Prove that $P(x,y|z)=P(x|z)P(y|x,z)$

Question

This is my attempt, but I'm stuck:

$P(x,y|z) = P(y|x,z)P(x)=P(y|x,z)\sum_{z}P(x,z)=P(y|x,z)\sum_{z}P(x|z)P(z)$

I think the steps I took thus far have been correct, although I'm not sure whether they have actually been goal oriented, I mean $\sum_{z}P(x|z)P(z) \neq P(x|z)$ isn't it?

Can anybody please give me a hint?

Answer 1

Actually intuitively this is obvious since it is well known that $$P(x,y)=P(y|x)P(x)$$ which of course will not change if you have additional information. i.e. if we write $$P(x,y|\cdot)=P(x|\cdot)P(y|x, \cdot)$$ Taking $z$ in the place of $"\cdot"$ you have the required equality.

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In your attempt the first equation is false, since you write $$P(x,y|z)=P(y|x,z)P(x)$$ which is not true since you should write $P(x|z)$ instead of $P(x)$.

Answer 2

Rigorously, this follows from the general definition $$ P(a,b,c,\ldots\mid u,v,w,\ldots)=\frac{P(a,b,c,\ldots, u,v,w,\ldots)}{P(u,v,w,\ldots)}. $$ Can you apply this identity to your setting? And possibly spot the faulty steps in your post? For example, $$ P(x,y|z) =\frac{P(x,y,z)}{P(z)}\quad\&\quad P(y|x,z)=\frac{P(x,y,z)}{P(x,z)}. $$