# Expectation regarding Brownian Motion

This is a formula regarding getting expectation under the topic of Brownian Motion.

\begin{align} E[W(s)W(t)] &= E[W(s)(W(t) - W(s)) + W(s)^2] \\ &= E[W (s)]E[W (t) - W (s)] + E[W(s)^2] \\ &= 0+s\\ &=\min(s,t) \end{align}

How does $E[W (s)]E[W (t) - W (s)]$ turn into 0?

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It's a product of independent increments. – Raskolnikov Nov 29 '12 at 23:17

$W(s)\sim N(0,s)$ and $W(t)-W(s)\sim N(0,t-s)$. So both expectations are $0$.

(The step that says $\mathbb E[W(s)(W(t)-W(s))]= \mathbb E[W(s)] \mathbb E[W(t)-W(s)]$ depends on an assumption that $t>s$.)

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thank you very much – user1486802 Nov 29 '12 at 23:51

Brownian motion has independent increments. This means the two random variables $W(t_1)$ and $W(t_2-t_1)$ are independent for every $t_1 < t_2$.

Independence for two random variables $X$ and $Y$ results into $E[X Y]=E[X] E[Y]$. This is zero if either $X$ or $Y$ has mean zero. Now, remember that for a Brownian motion $W(t)$ has a normal distribution with mean zero.

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Thanks a lot. This is very helpful. – user1486802 Nov 29 '12 at 23:50