It is my understanding that if we have an improper prior, we may either get a proper or a posterior distribution. My question is if there are cases where a proper prior distribution may, when combined with the likelihood function, result in an improper posterior. Are there any trivial cases? Thanks!
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$\begingroup$ This exact question was asked last year on Reddit: m.reddit.com/r/statistics/comments/2sncqo/… $\endgroup$– shadowtalkerCommented May 8, 2016 at 3:07
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$\begingroup$ Relevant discussion: stats.stackexchange.com/questions/87321/… $\endgroup$– KOECommented May 8, 2016 at 14:26
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I think that everything you need is contained in the several answers for this related but slighly different question (Does the Bayesian posterior need to be a proper distribution?). Here is a proposal of summary for your specific question:
From the sketch of proof https://stats.stackexchange.com/a/89157/14346: if the prior is proper and $p(x|\theta)$ none degenerate then the set of observations for which the posterior is improper is a Lesbegue null set. In other words: $$ \mbox{proper prior} \Rightarrow \mbox{proper posterior almost everywhere}. $$ So, there exists some observations $x$ (but a null set) leading to an improper $p(\theta|x)$. An example of such $x$ for a given $p(\theta|x)$ is given here https://stats.stackexchange.com/a/89150/14346.
Nevertheless, (as pointed out in the comment of the example below) density functions are equivalent (roughly, can be treated as the same) if they only differ on a null set. So the specific observations leading to improperness are complelety arbitrary depending on which specific forms of the density you choose from the equivalent sets.
