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?

  • 2
    $\begingroup$ Some people object to $p((a|b)|c)$. See this for example $\endgroup$
    – Henry
    Jul 28, 2012 at 23:50
  • $\begingroup$ 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. $\endgroup$
    – A.G.
    Oct 15, 2016 at 3:11

1 Answer 1


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

  • 3
    $\begingroup$ Thanks, I had failed to see the obvious $p(a,b,c)/p(c)=p(a,b|c)$ $\endgroup$
    – Jim
    Jul 29, 2012 at 0:20

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