$e^{X_t - \frac{t^3}{6}}$ is a martingale - show it

Question

I am trying to use Ito's integral properties to prove it is a martingale, but am getting stuck in the preliminaries.

More so, I wanted to confirm, do I use this formula?

Answer 1

You have $Y_t = f(X_t, t) = \exp(X_t - t^3/6)$. Assume $dX_t = m(t) dt + s(t) dW_t$ so $(dX_t)^2 = s(t)^2 dt$. Now apply Ito's lemma to $Y_t$: $$ \begin{split} dY_t &= \exp(X_t - t^3/6) \frac{-t^2dt}{2} + \exp(X_t - t^3/6) (dX_t) + \exp(X_t - t^3/6) (dX_t)^2/2 \\ &= Y_t \left[ \frac{-t^2dt}{2} + m(t) dt + s(t) dW_t + \frac{s(t)^2dt}{2} \right] \end{split} $$ which is a martingale when $$ s(t)^2/2 + m(t) - t^2/2 = 0. $$ The simplest example would be $m(t) = 0$ and $s(t) = t$, so the process $(X_t)$ such that $dX_t = t dW_t$ would make $Y_t = f(X_t,t)$ into a martingale.