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Suppose you have a particle which diffuses as:

$$\dot\theta(t)=\sqrt{2D_r}\xi_\theta$$

where $\xi_\theta$ is a random gaussian noise with:

$$<\xi_\theta(t)>=0 \\ <\xi_\theta(t)\xi_\theta(s)>=\delta(t-s) $$ Given this problem,

a) how can you prove that $<cos(\theta(t))\cdot cos(\theta(s))>=e^{-2D_r|t-s|}$

b) which will be the result of $<cos(\theta(t))>$? $e^{-D_rt}$?

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  • $\begingroup$ You know how to solve the SDE to get the marginal and joint distributions? After that you can probably go a long way by writing $\cos z = \Re e^{iz}$ $\endgroup$
    – Nadiels
    Mar 31 '18 at 16:33
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As I understand it you have a diffusion and you want to solve it and then calculate the moments of some functions of this diffusion, you haven't specified an initial condition so I'm taking it to be zero and we have $$ \theta(t)=\sqrt{2D}\int_0^t\xi(\tau)\operatorname{d}\tau, $$ that is to say $\theta(t)$ is a scalar multiple of the standard Weiner process and $$ \langle \theta(s),\theta(t)\rangle = 2D\min(s,t), $$ That is all the information you need from the diffusion. For the rest let's just consider a bivariate Gaussian random variable $X_1, X_2$ with $$ X_1,X_2 \sim\mathcal{N}\left(\begin{bmatrix}0 \\ 0 \end{bmatrix},\begin{bmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{12} & \sigma_{22} \end{bmatrix}\right) $$ then $$ \begin{align} \mathbb{E}\left[\cos(X_1)\cos(X_2)\right] &= \frac{1}{2}\mathbb{E}\left[ \Re \left(e^{i(X_1-X_2)} +e^{i(X_1+X_2)}\right)\right] \\ &=\frac{1}{2}\Re\left(\mathbb{E}e^{i(X_1-X_2)} + \mathbb{E} e^{i(X_1+X_2)}\right) \\ &=\frac{1}{2}\left(e^{-(\sigma_{11}-2\sigma_{12}+\sigma_{22})/2} +e^{-(\sigma_{11}+2\sigma_{12} + \sigma_{22})/2} \right). \end{align} $$ Now you can easily apply that to the situation you are interested in. Calculating $\mathbb{E}\cos X$ is straight forward using the same trick.

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  • $\begingroup$ Nice! Although I cannot see where do you get the |t-s| from... $\endgroup$ Apr 1 '18 at 11:59

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