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Thread: problem about discrete uniform distribution

  1. #1
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    problem about discrete uniform distribution

    Suppose that X and Y are independent discrete random variables with the same probability mass function

    pX(k) = pY(k) =pq^(k-1)

    where 0<p<1. Show that for any integer n greater than or equal to 2, the conditional probability mass function

    p(k) = P(X=k|X+Y=n)

    is a discrete uniform distribution

    So I do so in this way

    P(X=k|X+Y=n)
    =P(X=k|Y=n-k)
    =P(X=k, Y=n-k)/P(Y=n-k)
    = p(k,n-k)/P(Y=n-k)
    =p2qn-1+p2qn-1+..../pqn-k
    =np2qn-1/pqn-k
    =nqk-1
    which should be wrong in my opinion. So what should be the correct ans?
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  2. #2
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    Re: problem about discrete uniform distribution

    Sorry I dun know how to type in the summation sign summing up the probability from k=1 to n in step 3 ==,
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  3. #3
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    Re: problem about discrete uniform distribution

    $P[k] = P[X=k|X+Y=n] = \dfrac{P[X+Y=n|X=k]P[X=k]}{P[X+Y=n]}$

    $P[X+Y=n|X=k] = P[Y=n-k] = p q^{n-k}$

    $P[X=k] = p q^k$

    $P[X+Y=n] = \sum \limits_{j=1}^n~(p q^j)(p q^{n-j}) = p^2 n q^n$

    $P[k] = \dfrac{(p q^{n-k})(p q^k)}{p^2 n q^n} = \dfrac{1}{n}$
    Last edited by romsek; Nov 29th 2017 at 10:54 AM.
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