Autocorrelation of exponential to the power of Wiener-noise

Question

I want to find the autocorrelation $\langle x(t)x(t+\tau)\rangle$ of the following stochastic function

$$ x(t) = e^{-\beta W(t)}W(t) $$

I have so far

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

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?

Answer 1

Here is what I get based on @Tom's comments:

$$ \begin{align} \langle x(t)x(t+\tau) \rangle &= \langle (e^{-\beta W(t)}W(t))(e^{-\beta W(t+\tau)}W(t+\tau))\rangle \\ &= \langle (e^{-\beta W(t)}e^{-\beta W(t+\tau)})(W(t)W(t+\tau))\rangle \\ &= \langle (e^{-\beta W_t}e^{-\beta W_{t+\tau}})W_tW_{t+\tau}\rangle \\ &= \langle (e^{-\beta W_t}e^{-\beta (W_t + W_{t+\tau} - W_t)})W_t(W_t + W_{t+\tau} - W_t)\rangle \\ &= \langle (e^{-\beta W_t}e^{-\beta (W_t + \delta W_{t})})W_t(W_t + \delta W_{t})\rangle \\ &= \langle (e^{-\beta W_t}e^{-\beta W_t}e^{-\beta \delta W_{t}})W_t(W_t + \delta W_{t})\rangle \\ &= \langle e^{-\beta W_t}e^{-\beta W_t}e^{-\beta \delta W_{t}}W_tW_t\rangle + \langle e^{-\beta W_t}e^{-\beta W_t}e^{-\beta \delta W_{t}} W_t\delta W_{t}\rangle \\ &= \langle e^{-\beta W_t}e^{-\beta W_t}W_tW_t\rangle \langle e^{-\beta \delta W_{t}}\rangle + \langle e^{-\beta W_t}e^{-\beta W_t} W_t \rangle\langle e^{-\beta \delta W_{t}}\delta W_{t}\rangle \\ &= \int{e^{-\beta W_t}e^{-\beta W_t}W_tW_t \frac{1}{\sqrt{2\pi t}}e^{-W^2/2t}dW} \times \int{e^{-\beta \delta W_{t}} \frac{1}{\sqrt{2\pi \tau}}e^{-\delta W^2/2\tau}d(\delta W)} + \ldots \\ \end{align} $$

Looking at the first term

$$ \begin{align} \int{e^{-\beta W_t}e^{-\beta W_t}W_tW_t \frac{1}{\sqrt{2\pi t}}e^{-W^2/2t}dW} &= \frac{1}{\sqrt{2\pi t}}\int{W^2 e^{-W^2/2t -2\beta W}dW} \\ &= \frac{e^{-\beta \sqrt{2t}}}{\sqrt{2\pi t}}\int{W^2 e^{-(W/\sqrt{2t} + \beta\sqrt{2t})^2}dW} \end{align} $$