Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 1

Let $({\tau }_n)$ be a localizing sequence for $M$. For $s\le t$, the optional sampling theorem givesBoth sides converge almost surely to $M_t$ and $M_s$, and $|M_{u\wedge {\tau }_n}|\le Z$. Conditional dominated convergence therefore yields $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$, so $M$ is a martingale. Moreover, the family $(M_t{)}_{t\ge 0}$ is dominated by the integrable random variable $Z$, hence is uniformly integrable. Thus $M$ is a uniformly integrable martingale.

Put $A_t={\int }_0^t{H}_s^{2}ds$. It is continuous and tends to infinity almost surely, so $A_{T_{\sigma }}={\sigma }^2$. Define the right-continuous inverse $T(u)=\inf \{t:A_t>u\}$ andThe time-change theorem for local martingales shows that $W$ is a continuous local martingale in the time-changed filtration, andFor completeness, this proves the required case of the Dambis-Dubins-Schwarz theorem: for every $\theta \in \mathbb{R}$, Itô formula makes $\exp (i\theta W_u+{\theta }^{2}u/2)$ a local martingale; stopping and conditioning show that $W_v-W_u$ has conditional characteristic function $e^{-{\theta }^2(v-u)/2}$. Hence the increments are independent centered normal variables with the Brownian variances, and continuity makes $W$ a Brownian motion. ConsequentlyThus $X_{\sigma }$ is centered Gaussian with variance ${\sigma }^2$.

For $\theta >0$, the Exponential martingale for Brownian motion and the Doob maximal inequality for a nonnegative submartingale giveTaking $\theta =a/s$ gives $e^{-a^2/(2s)}$. Apply the same argument to $-B$ and use the union bound:

Itô formula givesso $M$ is a positive local martingale. On every finite interval $[0,T]$, the hypotheses implyThe Gaussian tail bound of the Brownian maximum has finite exponential moments of every subquadratic power. Therefore Novikov condition holds on $[0,T]$, and the stochastic exponential $M$ is a true martingale there. Since $T$ was arbitrary, $M$ is a true martingale.