# 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?

(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.

• What if $\int {\pi \left( \theta \right){\text{d}}\theta } = + \infty$, $\int {f\left( {\left. x \right|\theta } \right)\pi \left( \theta \right){\text{d}}\theta } = + \infty$ but $\frac{{\int {\theta f\left( {\left. x \right|\theta } \right)\pi \left( \theta \right){\text{d}}\theta } }}{{\int {f\left( {\left. x \right|\theta } \right)\pi \left( \theta \right){\text{d}}\theta } }} < + \infty$???
– user193726
Feb 2, 2020 at 21:25
• Something like a Feynman path integral maybe.
– user193726
Feb 3, 2020 at 20:19
• I finally answered the question below. Please tell me whether you accept my counterexample or not. BR
– user193726
Feb 4, 2020 at 8:36
• In fact, I'm struggling to interpret my own counterexample because it seems the situation is different a priori and a posteriori!? Indeed, suppose ${\mathbf{x}}$ has the same kind of prior distribution. By the same reasoning, we would conclude that in any case, it has a prior expectation. But this is NOT true. On the contrary, a priori we would rather say that it has NO expectation at all if, for instance, the covariance matrix is a singular matrix obtained from a finite difference scheme without (Dirichlet, Neumann) boundary conditions on the derivatives of function $x\left( t \right)$.
– user193726
Feb 4, 2020 at 9:09
• We want to constrain the derivatives only, not the function itself. So, a priori we say that ${\mathbf{x}}$ has no expectation at all, but a posteriori we want to estimate it by computing its posterior expectation as described below! So, it is difficult to interpret, but that's how it works, definitely: would the prior covariance matrix be positive definite, ${\mathbf{x}}$ would have zero prior expectation and this would be undesirable from the regularization point of view. But it is singular and everything works perfectly. But it hopefully admit a well-defined posterior expectation! Weird.
– user193726
Feb 4, 2020 at 9:16

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.

• Cool example, Tom!
– Zen
Mar 7, 2014 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? Mar 7, 2014 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. Mar 7, 2014 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. Dec 20, 2014 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. May 9, 2016 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 blows up if the prior is improper.

• You once wrote: "If we can start with a proper prior and get an improper posterior, then I will quit inference." Mar 7, 2014 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, 2014 at 15:39
• By the way, remarkable memory, Tom!
– Zen
Mar 7, 2014 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. Dec 20, 2014 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, 2014 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!

– whuber
Mar 7, 2014 at 21:20
• One way to interpret the OP question "does the posterior need to be a proper distribution?" is to ask if it is mathematically possible to start with a proper prior and end with an improper posterior. Minka's answer gives an explicit example in which it happens. I tried to complement it with my answer and point out that this can only happen inside a set of zero prior predictive probability.
– Zen
Mar 7, 2014 at 21:55
• @Zen It seems to me that a closely related interpretation is "if the posterior is not proper, what information can I get from it?" This accepted answer looks like it provides useful and correct advice related to that in a special circumstance (which is clearly described). The acceptance looks to me like a signal that eretmochelys struck home with a shrewd guess about the circumstances.
– whuber
Mar 7, 2014 at 22:03

Later is better than never. Here is a natural and useful counterexample I believe, arising from Bayesian nonparametrics.

Suppose $${\mathbf{x}} = \left( {{x_1},...,{x_i},...{x_n}} \right) \in {\mathbb{R}^n}$$ has posterior probability distribution

$$p\left( {\left. {\mathbf{x}} \right|D} \right) \propto {e^{ - \frac{1}{2}{{\mathbf{x}}^{\mathbf{T}}}{\mathbf{Ax}} + {{\mathbf{J}}^{\mathbf{T}}}{\mathbf{x}}}}$$

We want to evaluate the posterior expectation $$\mathbb{E}\left. {\mathbf{x}} \right|D$$. If $${\mathbf{A}}$$ is positive definite, then let

$$I \triangleq \int\limits_{{\mathbb{R}^n}} {{e^{ - \frac{1}{2}{{\mathbf{x}}^{\mathbf{T}}}{\mathbf{Ax}} + {{\mathbf{J}}^{\mathbf{T}}}{\mathbf{x}}}}{{\text{d}}^n}{\mathbf{x}}} = \sqrt {{{\left( {2\pi } \right)}^n}{{\left| {\mathbf{A}} \right|}^{ - 1}}} {e^{\frac{1}{2}{{\mathbf{J}}^{\mathbf{T}}}{{\mathbf{A}}^{ - 1}}{\mathbf{J}}}}$$

