posterior probability distribution of a 2D parameter

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?!

• I would have thought $p(\theta_1\mid Y) = \dfrac{\int p(\theta_1,\theta_2) L(\theta_1,\theta_2\mid Y)\,d\theta_2} {\int p(\theta) L(\theta\mid Y)\,d\theta}$ might be slightly clearer. In the denominator you integrate over the whole space while in the numerator just over the nuisance parameter Jun 25, 2020 at 9:57

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.

• Superb! well explained. thanks Jun 25, 2020 at 18:02