1. ## Likelihood confirmation question

Basic question but

$p(\theta | D) = \frac{p(D|\theta)p(\theta)}{p(D)} = \frac{p(D|\theta)p(\theta)}{\int p(D|\theta)p(\theta)}$

Why is $p(D) = \int p(D|\theta)p(\theta)$? I thought it should be the sum? Wouldn't the integral just be 1?

Thankyou

2. ## Re: Likelihood confirmation question

No. The probability of D occurring overall may be tiny across the entire range of $\theta$. It's an integral when $p(\theta)$ is a probability density function rather than a discrete probability distribution.

3. ## Re: Likelihood confirmation question

Originally Posted by romsek
No. The probability of D occurring overall may be tiny across the entire range of $\theta$. It's an integral when $p(\theta)$ is a probability density function rather than a discrete probability distribution.
Thankyou! So, it's just because it is a continuous distribution?
It looks similar to marginal likelihood, can I ask how this comes into play?
Thankyou

4. ## Re: Likelihood confirmation question

Originally Posted by AshleyCS
Thankyou! So, it's just because it is a continuous distribution?
It looks similar to marginal likelihood, can I ask how this comes into play?
Thankyou
Yes, because it's a continuous distribution.

If the distribution depended on parameters other than $\theta$ then yes this would be a marginal distribution, but since it doesn't you end up with just a value which is the probability of D occurring at all.