# Density of invariant measure of stochastic differential equation

I have a question: is it possible that an SDE has a "nice" density, but its invariant measure does not have a "nice" density?

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 (no further assumptions on uniform ellipticity of $$\sigma$$ or that $$b$$ is globally Lipschitz or that $$b$$ is bounded). Suppose that we know that this equation has a unique strong solution and that the transition kernel has a smooth density $$p_t(x,y)$$.

Assume that we know that this equation has a unique invariant measure $$\pi$$. Question: Is it true (without any further assumptions) 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?

EDIT: This answer is wrong. $$p_t$$ is not continuous at $$(0,0)$$ and thus not smooth. Further more $$p_t(0, \cdot)$$ is not a density instead $$p_t\left(y, dx\right) \to \delta_0(dx)$$ as $$y \to 0$$.

What's below shows instead that $$X_t$$ having a smooth density for all time $$t \geq 0$$ does not imply that $$X$$'s limiting distribution has a density. As well in what's below $$\nu P_t$$ is absolutely continuous with respect to the Lebesgue measure only if $$\nu\left(\left\{0\right\}\right) = 0$$.

A smooth transition density does not imply the invariant measure has a density even if the coefficients are globally Lipschitz. One counterexample is Geometric Brownian Motion with negative drift. Let $$\mu, \sigma > 0$$ and $$X$$ satisfy $$dX_t = -\mu X_t dt + \sigma X_t dW_t$$ The transition density of $$X$$ is $$p_t\left(x,y\right) = \frac{1}{\sqrt{2\pi}}\frac{1}{y\sigma \sqrt{t}}\exp{\left(-\frac{\left(\log{\left(\frac{y}{x}\right) + \left(\mu + \frac{1}{2}\sigma^2\right)t}\right)^2}{2\sigma^2 t}\right)}$$ which is smooth as $$\frac{\partial^n}{\partial y^n} p_t\left(x,y\right)$$ is a linear combination of terms of the form $$x^{-\alpha}\log^\beta(y)\exp\left(-\log^2(y)\right)$$. However the invariant density of $$X$$ is a dirac mass at zero. Further more for any initial distribution, $$\nu$$, let $$\nu P_t$$ denote the distribution of $$X_t$$ at time $$t$$, $$\nu P_t(A) = \int_A \left(\int_{\mathbb{R}} p_t(x,y)\nu(dx)\right)dy,$$ $$\nu P_t$$ converges to $$\delta_0$$ and exponential fast in the 1-Wasserstein metric, i.e. $$\mathcal{W}_1\left(\nu P_t, \delta_0 \right) \leq e^{-\mu t}\mathcal{W}_1\left(\nu, \delta_0 \right)$$.

• I do see why the invariant measure would be a dirac mass at zero, but isn't the transition density computed assuming that $y>0$? Sorry if I'm misunderstanding something Aug 1, 2021 at 5:31
• Dear Shiva, thank you for your reply. Unfortunately your counter-example won't work: indeed, in this case the process X does not have a density for $x=0$ and $P_t(0,A)=\delta_0(A)$. I'm interested in an SDE which has smooth density $p_t(x,y)$ for all $x,y$.
– Oleg
Aug 1, 2021 at 14:39
• @Snoop you are completely right: I would like to thank Shiva for his reply, but, unfortunately, his counterexample is not valid since the process $X$ does not have density if started at $0$.
– Oleg
Aug 1, 2021 at 14:40
• @Oleg Yes on second look it appears that I am mistaken and $p_t$ is does not have a density when $y = 0$, sorry. The example given shows that the distribution of $X$ at time $t$ having a smooth density does not imply that the limiting distribution of $X$ has a density. I had missed the subtle difference between this and what you were asking for. Aug 1, 2021 at 16:23