Q:

Assume $B_1$ and $B_2$ are two independent Brownian motions, what is the $E\left[e^{B_1(t_1)}e^{B_2(t_2)}\right]$?

Assume $B_1$ and $B_2$ are two independent Brownian motions, what is the $E[e^{B_1(t_1)}e^{B_2(t_2)}]$?
Since $B_1,B_2$ are independent, we can get $cov(B_1, B_2) = 0$ and $E[e^{B_1}e^{B_2}] = E[e^{B_1}]E[e^{B_2}]$.
How do you proceed from there? One possible idea I can think of is to use $e^x = \sum \frac{x^n}{n!}$, but I'm not sure if there is any special considerations about Brownian motion when applying the Taylor series.

A:

For $t > 0,$ since $B_i(t) \sim \mathcal{N}(0,t),$
$$\begin{align*}E\left[e^{B_i(t)}\right] &= \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty e^x e^{-x^2/(2t)}\,dx \\ &= \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty e^{t/2}e^{-(x-t)^2/(2t)}\,dx \\ &= e^{t/2} \cdot \left(\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty e^{-(x-t)^2/(2t)}\,dx\right) \\ &= e^{t/2}\end{align*}$$
Note that I'm assuming these are defined so that $B_i(0) = 0$ almost surely.  If not, you would have to deal with the distribution of $B_i(0)$ separately.  Also, while the proof requires $t \neq 0$, the result is immediately verifiable for $t=0.$

Using the above result, we conclude
$$E\left[e^{B_1(t_1)}e^{B_2(t_2)}\right] = E\left[e^{B_1(t_1)}\right]E\left[e^{B_2(t_2)}\right] = e^{t_1/2}e^{t_2/2} = e^{(t_1+t_2)/2}$$

