Did you know that if $X\sim\mathcal{E}(\lambda)$ and $\mathcal{L}(Y\,\vert\,X=x)=\mathcal{E}(x)$ for all $x\geq0$ then $Y$ follows a Pareto distribution with  probability density function $x\mapsto 1/(\lambda+x)^2$? Funny!

Consider now the kinetic diffusion process $(X_t,Y_t)_{t\geq0}$ on $\mathbb{R}^2$ where

$\displaystyle\begin{cases}dX_t&=dB_t-s(X_t)\lambda dt\\dY_t&=dW_t-s(Y_t)|X_t|dt\end{cases}$

where $(B_t)_{t\geq0}$ and $(W_t)_{t\geq0}$ are independent standard Brownian motions and $s$ is the sign function… Can you guess the invariant measure and control the speed of convergence?