3
$\begingroup$

I have a probability generating function

$$G_X(s) = \frac{p+ps}{1-s+p+ps}$$

and I need to find $P(X=r)$.

How do I get this from the probability generating function? I was thinking about finding the Mac Laurin expansion of $G_X(s)$ and finding the general formula for the coefficient of $s^r$, but that's proving very difficult.

Thanks

$\endgroup$
1
  • $\begingroup$ The derivatives get more and more complicated each time, so finding a general rth derivative is looking quite hard :( is there any other way? $\endgroup$
    – Taimur
    Dec 20 '13 at 15:22
2
$\begingroup$

The Maclaurin expansion idea is good. Note that if we find the expansion of $\frac{1}{1-s+p+ps}$, multiplying this by $p(1+s)$ will not be difficult.

Rewrite the bottom as $1+p-s(1-p)$, and then as $(1+p)\left(1-s\frac{1-p}{1+p}\right)$. To make things look nicer, let $a=\frac{1-p}{1+p}$. Then we have $$\frac{1}{1-s+p+ps}=\frac{1}{1+p}\cdot\frac{1}{1-as}.$$ But $$\frac{1}{1-as}=1+as+a^2s^2+a^3s^3+a^4s^4+\cdots$$ (geometric series).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.