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For a geometric distribution with $p_{x}(x)=p(1-p)^x, x=0,1,2,3,...$ I have been asked to find the probability generating function.

I know that the way to find this is by finding $E(s^x)$ (the expectation) but I've plugged in the probability mass function and summed it and I'm just not getting a proper answer (I roughly know what the end result should look like).

Can someone please help me with the steps of finding this?

Thanks

Gilles Bonnet
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George
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  • Please show "what the end result should look like" and what you found. – Did Nov 02 '13 at 13:26
  • I think it should be something like $\frac{p}{1-(1-p)s}$ but I have only got as far as $\sum_{x=0}^\infty s^x (1-p)^x p$ – George Nov 02 '13 at 13:31

1 Answers1

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Hint: For every $z$ in $\mathbb C$, $|z|\lt1$, $\sum\limits_{x=0}^\infty z^x=\frac1{1-z}$.

Did
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