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Math Help - Finding marginal pmf

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    Finding marginal pmf

    Here's the pmf:
    p_{X,Y}(i, j) = \frac{2}{n(n+1)} \mbox{ for } 1\leq j\leq i\leq n, n > 0

    I got the first part, which is verifying that this is a legitimate joint pmf, however I am stuck on finding the marginal pmf. I know that marginal pmf of X would be:

    p_{X}(i) = \sum_j p_{X,Y}(i, j)

    But wouldn't that just be (n)(\frac{2}{n(n+1)} in this case? But somehow I don't think it's that easy.
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    Hello,

    Actually, we have 1\leq j\leq i

    So we have :

    p_X(i)=\sum_{j=1}^i \frac{2}{n(n+1)}

    But \frac{2}{n(n+1)} is "counted" i times.

    So it's actually \frac{2i}{n(n+1)}
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    For the marginal pmf of X, that answer is correct. However, my TA told me the marginal pmf of Y isn't the same thing with j replacing i. How come?
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    Quote Originally Posted by jfz23 View Post
    For the marginal pmf of X, that answer is correct. However, my TA told me the marginal pmf of Y isn't the same thing with j replacing i. How come?
    Because i takes values from j to n.
    This means there are n-j+1 possible values for i.
    So you'll "count" \frac{2}{n(n+1)} n-j+1 times, which makes the pdf of Y equal to \frac{2(n-j+1)}{n(n+1)}

    Does it look clear to you ?
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    I see...thank you very much!
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    Quote Originally Posted by jfz23 View Post
    Here's the pmf:
    p_{X,Y}(i, j) = \frac{2}{n(n+1)} \mbox{ for } 1\leq j\leq i\leq n, n > 0

    I got the first part, which is verifying that this is a legitimate joint pmf, however I am stuck on finding the marginal pmf. I know that marginal pmf of X would be:

    p_{X}(i) = \sum_j p_{X,Y}(i, j)

    But wouldn't that just be (n)(\frac{2}{n(n+1)} in this case? But somehow I don't think it's that easy.
    A striking similarity to the question in this thread: http://www.mathhelpforum.com/math-he...tml#post373172
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