I have a question: is it possible that an SDE has a "nice" density, but its invariant measure does not have a "nice" density? I asked this question at math.stackexchange but without much success, so I'm asking it here.
More precisely, consider a stochastic differential equation $$ dX_t=b(X_t) dt + \sigma(X_t) dW_t, $$ where $W$ is a standard Brownian motion and $b,\sigma$ are continuous smooth functions. We can further assume, if needed, that $b$ is locally (but not globally!) Lipschitz and that the Hörmander condition holds (but $\sigma$ is not uniformly elliptic). Suppose that we know that this equation has a unique strong solution and that the transition kernel has a smooth density $p_t(x,y)$ for all $t>0$, $x,y\in\mathbb{R}^d$.
Assume that we know that this equation has a unique invariant measure $\pi$. Question: Is it true that $\pi$ has a smooth density $p$?
Indeed, it is immediate to see that for any measurable set $A$ one has $$ \pi(A)=\int_A\Bigl(\int_{\mathbb{R}^d} p_t(x,y) \pi(dx)\Bigr)dy, $$ which implies that $\pi$ has a density $p$ such that $$ p(y)=\int_{\mathbb{R}^d} p_t(x,y) p(x)dx. $$ However it is not clear to me why $p$ is finite everywhere or why it is differentiable everywhere. Can one construct a counter-example here?
