Expectations of a stochastic process

Question

Studying for a test and came across this question for practice

Given

$dX(t) = X(t)dW(t)$ where $W(t)$ is the Weiner process and X(t) is a process governed by dX(t) and X(0) = 2, I want to find the expected value $E[e^{-t}X^{2}(t)]$.

So far, I've been able to arrive at

$X(t) = X(0) + \int_{0}^{t}(X(u)dW(u))$

and

$X^2(t) = X(0)^2 + 2X(0)\int_{0}^{t}X(u)dW(u) + \int_{0}^{t}X^2(t)du$

So the

$E[e^{-t}X^2(t)] = e^{-t}E[4 + 4\int_{0}^{t}X(u)dW(u) + \int_{0}^{t}X^2(t)du]$

By linearity of expectation, this is simplified to :

$e^{-t}(4 + 4E[\int_{0}^{t}X(u)dW(u)] + E[\int_{0}^{t}X^2(t)du])$

Since $\int_{0}^{t}X(u)dW(u)$ is an ito integral, it is a martingale and it's expectated value will be $0$ so I finally arrived at

$e^{-t}(4 + E[\int_{0}^{t}X^2(t)du])$.

How can I proceed from here?

Answer 1

One simple approach is by noting that \begin{align*} X_t = X_0 e^{-\frac{1}{2}t + W_t}. \end{align*} Then \begin{align*} E\left(e^{-t}X_t^2 \right) &= E\left(4e^{-2t+2 W_t} \right)\\ &=4e^{-2t + \frac{4t}{2}} =4. \end{align*}