Is $W^3(t)$ a martingale if $W(t)$ is a Brownian motion

Question

Is $W^3(t)$ a martingale if $W(t)$ is a Brownian motion? The answer seems like no to me. Using Ito's lemma I can write $$W^3(t)=\frac{3}{2}W^2(t)+\int_0^t3W(u)dW(u)$$ The second piece on the LHS is an Ito integral and thus a martingale. However the first piece on the LHS in not a martingale and thus $W^3(t)$ is not a martingale. Can someone confirm or refute my argument (or perhaps make an argument of their own)?

Answer 1

You can in fact show that, for any $n>1$, $W(t)^n$ is not a (local) martingale: by applying Ito's lemma, you get:

$dW(t)^n = 0.5n(n-1)W(t)^{n-2}dt + nW(t)^{n-1}dW(t)$

$W(t)^n$ is a (local) martingale iff: $n=1$. Thus, for any $n>1$, $W(t)^n$ is not a (local) martingale.