Bayes theorem: normalisation denominator and likelihood I have been racking my brains trying to understand Bayes theorem. So, the way I have understood is that the likelihood is the probability of observing the particular outcome given a set of parameter values. The prior describes our belief probabilities over the different parameter values. The denominator acts like a normaliser ensuring that the posterior is a valid distribution.
I have a couple of things I am not clear about:
1: The likelihood is a probability and the prior is a probability as well. So, why is the numerator already not normalised between 0 and 1? Assume I have not ignored any multiplicative constants when computing the likelihood.
2: One way I think of this denominator now is that if we have a bunch of numbers and we would like to normalize them between 0 and 1, one thing we can do is divide each of these numbers by their sum. Is this what the denominator in the Bayes theorem is affectively doing? 
 A: From a technical point of view, here is the argument:
For densities (but the argument is analogous in the discrete case), we write
$$ \pi \left( \theta |y\right) =\frac{f\left( y|\theta \right) \pi \left(\theta \right) }{f(y)} 
$$
The norming constant can be obtained as, by writing a marginal density as a joint density and then writing the joint as conditional times marginal, with the other parameter integrated out,
\begin{align*}
f(y)&=\int f\left( y,\theta \right) d\theta\\
&=\int f\left( y|\theta \right) \pi \left(\theta \right)d\theta
\end{align*}
It ensures integration to 1 because
\begin{align*}
\int \pi \left( \theta |y\right) d\theta&=\int\frac{f\left( y|\theta \right) \pi \left(\theta \right) }{\int f\left( y|\theta \right) \pi \left(\theta \right)d\theta}d\theta\\ &=\frac{\int f\left( y|\theta \right) \pi \left(\theta \right) d\theta}{\int f\left( y|\theta \right) \pi \left(\theta \right)d\theta}\\
&=1,
\end{align*}
where we can "take out" the integral in the denominator because $\theta$ had already been integrated out there.
