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In the following integral, p(x) and q(x) are probability distributions. Can you help me to determine in what situation this integral is equal to infinity. For example, I think that such a situations is when only p(x) has an infinite peak.


Thank you very much!

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Thank you for the edit! – Marco Feb 8 '11 at 9:05

The integral that you have is the Kullback-Leibler divergence between distributions P and Q, $D_{KL}(P \parallel Q)$. This divergence is roughly a kind of a "distance" between the two distributions. The reason "distance" is in quotes is because this divergence is not symmetric and is hence not a metric. However, a useful way to think about the divergence $D_{KL}(P \parallel Q)$ is that it is the penalty paid for mistaking distribution $P$ as distribution $Q$. This statement can be made precise using information theory. If $D_{KL}(P \parallel Q)$ is infinity, it is because the two distributions are quite unlike each other that you incur an infinite penalty for mistaking $P$ as $Q$. This can happen for instance when distribution $P$ can produce values that distribution $Q$ can never do - in this case, mistaking $P$ as $Q$ is indeed a grievous error. I will leave it to you to interpret this in terms of the integral above to derive the condition under which the divergence is infinite.

Update: Adding more information in response to Marco's comment below. It got too unwieldy to be left as a comment: Given any $M>0$ and any distribution $P$, we can find $Q$ such that $D_{KL}(P \parallel Q) > M$. But note that as $M$ grows, we need to adaptively change $Q$ to make sure the divergence grows larger than $M$. This is different from saying $D_{KL}(P \parallel Q) = \infty$ for a given $P$, $Q$ which is what the question is stating. I think this can happen only if the support of $P$ includes points not in the support of $Q$.

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Thank you, that interpretation helps. However, let us assume that p is absolutely continuous with respect to q, meaning that the support of p is a subset of the support of q. So distribution p cannot produce values that distribution q can never do. However, I think it is not enough to ensure finite values... – Marco Feb 8 '11 at 10:04
What do you think of this? Suppose, for example, that p and q have the same support, (0, infinity). If q(x) --> 0 much more quickly than p(x) --> 0 then the integral is infinity. Thx – Marco Feb 8 '11 at 10:39
Thank you very much for the update. I still need to think on it, but I am more and more convinced! Thank you! – Marco Feb 9 '11 at 10:01
@Marco: See also mjqxxxx's explanation here:… – Dinesh Feb 9 '11 at 17:05

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