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Math Help - Joint Probability question

  1. #1
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    Joint Probability question

    Let X and Y each take on either the value 1 or -1. Let

    pr(1,1)=pr(X=1, Y=1)
    pr(1,-1)=pr(X=1,Y=-1)
    pr(-1,1)=pr(X=-1,Y=1)
    pr(-1,-1)=pr(X=-1,Y=-1)

    Suppose that E[X]=E[Y]=0. Show that

    (a)pr(1,1)=pr(-1,-1)
    (b)pr(1,-1)=pr(-1,1)

    Let p=2p(1,1), Find
    (c) Var[X]
    (d) Var[Y]
    (e) Cov[X,Y]

    Ok, I started by making this joint distribution table



    But I don't understand parts C, D, and E. What does p=2p(1,1) mean in relation to C, D, and E?

    If I ignore the part about p=2p(1,1) I can see that

    var[X]=var[Y]=0.5 and that cov[X,Y]=0 because X and Y are independent
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  2. #2
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    Re: Joint Probability question

    Quote Originally Posted by downthesun01 View Post
    Let X and Y each take on either the value 1 or -1. Let

    pr(1,1)=pr(X=1, Y=1)
    pr(1,-1)=pr(X=1,Y=-1)
    pr(-1,1)=pr(X=-1,Y=1)
    pr(-1,-1)=pr(X=-1,Y=-1)

    Suppose that E[X]=E[Y]=0. Show that

    (a)pr(1,1)=pr(-1,-1)
    (b)pr(1,-1)=pr(-1,1)

    Let p=2p(1,1), Find
    (c) Var[X]
    (d) Var[Y]
    (e) Cov[X,Y]

    Ok, I started by making this joint distribution table



    But I don't understand parts C, D, and E. What does p=2p(1,1) mean in relation to C, D, and E?

    If I ignore the part about p=2p(1,1) I can see that

    var[X]=var[Y]=0.5 and that cov[X,Y]=0 because X and Y are independent
    Your values 1/4, 1/4, 1/4 and 1/4 in your cells are only a particular solution. In fact, there a an infinite number of solutions: the general solution is p(-1,-1) = t, p(1, -1) = 0.5 - t, p(-1, 1) = 0.5 - t, p(1, 1) = t where t is a parameter and can have any value 0 \leq t \leq 0.5. Your particular solution corresponds to t = 0.25.

    So what "p=2p(1,1)" does is defines an alternative parameter to express the general solution. My t corresponds p/2.
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  3. #3
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    Re: Joint Probability question

    Ok. That makes more sense. Thanks
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