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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

1 Answers 1

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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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    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!2012-12-29
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    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.2012-12-29
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    ok, so is another way of doing the derivation to say $=$ since $W(t+\Delta t)=W(t)+W(\Delta t)$. Next we can say $=$ because $<\phi(t)\phi(\tau)>=0$2012-12-30
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    @AimForClarity What does "$<\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 $r \leq t$.2012-12-30
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    @AimForClarity Please do not forget to upvote solutions which satisfy you.2012-12-30
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    @ Saz, thanks you Saz, that makes sense now, great answers! (I did mean E() by <()>) @did Yep, just need 2 more rep points :)2012-12-30
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    @AimForClarity You are welcome. (By the way, you don't have to give away your reputation points if you upvote answers.)2012-12-30
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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