Let $X$ be an absolutely continuous (i.e. its law is absolutely continuous with respect to the Lebesgue measure) random variable with probability density $p$. Its differential entropy is given by
$$h(X) = - \int\_{\mathbb{R}} p(x) \log p(x) \mathrm{d} x$$
with the convention $0 \log 0 = 0$, as soon as the integral is absolutely convergent.


A random variable is infinitely divisible if, for any $n \geq 1$, $X$ can be decomposed as the sum of $n$ i.i.d. random variables.


**Question:** Are there infinitely divisible and absolutely continuous random variables for which the differential entropy does not exist?


*Comment:* It is possible to construct random variables for which the differential entropy does not exist. The constructions I could find are however handcrafted to make the differential entropy undefined. Since infinitely divisible random variables have a strong structure, I am wondering what can be said in this case.


It is moreover possible to find simple conditions so that the differential entropy is well-defined, for instance if $X$ admits some positive moments and $p$ is a bounded probability density. The former condition is however not always true for infinitely divisible laws, and I have no idea for the latter.


Any help would be appreciated.



Existence of the differential entropy for infinitely divisible laws
For real $t>0$, let
\begin{equation}
 p\_t:=e^{-t}e^{\*tf}\*g\_t:=e^{-t}\sum\_{n=0}^\infty\frac{t^n f^{\*n}}{n!}\*g\_t, \tag{0}
\end{equation}
where $f$ is the (bounded by $c:=1/e$) pdf given by  

\begin{equation}
 f(x)=\frac{1\{x\ge e\}}{x\ln^2 x}, \tag{0.5}
\end{equation}
$f^{\*n}:=f\*\cdots\*f$ ($n$ times, with $f^{\*0}$ defined as the Dirac delta function at $0$), and $g\_t$ is the normal pdf with mean $0$ and variance $t$. Then $p\_s\*p\_t=p\_{s+t}$ for all real $s,t>0$. So, $$p:=p\_1$$ is the pdf of an infinitely divisible distribution. Moreover, similarly to the main result in [this paper](https://link.springer.com/article/10.1007/BF00535504), we have
\begin{equation}
 p(x)\sim f(x) \tag{1}
\end{equation}
(the convergence everywhere here is as $x\to\infty$),
whence
\begin{equation}
 p(x)\ln p(x)\sim-\frac1{x\ln x}, \tag{1.5}
\end{equation}
so that the differential entropy does not exist.




---


Since the proof of (1) is a bit involved, let us make do with something weaker than (1), which however can be proved quickly. Indeed, note first here that, by (0) and (0.5), for $g:=g\_1$ and all real $x$
\begin{equation}
 p(x)\ge e^{-1}(f\*g)(x)\ge\frac1e\,\int\_{-1}^1f(x-y)g(y)\,dy\sim\frac c{x\ln^2 x}=:q(x), \tag{2}
\end{equation}
where $c:=\frac1e\,\int\_{-1}^1g(y)\,dy\in(0,\infty)$.
On the other hand, again by (0),
\begin{equation}
 p(x)=\frac1e\,\int\_{-\infty}^\infty g(x-y)\,(e^{\*f})(y)\,dy\to0, \tag{3}
\end{equation}
by dominated convergence.
Now note that the function $u\mapsto-u\ln u$ is positive and increasing in a right neighborhood of $0$. Hence, by (2) and (3), for all large enough $x>0$  

\begin{equation\*}
 -p(x)\ln p(x)\ge-\frac{q(x)}2\,\ln\frac{q(x)}2\sim\frac c{2x\ln x}
\end{equation\*}
(cf. (1.5)).
So, the differential entropy does not exist.


