# 泊松分布

| name       = Poisson Distribution

| type       = mass

| pdf_image  = 325px
| pdf_image  = 325px
| pdf_caption = The horizontal axis is the index k, the number of occurrences. λ is the expected rate of occurrences. The vertical axis is the probability of k occurrences given λ. The function is defined only at integer values of k; the connecting lines are only guides for the eye.
| pdf_caption = The horizontal axis is the index k, the number of occurrences. λ is the expected rate of occurrences. The vertical axis is the probability of k occurrences given λ. The function is defined only at integer values of k; the connecting lines are only guides for the eye.
| pdf _ caption = 横轴是索引 k，表示出现的次数。是预期发生率。垂直轴是给定的 k 发生概率。函数只定义在 k 的整数值上，连接线指示方向。
| cdf_image  = 325px
| cdf_image  = 325px
| cdf_caption = The horizontal axis is the index k, the number of occurrences. The CDF is discontinuous at the integers of k and flat everywhere else because a variable that is Poisson distributed takes on only integer values.
| cdf_caption = The horizontal axis is the index k, the number of occurrences. The CDF is discontinuous at the integers of k and flat everywhere else because a variable that is Poisson distributed takes on only integer values.
| cdf _ caption = 水平轴是索引 k，表示出现的次数。因为一个 {泊松分佈 Poisson distribution的变量只取整数值，所以 CDF 在 k 的整数和平坦的所有其他地方均不连续。
| notation   = $\displaystyle{ \operatorname{Pois}(\lambda) }$
| 表示法 = < math > operatorname { Pois }(lambda) </math >
| parameters = $\displaystyle{ \lambda\in (0, \infty) }$  (rate)
| support    = $\displaystyle{ k \in \mathbb{N}_0 }$ (Natural numbers starting from 0)
| support = < math > k in mathbb { n } _ 0 </math > (自然数从0开始)
| pdf        = $\displaystyle{ \frac{\lambda^k e^{-\lambda}}{k!} }$
| pdf = < math > frac { lambda ^ k e ^ {-lambda }{ k！{/math >
| cdf        = $\displaystyle{ \frac{\Gamma(\lfloor k+1\rfloor, \lambda)}{\lfloor k\rfloor !} }$, or $\displaystyle{ e^{-\lambda} \sum_{i=0}^{\lfloor k\rfloor} \frac{\lambda^i}{i!}\ }$, or $\displaystyle{ Q(\lfloor k+1\rfloor,\lambda) }$
| cdf = < math > frac { Gamma (lfloor k + 1 rfloor，lambda)}{ lfloor k rfloor！{ i = 0}{ lfloor k rfloor } frac { lambda ^ i }{ i！} </math > ，或者 < math > q (lfloor k + 1 rfloor，lambda) </math >

(for $\displaystyle{ k\ge 0 }$, where $\displaystyle{ \Gamma(x, y) }$ is the upper incomplete gamma function, $\displaystyle{ \lfloor k\rfloor }$ is the floor function, and Q is the regularized gamma function)

(for $\displaystyle{ k\ge 0 }$, where $\displaystyle{ \Gamma(x, y) }$ is the upper incomplete gamma function, $\displaystyle{ \lfloor k\rfloor }$ is the floor function, and Q is the regularized gamma function)

(对于 < math > k ge 0 </math > ，其中 < math > Gamma (x，y) </math > 是上面的不完全Γ函数，< math > lfloor k | | | | | | | | | | | | | | | | | | </math > 是下面的函数，q 是正则化的 Gamma 函数)

| mean       = $\displaystyle{ \lambda }$
| mean       = $\displaystyle{ \lambda }$

| mean = < math > > lambda </math >

| median     = $\displaystyle{ \approx\lfloor\lambda+1/3-0.02/\lambda\rfloor }$
| median     = $\displaystyle{ \approx\lfloor\lambda+1/3-0.02/\lambda\rfloor }$

| 中位数 = < math > > 大约1floor lambda + 1/3-0.02/lambda rfloor </math >

| mode       = $\displaystyle{ \lceil\lambda\rceil - 1, \lfloor\lambda\rfloor }$
| mode       = $\displaystyle{ \lceil\lambda\rceil - 1, \lfloor\lambda\rfloor }$

| mode = < math > lceil lambda rceil-1，lfloor lambda rfloor

| variance   = $\displaystyle{ \lambda }$
| variance   = $\displaystyle{ \lambda }$

| variance = < math > lambda </math >

| skewness   = $\displaystyle{ \lambda^{-1/2} }$
| skewness   = $\displaystyle{ \lambda^{-1/2} }$

| skewness = < math > lambda ^ {-1/2} </math >

| kurtosis   = $\displaystyle{ \lambda^{-1} }$
| kurtosis   = $\displaystyle{ \lambda^{-1} }$

| 峭度 = < math > lambda ^ {-1} </math >

| entropy    = $\displaystyle{ \lambda[1 - \log(\lambda)] + e^{-\lambda}\sum_{k=0}^\infty \frac{\lambda^k\log(k!)}{k!} }$
| entropy    = $\displaystyle{ \lambda[1 - \log(\lambda)] + e^{-\lambda}\sum_{k=0}^\infty \frac{\lambda^k\log(k!)}{k!} }$

| 熵 = < math > lambda [1-log (lambda)] + e ^ {-lambda } sum _ { k = 0} ^ infty frac { lambda ^ k log (k!)}{ k！{/math >

(for large $\displaystyle{ \lambda }$)

(for large $\displaystyle{ \lambda }$)

(对于大的 < math > > > lambda </math >)

$\displaystyle{ \frac{1}{2}\log(2 \pi e \lambda) - \frac{1}{12 \lambda} - \frac{1}{24 \lambda^2} -{} }$
$\displaystyle{ \qquad \frac{19}{360 \lambda^3} + O\left(\frac{1}{\lambda^4}\right) }$

$\displaystyle{ \frac{1}{2}\log(2 \pi e \lambda) - \frac{1}{12 \lambda} - \frac{1}{24 \lambda^2} -{} }$
$\displaystyle{ \qquad \frac{19}{360 \lambda^3} + O\left(\frac{1}{\lambda^4}\right) }$

12 lambda }-frac {1}{24 lambda ^ 2}-{{{} </math > < br > < math > qfrac {19}{360 lambda ^ 3} + o left (frac {1}{1}{ lambda ^ 4}右) </math > <

| pgf        = $\displaystyle{ \exp[\lambda(z - 1)] }$
| pgf        = $\displaystyle{ \exp[\lambda(z - 1)] }$

| pgf = < math > exp [ lambda (z-1)] </math >

| mgf        = $\displaystyle{ \exp[\lambda (e^{t} - 1)] }$
| mgf        = $\displaystyle{ \exp[\lambda (e^{t} - 1)] }$

| mgf = < math > exp [ lambda (e ^ { t }-1)] </math >

| char       = $\displaystyle{ \exp[\lambda (e^{it} - 1)] }$
| char       = $\displaystyle{ \exp[\lambda (e^{it} - 1)] }$

| char = < math > exp [ lambda (e ^ { it }-1)] </math >

| fisher     = $\displaystyle{ \frac{1}{\lambda} }$
| fisher     = $\displaystyle{ \frac{1}{\lambda} }$

| fisher = < math > frac {1}{ lambda } </math >

}}

}}

}}

In probability theory and statistics, the Poisson distribution (模板:IPAc-en; 模板:IPA-fr), named after French mathematician Siméon Denis Poisson, is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.模板:R The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume.

In probability theory and statistics, the Poisson distribution (; ), named after French mathematician Siméon Denis Poisson, is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume.

--fairywang(讨论)  【审校】“它表示在一个固定的时间段或空间中一定数量的事件的发生概率”改为“它表示在一个固定的时间段或空间中，一定数量的事件发生的概率”

For instance, an individual keeping track of the amount of mail they receive each day may notice that they receive an average number of 4 letters per day. If receiving any particular piece of mail does not affect the arrival times of future pieces of mail, i.e., if pieces of mail from a wide range of sources arrive independently of one another, then a reasonable assumption is that the number of pieces of mail received in a day obeys a Poisson distribution.模板:R Other examples that may follow a Poisson distribution include the number of phone calls received by a call center per hour and the number of decay events per second from a radioactive source.

For instance, an individual keeping track of the amount of mail they receive each day may notice that they receive an average number of 4 letters per day. If receiving any particular piece of mail does not affect the arrival times of future pieces of mail, i.e., if pieces of mail from a wide range of sources arrive independently of one another, then a reasonable assumption is that the number of pieces of mail received in a day obeys a Poisson distribution. Other examples that may follow a Poisson distribution include the number of phone calls received by a call center per hour and the number of decay events per second from a radioactive source.

--fairywang(讨论)  【审校】“呼叫中心每小时接到的电话数量和每秒从放射源衰变事件的数量。”改为“呼叫中心每小时接到的电话数量和每秒从放射源的衰变数。”

## Definitions定义

### Probability mass function概率分布函数

The Poisson distribution is popular for modeling the number of times an event occurs in an interval of time or space.

The Poisson distribution is popular for modeling the number of times an event occurs in an interval of time or space.

A discrete random variable X is said to have a Poisson distribution with parameter λ > 0, if, for k = 0, 1, 2, ..., the probability mass function of X is given by:模板:R

A discrete random variable X is said to have a Poisson distribution with parameter λ > 0, if, for k = 0, 1, 2, ..., the probability mass function of X is given by:

$\displaystyle{ \!f(k; \lambda)= \Pr(X = k)= \frac{\lambda^k e^{-\lambda}}{k!}, }$

where

The positive real number λ is equal to the expected value of X and also to its variance引用错误：没有找到与</ref>对应的<ref>标签

$\displaystyle{ \lambda=\operatorname{E}(X)=\operatorname{Var}(X). }$

The Poisson distribution can be applied to systems with a large number of possible events, each of which is rare. The number of such events that occur during a fixed time interval is, under the right circumstances, a random number with a Poisson distribution.

The Poisson distribution can be applied to systems with a large number of possible events, each of which is rare. The number of such events that occur during a fixed time interval is, under the right circumstances, a random number with a Poisson distribution.

### Example举例

The Poisson distribution may be useful to model events such as

The Poisson distribution may be useful to model events such as

• The number of meteorites greater than 1 meter diameter that strike Earth in a year
• The number of patients arriving in an emergency room between 10 and 11 pm
• The number of laser photons hitting a detector in a particular time interval
• 一年内撞击地球的直径大于1米的陨石数量
• 晚上10点到11点到达急诊室的病人人数
• 在特定时间间隔内撞击探测器的激光光子数

### Assumptions and validity假设与有效条件

The Poisson distribution is an appropriate model if the following assumptions are true:模板:R

The Poisson distribution is an appropriate model if the following assumptions are true:

• k is the number of times an event occurs in an interval and k can take values 0, 1, 2, ....
• The occurrence of one event does not affect the probability that a second event will occur. That is, events occur independently.
• The average rate at which events occur is independent of any occurrences. For simplicity, this is usually assumed to be constant, but may in practice vary with time.
• Two events cannot occur at exactly the same instant; instead, at each very small sub-interval exactly one event either occurs or does not occur.
• 事件在一个时间间隔内发生且{mvar | k}可以取值0，1，2，...；
• 一个事件的发生不影响第二个事件发生的概率，也就是时间发生相互独立；
• 事件发生的平均速率与任何事件无关。为简单起见，通常假定其为常数，但实际上可能随时间而变化；
• 两个事件不可能在完全相同的时刻发生，即在每一小段的时间内正好有一个事件发生或不发生。

If these conditions are true, then k is a Poisson random variable, and the distribution of k is a Poisson distribution.

If these conditions are true, then is a Poisson random variable, and the distribution of is a Poisson distribution.

The Poisson distribution is also the limit of a binomial distribution, for which the probability of success for each trial equals λ divided by the number of trials, as the number of trials approaches infinity (see Related distributions).

The Poisson distribution is also the limit of a binomial distribution, for which the probability of success for each trial equals divided by the number of trials, as the number of trials approaches infinity (see Related distributions). 每次试验的成功概率除以总试验次数，随着试验的数量趋于无穷大，泊松分布也是 二项式分布Binomial distribution的极限。（可参考相关分布）

### Probability of events for a Poisson distribution {泊松分佈 Poisson distribution的事件概率

An event can occur 0, 1, 2, ... times in an interval. The average number of events in an interval is designated $\displaystyle{ \lambda }$ (lambda). $\displaystyle{ \lambda }$ is the event rate, also called the rate parameter. The probability of observing k events in an interval is given by the equation

An event can occur 0, 1, 2, ... times in an interval. The average number of events in an interval is designated $\displaystyle{ \lambda }$ (lambda). $\displaystyle{ \lambda }$ is the event rate, also called the rate parameter. The probability of observing events in an interval is given by the equation

$\displaystyle{ P(k \text{ events in interval}) = \frac{\lambda^k e^{-\lambda}}{k!} }$

$\displaystyle{ P(k \text{ events in interval}) = \frac{\lambda^k e^{-\lambda}}{k!} }$

< math > p (k text { events in interval }) = frac { lambda ^ k e ^ {-lambda }{ k！{/math >

where

where

• $\displaystyle{ \lambda }$ is the average number of events per interval
• e is the number 2.71828... (Euler's number) the base of the natural logarithms
• k takes values 0, 1, 2, ...
• k! = k × (k − 1) × (k − 2) × ... × 2 × 1 is the factorial of k.

This equation is the probability mass function (PMF) for a Poisson distribution.

This equation is the probability mass function (PMF) for a Poisson distribution.

