# Conditional and joint probability manipulations when there are 3 variables

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

$$p(x, y \mid z)= p(x \mid y, z) p(y \mid z)$$

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

$$p(x, y \mid z) = p(z\mid x, y) p(x, y)/p(z)$$

However, this did not 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 chain rule that I can break it down to:

$$p(a, b, c)=p(a \mid b, c) p(b, c) = p(a \mid b, c) p(b \mid c) p(c)$$

Is it allowed to split by the second comma:

$$p(a, b, c) = p(a, b \mid c) p(c) ?$$

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?

• Some people object to $p((a|b)|c)$. See this for example Jul 28, 2012 at 23:50
• Actually $a$, $b$ and $c$ should refer to events rather than variables. $p(x,y|z)$ should be written as $P(x\bigcap y| z)$, etc.
– A.G.
Oct 15, 2016 at 3:11

$$\Pr(a,b,c)=\Pr(a,b\mid c)\Pr(c)$$ is allowed.
You are simply saying $$\Pr(d,c)=\Pr(d\mid c)\Pr(c)$$ where $$d = a \cap b$$.
Combine this with $$\Pr(a,b,c)=\Pr(a\mid b,c)\Pr(b,c)=\Pr(a\mid b,c)\Pr(b\mid c)\Pr(c)$$ and divide through by nonzero $$\Pr(c)$$ to get $$\Pr(a,b\mid c)=\Pr(a\mid b,c)\Pr(b\mid c)$$.
• Thanks, I had failed to see the obvious $p(a,b,c)/p(c)=p(a,b|c)$