# Autocorrelation of a wiener process

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

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chiro
ok, so can I say $<e^{\phi(t)+\phi(t+\tau)}>=<e^{\phi(t)+\phi(t)+\phi(\tau)}>$ since $W(t+\Delta t)=W(t)+W(\Delta t)$. Next we can say $<e^{\phi(t)+\phi(t)+\phi(\tau)}>=<e^{2\phi(t)}><e^{\phi(\tau)}>$ because $<\phi(t)\phi(\tau)>=0$