# W_t^3 martingale or not? two arguments puzzle me.

I want to study whether $$W_t^3$$ is a martingale or not? where $$W_t$$ is the standard Brownian motion.

I have method 1 argument, but I also got second argument which implies different conclusion. Please help me figure out.

Method 1: Using Ito formula, we have $$W_t^3 = \int_0^t 3 W_s^2~d W_s + \int_0^t 3 W_s~ds$$ The first term on the RHS is a Ito integral, thus a martingale. Consider the second term, notice that a stochastic process $$(X_t)_{t\geq 0}$$ is martingale if and only if for any bounded stopping time $$\tau$$, we have $$\mathbf{E}(X_\tau)=0$$ Back to the second term, $$\mathbf{E}\Big( \int_0^\tau W_s~ds \Big)=\int_0^T\mathbf{E}(W_{s\wedge \tau})~ds~=~0$$ where $$T$$ is a finite boundedness for $$\tau$$. Thus we proved that $$\int_0^t 3 W_s~ds$$ is also a martingale. This gives that $$W_t^3$$ is a martingale.

Method 2: Let $$\tau$$ be the first existing time of $$W_t$$ from the interval $$[-1, 2]$$, then from optional sampling theorem, we know $$\mathbf{P}(W_\tau = -1) = {2\over 3},~~\mathbf{P}(W_\tau = 2) = {1\over 3}$$. Suppose $$(W_t^3)_{t\geq 0}$$ is also a martingale, then we must have $$\mathbf{E}(W_\tau^3) = 0$$. However by computation, $$\mathbf{E}(W_\tau^3) = (-1)\times {2\over 3} + 8\times {1\over 3} \neq 0.$$ This contradiction implies that $$W_t^3$$ is not a martingale.

• Why not dierctly compute $E(W_{t+s}^{3}|\mathcal F_t)$ by writing $W_{t+s}^{3}$ as $(W_{t+s}-W_t)^{3}+3W_{t+s}W_t^{2}-3W_{t+s}^{2}W_t$. – Kavi Rama Murthy Jun 10 at 5:40

It's not a martingale. Basically, a process of bounded variation can never be a continuous-time martingale (unless it's constant), so intuitively you can see immediately that $$W_t^3$$ is not a martingale because the bounded variation part of the Ito decomposition doesn't vanish.
$$\mathbf{E}\Big( \int_0^\tau W_s~ds \Big)=\int_0^T\mathbf{E}(W_{s\wedge \tau})~ds$$
This isn't true. What is true is that $$\mathbf{E}\int_0^\tau W_s\,ds = \int_0^T \mathbf{E}[W_s 1_{\{s \le \tau\}}]\,ds$$ but the integrand on the right side is not the same as $$W_{s \wedge \tau}$$, which when $$s > \tau$$ yields $$W_\tau$$, not $$0$$. And there is no reason why $$E[W_s 1_{\{s \le \tau\}}]$$ should equal zero.