# Normalizing constant in Bayes theorem

I read that in Bayes rule, the denominator $\Pr(\textrm{data})$ of

$$\Pr(\text{parameters} \mid \text{data}) = \frac{\Pr(\textrm{data} \mid \textrm{parameters}) \Pr(\text{parameters})}{\Pr(\text{data})}$$

is called a normalizing constant. What exactly is it? What is its purpose? Why does it look like $\Pr(data)$? Why doesn't it depend on the parameters?

• When you integrate $f(\text{data}|\text{params})f(\text{params})$, you are integrating over the parameters and so the result has no term depending on the parameters, in the same way that $\int_{x=0}^{x=2}xy\;dx = 2y$ does not depend on $x$. – Henry Jun 20 '11 at 18:57

The denominator, $\Pr(\textrm{data})$, is obtained by integrating out the parameters from the join probability, $\Pr(\textrm{data}, \textrm{parameters})$. This is the marginal probability of the data and, of course, it does not depend on the parameters since these have been integrated out.

Now, since:

• $\Pr(\textrm{data})$ does not depend on the parameters for which one wants to make inference;
• $\Pr(\textrm{data})$ is generally difficult to calculate in a closed-form;

one often uses the following adaptation of Baye's formula:

$\Pr(\textrm{parameters} \mid \textrm{data}) \propto \Pr(\textrm{data} \mid \textrm{parameters}) \Pr(\textrm{parameters})$

Basically, $\Pr(\textrm{data})$ is nothing but a "normalising constant", i.e., a constant that makes the posterior density integrate to one.

• What do you exactly mean by "by integrating out the parameters"? What's the precise meaning of "integrating out" in this context? – nbro Mar 7 '18 at 17:51
• @nbro: I mean Pr(data) = integral over the parameters of Pr(data, parameters) – ocram Mar 8 '18 at 6:24

When applying Bayes' rule, we usually wish to infer the "parameters" and the "data" is already given. Thus, $\Pr(\textrm{data})$ is a constant and we can assume that it is just a normalizing factor.