Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 24 / 3

Use the independent increments of Brownian motion, rather than merely checking that an Itô formula drift vanishes. For $0\le s<t$, put $h=t-s$ and write $B_t=B_s+Z$, where $Z$ is independent of ${\mathcal{F}}_s$ and has normal distribution $N(0,h)$. Its first four moments are $0,h,0,3h^2$. ThereforeFor the cubic expression, the coefficient of $B_s$ after conditioning is $3(t-s)+\alpha (t)$, so $\alpha (t)=-3t$ makes it $\alpha (s)$. For the quartic expression, choose $\beta (t)=-6t$. Its conditioned coefficient of $B_s^2$ is then $6(t-s)-6t=-6s$. The constant term becomeswhich equals $3s^2$ when $\gamma (t)=3t^2$. Thus a standard choice isThe resulting stochastic processes are Hermite polynomial martingales $H_3(B_t,t)$ and $H_4(B_t,t)$. They are genuine integrable martingales, since Gaussian moments are finite at every finite time and the displayed conditional identities establish the martingale property directly.

The choice is not unique. Constants $c,d,e\in \mathbb{R}$ give the valid family $\alpha (t)=c-3t$, $\beta (t)=d-6t$, and $\gamma (t)=3t^2-dt+e$: these add $c{B}_t$ to the cubic martingale and $d(B_t^2-t)+e$ to the quartic one. The boxed choice sets these harmless additions to zero.

Write $\tau ={\tau }_x$ and $T_t=t\wedge \tau$. Path continuity gives $|B_{T_t}|\le x$. Stop the martingale $B_s^2-s$ at the bounded stopping time $T_t$ and apply the optional stopping theorem:By the monotone convergence theorem, $\mathbb{E}\tau \le x^2$, so $\tau <\infty$ almost surely. Path continuity then gives $B_{\tau }\in \{-x,x\}$. The dominated convergence theorem for the bounded variables $B_{T_t}^2$ shows

Next stop the quartic Hermite polynomial martingale from part (a), again only at $T_t$. Its expectation is zero, soMonotone convergence proves $\mathbb{E}{\tau }^2\le 2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_t{B}_{T_t}^2\le x^2\tau$ and $B_{T_t}^4\le x^4$, dominated convergence givesConsequently the Brownian symmetric interval-exit moments are

To obtain the Laplace transform of symmetric Brownian interval-exit time, put $a=\sqrt{2\lambda }$. The Exponential martingale for Brownian motion shows thatis a martingale with $H_0=1$. Bounded-time stopping gives $\mathbb{E}{H}_{T_t}=1$. Its stopped values are bounded by $\cosh (ax)$, so dominated convergence applies as $t\to \infty$. Since $\cosh (a{B}_{\tau })=\cosh (ax)$, it yieldsEvery use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.