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?
 A: 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. 
A: 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.
A: Most explanations of Bayes miss the mark.  Consider the following for the role of Pr(B).
The crux of Bayes is the "update factor" $[Pr(B|A) / Pr(B)]$.
This is the transformation applied to the prior.
If B always occurs in all states of the world, there is no information content & the update factor is 1.
In this case, $Pr(A|B) = Pr(A)$.
However, if B occurs frequently when A has occurred, but the overall probability of B occurring is very low, then there is high information content with respect to Pr(A).
The update factor will be HIGH and so $Pr(A|B) >> Pr(A)$.
For completeness, if B occurs rarely when A has occurred, but the overall probability of B occurring is very high, then there is also information content with respect to Pr(A), but in the opposite direction.
The update factor will be LOW and so $Pr(B|A) << Pr(A)$.
Purely mechanical explanations of Bayes seem to miss the genius of this simple equation.
