Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 29 / 1 / b / Solution

Use the representation from part (a), and put ${\Delta }_{j}M=M_{T_{j+1}}-M_{T_j}$, $U_j=h_j{\Delta }_{j}M$ and $N=H\cdot M$. The optional stopping theorem in the form needed here says that for a uniformly integrable martingale $L$ and stopping times $\sigma \le \tau$, including terminal infinite times, $\mathbb{E}[L_{\tau }\mid {\mathcal{F}}_{\sigma }]=L_{\sigma }$. An L2-bounded continuous martingale is uniformly integrable and has the required terminal value.

Consequently $\mathbb{E}[{\Delta }_{j}M\mid {\mathcal{F}}_{T_j}]=0$. Conditioning on ${\mathcal{F}}_t$ and distinguishing whether $T_j\le t$ givesThus $N_t=\mathbb{E}[{\sum }_j{U}_j\mid {\mathcal{F}}_t]$ is a martingale. Its paths are continuous, since each stopped increment is continuous and zero until its left endpoint.

For $i<j$, $U_i$ is ${\mathcal{F}}_{T_j}$-measurable, so the same conditional identity yields $\mathbb{E}(U_i{U}_j)=0$. These martingale-difference orthogonality relations also hold for the unweighted increments. ThereforeHere $N_{\infty }={\sum }_j{U}_j\in L^2$. By conditional Jensen, $\mathbb{E}{N}_t^2\le \mathbb{E}{N}_{\infty }^2$ for every $t$, soThis proves the norm bound without presupposing the Itô isometry requested next.