# 泊松分布

## 泊松分布的定义和性质 Definition and Properties of the Poisson Distributions

$$f(x|n=3600,p=0.00125)= \begin{cases} \begin{pmatrix} 3600\\x \end{pmatrix}p^x(1-p)^{3600-x}&\text{for }0\leq x\leq 3600\\ 0&\text{otherwise} \end{cases}$$

\begin{aligned} \frac{f(x+1)}{f(x)}&= \frac {\begin{pmatrix}n\\x+1\end{pmatrix}p^{x+1}(1-p)^{n-x-1}} {\begin{pmatrix}n\\x\end{pmatrix}p^{x+1}(1-p)^{n-x-1}}\\ &=\frac{(n-x)p}{(x+1)(1-p)}\\ &\approx\frac{np}{x+1} \end{aligned}

$$f(1)=f(0)\lambda\\ f(2)=f(1)\frac{\lambda}{2}=f(0)\frac{\lambda^2}{2}\\ f(3)=f(2)\frac{\lambda}{3}=f(0)\frac{\lambda^3}{6}\\ \vdots\\ f(n)=f(n-1)\frac{\lambda}{n}==f(0)\frac{\lambda^n}{n!}\\$$

$$\sum^{\infty}_{x=0}f(x)=1$$

$$\sum^{\infty}_{x=0}f(0)\frac{\lambda^n}{n!}=1\\ f(0)\sum^{\infty}_{x=0}\frac{\lambda^n}{n!}=1\\ \text{for :}\sum^{\infty}_{x=0}\frac{\lambda^n}{n!}=e^{\lambda}\\ \text{so :}f(0)=e^{-\lambda}$$

$$f(x|\lambda)= \begin{cases} \frac{e^{-\lambda}\lambda^x}{x!}&\text{for }x=1,2,3,\dots\\ 0&\text{otherwise} \end{cases}$$

Definition Poisson Distribution.Let $\lambda > 0$ .A random variable X has the Poisson Distribution with mean $\lambda$ if the p.f. of $X$ is as follow:
$$f(x|\lambda)= \begin{cases} \frac{e^{-\lambda}\lambda^x}{x!}&\text{for }x=1,2,3,\dots\\ 0&\text{otherwise} \end{cases}$$

### 泊松分布的均值 Mean

Theorem Mean. The mean of Poisson Distribution with p.f. equal to upside is $\lambda$ .

$$E(X)=\sum^{\infty}_{x=0}xf(x|\lambda)$$

\begin{aligned} E(X)&=\sum^{\infty}_{x=0}x\frac{e^{-\lambda}\lambda^x}{x!}\\ &=\sum^{\infty}_{x=1}\frac{e^{-\lambda}\lambda^x}{(x-1)!}\\ &=\lambda\sum^{\infty}_{x=1}\frac{e^{-\lambda}\lambda^{x-1}}{(x-1)!}\\ \text{if we set } y=x-1\\ &=\lambda\sum^{\infty}_{y=0}\frac{e^{-\lambda}\lambda^{y}}{y!} \end{aligned}

### 泊松分布的方差 Varaince

Theorem Variance.The variance of Poisson distribution with mean $\lambda$ is also $\lambda$

\begin{aligned} E[X(X-1)]&=\sum^{\infty}_{x=0}x(x-1)f(x|\lambda)\\ &=\sum^{\infty}_{x=2}x(x-1)f(x|\lambda)\\ &=\sum^{\infty}_{x=2}x(x-1)\frac{e^{-\lambda}\lambda^x}{x!}\\ &=\lambda^2\sum^{\infty}_{x=2}\frac{e^{-\lambda}\lambda^{x-2}}{x-2!}\\ \text{We set }y=x-2\\ E[X(X-1)]&=\lambda^2\sum^{\infty}_{y=0}\frac{e^{-\lambda}\lambda^y}{y!}\\ &=\lambda^2 \end{aligned}