By Leibniz rule/Feynman trick, we have

$$\frac{{\partial I}}{{\partial {J_j}}} = \int\limits_{{\mathbb{R}^n}} {\frac{{\partial {e^{ - \frac{1}{2}{{\mathbf{x}}^{\mathbf{T}}}{\mathbf{Ax}} + {{\mathbf{J}}^{\mathbf{T}}}{\mathbf{x}}}}}}{{\partial {J_j}}}{{\text{d}}^n}{\mathbf{x}}} = \int\limits_{\,{\mathbb{R}^n}} {{x_j}{e^{ - \frac{1}{2}{{\mathbf{x}}^{\mathbf{T}}}{\mathbf{Ax}} + {{\mathbf{J}}^{\mathbf{T}}}{\mathbf{x}}}}{{\text{d}}^n}{\mathbf{x}}} = \\ \frac{\partial }{{\partial {J_j}}}\sqrt {{{\left( {2\pi } \right)}^n}{{\left| {\mathbf{A}} \right|}^{ - 1}}} {e^{\frac{1}{2}{{\mathbf{J}}^{\mathbf{T}}}{{\mathbf{A}}^{ - 1}}{\mathbf{J}}}} = \sqrt {{{\left( {2\pi } \right)}^n}{{\left| {\mathbf{A}} \right|}^{ - 1}}} {e^{\frac{1}{2}{{\mathbf{J}}^{\mathbf{T}}}{{\mathbf{A}}^{ - 1}}{\mathbf{J}}}}\frac{\partial }{{\partial {J_j}}}\frac{1}{2}{{\mathbf{J}}^{\mathbf{T}}}{{\mathbf{A}}^{ - 1}}{\mathbf{J}} = \\ \frac{1}{2}I\frac{\partial }{{\partial {J_j}}}{{\mathbf{J}}^{\mathbf{T}}}{{\mathbf{A}}^{ - 1}}{\mathbf{J}} = I\sum\limits_{i = 1}^n {{\mathbf{A}}_{ij}^{ - 1}{{\mathbf{J}}_i}} \\$$

Therefore

$$\mathbb{E}\left. {{x_j}} \right|D = \frac{{\int\limits_{\,{\mathbb{R}^n}} {{x_j}{e^{ - \frac{1}{2}{{\mathbf{x}}^{\mathbf{T}}}{\mathbf{Ax}} + {{\mathbf{J}}^{\mathbf{T}}}{\mathbf{x}}}}{{\text{d}}^n}{\mathbf{x}}} }}{{\int\limits_{\,{\mathbb{R}^n}} {{e^{ - \frac{1}{2}{{\mathbf{x}}^{\mathbf{T}}}{\mathbf{Ax}} + {{\mathbf{J}}^{\mathbf{T}}}{\mathbf{x}}}}{{\text{d}}^n}{\mathbf{x}}} }} = \sum\limits_{i = 1}^n {{\mathbf{A}}_{ij}^{ - 1}{{\mathbf{J}}_i}}$$

and

$$\mathbb{E}\left. {\mathbf{x}} \right|D = {{\mathbf{A}}^{ - 1}}{\mathbf{J}}$$

Now, if $${\mathbf{A}}$$ is only positive semi-definite and singular, so that $$p\left( {\left. {\mathbf{x}} \right|D} \right)$$ is improper, degenerate and

$$\int\limits_{{\mathbb{R}^n}} {p\left( {\left. {\mathbf{x}} \right|D} \right){{\text{d}}^n}{\mathbf{x}}} = + \infty$$

it suffices to replace the matrix inverse $${{\mathbf{A}}^{ - 1}}$$ by its Moore-Penrose pseudoinverse $${{\mathbf{A}}^ + }$$ to get

$$\mathbb{E}\left. {\mathbf{x}} \right|D = {{\mathbf{A}}^ + }{\mathbf{J}}$$

IT WORKS. Same for higher moments. So, it seems that a Bayesian posterior does not need to be proper/non-degenerate in order to be proper, that is to yield legitimate and useful inferences.

• @Xi'an Thanks for the feedback. You should better ask Richard P. Feynman. This is routine calculation in QM. See for instance Path integral formulation, section Expectation values. The expectation $\left\langle F \right\rangle$ is defined as the quotient of two infinite path integrals. I agree this is wild maths but it DOES WORK in applications.
– user193726
Feb 4, 2020 at 9:51
• @Xi'an Surely, Mr Feynman is joking. :)
– user193726
Feb 4, 2020 at 11:01
• @Xi'an To recap: defining expectations and moments as ratios of two infinite quantities is routine in QM/QFT and Bayesian nonparametrics. This yield, concrete, accurate and useful inferences. This invalidate your claim (the proof is on your side, not Feynman's) of the neccessity of a finite (=1) posterior normalization constant. From my point of view, the real mystery is elsewhere: in Bayesian nonparametrics, those proper-improper posteriors arise from kind of improper priors. Therefore, by the same reasoning, they should be proper as well: they should have a well-defined expectation too.
– user193726
Feb 4, 2020 at 14:08
• But if this were true, this would damage regularization when constraining the underlying function derivatives (e.g. second-order smoothing prior): the prior would not only constrain the derivates as expected but also the function itself, which is undesirable. But this is NOT true, it seems the prior has actually no expectation because experiments show that there is no damage, no bias in the posterior expectation with a positive semi-definite, singular prior covariance regularization matrix. This, I don't understand, but that's an experimental fact.
– user193726
Feb 4, 2020 at 14:13

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.

• This answer is incorrect. See my answer. Mar 7, 2014 at 13:07