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Calculating Joint Distribution

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Let X and Y be two random variables, with joint distribution given by the following table:
    Code (Text):
                    a
    b       1       2       3       4
    --------------------------------------
    1       16/136  3/136   2/136   13/136
    2       5/136   10/136  11/136  8/136
    3       9/136   6/136   7/136   12/136
    4       4/136   15/136  14/136  1/136
     

    What is:

    a.) P(X = Y)
    b.) P(X + Y = 5)
    c.) P(1 < X <= 3, 1 < Y <= 3)
    d.) p((X,Y) [tex]\in[/tex] {1,4} x {1,4})


    2. Relevant equations



    3. The attempt at a solution

    To be honest, I'm not exactly sure what exactly these kinds of operators mean (e.g. what does it mean to say P(X = Y). But here's what I have so far:

    P(a = 1) = P(a = 2) = P(a = 3) = P(a = 4) = P(b = 1) = P(b = 2) = P(b = 3) = P(b = 4) = 34/136

    a.) P(X = Y) is [P(a = 1, b = 1) + ... + P(a = 4, b =4)] = 16/136 + 10/136 + 7/136 + 1/136

    b.) P(X + Y = 5) = [P(a = 1, b = 4) + P(a = 4, b = 1)] = 13/136 + 4/136

    c.) P(1 < X <= 3, 1 < Y <= 3) = c.) P(1 < a <= 3, 1 < b <= 3) =
    10/136 + 11/136 + 6/136 + 7/136

    d.) p((X,Y) [tex]\in[/tex] {1,4} x {1,4}) = ? I have no idea what this syntax means
     
  2. jcsd
  3. Oct 18, 2009 #2

    Mark44

    Staff: Mentor

    I don't know what you mean by the equations above. The probabilities in the table are joint probabilities.
    Looks good (above).
    What other ways can X + Y add to 5? You're showing just two of them. (above).
    Looks good (above).
    It's asking for P((X, Y)) where X can be 1, 2, 3, or 4 and Y can be 1, 2, 3, 4.
     
  4. Oct 18, 2009 #3

    whs

    User Avatar

    Part 'd' means the probability that the pair X,Y will be in the Cartesian product of {1,4} x {1,4}. Which means, P(X=1, Y=1) or P(X=2, Y=1) or ... and so on, for all pairs in that product.

    Seeing your table, it looks like that would cover all the possible values in that table, so it would be 1.
     
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