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泊松(Poisson)分布 $P(\lambda)$

Posted by haifeng on 2018-05-25 06:32:08 last update 2020-04-06 18:35:57 | Answers (3) | 收藏


泊松(Poisson)分布 $P(\lambda)$

若随机变量 $X$ 的概率分布为

\[
P\{X=k\}=\frac{\lambda^k}{k!}e^{-\lambda},\quad k=0,1,2,\ldots.
\]

其中 $\lambda > 0$, 则称 $X$ 服从参数为 $\lambda$ 的泊松分布, 记为 $X\sim P(\lambda)$.

历史上, 泊松分布是作为二项分布的近似而引入的.

所基于的理论是

 

Thm. (泊松定理) 设 $\lim\limits_{n\rightarrow+\infty}np_n=\lambda > 0$, 则

\[
\lim_{n\rightarrow+\infty}C_n^k p_n^k(1-p_n)^{n-k}=\frac{\lambda^k}{k!}e^{-\lambda},\quad k=0,1,2,\ldots,n.
\]

 

 


设 $X\sim P(\lambda)$, 则 $E(X)=\lambda$, $D(X)=\lambda$.

 


[English]

The Poisson Probability Distribution

[Def] A random variable $X$ is said to have a Poisson distribution if the pmf of $X$ is

\[
p(x;\lambda)=\frac{e^{-\lambda}\lambda^x}{x!},\quad x=0,1,2,\ldots
\]

for some $\lambda > 0$.


The rationale for using the Poisson distribution in many situation is provided by the following the above proposition. Also it can be stated as follows:

[Prop] Suppose that in the binomial pmf $b(x;n,p)$, we let $n\rightarrow\infty$ and $p\rightarrow 0$ in such a way that $np$ approaches a value $\lambda > 0$. Then $b(x;n,p)\rightarrow p(x;\lambda)$.

 

[Prop] If $X$ has a Poisson distribution with parameter $\lambda$, then $E(X)=V(X)=\lambda$.

 

References:

The above content in English is copied from the following book:

《Probability and Statistics For Engineering and The Sciences》(Fifth Edtion) P.131
Author: Jay L. Devore

Section 5 of Chapter 3.