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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 0 < p < q < r. Determine $E[W_p W_q W_r]$.

My attempt:

$0 = E[(W_r-W_q)(W_q-W_p)(W_p)]$

$\to E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 + W_qW_p^2]$

$\to E[W_p W_q W_r] = E[(W_r+W_q) W_p^2 + W_pW_q^2]$

$\to E[W_p W_q W_r] = E[E[(W_r+W_q) W_p^2 + W_pW_q^2]|\mathscr{F_p}]$

$\to E[W_p W_q W_r] = E[W_p^2E[(W_r+W_q)|\mathscr{F_p}] + E[W_pE[(W_q^2)|\mathscr{F_p}]]$

$\to E[W_p W_q W_r] = ...0$ ?

It looks like the stuff are $\mathscr{F_p}$-measurable? $E[(W_r+W_q)]=0=E[W_p]$

I don't know. Help please? :(

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3 Answers 3

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I think you are on the right track here.

You made a sign error in the first line, unfortunately: $$E[W_p W_q W_r] = E[W_r W_p^2 + W_pW_q^2 - W_qW_p^2]=\\ E[(W_r-W_q)W_p^2]+E[W_pW_q^2]= E[W_pW_q^2] $$ The first term is $0$ by independence (as $p<\text{min}(r,q)$ and the square does not affect independence).

To take care of the second term we do the standard expansion trick: $$E[W_pW_q^2] = E[W_p(W_q-W_p)^2]+2E[W_qW_p^2]-E[W_p^3]$$

Now, the first term is $0$ again, by independence. For the third term we use that the third central moment of a normal distibution is$ 0$. The second term is also $0$, as a simple calculation shows: $$ E[W_qW_p^2] = E[(W_q - W_p)W_p^2]+E[W_p^3] = 0$$ again, by independence of increments.

So it turns out that $$E[W_pW_qW_r]=0.$$

Hope I got everything right. I must admit, I expected a different result.

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  • $\begingroup$ We also have $E(W_t^3)=0$ which would aswell be consistent to $E(W_t^2)=t$ for $E(W_tW_p)=\min(t,p)=t,\,(t<p)$. $\endgroup$
    – emcor
    Commented Oct 6, 2014 at 15:54
  • $\begingroup$ May you pls explain the third central moment more? Thanks so much vanguard2k $\endgroup$
    – BCLC
    Commented Oct 6, 2014 at 19:06
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    $\begingroup$ For a normal distributed rv $X ~ N(\mu,\sigma)$, $E[(X-\mu)^3] = \mu^3+3\mu\sigma^2$ (just search for moment normal). In our case, $\mu=0$ and $\sigma = p$. Alternatively, you can calculate it by hand (several times integration by parts) or via the moment-generating function. $\endgroup$
    – vanguard2k
    Commented Oct 7, 2014 at 6:42
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We know that $(\tilde{W}_t) := (-W_t)$ is also a Wiener process so $$ E[W_pW_qW_r] = E[\tilde{W}_p\tilde{W}_q\tilde{W}_r] = (-1)^3E[W_pW_qW_r] $$ and that implies that $E[W_pW_qW_r] = 0$.

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  • $\begingroup$ Ybl how do you know that they are equal? Thanks $\endgroup$
    – BCLC
    Commented Oct 6, 2014 at 19:09
  • $\begingroup$ The product $W_pW_qW_r$ is not a Wiener process? $\endgroup$
    – emcor
    Commented Oct 6, 2014 at 19:24
  • $\begingroup$ Emcor, are wiener processes equal to each other? Thanks $\endgroup$
    – BCLC
    Commented Oct 7, 2014 at 2:22
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    $\begingroup$ @BCLC For the expectation they dont have to be equal (only identically distributed), YBL wants to show: $X=-X\Rightarrow X=0$. $\endgroup$
    – emcor
    Commented Oct 7, 2014 at 12:11
  • $\begingroup$ YBL, @emcor How do you know that they are identically distributed? I am just wondering how you know X = - X in the first place i.e. $E[WpWqWr] = E[-Wp*-Wq*-Wr]$ $\endgroup$
    – BCLC
    Commented Oct 9, 2014 at 13:37
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Note that $\{W_t \mid t \geq 0\}$ is a martingale. Then, for $0<p<q<r$, \begin{align*} E(W_pW_qW_r) &= E\Big( E(W_pW_qW_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q E(W_r \mid \mathcal{F}_q)\Big)\\ &=E\Big(W_pW_q^2\Big)\\ &=E\Big(W_p(W_q-W_p+W_p)^2\Big)\\ &=E\Big(W_p(W_q-W_p)^2+W_p^3+2W_p^2(W_q-W_p) \Big)\\ &=E(W_p)E\Big((W_q-W_p)^2\Big)+E(W_p^3)+2E\big(W_p^2\big)E(W_q-W_p)\\ &=0, \end{align*} by noting that \begin{align*} E(W_p) = E(W_p^3) = E(W_q-W_p) = 0. \end{align*}

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  • $\begingroup$ Are Wp and Wq-Wp independent? $\endgroup$
    – BCLC
    Commented Oct 10, 2014 at 10:40
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    $\begingroup$ For $p<q$, $W_p$ and $W_q-W_p$ are independent $\endgroup$
    – Gordon
    Commented Oct 10, 2014 at 13:08

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