Let $\phi(t)$ be a Brownian Walk (Wiener Process), where $\phi\in[0,2\pi)$. As such we work with the variable $z(t)=e^{i\phi(t)}$. I would like to calculate
$E(z(t)z(t+\tau)).$
This is equal to $E(e^{i\phi(t)+i\phi(t+\tau)})$ and I know that $E(e^{i\phi(t)})=e^{-\frac{1}{2}\sigma^{2}(t)}$, where the mean is $0$ and $\sigma^{2}(t)=2Dt$. However, I have been stuck a week on how to proceed, any thoughts?
Thank you :)
Aim For Clarity