Posison Distribution

泊松分布 (Poisson Distribution)

泊松分布是概率论中的一个重要离散分布,描述单位时间或单位空间内随机事件发生的次数,假设事件是独立的且平均发生率是已知的。


定义

泊松分布的概率质量函数 (PMF) 为:
P ( X = k ) = λ k e − λ k ! , k = 0 , 1 , 2 , … P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \ldots P(X=k)=k!λkeλ,k=0,1,2,

  • X X X:随机变量,表示单位时间或单位空间内事件发生的次数。
  • k k k:事件发生的具体次数(非负整数)。
  • λ \lambda λ:事件发生的平均次数(泊松分布的参数)。
  • e e e:自然对数的底数,约等于 2.718 2.718 2.718

性质

  1. 期望和方差
    E [ X ] = λ , V a r ( X ) = λ \mathbb{E}[X] = \lambda, \quad \mathrm{Var}(X) = \lambda E[X]=λ,Var(X)=λ
    泊松分布的期望值和方差均等于参数 (\lambda)。

  2. 稀疏事件建模:泊松分布常用于建模稀疏事件(事件发生概率低,但可能次数多)。

  3. 无上界性:虽然泊松分布是离散分布,但事件发生次数 (k) 没有上限(但概率会迅速趋近于 0)。

  4. 加法性:若 X 1 ∼ Poisson ( λ 1 ) X_1 \sim \text{Poisson}(\lambda_1) X1Poisson(λ1) X 2 ∼ Poisson ( λ 2 ) X_2 \sim \text{Poisson}(\lambda_2) X2Poisson(λ2),且 X 1 X_1 X1 X 2 X_2 X2 独立,则:
    X 1 + X 2 ∼ Poisson ( λ 1 + λ 2 ) X_1 + X_2 \sim \text{Poisson}(\lambda_1 + \lambda_2) X1+X2Poisson(λ1+λ2)


适用条件

  1. 独立性:单位时间/空间内事件的发生是独立的。
  2. 均匀性:事件发生的平均速率 (\lambda) 是固定的。
  3. 单事件性:在极小的时间间隔内,只能发生一次事件。

例子

例子 1:客户到达速率

某银行的客户到达速率为每分钟 2 2 2 ( λ = 2 (\lambda = 2 (λ=2)。假设客户到达服从泊松分布:

  • 问题:一分钟内没有客户到达的概率是多少?

    解答
    P ( X = 0 ) = λ 0 e − λ 0 ! = 2 0 e − 2 1 = e − 2 ≈ 0.1353 P(X = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = \frac{2^0 e^{-2}}{1} = e^{-2} \approx 0.1353 P(X=0)=0!λ0eλ=120e2=e20.1353

    结果:一分钟内没有客户到达的概率约为 (13.53%)。


例子 2:网页访问量

某网站的访问量平均每小时为 10 10 10 ( λ = 10 (\lambda = 10 (λ=10)。假设访问次数服从泊松分布:

  • 问题:一小时内正好有 (15) 次访问的概率是多少?

    解答
    P ( X = 15 ) = λ 15 e − λ 15 ! = 1 0 15 e − 10 15 ! P(X = 15) = \frac{\lambda^{15} e^{-\lambda}}{15!} = \frac{10^{15} e^{-10}}{15!} P(X=15)=15!λ15eλ=15!1015e10
    使用计算工具计算得:
    P ( X = 15 ) ≈ 0.0347 P(X = 15) \approx 0.0347 P(X=15)0.0347

    结果:一小时内正好有 15 15 15 次访问的概率约为 3.47 % 3.47\% 3.47%


例子 3:电话呼叫

某呼叫中心每小时接到的呼叫数平均为 6 6 6 l a m b d a = 6 lambda = 6 lambda=6。假设呼叫次数服从泊松分布:

  • 问题:一小时内接到超过 8 8 8 次呼叫的概率是多少?

    解答
    P ( X > 8 ) = 1 − P ( X ≤ 8 ) = 1 − ∑ k = 0 8 P ( X = k ) P(X > 8) = 1 - P(X \leq 8) = 1 - \sum_{k=0}^8 P(X = k) P(X>8)=1P(X8)=1k=08P(X=k)
    逐项计算 P ( X = k ) P(X = k) P(X=k),或直接使用计算工具得:
    P ( X > 8 ) ≈ 0.194 P(X > 8) \approx 0.194 P(X>8)0.194

    结果:一小时内接到超过 8 8 8 次呼叫的概率约为 19.4 % 19.4\% 19.4%


泊松分布与其他分布的关系

  1. 与二项分布的关系
    当二项分布的试验次数 n n n 很大,单次成功概率 p p p 很小,且 n p = λ np = \lambda np=λ 为常数时,二项分布可近似为泊松分布:
    P ( X = k ) = ( n k ) p k ( 1 − p ) n − k ≈ λ k e − λ k ! P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \approx \frac{\lambda^k e^{-\lambda}}{k!} P(X=k)=(kn)pk(1p)nkk!λkeλ

  2. 与指数分布的关系
    泊松过程中的事件间隔时间服从指数分布。如果事件发生的速率为 l a m b d a lambda lambda,则事件间隔时间 T T T 的概率密度函数为:
    f T ( t ) = λ e − λ t , t ≥ 0 f_T(t) = \lambda e^{-\lambda t}, \quad t \geq 0 fT(t)=λeλt,t0


总结

泊松分布在实际生活中应用广泛,包括:

  • 客户到达次数
  • 事故发生次数
  • 电话呼叫数量
  • 射线探测计数

它适合描述独立稀疏事件的发生次数,是统计学、工程学、管理科学等领域的重要工具。

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