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.