I have almost completed a question which is to find the mean and variance of the following:
$I_T = \int_0^T \sqrt{|W_t|}d W_t$
In order to do this, I know I need to find $\mathbb E [I_T]$ and $\mathbb E [I_T^2]$
For $\mathbb E [I_T]$, I have used the fact that $M_T$ is a martingale where $M_T = \int_0^T \sqrt{|W_t|}d W_t$, based on the properties of martingales we know that $ \mathbb E [M_t]= \mathbb E[M_0]=0 $ which is the mean.
Then I need to find the variance which is a bit more tricky, so I have used Ito Isometry: $var(I_T) = \mathbb E (I_T^2)- \mathbb E (I_T)^2 = \mathbb E [\int^T_0 |W_t|dt] $.
The part of this I need help with is to actually find a value for the variance, is it possible to simplify what I have so far and how?
All help is appreciated.
Best Answer
$W_t \sim N(0,t),$ so
\begin{align*} E|W_t| &= \int^{\infty}_{- \infty}|x| \frac{1}{\sqrt{2 \pi t}} e^{\frac{-x^2}{2t}}dx \\ &= \int^{\infty}_{0}2x \frac{1}{\sqrt{2 \pi t}} e^{\frac{-x^2}{2t}}dx \\ &= \sqrt{\frac{2t}{\pi}}. \end{align*} So $$var(I_T)= \int_0^T \sqrt{\frac{2t}{\pi}} dt$$