# Is differential entropy always less than infinity?

For an arbitrary continuous random variable, say $X$, is its differential entropy always less than $\infty$? (It’s ok if it’s $-\infty$.) If not, what’s the necessary and sufficient condition for it to be less than $\infty$?

I thought about this question some more and managed to find a counter-example, thanks also to the Piotr’s comments above. The answer to the first question is no – the differential entropy of a continuous random variable (RV) is not always less than $\infty$. For example, consider a continuous RV X whose pdf is

for $x > 2$.

It’s not hard to verify that its differential entropy is infinite. It grows quite slowly though (approx. logarithmically).

For the 2nd question, I am not aware of a simple necessary and sufficient condition. However, one partial answer is as follows. Categorize a continuous RV into one of the following 3 Types based on its support, i.e.

Type 1: a continuous RV whose support is bounded, i.e. contained in [a,b].
Type 2: a continuous RV whose support is half-bounded, i.e. contained in [a,$\infty$) or ($-\infty$,a]
Type 3: a continuous RV whose support is unbounded.

Then we have the following –

For a Type 1 RV, its entropy is always less than $\infty$, unconditionally.
For a Type 2 RV, its entropy is less than $\infty$, if its mean ($\mu$) is finite.
For a Type 3 RV, its entropy is less than $\infty$, if its variance ($\sigma^2$) is finite.

The differential entropy of a Type 1 RV is less than that of the corresponding uniform distribution, i.e. $log(b-a)$, a Type 2 RV, that of the exponential distribution, i.e. $1+log(|\mu-a|)$, and a Type 3 RV, that of the Gaussian distribution, i.e. $\frac{1}{2} log(2{\pi}e\sigma^2)$.

Note that for a Type 2 or 3 RV, the above condition is only a sufficient condition. For example, consider a Type 2 RV with
for $x > 3$. Clearly, its mean is infinite, but its entropy is 3.1 nats. Or consider a Type 3 RV with
for $|x| > 3$. Its variance is infinite, but its entropy is 2.6 nats. So it would be great, if someone can provide a complete or more elegant answer for this part.