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?!

  • $\begingroup$ 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 $\endgroup$
    – Henry
    Jun 25, 2020 at 9:57

1 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,


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.

  • $\begingroup$ Superb! well explained. thanks $\endgroup$
    – Alireza
    Jun 25, 2020 at 18:02

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