# Math Help - PGF of a branching process (very hard)

1. ## PGF of a branching process (very hard)

Let $X_0, X_1, ...$ be a branching process with $\mathbb{P} (X_0 = 1) = 1$, and offspring distribution $\mathbb{P} (Y= k ) = qp^k \ \mbox{for k} = 0,1,2,...$ where $q = 1- p \ \mbox{and} \ p \in (0,1)$.

(1) Prove that the pgf $G_n \ \mbox{of} \ X_n$ is

$G_n (s) = \begin{cases} \frac{n-(n-1)s}{n+1-ns} & p = q \\ \frac{q(p^n-q^n-ps(p^{n-1} - q^{n-1}))}{p^{n+1} - q^{n+1} - ps(p^n - q^n)} & p \neq p \end{cases}$

and

(2) Determine (explicitly) the pdf $H_n$ of $X_n$ conditioned on eventual extinction.

For part two, I have seen that the probability of eventual extension, $\beta$, is given by

$\beta = I_{\{p \leq \frac{1}{2} \}} + \alpha I_{\{p > \frac{1}{2}\}} \ \mbox{where} \ \alpha = \frac{q}{p}$

2. Originally Posted by RoyalFlush
Let $X_0, X_1, ...$ be a branching process with $\mathbb{P} (X_0 = 1) = 1$, and offspring distribution $\mathbb{P} (Y= k ) = qp^k \ \mbox{for k} = 0,1,2,...$ where $q = 1- p \ \mbox{and} \ p \in (0,1)$.

(1) Prove that the pgf $G_n \ \mbox{of} \ X_n$ is

$G_n (s) = \begin{cases} \frac{n-(n-1)s}{n+1-ns} & p = q \\ \frac{q(p^n-q^n-ps(p^{n-1} - q^{n-1}))}{p^{n+1} - q^{n+1} - ps(p^n - q^n)} & p \neq p \end{cases}$

and

(2) Determine (explicitly) the pdf $H_n$ of $X_n$ conditioned on eventual extinction.

For part two, I have seen that the probability of eventual extension, $\beta$, is given by

$\beta = I_{\{p \leq \frac{1}{2} \}} + \alpha I_{\{p > \frac{1}{2}\}} \ \mbox{where} \ \alpha = \frac{q}{p}$
The pgf of $X_1$ (or $Y$, equivalently) is easily found to be $f(s)=G_1(s)=\frac{q}{1-ps}$.
Then you probably know that the pgf of $X_2$ is $f(f(s))$, etc., in general $G_{n+1}(s)=f(G_n(s))$. The formulas from the text can then be verified by induction; it's just tedious computation.

As for (2), your computation of the extinction probability is correct. When $p\leq 1/2$, the probability is 1, hence conditioning to extinction is like doing nothing: $H_n=G_n$ ( $H_n$ being the pgf of $X_n$ conditioned to extinction). When $p>1/2$, this probability is positive ( $\alpha=\frac{q}{p}$). In this case, you can deduce the pgf of the offspring distribution of the process conditioned to extinction: from your lecture (very likely), it is $\widetilde{f}(s)=\widetilde{G}_1(s)=\frac{1}{\alph a}G_1(\alpha s) = \frac{p}{1-qs}$. As you notice, this is the same as $G_1$ except that $p$ and $q$ are swapped. The pgf $H_n$ is therefore obtained by swapping $p$ and $q$ in $G_n$. Notice that this swapping resumes to the situation when $p\leq 1/2$, where the extinction is almost sure: that's consistent with the fact that the process conditioned to extinction reaches extinction eventually.

3. Thank-you so very much Laurent.

4. Originally Posted by Laurent
The pgf of $X_1$ (or $Y$, equivalently) is easily found to be $f(s)=G_1(s)=\frac{q}{1-ps}$.
Then you probably know that the pgf of $X_2$ is $f(f(s))$, etc., in general $G_{n+1}(s)=f(G_n(s))$. The formulas from the text can then be verified by induction; it's just tedious computation.
I dont understand how you did this. Whenever I try I get something completely different.