Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 211 / 4

Localize when the predictable integrand first exceeds level $n$, just before that increment is taken. The resulting stopped integrand is bounded, so its martingale transform is a martingale by the stated result. The localization times increase to infinity because each finite collection of $H_t$ is finite almost surely. Hence $X$ is a local martingale.

On $\{H_T\ne 0,X_{T-1}\le 0\}$,Dividing by $\Vert H_T\Vert$ proves $\xi \ge 0$ there; outside that event $\xi =0$.

If $X_{T-1}<0$, then $X_T\ge 0$ forces $H_T\ne 0$ and a strictly positive increment in the $H_T$ direction. If $X_{T-1}=0$, that directional increment is strictly positive exactly when $X_T>0$. These disjoint cases give

If $X_{T-1}=0$ almost surely, the definition of $T$ gives $\mathbb{P}(X_T>0)>0$. Part c then implies $\mathbb{P}(\xi >0)>0$.

If $\mathbb{P}(X_{T-1}<0)>0$, the second term in the identity of part c is positive, so $\mathbb{P}(\xi >0)>0$.

The assumptions of this case imply $X_{T-1}\ge 0$ almost surely and $\mathbb{P}(X_{T-1}>0)>0$. This would satisfy the defining condition for $T$ one period earlier, contradicting the minimality of $T$. Thus the case is impossible.

The one-step predictable integrand $\theta$ is bounded by one. If $M$ were a martingale, its transform $\xi =\theta \cdot (M_T-M_{T-1})$ would be an integrable mean-zero random variable. Parts b–d show instead that it is nonnegative almost surely and strictly positive with positive probability, so its expectation is positive. This contradiction proves that $M$ cannot be a martingale.