posterior probability distribution of a 2D parameter

Question

In Little & Rubin's Statistical Analysis with Missing Data page 115, The posterior probability of a parameter $\theta_1$ as in $\theta = (\theta_1, \theta_2)$ given observations $Y$ has been given as:

$$ p(\theta_1|Y) = \frac{\int p(\theta) L(\theta|Y).d\theta_2} {\int p(\theta) L(\theta|Y).d\theta} $$

where $L(\theta|Y) \propto P(Y|\theta)$ is the likelihood function.

I readlly don't understand how this equation is true. the only clue I've got is to start from the simple product rule $P(\theta|Y) = P(\theta_1|Y)P(\theta_2|Y)$ (assume independence). but then how to proceed?!

Answer 1

The equation is nothing but the definition of conditional probability. If you ignore the integrals, and just think of summation instead, it looks quite straightforward. By definition,

$P(\theta_1|Y)=P(\theta_1,Y)/P(Y)$.

Now we will show that the numerator and denominator work out exactly to those above.

The numerator is just integrating over $\theta_2$, and note that $p(\theta_1,\theta_2) L(\theta_1,\theta_2|Y) = p(\theta_1,\theta_2,Y)$. So that the numerator can be written as

$\sum_{\theta_2} p(\theta_1,\theta_2,Y) = p(\theta_1,Y)$.

The denominator is summing over both $\theta_1$ and $\theta_2$, so can be written as

$\sum_{\theta_1,\theta_2} p(\theta_1,\theta_2,Y) = p(Y)$.

Thus we have shown that the numerator divided by the denominator equals the definition mentioned in the first equation.