Is this calculation of the expectation of Brownian motion correct?

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Consider the two Brownian motions $W(t_{1})$ and $W(t_{2})$ with $(t_{2}\geq t_{1})$. We are interested in the following process:

$\mathbb{E}[e^{W(t_{2})-W(t_{1})}]$

We know $W(t_{2})-W(t_{1})$ is distributed as $N(0,t_{2}-t_{1})$ can we hence rewrite this expectation as:

$\int_{-\infty }^{\infty }xe^{{\sqrt{t_{2}-t_{1}}}x}f(x)dx$

(where is the density of a standard normal random variable)?