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Math Help - poisson and negative binomial

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    poisson and negative binomial

    If X has Poisson distribution with the mean of 2.3, Y has Negative Binomial
    distribution with p = 0.62 and k = 4, and X and Y are independent,
    find Pr(X + Y > 5).
    Hint: Use probability generating function.

    I would usually fill in on what I have tried so far, but even with the hint here I can't seem to get started. Any help is appreciated
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    Quote Originally Posted by chrisc View Post
    If X has Poisson distribution with the mean of 2.3, Y has Negative Binomial
    distribution with p = 0.62 and k = 4, and X and Y are independent,
    find Pr(X + Y > 5).
    Hint: Use probability generating function.

    I would usually fill in on what I have tried so far, but even with the hint here I can't seem to get started. Any help is appreciated
    Let U = X + Y. Note that U = 4, 5, 6, \, ....

    \Pr(U > 5) = 1 - \Pr(U \leq 5) = 1 - \Pr(U = 4) - \Pr(U = 5).

    Let g_U(s) be the pgf of U.

    Then g_U(s) = g_X(s) \cdot g_Y(s) (since X and Y are independent) where g_X(s) is the pgf of X and g_Y(s) is the pgf of Y.

    And you should know that g_X^{(n)}(0) = n! \Pr(U = n).
    Last edited by mr fantastic; November 19th 2008 at 11:37 PM. Reason: Corrected a mistake in the inequality
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    Quote Originally Posted by mr fantastic View Post
    Let U = X + Y. Note that U = 4, 5, 6, \, ....

    \Pr(U > 5) = 1 - \Pr(U \leq 5) = 1 - \Pr(U = 4) - \Pr(U = 5).

    Let g_U(s) be the pgf of U.

    Then g_U(s) = g_X(s) \cdot g_Y(s) (since X and Y are independent) where g_X(s) is the pgf of X and g_Y(s) is the pgf of Y.

    And you should know that g_X^{(n)}(0) = n! \Pr(U = n).
    Perhaps an easier way (but not using the pgf) would be to note that

    \Pr(U = 4) = \Pr(X = 0) \cdot \Pr(Y = 4)

    \Pr(U = 5) = \Pr(X = 0) \cdot \Pr(Y = 5) + \Pr(X = 1) \cdot \Pr(Y = 4)

    since X and Y are independent.
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