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Thread: conditional independence

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    Exclamation conditional independence




    hello,

    I would like to ask the following:

    I have 3 random variables X , Y and Z.
    If I know that X is independent of Y given Z then does this imply that X is independent of Z?

    thanks !

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    Re: conditional independence

    (X ⊥ Y | Z) => (X|Z)

    I don't think so, because I think Z could be a D seperator (screen) between X and Y.

    So in causality it could be: X → Z → Y.

    Though I could be wrong, since I am just trying to learn these comcepts myself. I think this is in the first chapter of Judea Pearl's, Causality book - which I don't have access to.
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    Re: conditional independence

    Let X ~ N(0,1), E1 ~ N(0,1), E2 ~ N(0,1) all independent of one another. Now let Z = X + E1, and let Y = Z + E2

    X and Y are not independent but conditional on Z they are. X and Z are definitely not independent.

    So to answer your question - X being independent of Y given Z does not imply that X is independent of Z.
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    Re: conditional independence


    Thank for the confirmation Dason, but I expected something more elegant from you

    Your example could be written:

    Z = X + E1
    Y = Z + E2

    so, Y = (X+ E1) + E2
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