 Solution

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).