# Why Normalizing Factor is Required in Bayes Theorem?

Bayes theorem goes $$P(\textrm{model}|\textrm{data}) = \frac{P(\textrm{model}) \times P(\textrm{data}|\textrm{model})}{P(\textrm{data})}$$

This is all fine. But, I've read somewhere:

Basically, P(data) is nothing but a normalising constant, i.e., a constant that makes the posterior density integrate to one.

We know that $0 \leq P(\textrm{model}) \leq 1$ and $0 \leq P(\textrm{data}|\textrm{model}) \leq 1$.

Therefore, $P(\textrm{model}) \times P(\textrm{data}|\textrm{model})$ must be between 0 and 1 as well. In such a case, why do we need a normalizing constant to make the posterior integrate to one?

• When you are working with probability densities, as mentioned in this post, you can no longer conclude 0 <= P(model) <= 1 nor 0 <= P(data/model) <= 1, because either (or even both!) of those could exceed $1$ (and even be infinite). See stats.stackexchange.com/questions/4220. – whuber Dec 18 '14 at 22:35
• It is not the case that $$P(\textrm{data}|\textrm{model})\le 1$$ because this vague notation represents the integrated likelihood of the data, not a probability. – Xi'an Dec 19 '14 at 11:29

First, the integral of "likelihood x prior" is not necessarily 1.

It is not true that if:

$$0 \leq P(\textrm{model}) \leq 1$$ and $$0 \leq P(\textrm{data}|\textrm{model}) \leq 1$$

then the integral of this product with respect to the model (to the parameters of the model, indeed) is 1.

Demonstration. Imagine two discrete densities: $$P(\textrm{model}) = [0.5, 0.5] \text{ (this is called "prior")}\\ P(\textrm{data | model}) = [0.80, 0.2] \text{ (this is called "likelihood")}\\$$

If you multiply them both you get: $$[0.40, 0.25]$$ which is not a valid density since it does not integrate to one: $$0.40 + 0.25 = 0.65$$

So, what should we do to force the integral to be 1? Use the normalizing factor, which is: $$\sum_{\text{model_params}} P(\text{model}) P(\text{data | model}) = \sum_\text{model_params} P(\text{model, data}) = P(\text{data}) = 0.65$$

(sorry about the poor notation. I wrote three different expressions for the same thing since you might see them all in the literature)

Second, the "likelihood" can be anything, and even if it is a density, it can have values higher than 1.

As @whuber said this factors do not need to be between 0 and 1. They need that their integral (or sum) be 1.

You will often see: $$P(\textrm{model}|\textrm{data}) \propto P(\textrm{data}|\textrm{model}) P(\text{model})$$ because the missing denominator can be easily get by integrating this product. Note that this integration will have one well known result if the prior and the likelihood are conjugate.

• +1. This is the only answer that actually addresses the original question of why the normalization constant is needed to make the posterior integrate to one. What you do with the posterior later (e.g. MCMC inference or calculating absolute probabilities) is a different matter. – Pedro Mediano May 30 '17 at 13:07
• I have difficulties following the notation. What does it mean $P(model)=[0.5,0.5]$? Suppose we use a Gaussian model with given variance $\sigma^2=1$, and we wish to find the mean $\mu$. What does it mean $P(\mu) = [0.5, 0.5]$? – raffaem Apr 27 '19 at 9:57
• I tried to follow the notation of the OP. I mean that you have two possible values for your model parameter, and that both are equally likely. For instance, is there a teapot revolving around the Earth? yes/no? In your case it does not make sense because your parameter $\mu$ is continuous and can take infinite values. – alberto May 4 '19 at 19:01

The short answer to your question is that without the denominator, the expression on the right-hand side is merely a likelihood, not a probability, which can only range from 0 to 1. The "normalizing constant" allows us to get the probability for the occurrence of an event, rather than merely the relative likelihood of that event compared to another.

$$P(\text{model}~ | ~\text{data}) \propto P(\text{model}) \times P(\text{data}~|~\text{model})$$