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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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Let $(\varphi_t)_t$ a Brownian motion, then

$\begin{align} \mathbb{E} \left(e^{\imath \, \varphi(t)+\imath \, \varphi(t+r)} \right) &= \mathbb{E} \left( e^{\imath \, (\varphi(t+r)-\varphi(t))+2 \imath \varphi(t))} \right) = \underbrace{\mathbb{E}\left(e^{\imath \, (\varphi(t+r)-\varphi(t))} \right)}_{\mathbb{E}\left(e^{\imath \varphi(r)} \right)} \cdot \mathbb{E}\left(e^{2\imath \, \varphi(t)} \right) \\ &=\mathbb{E}\left(e^{\imath \varphi(r)} \right) \cdot \mathbb{E}\left(e^{2\imath \, \varphi(t)} \right) \end{align}$

where we used the independence (thus $\varphi(t+r)-\varphi(t)$,$\varphi(t)$ are independent) and stationarity (thus $\varphi(t+r) - \varphi(t) \sim \varphi((t+r)-r)=\varphi(r)$) of the increments. The remaining expectation values you can calculate using the formula you mentioned.

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    @saz I have 13 rep and need 15 to upvote in general, it wont let me do it yet :)2012-12-30