$$Var(X)=E[X^2]-E^2[x]=\lambda^2+\lambda-\lambda^2=\lambda$$

### 泊松分布的距生成函数 m.g.f.

Theorem Moment Generating Function.The m.g.f. of the Poisson distribution with mean $\lambda$ is
$$\psi(t)=e^{\lambda(e^t-1)}$$
for all real $t$

$$\psi(t)=E(e^{tX})=\sum^{\infty}_{x=0}\frac{e^{tx}e^{-\lambda}\lambda^x}{x!}=e^{-\lambda}\sum^{\infty}_{x=0}\frac{(\lambda e^t)^x}{x!}$$

$$\sum^{\infty}_{x=0}\frac{(\lambda e^t)^x}{x!}=e^{\lambda e^t}$$

$$\psi(t)=e^{-\lambda}e^{\lambda e^t}=e^{\lambda(e^t-1)}$$

### 泊松分布随机变量相加

Theorem If the random variable $X_1,\dots,X_k$ are independent and if $X_i$ has Poisson distribution with mean $\lambda_i(i=1,\dots,k)$ ,then the sum $X_1+\dots+X_k$ has the Poisson distribution with mean $\lambda_1+\dots+\lambda_k$

$$\psi(t)=\Pi^k_{i=1}\psi_i(t)=\Pi^k_{i=1}e^{\lambda_i(e^t-1)}=e^{(\lambda_1+\dots+\lambda_k)(e^t-1)}$$

## 二项分布的泊松近似 The Poisson Approximation to Binomial Distributions

Theorem Closeness of Binomial and Pisson Distribution.For each integer n and each $0 < p < 1$ ,let $f(x|n,p)$ denote the p.f. of the binomial distribtuion with parameters $n$ and $p$ .Let $f(x|\lambda)$ denote the p.f. of the Poisson distribution with mean $\lambda$ .Let ${{P_n}}^{\infty}{n=1}$ be a sequence of numbers between 0 and 1 such that $lim{n\to \infty}np_n=\lambda$ . Then
$$lim_{n\to \infty}f(x|n,p_n)=f(x|\lambda)$$
for all $x=0,1\dots$

$$f(x|n,p_n)=\frac{n(n-1)\dots(n-x+1)}{x!}p_n^x(1-p_n)^{n-x}$$

$$f(x|n,p_n)=\frac{\lambda_n^x}{x!}\frac{n}{n}\cdot\frac{n-1}{n}\dots \frac{n-x+1}{n}(1-\frac{\lambda_n}{n})^n(1-\frac{\lambda_n}{n})^{-x}$$

$$lim_{n\to \infty}\frac{n}{n}\cdot\frac{n-1}{n}\dots \frac{n-x+1}{n}(1-\frac{\lambda_n}{n})^{-x}=1$$

$$lim_{n\to \infty}(1-\frac{\lambda_n}{n})^{n}=e^{-\lambda}$$

$$lim_{n\to \infty}f(x|n,p_n)=\frac{e^{-\lambda}\lambda^x}{x!}=f(x|\lambda)$$

Theorem Closeness of Hypergeometric and Poisson Distribution.Let $\lambda>0$ .Let $Y$ have the Poisson distribution with mean $\lambda$ .For each postive integer $T$ ,let $A_T,B_T$ ,and $n_T$ be integers such that $lim_{T\to \infty}n_TA_T/(A_T+B_T)=\lambda$ .Let $X_T$ have the hypergeometric distribution with parameters $A_T,B_T$ and $n_T$ .Tor each fixed $x=0,1,\dots$ ，
$$lim_{T\to \infty}\frac{Pr(Y=x)}{Pr(X_t=x)}=1$$

## 泊松过程 Poisson Processes

Definition Poisson Process.A Poisson process with rate $\lambda$ per unit time is a process that satisfies the following two properties:
i: The number of arrivals in every fixed interval of time of length $t$ has the Poisson distribution with mean $\lambda t$
ii: The numbers of arrivals in every collection of disjoint time intervals are independent

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