Branching process, using martingale, in Durrett

martingalesprobabilityprobability theory

I'm trying to solve a problem in Durrett, 5th edition.

The problem is:

Let $Z_n$ be a branching process with offspring distribution $p_k$, and let $\phi(\theta) = \sum p_k \theta^k$. Suppose $\rho < 1$ has $\phi(\rho) = \rho$. Show that $\rho^{Z_n}$ is martingale and use this to conclude $P(Z_n = 0 \mbox{ for some } n \ge 1 \ | \ Z_0 = x) = \rho^x$.

It was easy to show that $\rho^{Z_n}$ is martingale, but I don't know how to use it to prove the desired equality.

It seems to be use $\mathbb{E}(\rho^{Z_n}) = \mathbb{E}(\rho^{Z_0}) = \rho^x$, but I don't know whether this quantity equals to the desired probability or not.

Could anyone help?

Best Answer

I think that the question is same as this.

I answered there as:

Here, you can prove that $$\{\lim Z_n / \mu^n > 0 \} = \{Z_n > 0 \mbox{ for all } n\} a.s.$$

With this note, let $N = \inf \{ n : Z_n = 0\}$ be a stopping time. Since $\rho^{Z_n}$ is martingale, $\rho^{Z_{n \wedge N}}$ is martingale, and thus $$\rho^x = \mathbb{E}[\rho^{Z_{0 \wedge N}}] = \mathbb{E}[\rho^{Z_{n \wedge N}}].$$ Since $\rho < 1$, we can apply the dominated convergence theorem, to get $$\rho^x = \rho^0 \mathbb{P}(N < \infty) + \rho^{\infty}\mathbb{P}(N = \infty) = \mathbb{P}(N < \infty)$$ where the first equality from the first notification that I gave above.

EDIT:

Above almost sure equality may hold under the condition that $\mathbb{P}(\lim Z_{n}/\mu^n = 0) < 1$ and I'm not sure whether it holds or not.

First, we may assume that $p_0 > 0$ since otherwise it becomes trivial.

Now, the following lemma is helpful:

Lemma let $X_n$'s be random variables taking values in $[0, \infty)$, and $D = \{ X_n = 0 \mbox{ for some } n \ge 1\}$. Assume that $\mathbb{P}(D | X_1, \cdots, X_n) \ge \delta(x) > 0$ a.s. on $\{X_n \le x\}$, $$\mathbb{P}(D \cup \{\lim X_n = \infty\}) = 1$$

Proof On $\{\liminf X_n \le M \}$, $X_n \le < M + 1$ infinitely often so $$\mathbb{P}(D | X_1, \cdots, X_n) \ge \delta(M+1) > 0$$ i.o.

By Levy's 0-1 law, LHS converges to $1_D$ so we have $\{\liminf X_n \le M \} \subseteq D$. Taking $M \to \infty$, $\{\liminf X_n < \infty \} \subseteq D$ a.s. and the result follows.

Now, from $p_0 > 0$, $\mathbb{P}(Z_{n+1} = 0 | Z_1, \cdots, Z_n) \ge p_0 ^k$ on $\{Z_n \le k\}$. Thus by the lemma, $$\mathbb{P}(\{Z_n = 0 \mbox{ for some } n\} \cup \{\lim Z_n = \infty\}) = 1$$ which allows us to do the calculation in the original answer.

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