# Autocorrelation of wrapped 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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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.
This is great! And my question may be of rather simple nature, but why is $E(e^{i\phi(t+\tau)-i\phi(t)}) = e^{i\phi(r)}$, ie why r in particular, i understand the increments are stationary and independent, but why not $e^{i\phi(t+r)}$? thank you! – AimForClarity Dec 29 '12 at 22:27
Stationary increments means that $\phi(t+r)-\phi(t)$ has the same distribution as $\phi(r)-\phi(0)$. Since for a Brownian motion $\phi$ we have $\phi(0)=0$ a.s., it follows that $\phi(t+r)-\phi(t)$ has the same distribution as $\phi(r)$; so the expected values of the exponentials are the same. – Eckhard Dec 29 '12 at 23:36
ok, so is another way of doing the derivation to 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 $<\phi(t)\phi(\tau)>=0$ – AimForClarity Dec 30 '12 at 4:57
@AimForClarity What does "$<\ldots>$" mean? Is it abbrevation for expectation value? In this case, the answer is no: The "equality" $$\mathbb{E}(f(\varphi(t),\varphi(t+r))) = \mathbb{E}(f(\varphi(t),\varphi(t)+\varphi(r)))$$ is in general not correct (you applied it to $f(x,y) := e^{\imath \, (x+y)}$ as far as I can see), because the joint distributions are not the same. You can see this if you take for example $f(x,y):=x \cdot y$. Then lefthandside is equal to $t$ whereas the righthandside is equal to $t+r$ for all $r \leq t$. – saz Dec 30 '12 at 9:11