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

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.