I see that the Radon-Nikodym derivative is the ratio of probability measures, $dP/dQ$. How is this different, in general, from a likelihood ratio of two continuous distributions? I understand the RN-definition broadly applies for discrete/continuous/mixture densities, but beyond that is there any difference?


1 Answer 1


If $dx$ is Lebesgue measure, then it dominates both measures because they correspond with continuous random variables, and one of the properties of RN derivatives is $$ \frac{dP}{dQ} = \frac{\frac{dP}{dx}}{\frac{dQ}{dx}}. $$ The numerator is the density of $P$, and the denominator is the density of $Q$. This is the second property on wikipedia.

So yes, the likelihood ratio is just a particular case. If these two measures were for discrete random variables, then you would replace $dx$ with the counting measure, and you would get a ratio of probability mass functions.

  • $\begingroup$ In order for $dP/dQ$ to make sense, we need $P \ll Q$. Why is this the case? All we seem to have is that $P\ll dx$ and $Q\ll dx$, where $dx$ is Lebesgue measure. $\endgroup$
    – Satana
    Feb 22, 2023 at 2:09
  • 1
    $\begingroup$ Because that is the Radon-Nikodym theorem: density (i.e. $dP/dQ$) existence iff absolute-continuity (sometimes called domination in certain textbooks). This is the foundational assumption, and the Lebesgue bit is an added/secondary assumption. $\endgroup$
    – Taylor
    Feb 22, 2023 at 3:52

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