Given prob space $(\Omega, \mathscr{F}, P)$ and a Wiener process $(W_t)_{t \geq 0}$, define filtration $\mathscr{F}_t = \sigma(W_u : u \leq t)$

Let $(B_t)_{t \geq 0}$ where $B_t = W_t^3 - 3tW_t$. Show that $E[B_t|\mathscr{F}_s] = B_s$ whenever $s < t$.

I think this all comes down to manipulation since there are martingales somewhere

My attempt:

Splitting up into $E[W_t^3|\mathscr{F}_s] - 3E[tW_t|\mathscr{F}_s]$ doesn't do anything since those guys aren't martingales? So, I tried splitting it up into:

$E[W_t(W_t^2 - 3t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t -2 t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t) -2 tW_t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t)|\mathscr{F}_s] -2E[ tW_t|\mathscr{F}_s]$

$W_t$ is not $\mathscr{F}_s$-measurable, so we can't take that out...

$tW_{1/t}$ is Brownian and thus a martingale, but I don't know about $tW_t$...

$cW_{t/c^2}$ is Brownian and thus a martingale, but I don't think we can set c = t...

Help please?


3 Answers 3


\begin{align*} E\Big(W_t^3-3tW_t \mid \mathcal{F}_s\Big) &= E\Big((W_t-W_s+W_s)^3-3t(W_t-W_s+W_s) \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3+W_s^3+3(W_t-W_s)^2W_s + 3 (W_t-W_s)W_s^2\\ &\qquad \qquad -3t(W_t-W_s)-3tW_s \mid \mathcal{F}_s\Big) \\ &=E\Big((W_t-W_s)^3\Big) + W_s^3+3W_sE\Big((W_t-W_s)^2\Big)\\ &\qquad \qquad + 3W_s^2 E(W_t-W_s)-3tE(W_t-W_s)-3tW_s\\ &=W_s^3+3W_s (t-s)-3tW_s\\ &=W_s^3 -3sW_s, \end{align*} by noting that \begin{align*} E\Big((W_t-W_s)^3\Big) = E(W_t-W_s) =0, \end{align*} and \begin{align*} E\Big((W_t-W_s)^2\Big) = t-s. \end{align*}


The trick is to start with the highest power, rewrite it as something you know (a third order moment) and then work backwards on the remaining terms. By that I mean you can complete the cube as follows:

$$E[W_t^3 - 3tW_t|\mathcal{F}_s] = E[(W_t-W_s)^3 - C -3tW_t|\mathcal{F}_s]$$

where you'll need to find $C$ such that the equality holds (i.e. $C=W_s^3 + 3W_s^2 W_t + ...$). Next, you can use that

$$ (W_t-W_s)$$

is normal w.r.t. $\mathcal{F}_s$ with mean zero ($E[(W_t-W_s) |\mathcal{F}_s] = 0$) and variance $E[(W_t-W_s)^2 |\mathcal{F}_s] = (t-s)$, so the third order moment is zero, i.e. $E[(W_t-W_s)^3|\mathcal{F}_s] = 0$. What we accomplished here is that we got rid of the highest order of $W_t$ in the expectation value (the cube term, $W_t^3$). The remaining terms are either independent, linear or quadratic in $W_t$. So we can apply the same trick to these terms, that is, "completing the difference". I.e. if you encounter a term like:

$$E[W_s W_t^2|\mathcal{F_s}]$$

then you again complete the square of $W_t$ such that you end up with an expression that involves the difference $W_t - W_s$. So find a $C'$ such that

$$E[W_s W_t^2|\mathcal{F_s}] = E[W_s (W_t - W_s)^2 - C'|\mathcal{F_s}]$$

holds. You then recognise the variance of $W_t - W_s$ in the first term, and $C'$ will contain terms at most linear in $W_t$. You keep doing that for all the terms, and you should end up with the identity that you set out to prove.

I've been intentionally a bit vague, because its a good exercise to complete. But if things are not clear, then let me know.

  • $\begingroup$ I appreciate your attempt, but its not a full answer as you mentioned. Since $W^3$ is a cubed term, it may be that the expectation of $C$ is still not directly calculable, so it may not lead to a solution. $\endgroup$
    – emcor
    Commented Oct 7, 2014 at 21:31
  • $\begingroup$ Not sure I follow, so please correct me if I'm wrong. But I think that $E[C]$ can again be be completely rewritten in terms of $E[(W_t - W_s)^2|\mathcal{F}_s]$, $E[(W_t - W_s)|\mathcal{F}_s]$ and terms that only depend on $W_s$. And these are all known. $\endgroup$
    – Olaf
    Commented Oct 8, 2014 at 6:43
  • $\begingroup$ Ok, but then how do you compute $E(tW_t|F_s)$? $\endgroup$
    – emcor
    Commented Oct 8, 2014 at 7:02
  • 2
    $\begingroup$ It's better to work on the higher power first. So first eliminate the quadratic term in $W_t$, as this will generate more linear terms in $W_t$. However, if you just want to compute the term you mentioned, then: $E[tW_t|F_s] = t E[(W_t - W_s) + W_s|F_s] = t E[(W_t - W_s)|F_s] + tE[W_s|F_s] = t W_s$. This term alone is of course not a martingale. $\endgroup$
    – Olaf
    Commented Oct 8, 2014 at 7:05
  • 1
    $\begingroup$ Thanks Olaf. It looks like you've been using that username since before Frozen. +1 $\endgroup$
    – BCLC
    Commented Oct 9, 2014 at 13:35

You can use that $f(t,W_t)\in C^2$ is Martingale iff:$$\partial_t f+\frac{1}{2}\partial_{WW}f= 0$$

We get:$$\partial_t f=-3W_t$$$$\partial_{WW}f=6W_t$$


$$-3W_t+3W_t= 0$$


The proof of theorem follows by writing out $f(t,W_t)$ via Ito formula. Proof of theorem: enter image description here

  • $\begingroup$ Thanks, but none of these were yet discussed in our Stochastic Calculus class. We did learn about some in Derivatives Pricing. Is there any other way to do this? $\endgroup$
    – BCLC
    Commented Oct 7, 2014 at 17:53
  • $\begingroup$ Technical point: the pde condition on f only guarantees that it will be a local martingale. Integrability has to be checked separately. $\endgroup$
    – quasi
    Commented Oct 9, 2014 at 17:58
  • $\begingroup$ @quasi Yes, the total proof goes a bit longer to check on this issue requires some bounded derivatives, but for the normal distribution we know that all the terms in $f$ are integrable. $\endgroup$
    – emcor
    Commented Oct 9, 2014 at 19:52
  • $\begingroup$ Yep, totally correct. $\endgroup$
    – quasi
    Commented Oct 9, 2014 at 20:32

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