5
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$
0
$\begingroup$

In Probability Theory, density functions are generally defined as Radon-Nikodym derivatives themselves, $\frac{dP}{dQ}$.

The likelihood function interprets these densities (R-N derivatives) as a function of the parameters, given some observed outcome. More explicitly, let $X$ be an absolutely continuous random variable. Then, $$\mathcal{L}(\theta|x\in X) = f(x|\theta) = \mathbb{P}(x\in X|\theta)$$ In other words, the likelihood function measures the probability of observing $x$ given the parameters $\theta$.

The likelihood ratio is meant to assess the goodness-of-fit of two statistical models (with different parameters) given the same set of observations $x$, not two entirely different distributions. More explicitly, let $\Theta$ be the set of all possible parameters, and consider some subsets $\Theta_0, \Theta_1 \in \Theta$. The likelihood ratio is then, $$\mathcal{L(\Theta_0,\Theta_1)} = -2\log\frac{\sup_{\Theta_0\in\Theta} \mathcal{L}(\theta)}{\sup_{\Theta_1\in\Theta}\mathcal{L}(\theta)}$$ which tests for the null hypothesis $\theta\in\Theta_0$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ "Radon-Nikodym derivatives themselves" yes but then you should switch the notation instead of implicitly calling $Q$ Lebesgue measure $\endgroup$ – Taylor Apr 13 at 16:14
  • $\begingroup$ Also, densities are not probability mass functions. They cannot be interepreted that way $\endgroup$ – Taylor Apr 13 at 16:14
  • $\begingroup$ finally, the last expression is not the likelihood ratio, but it is a function of the likelihood ratio. The reason it is transformed is that when it is written in thsi way, it's asymptotically $\chi^2$ distributed--but this is irrelevant at the moment. $\endgroup$ – Taylor Apr 13 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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