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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 $t>s$,

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

Brian M. Scott
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user1486802
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    Try rewriting one of the processes in $E[W(s)W(t)]$ and think of why doing so is useful. – Patrick Dec 02 '12 at 03:53
  • got it... i feel so dumb. LOL thanks. – user1486802 Dec 02 '12 at 04:09
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    Don't feel dumb, just take the new knowledge you have and keep on learning. – Patrick Dec 02 '12 at 04:22
  • @Patrick, Just to be clear, this doesn't necessarily mean that $W(t)$ is independent of $W(s)$, right? If it did then the expected value would just be zero. But, since $W(s)$ is independent of $W(t)-W(s)$, then we are able to take out the $W(s)$, right? and if so, why? – makansij Feb 03 '19 at 19:53

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