Auto-correlation of a stochastic equation

[Tyler_D]

Homework Statement

Calculate the auto-correlation of $x(t)=e^{-\beta W(t)}W(t)$

Relevant Equations

$\langle x(t)x(t+\tau)\rangle$, $W(t)=\int_0^t dW(s)$

Here is my attempt:

$$
\begin{align}
\langle x(t)x(t+\tau) \rangle &= \langle (e^{-\beta W(t)}W(t))(e^{-\beta W(t+\tau)}W(t+\tau))\rangle \\
&= \langle (e^{-\beta W(t)}e^{-\beta W(t+\tau)})(W(t)W(t+\tau))\rangle \\
&= \langle (e^{-\beta W(t)}e^{-\beta W(t+\tau)})(\int_0^t dW(t)\int_0^{t+\tau}dW(t))\rangle \\
&= \langle (e^{-\beta W(t)}e^{-\beta W(t+\tau)})\int_0^t dW(t)(\int_0^{t}dW(t)+\int_t^{\tau}dW(t))\rangle
\end{align}
$$

From here on I am quite lost on what to do with the exponentials. Do I Taylor-expand them, or what is the best thing to do?

 

[mfb]

Is W(t) some special function? The Lambert W function maybe? That would make nice integrals.

 

[Tyler_D]

It is the Wiener-integral/Brownian motion (my bad, should've written it in the OP):

$W(t) = \int_0^t dW(s)$

 

[MathematicalPhysicist]

I don't understand something, you wrote in Eq. (4) $\int_0^{t+\tau}dW(s) = \int_0^t dW(s)+\int_t^{\tau}dW(s)$, but obviously this means that: $\int_0^{t+\tau}dW(s)=\int_0^\tau dW(s)$.
Does this mean that Weiner integral is periodic with period $t$, since then $W(t+\tau)=W(\tau)$?

 

[Tyler_D]

That's also a typo, thanks for spotting it. The upper limit in the last integral should naturally be $\tau+t$, not just $\tau$.

 

[mfb]

The integral from 0 to t and the integral from t to t+tau are independent, <AB>=<A><B> for them. The expectation value for the latter should have some nice expression I think.
Wikipedia has some relations that can be useful