Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let two continuous random variables, where the one is a function of the other: $X\, $ and $\, Y=g\left(X\right)$. Their mutual information is defined as $$I\left(X,Y\right)\,=\,h\left(X\right)\,-\,h\left(X|Y\right)=\,h\left(Y\right)\,-\,h\left(Y|X\right)$$ where lowercase h denotes differential entropy, the entropy concept for continuous rv's. It is a proven fact that the mutual information between two variables, for discrete as well as for continuous rv's, is non-negative, and becomes zero only when the two rv's are independent (clearly not our case). Using the fact that Y is a function of X we have $$I\left(X,g\left(X\right)\right)\,=\,h\left(g\left(X\right)\right)\,-\,h\left(g\left(X\right)|X\right) \gt\,0\,\Rightarrow \,h\left(g\left(X\right)\right)\,\gt \,h\left(g\left(X\right)|X\right)$$ Now, differential entropy (unlike entropy for discrete rv's) can take negative values. Assume that it so happens that $h\left(g\left(X\right)\right)\lt\,0$. Then from the positivity of mutual information we obtain $$0\,\gt \,h\left(g\left(X\right)\right)\,\gt \,h\left(g\left(X\right)|X\right) \Rightarrow\; h\left(g\left(X\right)|X\right)\neq\,0$$ And this is the counter-intuitive puzzle: for any discrete random variable Z we always have $h\left(g\left(Z\right)|Z\right)\,=\,0$. This is intuitive: if Z is known, then any function of Z is completely determined -no entropy, no uncertainty remains, and so the conditional entropy measure is zero. But we just saw that, when dealing with continuous rv's where the one is a function of the other, their conditional differential entropy may be non-zero (it doesn't matter whether it is positive or negative), which is not intuitive at all. Because, even in this strange world of continuous rv's, knowing X, completely determines Y=g(X). I have searched high and low to find any discussion, comment or exposition of the matter, but I found nothing. Cover & Thomas book does not mention it, other books do not mention it, a myriad of scientific papers or web sites do not mention it.

My motives: a) Scientific curiosity. b) I want to use the concept of mutual information for continuous rv's in an econometrics paper I am writing, and I feel very uncomfortable to just mention the "non-zero conditional differential entropy" case without being able to discuss it a bit. So any intuition, reference, suggestion, idea, or full answer of course, would be greatly appreciated. Thanks.

share|cite|improve this question
Although conditional differential entropy changes, MI does not change for invertible transformations, and is well defined. If you want more interpretable entropies for continuous variables, perhaps you should look into relative entropy. – Memming Jul 28 '13 at 13:34
Thanks for the suggestion, I will look into it... although the symmetric nature of MI is much more appealing. Still, it is strange that I couldn't find any mention of the case I presented in my question. – Alecos Papadopoulos Jul 28 '13 at 14:15
@AlecosPapadopoulos I think it returns to the question that why differential entropy is called differential entropy! I remember our professor told us the answer of your question, but unfortunately I don't remember the answer! – Mahdi Khosravi Jul 28 '13 at 17:09
up vote 2 down vote accepted

The paradox can be stated in a simpler form:

We know that $I(X;Y)=h(X)-h(X|Y)\ge 0$ holds, also for continuous variables. Take the particular case $Y=X$; then second term vanishes ($h(X|X)=0$) and we get

$$ I(X;X)= h(X)-h(X|X)=h(X) \ge 0$$

But this is not right. The differential entropy can be negative. So what?

when dealing with continuous rv's where the one is a function of the other, their conditional differential entropy may be non-zero (it doesn't matter whether it is positive or negative), which is not intuitive at all.

Your problem (and the problem with the above paradox) is to implicitly assume that the concept "zero entropy means no uncertainty" applies also to differential entropies. That's false. It's false that $h(g(X)|X)=0$, as is $h(X|X)=0$, as is the assumption that $h(X)=0$ implies zero uncertainty for differential entropies. The fact that (by a mere change of scale) a differential entropy can be negative, suggests by itself that here zero entropy (conditional or not) has no special meaning. In particular, a uniform variable in $[0,1]$ has $h(X)=0$.

share|cite|improve this answer
Thank you very much for reaching back two years. I will read your answer carefully (I am very much still interested in the matter), and maybe I will have a question or two in order to understand it fully. – Alecos Papadopoulos May 27 '15 at 23:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.