问题及解答

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.

Poisson 定理的证明

\[
\begin{split}
&C_n^k p_n^k(1-p_n)^{n-k}\\
=&\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}p_n^k(1-p_n)^{n-k}\\
=&\frac{(np_n)^k}{k!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k-1}{n})(1-p_n)^{n-k}.
\end{split}
\]

由于 $\lim\limits_{n\rightarrow+\infty}np_n=\lambda > 0$, 故有 $\lim\limits_{n\rightarrow+\infty}p_n=0$.  注意到 $k$ 是固定的某个正整数. 于是

\[
\begin{split}
&\lim_{n\rightarrow+\infty}C_n^k p_n^k(1-p_n)^{n-k}\\
=&\lim_{n\rightarrow+\infty}\frac{(np_n)^k}{k!}(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k-1}{n})(1-p_n)^{n-k}\\
=&\frac{\lambda^k}{k!}\lim_{n\rightarrow+\infty}(1-p_n)^{n-k}\\
=&\frac{\lambda^k}{k!}\lim_{n\rightarrow+\infty}(1-p_n)^{n},\\
\end{split}
\]

而

\[
\begin{split}
&\lim_{n\rightarrow+\infty}(1-p_n)^{n}\\
=&\lim_{n\rightarrow+\infty}(1+\frac{1}{-1/p_n})^{(-1/p_n)\cdot(-np_n)}\\
=&\lim_{n\rightarrow+\infty}e^{-\frac{1}{p_n}(-np_n)\ln(1+\frac{1}{-1/p_n})}\\
=&e^{\lim\limits_{n\rightarrow+\infty}(-np_n)\ln(1+\frac{1}{-1/p_n})^{-1/p_n}}\\
=&e^{-\lambda}.
\end{split}
\]

因此, 我们证明了

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

根据离散型随机变量之期望的定义(参见问题2132)

\[
\begin{split}
E(X)&=\sum_{k=0}^{+\infty}x_k p_k\\
&=\sum_{k=0}^{+\infty}k\cdot\frac{\lambda^k}{k!}e^{-\lambda}\\
&=\lambda e^{-\lambda}\sum_{k=1}^{+\infty}\frac{\lambda^{k-1}}{(k-1)!}\\
&\xlongequal{m=k-1}\lambda e^{-\lambda}\sum_{m=0}^{+\infty}\frac{\lambda^m}{m!}\\
&=\lambda e^{-\lambda}e^{\lambda}\\
&=\lambda.
\end{split}
\]

我们利用公式 $D(X)=E(X^2)-(E(X))^2$ 来计算方差.

\[
\begin{split}
E(X^2)&=E[X+X(X-1)]\\
&=E(X)+E(X(X-1))\\
&=\lambda+\sum_{k=0}^{+\infty}k(k-1)\frac{\lambda^k}{k!}e^{-\lambda}\\
&=\lambda+\sum_{k=2}^{+\infty}k(k-1)\frac{\lambda^k}{k!}e^{-\lambda}\\
&=\lambda+\lambda^2 e^{-\lambda}\sum_{k=2}^{+\infty}\frac{\lambda^{k-2}}{(k-2)!}\\
&\xlongequal{j=k-2}\lambda+\lambda^2 e^{-\lambda}\sum_{j=0}^{+\infty}\frac{\lambda^j}{j!}\\
&=\lambda+\lambda^2 e^{-\lambda}e^{\lambda}\\
&=\lambda+\lambda^2,
\end{split}
\]

因此

\[
D(X)=E(X^2)-(E(X))^2=\lambda+\lambda^2-\lambda^2=\lambda.
\]