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Suppose $\{W_t, t>=0\}$ is a Standard Brownian Motion. How to compute $ \mathbb{E} \left[ W_2 W_3 \vert W_1 =0 \right]$? We know $ W_2 \vert W_1 = 0 \sim N(0,1)$ and $ W_3 \vert W_1 = 0 \sim N(0,2)$. Thank you so much.

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closed as off-topic by LocalVolatility, byouness, skoestlmeier, Helin, lehalle Feb 22 at 19:32

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    $\begingroup$ Hint: $\mathbb{E} \left[ W_2 W_3 \vert W_1 = 0 \right] = \mathbb{E} \left[ W_1 W_2 \right]$. Then use $W_2 = W_1 + \left( W_2 - W_1 \right)$. $\endgroup$ – LocalVolatility Feb 5 at 21:56
  • $\begingroup$ Alternatively, $E(W_2W_3\mid W_1=0) = E(W_2^2\mid W_1=0)$. $\endgroup$ – Gordon Feb 6 at 14:01
  • $\begingroup$ You are asking about very basic properties of Brownian motion and conditional expectation. Have a look at e.g. the first few chapters in Shreve's "Stochastic Calculus for Finance II". $\endgroup$ – LocalVolatility Feb 6 at 16:09
  • $\begingroup$ $\bf{E}[W_t W_u | W_s = 0] =\bf{E}[W_{t-s} W_{u-s}] $ since $W_0=0$ It is a time shift by $-s$ on all three subscripts. $\endgroup$ – Alex C Feb 6 at 17:34
  • $\begingroup$ To Alex C: How do you show E[WtWu|Ws=0] = E[Wt-sWu-s]? Which source ( book or website ) to get the proof of your statement? Thank you so much. $\endgroup$ – MathMan12 Feb 6 at 23:16
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The conditional expectation with respect to $W_1=0$ can be treated as the conditional expectation with respect to $W_1$ and then set $W_1$ to 0. See more discussions in this question.

Note that $W_3 = W_3-W_2+W_2$, and $W_2 = W_2-W_1+W_1$. Then \begin{align*} E\big(W_2W_3\mid W_1\big) &= E\Big(\big(W_3-W_2\big)\big(W_2-W_1\big)\\ &\qquad+ \big(W_3-W_2\big)W_1 + \big(W_2-W_1\big) W_2 + W_2 W_1\mid W_1\Big)\\ &=E\Big(\big(W_2-W_1\big) W_2 + W_2 W_1\mid W_1\Big)\\ &=E\Big(W_2^2\mid W_1\Big)\\ &=E\Big(\big(W_2-W_1\big)^2 + 2\big(W_2-W_1\big) W_1 + W_1^2\mid W_1\Big)\\ &=1+ W_1^2. \end{align*} That is, \begin{align*} E\big(W_2W_3\mid W_1=0\big)=1. \end{align*}

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