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Regarding Brownian Motion formula below, how does $E[W(s)W(t)]$ turn into $$E\left[W(s)\big(W(t)−W(s)\big)+W(s)^2\right]\;??$$

I have asked a question using the formula below, but this and that are totally different questions. Thanks for all the help!!

Assuming $t>s$,

$$\begin{align*} E[W(s)W(t)]&=E\left[W(s)\big(W(t)−W(s)\big)+W(s)^2\right]\\ &=E[W(s)]E[W(t)−W(s)]+E\left[W(s)^2\right]\\ &=0+s\\ &=\min(s,t)\;. \end{align*}$$

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2  
Try rewriting one of the processes in $E[W(s)W(t)]$ and think of why doing so is useful. –  Patrick Dec 2 '12 at 3:53
    
got it... i feel so dumb. LOL thanks. –  user1486802 Dec 2 '12 at 4:09
3  
Don't feel dumb, just take the new knowledge you have and keep on learning. –  Patrick Dec 2 '12 at 4:22

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