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Math Help - Random Variables

  1. #1
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    Random Variables

    Given (\Omega, \mathcal{A}, P), suppose X is a random variable with X\geq0 and E{X}=1. Definite Q:\mathcal{A}->R by Q(A)= E{(X1_A)}. Show that if P(A) = 0, then Q(A)=0. Give an example that shows that Q(A)=0 does not in general imply P(A)=0.

    I'm pretty sure that E{(1_A)}=P(A), so I was thinking that maybe we could break Q(A) up into E{(X)}E{(1_A)}. However, this would be assuming that X and 1_A are independent random variables? Would this be true? If so, then it's very obvious why, if P(A)=0, then Q(A)=0. Anyone have any hints? Thanks!
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  2. #2
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    Quote Originally Posted by azdang View Post
    Given (\Omega, \mathcal{A}, P), suppose X is a random variable with X\geq0 and E{X}=1. Definite Q:\mathcal{A}->R by Q(A)= E{(X1_A)}. Show that if P(A) = 0, then Q(A)=0. Give an example that shows that Q(A)=0 does not in general imply P(A)=0.

    I'm pretty sure that E{(1_A)}=P(A), so I was thinking that maybe we could break Q(A) up into E{(X)}E{(1_A)}. However, this would be assuming that X and 1_A are independent random variables? Would this be true? If so, then it's very obvious why, if P(A)=0, then Q(A)=0. Anyone have any hints? Thanks!
    It is false (without further assumption) that Q(A)=E{(X)}E{(1_A)}. However, if P(A)=0, then X 1_A=0 almost-everywhere (it is 0 outside A, and A is negligible), hence Q(A)=0.

    As for the example, you must choose X in such a way that X 1_A is zero everywhere, i.e. X is 0 on A, and it is chosen outside A in such a way that E[X](=E[X 1_{A^c}])=1).
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  3. #3
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    Ooh, the first part seems so obvious. Thank you, Laurent.

    As for the second part, I'm still a little lost. My first question is: Is it correct that X1_A = X if x is in A and 0 if x is not in A?
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  4. #4
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    Quote Originally Posted by azdang View Post
    Ooh, the first part seems so obvious. Thank you, Laurent.

    As for the second part, I'm still a little lost. My first question is: Is it correct that X1_A = X if x is in A and 0 if x is not in A?
    No. What is true is that X(\omega) 1_A(\omega)=X(\omega) if \omega\in A and X(\omega) 1_A(\omega)=0 else. Remember A is an event, not a subset of \mathbb{R}.

    For instance (you should write this more formally), suppose we toss a fair coin, and define the event A=\{\text{the outcome is heads}\} and X is defined to be 0 in case heads shows up, and to be 2 if tails shows up. Then what are P(A), E[X] and E[X 1_A]?
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  5. #5
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    Well, P(A) should be .5, correct? So, wouldn't EX1_A be 0 in either case (if we roll a heads or a tales)? So in this case, Q(A) = 0, but P(A) does not equal 0.
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  6. #6
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    Quote Originally Posted by azdang View Post
    Well, P(A) should be .5, correct? So, wouldn't EX1_A be 0 in either case (if we roll a heads or a tales)? So in this case, Q(A) = 0, but P(A) does not equal 0.
    Yes. (More correctly, X1_A is 0 in both cases, hence E(X1_A)=0. )
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