泊松分布
概率质量函数
k ，发生次数。该函数只定义在k 为整数的时候。连接线是只为了指导视觉。
累积分布函数
k ，发生次数。CDF在整数k 处不连续，且在其他任何地方都是水平的，因为服从泊松分布的变量只针对整数值。 参数
λ > 0（实数 ） 值域
k
∈
{
0
,
1
,
2
,
3
,
⋯
}
{\displaystyle k\in \{0,1,2,3,\cdots \}}
概率质量函数
λ
k
k
!
e
−
λ
{\displaystyle {\frac {\lambda ^{k}}{k!}}e^{-\lambda }}
累积分布函数
Γ
(
⌊
k
+
1
⌋
,
λ
)
⌊
k
⌋
!
{\displaystyle {\frac {\Gamma (\lfloor k+1\rfloor ,\lambda )}{\lfloor k\rfloor !}}}
e
−
λ
∑
i
=
0
⌊
k
⌋
λ
i
i
!
{\displaystyle e^{-\lambda }\sum _{i=0}^{\lfloor k\rfloor }{\frac {\lambda ^{i}}{i!}}\ }
Q
(
⌊
k
+
1
⌋
,
λ
)
{\displaystyle Q(\lfloor k+1\rfloor ,\lambda )}
(对于
k
≥
0
{\displaystyle k\geq 0}
Γ
(
x
,
y
)
{\displaystyle \Gamma (x,y)}
不完全Γ函数 ，
⌊
k
⌋
{\displaystyle \lfloor k\rfloor }
高斯符号 ，Q是规则化Γ函数) 期望值
λ
{\displaystyle \lambda }
中位数
≈
⌊
λ
+
1
/
3
−
0.02
/
λ
⌋
{\displaystyle \approx \lfloor \lambda +1/3-0.02/\lambda \rfloor }
众数
⌈
λ
⌉
−
1
,
⌊
λ
⌋
{\displaystyle \lceil \lambda \rceil -1,\lfloor \lambda \rfloor }
方差
λ
{\displaystyle \lambda }
偏度
λ
−
1
/
2
{\displaystyle \lambda ^{-1/2}}
峰度
λ
−
1
{\displaystyle \lambda ^{-1}}
熵
λ
[
1
−
log
(
λ
)
]
+
e
−
λ
∑
k
=
0
∞
λ
k
log
(
k
!
)
k
!
{\displaystyle \lambda [1-\log(\lambda )]+e^{-\lambda }\sum _{k=0}^{\infty }{\frac {\lambda ^{k}\log(k!)}{k!}}}
λ
{\displaystyle \lambda }
1
2
log
(
2
π
e
λ
)
−
1
12
λ
−
1
24
λ
2
−
{\displaystyle {\frac {1}{2}}\log(2\pi e\lambda )-{\frac {1}{12\lambda }}-{\frac {1}{24\lambda ^{2}}}-}
19
360
λ
3
+
O
(
1
λ
4
)
{\displaystyle \qquad {\frac {19}{360\lambda ^{3}}}+O\left({\frac {1}{\lambda ^{4}}}\right)}
矩生成函数
exp
(
λ
(
e
t
−
1
)
)
{\displaystyle \exp(\lambda (e^{t}-1))}
特征函数
exp
(
λ
(
e
i
t
−
1
)
)
{\displaystyle \exp(\lambda (e^{it}-1))}
概率母函数
exp
(
λ
(
z
−
1
)
)
{\displaystyle \exp(\lambda (z-1))}
泊松分布 （法语：loi de Poisson ；英语：Poisson distribution ）又称Poisson分布 、帕松分布 、布瓦松分布 、布阿松分布 、普阿松分布 、波以松分布 、卜氏分布 、帕松小数法则 （Poisson law of small numbers），是一种统计 与概率 学里常见到的离散机率分布 ，由法国 数学家西莫恩·德尼·泊松 在1838年时发表。
泊松分布适合于描述单位时间内随机事件发生的次数的概率分布。如某一服务设施在一定时间内受到的服务请求的次数，电话 交换机 接到呼叫的次数、汽车站台的候客人数、机器出现的故障数、自然灾害 发生的次数、DNA序列的变异数、放射性原子核的衰变数、雷射 的光子数分布等等。（单位时间内发生的次数，可以看作事件发生的频率，类似物理的频率
f
{\displaystyle f}
泊松分布的机率质量函数 为：
P
(
X
=
k
)
=
e
−
λ
λ
k
k
!
