I wish to calculate the following expectation:
$\mathbb{E}[e^{-t}W(e^{2t})]$
Where $W(t)$ is a Brownian motion.
Can I rewrite this as follows?
$\mathbb{E}[e^{-t}W(e^{2t})]=\int_{-\infty }^{\infty }e^{-x}\sqrt{e^{2x}}xf(x)dx$
where $f$ is the density of a standard normal random variable and where I have used that:
$W(e^{2t})\sim \sqrt{e^{2t}}N(0,1)$
On
Yes, $W(t) \sim N(0,t)$, hence $W(e^{2t}) \sim \sqrt{e^{2t}}W(1) = e^t N(0,1)$.
Thus the answer is $Ee^{-t}W(e^{2t}) = EN(0,1) = 0$ because $\int_{\mathbb{R}} x f(x) dx = 0$ because of the fact that $f$ is odd.
The expression $\mathbb{E}[e^{-t}W(e^{2t})]=\int_{-\infty }^{\infty }e^{-t}\sqrt{e^{2t}}xf(x)dx$ is also correct.
$e^{-t}$ is not random, so $\mathbb{E}[e^{-t}W(e^{2t})] = e^{-t} \mathbb{E}[W(e^{2t})] = 0$.