How do you find the expectation of a random variable involving a function of Brownian motions?

Question

I understand how to calculate the expectation of simple random variables like:

$Y(t)=W(t+t_{0})-W(t_{0})$

But what if you had something more involved like:

$\mathbb{E}[e^{\alpha W(t)}W(t)]$

Where $W(t)$ is a Brownian motion and $\alpha$ is a real number.

Would you have to use Ito's equations for these types of questions?

Answer 1

All you need to use here is that $W(t)$ has the same distribution as $\sqrt t N$, where $N$ has a standard normal distribution. Therefore, $e^{\alpha W(t)}W(t)$ has the same distribution as $e^{\alpha \sqrt t N}\sqrt{t}N$ and you are reduced to compute the integral $$ \int_{-\infty}^\infty e^{\alpha \sqrt{t} x}\sqrt{t}xf(x)\mathrm{d}x, $$ where $f$ is the density of a standard normal random variable. It will be useful to complete the squares in the exponential to get the final result.

Answer 2

Since $W(t) = Z \sim \mathcal{N}(0,t)$ you can compute the probability density of any function $f(Z)$ using variable transformation. From there, it is straightforward to compute moments of this new random variable.

There are of course some special cases where we know that $f(W(t))$ (as a stochastic process) is a martingale. An example would be $f(W(t)) = W(t)^2 - t$ and Ito's Lemma can indeed help prove this.

Then, we get the expectation directly using the tower property.