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\u00a0 probability density function $x\\mapsto 1\/(\\lambda+x)^2$? Funny!<\/p>\n
Consider now the kinetic diffusion process $(X_t,Y_t)_{t\\geq0}$ on $\\mathbb{R}^2$ where<\/p>\n
$$\\begin{cases}dX_t&=dB_t-s(X_t)\\lambda dt\\\\dY_t&=dW_t-s(Y_t)|X_t|dt\\end{cases}$$ 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?<\/p>\n","protected":false},"excerpt":{"rendered":"
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…<\/p>\n