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募捐到 100K：所需人数的分布

Organize a charity event

专题
Probability / 概率
难度
L4

题目详情

To organize a charity event that costs $100K, an organization raises funds.
Independent of each other, one donor after another donates some amount of money that is exponentially distributed with a mean of $20K.

The process is stopped as soon as $100K or more has been collected.

Find the distribution, mean, and variance of the number of donors needed until at least $100K has been collected.

Mathematical formulation

  • Let each donation Xi∼Exp(λ)X_i \sim \mathrm{Exp}(\lambda) with mean 1λ=20K\frac{1}{\lambda} = 20\text{K}.
    So λ=120K\lambda = \frac{1}{20\text{K}}.

  • Let Sn=∑i=1nXiS_n = \sum_{i=1}^n X_i.

  • We want the random variable

    N=min⁡{n:Sn≥100K}.N = \min\{n : S_n \ge 100\text{K}\}.

Find the distribution, mean E[N]E[N], and variance Var(N)\mathrm{Var}(N).

解析

每位捐款 Xi∼Exp(mean 20K)X_i\sim\mathrm{Exp}(\text{mean }20\text{K})，即速率 λ=1/20K\lambda=1/20\text{K}。令累计 Sn=∑i=1nXiS_n=\sum_{i=1}^n X_i，停止规则为 N=min⁡{n:Sn≥100K}N=\min\{n:S_n\ge 100\text{K}\}。

指数分布的等待时间之和对应泊松过程的到达时刻：令 NtN_t 为速率 λ\lambda 的泊松过程，则 SnS_n 是第 nn 次到达时间，且

N=N100K+1.N = N_{100\text{K}}+1.

因为 N100K∼Poisson(λ⋅100K)=Poisson(5)N_{100\text{K}}\sim\mathrm{Poisson}(\lambda\cdot 100\text{K})=\mathrm{Poisson}(5)，所以

N=d1+Poisson(5).\boxed{N\stackrel{d}{=}1+\mathrm{Poisson}(5)}.

即对 n=1,2,…n=1,2,\ldots

P(N=n)=e−55n−1(n−1)!.\boxed{\mathbb{P}(N=n)=e^{-5}\frac{5^{n-1}}{(n-1)!}}.

均值与方差：

E[N]=6,Var⁡(N)=5.\boxed{\mathbb{E}[N]=6},\qquad \boxed{\operatorname{Var}(N)=5}.