This equation can be adapted if, instead of the average number of events $\displaystyle{ \lambda }$, we are given a time rate $\displaystyle{ r }$ for the events to happen. Then $\displaystyle{ \lambda = r t }$ (with $\displaystyle{ r }$ in units of 1/time), and

This equation can be adapted if, instead of the average number of events $\displaystyle{ \lambda }$, we are given a time rate $\displaystyle{ r }$ for the events to happen. Then $\displaystyle{ \lambda = r t }$ (with $\displaystyle{ r }$ in units of 1/time), and

--fairywang(讨论)  【审校】“那么这个方程就可以适用”改为“那么这个方程就是适用的”
$\displaystyle{ P(k \text{ events in interval } t) = \frac{(r t)^k e^{-r t}}{k!} }$

< math > p (k text { events in interval } t) = frac {(r t) ^ k e ^ {-r t }{ k！{/math >

#### Examples of probability for Poisson distributions {泊松分佈 Poisson distribution概率的示例

On a particular river, overflow floods occur once every 100 years on average. Calculate the probability of k = 0, 1, 2, 3, 4, 5, or 6 overflow floods in a 100-year interval, assuming the Poisson model is appropriate.

On a particular river, overflow floods occur once every 100 years on average. Calculate the probability of = 0, 1, 2, 3, 4, 5, or 6 overflow floods in a 100-year interval, assuming the Poisson model is appropriate.

--fairywang(讨论)  【审校】“计算在100年间洪水泛滥次数= 0,1,2,3,4,5或6次的概率”

Because the average event rate is one overflow flood per 100 years, λ = 1

Because the average event rate is one overflow flood per 100 years, λ = 1

$\displaystyle{ P(k \text{ overflow floods in 100 years}) = \frac{\lambda^k e^{-\lambda}}{k!} = \frac{1^k e^{-1}}{k!} }$
$\displaystyle{ P(k \text{ overflow floods in 100 years}) = \frac{\lambda^k e^{-\lambda}}{k!} = \frac{1^k e^{-1}}{k!} }$

< math > p (k text { overflow in 100 years }) = frac { lambda ^ k e ^ {-lambda }}{ k! }= frac {1 ^ k e ^ {-1}{ k！{/math >

$\displaystyle{ P(k = 0 \text{ overflow floods in 100 years}) = \frac{1^0 e^{-1}}{0!} = \frac{e^{-1}}{1} \approx 0.368 }$
$\displaystyle{ P(k = 0 \text{ overflow floods in 100 years}) = \frac{1^0 e^{-1}}{0!} = \frac{e^{-1}}{1} \approx 0.368 }$

< math > p (k = 0 text { overflow flood in 100 years }) = frac {1 ^ 0 e ^ {-1}{0! }= frac { e ^ {-1}{1}约0.368 </math >

$\displaystyle{ P(k = 1 \text{ overflow flood in 100 years}) = \frac{1^1 e^{-1}}{1!} = \frac{e^{-1}}{1} \approx 0.368 }$
$\displaystyle{ P(k = 1 \text{ overflow flood in 100 years}) = \frac{1^1 e^{-1}}{1!} = \frac{e^{-1}}{1} \approx 0.368 }$

< math > p (k = 1 text { overflow flood in 100 years }) = frac {1 ^ 1 e ^ {-1}{1! }= frac { e ^ {-1}{1}约0.368 </math >

$\displaystyle{ P(k = 2 \text{ overflow floods in 100 years}) = \frac{1^2 e^{-1}}{2!} = \frac{e^{-1}}{2} \approx 0.184 }$
$\displaystyle{ P(k = 2 \text{ overflow floods in 100 years}) = \frac{1^2 e^{-1}}{2!} = \frac{e^{-1}}{2} \approx 0.184 }$

< math > p (k = 2 text { overflow flood in 100 years }) = frac {1 ^ 2 e ^ {-1}{2! }= frac { e ^ {-1}{2} approx 0.184 </math >

The table below gives the probability for 0 to 6 overflow floods in a 100-year period.

The table below gives the probability for 0 to 6 overflow floods in a 100-year period.

{ | class = “ wikitable”
k P(k overflow floods in 100 years) P( overflow floods in 100 years) 100年内泛滥成灾
0 0.368 0 0.368 0 0.368
1 0.368 1 0.368 1 0.368
2 0.184 2 0.184 2 0.184
3 0.061 3 0.061 3 0.061
4 0.015 4 0.015 4 0.015
5 0.003 5 0.003 5 0.003
6 0.0005 6 0.0005 6 0.0005

|}

Ugarte and colleagues report that the average number of goals in a World Cup soccer match is approximately 2.5 and the Poisson model is appropriate.模板:R

Ugarte and colleagues report that the average number of goals in a World Cup soccer match is approximately 2.5 and the Poisson model is appropriate.

Because the average event rate is 2.5 goals per match, λ = 2.5.

Because the average event rate is 2.5 goals per match, λ = 2.5.

$\displaystyle{ P(k \text{ goals in a match}) = \frac{2.5^k e^{-2.5}}{k!} }$
$\displaystyle{ P(k \text{ goals in a match}) = \frac{2.5^k e^{-2.5}}{k!} }$

< math > p (k text { goals in a match }) = frac {2.5 ^ k e ^ {-2.5}{ k！{/math >

$\displaystyle{ P(k = 0 \text{ goals in a match}) = \frac{2.5^0 e^{-2.5}}{0!} = \frac{e^{-2.5}}{1} \approx 0.082 }$
$\displaystyle{ P(k = 0 \text{ goals in a match}) = \frac{2.5^0 e^{-2.5}}{0!} = \frac{e^{-2.5}}{1} \approx 0.082 }$

= frac {2.5 ^ 0 e ^ {-2.5}{0! }= frac { e ^ {-2.5}{1}大约0.082

$\displaystyle{ P(k = 1 \text{ goal in a match}) = \frac{2.5^1 e^{-2.5}}{1!} = \frac{2.5 e^{-2.5}}{1} \approx 0.205 }$
$\displaystyle{ P(k = 1 \text{ goal in a match}) = \frac{2.5^1 e^{-2.5}}{1!} = \frac{2.5 e^{-2.5}}{1} \approx 0.205 }$

= frac {2.5 ^ 1 e ^ {-2.5}{1! }= frac {2.5 e ^ {-2.5}{1}约0.205

$\displaystyle{ P(k = 2 \text{ goals in a match}) = \frac{2.5^2 e^{-2.5}}{2!} = \frac{6.25 e^{-2.5}}{2} \approx 0.257 }$
$\displaystyle{ P(k = 2 \text{ goals in a match}) = \frac{2.5^2 e^{-2.5}}{2!} = \frac{6.25 e^{-2.5}}{2} \approx 0.257 }$

< math > p (k = 2 text { goals in a match }) = frac {2.5 ^ 2 e ^ {-2.5}{2! }= frac {6.25 e ^ {-2.5}{2}{2}{约0.257 </math >

The table below gives the probability for 0 to 7 goals in a match.

The table below gives the probability for 0 to 7 goals in a match.

{ | class = “ wikitable”
k P(k goals in a World Cup soccer match) P( goals in a World Cup soccer match) P (世界杯足球赛进球)
0 0.082 0 0.082 0 0.082
1 0.205 1 0.205 1 0.205
2 0.257 2 0.257 2 0.257
3 0.213 3 0.213 3 0.213
4 0.133 4 0.133 4 0.133
5 0.067 5 0.067 5 0.067
6 0.028 6 0.028 6 0.028
7 0.010 7 0.010 7 0.010

|}

#### Once in an interval events: The special case of λ = 1 and k = 0 事件唯一发生：λ = 1 与 k = 0的特殊情形

Suppose that astronomers estimate that large meteorites (above a certain size) hit the earth on average once every 100 years (λ = 1 event per 100 years), and that the number of meteorite hits follows a Poisson distribution. What is the probability of k = 0 meteorite hits in the next 100 years?

Suppose that astronomers estimate that large meteorites (above a certain size) hit the earth on average once every 100 years (λ = 1 event per 100 years), and that the number of meteorite hits follows a Poisson distribution. What is the probability of = 0 meteorite hits in the next 100 years?

--fairywang(讨论)  【审校】“而且陨石撞击的次数紧随泊松分布之后”改为“而且陨石撞击的次数紧随泊松分布之后”
$\displaystyle{ P(k = \text{0 meteorites hit in next 100 years}) = \frac{1^0 e^{-1}}{0!} = \frac{1}{e} \approx 0.37 }$
$\displaystyle{ P(k = \text{0 meteorites hit in next 100 years}) = \frac{1^0 e^{-1}}{0!} = \frac{1}{e} \approx 0.37 }$

< math > p (k = text {0 meteorites hit in next 100 years }) = frac {1 ^ 0 e ^ {-1}{0! }= frac {1}{ e }大约0.37 </math >

Under these assumptions, the probability that no large meteorites hit the earth in the next 100 years is roughly 0.37. The remaining 1 − 0.37 = 0.63 is the probability of 1, 2, 3, or more large meteorite hits in the next 100 years.

Under these assumptions, the probability that no large meteorites hit the earth in the next 100 years is roughly 0.37. The remaining 1 − 0.37 = 0.63 is the probability of 1, 2, 3, or more large meteorite hits in the next 100 years.

In an example above, an overflow flood occurred once every 100 years (λ = 1). The probability of no overflow floods in 100 years was roughly 0.37, by the same calculation.

In an example above, an overflow flood occurred once every 100 years (λ = 1). The probability of no overflow floods in 100 years was roughly 0.37, by the same calculation.

In general, if an event occurs on average once per interval (λ = 1), and the events follow a Poisson distribution, then P(0 events in next interval) = 0.37. In addition, P(exactly one event in next interval) = 0.37, as shown in the table for overflow floods.

In general, if an event occurs on average once per interval (λ = 1), and the events follow a Poisson distribution, then . In addition, P(exactly one event in next interval) = 0.37, as shown in the table for overflow floods.

### Examples that violate the Poisson assumptions 违反泊松假设的例子

The number of students who arrive at the student union per minute will likely not follow a Poisson distribution, because the rate is not constant (low rate during class time, high rate between class times) and the arrivals of individual students are not independent (students tend to come in groups).

The number of students who arrive at the student union per minute will likely not follow a Poisson distribution, because the rate is not constant (low rate during class time, high rate between class times) and the arrivals of individual students are not independent (students tend to come in groups).

The number of magnitude 5 earthquakes per year in a country may not follow a Poisson distribution if one large earthquake increases the probability of aftershocks of similar magnitude.

The number of magnitude 5 earthquakes per year in a country may not follow a Poisson distribution if one large earthquake increases the probability of aftershocks of similar magnitude.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

Examples in which at least one event is guaranteed are not Poission distributed; but may be modeled using a Zero-truncated Poisson distribution.

Examples in which at least one event is guaranteed are not Poission distributed; but may be modeled using a Zero-truncated Poisson distribution.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

Count distributions in which the number of intervals with zero events is higher than predicted by a Poisson model may be modeled using a Zero-inflated model.

Count distributions in which the number of intervals with zero events is higher than predicted by a Poisson model may be modeled using a Zero-inflated model.

## Properties 性能

### 描述统计学Descriptive statistics

• 一个泊松分布随机变量的期望值和方差均等于λ
$\displaystyle{ \operatorname{E}[|X-\lambda|]= \frac{2 \lambda^{\lfloor\lambda\rfloor + 1} e^{-\lambda}}{\lfloor\lambda\rfloor!}. }$

$\displaystyle{ \operatorname{E}[|X-\lambda|]= \frac{2 \lambda^{\lfloor\lambda\rfloor + 1} e^{-\lambda}}{\lfloor\lambda\rfloor!}. }$

[ | x-lambda | ] = frac {2 lambda ^ { lfloor lambda rfloor + 1} e ^ {-lambda }{ lfloor lambda rfloor！} . </math >

• The mode of a Poisson-distributed random variable with non-integer λ is equal to $\displaystyle{ \scriptstyle\lfloor \lambda \rfloor }$, which is the largest integer less than or equal to λ. This is also written as floor(λ). When λ is a positive integer, the modes are λ and λ − 1.
• All of the cumulants of the Poisson distribution are equal to the expected value λ. The nth factorial moment of the Poisson distribution is λn.
• The expected value of a Poisson process is sometimes decomposed into the product of intensity and exposure (or more generally expressed as the integral of an "intensity function" over time or space, sometimes described as “exposure”).模板:R
• 一个具有非整值λ 的泊松分布随机变量的统计值等于$\displaystyle{ \scriptstyle\lfloor \lambda \rfloor }$, 小于λ的最大整数。它也写作floor(λ). λ为正整数时，取值为λ 以及 λ − 1。
• 所有泊松分布的积均等于期望值 λ。泊松分布的n阶指数积为λn
• 期望值与泊松过程有时分解为“强度”与“面积”的乘积（或更一般地表示为强度函数随时间或空间的积分，有时描述为“暴露”“exposure”。）模板:R

### Median 中值

Bounds for the median ($\displaystyle{ \nu }$) of the distribution are known and are sharp:模板:R

Bounds for the median ($\displaystyle{ \nu }$) of the distribution are known and are sharp:

$\displaystyle{ \lambda - \ln 2 \le \nu \lt \lambda + \frac{1}{3}. }$
$\displaystyle{ \lambda - \ln 2 \le \nu \lt \lambda + \frac{1}{3}. }$

2 le nu < lambda + frac {1}{3}.数学

### Higher moments 高阶矩

$\displaystyle{ m_k = \sum_{i=0}^k \lambda^i \left\{\begin{matrix} k \\ i \end{matrix}\right\}, }$
$\displaystyle{ m_k = \sum_{i=0}^k \lambda^i \left\{\begin{matrix} k \\ i \end{matrix}\right\}, }$

< math > m _ k = sum { i = 0} ^ k lambda ^ i left { begin { matrix } k i end { matrix } right } ，</math >

• 泊松分布原点的高阶矩moments mk是同余多项式，λ中：
where the {braces} denote Stirling numbers of the second kind.模板:R模板:R The coefficients of the polynomials have a combinatorial meaning. In fact, when the expected value of the Poisson distribution is 1, then Dobinski's formula says that the nth moment equals the number of partitions of a set of size n.
where the {braces} denote Stirling numbers of the second kind. The coefficients of the polynomials have a combinatorial meaning. In fact, when the expected value of the Poisson distribution is 1, then Dobinski's formula says that the nth moment equals the number of partitions of a set of size n.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

For the non-centered moments we define $\displaystyle{ B=k/\lambda }$, then模板:R

For the non-centered moments we define $\displaystyle{ B=k/\lambda }$, then

$\displaystyle{ \lt math\gt 《数学》 E[X^k]^{1/k} \le C\cdot E[X^k]^{1/k} \le C\cdot E [ x ^ k ] ^ {1/k } le c dot \begin{cases} \begin{cases} 开始{ cases } k/B & \text{if}\quad B \lt e \\ k/B & \text{if}\quad B \lt e \\ k/B & text { if }方 b \lt e k/\log B & \text{if}\quad B\ge e k/\log B & \text{if}\quad B\ge e 如果你想要一个更好的工具，你需要一个更好的工具 \end{cases} \end{cases} 结束{ cases } }$

[/itex]

where $\displaystyle{ C }$ is some absolute constant greater than 0.

where $\displaystyle{ C }$ is some absolute constant greater than 0.

