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Let $1 = Z_0,Z_1,Z_2,\ldots$ be a Galton-Watson branching process with offspring distribution $p_0,p_1,p_2,\ldots$. That is, $p_k$ is the probability that an individual will have $k$ offspring. Suppose that $p_0 = 2/3$ and $p_2 = 1/3$. Let $V = Z_0 + Z_1 + Z_2 +\cdots$.

Why is it that $P(V < \infty) = 1$, and what is the probability generating function of $V$?

I note that we can try to use a transition matrix, if it helps?

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see karlin & taylor v.1 sec 8.2 or Ross' probability models –  mike Apr 13 '12 at 12:01
    
mary: What part(s) of the answer below is (are) not clear to you? –  Did Apr 20 '12 at 16:01

1 Answer 1

up vote 2 down vote accepted

If $Z_1=0$, $V=1$. If $Z_1=2$, $V=1+V'+V''$ where $V'$ and $V''$ are independent copies of $V$. In the first case, which has probability $\frac23$, $V$ is finite. In the second case, which has probability $\frac13$, $V$ is finite if and only if $V'$ and $V''$ are both finite.

Introducing the probability of extinction $q=\mathrm P(V\ \text{finite})$ and the probability generating function $u:s\mapsto\mathrm E(s^V)$, one gets $$ q=\tfrac23+\tfrac13q^2,\qquad u(s)=\tfrac23s+\tfrac13su(s)^2. $$ The only solution $q$ in $[0,1]$ of the first equation is $q=1$ hence $V$ is finite almost surely.

The only solution $u$ defined on $[0,1]$ of the second equation is $u(s)=\frac1{2s}(3-\sqrt{9-4s^2})$.

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