Let $(B_t)$ be a Brownian motion and set $X_t = \int_0^t B_s^2 dB_s$. Is $X_t$ martingale?
My idea is to rewrite $X_t$ in terms of Ito's Formula $(f(x) = \frac{1}{3}x^3)$
$X_t = \int_0^t B_s^2 dB_s = 2 \int_0^t B_s d \langle B \rangle_s – \frac{1}{3} B_t^3$
and since $-f$ for $x \geq 0$ is a concave function, $- \frac{1}{3} B_s^3 $ is a supermartingale, thus a martingale. So for $X_t$ to be a martingale, $\int_0^t B_t^2 dB_s$ needs to be martingale. But since it is locally of bounded variation, if we assume it is a martingale it is constant in t, which can not be true (a brownian motion is not equal to zero for all t).
Best Answer
Yes, the process $(X_t)$ defined by $X_t=\displaystyle\int_0^tB_s^2\mathrm dB_s$ is a martingale, as every square integrable stochastic integral $\displaystyle\int_0^tu(s,B_s)\mathrm dB_s$, but, no, going back to Itô's formula to show this is not useful since this is going backwards because one already knows that $\mathrm dX=B^2\mathrm dB$, which is the kind of conclusion Itô's formula can help to reach.
No, the process $(Y_t)$ defined by $Y_t=B_t^3$ is not a martingale since, for every $t\gt s\gt0$, the decomposition $B_t=B_s+(B_t-B_s)$ with $B_s$ measurable with respect to $\mathcal F_s$ and $B_t-B_s$ independent of $\mathcal F_s$ yields the almost sure identity $E(B_t^3\mid\mathcal F^B_s)=B_s^3+3(t-s)B_s\ne B_s^3$.
No, a process being a supermartingale does not imply that it is a martingale, and no, the process $-\frac13B^3$ is not a supermartingale (and not a submartingale either).