# Does the Bayesian posterior need to be a proper distribution?

I know that priors need not be proper and that the likelihood function does not integrate to 1 either. But does the posterior need to be a proper distribution? What are the implications if it is/is not?

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(It is somewhat of a surprise to read the previous answers, which focus on the potential impropriety of the posterior when the prior is proper, since, as far as I can tell, the question is whether or not the posterior has to be proper (i.e., integrable to one) to be a proper (i.e., acceptable for Bayesian inference) posterior.)

In Bayesian statistics, the posterior distribution has to be a probability distribution, from which one can derive moments like the posterior mean $\mathbb{E}^\pi[h(\theta)|x]$ and probability statements like the coverage of a credible region, $\mathbb{P}(\pi(\theta|x)>\kappa|x)$. If $$\int f(x|\theta)\,\pi(\theta)\,\text{d}\theta = +\infty\,,\qquad (1)$$ the posterior $\pi(\theta|x)$ cannot be normalised into a probability density and Bayesian inference simply cannot be conducted. The posterior simply does not exist in such cases.

Actually, (1) must hold for all $x$'s in the sample space and not only for the observed $x$ for, otherwise, selecting the prior would depend on the data. This means that priors like Haldane's prior, $\pi(p)\propto \{1/p(1-p)\}$, on the probability $p$ of a Binomial or a Negative Binomial variable $X$ cannot be used, since the posterior is not defined for $x=0$.

I know of one exception when one can consider "improper posteriors": it is found in "The Art of Data Augmentation" by David van Dyk and Xiao-Li Meng. The improper measure is over a so-called working parameter $\alpha$ such that the observation is produced by the marginal of an augmented distribution $$f(x|\theta)=\int_{T(x^\text{aug})=x} f(x^\text{aug}|\theta,\alpha)\,\text{d}x^\text{aug}$$ and van Dyk and Meng put an improper prior $p(\alpha)$ on this working parameter $\alpha$ in order to speed up the simulation of $\pi(\theta|x)$ (which remains well-defined as a probability density) by MCMC.

In another perspective, somewhat related to the answer by eretmochelys, namely a perspective of Bayesian decision theory, a setting where (1) occurs could still be acceptable if it led to optimal decisions. Namely, if $L(\delta,\theta)\ge 0$ is a loss function evaluating the impact of using the decision $\delta$, a Bayesian optimal decision under the prior $\pi$ is given by $$\delta^\star(x)=\arg\min_\delta \int L(\delta,\theta) f(x|\theta)\,\pi(\theta)\,\text{d}\theta$$ and all that matters is that this integral is not everywhere (in $\delta$) infinite. Whether or not (1) holds is secondary for the derivation of $\delta^\star(x)$, even though properties like admissibility are only guaranteed when (1) holds.

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The posterior distribution need not be proper even if the prior is proper. For example, suppose $v$ has a Gamma prior with shape 0.25 (which is proper), and we model our datum $x$ as drawn from a Gaussian distribution with mean zero and variance $v$. Suppose $x$ is observed to be zero. Then the likelihood $p(x|v)$ is proportional to $v^{-0.5}$, which makes the posterior distribution for $v$ improper, since it is proportional to $v^{-1.25} e^{-v}$. This problem arises because of the wacky nature of continuous variables.

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Cool example, Tom! – Zen Mar 7 '14 at 13:41
+1, though could you expand the answer to the OP's last sentence? Is this whacky posterior meaningful (can you do the kinds of things you'd usually do with a posterior), or is it more analogous to getting a NaN or Inf from some calculations? Is it a sign that something's wrong with your model? – Wayne Mar 7 '14 at 14:48
There's nothing wrong with the model. This posterior is meaningful in the sense that if you receive another observation, you can multiply it in and possibly get back to a proper posterior. So it's not like a NaN, upon which all further operations are NaN. – Tom Minka Mar 7 '14 at 14:59
Although this is probably too late to matter,I do not think using such "counter-examples" help beginners: the problem arises because you use a specific version of the Gaussian density at $x=0$, when it can be arbitrarily defined on this set of measure zero. And hence make the posterior proper or improper depending on the chosen version. – Xi'an Dec 20 '14 at 22:06
Interesting - if you take the general $x$, then the posterior is a generalised inverse gaussian with parameters $-0.25,1,x^2$. @Xi'an - it would be good to see the alternative way to get a proper posterior out of this. – probabilityislogic May 9 at 11:11

Defining the set $$\text{Bogus Data} = \left\{ x:\int f(x\mid \theta)\,\pi(\theta)\,d\theta = \infty \right\} \, ,$$ we have $$\mathrm{Pr}\left(X\in\text{Bogus Data}\right) = \int_\text{Bogus Data} \int f(x\mid \theta)\,\pi(\theta)\,d\theta\,dx = \int_\text{Bogus Data} \infty\,dx \, .$$ The last integral will be equal to $\infty$ if the Lebesgue measure of $\text{Bogus Data}$ is positive. But this is impossible, because this integral gives you a probability (a real number between $0$ and $1$). Hence, it follows that the Lebesgue measure of $\text{Bogus Data}$ is equal to $0$, and, of course, it also follows that $\mathrm{Pr}\left(X\in\text{Bogus Data}\right)=0$.

In words: the prior predictive probability of those sample values that make the posterior improper is equal to zero.

Moral of the story: beware of null sets, they may bite, however improbable it may be.

P.S. As pointed out by Prof. Robert in the comments, this reasoning breaks down if the prior is improper.

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You once wrote: "If we can start with a proper prior and get an improper posterior, then I will quit inference." – Tom Minka Mar 7 '14 at 14:23
A little bit tongue in cheek, there was an implicit quantifier: If we can start with a proper prior and get an improper posterior, for every possible sample value, then I will quit inference. ;-) – Zen Mar 7 '14 at 15:39
By the way, remarkable memory, Tom! – Zen Mar 7 '14 at 22:01
@Zen: I think there is a problem with your reasoning in that you assume that $\mathrm{Pr}\left(X\in\text{Bogus Data}\right)$ is a probability, hence that the joint measure on $(\theta,x)$ is a probability measure, which implies that the prior has to be a (proper) probability measure. – Xi'an Dec 20 '14 at 22:15
You're correct. The reasoning in the answer only works with proper priors. Good point. I'll add a note. – Zen Dec 20 '14 at 22:57

Any "distribution" must sum (or integrate) to 1. I can think a few examples where one might work with un-normalized distributions, but I am uncomfortable ever calling anything which marginalizes to anything but 1 a "distribution".

Given that you mentioned Bayesian posterior, I bet your question might come from a classification problem of searching for the optimal estimate of $x$ given some feature vector $d$

\begin{align} \hat{x} &= \arg \max_x P_{X|D}(x|d) \\ &= \arg \max_x \frac{P_{D|X}(d|x) P_X(x)}{P_D(d)} \\ &= \arg \max_x {P_{D|X}(d|x) P_X(x)} \end{align}

where the last equality comes from the fact that $P_D$ doesn't depend on $x$. We can then choose our $\hat{x}$ exclusively based on the value $P_{D|X}(d|x) P_X(x)$ which is proportional to our Bayesian posterior, but do not confuse it for a probability!

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Improper posterior distribution only arises when you're having an improper prior distribution. The implication of this is that the asymptotic results do not hold. As an example, consider a binomial data consisting of $n$ success and 0 failures, if using $Beta(0,0)$ as the prior distribution, then the posterior will be improper. In this situation, the best is to think of a proper prior distribution to substitute your improper prior.