# Proof of semi-graphoid symmetry axiom

• Sep 22nd 2012, 01:20 AM
Lepzed
Proof of semi-graphoid symmetry axiom
I have some trouble understanding why this proof is a proof:
Lemma:
$\displaystyle I_p(X, Z, Y) \iff I_p(Y, Z, X)$

Proof:
$\displaystyle I_p(X, Z, Y) \iff P(C_x | C_y \land C_z) = P(C_x | C_z)$
Ok, this is just conditional independence, x independent of y given z.

$\displaystyle \iff \frac{P(C_x \land C_y \land C_z)}{P(C_y \land C_z)} = \frac{P(C_x \land C_z)}{P(C_z)}$
This is rewritten according to the rule of conditional independence P(A|B) = P(A & B) / P(B).

Bottom up the same is done for the right hand side of the lemma, but I don't get how the following and previous step are equal.. Why does this prove anything?
$\displaystyle \iff \frac{P(C_x \land C_y \land C_z)}{P(C_x \land C_z)} = \frac{P(C_y \land C_z)}{P(C_z)}$

$\displaystyle \iff P(C_y | C_x \land C_z) = P(C_y | C_z)$

$\displaystyle \iff I_p(Y, Z, X)$
• Sep 22nd 2012, 11:31 PM
chiro
Re: Proof of semi-graphoid symmetry axiom
Hey Lepzed.

Is the random variable X independent to Z?
• Sep 23rd 2012, 09:46 AM
Lepzed
Re: Proof of semi-graphoid symmetry axiom
The independence relation $\displaystyle I_p(X, Z, Y)$ expresses that in the context of information about Z, information about Y is irrelevant with respect to Y or put differently: X is condtionally independent of Y given Z.