0
$\begingroup$

Let $Z_n$ be the Galton Watson Branching Process. Let $Z_n=\sum_{k=1}^{Z_{n-1}}X_{k,n}$ where $X_{k,n}\sim X$ are iid progeny distribution. If $p_0=P(X=0)>0$ then show that $\forall s\in(0,1)$ we have $\phi_n'(s)\to0$ where $\phi_n(s)$ is the generating function of $Z_n$, evaluated at $s$.

I tried to use $\phi_n(s)=\phi_X(\phi_{n-1}(s))$ and then differentiate both sides to get $\phi_n'(s)=\phi_X'(\phi_{n-1}(s))\phi_{n-1}(s)$ and iterate but going nowhere.

Actually, I am trying to prove that $\phi_n(s)\to\eta$ for every $s\in[0,1)$ where $\eta$ is the extinction probability. I know that $\phi_n(0)\to\eta$ as $n\to\infty$, so I can write $\\phi_n(s)=\phi_n(0)+s\phi_n'(s_n)$ where $s_n\in(0,s)$. I will be done if I show $\phi_n'(s_n)\to0$. But note that $0<s_n<s$ so $\phi_n'(s_n)\leq \phi_n(s)$ so it suffices to show $\phi_n'(s)\to0$ for every $s\in(0,1)$.

$\endgroup$
5
  • $\begingroup$ Hint: Fix some $s$ in $(0,1)$, then $s\varphi'_n(s)=E(Z_ns^{Z_n})$. Almost surely, $Z_n\to0$ or $Z_n\to\infty$, hence $Z_ns^{Z_n}\to0$. For every nonnegative $u$, $us^u\leqslant-1/(e\ln s)$, thus, by dominated convergence $E(Z_ns^{Z_n})\to0$, qed. (The condition that $p_0\ne0$ is not needed, only that $p_1\ne1$.) $\endgroup$
    – Did
    Commented Mar 26, 2016 at 15:22
  • $\begingroup$ Oooh, how do you get these Did? How do you see? $\endgroup$ Commented Mar 26, 2016 at 16:21
  • $\begingroup$ By the way, I don't see why $Z_n\to0$ or $\infty$ a.s.. Why cannot it converge to something finite? $\endgroup$ Commented Mar 27, 2016 at 3:37
  • 1
    $\begingroup$ This is a general result about Markov chains with an absorbing state $\partial$. Roughly speaking, each time the chain visits some fixed finite set $F$ of states, it has some positive chance to hit $\partial$ (hence to stay there forever afterwards), thus, either it does hit $\partial$ or, after a while, it leaves $F$ forever. This holds for every finite $F$, that is, the chain either hits $\partial$ eventually or it goes to infinity. $\endgroup$
    – Did
    Commented Mar 27, 2016 at 7:41
  • $\begingroup$ @LandonCarter Was my answer helpful? $\endgroup$
    – Math1000
    Commented Jan 3, 2020 at 4:03

1 Answer 1

0
$\begingroup$

Expanding on @Did's comment, we have $$\varphi_n(s)=\sum_{j=0}^\infty \mathbb P(Z_n=j)s^j $$ and so $$ s\varphi'_n(s)=\sum_{j=0}^\infty j\ \mathbb P(Z_n=j)s^j = \mathbb E\left[Z_n s^{Z_n}\right].$$ Now, $\{Z_n:n=0,1,2,\ldots\}$ is a Markov chain on $\{0,1,2,\ldots\}$ with $\mathbb P(Z_0=1)=1$ and transition probabilities according to the branching process. Since $P_{00}=1$, $\{0\}$ is an absorbing state. Let $\tau = \inf\{n>0:Z_n=0\}$. Conditional on $\{\tau<\infty\}$, we have that $$\mathbb P\left( \liminf_{n\to\infty} \{Z_n=0\}\right)=1, $$ so $Z_n\stackrel{a.s.}\longrightarrow0$, hence $s^{Z_n}\stackrel{a.s.}\longrightarrow 1$, so that $Z_ns^{Z_n}\stackrel{a.s.}\longrightarrow0$. Conditional on $\{\tau=\infty\}$, we have $$\mathbb P\left(\bigcap_{m=1}^\infty \liminf_{n\to\infty} \{Z_n>m\}\right)=1, $$ so $Z_n\stackrel{a.s.}\longrightarrow\infty$, hence $s^{Z_n}\stackrel{a.s.}\longrightarrow0$ and $Z_ns^{Z_n}\stackrel{a.s.}\longrightarrow0$. Then from the inequality $$us^u \leqslant -(e\log s)^{-1} \tag1,\ u\geqslant 0 $$ we conclude from dominated convergence that $$ \lim_{n\to\infty}\varphi'_n(s) = \lim_{n\to\infty}s^{-1}\mathbb E\left[ Z_ns^{Z_n}\right]=s^{-1}\mathbb E\left[\lim_{n\to\infty} Z_ns^{Z_n}\right]=0.$$

P.S. If someone can help me justify $(1)$ I would be very grateful.

$\endgroup$

You must log in to answer this question.

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