### Sums of Poisson-distributed random variables 泊松分布随机变量和

If $\displaystyle{ X_i \sim \operatorname{Pois}(\lambda_i) }$ for $\displaystyle{ i=1,\dotsc,n }$ are independent, then $\displaystyle{ \sum_{i=1}^n X_i \sim \operatorname{Pois}\left(\sum_{i=1}^n \lambda_i\right) }$.模板:R A converse is Raikov's theorem, which says that if the sum of two independent random variables is Poisson-distributed, then so are each of those two independent random variables.模板:R模板:R
If $\displaystyle{ X_i \sim \operatorname{Pois}(\lambda_i) }$ for $\displaystyle{ i=1,\dotsc,n }$ are independent, then $\displaystyle{ \sum_{i=1}^n X_i \sim \operatorname{Pois}\left(\sum_{i=1}^n \lambda_i\right) }$. A converse is Raikov's theorem, which says that if the sum of two independent random variables is Poisson-distributed, then so are each of those two independent random variables.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

### Other properties 其他特性

• The directed Kullback–Leibler divergence of $\displaystyle{ \operatorname{Pois}(\lambda_0) }$ from $\displaystyle{ \operatorname{Pois}(\lambda) }$ is given by
$\displaystyle{ \operatorname{D}_{\text{KL}}(\lambda\mid\lambda_0) = \lambda_0 - \lambda + \lambda \log \frac{\lambda}{\lambda_0}. }$
$\displaystyle{ \operatorname{D}_{\text{KL}}(\lambda\mid\lambda_0) = \lambda_0 - \lambda + \lambda \log \frac{\lambda}{\lambda_0}. }$

< math > operatorname { d }{ text { KL }}(lambda mid lambda _ 0) = lambda _ 0-lambda + lambda log frac { lambda }{ lambda _ 0} . </math >

• Bounds for the tail probabilities of a Poisson random variable $\displaystyle{ X \sim \operatorname{Pois}(\lambda) }$ can be derived using a Chernoff bound argument.模板:R
• 泊松随机变量尾概率的界$\displaystyle{ X \sim \operatorname{Pois}(\lambda) }$ 可以用[[ 切诺夫界Chernoff bound]]参数派生模板:R
$\displaystyle{ P(X \geq x) \leq \frac{(e \lambda)^x e^{-\lambda}}{x^x}, \text{ for } x \gt \lambda }$,
$\displaystyle{ P(X \geq x) \leq \frac{(e \lambda)^x e^{-\lambda}}{x^x}, \text{ for } x \gt \lambda }$,

(x x x) leq frac {(e lambda) ^ x e ^ {-lambda }{ x ^ x } ，text { for } x > lambda </math > ,

$\displaystyle{ P(X \leq x) \leq \frac{(e \lambda)^x e^{-\lambda} }{x^x}, \text{ for } x \lt \lambda. }$
$\displaystyle{ P(X \leq x) \leq \frac{(e \lambda)^x e^{-\lambda} }{x^x}, \text{ for } x \lt \lambda. }$

< math > p (x leq x) leq frac {(e lambda) ^ x e ^ {-lambda }{ x ^ x } ，text { for } x < lambda. </math >

• The upper tail probability can be tightened (by a factor of at least two) as follows:模板:R
• 长尾概率可被收紧（至少两倍）如下：模板:R
$\displaystyle{ P(X \geq x) \leq \frac{e^{-\operatorname{D}_{\text{KL}}(x\mid\lambda)}}{\max{(2, \sqrt{4\pi\operatorname{D}_{\text{KL}}(x\mid\lambda)}})}, \text{ for } x \gt \lambda, }$
$\displaystyle{ P(X \geq x) \leq \frac{e^{-\operatorname{D}_{\text{KL}}(x\mid\lambda)}}{\max{(2, \sqrt{4\pi\operatorname{D}_{\text{KL}}(x\mid\lambda)}})}, \text{ for } x \gt \lambda, }$

< math > p (x geq x) leq frac { e ^ {-operatorname { d }{ text { KL }(x mid lambda)}}{ max {(2，sqrt {4 pi operatorname { d }{ text { KL }(x mid lambda)}}}}) ，text { for } x > lambda，</math >

where $\displaystyle{ \operatorname{D}_{\text{KL}}(x\mid\lambda) }$ is the directed Kullback–Leibler divergence, as described above.
where $\displaystyle{ \operatorname{D}_{\text{KL}}(x\mid\lambda) }$ is the directed Kullback–Leibler divergence, as described above.

• Inequalities that relate the distribution function of a Poisson random variable $\displaystyle{ X \sim \operatorname{Pois}(\lambda) }$ to the Standard normal distribution function $\displaystyle{ \Phi(x) }$ are as follows:模板:R
• 关于泊松随机变量分布函数的不等式 $\displaystyle{ X \sim \operatorname{Pois}(\lambda) }$与 标准正态分布函数$\displaystyle{ \Phi(x) }$ 如下:模板:R
$\displaystyle{ \Phi\left(\operatorname{sign}(k-\lambda)\sqrt{2\operatorname{D}_{\text{KL}}(k\mid\lambda)}\right) \lt P(X \leq k) \lt \Phi\left(\operatorname{sign}(k-\lambda+1)\sqrt{2\operatorname{D}_{\text{KL}}(k+1\mid\lambda)}\right), \text{ for } k \gt 0, }$
$\displaystyle{ \Phi\left(\operatorname{sign}(k-\lambda)\sqrt{2\operatorname{D}_{\text{KL}}(k\mid\lambda)}\right) \lt P(X \leq k) \lt \Phi\left(\operatorname{sign}(k-\lambda+1)\sqrt{2\operatorname{D}_{\text{KL}}(k+1\mid\lambda)}\right), \text{ for } k \gt 0, }$

< math > Phi left (operatorname { sign }(k-lambda) sqrt {2 operatorname { d }{ text { KL }(k mid lambda)}右) < p (x leq k) < left (operatorname { sign }(k-lambda + 1) sqrt {2 operatorname { d }{ text { KL }}}(k + 1 mid lambda)}右) ，text { for } k > 0，</math >

where $\displaystyle{ \operatorname{D}_{\text{KL}}(k\mid\lambda) }$ is again the directed Kullback–Leibler divergence.
where $\displaystyle{ \operatorname{D}_{\text{KL}}(k\mid\lambda) }$ is again the directed Kullback–Leibler divergence.

### Poisson races 泊松族群

Let $\displaystyle{ X \sim \operatorname{Pois}(\lambda) }$ and $\displaystyle{ Y \sim \operatorname{Pois}(\mu) }$ be independent random variables, with $\displaystyle{ \lambda \lt \mu }$, then we have that

Let $\displaystyle{ X \sim \operatorname{Pois}(\lambda) }$ and $\displaystyle{ Y \sim \operatorname{Pois}(\mu) }$ be independent random variables, with $\displaystyle{ \lambda \lt \mu }$, then we have that

$\displaystyle{ \lt math\gt 《数学》 \frac{e^{-(\sqrt{\mu} -\sqrt{\lambda})^2 }}{(\lambda + \mu)^2} - \frac{e^{-(\lambda + \mu)}}{2\sqrt{\lambda \mu}} - \frac{e^{-(\lambda + \mu)}}{4\lambda \mu} \leq P(X - Y \geq 0) \leq e^{- (\sqrt{\mu} -\sqrt{\lambda})^2} \frac{e^{-(\sqrt{\mu} -\sqrt{\lambda})^2 }}{(\lambda + \mu)^2} - \frac{e^{-(\lambda + \mu)}}{2\sqrt{\lambda \mu}} - \frac{e^{-(\lambda + \mu)}}{4\lambda \mu} \leq P(X - Y \geq 0) \leq e^{- (\sqrt{\mu} -\sqrt{\lambda})^2} Frac { e ^ {-(sqrt { mu }-sqrt { lambda }) ^ 2}{(lambda + mu) ^ 2}-frac { e ^ {-(lambda + mu)}}{2 sqrt { lambda mu }-frac { e ^ { e ^ {-(lambda + mu)}-(lambda + mu)}}}{4 lambda } leq p (x-y geq 0) leq ^ e ^ {-(sqrt { mu }-sqrt { lambda }) ^ 2} }$

[/itex]

The upper bound is proved using a standard Chernoff bound.

The upper bound is proved using a standard Chernoff bound.

The lower bound can be proved by noting that $\displaystyle{ P(X-Y\geq0\mid X+Y=i) }$ is the probability that $\displaystyle{ Z \geq \frac{i}{2} }$, where $\displaystyle{ Z \sim \operatorname{Bin}\left(i, \frac{\lambda}{\lambda+\mu}\right) }$, which is bounded below by $\displaystyle{ \frac{1}{(i+1)^2} e^{\left(-iD\left(0.5 \| \frac{\lambda}{\lambda+\mu}\right)\right)} }$, where $\displaystyle{ D }$ is relative entropy (See the entry on bounds on tails of binomial distributions for details). Further noting that $\displaystyle{ X+Y \sim \operatorname{Pois}(\lambda+\mu) }$, and computing a lower bound on the unconditional probability gives the result. More details can be found in the appendix of Kamath et al..模板:R

The lower bound can be proved by noting that $\displaystyle{ P(X-Y\geq0\mid X+Y=i) }$ is the probability that $\displaystyle{ Z \geq \frac{i}{2} }$, where $\displaystyle{ Z \sim \operatorname{Bin}\left(i, \frac{\lambda}{\lambda+\mu}\right) }$, which is bounded below by $\displaystyle{ \frac{1}{(i+1)^2} e^{\left(-iD\left(0.5 \| \frac{\lambda}{\lambda+\mu}\right)\right)} }$, where $\displaystyle{ D }$ is relative entropy (See the entry on bounds on tails of binomial distributions for details). Further noting that $\displaystyle{ X+Y \sim \operatorname{Pois}(\lambda+\mu) }$, and computing a lower bound on the unconditional probability gives the result. More details can be found in the appendix of Kamath et al..

## ' 相关分布'Related distributions

### Genera通常l

--fairywang(讨论)  【审校】“Genera通常l”改为“General通常”
• If $\displaystyle{ X_1 \sim \mathrm{Pois}(\lambda_1)\, }$ and $\displaystyle{ X_2 \sim \mathrm{Pois}(\lambda_2)\, }$ are independent, then the difference $\displaystyle{ Y = X_1 - X_2 }$ follows a Skellam distribution.
• If $\displaystyle{ X_1 \sim \mathrm{Pois}(\lambda_1)\, }$ and $\displaystyle{ X_2 \sim \mathrm{Pois}(\lambda_2)\, }$ are independent, then the distribution of $\displaystyle{ X_1 }$ conditional on $\displaystyle{ X_1+X_2 }$ is a binomial distribution.
Specifically, if $\displaystyle{ X_1+X_2=k }$, then $\displaystyle{ \!X_1\sim \mathrm{Binom}(k, \lambda_1/(\lambda_1+\lambda_2)) }$.

Specifically, if $\displaystyle{ X_1+X_2=k }$, then $\displaystyle{ \!X_1\sim \mathrm{Binom}(k, \lambda_1/(\lambda_1+\lambda_2)) }$.

More generally, if X1, X2,..., Xn are independent Poisson random variables with parameters λ1, λ2,..., λn then

More generally, if X1, X2,..., Xn are independent Poisson random variables with parameters λ1, λ2,..., λn then

given $\displaystyle{ \sum_{j=1}^n X_j=k, }$ $\displaystyle{ X_i \sim \mathrm{Binom}\left(k, \frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right) }$. In fact, $\displaystyle{ \{X_i\} \sim \mathrm{Multinom}\left(k, \left\{\frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right\}\right) }$.
given $\displaystyle{ \sum_{j=1}^n X_j=k, }$ $\displaystyle{ X_i \sim \mathrm{Binom}\left(k, \frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right) }$. In fact, $\displaystyle{ \{X_i\} \sim \mathrm{Multinom}\left(k, \left\{\frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right\}\right) }$.

