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Math Help - Iterated expectations

  1. #1
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    Iterated expectations

    Let I_A be the indicator function of the event A

    E[I_A]=E[E[I_A|A]]=E[E[1]]=1

    E[I_A]=E[E[I_A|A^C]]=E[E[0]]=0

    :S

    What's going on here?
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  2. #2
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    Quote Originally Posted by HD09 View Post
    Let I_A be the indicator function of the event A

    E[I_A]=E[E[I_A|A]]=E[E[1]]=1

    E[I_A]=E[E[I_A|A^C]]=E[E[0]]=0

    :S

    What's going on here?
    With a complete system of events {A_1, A_2, ..., A_n}, we have formula:
     EX =  \sum_{i=1}^n P(A_i) E(X | A_i)
    So I think
    E[I_A]=P(A) E[I_A|A]+ P(A^c) E[I_A|A^C]
    =P(A) E[1|A]+ P(A^c) E[0|A^C]=p(A)
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  3. #3
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    I agree but I'm not really after THE answer so much as the reason why taking iterated expectations is giving me the wrong answer
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  4. #4
    Moo
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    Hello,
    Quote Originally Posted by HD09 View Post
    Let I_A be the indicator function of the event A

    E[I_A]=E[E[I_A|A]]=E[E[1]]=1

    E[I_A]=E[E[I_A|A^C]]=E[E[0]]=0

    :S

    What's going on here?
    I guess the mistake is that in the iterated expectations, you have \mathbb{E}[X]=\mathbb{E}[\mathbb{E}[X|Y]], where Y is a random variable, and not an event. And X has to be a random variable too.
    Here, I_A is indeed a random variable. But A is defined as an event.

    A few facts :
    X|A is a random variable. E(X|A) is a number.
    X|Y is a random variable. E(X|Y) is a random variable. E(E(X|Y)) is a number.
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  5. #5
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    Thanks! That's made it much clearer
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  6. #6
    MHF Contributor matheagle's Avatar
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    e(i_a)=p(a) won't stay upper case for some reason ....


    E(I_A)=P(A)
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