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

First justify optional sampling for the square-minus-bracket process. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_{t\ge 0}|M_t{|}^2\le 4\mathbb{E}|M_{\infty }{|}^2<\infty$. Stop the local martingale $M^2-[M]$ using a localizing sequence and bounded bracket levels. The stopped expectation identity, followed by Fatou, bounds $\mathbb{E}[M{]}_{\infty }$ by the expected squared path supremum. Hence $M^2-[M]$ has an integrable absolute supremum, and localization with conditional dominated convergence makes it a uniformly integrable martingale. The optional stopping theorem stated in part (b) is therefore applicable to both $M$ and $M^2-[M]$.

For ordered stopping times $\sigma \le \tau$, expand the squared increment and condition on ${\mathcal{F}}_{\sigma }$. The mean increment is zero, soThis is the conditional bracket isometry for stopped martingale increments. Apply it with $\sigma =T_j$, $\tau =T_{j+1}$ and multiply by the bounded ${\mathcal{F}}_{T_j}$-measurable variable $h_j^2$. The orthogonality proved in part (b) gives the Itô isometry

To identify the whole bracket process, let $N=H\cdot M$ and $C_t={\int }_0^t{H}_s^{2}d[M{]}_s$. For deterministic $s<t$ and $E\in {\mathcal{F}}_s$, apply the proved isometry to the simple integrand $1_E{1}_{(s,t]}H$. It yieldsThus $\mathbb{E}[(N_t-N_s{)}^2\mid {\mathcal{F}}_s]=\mathbb{E}[C_t-C_s\mid {\mathcal{F}}_s]$. Since $N$ is a martingale, the cross term $2N_s(N_t-N_s)$ has conditional mean zero. Consequently $N^2-C$ is a martingale. The process $C$ is continuous, adapted, nondecreasing and starts at zero, so uniqueness in the definition of quadratic variation provesThis deduction uses only the proved elementary isometry, not a previously assumed formula for the quadratic variation of stochastic integrals.