Suppose that $W$ is standard Brownian motion and that process $X$ is defined via $X_t = W_t^3 – 3tW_t$. I need to show $X$ is a martingale by calculating $\mathbb E[X_t|F_s]$, where $s<t$. It's all going fine except I'm not sure how to calculate $$\mathbb E[(W_t-W_s)^3|F_s] = \mathbb E[(W_t-W_s)^3].$$
Can anyone give me a hint?
Best Answer
You want to prove that $\mathbb{E}[ W_t^3 - 3tW_t |\mathcal{F}_s ] \overset{d}= W_s^3 - 3sW_s $, right?
It's not so clear to me where you've got the $\mathbb{E}[ (W_t - W_s)^3|\mathcal{F}_s] $ term from (maybe you could explain this if you still need more help), but you could try expanding the cubed expression into $W_t^3 + 3W_t^2 W_s + 3W_tW_s^2 + W_s^3$, and use standard properties of the normal distribution, and the fact that $W_s$ is just a fixed number when you're using the conditioning on knowing $\mathcal{F}_s$.
I hope this helps - good luck! :)