How to simplify $\int_0^T\int_0^T\mathbb{E}_{\pi}\left(\bar{f}(X_t)\bar{f}(X_s)\right)\mathrm{d}t\mathrm{d}s$?

Question

I would like to get the result

$$\int_0^T\int_0^T\mathbb{E}_{\pi}\left(\bar{f}(X_t)\bar{f}(X_s)\right)\mathrm{d}t\mathrm{d}s = 2\int_0^T(T-s)\mathbb{E}_{\pi}\left(\bar{f}(X_s)\bar{f}(X_0)\right)\mathrm{d}s.$$

$X_t$ is a continuous time real-valued stochastic process, $$\bar{f}:=f - \mathbb{E}_{\pi}f,$$ and $\mathbb{E}_{\pi}$ is defined to be such that $$\mathbb{E}_{\pi}f = \int_{\chi} f(x)\pi(x)\mathrm{d}x,$$ where $\pi$ is the stationary distribution of $X_t$ and $\chi$ is the state space.

Answer 1

As noted by @Did in the comments, $\pi$ is the stationary distribution of $X_t$, and hence $\mathbb{E}_{\pi}\left(\bar{f}(X_t),\bar{f}(X_s)\right)=R_{\bar{f}}(t-s)$.

To evaluate the integral, let's start with the transformation $(t,s)\mapsto(t-s,t+s)$. Then, instead of integrating over the square $(0,0), (0,T), (T,0), (T,T)$, we're integrating over the square $(0,0), (-T,T), (T,T), (0,2T)$. Seeing as $f$ is real-valued, the autocorrelation function is even, and all we need to do is find the integral over the triangle with vertices $(0,0), (T,T), (0,2T)$ and multiply it by $2$.

Let's find the Jacobian of the transformation. Setting $(u,v)=(t-s,t+s)$ (and hence $t=\frac{u+v}{2}$, $s=\frac{u-v}{2}$), we get $$ \begin{vmatrix} \frac{1}{2} &\frac{1}{2} \\ \frac{1}{2} &-\frac{1}{2} \end{vmatrix}=-\frac{1}{2}. $$ Next, note that the cross-section with the line $x=s$ is $2(T-s)$. Putting all this together, our integral becomes $$2\int_0^T 2(T-s)\left|-\frac{1}{2}\right|R_{\bar{f}}(s)\mathrm{d}s=2\int_0^T(T-S)\mathbb{E}_{\pi}\left(\bar{f}(X_s),\bar{f}(X_0)\right)\mathrm{d}s,$$ which is what we wanted to show.

Answer 2

Since $\pi$ is the stationary distribution, then $\mathsf{E}_{\pi}[\bar{f}(X_t)\bar{f}(X_s)]=\mathsf{E}_{\pi}[\bar{f}(X_{|t-s|})\bar{f}(X_0)]$, and \begin{align}\int_0^T&\int_0^T\mathsf{E}_{\pi}[\bar{f}(X_t)\bar{f}(X_s)]\,dtds =2\iint_{0\le s\le t\le T}\mathsf{E}_{\pi}[\bar{f}(X_{|t-s|})\bar{f}(X_0)]\,dtds\\ &=2\iint_{0\le s\le s+\tau\le T}\mathsf{E}_{\pi}[\bar{f}(X_{\tau})\bar{f}(X_0)]\,d\tau ds=2\iint_{0\le s\le T-\tau,0\le \tau\le T}\mathsf{E}_{\pi}[\bar{f}(X_{\tau})\bar{f}(X_0)]\,dsd\tau\\ &=2\int_0^T(T-\tau)\mathsf{E}_{\pi}[\bar{f}(X_{\tau})\bar{f}(X_0)]\,d\tau =2\int_0^T(T-s)\mathsf{E}_{\pi}[\bar{f}(X_s)\bar{f}(X_0)]\,ds \end{align}