how do you get the Expectation of a Brownian process $W_s$ times the exponential of it? Where 0$\le$s$\lt$t and $W_0$=0?
I know that $E[e^{W_s-W_t}]$ is $\frac 12$$e^{(t-s)}$
So what is $E[W_se^{W_s-W_t}]$?
2026-04-04 15:08:01.1775315281
Expectation of a Brownian process $W_s$ times its exponential
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Note that $\sigma(W_t-W_s)$ and $\sigma(W_u:u\leq s)$ are independent and that $W_s$ is measurable w.r.t. the latter.