• If $\displaystyle{ X \sim \mathrm{Pois}(\lambda)\, }$ and the distribution of $\displaystyle{ Y }$, conditional on X = k, is a binomial distribution, $\displaystyle{ Y \mid (X = k) \sim \mathrm{Binom}(k, p) }$, then the distribution of Y follows a Poisson distribution $\displaystyle{ Y \sim \mathrm{Pois}(\lambda \cdot p)\, }$. In fact, if $\displaystyle{ \{Y_i\} }$, conditional on X = k, follows a multinomial distribution, $\displaystyle{ \{Y_i\} \mid (X = k) \sim \mathrm{Multinom}\left(k, p_i\right) }$, then each $\displaystyle{ Y_i }$ follows an independent Poisson distribution $\displaystyle{ Y_i \sim \mathrm{Pois}(\lambda \cdot p_i), \rho(Y_i, Y_j) = 0 }$.
• The Poisson distribution can be derived as a limiting case to the binomial distribution as the number of trials goes to infinity and the expected number of successes remains fixed — see law of rare events below. Therefore, it can be used as an approximation of the binomial distribution if n is sufficiently large and p is sufficiently small. There is a rule of thumb stating that the Poisson distribution is a good approximation of the binomial distribution if n is at least 20 and p is smaller than or equal to 0.05, and an excellent approximation if n ≥ 100 and np ≤ 10.模板:R
$\displaystyle{ F_\mathrm{Binomial}(k;n, p) \approx F_\mathrm{Poisson}(k;\lambda=np)\, }$
$\displaystyle{ F_\mathrm{Binomial}(k;n, p) \approx F_\mathrm{Poisson}(k;\lambda=np)\, }$

(k; n，p)接近 f _ mathrm { Poisson }(k; lambda = np) ，</math >

• The Poisson distribution is a special case of the discrete compound Poisson distribution (or stuttering Poisson distribution) with only a parameter.模板:R The discrete compound Poisson distribution can be deduced from the limiting distribution of univariate multinomial distribution. It is also a special case of a compound Poisson distribution.
• 这一泊松分布是离散复合泊松分布（或断续泊松分布）在只有一个参数情况下的特殊情形模板:R离散复合泊松分布可由一元多项式分布的极限分布导出。同时它也是复合泊松分布#特殊情况 复合泊松分布的一个特例。
• For sufficiently large values of λ, (say λ>1000), the normal distribution with mean λ and variance λ (standard deviation $\displaystyle{ \sqrt{\lambda} }$) is an excellent approximation to the Poisson distribution. If λ is greater than about 10, then the normal distribution is a good approximation if an appropriate continuity correction is performed, i.e., if P(X ≤ x), where x is a non-negative integer, is replaced by P(X ≤ x + 0.5).
• 对于足够大的值λ，（如 λ>1000)，具有均值 λ 的正态分布与变量 λ （标准差 $\displaystyle{ \sqrt{\lambda} }$），是泊松分布的完美近似。如果 λ 大于10，则正态分布在适当的校正下可近似模拟，例如如果P(X ≤ x)，x 为非负整数，则将其改为P(X ≤ x + 0.5)。
$\displaystyle{ F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)\, }$
$\displaystyle{ F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)\, }$

(x; λ) approx f _ mathrm { normal }(x; mu = lambda，sigma ^ 2 = lambda) ，</math >

$\displaystyle{ Y = 2 \sqrt{X} \approx \mathcal{N}(2\sqrt{\lambda};1) }$,模板:R

$\displaystyle{ Y = 2 \sqrt{X} \approx \mathcal{N}(2\sqrt{\lambda};1) }$,

(2 sqrt { lambda } ; 1) </math > ,

and

and

$\displaystyle{ Y = \sqrt{X} \approx \mathcal{N}(\sqrt{\lambda};1/4) }$.模板:R

$\displaystyle{ Y = \sqrt{X} \approx \mathcal{N}(\sqrt{\lambda};1/4) }$.

(sqrt { lambda } ; 1/4) </math > .

Under this transformation, the convergence to normality (as $\displaystyle{ \lambda }$ increases) is far faster than the untransformed variable.[citation needed] Other, slightly more complicated, variance stabilizing transformations are available,模板:R one of which is Anscombe transform.模板:R See Data transformation (statistics) for more general uses of transformations.

Under this transformation, the convergence to normality (as $\displaystyle{ \lambda }$ increases) is far faster than the untransformed variable. Other, slightly more complicated, variance stabilizing transformations are available, one of which is Anscombe transform. See Data transformation (statistics) for more general uses of transformations.

• If for every t > 0 the number of arrivals in the time interval [0, t] follows the Poisson distribution with mean λt, then the sequence of inter-arrival times are independent and identically distributed exponential random variables having mean 1/λ.模板:R
$\displaystyle{ F_\text{Poisson}(k;\lambda) = 1-F_{\chi^2}(2\lambda;2(k+1)) \quad\quad \text{ integer } k, }$

$\displaystyle{ F_\text{Poisson}(k;\lambda) = 1-F_{\chi^2}(2\lambda;2(k+1)) \quad\quad \text{ integer } k, }$

1-F _ { chi ^ 2}(2 lambda; 2(k + 1)) quad text { integer } k，</math >

and模板:R

and

$\displaystyle{ \Pr(X=k)=F_{\chi^2}(2\lambda;2(k+1)) -F_{\chi^2}(2\lambda;2k) . \lt math\gt \Pr(X=k)=F_{\chi^2}(2\lambda;2(k+1)) -F_{\chi^2}(2\lambda;2k) . \lt math \gt Pr (x = k) = f { chi ^ 2}(2 lambda; 2(k + 1))-f { chi ^ 2}(2 lambda; 2k). }$

[/itex]

### Poisson Approximation 泊松近似

Assume $\displaystyle{ X_1\sim\operatorname{Pois}(\lambda_1), X_2\sim\operatorname{Pois}(\lambda_2), \dots, X_n\sim\operatorname{Pois}(\lambda_n) }$ where $\displaystyle{ \lambda_1 + \lambda_2 + \dots + \lambda_n=1 }$, then[3] $\displaystyle{ (X_1, X_2, \dots, X_n) }$ is multinomially distributed

Assume $\displaystyle{ X_1\sim\operatorname{Pois}(\lambda_1), X_2\sim\operatorname{Pois}(\lambda_2), \dots, X_n\sim\operatorname{Pois}(\lambda_n) }$ where $\displaystyle{ \lambda_1 + \lambda_2 + \dots + \lambda_n=1 }$, then $\displaystyle{ (X_1, X_2, \dots, X_n) }$ is multinomially distributed

$\displaystyle{ (X_1, X_2, \dots, X_n) \sim \operatorname{Mult}(N, \lambda_1, \lambda_2, \dots, \lambda_n) }$ conditioned on $\displaystyle{ N = X_1 + X_2 + \dots X_n }$.

$\displaystyle{ (X_1, X_2, \dots, X_n) \sim \operatorname{Mult}(N, \lambda_1, \lambda_2, \dots, \lambda_n) }$ conditioned on $\displaystyle{ N = X_1 + X_2 + \dots X_n }$.

(x _ 1，x _ 2，dots，x _ n) sim 操作员名称{ Mult }(n，λ _ 1，λ _ 2，dots，λ _ n) </math > 取决于 < math > n = x1 + x _ 2 + dots x _ n </math > 。

This means模板:R, among other things, that for any nonnegative function $\displaystyle{ f(x_1,x_2,\dots,x_n) }$,

This means, among other things, that for any nonnegative function $\displaystyle{ f(x_1,x_2,\dots,x_n) }$,

if $\displaystyle{ (Y_1, Y_2, \dots, Y_n)\sim\operatorname{Mult}(m, \mathbf{p}) }$ is multinomially distributed, then

if $\displaystyle{ (Y_1, Y_2, \dots, Y_n)\sim\operatorname{Mult}(m, \mathbf{p}) }$ is multinomially distributed, then

$\displaystyle{ \lt math\gt 《数学》 \operatorname{E}[f(Y_1, Y_2, \dots, Y_n)] \le e\sqrt{m}\operatorname{E}[f(X_1, X_2, \dots, X_n)] \operatorname{E}[f(Y_1, Y_2, \dots, Y_n)] \le e\sqrt{m}\operatorname{E}[f(X_1, X_2, \dots, X_n)] 操作员名称{ e }[ f (y _ 1，y _ 2，dots，y _ n)] le e sqrt { m }操作员名称{ e }[ f (x _ 1，x _ 2，dots，x _ n)] }$

[/itex]

where $\displaystyle{ (X_1, X_2, \dots, X_n)\sim\operatorname{Pois}(\mathbf{p}) }$.

where $\displaystyle{ (X_1, X_2, \dots, X_n)\sim\operatorname{Pois}(\mathbf{p}) }$.

The factor of $\displaystyle{ e\sqrt{m} }$ can be removed if $\displaystyle{ f }$ is further assumed to be monotonically increasing or decreasing.

The factor of $\displaystyle{ e\sqrt{m} }$ can be removed if $\displaystyle{ f }$ is further assumed to be monotonically increasing or decreasing.

### 二元泊松分布Bivariate Poisson distribution

This distribution has been extended to the bivariate case.模板:R The generating function for this distribution is

This distribution has been extended to the bivariate case. The generating function for this distribution is

$\displaystyle{ g( u, v ) = \exp[ ( \theta_1 - \theta_{12} )( u - 1 ) + ( \theta_2 - \theta_{12} )(v - 1) + \theta_{12} ( uv - 1 ) ] }$
$\displaystyle{ g( u, v ) = \exp[ ( \theta_1 - \theta_{12} )( u - 1 ) + ( \theta_2 - \theta_{12} )(v - 1) + \theta_{12} ( uv - 1 ) ] }$

< math > g (u，v) = exp [(theta _ 1-theta _ {12})(u-1) + (theta _ 2-theta _ {12})(v-1) + theta _ {12}(uv-1)] </math >

with

with

$\displaystyle{ \theta_1, \theta_2 \gt \theta_{ 12 } \gt 0 \, }$
$\displaystyle{ \theta_1, \theta_2 \gt \theta_{ 12 } \gt 0 \, }$

1，theta 2，theta {12} > 0，</math >

The marginal distributions are Poisson(θ1) and Poisson(θ2) and the correlation coefficient is limited to the range

The marginal distributions are Poisson(θ1) and Poisson(θ2) and the correlation coefficient is limited to the range

$\displaystyle{ 0 \le \rho \le \min\left\{ \frac{ \theta_1 }{ \theta_2 }, \frac{ \theta_2 }{ \theta_1 } \right\} }$
$\displaystyle{ 0 \le \rho \le \min\left\{ \frac{ \theta_1 }{ \theta_2 }, \frac{ \theta_2 }{ \theta_1 } \right\} }$

[数学][数学][数学][数学]

A simple way to generate a bivariate Poisson distribution $\displaystyle{ X_1,X_2 }$ is to take three independent Poisson distributions $\displaystyle{ Y_1,Y_2,Y_3 }$ with means $\displaystyle{ \lambda_1,\lambda_2,\lambda_3 }$ and then set $\displaystyle{ X_1 = Y_1 + Y_3,X_2 = Y_2 + Y_3 }$. The probability function of the bivariate Poisson distribution is

A simple way to generate a bivariate Poisson distribution $\displaystyle{ X_1,X_2 }$ is to take three independent Poisson distributions $\displaystyle{ Y_1,Y_2,Y_3 }$ with means $\displaystyle{ \lambda_1,\lambda_2,\lambda_3 }$ and then set $\displaystyle{ X_1 = Y_1 + Y_3,X_2 = Y_2 + Y_3 }$. The probability function of the bivariate Poisson distribution is

--fairywang(讨论)  【审校】“泊松分佈分布”改为“泊松分布”，“泊松分佈变量”改为“泊松分布变量”
\displaystyle{ \lt math\gt 《数学》 \begin{align} \begin{align} 开始{ align } & \Pr(X_1=k_1,X_2=k_2) \\ & \Pr(X_1=k_1,X_2=k_2) \\ & Pr (x _ 1 = k _ 1，x _ 2 = k _ 2) = {} & \exp\left(-\lambda_1-\lambda_2-\lambda_3\right) \frac{\lambda_1^{k_1}}{k_1!} \frac{\lambda_2^{k_2}}{k_2!} \sum_{k=0}^{\min(k_1,k_2)} \binom{k_1}{k} \binom{k_2}{k} k! \left( \frac{\lambda_3}{\lambda_1\lambda_2}\right)^k = {} & \exp\left(-\lambda_1-\lambda_2-\lambda_3\right) \frac{\lambda_1^{k_1}}{k_1!} \frac{\lambda_2^{k_2}}{k_2!} \sum_{k=0}^{\min(k_1,k_2)} \binom{k_1}{k} \binom{k_2}{k} k! \left( \frac{\lambda_3}{\lambda_1\lambda_2}\right)^k = {} & exp left (- lambda _ 1-lambda _ 2-lambda _ 3 right) frac { lambda _ 1 ^ { k _ 1}{ k _ 1! }2 ^ { k2}{ k2! }{ k = 0} ^ { min (k _ 1，k _ 2)} binom { k _ 1}{ k } binom { k _ 2}{ k } ！左(frac { lambda _ 3}{ lambda _ 1 lambda _ 2}右) ^ k \end{align} \end{align} 结束{ align } }

[/itex]

### 自由泊松分布Free Poisson distribution

The free Poisson distribution[4] with jump size $\displaystyle{ \alpha }$ and rate $\displaystyle{ \lambda }$ arises in free probability theory as the limit of repeated free convolution

The free Poisson distribution with jump size $\displaystyle{ \alpha }$ and rate $\displaystyle{ \lambda }$ arises in free probability theory as the limit of repeated free convolution

$\displaystyle{ \lt math\gt 《数学》 \left( \left(1-\frac{\lambda}{N}\right)\delta_0 + \frac{\lambda}{N}\delta_\alpha\right)^{\boxplus N} }$

\left( \left(1-\frac{\lambda}{N}\right)\delta_0 + \frac{\lambda}{N}\delta_\alpha\right)^{\boxplus N}[/itex]

as N → ∞.

as N → ∞.

as N → ∞.

In other words, let $\displaystyle{ X_N }$ be random variables so that $\displaystyle{ X_N }$ has value $\displaystyle{ \alpha }$ with probability $\displaystyle{ \frac{\lambda}{N} }$ and value 0 with the remaining probability. Assume also that the family $\displaystyle{ X_1,X_2,\ldots }$ are freely independent. Then the limit as $\displaystyle{ N\to\infty }$ of the law of $\displaystyle{ X_1+\cdots +X_N }$

In other words, let $\displaystyle{ X_N }$ be random variables so that $\displaystyle{ X_N }$ has value $\displaystyle{ \alpha }$ with probability $\displaystyle{ \frac{\lambda}{N} }$ and value 0 with the remaining probability. Assume also that the family $\displaystyle{ X_1,X_2,\ldots }$ are freely independent. Then the limit as $\displaystyle{ N\to\infty }$ of the law of $\displaystyle{ X_1+\cdots +X_N }$

--fairywang(讨论)  【审校】“假设家庭成员 ”改为“假设集合 ”

is given by the Free Poisson law with parameters $\displaystyle{ \lambda,\alpha }$.

is given by the Free Poisson law with parameters $\displaystyle{ \lambda,\alpha }$.

