Past exam of the martingales and financial mathematics course of the University of Oxford / 2010 / B / 5

For the time integral $I_t={\int }_0^t{W}_{s}ds$, Fubini's theorem and $\mathbb{E}{W}_s=0$ give $\mathbb{E}{I}_t=0$. Brownian covariance is $\mathbb{E}[W_u{W}_v]=\min (u,v)$, soConsequently

For geometric Brownian motion, $\log S_u=\log S_0+(\mu -{\sigma }^2/2)u+\sigma W_u$. Averaging in time givesThus $\log A_t$ is normal with mean $\log S_0+(\mu -{\sigma }^2/2)t/2$ and variance ${\sigma }^{2}t/3$. The moment-generating function of a normal distribution gives

For $v\le t$, the two stock paths agree, so their geometric averages agree. For $v>t$, the future log-price isSplit the average's log-integral into $[0,t]$ and $[t,v]$. The first part is $t\log A_t$, while integration of the displayed formula givesDividing by $v$ and exponentiating proves

Put $\tau =T-t$. Conditional on present information, $J={\int }_t^T(W_u-W_t)du$ is independent of the past and is normal with mean zero and variance ${\tau }^3/3$. Part (c) givesThe future drift $r$ in this auxiliary stock is precisely the risk-neutral drift. Discounting the conditional expectation and applying the normal exponential-moment formula yieldsAt expiry this reduces to $A_T^2$; at time zero it gives $S_0^2{e}^{{\sigma }^{2}T/6}$. This is a special case of conditional power moments of a continuous geometric Asian average.