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第169行： − 在概率论和统计学中，泊松分佈是以法国数学家西莫恩·德尼·泊松命名的，是一个离散的概率分布，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，这些事件以一个已知的常数发生，并且独立于与上一个事件发生的间隔时间。泊松分佈还可以用来表示其他特定间隔的事件数量，如距离、面积或体积。+ 第177行：
第177行： − 例如，记录每天收到邮件数量的个人可能会注意到，他们平均每天收到4封信。如果收到任何特定的邮件并不影响未来邮件的到达时间，也就是说，如果来自不同来源的邮件彼此独立地到达，那么一个合理的假设是，每天收到的邮件数量服从一个泊松分佈。其他例子可能遵循一个泊松分佈包括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。+ 第193行：
第193行： − 泊松分佈模型用来模拟一个事件在一段时间或空间内发生的次数。+ 第201行：
第201行： − + 第245行：
第245行： − 泊松分佈可以应用于包括大量罕见的可能事件的系统。在正确的条件下，在一个固定的时间间隔内发生的这类事件的数量是一个带有泊松分佈的随机数。+ 第256行：
第256行： − 泊松分佈模型可以用来模拟事件，比如+ 第272行：
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第278行： − 如果下面的假设是正确的，泊松分佈是一个合适的模型:+ 第292行：
第292行： − + − + 第300行：
第300行： − 如果这些条件成立，那么泊松是一个随机变量，其分布是一个泊松分佈。+ 第307行：
第307行： − + − 随着试验的数量趋于无穷大，每次试验的成功概率除以试验的数量，泊松分佈也是二项分布的极限。 − 泊松分布的事件概率+ − + 第348行：
第347行： − 这个方程就是概率质量函数的泊松分佈。+ 第369行：
第368行： − + 第380行：
第379行： − 在一条特定的河流上，泛滥的洪水平均每100年发生一次。计算概率 = 0,1,2,3,4,5，或6溢出洪水在100年的间隔，假设泊松模型是适当的。+ 第388行：
第387行： − 因为平均事件率是每100年溢出一次洪水，= 1+ 第554行：
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第759行： − 在上面的一个例子中，溢流洪水每100年发生一次(= 1)。根据同样的计算，100年内不会有洪水泛滥的概率大约是0.37。+ 第768行：
第767行： − 一般来说，如果一个事件平均每个间隔发生一次(= 1) ，并且事件遵循一个泊松分佈，那么。此外，p (下一个间隔中正好有一个事件) = 0.37，如溢出洪水的表所示。+ − + 第780行：
第779行： − 每分钟抵达学生会的学生人数可能不会遵循一个泊松分佈，因为这个比率不是恒定的(上课时间的低比率，上课时间之间的高比率) ，而且个别学生的到达也不是独立的(学生往往是成群结队来的)。+ 第788行：
第787行： − 如果一个国家每年发生5级地震的次数增加了发生类似震级余震的可能性，那么这个国家每年发生5级地震的次数可能不会超过一个泊松分佈。+ 第796行：
第795行： − 至少有一个事件得到保证的例子不是 Poission 分布式的，而是可以使用零截断泊松分佈进行建模。+ 第809行：
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第1,403行： − 在哪里+ 第1,433行：
第1,432行： − + 第1,662行：
第1,661行： − + 第1,686行：
第1,685行： − 要保持这个等式，< math > g (t) </math > 必须为0。这源于这样一个事实: 对于所有 < math > t </math > 的和和以及 < math > > lambda </math > 的所有可能值，其他项都不会为0。因此，e (g (t)) = 0 </math > > lambda </math > 意味着 < math > p _ lambda (g (t) = 0 = 1 </math > ，统计已被证明是完整的。+ − + − 置信区间的平均泊松分佈可以用泊松分布和卡方分布的累积分布函数之间的关系来表示。卡方分布本身与伽玛分布密切相关，这导致了另一种表达方式。给定一个来自平均泊松分佈的观测值 k，一个带有置信水平的置信区间是+
无编辑摘要
{{Probability distribution
{{Probability distribution
{概率分布
'''<font color="#ff8000"> {概率分布Probability distribution</font>'''
| name = Poisson Distribution
| name = Poisson Distribution
| name = Poisson Distribution
| name = Poisson Distribution
泊松分佈
'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''
| type = mass
| type = mass
| 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。因为一个泊松分布的变量只取整数值，所以 CDF 在 k 的整数和平坦的所有其他地方均不连续。
| cdf _ caption = 水平轴是索引 k，表示出现的次数。因为一个'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''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 = <math>\operatorname{Pois}(\lambda)</math>
| notation = <math>\operatorname{Pois}(\lambda)</math>
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.
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.
