4
$\begingroup$

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?

| improve this question | |
$\endgroup$
3
$\begingroup$

\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*}

| improve this answer | |
$\endgroup$
7
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Oct 7 '14 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 Oct 8 '14 at 6:43
  • $\begingroup$ Ok, but then how do you compute $E(tW_t|F_s)$? $\endgroup$ – emcor Oct 8 '14 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 Oct 8 '14 at 7:05
  • 1
    $\begingroup$ Thanks Olaf. It looks like you've been using that username since before Frozen. +1 $\endgroup$ – BCLC Oct 9 '14 at 13:35
1
$\begingroup$

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

Finally:

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

q.e.d.

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Oct 7 '14 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 Oct 9 '14 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 Oct 9 '14 at 19:52
  • $\begingroup$ Yep, totally correct. $\endgroup$ – quasi Oct 9 '14 at 20:32

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.