{\displaystyle P(X=k)={\frac {e^{-\lambda }\lambda ^{k}}{k!}}}
泊松分布的参数
λ
{\displaystyle \lambda }
若
X
{\displaystyle X}
λ
{\displaystyle \lambda }
X
∼
π
(
λ
)
{\displaystyle X\sim \pi (\lambda )}
X
∼
P
o
i
s
s
o
n
(
λ
)
{\displaystyle X\sim Poisson(\lambda )}
1、服从泊松分布的随机变量 ，其数学期望 与方差 相等，同为参数
λ
{\displaystyle \lambda }
E
(
X
)
=
V
(
X
)
=
λ
{\displaystyle E(X)=V(X)=\lambda }
2、两个独立且服从泊松分布的随机变量 ，其和仍然服从泊松分布。更精确地说，若
X
∼
P
o
i
s
s
o
n
(
λ
1
)
{\displaystyle X\sim Poisson(\lambda _{1})}
Y
∼
P
o
i
s
s
o
n
(
λ
2
)
{\displaystyle Y\sim Poisson(\lambda _{2})}
X
+
Y
∼
P
o
i
s
s
o
n
(
λ
1
+
λ
2
)
{\displaystyle X+Y\sim Poisson(\lambda _{1}+\lambda _{2})}
Raikov定理
3、其动差母函数 为：
M
X
(
t
)
=
E
[
e
t
X
]
=
∑
x
=
0
∞
e
t
x
e
−
λ
λ
x
x
!
=
e
−
λ
∑
x
=
0
∞
(
e
t
λ
)
x
x
!
=
e
λ
(
e
t
−
1
)
{\displaystyle M_{X}(t)=E[e^{tX}]=\sum _{x=0}^{\infty }e^{tx}{\frac {e^{-\lambda }\lambda ^{x}}{x!}}=e^{-\lambda }\sum _{x=0}^{\infty }{\frac {({e^{t}}\lambda )^{x}}{x!}}=e^{{\lambda }(e^{t}-1)}}
期望值：(倒数第三至第二是使用泰勒展开式 )
E
(
X
)
=
∑
i
=
0
∞
i
P
(
X
=
i
)
=
∑
i
=
1
∞
i
e
−
λ
λ
i
i
!
=
λ
e
−
λ
∑
i
=
1
∞
λ
i
−
1
(
i
−
1
)
!
=
λ
e
−
λ
∑
i
=
0
∞
λ
i
i
!
=
λ
e
−
λ
e
λ
=
λ
{\displaystyle {\begin{aligned}\mathrm {E} (X)&=\sum _{i=0}^{\infty }\displaystyle iP(X=i)\\&=\sum _{i=1}^{\infty }\displaystyle i{e^{-\lambda }\lambda ^{i} \over i!}\\&=\lambda e^{-\lambda }\sum _{i=1}^{\infty }\displaystyle {\lambda ^{i-1} \over (i-1)!}\\&=\lambda e^{-\lambda }\sum _{i=0}^{\infty }\displaystyle {\lambda ^{i} \over i!}\\&=\lambda e^{-\lambda }e^{\lambda }\\&=\lambda \end{aligned}}}
E
(
X
2
)
=
∑
i
=
0
∞
i
2
P
(
X
=
i
)
=
∑
i
=
1
∞
i
2
e
−
λ
λ
i
i
!
=
λ
e
−
λ
∑
i
=
1
∞
i
λ
i
−
1
(
i
−
1
)
!
