Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Some people object to $p((a|b)|c)$. See this for example –  Henry Jul 28 '12 at 23:50

1 Answer 1

up vote 7 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 $\Pr(a,b|c)=\Pr(a|b,c)\Pr(b|c)$.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.