$W_t$ is a Brownian motion. How do we calculate this expectation? there are two cases:
- $s < t$
- $t < s$
Do we have to distinguish the two cases or there is a unified way of calculating it
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For $s<t$, then \begin{align*} E\big(W_sW_t \,|\, W_s\big) &= W_sE\big((W_t-W_s + W_s)\,|\,W_s\big) \\ &=W_s^2. \end{align*}
For $0 < t < s$, then $$E\left(W_s \Big(W_t-\frac{t}{s}W_s\Big) \right)= 0,$$ and, given their joint normality, $W_s$ and $W_t-\frac{t}{s}W_s$ are independent. Therefore, \begin{align*} E\big(W_sW_t \,|\, W_s\big) &= W_sE\left(\Big(W_t-\frac{t}{s}W_s +\frac{t}{s} W_s\Big)\,|\,W_s\right) \\ &= \frac{t}{s} W_s^2. \end{align*}
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1$\begingroup$ I thought $W_t$ is $F_s$ measurable if $s>t$. Is that not the case? $\endgroup$– SanjayNov 1, 2019 at 20:08
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2$\begingroup$ @Sanjay: $W_t$ is $\mathscr{F}_s$ measurable if $s>t$, but not $\sigma(W_s)$ measurable. That is, $E(W_t \,|\, W_s) \ne W_t$. $\endgroup$– GordonNov 1, 2019 at 20:11
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$\begingroup$ Ok. I have learned stochastic calculus before proper training in measure theory (and sigma algebra). Sometimes this fact bites me in the a** :D Thanks for clarification $\endgroup$– SanjayNov 2, 2019 at 0:12