Let $X$ be a real valued random variable. Of course, the integer part $\lfloor X \rfloor$ of $X$ is a discrete random variable taking values in $\mathbb{Z}$. We can therefore define its discrete entropy \begin{equation} H(\lfloor X \rfloor) = - \sum_{n\in\mathbb{Z}} \mathbb{P}( \lfloor X \rfloor = n ) \log( \mathbb{P}( \lfloor X \rfloor = n ) ), \end{equation} which is in $[0,\infty]$ as a sum of nonnegative terms, since $- x \log x \geq 0$ for any $0 \leq x \leq 1$ (with the convention $0\log 0 = 0$).

I am looking for sufficient conditions such that $H(\lfloor X \rfloor ) < \infty$. For instance, is it sufficient to know that $X$ has a finite absolute moment $\mathbb{E}[|X|^p] < \infty$ for some $p>0$? Any condition of this type, possibly weaker, is welcome.

Motivation: There are strong connection between the differential entropy of $X$ (assuming $X$ has a pdf whose differential entropy is well-defined) and the discrete entropy of $\lfloor nX \rfloor$ when $n\rightarrow0$. This was the main topic of the 1959 paper from Alfred Rényi intitled On the dimension and entropy of probability distributions: I am questioning the assumptions under which the discrete entropy is well-defined.


3 Answers 3


Since $\lfloor X\rfloor$ has finite entropy if and only if $|\lfloor X\rfloor|$ has finite entropy, it suffices to consider random variables taking values in the natural numbers. Write $p_n$ for $\mathbb P(X=n)$ (so that $\sum_n p_n=1$). We have $X\in L^q$ if and only if $\sum p_n n^q<\infty$.

Suppose $X\in L^q$ so that $\sum p_n n^q<\infty$. Then let $S_1=\{n\colon p_n<\frac{1}{n^2}\}$ and $S_2=\{n\colon p_n\ge \frac 1{n^2}\}$. We have $$ H(X)=\sum_n -p_n\log p_n=-\sum_{n\in S_1}p_n\log p_n-\sum_{n\in S_2}p_n\log p_n. $$ Since $-t\log t$ is increasing on $[0,\frac 1e]$, the first sum is bounded above by $$ \sum_{n\in S_1}\frac{2\log n}{n^2}<\infty. $$ There exists an $n_0$ so that for $n\ge n_0$, $2\log n<n^q$. For $n\in S_2$ such that $n\ge n_0$, $-\log p_n<2\log n<n^q$, so that $$ -\sum_{n\in S_2,\,n\ge n_0}p_n\log p_n\le \sum_{n\in S_2,\,n\ge n_0}p_n n^q<\infty. $$ Hence $H(X)<\infty$. (This trick appears in a couple of papers of mine: one with Ciprian Demeter in NYJM and another more recent preprint with Tamara Kucherenko and Christian Wolf).

  • $\begingroup$ What are some examples of distributions where the entropy is not well-defined? $\endgroup$
    – user44143
    Aug 4, 2020 at 10:53
  • $\begingroup$ That's a cute trick effectively, for a neat proof. Thanks! $\endgroup$
    – Goulifet
    Aug 4, 2020 at 23:21
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    $\begingroup$ @MattF: are you asking for examples where the entropy is infinite? My go-to example is $P(X=n)=c/[n(\log n)^2]$ for $n\ge2$ and with suitable normalization. $\endgroup$ Aug 5, 2020 at 1:20

Using (say) decimal notation, ASCII encoding, and a delimiter symbol such as a space or comma, as well as the law of large numbers, one can almost surely encode $N$ independent copies of $\lfloor X \rfloor$ using $O( N {\bf E} \log( 2 + |X| ) ) + o(N)$ bits. Applying the Shannon source coding theorem, we conclude that

$$ {\bf H}( \lfloor X \rfloor ) \ll {\bf E} \log(2 + |X| )$$

which by Jensen's inequality also gives

$$ {\bf H}( \lfloor X \rfloor ) \ll_p \log(2 + {\bf E} |X|^p)$$

for any $0 < p < \infty$.


