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We consider here a Galton–Watson process with an offspring distribution $X$, where $\mathbb{E}X = \mu$ and $\operatorname{Var} X = \sigma^{2} < \infty$ and $q = \mathbb{P}(\text{extinction})$, i.e., the extinction probability.

I want to prove that the general upper bound of the extinction probability is given by: $$q \leq 1 - \frac{\mu - 1}{\sigma^{2} + \mu^{2} - \mu}.$$

So if this helps, I can rewrite the above form as follows: $$q \leq 1 - \frac{G_X'(1) - G_X(1)}{G_X''(1)},$$ where $G_X(s)$ is the probability generating function of the random variable $X$.

But here I am stuck.

I know the question is not well-asked, but hope that anyone can give me some hints.

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    $\begingroup$ I think you should change your username to a more neutral one. $\endgroup$ Commented May 1, 2020 at 23:19
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    $\begingroup$ Cross-posted at math.stackexchange.com/questions/3649820/… . You should disclose your cross-posts to prevent duplication of effort. $\endgroup$ Commented May 3, 2020 at 16:41
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    $\begingroup$ @YCor Sure I'll change as soon as I can. Since I already changed my username around a month ago, I am currently not able to change, but 4 days later :) $\endgroup$ Commented May 3, 2020 at 19:33

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We have $$G(s):=G_X(s)=Es^X,$$ with the convention $0^0:=1$, where $X$ is a random variable with values in $\{0,1,\dots\}$ and $EX^2<\infty$. So, $G$ is a nonnegative nondecreasing convex function from $[0,1]$ to $[0,1]$ with nondecreasing $G''$. Also, $G(0)=P(X=0)$ and $G'(0)=P(X=1)$. So, excepting the case when $P(X=0)=0$ and $P(X=1)=1$ (and hence $G''=0$, which makes your inequality devoid of meaning), the extinction probability is the smallest root of the equation $$G(q)=q. \tag{1}$$ Also, $G(1)=1$. So, by the convexity of $G$ and (1), we have $G(s)\le s$ for $s\in[q,1]$ and $G(s)\ge s$ for $s\in[0,q]$. So,
$$G'(q)\le1.$$ So, $$G'(1)-G(1)=G'(1)-1\le G'(1)-G'(q)=\int_q^1 G''(s)\,ds \le \int_q^1 G''(1)\,ds=(1-q)G''(1),$$ which implies that indeed $$q\le1-\frac{G'(1)-G(1)}{G''(1)},$$ as desired.

To illustrate this, here are the graphs $\{(s,s)\colon0\le s\le1\}$ (blue) and $\{(s,G(s))\colon0\le s\le1\}$ (gold) for the case when $X$ takes values $0,1,2$ with probabilities $\frac2{10},\frac1{10},\frac7{10}$, respectively, so that here $q=\frac27$.

enter image description here

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  • $\begingroup$ How does $G''(s) \leq G''(1)$ inequality trivially hold? Is it a property of convexity? $\endgroup$ Commented May 3, 2020 at 19:37
  • $\begingroup$ @AsianbutnoChinese : $G''$ is nondecreasing, because $G'''\ge0$ (and actually $G^{(k)}\ge0$ for all $k=0,1,\dots$) on the interval $[0,1)$. $\endgroup$ Commented May 3, 2020 at 23:35
  • $\begingroup$ Ahh right it was not clever to ask. Thank you! $\endgroup$ Commented May 4, 2020 at 19:48

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