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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 $t>s$,

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

  • 2
    Try rewriting one of the processes in $E[W(s)W(t)]$ and think of why doing so is useful.2012-12-02
  • 0
    got it... i feel so dumb. LOL thanks.2012-12-02
  • 7
    Don't feel dumb, just take the new knowledge you have and keep on learning.2012-12-02

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