poisson and negative binomial

**chrisc** · November 19th 2008, 08:01 AM

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

**mr fantastic** · November 19th 2008, 02:31 PM

Originally Posted by chrisc

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 $g_X^{(n)}(0) = n! \Pr(U = n)$ .

**mr fantastic** · November 19th 2008, 11:41 PM

Originally Posted by mr fantastic

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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