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I am stuck on exercise 11.2 From Grimmett's probability on graphs. Here is a link to the pdf on his website.

Consider a branching process whose family-sizes have the binomial distribution bin$(n, \frac{\lambda}{n})$. Show that the extinction probability converges to $\eta(\lambda)$ as $n \rightarrow \infty$, where $\eta(\lambda)$ is the extinction probability of a branching process with family-sizes distributed as $\text{Po}(\lambda)$.

To solve this exercise, we use a theorem from his other book, which states that if you have a branching process whose family sizes are $X$ distributed then the extinction probability of this branching process is the smallest non-negative fixed point of the equation $s = G(s)$ where $G$ is the probability generating function $G(s) = \sum_{k = 0}^\infty \mathbb{P}(X = k)s^k$.

My next step was computing these probability generating functions for $\text{bin}(n,\lambda)$, these are $G_n(s) = (1+\frac{\lambda( s -1)}{n})^n$. When the family sizes are Poisson distributed with parameter $\lambda$ we get $G(s) = e^{\lambda(s-1)}$. and so clearly $G_n$ converges pointwise to $G$.

This however is not enough, as we want to show that the extinction probabilities also converge, so we have $s_n$ extinction probabilities of the $n$-th process, i.e. $s_n$ is the smallest non-negative solution to $s = G_n(s)$, and we want to show that this converges the smallest non-negative solution of $s = G(s)$. I hoped I could use some theorem which states that if you have pointwise convergence then the fixed points also converge, but this is false, see this counterexample.

During writing this I thought of an attempt to solve this using Hurwitz theorem, namely, show that the smallest non-negative fixed point of $G(s)$ is smaller than 1, then show that the function $G'$ is complex differentiable with non-zero derrivative, use inverse function theorem, get an open neighbourhood of our fixed point, find sequence of fixed points which converge to the smallest non-negative fixed point $\eta(\lambda)$ of $G$. Here I am stuck trying to show that $s_n$ are the smallest non-negative fixed points of $G_n$.

How to proceed?

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Disclaimer: This is, so far, one of my most downvoted answers on the site. Needless to say, it is perfectly correct, and it answers the question. The downvotes might be due to extra-mathematical reasons. Happy reading!

For each $n$ large enough, let $q_n(\lambda)$ denote the extinction probability of the binomial branching process with parameters $(n,\lambda/n)$. Let $q(\lambda)$ denote the extinction probability of the Poisson branching process with parameter $\lambda$ (this is your $\eta(\lambda)$).

For every fixed $s$ in $[0,1)$, the sequence $(G_n(s))$ increases to $G(s)$, each $q_n(\lambda)$ is the smallest solution in $[0,1]$ of the equation $s=G_n(s)$, and $q(\lambda)$ is the smallest solution in $[0,1]$ of the equation $s=G(s)$, hence the sequence $(q_n(\lambda))_n$ is increasing and $q_n(\lambda)<q(\lambda)$ for every $n$ (to prove this, use that every $G_n$ is increasing on $[0,1]$).

Let $r(\lambda)$ denote the limit of the sequence $(q_n(\lambda))_n$. Then $q_n(\lambda)<r(\lambda)\leqslant q(\lambda)$ for every $n$, in particular, $G_n(r(\lambda))<r(\lambda)$, which implies that $G(r(\lambda))=\lim G_n(r(\lambda))\leqslant r(\lambda)$. Now, $G(r(\lambda))\leqslant r(\lambda)$ is equivalent to $r(\lambda)\geqslant q(\lambda)$, hence all this proves that $\lim q_n(\lambda)=r(\lambda)=q(\lambda)$, as desired.

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  • $\begingroup$ Note that $G_n$ becomes eventually increasing, for example when $n = 2, \lambda > 2$ we get that $G_n$ is actually strictly decreasing on $[0,1]$ as you then get $G_2(s) = (1 + 1.5( s - 1))^2$ has a decreasing part on $[0,1]$. Further, to conclude that q_n is increasing you need something more than that G_n is increasing, namely that $G_n'$ is also increasing on $[0,b]$ for some $b<1$, which again, holds eventually. I dont yet see how to pick b uniformly over all n such that we can show this result for all q_n at once. $\endgroup$ – S. Franssen Jan 15 '17 at 14:16
  • $\begingroup$ You have some serious misconceptions about generating functions, it seems... Every generating function $g(s)=E(s^X)$ is increasing (well, except if $P(X=0)=1$, then it is only nondecreasing) and strictly convex (well, except if $P(X=0)+P(X=1)=1$, then it is only convex) on $[0,1]$. The function $G(s)=(1+1.5(s-1))^2$ is not a generating function. Every generating function $g$ is such that $g'$ is increasing on $[0,1)$. Hmmm... $\endgroup$ – Did Jan 15 '17 at 14:28
  • $\begingroup$ By the way, the binomial distribution $(n,\lambda)$ simply does not exist if $\lambda>1$ and $G(s)=(1+\lambda(s-1)/n)^n$ with $n$ some positive integer, defines a generating function only if $0\leqslant\lambda\leqslant n$. $\endgroup$ – Did Jan 15 '17 at 14:30
  • $\begingroup$ In the exercise we use $(n, \frac{\lambda}{n})$ and in the book Grimmett uses it for $\lambda \in (0, \infty)$, see page 217 of the pdf, 207 of the book. This is my first time using generating functions so I have no clue what properties they have. For now I am stuck showing that $q_n$ gives an (eventually) increasing sequence. $\endgroup$ – S. Franssen Jan 15 '17 at 14:45
  • $\begingroup$ To sum up, no binomial $(n,\lambda)$ is involved and binomial $(n,\lambda/n)$ is allright since for every fixed $\lambda$, indeed $\lambda/n\leqslant1$ for every $n$ large enough. To show that $q_n\leqslant q_{n+1}$, note that $q_n=G_n(q_n)<G_{n+1}(q_n)$ and identify the set of points $s$ such that $s<G_{n+1}(s)$. // "This is my first time using generating functions so I have no clue what properties they have." If really you mean it then we have a clear case of putting the cart before the horse. One simply cannot study branching processes without some serious knowledge of generating functions. $\endgroup$ – Did Jan 15 '17 at 14:52

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