Proving that $p(x,y|z) = p(x|z) p(y|z) \iff p(x|y,z) = p(x|z)$

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How can I prove the equivalence of these two conditional independence definitions?
Definition 1: $p(x,y|z) = p(x|z) * p(y|z)$
Definition 2: $p(x|y,z) = p(x|z)$

I tried substituting Definition 2 into Definition 1: $$p(x,y|z) = p(x|y,z) * p(y|z)\\ = p(x|y) * p(x|z) * p(y|z)\\ = p(x|y) * p(y|z) * p(x|z)$$

However, I am stuck after this step. I may be wrong from the first step.

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I would suggesting using the definition of conditional probability (or density):

$$p(u|v):=\frac{p(u,v)}{p(v)}.$$

So to show definition 2 implies definition 1, we have

$$p(x,y|z)=\frac{p(x,y,z)}{p(z)}=\frac{p(x|y,z)p(y,z)}{p(z)}=p(x|y,z)p(y|z)=p(x|z)p(y|z),$$

where the last step uses definition 2. Now try the other direction.