To show martingale, what other conditions do you need to check if the process has zero drift?

Question

Given a Brownian motion $(W_t)_t,$ to show that $W_t^3$ is not a martingale, one usually applies Ito's lemma and show that the SDE satisfies by $W_t^3$ has no drift term, that is, no $dt$ term.

In Mark Joshi et al. Quant Jon Interview, they ask What other conditions do you need to check if the process has zero drift?

I have no idea how to answer this question.

Answer 1

If an Ito process has no drift, then we have a process of the type $I_t = \int_0^t H_u dW_u,$ (i.e the stochastic differential is $dI_u = H_u dW_u$). The previous integral is defined for integrand processes $H$ such that $H$ is $\{ \mathscr{F}_t \}$-adapted and $\int_0^t |H_u|^2 du < \infty $ a.s for every $t>0.$ But $I_t$ defined like that, is ("only") a local martingale. For $I_t$ to be a (true) martingale we have to check that $E \int_0^t |H_u|^2 du < \infty $ a.s for every $t>0.$

Answer 2

But, by Ito, $W_t^3$ isn't even a local martingale: $$ W_t^3=\int_0^t 3W_s\,dW_s + \int_0^t 3W_s\,ds. $$

Answer 3

One necessary and sufficient condition for a continuous local martingale $X_t$ to be a martingale is if $\{X_T : T \text{ is a stopping time}, T \le a \}$ is uniformly integrable for all $a > 0.$ That's a pretty difficult condition to check, so a sufficient condition to check is if $\mathbb{E}[\langle X,X \rangle_t] < \infty$ for all $t > 0$. If $X_t = \int_0^t H_s dW_s$ for some predictable process $H_s$, this is to the condition that UBM posted.