Generating function of random variable $X$

Question

Suppose random variable $X$ with values ${2,3,...}$ has distribution $P(X=k) = (k-1)p^2(1-p)^{k-2}$. Show that $$G_x(s) = \left(\frac{ps}{1-s(1-p)}\right)^2$$ Is this easy to show by definition or how do I solve it?

Answer 1

Hint Remember the definition of $G_X$ :

$$ G_X(s) = \sum_{n=0}^\infty \mathbb P (X=k)s^k $$

which is defined on $[-1,1]$. So with that definition, and by making appear a geometric sum (or something close to it) you might be able to find the answer.

Answer 2

Fot $0<t<1$, $\sum_{k=0}^{\infty} t^{k}= \frac 1 {1-t}$ and $\sum_{k=0}^{\infty} kt^{k-1}= -\frac 1 {(1-t)^{2}}$. This gives $\sum_{k=0}^{\infty} kt^{k}= -\frac t {(1-t)^{2}}$. Now write down $\sum t^{k}P\{X=k\}$ and use these formulas to compute $G_X(t)$.