Autocorrelation of a wiener process

[aimforclarity]

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

 

[chiro]

Hey aimforclarity.

Have you made an attempt to partition this whole thing in terms of independent processes so that you can calculate E[XY] = E[X]E[Y]?

 

[aimforclarity]

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