5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why not dierctly compute $E(W_{t+s}^{3}|\mathcal F_t)$ by writing $W_{t+s}^{3}$ as $(W_{t+s}-W_t)^{3}+3W_{t+s}W_t^{2}-3W_{t+s}^{2}W_t$. $\endgroup$ – Kavi Rama Murthy Jun 10 at 5:40
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.