=
λ
e
−
λ
∑
i
=
1
∞
1
(
i
−
1
)
!
d
d
λ
(
λ
i
)
=
λ
e
−
λ
d
d
λ
[
∑
i
=
1
∞
λ
i
(
i
−
1
)
!
]
=
λ
e
−
λ
d
d
λ
[
λ
∑
i
=
1
∞
λ
i
−
1
(
i
−
1
)
!
]
=
λ
e
−
λ
d
d
λ
(
λ
e
λ
)
=
λ
e
−
λ
(
e
λ
+
λ
e
λ
)
=
λ
+
λ
2
{\displaystyle {\begin{aligned}\mathrm {E} (X^{2})&=\sum _{i=0}^{\infty }\displaystyle i^{2}P(X=i)\\&=\sum _{i=1}^{\infty }\displaystyle i^{2}{e^{-\lambda }\lambda ^{i} \over i!}\\&=\lambda e^{-\lambda }\sum _{i=1}^{\infty }\displaystyle {i\lambda ^{i-1} \over (i-1)!}\\&=\lambda e^{-\lambda }\sum _{i=1}^{\infty }\displaystyle {1 \over (i-1)!}{d \over d\lambda }(\lambda ^{i})\\&=\lambda e^{-\lambda }{d \over d\lambda }\left[\sum _{i=1}^{\infty }\displaystyle {\lambda ^{i} \over (i-1)!}\right]\\&=\lambda e^{-\lambda }{d \over d\lambda }\left[\lambda \sum _{i=1}^{\infty }\displaystyle {\lambda ^{i-1} \over (i-1)!}\right]\\&=\lambda e^{-\lambda }{d \over d\lambda }(\lambda e^{\lambda })=\lambda e^{-\lambda }(e^{\lambda }+\lambda e^{\lambda })=\lambda +\lambda ^{2}\end{aligned}}}
我们可以得到：
V
a
r
(
X
)
=
(
λ
+
λ
2
)
−
λ
2
=
λ
{\displaystyle Var(X)=(\lambda +\lambda ^{2})-\lambda ^{2}=\lambda }
如同性质：
E
(
X
)
=
V
a
r
(
X
)
=
λ
{\displaystyle E(X)=Var(X)=\lambda }
σ
X
=
λ
{\displaystyle \sigma _{X}={\sqrt {\lambda }}}
相互独立的卜瓦松分布随机变数之和仍服从卜瓦松分布：
X
∼
P
o
i
s
s
o
n
(
λ
1
)
,
Y
∼
P
o
i
s
s
o
n
(
λ
2
)
.
{\displaystyle X\sim Poisson(\lambda _{1}),Y\sim Poisson(\lambda _{2}).}
P
(
X
=
k
1
)
=
λ
1
k
1
e
−
λ
1
k
1
!
,
P
(
Y
=
k
2
)
=
λ
2
k
2
e
−
λ
2
k
2
!
.
{\displaystyle P(X=k_{1})={\dfrac {\lambda _{1}^{k_{1}}e^{-\lambda _{1}}}{k_{1}!}},P(Y=k_{2})={\dfrac {\lambda _{2}^{k_{2}}e^{-\lambda _{2}}}{k_{2}!}}.}
P
(
X
+
Y
=
k
)
=
∑
i
=
0
k
P
(
X
=
i
)
P
(
Y
=
k
−
i
)
=
∑
i
=
0
k
λ
1
i
λ
2
k
−
i
e
−
(
λ
1
+
λ
2
)
i
!
(
k
−
i
)
!
=
e
−
(
λ
1
+
λ
2
)
k
!
∑
i
=
0
k
C
k
i
λ
1
i
λ
2
k
−
i
=
e
−
(
λ
1
+
λ
2
)
(
λ
1
+
λ
2
)
k
k
!
