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SolidSnake
Oct12-09, 11:28 AM
1. The problem statement, all variables and given/known data
The probabilty generating funtion G is definied for random varibles whos range are \subset {0,1,2,3,......}. If Y is such a random variable we will call it a counting random varible. Its probabiltiy generating function is G(s) = E(s^{y}) for those s's such that E(|s|^{y})) < \infty.

2. Relevant equations

binomial distribution = \left(\stackrel{n}{y}\right)p^{y}q^{n-y} , y = 0,1,2,3,....n and 0 \leq p \leq 1

3. The attempt at a solution

What i have so far is...

G(s) = E(s^{y}) = \sum s^{y}\left(\stackrel{n}{y}\right)p^{y}q^{n-y}

G(s) = E(s^{y}) = \sum \left(\stackrel{n}{y}\right)(sp)^{y}q^{n-y}

not sure where to go from that. i managed to do it for the geometric random variable distribution b/c there was no "n choose y". Thanks to wiki, I know what the answer should be. The answer is G(s) = [(1-p) + ps]^{n}. I can't see how they went from what i have above to that.

SolidSnake
Oct12-09, 12:36 PM
well i just realized that G(s) = E(s^{y}) = \sum \left(\stackrel{n}{y}\right)(sp)^{y}q^{n-y}

is the same thing as (q + sp)^{n} .

Also by definition p + q = 1 \Rightarrow q = 1-p which means...

G(s) = E(s^{y}) = [(1-p) + ps]^{n}