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Math Help - Variation of a RV conditional on another - help pls

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    Question Variation of a RV conditional on another - help pls

    Hello. I have the following problem to solve: prove that

    Var(X)=E(Var(XY)) + Var(E(XY))

    where denotes conditional probability. Sorry, I do not know how write on this LaTex thing
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by dely84 View Post
    Hello. I have the following problem to solve: prove that

    Var(X)=E(Var(XY)) + Var(E(XY))

    where denotes conditional probability. Sorry, I do not know how write on this LaTex thing
    Sorry I won't write it in Latex, I hope it will be enough.

    The mean of a variable is not a variable anymore. It is a constant.
    And so is the variance of a variable.

    Now there are 2 properties of the mean and the variance. If a is a constant, then :
    E{a}=a and Var{a}=0.

    Hence E{Var(X|Y)}+Var{E(X|Y)}=Var(X|Y)+0=Var(X|Y)

    So... is there some missing information or a mistake somewhere ? :s
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    Quote Originally Posted by Moo View Post
    Hello,

    Sorry I won't write it in Latex, I hope it will be enough.

    The mean of a variable is not a variable anymore. It is a constant.
    True, but the conditional mean given a random variable is a random variable (there probably are another few chapters before you get to that cool notion ).

    To prove the equality, one should make the definitions of the right-hand side explicit: since {\rm Var}(Z)=E[Z^2]-E[Z]^2,

    E[{\rm Var}(X|Y)]+{\rm Var}(E[X|Y])=E[E[X^2|Y]-E[X|Y]^2]+E[E[X|Y]^2]-E[E[X|Y]]^2.

    Now, you should notice that two terms simplify, and if you remember that E[E[Z|Y]]=E[Z] for any integrable Z, you'll notice that you're done.
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    Moo
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    Quote Originally Posted by Laurent View Post
    True, but the conditional mean given a random variable is a random variable (there probably are another few chapters before you get to that cool notion ).
    Sorry for the "bump" or "up"...

    We got into it - not in the class I expected though -, and it's just as you said, cool
    Along with some introductory Markov chains
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