{\displaystyle {\begin{aligned}P(X+Y=k)&=\sum _{i=0}^{k}P(X=i)P(Y=k-i)\\&=\sum _{i=0}^{k}{\frac {\lambda _{1}^{i}\lambda _{2}^{k-i}e^{-(\lambda _{1}+\lambda _{2})}}{i!(k-i)!}}\\&={\frac {e^{-(\lambda _{1}+\lambda _{2})}}{k!}}\sum _{i=0}^{k}C_{k}^{i}\lambda _{1}^{i}\lambda _{2}^{k-i}\\&={\frac {e^{-(\lambda _{1}+\lambda _{2})}(\lambda _{1}+\lambda _{2})^{k}}{k!}}\end{aligned}}}
X
+
Y
∼
P
o
i
s
s
o
n
(
λ
1
+
λ
2
)
{\displaystyle X+Y\sim Poisson(\lambda _{1}+\lambda _{2})}
在二项分布 的伯努利试验 中，如果试验次数
n
{\displaystyle n}
p
{\displaystyle p}
λ
=
n
p
{\displaystyle \lambda =np}
证明如下。首先，回顾自然对数
e
{\displaystyle e}
lim
n
→
∞
(
1
−
λ
n
)
n
=
e
−
λ
,
{\displaystyle \lim _{n\to \infty }\left(1-{\lambda \over n}\right)^{n}=e^{-\lambda },}
二项分布的定义：
P
(
X
=
k
)
=
(
n
k
)
p
k
(
1
−
p
)
n
−
k
{\displaystyle P(X=k)={n \choose k}p^{k}(1-p)^{n-k}}
如果令
p
=
λ
n
{\displaystyle p={\frac {\lambda }{n}}}
n
{\displaystyle n}
P
{\displaystyle P}
lim
n
→
∞
P
(
X
=
k
)
=
lim
n
→
∞
(
n
k
)
p
k
(
1
−
p
)
n
−
k
=
lim
n
→
∞
n
!
(
n
−
k
)
!
k
!
(
λ
n
)
k
(
1
−
λ
n
)
n
−
k
=
lim
n
→
∞
[
n
!
n
k
(
n
−
k
)
!
]
⏟
F
(
λ
k
k
!
)
(
1
−
λ
n
)
n
⏟
→
exp
(
−
λ
)
(
1
−
λ
n
)
−
k
⏟
→
1
=
lim
n
→
∞
[
(
1
−
1
n
)
(
1
−
2
n
)
…
(
1
−
k
−
1
n
)
]
⏟
→
1
(
λ
k
k
!
)
(
1
−
λ
n
)
n
⏟
→
exp
(
−
λ
)
(
1
−
λ
n
)
−
k
⏟
→
1
=
(
λ
k
k
!
)
exp
(
−
λ
)
{\displaystyle {\begin{aligned}\lim _{n\to \infty }P(X=k)&=\lim _{n\to \infty }{n \choose k}p^{k}(1-p)^{n-k}\\&=\lim _{n\to \infty }{n! \over (n-k)!k!}\left({\lambda \over n}\right)^{k}\left(1-{\lambda \over n}\right)^{n-k}\\&=\lim _{n\to \infty }\underbrace {\left[{\frac {n!}{n^{k}\left(n-k\right)!}}\right]} _{F}\left({\frac {\lambda ^{k}}{k!}}\right)\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{n}} _{\to \exp \left(-\lambda \right)}\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{-k}} _{\to 1}\\&=\lim _{n\to \infty }\underbrace {\left[\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)\ldots \left(1-{\frac {k-1}{n}}\right)\right]} _{\to 1}\left({\frac {\lambda ^{k}}{k!}}\right)\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{n}} _{\to \exp \left(-\lambda \right)}\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{-k}} _{\to 1}\\&=\left({\frac {\lambda ^{k}}{k!}}\right)\exp \left(-\lambda \right)\end{aligned}}}
给定
n
{\displaystyle n}
k
i
{\displaystyle k_{i}}
λ
{\displaystyle \lambda }
最大似然估计 值，列出对数似然函数：
L
(
λ
)
=
ln
∏
i
=
1
n
f
(
k
i
∣
λ
)
=
∑
i
=
1
n
ln
(
e
−
λ
λ
k
i
k
i
!
