S=\Phi (d_1)$. Therefore, in units of the two traded assets, the replicating strategy isIts value is $h_t^B{B}_t+h_t^S{S}_t=V(t,S_t)$. To check self-financing explicitly, differentiation also givesHence $V_t+rS{V}_S+\frac{1}{2}{\sigma }^2{S}^2{V}_{SS}=rV$. Applying Itô formula under the physical measure givesThus the holdings form a self-financing strategy and their terminal value is the required payout. Holdings at the single maturity instant can be assigned by their limiting values; the continuous-time trading gains do not depend on that assignment. If ${\pi }_t$ denotes wealth fractions instead, the stock fraction is $S_t\Phi (d_1)/V(t,S_t)$ and the bank fraction is $K{e}^{-r\tau }\Phi (-d_2)/V(t,S_t)$. For $K\le 0$, hold one stock and no bank-account units.