= Solution
Use $V=\langle X\rangle$. Applying the <Itô formula> and the <Itô product rule> gives
$$
d(X^4)=4X^3\,dX+6X^2\,dV,\qquad d(X^2V)=2XV\,dX+(V+X^2)\,dV,\qquad d(V^2)=2V\,dV.
$$
The finite-variation terms cancel in the prescribed combination, leaving
$$
dY_t=(4X_t^3-12X_tV_t)\,dX_t.
$$
Thus $Y$ is initially a <continuous local martingale>. It is a true <martingale>, not merely local. On each fixed horizon $T$,
$$
\sup_{s\leq T}|Y_s|\leq S_T^4+6S_T^2V_T+3V_T^2.
$$
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 $\mathbb EV_t=\mathbb EX_t^2=t$.
The zero <covariance> now means $\mathbb E(X_t^2V_t)=t^2$. Since $Y_0=0$ and $\mathbb EY_t=0$,
$$
\boxed{\mathbb EX_t^4=6t^2-3\mathbb EV_t^2=3t^2-3\operatorname{Var}(V_t)\leq3t^2.}
$$
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 \b[$X$ is <Brownian motion> in its given filtration]. A proof of that characterization by conditional characteristic functions is included in Question 2(a).
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