)
=
−
n
λ
+
(
∑
i
=
1
n
k
i
)
ln
(
λ
)
−
∑
i
=
1
n
ln
(
k
i
!
)
.
{\displaystyle {\begin{aligned}L(\lambda )&=\ln \prod _{i=1}^{n}f(k_{i}\mid \lambda )\\&=\sum _{i=1}^{n}\ln \!\left({\frac {e^{-\lambda }\lambda ^{k_{i}}}{k_{i}!}}\right)\\&=-n\lambda +\left(\sum _{i=1}^{n}k_{i}\right)\ln(\lambda )-\sum _{i=1}^{n}\ln(k_{i}!).\end{aligned}}}
d
d
λ
L
(
λ
)
=
0
⟺
−
n
+
(
∑
i
=
1
n
k
i
)
1
λ
=
0.
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} \lambda }}L(\lambda )=0\iff -n+\left(\sum _{i=1}^{n}k_{i}\right){\frac {1}{\lambda }}=0.\!}
解得λ 从而得到一个驻点 （stationary point）：
λ
^
M
L
E
=
1
n
∑
i
=
1
n
k
i
.
{\displaystyle {\widehat {\lambda }}_{\mathrm {MLE} }={\frac {1}{n}}\sum _{i=1}^{n}k_{i}.\!}
检查函数
L
{\displaystyle L}
λ
{\displaystyle \lambda }
k
i
{\displaystyle k_{i}}
L
{\displaystyle L}
∂
2
L
∂
λ
2
=
∑
i
=
1
n
−
λ
−
2
k
i
{\displaystyle {\frac {\partial ^{2}L}{\partial \lambda ^{2}}}=\sum _{i=1}^{n}-\lambda ^{-2}k_{i}}
对某公共汽车站的客流做调查，统计了某天上午10:30到11:47来到候车的乘客情况。假定来到候车的乘客各批（每批可以是1人也可以是多人）是互相独立发生的。观察每20秒区间来到候车的乘客批次，共观察77分钟*3=231次，共得到230个观察记录。其中来到0批、1批、2批、3批、4批及4批以上的观察记录分别是100次、81次、34次、9次、6次。使用极大似真估计（MLE），得到
λ
{\displaystyle \lambda }
81
×
1
+
34
×
2
+
9
×
3
+
6
×
4
230
≈
0.87
{\displaystyle {\frac {81\times 1+34\times 2+9\times 3+6\times 4}{230}}\approx 0.87}
一个用来生成随机泊松分布的数字（伪随机数抽样）的简单算法，已经由高德纳 给出（见下文参考）：
algorithm poisson random number (Knuth) :
init :
Let L ← e −λ , k ← 0 and p ← 1.
do :
k ← k + 1.
Generate uniform random number u in [0,1] and let p ← p×u.
while p > L.
return k − 1.
尽管简单，但复杂度是线性的，在返回的值
k
{\displaystyle k}
λ
{\displaystyle \lambda }
λ
{\displaystyle \lambda }
e
−
λ
{\displaystyle e^{-\lambda }}
λ
{\displaystyle \lambda }
拒绝采样 ，另一种是采用泊松分布的高斯近似。
对于很小的
λ
{\displaystyle \lambda }
u
{\displaystyle u}
algorithm Poisson generator based upon the inversion by sequential search :[ 1] init :
Let x ← 0, p ← e −λ , s ← p.
Generate uniform random number u in [0,1].
do :
x ← x + 1.
p ← p * λ / x.
s ← s + p.
while u > s.
return x.
