Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 27 / 1

Put $V_t=⟨X{⟩}_t$ and $S_t={\sup }_{s\le t}|X_s|$. Suppose first that $|X|\le K$. The Itô formula for $|x{|}^p$, which is twice continuously differentiable for $p\ge 2$, givesFor $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb{E}{V}_t<\infty$ makes it a square-integrable martingale. ConsequentlyThis is the first required estimate.

A bounded local martingale is a true martingale. Apply the allowed Doob Lp maximal inequality, and then Hölder's inequality with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1){)}^p$, this yieldsIf the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb{E}{X}_t^2=\mathbb{E}{V}_t$ and Doob's inequality. A usable constant is thereforeIn particular $C_2=4$. This is the upper maximal moment bound for a continuous local martingale.

For an unbounded continuous local martingale, stop at ${\tau }_n=\inf \{s:|X_s|\ge n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge {\tau }_n}\le V_t$. The proved inequality gives a bound by $C_p\mathbb{E}{V}_t^{p/2}$, independent of $n$. Continuity makes ${\tau }_n\uparrow \infty$, and the stopped maxima increase to $S_t$. Monotone convergence proves the same inequality for the original unbounded process, with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.

Use $V=⟨X⟩$. Applying the Itô formula and the Itô product rule givesThe finite-variation terms cancel in the prescribed combination, leavingThus $Y$ is initially a continuous local martingale. It is a true martingale, not merely local. On each fixed horizon $T$,Part (a) with $p=4$, the assumed bracket moments and Cauchy-Schwarz inequality make this bound integrable. The integrable-supremum martingale criterion now applies, by localization and dominated conditional expectations. The same reasoning makes $X^2-V$ a martingale, so $\mathbb{E}{V}_t=\mathbb{E}{X}_t^2=t$.

The zero covariance now means $\mathbb{E}(X_t^2{V}_t)=t^2$. Since $Y_0=0$ and $\mathbb{E}{Y}_t=0$,This is the fourth-moment deficit and bracket variance identity. If equality holds for every $t$, then $V_t=t$ almost surely for each $t$. Take a single probability-one event for all rational times and use continuity to obtain $V_t=t$ simultaneously for all times. The Lévy characterization of Brownian motion then proves $X$ is Brownian motion in its given filtration. A proof of that characterization by conditional characteristic functions is included in Question 2(a).