Each year a tree of a particular type flowers once and produces a random number $N$ of flowers, where $\mathbb{P}(N=n)=(1-p)p^n$, $n=0,1,2,\dots $ and $0<p<1$. Each flower has probability $1/2$ of producing a ripe fruit, independently of all other flowers. Find the probability generating function for the number $R$ of ripe fruits produced by the tree. Hence, find the probability that in a given year

  1. the tree produces $r$ ripe fruits
  2. the tree has $n$ flowers if it produces $r$ ripe fruits

For clarity, reform $\mathbb{P}(N=n)=(1-p)p^n$ into $\mathbb{P}(X=k)=q(1-q)^{k-1}$, where $q=1-p$ and $n=k-1$. Therefore, $N$~$Geo(q)$ and $X_i$~$Ber(1/2)$. The p.g.f. of these RVs are

$G_N(s)=\frac{qs}{1-(1-q)s}$ and $G_{X_1}(s)=\frac{1}{2}(1+s)$

Since the p.g.f. of $\sum_{i=1}^N X_i$ is $G_N(G_X(s))=G_N(\frac{1}{2}(1+s))=\frac{q\frac{1}{2}(1+s)}{1-(1-q)\frac{1}{2}(1+s)}=$$\frac{1}{2}\frac{q+qs}{1-\frac{1}{2}(1+s-q-qs)}=\frac{q+qs}{1-s+q+qs}$

I am unsure of how to interpret this, or where I went wrong.

More generally, how do we interpret a general pgf of a geometric RV?

  • $\begingroup$ What you have seems right to me. What do you think is wrong? $\endgroup$
    – Math1000
    Commented May 29, 2015 at 21:59
  • $\begingroup$ How do interpret the result that I have? In other words, what is $\mathbb{P}(R=r)$ and $\mathbb{P}(N=n| R=r)$? $\endgroup$ Commented May 29, 2015 at 22:40

1 Answer 1


Let $G_R(s)$ be the generating function of $R$. Then

$$ \begin{align*} G_R(s) &= \frac{q+qs}{1-s+q+qs}\\ &= \frac{q+qs}{1+q-(1-q)s}\\ &= \frac q{1+q}(1+s)\left(\frac 1{1+\left(\frac{1-q}{1+q}\right)s}\right)\\ &= \frac q{1+q}(1+s)\sum_{n=0}^\infty \left(\frac{1-q}{1+q}\right)^n s^n\\ &= \frac q{1+q}\left(1 + \sum_{n=1}^\infty\left(\frac{1-q}{1+q}\right)^n s^n + \sum_{n=0}^\infty\left(\frac{1-q}{1+q}\right)^n s^{n+1}\right)\\ &= \frac q{1-q}\left(1 + \sum_{n=1}^\infty \frac2{1-q}\left(\frac{1-q}{1+q}\right)^n s^n \right)\\ &= \frac q{1+q} + \sum_{n=1}^\infty \left(\frac 2{1-q^2}\right)\left(\frac{1-q}{1+q}\right)^n s^n \end{align*} $$ Hence $$\mathbb P(R=r) = \begin{cases} \frac q{1+q},& r=0\\ \left(\frac2{1-q^2}\right)\left(\frac{1-q}{1+q}\right)^r,& r>0\end{cases}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .