How to prove that $r_n=r_{n-1} - \frac12 b r_{n-1}^2 $ is asymptotic to $\frac{1}{n}$?

Question

So my question is:

How do we prove that $r_n=r_{n-1} -\frac12 b r_{n-1}^2 $ is asymptotic to $\frac{1}{n}$ when we have an offspring distribution $p_i:=P(\xi=i)$ and b is the variance $$b=\sum_{i\geq 2}i(i-1) p_i< \infty$$ and $r_n:=1-q^n(0)$ for $$q^n(x):=q(q^{n-1}(x)), \quad q(x):=p_0+p_1x+p_2x^2+\ldots,$$ what gives us $$r_1=1-p_0.$$ Also we have $$\sum_{i\geq 1}i p_i=1.$$

Best regards!

Edit: In fact it is asymptotic to $\frac{2}{bn}$.

Answer 1

Write $q(x) = \mathbf{E}[x^{\xi}]$. If $\mathbf{E}[\xi] = 1$ and $b = \mathbf{Var}(\xi) \in (0, \infty)$, then

• $r_n = 1 - q^{\circ n}(0)$ solves $r_n = 1 - q(1 - r_{n-1})$, hence decreases to $0$ as $n\to\infty$.

• By the Taylor's theorem, $1-q(1-h) = h - \left(\frac{b}{2}+o(1)\right)h^2$.

So, if we write $a_n = \frac{1}{r_n}$, then

$$a_{n} = a_{n-1} \left( 1 - \frac{\frac{b}{2}+o(1)}{a_{n-1}} \right)^{-1} = a_{n-1} + \frac{b}{2} + o(1). $$

Using this, we can prove that $a_n = \left( \frac{b}{2} + o(1) \right) n$ as $n\to\infty$.