慈善募款人数

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英文解析

Denote by $a_{i}$ the amount of money donated by donor $i$ , that is exponentially distributed with mean $1 / \lambda$ . Let $s_{n} = \sum_{i = 1}^{n} a_{i}$ be the total amount of money donated by donors $1, \ldots , n$ , and let

$$N = \min_{n \geq 1} \{n \text{such that} s_{n} \geq a\}$$

be the discrete random variable denoting the smallest index $n$ such that $s_{n}$ is at least $a$ .

Denote by $P(n \mid a), n \geq 1$ , the probability mass function of $N$ , that is, the probability that $N = n$ when a total of $a$ needs to be raised. Note that

$$P(1 \mid a) = P(a_{1} \geq a) = e^{-\lambda a}.$$

We find $P(n \mid a), n > 1$ , by conditioning on $a_{1}$ . Given that the first donor donated $a_{1} = x < a, N$ is equal to $n$ if and only if the remaining amount $a - x$ is raised by the next $n - 1$ donors (and not by fewer than the next $n - 1$ donors), an event that by definition has probability $P(n - 1 \mid a - x)$ . Since the probability density function of $a_{1}$ is $f_{a_{1}}(x) = \lambda e^{- \lambda x}$ , then, for $n > 1$ , the law of total probability yields

$$\begin{array}{c}{{P(n\mid a)=\int_{0}^{a}P(n-1\mid a-x)f_{a_{1}}(x)d x}}\\ {{=\int_{0}^{a}\lambda e^{-\lambda x}P(n-1\mid a-x)d x}}\end{array}$$

We will prove that

$$P(n \mid a) = \frac{(\lambda a)^{n - 1}}{(n - 1)!} e^{-\lambda a}, \quad \forall a \geq 0,$$

by induction on $n$ . The base case $n = 1$ was already established; see (3.219). Assume that (3.221) holds for $n > 1$ ; we will show that it also holds for $n + 1$ .

From the induction hypothesis, we obtain that

$$P(n \mid a - x) = \frac{(\lambda(a - x))^{n - 1}}{(n - 1)!} e^{-\lambda (a - x)}$$

for all $0 \leq x \leq a$ . From (3.220), it follows that

$$P(n + 1 \mid a) = \int_{0}^{a} \lambda e^{-\lambda x} P(n \mid a - x) dx.$$

From (3.222) and (3.223), we find that

$$\begin{array}{l}{P(n + 1\mid a)}\\ {= \int_{0}^{a}\lambda e^{-\lambda x}\cdot \frac{(\lambda(a - x))^{n - 1}}{(n - 1)!} e^{-\lambda (a - x)}d x}\\ {= \frac{\lambda^{n}e^{-\lambda a}}{(n - 1)!}\int_{0}^{a}(a - x)^{n - 1}d x}\\ {= \frac{(\lambda a)^{n}}{n!} e^{-\lambda a}.} \end{array}$$

We conclude that (3.221) holds for $n + 1$ , and therefore (3.221) is proved by induction.

From (3.221), it follows that $N$ has the same distribution as $1 + M$ , where $M$ has a Poisson distribution with mean $\lambda a$ . Then,

$$E[N] = 1 + \lambda a; \quad \operatorname {Var}(N) = \lambda a.$$

For our problem, $1 / \lambda =$ 20 $\mathrm{K}$ and $a =$ 100 $\mathrm{K}$ . Thus, $E[N] = 1 + \lambda a = 6$ and $\operatorname {Var}(N) = \lambda a = 5$

We conclude that the number of donors needed until at least $$100 \mathrm{K}$ is collected has mean 6 and variance 5.