For most of my time in stats, I have been able to ignore the marginal distribution that is usually present at the denominator of any bayesian posterior distribution.

For example, if we write down $L_x(\theta)\pi(\theta)$ and recognize that this function of $\theta$ looks like a distribution of $\theta$ but with an incorrect normalizing constant, I usually just mix and match till I get it since

$$ \pi(x|\theta) \propto L_x(\theta)\pi(\theta) $$


why can I do this, are there any cases where this breaks down?

  • 1
    $\begingroup$ Draw a graph with $\theta $ on horizontal and $ L_x (\theta)\pi (\theta) $ on the vertical. Now what happens when you plot $\pi (\theta|x) $ on the vertical instead? The scale of the vertical axis changes - and nothing else. $\endgroup$ Feb 5, 2014 at 10:04

4 Answers 4


Theoretically, why do we not need to compute a marginal distribution constant for finding a Bayesian posterior?

Generally speaking, you do need to - it's just that sometimes it's so easy that you might not notice you did it.

With 'textbook' problems you can often take $\pi(x|\theta) \propto L_x(\theta)\pi(\theta)$, then play about with the result and recognize the density function, at which point you've computed what the normalizing constant must have been - the thing required to scale your $L_x(\theta)\pi(\theta)$ so it integrates to 1. Since it's a pdf you know it integrates to 1, and since it's proportional to $ L_x(\theta)\pi(\theta)$, you know you have divided by the integral of that.

With cases where that doesn't work there are often a few choices.

One is numerical integration - you can integrate $ L_x(\theta)\pi(\theta)$ to work out the normalizing constant. So then you can compute expectations, and so on.

Another is sampling; maybe you can't find the integral but you can bound it and use rejection sampling, or approximate it and use Metropolis-Hastings etc. With a sample from the posterior, you can again find means or other quantities as needed, or get a good approximation to the density or the cdf.

There are other approaches.

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    $\begingroup$ My main concern is that there appears to be no rigorous proof I have seen as to why if the likelihood times the prior has "looks" like a certain distribution, we can just add on the constants so that it "is" the distribution. The only "proof" I have seen just comes from textbooks telling you that since its proportional, we can just add on constants. Any thoughts to how I can understand it at a deeper level or a more rigorous manner? Meaning, is there a case where, for example, say the product of the likelihood and prior HAS a certain form, but ISNT even as we try to add constants? Thanks! $\endgroup$
    – user123276
    Feb 5, 2014 at 5:40
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    $\begingroup$ It follows immediately from the fact that integration is linear: $\int af(x) dx = a \int f(x) dx$. I assume you're already aware of that. $\endgroup$
    – Glen_b
    Feb 5, 2014 at 10:56

Your problem is equivalent to the following: Suppose you have a function $f(x)$ such that $\int f(x) dx < \infty$, and you are looking for a constant $c$ such that $\int cf(x) dx = 1$. Clearly, $c = 1/ \int f(x) dx$ would work, but perhaps it isn't easy to compute $\int f(x) dx$. In your quest to find $c$, perhaps you find a probability density function $g(x)$ (integrates to $1$) that shares the same "form" as $f(x)$. That is $g(x) = d f(x)$ for some constant $d$. Then $1 = \int g(x) dx = \int d f(x) dx$ and hence $c = d$. In other words, multiplying $f(x)$ by $c$ gives us the density function $g(x)$. This is the same logic that allows us to find the posterior distribution by recognizing the functional form of the density.


The "formal proof" you are looking for is called Bayes' Theorem (see also Posterior probability) which states that:

$$\pi(\theta\vert{\bf x}) = \dfrac{f({\bf x}\vert\theta)\pi(\theta)}{\pi({\bf x})}.$$

The left-hand side represents the posterior distribution and IT IS a distribution as long as the prior is proper. From this expression you can identify $f({\bf x}\vert\theta)$ as the likelihood function $L(\theta;{\bf x})$ and you can also see that $\pi({\bf x})$ does not depend upon $\theta$. Therefore

$$\pi(\theta\vert{\bf x}) \propto L(\theta;{\bf x})\pi(\theta).$$

Also, note that it should be $\theta\vert{\bf x}$ and not the other way round. $f({\bf x}\vert \theta)$ is the likelihood function, which is not a distribution as a function of $\theta$, tipically.

Discussion: In Bayesian statistics it is impossible, for non trivial examples, to identify what sort of distribution is this (e.g. normal, student-t ...). Then, the use of MCMC methods is often necessary to sample from the posterior and to conduct a Bayesian data analysis. MCMC methods require the evaluation of the posterior up to a proportionality constant. For this reason, it is not necessary to calculate $\pi({\bf x})$. However, for Bayesian model comparison you need to obtain a numerical approximation of this quantity, given that the Bayes factors are defined in terms of the normalising constant.



In simple terms, the denominator, or the marginal distribution of the RHS of your Bayes theorem is just a constant that is used to make the RHS numerator a pdf. If you know what kind of distribution your RHS numerator, i.e, the Likelihood function * prior distribution follows, then you can find out the denominator(marginal) easily. For example, if your prior is uniform and your likelihood function is a Binomial, then your posterior will be proportional to a Beta distribution. You can now easily find out the constants for a Beta Distribution.

  • $\begingroup$ You can add Latex typesetting to your math using the dollar sign, for instance $x_{abc}^{23}$ produces $x_{abc}^{23}$; doubling up the dollar signs also centers the equation. $\endgroup$
    – Silverfish
    Sep 23, 2016 at 22:57

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