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Is the KL divergence not defined because uniform has bounded support and gaussian has unbounded support?

How else can I calculate the distance of my gaussian to a 'maximum entropy' distribution if I can't use the uniform distribution?

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    $\begingroup$ The distance is infinite. $\endgroup$ – Xi'an May 21 at 6:28
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The KL divergence $$ KL\left(P \middle\| Q\right) = \int \log \frac{d P}{d Q }dP $$ is only defined if the Radon-Nikodym derivative exists, which is when $P$ is absolutely continuous with respect to $Q$ (written $P \ll Q$). This means that there can't be any sets $A$ where $P(A) > 0$ and $Q(A) = 0$, otherwise we would be dividing by zero.

In your case, $p$ is the density of the uniform random variable, and $q$ is the density of the normal random variable (they are both dominated by the Lebesgue measure), so you could calculate $$ KL\left(P \middle\| Q\right) = \int \log \frac{p(x)}{q(x)}p(x)dx, $$ but you couldn't calculate $KL\left(Q \middle\| P\right)$. You can calculate $KL\left(P \middle\| Q\right)$ because there are no sets $A$ such that $\int_A p(x) dx > 0$ and $\int_A q(x) dx = 0$.

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  • $\begingroup$ Yes, unfortunately I need KL(Q||P). j $\endgroup$ – user3180 May 21 at 4:11

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