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I can compute $E(W_tW_s)$, but I cannot compute $E[(W_t)^2(W_s)^2] $. please help me! Thanks a lot!

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Hints:

  1. For $t>s$, $f(W_t-W_s)$ and $f(W_s)$ are independent random variables with some measurable function $f$ since $W_t-W_s$ and $W_s$ are independent (see here).
  2. $W_t-W_s\sim\mathcal N(0,t-s)$ for $0\le s< t$.
  3. The fourth moment of the Gaussian random variable $\xi\sim\mathcal N(0,\sigma^2)$ is equal to $3\sigma^4$ (see here).

Edit:

Without loss of generality, let us assume that $t>s$. Then \begin{align*} \operatorname E[W_t^2W_s^2] &=\operatorname E[(W_t-W_s+W_s)^2W_s^2]\\ &=\operatorname E[((W_t-W_s)^2+2(W_t-W_s)W_s+W_s^2)W_s^2]\\ &=\operatorname E[(W_t-W_s)^2W_s^2]+2\operatorname E[(W_t-W_s)W_s^3]+\operatorname EW_s^4\\ &=\operatorname E(W_t-W_s)^2\operatorname EW_s^2+2\operatorname E(W_t-W_s)\operatorname EW_s^3+\operatorname EW_s^4. \end{align*} Can you take it from here?

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  • $\begingroup$ Thank you Cm7F7Bb but I still cannot compute. $\endgroup$
    – Ooshi Osoto
    May 8, 2017 at 10:21
  • $\begingroup$ Can you help me compute specifically. $\endgroup$
    – Ooshi Osoto
    May 8, 2017 at 10:22
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    $\begingroup$ @OoshiOsoto You should include your attempt and where you get stuck in the question. Then we will know what kind of help you need. I updated my answer with more details. $\endgroup$
    – Cm7F7Bb
    May 8, 2017 at 10:25
  • $\begingroup$ I got it. Thank you so much, Cm7F7Bb. $\endgroup$
    – Ooshi Osoto
    May 8, 2017 at 13:08
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    $\begingroup$ @OoshiOsoto You're welcome! Since you're new to the site, I'd like to remind you that you can accept this answer if it helped you. But you can also wait until a better answer comes along. $\endgroup$
    – Cm7F7Bb
    May 8, 2017 at 13:11

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