Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 36 / 3 / b

Fix a finite horizon $T$ and define ${\tau }_n=\inf \{s\ge 0:|B_s|\ge n\}$. These stopping times increase to infinity almost surely because continuous paths are bounded on each finite interval. Applying the Itô formula to $B_{t\wedge {\tau }_n}^k$, for $k\ge 2$, givesThe first integrand is bounded by $k{n}^{k-1}$, so its integral is square-integrable and has expectation zero. ThusTo justify both limits explicitly, let $S_T={\sup }_{s\le T}|B_s|$ and choose an integer $q>\max \{k,1\}$. The Gaussian density gives $\mathbb{E}|B_T{|}^q<\infty$ by its exponential tail. The Doob Lp maximal inequality statesTherefore $S_T^k$ and $1+S_T^{k-2}$ are integrable. The stopped terminal powers converge almost surely to $B_t^k$ and are dominated in absolute value by $S_T^k$. The integrands in the time integral converge almost surely and are bounded by $1+S_T^{k-2}$, so their integrals are dominated by $T(1+S_T^{k-2})$. Dominated convergence theorem on both sides, and then Fubini justified by the same bound, give the Brownian moment recursionThis uses only finiteness of Gaussian moments, not the closed moment formula requested in the next part.