# The conditional mean of a product of standard Brownian motions [closed]

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.

• Hint: $\mathbb{E} \left[ W_2 W_3 \vert W_1 = 0 \right] = \mathbb{E} \left[ W_1 W_2 \right]$. Then use $W_2 = W_1 + \left( W_2 - W_1 \right)$. – LocalVolatility Feb 5 '19 at 21:56
• Alternatively, $E(W_2W_3\mid W_1=0) = E(W_2^2\mid W_1=0)$. – Gordon Feb 6 '19 at 14:01
• 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". – LocalVolatility Feb 6 '19 at 16:09
• $\bf{E}[W_t W_u | W_s = 0] =\bf{E}[W_{t-s} W_{u-s}]$ since $W_0=0$ It is a time shift by $-s$ on all three subscripts. – Alex C Feb 6 '19 at 17:34
• 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. – MathMan12 Feb 6 '19 at 23:16

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*}