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?


1 Answer 1


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. $\mathcal{W}_1\left(\nu P_t, \delta_0 \right) \leq e^{-\mu t}\mathcal{W}_1\left(\nu, \delta_0 \right)$.

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    $\begingroup$ 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 $\endgroup$
    – Snoop
    Aug 1, 2021 at 5:31
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    $\begingroup$ 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 $x,y$. $\endgroup$
    – Oleg
    Aug 1, 2021 at 14:39
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    $\begingroup$ @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 $0$. $\endgroup$
    – Oleg
    Aug 1, 2021 at 14:40
  • $\begingroup$ @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. $\endgroup$
    – Shiva
    Aug 1, 2021 at 16:23

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