This definition is analogous to one of the ways in which the classical Poisson distribution is obtained from a (classical) Poisson process.

This definition is analogous to one of the ways in which the classical Poisson distribution is obtained from a (classical) Poisson process.

--fairywang(讨论)  【审校】“泊松分佈 ”改为“泊松分布 ”

The measure associated to the free Poisson law is given by[5]

The measure associated to the free Poisson law is given by 与自由泊松定律相关的测度由以下给出

$\displaystyle{ \mu=\begin{cases} (1-\lambda) \delta_0 + \lambda \nu,& \text{if } 0\leq \lambda \leq 1 \\ \lt math\gt \mu=\begin{cases} (1-\lambda) \delta_0 + \lambda \nu,& \text{if } 0\leq \lambda \leq 1 \\ \lt math \gt mu = begin { cases }(1-lambda) delta _ 0 + lambda nu，& text { if }0 leq lambda leq 1 \nu, & \text{if }\lambda \gt 1, \nu, & \text{if }\lambda \gt 1, 1，& text { if } lambda \gt 1, \end{cases} \end{cases} 结束{ cases } }$

[/itex]

where

where

$\displaystyle{ \nu = \frac{1}{2\pi\alpha t}\sqrt{4\lambda \alpha^2 - ( t - \alpha (1+\lambda))^2} \, dt }$
$\displaystyle{ \nu = \frac{1}{2\pi\alpha t}\sqrt{4\lambda \alpha^2 - ( t - \alpha (1+\lambda))^2} \, dt }$

4 lambda alpha ^ 2-(t-alpha (1 + lambda)) ^ 2} ，dt </math >

and has support $\displaystyle{ [\alpha (1-\sqrt{\lambda})^2,\alpha (1+\sqrt{\lambda})^2] }$.

and has support $\displaystyle{ [\alpha (1-\sqrt{\lambda})^2,\alpha (1+\sqrt{\lambda})^2] }$.

This law also arises in random matrix theory as the Marchenko–Pastur law. Its free cumulants are equal to $\displaystyle{ \kappa_n=\lambda\alpha^n }$.

This law also arises in random matrix theory as the Marchenko–Pastur law. Its free cumulants are equal to $\displaystyle{ \kappa_n=\lambda\alpha^n }$.

#### Some transforms of this law这一定律的一些变换

We give values of some important transforms of the free Poisson law; the computation can be found in e.g. in the book Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher[6]

We give values of some important transforms of the free Poisson law; the computation can be found in e.g. in the book Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher

The R-transform of the free Poisson law is given by

The R-transform of the free Poisson law is given by

$\displaystyle{ R(z)=\frac{\lambda \alpha}{1-\alpha z}. }$
$\displaystyle{ R(z)=\frac{\lambda \alpha}{1-\alpha z}. }$

1-alpha z }.数学

The Cauchy transform (which is the negative of the Stieltjes transformation) is given by

The Cauchy transform (which is the negative of the Stieltjes transformation) is given by

$\displaystyle{ \lt math\gt 《数学》 G(z) = \frac{ z + \alpha - \lambda \alpha - \sqrt{ (z-\alpha (1+\lambda))^2 - 4 \lambda \alpha^2}}{2\alpha z} G(z) = \frac{ z + \alpha - \lambda \alpha - \sqrt{ (z-\alpha (1+\lambda))^2 - 4 \lambda \alpha^2}}{2\alpha z} G (z) = frac { z + alpha-lambda alpha-sqrt {(z-alpha (1 + lambda)))) ^ 2-4 lambda alpha ^ 2}{2 alpha z } }$

[/itex]

The S-transform is given by

The S-transform is given by

$\displaystyle{ \lt math\gt 《数学》 S(z) = \frac{1}{z+\lambda} S(z) = \frac{1}{z+\lambda} (z) = frac {1}{ z + lambda } }$

[/itex]

in the case that $\displaystyle{ \alpha=1 }$.

in the case that $\displaystyle{ \alpha=1 }$.

## Statistical Inference 统计学推论

### Parameter estimation参数估计

Given a sample of n measured values $\displaystyle{ k_i \in \{0,1,...\} }$, for i = 1, ..., n, we wish to estimate the value of the parameter λ of the Poisson population from which the sample was drawn. The maximum likelihood estimate is [7]

Given a sample of n measured values $\displaystyle{ k_i \in \{0,1,...\} }$, for i = 1, ..., n, we wish to estimate the value of the parameter λ of the Poisson population from which the sample was drawn. The maximum likelihood estimate is

$\displaystyle{ \widehat{\lambda}_\mathrm{MLE}=\frac{1}{n}\sum_{i=1}^n k_i. \! }$
$\displaystyle{ \widehat{\lambda}_\mathrm{MLE}=\frac{1}{n}\sum_{i=1}^n k_i. \! }$

1}{ n } sum { i = 1} ^ n ki.! 数学

Since each observation has expectation λ so does the sample mean. Therefore, the maximum likelihood estimate is an unbiased estimator of λ. It is also an efficient estimator since its variance achieves the Cramér–Rao lower bound (CRLB).[citation needed] Hence it is minimum-variance unbiased. Also it can be proven that the sum (and hence the sample mean as it is a one-to-one function of the sum) is a complete and sufficient statistic for λ.

Since each observation has expectation λ so does the sample mean. Therefore, the maximum likelihood estimate is an unbiased estimator of λ. It is also an efficient estimator since its variance achieves the Cramér–Rao lower bound (CRLB). Hence it is minimum-variance unbiased. Also it can be proven that the sum (and hence the sample mean as it is a one-to-one function of the sum) is a complete and sufficient statistic for λ.

To prove sufficiency we may use the factorization theorem. Consider partitioning the probability mass function of the joint Poisson distribution for the sample into two parts: one that depends solely on the sample $\displaystyle{ \mathbf{x} }$ (called $\displaystyle{ h(\mathbf{x}) }$) and one that depends on the parameter $\displaystyle{ \lambda }$ and the sample $\displaystyle{ \mathbf{x} }$ only through the function $\displaystyle{ T(\mathbf{x}) }$. Then $\displaystyle{ T(\mathbf{x}) }$ is a sufficient statistic for $\displaystyle{ \lambda }$.

To prove sufficiency we may use the factorization theorem. Consider partitioning the probability mass function of the joint Poisson distribution for the sample into two parts: one that depends solely on the sample $\displaystyle{ \mathbf{x} }$ (called $\displaystyle{ h(\mathbf{x}) }$) and one that depends on the parameter $\displaystyle{ \lambda }$ and the sample $\displaystyle{ \mathbf{x} }$ only through the function $\displaystyle{ T(\mathbf{x}) }$. Then $\displaystyle{ T(\mathbf{x}) }$ is a sufficient statistic for $\displaystyle{ \lambda }$.

$\displaystyle{ P(\mathbf{x})=\prod_{i=1}^n\frac{\lambda^{x_i} e^{-\lambda}}{x_i!}=\frac{1}{\prod_{i=1}^n x_i!} \times \lambda^{\sum_{i=1}^n x_i}e^{-n\lambda} }$
$\displaystyle{ P(\mathbf{x})=\prod_{i=1}^n\frac{\lambda^{x_i} e^{-\lambda}}{x_i!}=\frac{1}{\prod_{i=1}^n x_i!} \times \lambda^{\sum_{i=1}^n x_i}e^{-n\lambda} }$

= prod _ { i = 1} ^ n frac { lambda ^ { x _ i } e ^ {-lambda }{ x _ i！1}{ prod { i = 1} ^ n x i! }乘以 lambda ^ { sum { i = 1} ^ n xi } e ^ {-n lambda } </math >

The first term, $\displaystyle{ h(\mathbf{x}) }$, depends only on $\displaystyle{ \mathbf{x} }$. The second term, $\displaystyle{ g(T(\mathbf{x})|\lambda) }$, depends on the sample only through $\displaystyle{ T(\mathbf{x})=\sum_{i=1}^nx_i }$. Thus, $\displaystyle{ T(\mathbf{x}) }$ is sufficient.

The first term, $\displaystyle{ h(\mathbf{x}) }$, depends only on $\displaystyle{ \mathbf{x} }$. The second term, $\displaystyle{ g(T(\mathbf{x})|\lambda) }$, depends on the sample only through $\displaystyle{ T(\mathbf{x})=\sum_{i=1}^nx_i }$. Thus, $\displaystyle{ T(\mathbf{x}) }$ is sufficient.

To find the parameter λ that maximizes the probability function for the Poisson population, we can use the logarithm of the likelihood function:

To find the parameter λ that maximizes the probability function for the Poisson population, we can use the logarithm of the likelihood function:

\displaystyle{ \begin{align} \ell(\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{align} }
\displaystyle{ \begin{align} \ell(\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{align} }

{ align } ell (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!).结束{ align } </math >

We take the derivative of $\displaystyle{ \ell }$ with respect to λ and compare it to zero:

We take the derivative of $\displaystyle{ \ell }$ with respect to λ and compare it to zero:

$\displaystyle{ \frac{\mathrm{d}}{\mathrm{d}\lambda} \ell(\lambda) = 0 \iff -n + \left(\sum_{i=1}^n k_i\right) \frac{1}{\lambda} = 0. \! }$
$\displaystyle{ \frac{\mathrm{d}}{\mathrm{d}\lambda} \ell(\lambda) = 0 \iff -n + \left(\sum_{i=1}^n k_i\right) \frac{1}{\lambda} = 0. \! }$

1}{ lambda } = 0 iff-n + left (sum { i = 1} ^ n k i right) frac {1}{ lambda } = 0.! 数学

Solving for λ gives a stationary point.

Solving for λ gives a stationary point.

$\displaystyle{ \lambda = \frac{\sum_{i=1}^n k_i}{n} }$
$\displaystyle{ \lambda = \frac{\sum_{i=1}^n k_i}{n} }$

{ math > lambda = frac { sum { i = 1} ^ n k _ i }{ n } </math >

So λ is the average of the ki values. Obtaining the sign of the second derivative of L at the stationary point will determine what kind of extreme value λ is.

So λ is the average of the ki values. Obtaining the sign of the second derivative of L at the stationary point will determine what kind of extreme value λ is.

K < sub > i 值的平均值也是如此。在驻点得到 l 的二阶导数的符号将决定什么是极值。

$\displaystyle{ \frac{\partial^2 \ell}{\partial \lambda^2} = -\lambda^{-2}\sum_{i=1}^n k_i }$
$\displaystyle{ \frac{\partial^2 \ell}{\partial \lambda^2} = -\lambda^{-2}\sum_{i=1}^n k_i }$

{ partial ^ 2 ell }{ partial lambda ^ 2} =-lambda ^ {-2} sum { i = 1} ^ n k i </math >

Evaluating the second derivative at the stationary point gives:

Evaluating the second derivative at the stationary point gives:

$\displaystyle{ \frac{\partial^2 \ell}{\partial \lambda^2} = - \frac{n^2}{\sum_{i=1}^n k_i} }$
$\displaystyle{ \frac{\partial^2 \ell}{\partial \lambda^2} = - \frac{n^2}{\sum_{i=1}^n k_i} }$

{ partial ^ 2 ell }{ partial lambda ^ 2} =-frac { n ^ 2}{ sum { i = 1} ^ n ki } </math >

which is the negative of n times the reciprocal of the average of the ki. This expression is negative when the average is positive. If this is satisfied, then the stationary point maximizes the probability function.

which is the negative of n times the reciprocal of the average of the ki. This expression is negative when the average is positive. If this is satisfied, then the stationary point maximizes the probability function.

For completeness, a family of distributions is said to be complete if and only if $\displaystyle{ E(g(T)) = 0 }$ implies that $\displaystyle{ P_\lambda(g(T) = 0) = 1 }$ for all $\displaystyle{ \lambda }$. If the individual $\displaystyle{ X_i }$ are iid $\displaystyle{ \mathrm{Po}(\lambda) }$, then $\displaystyle{ T(\mathbf{x})=\sum_{i=1}^nX_i\sim \mathrm{Po}(n\lambda) }$. Knowing the distribution we want to investigate, it is easy to see that the statistic is complete.

For completeness, a family of distributions is said to be complete if and only if $\displaystyle{ E(g(T)) = 0 }$ implies that $\displaystyle{ P_\lambda(g(T) = 0) = 1 }$ for all $\displaystyle{ \lambda }$. If the individual $\displaystyle{ X_i }$ are iid $\displaystyle{ \mathrm{Po}(\lambda) }$, then $\displaystyle{ T(\mathbf{x})=\sum_{i=1}^nX_i\sim \mathrm{Po}(n\lambda) }$. Knowing the distribution we want to investigate, it is easy to see that the statistic is complete.

$\displaystyle{ E(g(T))=\sum_{t=0}^\infty g(t)\frac{(n\lambda)^te^{-n\lambda}}{t!}=0 }$
$\displaystyle{ E(g(T))=\sum_{t=0}^\infty g(t)\frac{(n\lambda)^te^{-n\lambda}}{t!}=0 }$

< math > e (g (t)) = sum { t = 0} ^ infty g (t) frac {(n lambda) ^ te ^ {-n lambda }{ t！0 </math >

For this equality to hold, $\displaystyle{ g(t) }$ must be 0. This follows from the fact that none of the other terms will be 0 for all $\displaystyle{ t }$ in the sum and for all possible values of $\displaystyle{ \lambda }$. Hence, $\displaystyle{ E(g(T)) = 0 }$ for all $\displaystyle{ \lambda }$ implies that $\displaystyle{ P_\lambda(g(T) = 0) = 1 }$, and the statistic has been shown to be complete.

