# conditional and joint probability (3 variables)

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

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

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

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