# Galton-Watson process extinction probability

A similar version of this question has been posted before but it was never answered and I need help with it.

Let $$S_n$$ be a Galton Watson Process with offspring distribution $$p_k$$. We assume that $$p_0 > 0$$ and that $$\sum_{i=0}^{\infty} k p_k > 1$$. Also $$S_0 = 1$$.

We define $$T_0 := \inf \{n > 0 : S_n = 0\}$$. Let $$d := P(T_0 < \infty$$). Also

$$G(z) := \sum_{i=0}^{\infty} z^k P(S_1 = k)$$

is the generating function of $$S_1$$.

I have shown that $$P(T_0 < \infty | S_1 = k) = d^k$$ and $$r_k := P(S_1 = k \vert T_0 < \infty) = d^{k-1}$$ P($$S_1 = k)$$

Now I have to show that if $$Z_n$$ is a Galton Watson Process with offspring distribution $$r_k$$, then $$Z_n$$ is going to die out almost surely. For this I want to show that $$\frac{1}{d}\sum_{k=0}^\infty kd^kp_k \le 1$$

but I have no idea how to show this. I am not given any upper bound for $$d$$ or $$p_k$$ (other than $$1$$ obviously). Can anyone help?

1. What does the condition $$\sum kp_k>1$$ mean for the graph of $$G$$ ?

2. How is $$d$$ determined in terms of $$G$$ ?

3. Denote $$H$$ the generating function of $$Z_1$$. How can you rewrite $$H$$ in terms of $$G$$ and $$d$$ ? What can you say now about the new Galton-Watson process ?

• 1. It means G has another fixed point $<1$. And for 3: $H(t) = \frac{1}{d}G(td)$ Commented Nov 22, 2019 at 9:07
• Ok I think I worked it out now with your hint. Since $H(t) = \frac{1}{d}G(td)$ and I know that $d$ is the smallest non-negative fixed point of $G$ I can argue that for $t \lt 1$ we have $H(t) = \frac{1}{d}G(td) \ne \frac{td}{d} = t$, so only $t=1$ can be a fixed point of $H$. Commented Nov 22, 2019 at 9:24
• yes! glad you could work it out. You can further extract that $E[Z_1] < 1$, so $Z$ is actually subcritical. Commented Nov 27, 2019 at 20:39

Let's postpone the consideration of $$Z_n$$, and focus on $$S_n$$ for a moment.

We can further have $$d:=P(T_0 < \infty)=\sum_{k=0}^\infty P(T_0 < \infty | S_1 = k)P(S_1 = k) = \sum_{k=0}^\infty d^kP(S_1 = k),$$ i.e. $$d= \sum_{k=0}^\infty d^kp_k.\quad (1)$$

Note the process $$Z_n$$ is the same as the process $$S_n$$ except that the $$p_k$$ is replaced with $$r_k=d^{k-1}p_k$$. Let $$N_0 := \inf \{n > 0 : Z_n = 0\}, e:=P(N_0 < \infty).$$ Following the same arguement, in the context of $$Z_n$$, we have $$e= \sum_{k=0}^\infty e^kr_k =\sum_{k=0}^\infty e^kd^{k-1}p_k$$ $$\Leftrightarrow ed = \sum_{k=0}^\infty {(ed)}^{k}p_k. \quad (2)$$

For $$x \in (0, 1)$$, the equation $$\sum_{k=0}^\infty p_k x^k -x = 0 \Leftrightarrow \sum_{k=0}^\infty p_k (x^k -x) = 0$$ $$\Leftrightarrow (1-x)p_0 = \sum_{k=2}^\infty p_k x(1 - x^{k-1})$$ $$\Leftrightarrow p_0 = \sum_{k=2}^\infty p_k x\frac{1 - x^{k-1}}{1-x}.$$ Note that $$x\frac{1 - x^{k-1}}{1-x}$$ is a strictly increasing function of $$x$$ in (0, 1). Hence the equation has at most one solution in (0, 1).

We further note $$e > 0$$ since $$r_0 > 0$$. Hence $$d$$ and $$ed$$ are both solutions to the above equation in (0, 1). It follows that $$d = ed$$, and $$e = 1$$ i.e. the $$Z_n$$ dies a.s.

• I don't see why you can just leave out $k=0$ from that last sum. If we plug in $k=0$ we get $(e-1)p_0$ Commented Nov 22, 2019 at 8:49
• Thanks for pointing out the miscalculation. Hopefully it is now corrected. Commented Nov 22, 2019 at 9:47