For this equality to hold, $\displaystyle{ g(t) }$ must be 0. This follows from the fact that none of the other terms will be 0 for all $\displaystyle{ t }$ in the sum and for all possible values of $\displaystyle{ \lambda }$. Hence, $\displaystyle{ E(g(T)) = 0 }$ for all $\displaystyle{ \lambda }$ implies that $\displaystyle{ P_\lambda(g(T) = 0) = 1 }$, and the statistic has been shown to be complete.

### Confidence interval 置信区间Confidence interval

The confidence interval for the mean of a Poisson distribution can be expressed using the relationship between the cumulative distribution functions of the Poisson and chi-squared distributions. The chi-squared distribution is itself closely related to the gamma distribution, and this leads to an alternative expression. Given an observation k from a Poisson distribution with mean μ, a confidence interval for μ with confidence level 1 – α is

The confidence interval for the mean of a Poisson distribution can be expressed using the relationship between the cumulative distribution functions of the Poisson and chi-squared distributions. The chi-squared distribution is itself closely related to the gamma distribution, and this leads to an alternative expression. Given an observation k from a Poisson distribution with mean μ, a confidence interval for μ with confidence level is

--fairywang(讨论)  【审校】“泊松分佈 ”改为“泊松分布 ”
$\displaystyle{ \tfrac 12\chi^{2}(\alpha/2; 2k) \le \mu \le \tfrac 12 \chi^{2}(1-\alpha/2; 2k+2), }$

$\displaystyle{ \tfrac 12\chi^{2}(\alpha/2; 2k) \le \mu \le \tfrac 12 \chi^{2}(1-\alpha/2; 2k+2), }$

12 chi ^ {2}(alpha/2; 2k) le mu le tfrac 12 chi ^ {2}(1-alpha/2; 2k + 2) ，</math >

or equivalently,

or equivalently,

$\displaystyle{ F^{-1}(\alpha/2; k,1) \le \mu \le F^{-1}(1-\alpha/2; k+1,1), }$

$\displaystyle{ F^{-1}(\alpha/2; k,1) \le \mu \le F^{-1}(1-\alpha/2; k+1,1), }$

(1-alpha/2; k，1) le mu le f ^ {-1}(1-alpha/2; k + 1,1) ，</math >

where $\displaystyle{ \chi^{2}(p;n) }$ is the quantile function (corresponding to a lower tail area p) of the chi-squared distribution with n degrees of freedom and $\displaystyle{ F^{-1}(p;n,1) }$ is the quantile function of a gamma distribution with shape parameter n and scale parameter 1.模板:R This interval is 'exact' in the sense that its coverage probability is never less than the nominal 1 – α.

where $\displaystyle{ \chi^{2}(p;n) }$ is the quantile function (corresponding to a lower tail area p) of the chi-squared distribution with n degrees of freedom and $\displaystyle{ F^{-1}(p;n,1) }$ is the quantile function of a gamma distribution with shape parameter n and scale parameter 1. This interval is 'exact' in the sense that its coverage probability is never less than the nominal .

When quantiles of the gamma distribution are not available, an accurate approximation to this exact interval has been proposed (based on the Wilson–Hilferty transformation):模板:R

When quantiles of the gamma distribution are not available, an accurate approximation to this exact interval has been proposed (based on the Wilson–Hilferty transformation):

$\displaystyle{ k \left( 1 - \frac{1}{9k} - \frac{z_{\alpha/2}}{3\sqrt{k}}\right)^3 \le \mu \le (k+1) \left( 1 - \frac{1}{9(k+1)} + \frac{z_{\alpha/2}}{3\sqrt{k+1}}\right)^3, }$

$\displaystyle{ k \left( 1 - \frac{1}{9k} - \frac{z_{\alpha/2}}{3\sqrt{k}}\right)^3 \le \mu \le (k+1) \left( 1 - \frac{1}{9(k+1)} + \frac{z_{\alpha/2}}{3\sqrt{k+1}}\right)^3, }$

< math > k left (1-frac {1}{9k }-frac { z _ { alpha/2}{3 sqrt { k }右) ^ 3 le mu le (k + 1) left (1-frac {1}{9(k + 1)}} + frac { z _ { alpha/2}{3 sqrt { k + 1}右) ^ 3，</math >

where $\displaystyle{ z_{\alpha/2} }$ denotes the standard normal deviate with upper tail area α / 2.

where $\displaystyle{ z_{\alpha/2} }$ denotes the standard normal deviate with upper tail area .

For application of these formulae in the same context as above (given a sample of n measured values ki each drawn from a Poisson distribution with mean λ), one would set

For application of these formulae in the same context as above (given a sample of n measured values ki each drawn from a Poisson distribution with mean λ), one would set

--fairywang(讨论)  【审校】“泊松分佈 ”改为“泊松分布 ”
$\displaystyle{ k=\sum_{i=1}^n k_i ,\! }$

$\displaystyle{ k=\sum_{i=1}^n k_i ,\! }$

[ math > k = sum { i = 1} ^ n k _ i，

calculate an interval for μ = , and then derive the interval for λ.

calculate an interval for μ = nλ, and then derive the interval for λ.

### Bayesian inference 贝叶斯推理

In Bayesian inference, the conjugate prior for the rate parameter λ of the Poisson distribution is the gamma distribution.模板:R Let

In Bayesian inference, the conjugate prior for the rate parameter λ of the Poisson distribution is the gamma distribution. Let

Bayesian inference 贝叶斯推理中，泊松分佈的速率参数的 共轭先验Conjugate prior是伽玛分布。让

--fairywang(讨论)  【审校】“泊松分佈 ”改为“泊松分布 ”
$\displaystyle{ \lambda \sim \mathrm{Gamma}(\alpha, \beta) \! }$

$\displaystyle{ \lambda \sim \mathrm{Gamma}(\alpha, \beta) \! }$

{ Gamma }(alpha，beta)

denote that λ is distributed according to the gamma density g parameterized in terms of a shape parameter α and an inverse scale parameter β:

denote that λ is distributed according to the gamma density g parameterized in terms of a shape parameter α and an inverse scale parameter β:

$\displaystyle{ g(\lambda \mid \alpha,\beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \; \lambda^{\alpha-1} \; e^{-\beta\,\lambda} \qquad \text{ for } \lambda\gt 0 \,\!. }$

$\displaystyle{ g(\lambda \mid \alpha,\beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)} \; \lambda^{\alpha-1} \; e^{-\beta\,\lambda} \qquad \text{ for } \lambda\gt 0 \,\!. }$

G (lambda mid alpha，beta) = frac { beta ^ { alpha }{ Gamma (alpha)} ; lambda ^ { alpha-1} ; e ^ {-beta，lambda } qquad text { for } lambda > 0，! </math >

Then, given the same sample of n measured values ki as before, and a prior of Gamma(α, β), the posterior distribution is

Then, given the same sample of n measured values ki as before, and a prior of Gamma(α, β), the posterior distribution is

$\displaystyle{ \lambda \sim \mathrm{Gamma}\left(\alpha + \sum_{i=1}^n k_i, \beta + n\right). \! }$

$\displaystyle{ \lambda \sim \mathrm{Gamma}\left(\alpha + \sum_{i=1}^n k_i, \beta + n\right). \! }$

The posterior mean E[λ] approaches the maximum likelihood estimate $\displaystyle{ \widehat{\lambda}_\mathrm{MLE} }$ in the limit as $\displaystyle{ \alpha\to 0,\ \beta\to 0 }$, which follows immediately from the general expression of the mean of the gamma distribution.

The posterior mean E[λ] approaches the maximum likelihood estimate $\displaystyle{ \widehat{\lambda}_\mathrm{MLE} }$ in the limit as $\displaystyle{ \alpha\to 0,\ \beta\to 0 }$, which follows immediately from the general expression of the mean of the gamma distribution.

The posterior predictive distribution for a single additional observation is a negative binomial distribution,模板:R sometimes called a gamma–Poisson distribution.

The posterior predictive distribution for a single additional observation is a negative binomial distribution, sometimes called a gamma–Poisson distribution.

--fairywang(讨论)  【审校】“泊松分佈 ”改为“泊松分布 ”

### Simultaneous estimation of multiple Poisson means 多重泊松均值的同步估计

Suppose $\displaystyle{ X_1, X_2, \dots, X_p }$ is a set of independent random variables from a set of $\displaystyle{ p }$ Poisson distributions, each with a parameter $\displaystyle{ \lambda_i }$, $\displaystyle{ i=1,\dots,p }$, and we would like to estimate these parameters. Then, Clevenson and Zidek show that under the normalized squared error loss $\displaystyle{ L(\lambda,{\hat \lambda})=\sum_{i=1}^p \lambda_i^{-1} ({\hat \lambda}_i-\lambda_i)^2 }$, when $\displaystyle{ p\gt 1 }$, then, similar as in Stein's example for the Normal means, the MLE estimator $\displaystyle{ {\hat \lambda}_i = X_i }$ is inadmissible. 模板:R

Suppose $\displaystyle{ X_1, X_2, \dots, X_p }$ is a set of independent random variables from a set of $\displaystyle{ p }$ Poisson distributions, each with a parameter $\displaystyle{ \lambda_i }$, $\displaystyle{ i=1,\dots,p }$, and we would like to estimate these parameters. Then, Clevenson and Zidek show that under the normalized squared error loss $\displaystyle{ L(\lambda,{\hat \lambda})=\sum_{i=1}^p \lambda_i^{-1} ({\hat \lambda}_i-\lambda_i)^2 }$, when $\displaystyle{ p\gt 1 }$, then, similar as in Stein's example for the Normal means, the MLE estimator $\displaystyle{ {\hat \lambda}_i = X_i }$ is inadmissible.

In this case, a family of minimax estimators is given for any $\displaystyle{ 0 \lt c \leq 2(p-1) }$ and $\displaystyle{ b \geq (p-2+p^{-1}) }$ as模板:R

In this case, a family of minimax estimators is given for any $\displaystyle{ 0 \lt c \leq 2(p-1) }$ and $\displaystyle{ b \geq (p-2+p^{-1}) }$ as

$\displaystyle{ {\hat \lambda}_i = \left(1 - \frac{c}{b + \sum_{i=1}^p X_i}\right) X_i, \qquad i=1,\dots,p. }$

$\displaystyle{ {\hat \lambda}_i = \left(1 - \frac{c}{b + \sum_{i=1}^p X_i}\right) X_i, \qquad i=1,\dots,p. }$

{ hat lambda } _ i = left (1-frac { c }{ b + sum { i = 1} ^ p x _ i } right) xi，qquad i = 1，dots，p </math >

## Occurrence and applications 发生与应用

Applications of the Poisson distribution can be found in many fields including:模板:R

Applications of the Poisson distribution can be found in many fields including:

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”
--fairywang(讨论)  【审校】“到达系统的电话”改为“一个系统中的来电数 ”

The Poisson distribution arises in connection with Poisson processes. It applies to various phenomena of discrete properties (that is, those that may happen 0, 1, 2, 3, ... times during a given period of time or in a given area) whenever the probability of the phenomenon happening is constant in time or space. Examples of events that may be modelled as a Poisson distribution include:

The Poisson distribution arises in connection with Poisson processes. It applies to various phenomena of discrete properties (that is, those that may happen 0, 1, 2, 3, ... times during a given period of time or in a given area) whenever the probability of the phenomenon happening is constant in time or space. Examples of events that may be modelled as a Poisson distribution include:

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

-->
-->
• The number of soldiers killed by horse-kicks each year in each corps in the Prussian cavalry. This example was used in a book by Ladislaus Bortkiewicz (1868–1931).模板:R
• The number of yeast cells used when brewing Guinness beer. This example was used by William Sealy Gosset (1876–1937).模板:R模板:R
• 酿造吉尼斯啤酒时使用的酵母细胞数量。这个例子被William sely Gosset（1876-1937）使用。模板:R模板:R
• The number of phone calls arriving at a call centre within a minute. This example was described by A.K. Erlang (1878–1929).模板:R
• 一分钟内到达呼叫中心的电话数。这个例子由 A.K.Erlang（1878-1929）描述模板:R
• Internet traffic.
• 互联网堵塞。
• The number of goals in sports involving two competing teams.模板:R
• 两支参赛队伍在运动中的进球数。模板:R
• The number of deaths per year in a given age group.
• 特定年龄组每年的死亡人数。
• The number of jumps in a stock price in a given time interval.
• 股票价格在给定时间间隔内的波动次数。
• Under an assumption of homogeneity, the number of times a web server is accessed per minute.
• 在[[泊松过程#齐次|均匀性]假设下，每分钟访问web服务器的次数。
• The number of mutations in a given stretch of DNA after a certain amount of radiation.
• 在一定量的辐射之后，在给定的DNA段中突变的数目。
• The proportion of cells that will be infected at a given multiplicity of infection.
• 在给定的时间内被感染的细胞的比例。
• The number of bacteria in a certain amount of liquid.模板:R
• 一定量液体中细菌的数量。模板:R
• The arrival of photons on a pixel circuit at a given illumination and over a given time period.
• 在给定的光照和时间间隔，到达像素电路的光子
• The targeting of V-1 flying bombs on London during World War II investigated by R. D. Clarke in 1946.模板:R
• 二战期间伦敦对V-1飞弹的目标调查||由R. D. Clarke 1946年调查。模板:R

Gallagher showed in 1976 that the counts of prime numbers in short intervals obey a Poisson distribution模板:R provided a certain version of the unproved prime r-tuple conjecture of Hardy-Littlewood模板:R is true.

Gallagher showed in 1976 that the counts of prime numbers in short intervals obey a Poisson distribution provided a certain version of the unproved prime r-tuple conjecture of Hardy-Littlewood is true.

