Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 37 / 6

Put $R_{\epsilon }={\int }_0^{\epsilon }(H_u-H_0)d{B}_u$ and $d_{\epsilon }={\epsilon }^{-1}{\int }_0^{\epsilon }\mathbb{E}|H_u-H_0{|}^{2}du$. Boundedness and path continuity of $H$ make $\mathbb{E}|H_u-H_0{|}^2\to 0$ by the dominated convergence theorem, so $d_{\epsilon }\to 0$. The Itô isometry gives $\mathbb{E}{R}_{\epsilon }^2=\epsilon d_{\epsilon }$.

We must not assume independence between the integral and its Brownian denominator. Instead, the Cauchy-Schwarz inequality givesConcavity, or Jensen inequality, bounds the first factor by $(\mathbb{E}{R}_{\epsilon }^2{)}^{1/8}$. Since $B_{\epsilon }\stackrel{d}{=}\sqrt{\epsilon }N$, the second factor is ${\epsilon }^{-1/8}(\mathbb{E}|N{|}^{-1/2}{)}^{1/2}$. This negative normal moment is finite: the density is bounded near zero, where ${\int }_0^1{u}^{-1/2}du<\infty$, and has Gaussian decay at infinity. In fact $\mathbb{E}|N{|}^{-1/2}=2^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The powers of $\epsilon$ cancel, yielding the fractional-moment control of Brownian increment ratiosThe ratio may be assigned any value on the null event $B_{\epsilon }=0$.

Fix a deterministic $t>0$ and define ${\hat{B}}_u=B_{t+u}-B_t$, ${\hat{H}}_u=H_{t+u}$ and ${\hat{\mathcal{F}}}_u={\mathcal{F}}_{t+u}$. The increment process is a standard Brownian motion in the shifted filtration, and $\hat{H}$ remains bounded, continuous and adapted. Thus part (a) applies even though the initial value ${\hat{H}}_0=H_t$ is random. For $\epsilon >0$,Here the constant-in-time integrand $H_t$ is ${\mathcal{F}}_t$-measurable, so its stochastic integral equals $H_t(B_{t+\epsilon }-B_t)$. The expectation of the absolute value of the right-hand side to the power $1/4$ tends to zero by part (a). For every $\eta >0$, Markov inequality consequently givesThereforeThe stated convergence in probability holds for each fixed deterministic time.