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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.

Please help me figure out what's wrong with the arguments.

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  • $\begingroup$ Why not dierctly compute $E(W_{t+s}^{3}|\mathcal F_t)$ by writing $W_{t+s}^{3}$ as $(W_{t+s}-W_t)^{3}+3W_{t+s}W_t^{2}-3W_{t+s}^{2}W_t$. $\endgroup$ Jun 10, 2019 at 5:40

1 Answer 1

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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.

The error in your reasoning is here:

$$\mathbf{E}\Big( \int_0^\tau W_s~ds \Big)=\int_0^T\mathbf{E}(W_{s\wedge \tau})~ds$$

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.

Your reasoning in Method 2 is correct.

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