$\newcommand{\fx}{\lfloor X\rfloor}$ $\newcommand\Z{\mathbb{Z}}$ We shall prove more than requested: that $H(\fx)<\infty$ if $E\ln(1+|X|)<\infty$.

Indeed, let $$p_n:=P(\fx=n),$$ so that $$H(\fx)=-\sum_{n\in\Z}p_n\ln p_n.$$ Let $q\colon\mathbb R\to(0,\infty)$ be any function such that $$\sum_{n\in\Z}q(n)=1\tag{1}$$ and $$q(x)\le cq(\lfloor x\rfloor)\tag{2}$$ for some real $c>0$ and all real $x$.

Then by the Gibbs inequality for the Kullback–Leibler divergence between $(p_n)_{n\in\Z}$ and $(q(n))_{n\in\Z}$ we have $$0\le KL((p_n)_{n\in\Z}||(q(n))_{n\in\Z})=\sum_{n\in\Z}p_n\ln\frac{p_n}{q(n)}=-H(\fx)+\sum_{n\in\Z}p_n\ln\frac1{q(n)},$$ whence, in view of (2), $$H(\fx)\le\sum_{n\in\Z}p_n\ln\frac1{q(n)} \\ =\sum_{n\in\Z}\int_{[n,n+1)}P(X\in dx)\ln\frac1{q(n)} \\ \le\sum_{n\in\Z}\int_{[n,n+1)}P(X\in dx)\ln\frac c{q(x)} \\ =E\ln\frac c{q(X)}=\ln c+E\ln\frac1{q(X)}.$$ So, $$H(\fx)<\infty\quad\text{if}\quad E\ln\frac1{q(X)}<\infty.$$ Taking here e.g. $q(x)=\frac C{(1+|x|)^2}$, where $C:=1/\sum_{n\in\Z}\frac1{(1+|x|)^2}[=\frac3{\pi ^2-3}]$, we have conditions (1) and (2) satisfied. So, $$H(\lfloor X\rfloor)<\infty\quad\text{if}\quad E\ln(1+|X|)<\infty.$$ It follows that for any real $a>0$ $$H(\lfloor X\rfloor)<\infty\quad\text{if}\quad E|X|^a<\infty,$$ as was initially desired.

  • $\begingroup$ Is there any variable $X$ for which the $\alpha$-moment is undefined for all $\alpha>0$? Or does this imply that the entropy is in fact finite for all variables? $\endgroup$
    – user44143
    Aug 4, 2020 at 16:28
  • $\begingroup$ @MattF. If you restrict $\alpha$ to integers the answer is yes (Cauchy distribution). If you allow for fractional moments then this paper makes the answer seem to be "no". $\endgroup$
    – Mark
    Aug 4, 2020 at 16:57
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    $\begingroup$ @MattF. : If e.g. a random variable $X$ has the pdf $f$ given by $f(x)=\frac1{2|x|\ln^2|x|}\,1\{|x|>e\}$, then $EX_+^a=EX_-^a=\infty$ for all $a>0$, where $X_+:=\max(0,X)$ and $X_-:=\max(0,-X)$. Is this what your question is about? $\endgroup$ Aug 4, 2020 at 17:30
  • $\begingroup$ Very nice. The question about the finiteness of the entropy for measures on $\mathbb Z$ with a finite logarithmic moment is more or less equivalent to that about the measures with a finite first moment on the set of words in a finite alphabet (which naturally arises in symbolic dynamics). The standard argument in the literature is the one quoted by Anthony - which I have always found somewhat artificial. $\endgroup$
    – R W
    Aug 4, 2020 at 22:57
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    $\begingroup$ @Goulifet : The Kullback--Leibler divergence is well-defined (and nonnegative), because $p_n\ln\frac{p_n}{q_n}\ge p_n-q_n$ and $\sum_n(p_n-q_n)=0$, where $q_n:=q(n)$. Cf. en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/Gibbs%27_inequality. $\endgroup$ Aug 5, 2020 at 0:03

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