Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 4

For $0\le s\le t$, write $B_t=B_s+(B_t-B_s)$. The increment is independent of ${\mathcal{F}}_s$ and is normally distributed with variance $t-s$. Its moment generating function givesThe process is integrable for every real $\lambda$, so it is the exponential Brownian martingale.

Differentiate the conditional identity from part a. To justify doing so, fix a compact parameter interval $|\lambda |\le L$. Every $n$th derivative of $M_{\lambda }(t)$ is a polynomial in $B_t,t,$ and $\lambda$ times $M_{\lambda }(t)$, and its absolute value is bounded byThis bound is integrable because a Gaussian random variable has every polynomially weighted exponential moment. Dominated differentiation of conditional expectation therefore givesThus every parameter derivative of the exponential Brownian martingale is itself a martingale.

Let $T={\tau }_b\wedge {\tau }_{-a}$ and $A=\{{\tau }_b<{\tau }_{-a}\}$. The Brownian exit time $T$ is finite almost surely. Optional stopping of the bounded martingale $B_{t\wedge T}$ givesFor $T_n=T\wedge n$, optional stopping of $B_t^2-t$ gives $\mathbb{E}{T}_n=\mathbb{E}[B_{T_n}^2]\le \max (a^2,b^2)$. Letting $n\to \infty$ by monotone and bounded convergence proves $\mathbb{E}T=\mathbb{E}[B_T^2]=ab$.

The third derivative in part b at $\lambda =0$ is the cubic martingale $B_t^3-3t{B}_t$. Optional stopping at $T_n$ is valid because $T_n$ is bounded. Since $B_{T_n}$ is bounded and $T_n\to T$ in $L^1$, its stopped identity passes to the limit and givesPut $x=\mathbb{E}[T{1}_A]$ and $y=\mathbb{E}[T{1}_{A^c}]$. Then $x+y=ab$, whileSolving gives $x=ab(2a+b)/(3(a+b))$. Dividing by $\mathbb{P}(A)=a/(a+b)$ proves the conditional Brownian interval-exit time formula