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变形虫分裂过程：最终灭绝概率

Start off with one

专题
Probability / 概率
难度
L4

题目详情

You start off with one amoeba. Every minute, this amoeba can either die, do nothing, split into two amoebas, or split into three amoebas; all these scenarios being equally likely to happen. All further amoebas behave the same way. What is the probability that the amoebas eventually die off?

解析

这是 Galton–Watson 分枝过程。每分钟每个个体的子代数为 0,1,2,30,1,2,3，各概率 1/41/4。

令 qq 为最终灭绝概率，则满足生成函数不动点方程

q=1+q+q2+q34.q=\frac{1+q+q^2+q^3}{4}.

化简得

q3+q2−3q+1=0=(q−1)(q2+2q−1).q^3+q^2-3q+1=0=(q-1)(q^2+2q-1).

在 [0,1][0,1] 内除 q=1q=1 外还有根 q=2−1q=\sqrt2-1。

由于该过程均值 E[子代]=0+1+2+34=32>1\mathbb{E}[\text{子代}]=\frac{0+1+2+3}{4}=\frac32>1，灭绝概率是 [0,1][0,1] 内小于 1 的不动点，因此

q=2−1≈0.4142.\boxed{q=\sqrt2-1}\approx 0.4142.