1976年，加拉格尔指出，只要Hardy-Littlewo素数r-元组猜想的一个版本为正确，短时间间隔内质数的计数即服从泊松分布。

### Law of rare events稀有事件定律

Comparison of the Poisson distribution (black lines) and the binomial distribution with n = 10 (red circles), n = 20 (blue circles), n = 1000 (green circles). All distributions have a mean of 5. The horizontal axis shows the number of events k. As n gets larger, the Poisson distribution becomes an increasingly better approximation for the binomial distribution with the same mean.

Comparison of the Poisson distribution (black lines) and the [[binomial distribution with n = 10 (red circles), n = 20 (blue circles), n = 1000 (green circles). All distributions have a mean of 5. The horizontal axis shows the number of events k. As n gets larger, the Poisson distribution becomes an increasingly better approximation for the binomial distribution with the same mean.]]

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

The rate of an event is related to the probability of an event occurring in some small subinterval (of time, space or otherwise). In the case of the Poisson distribution, one assumes that there exists a small enough subinterval for which the probability of an event occurring twice is "negligible". With this assumption one can derive the Poisson distribution from the Binomial one, given only the information of expected number of total events in the whole interval. Let this total number be $\displaystyle{ \lambda }$. Divide the whole interval into $\displaystyle{ n }$ subintervals $\displaystyle{ I_1,\dots,I_n }$ of equal size, such that $\displaystyle{ n }$ > $\displaystyle{ \lambda }$ (since we are interested in only very small portions of the interval this assumption is meaningful). This means that the expected number of events in an interval $\displaystyle{ I_i }$ for each $\displaystyle{ i }$ is equal to $\displaystyle{ \lambda/n }$. Now we assume that the occurrence of an event in the whole interval can be seen as a Bernoulli trial, where the $\displaystyle{ i^{th} }$ trial corresponds to looking whether an event happens at the subinterval $\displaystyle{ I_i }$ with probability $\displaystyle{ \lambda/n }$. The expected number of total events in $\displaystyle{ n }$ such trials would be $\displaystyle{ \lambda }$, the expected number of total events in the whole interval. Hence for each subdivision of the interval we have approximated the occurrence of the event as a Bernoulli process of the form $\displaystyle{ \textrm{B}(n,\lambda/n) }$. As we have noted before we want to consider only very small subintervals. Therefore, we take the limit as $\displaystyle{ n }$ goes to infinity.

The rate of an event is related to the probability of an event occurring in some small subinterval (of time, space or otherwise). In the case of the Poisson distribution, one assumes that there exists a small enough subinterval for which the probability of an event occurring twice is "negligible". With this assumption one can derive the Poisson distribution from the Binomial one, given only the information of expected number of total events in the whole interval. Let this total number be $\displaystyle{ \lambda }$. Divide the whole interval into $\displaystyle{ n }$ subintervals $\displaystyle{ I_1,\dots,I_n }$ of equal size, such that $\displaystyle{ n }$ > $\displaystyle{ \lambda }$ (since we are interested in only very small portions of the interval this assumption is meaningful). This means that the expected number of events in an interval $\displaystyle{ I_i }$ for each $\displaystyle{ i }$ is equal to $\displaystyle{ \lambda/n }$. Now we assume that the occurrence of an event in the whole interval can be seen as a Bernoulli trial, where the $\displaystyle{ i^{th} }$ trial corresponds to looking whether an event happens at the subinterval $\displaystyle{ I_i }$ with probability $\displaystyle{ \lambda/n }$. The expected number of total events in $\displaystyle{ n }$ such trials would be $\displaystyle{ \lambda }$, the expected number of total events in the whole interval. Hence for each subdivision of the interval we have approximated the occurrence of the event as a Bernoulli process of the form $\displaystyle{ \textrm{B}(n,\lambda/n) }$. As we have noted before we want to consider only very small subintervals. Therefore, we take the limit as $\displaystyle{ n }$ goes to infinity.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

In this case the binomial distribution converges to what is known as the Poisson distribution by the Poisson limit theorem.

In this case the binomial distribution converges to what is known as the Poisson distribution by the Poisson limit theorem.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

In several of the above examples—such as, the number of mutations in a given sequence of DNA—the events being counted are actually the outcomes of discrete trials, and would more precisely be modelled using the binomial distribution, that is

In several of the above examples—such as, the number of mutations in a given sequence of DNA—the events being counted are actually the outcomes of discrete trials, and would more precisely be modelled using the binomial distribution, that is

$\displaystyle{ X \sim \textrm{B}(n,p). \, }$

$\displaystyle{ X \sim \textrm{B}(n,p). \, }$

X sim textrm { b }(n，p).，math

In such cases n is very large and p is very small (and so the expectation np is of intermediate magnitude). Then the distribution may be approximated by the less cumbersome Poisson distribution[citation needed]

In such cases n is very large and p is very small (and so the expectation np is of intermediate magnitude). Then the distribution may be approximated by the less cumbersome Poisson distribution

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”
$\displaystyle{ X \sim \textrm{Pois}(np). \, }$

$\displaystyle{ X \sim \textrm{Pois}(np). \, }$

$\displaystyle{ X \sim \textrm{Pois}(np).，math This approximation is sometimes known as the ''law of rare events'',{{r|Cameron1998|p=5}}since each of the ''n'' individual [[Bernoulli distribution|Bernoulli events]] rarely occurs. The name may be misleading because the total count of success events in a Poisson process need not be rare if the parameter ''np'' is not small. For example, the number of telephone calls to a busy switchboard in one hour follows a Poisson distribution with the events appearing frequent to the operator, but they are rare from the point of view of the average member of the population who is very unlikely to make a call to that switchboard in that hour. This approximation is sometimes known as the law of rare events,since each of the n individual Bernoulli events rarely occurs. The name may be misleading because the total count of success events in a Poisson process need not be rare if the parameter np is not small. For example, the number of telephone calls to a busy switchboard in one hour follows a Poisson distribution with the events appearing frequent to the operator, but they are rare from the point of view of the average member of the population who is very unlikely to make a call to that switchboard in that hour. 这种近似有时被称为稀有事件定律，因为 n 个伯努利事件中的每一个很少发生。这个名称可能有误导性，因为如果参数 np 不小，那么 Poisson 过程中成功事件的总计数就不会很少。例如，一个小时内打给忙碌总机的电话数量跟随着一个泊松分佈，这些事件在接线员看来是频繁的，但是从普通人的角度来看，这些事件很少发生，因为他们不太可能在那个小时内打电话给总机。 --[[用户:fairywang|fairywang]]([[用户讨论:fairywang|讨论]]) 【审校】“泊松分佈”改为“泊松分布” The word ''law'' is sometimes used as a synonym of [[probability distribution]], and ''convergence in law'' means ''convergence in distribution''. Accordingly, the Poisson distribution is sometimes called the "law of small numbers" because it is the probability distribution of the number of occurrences of an event that happens rarely but has very many opportunities to happen. ''The Law of Small Numbers'' is a book by Ladislaus Bortkiewicz about the Poisson distribution, published in 1898.{{r|vonBortkiewitsch1898}}{{r|Edgeworth1913}} The word law is sometimes used as a synonym of probability distribution, and convergence in law means convergence in distribution. Accordingly, the Poisson distribution is sometimes called the "law of small numbers" because it is the probability distribution of the number of occurrences of an event that happens rarely but has very many opportunities to happen. The Law of Small Numbers is a book by Ladislaus Bortkiewicz about the Poisson distribution, published in 1898. 法律一词有时被用作概率分布的同义词，法律的趋同意味着分配的趋同。因此，泊松分佈有时被称为“小数定律” ，因为它是一个事件发生次数的概率分布，这个事件很少发生，但却有很多机会发生。小数定律》是拉迪斯劳斯·博特基威茨的一本关于泊松分佈的书，出版于1898年。 --[[用户:fairywang|fairywang]]([[用户讨论:fairywang|讨论]]) 【审校】“泊松分佈”改为“泊松分布” ==='''\lt font color="#ff8000"\gt Poisson point process 泊松点过程\lt /font\gt '''=== {{Main|Poisson point process}} The Poisson distribution arises as the number of points of a [[Poisson point process]] located in some finite region. More specifically, if ''D'' is some region space, for example Euclidean space '''R'''\lt sup\gt ''d''\lt /sup\gt , for which |''D''|, the area, volume or, more generally, the Lebesgue measure of the region is finite, and if {{nowrap|''N''(''D'')}} denotes the number of points in ''D'', then The Poisson distribution arises as the number of points of a Poisson point process located in some finite region. More specifically, if D is some region space, for example Euclidean space R\lt sup\gt d\lt /sup\gt , for which |D|, the area, volume or, more generally, the Lebesgue measure of the region is finite, and if denotes the number of points in D, then 泊松分佈是位于某个有限区域的泊松过程的点数。更具体地说，如果 d 是某个区域空间，例如欧几里德空间 r \lt sup \gt d \lt /sup \gt ，对于这个区域 | d | ，区域的面积、体积或者更一般地说，区域的勒贝格测度是有限的，如果表示 d 中的点数，那么 --[[用户:fairywang|fairywang]]([[用户讨论:fairywang|讨论]]) 【审校】“泊松分佈”改为“泊松分布” : \lt math\gt P(N(D)=k)=\frac{(\lambda|D|)^k e^{-\lambda|D|}}{k!} . }$

$\displaystyle{ P(N(D)=k)=\frac{(\lambda|D|)^k e^{-\lambda|D|}}{k!} . }$

< math > p (n (d) = k) = frac {(lambda | d |) ^ k e ^ {-lambda | d | }{ k! }. math

### Poisson regression and negative binomial regression 泊松回归与负二项回归

Poisson regression and negative binomial regression are useful for analyses where the dependent (response) variable is the count (0, 1, 2, ...) of the number of events or occurrences in an interval.

Poisson regression and negative binomial regression are useful for analyses where the dependent (response) variable is the count (0, 1, 2, ...) of the number of events or occurrences in an interval.

Poisson regression and negative binomial regression 泊松回归与负二项回归分析是有用的，其中依赖(响应)变量是计数(0,1,2，...)的在一个区间内事件发生的数量。

### Other applications in science 科学上的其他应用

In a Poisson process, the number of observed occurrences fluctuates about its mean λ with a standard deviation $\displaystyle{ \sigma_k =\sqrt{\lambda} }$. These fluctuations are denoted as Poisson noise or (particularly in electronics) as shot noise.

In a Poisson process, the number of observed occurrences fluctuates about its mean λ with a standard deviation $\displaystyle{ \sigma_k =\sqrt{\lambda} }$. These fluctuations are denoted as Poisson noise or (particularly in electronics) as shot noise.

The correlation of the mean and standard deviation in counting independent discrete occurrences is useful scientifically. By monitoring how the fluctuations vary with the mean signal, one can estimate the contribution of a single occurrence, even if that contribution is too small to be detected directly. For example, the charge e on an electron can be estimated by correlating the magnitude of an electric current with its shot noise. If N electrons pass a point in a given time t on the average, the mean current is $\displaystyle{ I=eN/t }$; since the current fluctuations should be of the order $\displaystyle{ \sigma_I=e\sqrt{N}/t }$ (i.e., the standard deviation of the Poisson process), the charge $\displaystyle{ e }$ can be estimated from the ratio $\displaystyle{ t\sigma_I^2/I }$.[citation needed]

The correlation of the mean and standard deviation in counting independent discrete occurrences is useful scientifically. By monitoring how the fluctuations vary with the mean signal, one can estimate the contribution of a single occurrence, even if that contribution is too small to be detected directly. For example, the charge e on an electron can be estimated by correlating the magnitude of an electric current with its shot noise. If N electrons pass a point in a given time t on the average, the mean current is $\displaystyle{ I=eN/t }$; since the current fluctuations should be of the order $\displaystyle{ \sigma_I=e\sqrt{N}/t }$ (i.e., the standard deviation of the Poisson process), the charge $\displaystyle{ e }$ can be estimated from the ratio $\displaystyle{ t\sigma_I^2/I }$.

An everyday example is the graininess that appears as photographs are enlarged; the graininess is due to Poisson fluctuations in the number of reduced silver grains, not to the individual grains themselves. By correlating the graininess with the degree of enlargement, one can estimate the contribution of an individual grain (which is otherwise too small to be seen unaided).[citation needed] Many other molecular applications of Poisson noise have been developed, e.g., estimating the number density of receptor molecules in a cell membrane.

An everyday example is the graininess that appears as photographs are enlarged; the graininess is due to Poisson fluctuations in the number of reduced silver grains, not to the individual grains themselves. By correlating the graininess with the degree of enlargement, one can estimate the contribution of an individual grain (which is otherwise too small to be seen unaided). Many other molecular applications of Poisson noise have been developed, e.g., estimating the number density of receptor molecules in a cell membrane.

$\displaystyle{ \lt math\gt 《数学》 \Pr(N_t=k) = f(k;\lambda t) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}. }$
\Pr(N_t=k) = f(k;\lambda t) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}.[/itex]

Pr (n _ t = k) = f (k; lambda t) = frac {((lambda t) ^ k e ^ {-lambda t }}{ k！} . </math >

In Causal Set theory the discrete elements of spacetime follow a Poisson distribution in the volume.

In Causal Set theory the discrete elements of spacetime follow a Poisson distribution in the volume.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

## Computational methods计算方法

The Poisson distribution poses two different tasks for dedicated software libraries: Evaluating the distribution $\displaystyle{ P(k;\lambda) }$, and drawing random numbers according to that distribution.

The Poisson distribution poses two different tasks for dedicated software libraries: Evaluating the distribution $\displaystyle{ P(k;\lambda) }$, and drawing random numbers according to that distribution.

