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Let $(B_t)_{t \geq0} $ be a Brownian Motion. Calculate $Cov(e^ {B_t} ,e^{B_s})$ I would verify the following solution which the result looks a bit weird.

My solution: let $0 \leq s \leq t$. $$Cov(e^ {B_t} ,e^{B_s})=E[e^{B_t + B_s}]-E[e^{B_t}]E[e^{B_s}] \\=E[e^{B_t -B_s} e^{2B_s}]-e^{t/2}e^{s/2} \\ =E[e^{B_t -B_s}]\ E[e^{2B_s}]-e^{t/2}e^{s/2} (\because independence) \\= e^{(t-s)/2}e^{2s} - e^{t/2}e^{s/2} \\=e^{(t+2s)/2} (e^{s/2}-e^{-s/2})\\ =2e^{t/2+s} \sinh(s/2) $$

Here I have extensively used the fact $ E[e^{X}]=e^{E[X]+Var(X)/2}$ whenever $X$ is normal.

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Your result seems correct.

If you were worried about the symmetry of $s$ and $t$, you may simply put, provided that $0\le s\le t$, \begin{align} \text{Cov}\left(e^{B_t},e^{B_s}\right)&=e^{\left(t-s\right)/2}e^{2s}-e^{t/2}e^{s/2}&&\text{(your result)}\\ &=e^{t/2}e^{3s/2}-e^{t/2}e^{s/2}\\ &=e^{t/2}e^{s/2}\left(e^s-1\right). \end{align} Thus for general $t\ge 0$ and $s\ge 0$, $$ \text{Cov}\left(e^{B_t},e^{B_s}\right)=e^{t/2}e^{s/2}\left(e^{t\wedge s}-1\right), $$ where $t\wedge s=\min\left\{t,s\right\}$.

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