Binomial branching process ($f(s)=(1-p+ps{)}^n$)

A Galton-Watson process whose offspring law is a binomial distribution $\mathrm{Bin}(n,p)$ has mean $np$ and the displayed offspring probability generating function. Its branching extinction probability is the smallest root of $q=f(q)$ in $[0,1]$. It dominates a breadth-first exploration of a binomial random graph by allowing each explored vertex up to $n$ independent possible children, including previously exposed or already discovered vertices as phantom children.

For $p=(1+\epsilon )/n$, the branching survival probability satisfies $1-\rho =(1-p\rho {)}^n$. If $(1+\epsilon {)}^2<n$, comparison with a Poisson branching process and expansion of the logarithm giveFor the upper bound use $\epsilon ={\sum }_{j\ge 1}(1-n{p}^{j+1}){\rho }^j/(j+1)$ and retain its first nonnegative term. For fixed $n>1$, the leading term is instead $2n\epsilon /(n-1)$. In particular $n=2$ gives $\rho =4\epsilon /(1+\epsilon {)}^2$, showing that the uncorrected exact bound $\rho \le 2\epsilon$ is false for small positive $\epsilon$.

For $0<\epsilon \le 1/4$ and $n\epsilon \ge 2$, the uncorrected exact upper bound does hold for the binomial branching process. Here $n{p}^2\le \epsilon$ and $n{p}^3\le 1/4$, so at $x=2\epsilon$ the first two terms of the nonnegative series give $\epsilon (1-n{p}^2)+(4/3){\epsilon }^2(1-n{p}^3)\ge \epsilon$. Monotonicity of that series yields $\rho \le 2\epsilon$. This recovers the intended large-$n$ bracket without claiming it for every finite reproduction law.