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?


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! :)

  • $\begingroup$ Well, $\mathbb E[W_t^3 - 3tW_t|F_s] = \mathbb E[W_t^3 |F_s] - \mathbb E[3tW_t|F_s]$ and to deal with the first term I've written $\mathbb E[W_t^3 |F_s] = \mathbb E[(W_t - W_s + W_s)^3 |F_s] $ and proceeded as follows. Is this the wrong approach? $\endgroup$ – Vladimir Nabokov Apr 3 '17 at 10:04
  • $\begingroup$ That sounds good. $\endgroup$ – ADB Apr 3 '17 at 10:05
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    $\begingroup$ So from here, can you expand the $((W_t - W_s) + W_s)^3$ terms and then use the fact that $\mathbb{E}[W_s^n|F_n]=W_s^n$? $\endgroup$ – ADB Apr 3 '17 at 10:06
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    $\begingroup$ Actually, it's probably more helpful to use that $W_t \overset{d}= W_s + W_{t-s}^\prime$, where $W^\prime$ is an independent copy of $W$. Then it should work out, as W and W' are independent. Does this help? $\endgroup$ – ADB Apr 3 '17 at 10:09
  • $\begingroup$ Yes - that's very helpful! Thanks a lot $\endgroup$ – Vladimir Nabokov Apr 3 '17 at 10:10

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