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I'm studying the use of Brownian motion (aka Wiener Process) in the context of the Doob's optional stopping theorem for continuous cases. I do remember the basic properties of a Brownian motion (stationarity, independence or the increments, the fact that $W(t) \sim N(0, t)$), but I struggle with a "trivial" point on the proof of the fact that $Cov (W(t_1), W(t_2)) = \min(t_1, t_2)$.

By using the different properties I reach the point where I can say that, for $t_1 < t_2$, $Cov (W(t_1), W(t_2)) = E((W(t_1))^2) = t_1$.

Maybe my brain used all its energy but I can't find why $E(W(t_1)^2) = t_1$, knowing that the first moment is $0$.

endlessend2525
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  • Do you mean $\mathbb{E}(W_{\color{red}{t_1}}^2) = t_1$....? ($\mathbb{E}(W_1^2) = t_1$ is not correct unless $t_1 = 1$.) – saz Mar 11 '17 at 09:08
  • What do you mean by $W_1$? – RandomGuy Mar 11 '17 at 09:08
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    If you mean $E(W_{t_1}^2)=t_1$, then the answer is because the variance $Var(W_{t_1})=t_1$ for every BM and since $Var(W_{t_1}) = E(W_{t_1}^2)-E(W_{t_1})^2=E(W_{t_1}^2) $, since $E(W_{t_1})=0$ for a standard Brownian motion. – RandomGuy Mar 11 '17 at 09:14
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    Sorry for the typos guys. Of course, the answer is bigger than an elephant in front of me and I missed it. Thanks a lot @RandomGuy – endlessend2525 Mar 11 '17 at 09:23

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