1
$\begingroup$

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

Three intermediate questions could help you:

  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 ?

$\endgroup$
3
  • 1
    $\begingroup$ 1. It means G has another fixed point $<1$. And for 3: $H(t) = \frac{1}{d}G(td)$ $\endgroup$
    – yagod
    Commented Nov 22, 2019 at 9:07
  • 1
    $\begingroup$ 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 $H$. $\endgroup$
    – yagod
    Commented Nov 22, 2019 at 9:24
  • $\begingroup$ yes! glad you could work it out. You can further extract that $E[Z_1] < 1$, so $Z$ is actually subcritical. $\endgroup$
    – justt
    Commented Nov 27, 2019 at 20:39
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I don't see why you can just leave out $k=0$ from that last sum. If we plug in $k=0$ we get $(e-1)p_0$ $\endgroup$
    – yagod
    Commented Nov 22, 2019 at 8:49
  • $\begingroup$ Thanks for pointing out the miscalculation. Hopefully it is now corrected. $\endgroup$ Commented Nov 22, 2019 at 9:47

You must log in to answer this question.

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