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Math Help - Bivariate RV Probability

  1. #1
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    Bivariate RV Probability

    Let X=R(0,1) and Y=Bi(1,0.5) be independent random variables, Z=X^Y.

    1. Compute the function n(y)=E(Z|Y=y) and hence specify the rv E(Z|Y).

    2. Compute the function S(y)=var(Z|Y=y) and hence specify the rc var(Z|Y).

    3. Compute E(Z).

    How do I do the questions?
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  2. #2
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    Quote Originally Posted by maibs89 View Post
    Let X=R(0,1) and Y=Bi(1,0.5) be independent random variables, Z=X^Y.

    1. Compute the function n(y)=E(Z|Y=y) and hence specify the rv E(Z|Y).

    2. Compute the function S(y)=var(Z|Y=y) and hence specify the rc var(Z|Y).

    3. Compute E(Z).

    How do I do the questions?
    1. Is R(0, 1) a uniform pdf?
    2. Do I understand you correctly ....... n = 1 in the pmf for Y, that is, there's only 1 trial?
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  3. #3
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    Yes, that is precisely right.

    I have no idea where to get the answer.

    Help!
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  4. #4
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    Quote Originally Posted by maibs89 View Post
    Let X=R(0,1) and Y=Bi(1,0.5) be independent random variables, Z=X^Y.

    1. Compute the function n(y)=E(Z|Y=y) and hence specify the rv E(Z|Y).

    2. Compute the function S(y)=var(Z|Y=y) and hence specify the rc var(Z|Y).

    3. Compute E(Z).

    How do I do the questions?
    Note that Y = 0 or 1 only. So perhaps it's this simple:

    1. When Y = 0, Z = 1 and E(Z) = 1. So n(0) = 1.
    When Y = 1, Z = X and E(Z) = E(X) = 1/2.

    2. When Y = 0, Z = 1 and Var(Z) = 0. So S(0) = 0.
    When Y = 1, Z = X and Var(Z) = Var(X) = 1/12. So S(1) = 1/12.

    3. I think it will be

    (1) \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) = .........

    Someone might jump in to confirm or deny this ......

    By the way, there's an interesting question if X = R(0,1) and Y = R(0,1) ......
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  5. #5
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    What are the rv's of E(Z|Y) and var(Z|Y)?

    And could you please explain part 3?
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  6. #6
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    Quote Originally Posted by maibs89 View Post
    What are the rv's of E(Z|Y) and var(Z|Y)?

    And could you please explain part 3?
    I have already done parts 1. and 2.

    Part 3: Clearly either Z = Z^0 = 1 or Z = X^1 = X. Each obviously occurs with probability 1/2. Obviously E(1) = 1. When Z = X, E(Z) = E(X) = 1/2, as stated in my answer to part 1. Remember that X ~ R(0, 1).

    It should be clear from all this where my answer has come from.

    You started off with "I have no idea where to get the answer". I've given you a lot of food for thought. It's time now for you to step up to the plate.
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