I need help with this problem. Does anyone know how to approach it?
The distribution of the nonnegative, integer-valued random variable X has the following properties: For every n>=1:
$P(X=2n)=\frac{1}{2}P(X=2n-1)=\frac{2}{3}P(X=2n+1)$
Moreover, $P(X=0)=\frac{2}{3}P(X=1)$. Compute the generating function of X.
Thanks
I think the idea is to try to express everything in terms of $P(X=0)$. Let $p=P(X=0)$. Then iterating you get $$p=\frac{2}{3}P(X=1)=\frac{4}{3}P(X=2)=\frac{8}{9}P(X=3)=\frac{16}{9}P(X=4)=\cdots$$
Rearranging each term successively and using the definition of the generating function you end up with \begin{align} G(x)&=px^0+\frac{3}{2}px^1+\frac{3}{4}px^2+\frac{9}{8}px^3+\frac{9}{16}px^4+\cdots\\ &=p\left[\sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^{n}\left(x^{2n}+\frac{1}{2}x^{2n+1}\right)\right] \end{align}