What is the Expectation of Square of Wiener Process?

Question

I am trying to find $E[W(t)^2\cdot e^{-2at}]$, where $a$ is a constant and $W(t)$ is the Wiener Process. This arises when finding the variance of the Vasicek interest rate model. I think it should be $t$ since $dW^2=dt$ but not sure. The $e^{-2at}$ term should come out of the expectation but I dont know how to calculate $E[W(t)^2]$. Kindly help.

Answer 1

The Vasicek Model follows the SDE

$$ dr_t = k[\theta - r_t]dt + \sigma dW_t $$

and has solution $r_t = r(s)e^{-k(t-s)} +\theta (1-e^{-k(t-s)})+\sigma \int_{s}^{t}e^{-k(t-u)}dW_u$.

To compute the variance, we can notice that the first two terms are deterministic and so give no contribution and we are left with

$$ \text{Var}(r_t \mid \mathcal{F}_s) = \sigma^2 \mathbb{E}\bigg[\bigg(\int_{s}^{t}e^{-k(t-u)}dW_u\bigg)^2\bigg]=\text{Ito Isometry} \\ = \sigma^2 \mathbb{E}\bigg[\int_{s}^{t}e^{-2k(t-u)}du\bigg] = \frac{\sigma^2}{2k}(1-e^{-2k(t-s)}) $$

I believe the mistake was to "take out" the exponential in the stochastic integral. This is not allowed since it depends on time and is not constant.