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Math Help - Proof of semi-graphoid symmetry axiom

  1. #1
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    Proof of semi-graphoid symmetry axiom

    I have some trouble understanding why this proof is a proof:
    Lemma:
    I_p(X, Z, Y) \iff I_p(Y, Z, X)

    Proof:
    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.

    \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?
    \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)}

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

    \iff I_p(Y, Z, X)
    Last edited by Lepzed; September 22nd 2012 at 01:39 AM.
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  2. #2
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    Re: Proof of semi-graphoid symmetry axiom

    Hey Lepzed.

    Is the random variable X independent to Z?
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  3. #3
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    Re: Proof of semi-graphoid symmetry axiom

    The independence relation 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.
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