# Probability generating function (binomial distribution)

1. Oct 12, 2009

### SolidSnake

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.

2. Oct 12, 2009

### SolidSnake

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...

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

Last edited: Oct 12, 2009