# General upper bound of extinction probability

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.

• I think you should change your username to a more neutral one. – YCor May 1 '20 at 23:19
• Cross-posted at math.stackexchange.com/questions/3649820/… . You should disclose your cross-posts to prevent duplication of effort. – S. Carnahan May 3 '20 at 16:41
• @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 :) – Math is like Friday May 3 '20 at 19:33

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 $$q=\frac27$$.
• How does $G''(s) \leq G''(1)$ inequality trivially hold? Is it a property of convexity? – Math is like Friday May 3 '20 at 19:37
• @AsianbutnoChinese : $G''$ is nondecreasing, because $G'''\ge0$ (and actually $G^{(k)}\ge0$ for all $k=0,1,\dots$) on the interval $[0,1)$. – Iosif Pinelis May 3 '20 at 23:35