### Evaluating the Poisson distribution 计算泊松分布

Computing $\displaystyle{ P(k;\lambda) }$ for given $\displaystyle{ k }$ and $\displaystyle{ \lambda }$ is a trivial task that can be accomplished by using the standard definition of $\displaystyle{ P(k;\lambda) }$ in terms of exponential, power, and factorial functions. However, the conventional definition of the Poisson distribution contains two terms that can easily overflow on computers: λk and k!. The fraction of λk to k! can also produce a rounding error that is very large compared to e−λ, and therefore give an erroneous result. For numerical stability the Poisson probability mass function should therefore be evaluated as

Computing $\displaystyle{ P(k;\lambda) }$ for given $\displaystyle{ k }$ and $\displaystyle{ \lambda }$ is a trivial task that can be accomplished by using the standard definition of $\displaystyle{ P(k;\lambda) }$ in terms of exponential, power, and factorial functions. However, the conventional definition of the Poisson distribution contains two terms that can easily overflow on computers: λk and k!. The fraction of λk to k! can also produce a rounding error that is very large compared to e−λ, and therefore give an erroneous result. For numerical stability the Poisson probability mass function should therefore be evaluated as

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”
$\displaystyle{ \!f(k; \lambda)= \exp \left[ k\ln \lambda - \lambda - \ln \Gamma (k+1) \right], }$

$\displaystyle{ \!f(k; \lambda)= \exp \left[ k\ln \lambda - \lambda - \ln \Gamma (k+1) \right], }$

= exp left [ k ln lambda-lambda-ln Gamma (k + 1) right ] ，</math >

which is mathematically equivalent but numerically stable. The natural logarithm of the Gamma function can be obtained using the lgamma function in the C standard library (C99 version) or R, the gammaln function in MATLAB or SciPy, or the log_gamma function in Fortran 2008 and later.

which is mathematically equivalent but numerically stable. The natural logarithm of the Gamma function can be obtained using the lgamma function in the C standard library (C99 version) or R, the gammaln function in MATLAB or SciPy, or the log_gamma function in Fortran 2008 and later.

Some computing languages provide built-in functions to evaluate the Poisson distribution, namely

Some computing languages provide built-in functions to evaluate the Poisson distribution, namely

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”
• R: function dpois(x, lambda);
• Excel: function POISSON( x, mean, cumulative), with a flag to specify the cumulative distribution;
• Mathematica: univariate Poisson distribution as PoissonDistribution[$\displaystyle{ \lambda }$],[8] bivariate Poisson distribution as MultivariatePoissonDistribution[$\displaystyle{ \theta_{12} }$,{ $\displaystyle{ \theta_1 - \theta_{12} }$, $\displaystyle{ \theta_2 - \theta_{12} }$}],.[9]

### Random drawing from the Poisson distribution 从泊松分布中抽取随机量

The less trivial task is to draw random integers from the Poisson distribution with given $\displaystyle{ \lambda }$.

The less trivial task is to draw random integers from the Poisson distribution with given $\displaystyle{ \lambda }$.

Solutions are provided by:

Solutions are provided by:

• R: function rpois(n, lambda);

### Generating Poisson-distributed random variables 生成泊松分布随机变量

A simple algorithm to generate random Poisson-distributed numbers (pseudo-random number sampling) has been given by Knuth:模板:R

A simple algorithm to generate random Poisson-distributed numbers (pseudo-random number sampling) has been given by Knuth:

algorithm poisson random number (Knuth):
algorithm poisson random number (Knuth):

init:
init:

Let L ← e−λ, k ← 0 and p ← 1.
Let L ← e−λ, k ← 0 and p ← 1.
Let L ← e−λ, k ← 0 and p ← 1.
do:
do:

k ← k + 1.
k ← k + 1.
k ← k + 1.
Generate uniform random number u in [0,1] and let p ← p × u.
Generate uniform random number u in [0,1] and let p ← p × u.

while p > L.
while p > L.

return k − 1.
return k − 1.
return k − 1.

The complexity is linear in the returned value k, which is λ on average. There are many other algorithms to improve this. Some are given in Ahrens & Dieter, see 模板:Slink below.

The complexity is linear in the returned value k, which is λ on average. There are many other algorithms to improve this. Some are given in Ahrens & Dieter, see below.

For large values of λ, the value of L = e−λ may be so small that it is hard to represent. This can be solved by a change to the algorithm which uses an additional parameter STEP such that e−STEP does not underflow: [citation needed]

For large values of λ, the value of L = e−λ may be so small that it is hard to represent. This can be solved by a change to the algorithm which uses an additional parameter STEP such that e−STEP does not underflow:

--fairywang(讨论)  【审校】“底流”改为“下溢”
algorithm poisson random number (Junhao, based on Knuth):
algorithm poisson random number (Junhao, based on Knuth):

init:
init:

Let λLeft ← λ, k ← 0 and p ← 1.
Let λLeft ← λ, k ← 0 and p ← 1.
Let λLeft ← λ, k ← 0 and p ← 1.
do:
do:

k ← k + 1.
k ← k + 1.
k ← k + 1.
Generate uniform random number u in (0,1) and let p ← p × u.
Generate uniform random number u in (0,1) and let p ← p × u.

while p < 1 and λLeft > 0:
while p < 1 and λLeft > 0:
while p < 1 and λLeft > 0:
if λLeft > STEP:
if λLeft > STEP:
if λLeft > STEP:
p ← p × eSTEP
p ← p × eSTEP
p ← p × eSTEP
λLeft ← λLeft − STEP
λLeft ← λLeft − STEP
λLeft ← λLeft − STEP
else:
else:

p ← p × eλLeft
p ← p × eλLeft
p ← p × eλLeft
λLeft ← 0
λLeft ← 0
λLeft ← 0
while p > 1.
while p > 1.

return k − 1.
return k − 1.
return k − 1.

The choice of STEP depends on the threshold of overflow. For double precision floating point format, the threshold is near e700, so 500 shall be a safe STEP.

The choice of STEP depends on the threshold of overflow. For double precision floating point format, the threshold is near e700, so 500 shall be a safe STEP.

STEP 的选择取决于溢出 阈值Threshold。对于双精度浮点格式， 阈值Threshold接近 e < sup > 700 ，因此500应该是一个安全的 STEP。

Other solutions for large values of λ include rejection sampling and using Gaussian approximation.

Other solutions for large values of λ include rejection sampling and using Gaussian approximation.

Inverse transform sampling is simple and efficient for small values of λ, and requires only one uniform random number u per sample. Cumulative probabilities are examined in turn until one exceeds u.

Inverse transform sampling is simple and efficient for small values of λ, and requires only one uniform random number u per sample. Cumulative probabilities are examined in turn until one exceeds u.

algorithm Poisson generator based upon the inversion by sequential search:模板:R
algorithm Poisson generator based upon the inversion by sequential search:

init:
init:

Let x ← 0, p ← e−λ, s ← p.
Let x ← 0, p ← e−λ, s ← p.
Let x ← 0, p ← e−λ, s ← p.
Generate uniform random number u in [0,1].
Generate uniform random number u in [0,1].

while u > s do:
while u > s do:

x ← x + 1.
x ← x + 1.
x ← x + 1.
p ← p × λ / x.
p ← p × λ / x.
p ← p × λ / x.
s ← s + p.
s ← s + p.
s ← s + p.
return x.
return x.

## History历史

The distribution was first introduced by Siméon Denis Poisson (1781–1840) and published together with his probability theory in his work Recherches sur la probabilité des jugements en matière criminelle et en matière civile(1837).模板:R The work theorized about the number of wrongful convictions in a given country by focusing on certain random variables N that count, among other things, the number of discrete occurrences (sometimes called "events" or "arrivals") that take place during a time-interval of given length. The result had already been given in 1711 by Abraham de Moivre in De Mensura Sortis seu; de Probabilitate Eventuum in Ludis a Casu Fortuito Pendentibus .模板:R模板:R模板:R模板:R This makes it an example of Stigler's law and it has prompted some authors to argue that the Poisson distribution should bear the name of de Moivre.模板:R

The distribution was first introduced by Siméon Denis Poisson (1781–1840) and published together with his probability theory in his work Recherches sur la probabilité des jugements en matière criminelle et en matière civile(1837). The work theorized about the number of wrongful convictions in a given country by focusing on certain random variables N that count, among other things, the number of discrete occurrences (sometimes called "events" or "arrivals") that take place during a time-interval of given length. The result had already been given in 1711 by Abraham de Moivre in De Mensura Sortis seu; de Probabilitate Eventuum in Ludis a Casu Fortuito Pendentibus . This makes it an example of Stigler's law and it has prompted some authors to argue that the Poisson distribution should bear the name of de Moivre.

The distribution was first introduced by Siméon Denis Poisson (1781–1840) and published together with his probability theory in his work Recherches sur la probabilité des jugements en matière criminelle et en matière civile(1837). 这种分布最早由西蒙·丹尼斯·泊松（1781-1840）提出，与他的概率理论一起发表在他的著作《苏拉河畔的调查》中，sur la probabilité des jugements en matière criminelle et en matière civile(1837) 这项工作通过关注某些随机变量 n (其中包括在给定时间间隔内发生的离散事件(有时称为“事件”或“到达事件”)的数量)来推断某一国家的错误定罪数量。这个结果早在1711年就已经在《亚伯拉罕·棣莫弗》给出了。在 Ludis 举行的猜测活动。这使它成为斯蒂格勒定律的一个例子，也使一些作者提出， 泊松分佈Poisson distribution 应该以德莫伊弗雷的名字命名。

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

In 1860, Simon Newcomb fitted the Poisson distribution to the number of stars found in a unit of space.模板:R

In 1860, Simon Newcomb fitted the Poisson distribution to the number of stars found in a unit of space.

1860年，Simon Newcomb 将 泊松分佈Poisson distribution 与天文台一个空间单位中发现的恒星数量进行了比较。

A further practical application of this distribution was made by Ladislaus Bortkiewicz in 1898 when he was given the task of investigating the number of soldiers in the Prussian army killed accidentally by horse kicks;模板:R this experiment introduced the Poisson distribution to the field of reliability engineering.

A further practical application of this distribution was made by Ladislaus Bortkiewicz in 1898 when he was given the task of investigating the number of soldiers in the Prussian army killed accidentally by horse kicks; this experiment introduced the Poisson distribution to the field of reliability engineering.

--fairywang(讨论)  【审校】“泊松分佈”改为“泊松分布”

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|location=Lyon, France
|location=Lyon, France

| 地点: 法国里昂

|url=http://www.iarc.fr/en/publications/pdfs-online/stat/sp82/index.php
|url=http://www.iarc.fr/en/publications/pdfs-online/stat/sp82/index.php
|isbn=978-92-832-0182-3
|isbn=978-92-832-0182-3

| isbn = 978-92-832-0182-3

|access-date=2012-03-11
|access-date=2012-03-11

| access-date = 2012-03-11

|archive-url=https://web.archive.org/web/20180808161401/http://www.iarc.fr/en/publications/pdfs-online/stat/sp82/index.php
|archive-url=https://web.archive.org/web/20180808161401/http://www.iarc.fr/en/publications/pdfs-online/stat/sp82/index.php
|archive-date=2018-08-08
|archive-date=2018-08-08

| archive-date = 2018-08-08

}}</ref>
}}</ref>

} </ref >

}}</ref>

} </ref >

}}</ref>

} </ref >

}}</ref>

} </ref >

}}</ref>

} </ref >

}}</ref>

} </ref >

}}

}}

}}

### Sources资源

• Ahrens, Joachim h.; Dieter, Ulrich (1974

1974年), "从伽马分布、贝塔分布、泊松分布和二项分布抽样的计算机方法", Computing 日志 = 计算, 12

12 (3

}}

}}

• Ahrens, Joachim h.; Dieter, Ulrich (1982

1982年), "计算机生成的泊松偏离", ACM 数学软件汇刊, 8

8 (2

2): 163–179, doi:[//doi.org/10.1145%2F355993.355997%0A%0A10.1145%2F355993.355997 10.1145/355993.355997 10.1145/355993.355997] Check |doi= value (help) Unknown parameter |页= ignored (help); line feed character in |doi= at position 22 (help); line feed character in |volume= at position 2 (help); line feed character in |year= at position 5 (help); line feed character in |issue= at position 2 (help); Check date values in: |year= (help)

}}

}}

• Evans, Ronald j.; Boersma, j.; Blachman, n.; Jagers, a.答:。 (1988

Https://research.tue.nl/nl/publications/solution-to-problem-876--the-entropy-of-a-poisson-distribution(94cf6dd2-b35e-41c8-9da7-6ec69ca391a0).html "泊松分佈的熵: 问题87-6"] Check |url= value (help), SIAM Review, 30

30 (2

2): 314–317, doi:[//doi.org/10.1137%2F1030059%0A%0A10.1137%2F1030059 10.1137/1030059 10.1137/1030059] Check |doi= value (help) Unknown parameter |页数= ignored (help); line feed character in |doi= at position 16 (help); line feed character in |volume= at position 3 (help); line feed character in |issue= at position 2 (help); line feed character in |url= at position 146 (help); line feed character in |year= at position 5 (help); Check date values in: |year= (help)

}}
}}

Category:Articles with example pseudocode

Category:Conjugate prior distributions

Category:Factorial and binomial topics

Category:Infinitely divisible probability distributions

This page was moved from wikipedia:en:Poisson distribution. Its edit history can be viewed at 泊松分布/edithistory

1. [ http://www.proofwiki.org/wiki/expectation_of_poisson_distribution 证明 wiki: 期望]和[ http://www.proofwiki.org/wiki/variance_of_poisson_distribution 证明 wiki: variance ]
2. Free Random Variables by D. Voiculescu, K. Dykema, A. Nica, CRM Monograph Series, American Mathematical Society, Providence RI, 1992
3. James A. Mingo, Roland Speicher: Free Probability and Random Matrices. Fields Institute Monographs, Vol. 35, Springer, New York, 2017.
4. Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher, pp. 203–204, Cambridge Univ. Press 2006
5. Paszek, Ewa. "Maximum Likelihood Estimation – Examples".
6. "Wolfram Language: PoissonDistribution reference page". wolfram.com. Retrieved 2016-04-08.
7. "Wolfram Language: MultivariatePoissonDistribution reference page". wolfram.com. Retrieved 2016-04-08.
8. Empty citation (help) 引用错误：无效<ref>标签；name属性“Breslow1987”使用不同内容定义了多次