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I'm having troubles verifying why the following is correct.

$$p(x,y|z)= p(x|y,z)p(y|z)$$

I tried grouping the (x,y) together and split by the conditional which gives me


However this didn't bring me any closer. I'm uncertain about what kind of manipulations are allowed given more than 2 variables.

Say an expression like: $$p(a,b,c)$$ Then I know from the chainrule that I can break it down to: $$p(a,b,c)=p(a|b,c)p(b,c)=p(a|b,c)p(b|c)p(c)$$

Is it allowed to split by the second comma: $$p(a,b,c)=p(a,b|c)*p(c) ?$$

And even more complicated and expression like: $$p(a|b,c)$$

Am I allowed to rewrite this expression by grouping (a|b) together to give me something like $$p(a|b,c)=p((a|b)|c)p(c)$$ And does this expression even make sense?

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Some people object to $p((a|b)|c)$. See this for example – Henry Jul 28 '12 at 23:50
up vote 14 down vote accepted

$\Pr(a,b,c)=\Pr(a,b|c)\Pr(c)$ is allowed.

You are simply saying $\Pr(d,c)=\Pr(d|c)\Pr(c)$ where $d = a \cap b$.

Combine this with $\Pr(a,b,c)=\Pr(a|b,c)\Pr(b,c)=\Pr(a|b,c)\Pr(b|c)\Pr(c)$ and divide through by nonzero $\Pr(c)$ to get $\Pr(a,b|c)=\Pr(a|b,c)\Pr(b|c)$.

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Thanks, I had failed to see the obvious $p(a,b,c)/p(c)=p(a,b|c)$ – Jim Jul 29 '12 at 0:20

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