Let $W$ be a standard Brownian motion, and let $X$ be the solution to the one dimensional SDE


$$dX\_t = \sigma(t, X\_t) \, dW\_t$$


with initial condition $X\_0 = x\_0$ a.s. for some $x\_0 \in \mathbb R$. We assume $\sigma$ is Lipschitz continuous and uniformly bounded away from $0$.


Suppose that $X\_t$ admits a density $f\_t$ for all $t > 0$.


**Question:** Is it true that we have the entropy inequality


$$\mathbb E[-\log f\_t (X\_t)] > \mathbb E[-\log f\_s (X\_s)]$$


for all $t > s$ in $\mathbb R\_+$?



Does the entropy of a SDE with nondegenerate noise always increase?
$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb R}\newcommand{\pa}{\partial}$The answer is no. The idea is to get a diffusion version of my [two-state Markov chain example](https://mathoverflow.net/questions/447513/does-the-entropy-of-a-sde-with-nondegenerate-noise-always-increase#comment1156881_447513).


Indeed, for $t\in(0,\infty)$ and real $x$, let
\begin{equation\*}
 b(x,t):=e^{(t+1)^2 x^2/(2 t)}+\frac{1-t}{2 (t+1)^3}\ge1+\frac{1-t}{2 (t+1)^3}\ge\frac{53}{54}>\frac12, \tag{-1}\label{-1}
\end{equation\*}
so that
\begin{equation\*}
 \si(x,t):=\sqrt{2b(x,t)}\ge1. 
\end{equation\*}
Moreover, letting
\begin{equation\*}
 f\_t(x):=f(x,t):=g\_{0,t/(t+1)^2}(x), \tag{0}\label{0}
\end{equation\*}
where $g\_{a,s^2}$ is the density of the normal distribution with mean $a$ and variance $s^2$, we see that $f$ is a solution of the [Fokker–Planck equation](https://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation)
\begin{equation\*}
\pa\_t f(x,t)=\pa\_x^2(b(x,t)f(x,t)). \tag{1}\label{1}
\end{equation\*}
So, $f\_t$ is the density of $X\_t$ given the SDE
\begin{equation\*}
 dX\_t=\si(X\_t,t)\,dW\_t
\end{equation\*}
with the initial condition $X\_0=0$ (since the $EX\_t^2=t/(t+1)^2\to0$ as $t\downarrow0$).


However, the entropy
\begin{equation\*}
 \int\_\R f\_t\ln\frac1{f\_t}=\frac{1}{2} (\ln (2 \pi t)-2 \ln (t+1)+1)
\end{equation\*}
decreases in $t\ge1$. $\quad\Box$




---


*Discussion:* The example above may seem counterintuitive. Indeed, if $\si(x,t)$ does not depend on $x$, then $X\_t$ will be normally distributed for each $t$ with variance increasing with $t$, and hence with the entropy increasing with $t$.


In our example, $X\_t$ is still normally distributed for each $t$, but the variance $t/(t+1)^2$ of $X\_t$ is decreasing in $t\ge1$. As noted in the last paragraph, this can only happen if the diffusion coefficient $b(x,t)=\si(x,t)^2/2$ depends on $x$.


We wanted the variance of $X\_t$ to be decreasing in (say) $t\ge1$. It may then seem counterintuitive that the diffusion coefficient $b(x,t)$ in our example increases very fast in $|x|$, especially for large $t$. The Fokker–Planck equation \eqref{1} may help shed some light here. Indeed, suppose first that we are looking for a solution $f$ of \eqref{1} stationary in $t$. Then \eqref{1} implies that $b(x,t)f(x,t)$ is affine in $x$. If $b(x,t)$ and $f(x,t)$ are also even in $x$, then $b(x,t)f(x,t)$ must be constant in $x$. So then, if $f\_t$ is a normal density and hence $f(x,t)$ is decreasing fast in $|x|$, then $b(x,t)$ must be increasing fast in $|x|$. If now the variance of $X\_t$ is decreasing somewhat slowly for large $t$, then we may expect that $b(x,t)$ must still be increasing fast in $|x|$, as is the case in our example.


Actually, the way $b(x,t)$ was found in our example is as follows. We **want** \eqref{0} to hold. With $f$ so prescribed, for each $t$ equation \eqref{1} becomes a simple ODE (with respect to $x$) for the function $b\_t:=b(\cdot,t)$. Thus we get the expression for $b(x,t)$ in \eqref{-1}, with a certain choice of the integration constants.