Guerriero V. Power Law Distribution: Method of Multi-scale Inferential Statistics . Journal of Modern Mathematics Frontier (JMMF). 2012, 1 : 21–28 [2017-10-30 ] . （原始内容 存档于2018-02-21）. Joachim H. Ahrens, Ulrich Dieter. Computer Methods for Sampling from Gamma, Beta, Poisson and Binomial Distributions. Computing. 1974, 12 (3): 223–246. doi:10.1007/BF02293108 Joachim H. Ahrens, Ulrich Dieter. Computer Generation of Poisson Deviates . ACM Transactions on Mathematical Software. 1982, 8 (2): 163–179. doi:10.1145/355993.355997 Ronald J. Evans, J. Boersma, N. M. Blachman, A. A. Jagers. The Entropy of a Poisson Distribution: Problem 87-6. SIAM Review. 1988, 30 (2): 314–317. doi:10.1137/1030059 Donald E. Knuth. Seminumerical Algorithms. The Art of Computer Programming. Volume 2. Addison Wesley . 1969.
![{\displaystyle \lambda [1-\log(\lambda )]+e^{-\lambda }\sum _{k=0}^{\infty }{\frac {\lambda ^{k}\log(k!)}{k!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf6cf37058d59e89453fd5bf9a1ece59a8c81d1a)
![{\displaystyle M_{X}(t)=E[e^{tX}]=\sum _{x=0}^{\infty }e^{tx}{\frac {e^{-\lambda }\lambda ^{x}}{x!}}=e^{-\lambda }\sum _{x=0}^{\infty }{\frac {({e^{t}}\lambda )^{x}}{x!}}=e^{{\lambda }(e^{t}-1)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07027a2d97d52986d99b0a73c00430e95cdbdc8a)
![{\displaystyle {\begin{aligned}\mathrm {E} (X^{2})&=\sum _{i=0}^{\infty }\displaystyle i^{2}P(X=i)\\&=\sum _{i=1}^{\infty }\displaystyle i^{2}{e^{-\lambda }\lambda ^{i} \over i!}\\&=\lambda e^{-\lambda }\sum _{i=1}^{\infty }\displaystyle {i\lambda ^{i-1} \over (i-1)!}\\&=\lambda e^{-\lambda }\sum _{i=1}^{\infty }\displaystyle {1 \over (i-1)!}{d \over d\lambda }(\lambda ^{i})\\&=\lambda e^{-\lambda }{d \over d\lambda }\left[\sum _{i=1}^{\infty }\displaystyle {\lambda ^{i} \over (i-1)!}\right]\\&=\lambda e^{-\lambda }{d \over d\lambda }\left[\lambda \sum _{i=1}^{\infty }\displaystyle {\lambda ^{i-1} \over (i-1)!}\right]\\&=\lambda e^{-\lambda }{d \over d\lambda }(\lambda e^{\lambda })=\lambda e^{-\lambda }(e^{\lambda }+\lambda e^{\lambda })=\lambda +\lambda ^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67cfdcfdefe078c4ae07a091d4ec9cc56b007d67)
![{\displaystyle {\begin{aligned}\lim _{n\to \infty }P(X=k)&=\lim _{n\to \infty }{n \choose k}p^{k}(1-p)^{n-k}\\&=\lim _{n\to \infty }{n! \over (n-k)!k!}\left({\lambda \over n}\right)^{k}\left(1-{\lambda \over n}\right)^{n-k}\\&=\lim _{n\to \infty }\underbrace {\left[{\frac {n!}{n^{k}\left(n-k\right)!}}\right]} _{F}\left({\frac {\lambda ^{k}}{k!}}\right)\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{n}} _{\to \exp \left(-\lambda \right)}\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{-k}} _{\to 1}\\&=\lim _{n\to \infty }\underbrace {\left[\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)\ldots \left(1-{\frac {k-1}{n}}\right)\right]} _{\to 1}\left({\frac {\lambda ^{k}}{k!}}\right)\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{n}} _{\to \exp \left(-\lambda \right)}\underbrace {\left(1-{\frac {\lambda }{n}}\right)^{-k}} _{\to 1}\\&=\left({\frac {\lambda ^{k}}{k!}}\right)\exp \left(-\lambda \right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f3305a795ed8b2ce32f86487f67b44ecd5f69cb)
