combining conditional probabilities

Question

I've come across a physics paper in which pdf $$ p(a|b) $$ is desired, but only $$ p(a|c)\\ p(c|b) $$ are known. It is claimed that $$ p(a|b)=\int p(a|c)p(c|b) dc. $$ Is this correct wlog? I can't prove it. I think it is in fact a conditional expectation for $p(a|c)$. Am I right? Is it possible to find $p(a|b)$ exactly?

Answer 1

This made me very confused a couple of months ago. I struggled a lot trying to rewrite in all possible ways I could think of. In my case, I was interested in the posterior predictive distribution. Using the same notation as Wikipedia (but ignoring the hyperparameters), it is defined as:

$$ p(\tilde{x}|\mathbf{X})=\int_\theta p(\tilde{x}|\theta)p(\theta|\mathbf{X})d\theta $$ and it is just the same thing as in your example. As has been pointed out in the comments, more information is used. It is assumed that $p(\tilde{x}|\theta)$ is the same as $p(\tilde{x}|\theta, \mathbf{X})$ -- that is, conditioning on $\mathbf{X}$ is redundant. This means that $\tilde{x}$ and $\mathbf{X}$ - or $a$ and $b$ in your case - are independent conditional on $\theta$. Then we can rewrite it as:

$$ p(\tilde{x}|\mathbf{X})=\int_\theta p(\tilde{x}|\theta, \mathbf{X})p(\theta|\mathbf{X})d\theta=\int_\theta \frac{p(\tilde{x},\theta, \mathbf{X})}{p(\theta, \mathbf{X})}\frac{p(\theta,\mathbf{X})}{p(\mathbf{X})}d\theta\\ =\int_\theta \frac{p(\tilde{x},\theta, \mathbf{X})}{p(\mathbf{X})}d\theta= \frac{p(\tilde{x}, \mathbf{X})}{p(\mathbf{X})}=p(\tilde{x}|\mathbf{X}) $$ which is what we wanted to show.