# Relative entropy between singular measures

Usually, to define relative entropy between two probability measures, one assumes absolute continuity. Is it possible to extend the usual definition in the non absolutely continuous case?

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+1, Interesting question! FYI, Kullback-Leibler divergence wiki-page explicitly mentions the case of discrete random variates. –  Sasha Nov 7 '11 at 15:36

Relative entropy between two probability measures $P$ and $Q$ can be defined even if $P$ is not absolutely continuous w.r.to $Q$. In any case, $P$ and $Q$ are absolutely continuous w.r.to a common measure $\mu$ (one can take $\mu$ to be $\frac{P+Q}{2}$). Then relative entropy between $P$ and $Q$ is defined as $$D(P\|Q)=\int p\log\frac{p}{q}d\mu,$$ where $p=dP/d\mu$ and $q=dQ/d\mu$.
It's a bit unclear how this integral is defined: for some cases mutual singularity can lead to existence of $A$ s.t. $P(A) = 0$ and $Q(A) = 0$ (like with point mass and Lebesgue measure). Then $p,q$ are never non-zero together and the $\log\frac pq$ is undefined. Moreover, for any measurable function $f$ we have $\int pfd\mu = \frac12\int f\,dP$ so even if $\log\frac pq$ can be defined, we'll have to integrate it over values of $x$ for which $q=0$, so $D(P||Q) = \infty$ at least informally. Though it makes sense for mutually singular measure to have infinite relative entropy. –  Ilya Nov 11 '11 at 9:24
Yes you are right. But, while defining the above said formula, we always assume the following convention. $\log 0=-\infty$, $\log\frac{a}{0}=+\infty$, for any $a\ge 0$ and $0\cdot (\pm \infty)=0$. I hope, it makes sense now. –  Ashok Nov 11 '11 at 10:20
I had a typo in my first comment: there may exist $A$ s.t. $P(A) = 0$ and $Q(A^c) =0$. Does it mean then that $D(P\| Q) = +\infty$ in such case? –  Ilya Nov 11 '11 at 10:48
I think, what you have in mind is that $P$ and $Q$ have disjoint support. In such case $D(P\|Q)=\infty$. But in the case $P(A)=0$ and $Q(A^c)=0$, $D(P\|Q)$ may or may not be $\infty$. If for some $A$, $P(A)\neq 0$ but $Q(A)=0$ then we will definitely have $D(P\|Q)=\infty$. –  Ashok Nov 11 '11 at 11:02