在概率论和统计学中，泊松分佈是以法国数学家西莫恩·德尼·泊松命名的，是一个离散的概率分布，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，这些事件以一个已知的常数平均速率发生，并且独立于与上一个事件的间隔发生时间。'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''还可以用来表示其他有特定间隔的事件数量，如距离、面积或体积。
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.
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.
例如，记录每天收到邮件数量的个人可能会注意到，他们平均每天收到4封信。如果收到任何邮件都并不影响未来邮件的到达时间，也就是说，如果不同来源的邮件彼此独立地到达，那么一个合理的假设是，每天收到的邮件数量服从一个'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''。其他可能遵循一个'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''的例子包括：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。
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.
'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''模型用来模拟一个事件在一段时间或空间内发生的次数。
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:
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:
一个离散的随机变量 x 被称为具有参数 > 0的泊松分佈，如果，对于 k = 0,1,2，... ，x 的概率分布函数是:
一个离散的随机变量 x 被称为具有参数 > 0的'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''，如果，对于 k = 0,1,2，... ，x 的概率分布函数是:
:<math>\!f(k; \lambda)= \Pr(X = k)= \frac{\lambda^k e^{-\lambda}}{k!},</math>
:<math>\!f(k; \lambda)= \Pr(X = k)= \frac{\lambda^k e^{-\lambda}}{k!},</math>
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.
'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''可以应用于包括大量罕见可能事件的系统。在正确的条件下，在一个固定的时间间隔内发生的这类事件的数量是一个具有'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''的随机数。
The Poisson distribution may be useful to model events such as
The Poisson distribution may be useful to model events such as
'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''模型可以用来模拟事件，比如
* The number of meteorites greater than 1 meter diameter that strike Earth in a year
* The number of meteorites greater than 1 meter diameter that strike Earth in a year
===Assumptions and validity===
===Assumptions and validity===
假设与有效条件
==假设与有效条件==
The Poisson distribution is an appropriate model if the following assumptions are true:{{r|Koehrsen2019}}
The Poisson distribution is an appropriate model if the following assumptions are true:{{r|Koehrsen2019}}
The Poisson distribution is an appropriate model if the following assumptions are true:
The Poisson distribution is an appropriate model if the following assumptions are true:
以下假设成立时，'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''模型适用:
* {{mvar|k}} is the number of times an event occurs in an interval and {{mvar|k}} can take values 0, 1, 2, ....
* {{mvar|k}} is the number of times an event occurs in an interval and {{mvar|k}} can take values 0, 1, 2, ....
*一个事件的发生不影响第二个事件发生的概率
*一个事件的发生不影响第二个事件发生的概率
*事件发生的平均速率与任何事件无关。为简单起见，通常假定为常数，但实际上可能随时间而变化
*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，而是在每一小的时间正好有一个事件发生或不发生
*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 {{mvar|k}} is a Poisson random variable, and the distribution of {{mvar|k}} is a Poisson distribution.
If these conditions are true, then {{mvar|k}} is a Poisson random variable, and the distribution of {{mvar|k}} is a Poisson distribution.
If these conditions are true, then is a Poisson random variable, and the distribution of is a Poisson distribution.
If these conditions are true, then is a Poisson random variable, and the distribution of is a Poisson distribution.
如果这些条件成立，那么它是一个泊松随机变量，其分布是一个'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''。
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).
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，（可得二项式分布），随着试验的数量趋于无穷大，'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''也是二项式分布的极限。
===Probability of events for a Poisson distribution===
===Probability of events for a Poisson distribution===
==='''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''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 <math> \lambda </math> (lambda). <math> \lambda </math> is the event rate, also called the rate parameter. The probability of observing {{mvar|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 <math> \lambda </math> (lambda). <math> \lambda </math> is the event rate, also called the rate parameter. The probability of observing {{mvar|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 <math> \lambda </math> (lambda). <math> \lambda </math> is the event rate, also called the rate parameter. The probability of observing 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 <math> \lambda </math> (lambda). <math> \lambda </math> is the event rate, also called the rate parameter. The probability of observing events in an interval is given by the equation
一个事件可以在一个间隔内发生0,1,2，... 次。区间内的平均事件数被指定为 < math > lambda </math > (lambda)。Lambda </math > 是事件速率，也称为速率参数。该方程给出了在一个区间内观测事件的概率
一个事件可以在一个间隔内发生0,1,2，... 次。区间内的平均事件数被指定为 < math > lambda </math > (lambda)。Lambda </math > 是事件速率，也称为速率参数。以下方程给出了在一个区间内观测事件的概率
This equation is the probability mass function (PMF) for a Poisson distribution.
This equation is the probability mass function (PMF) for a Poisson distribution.
这个方程就是概率质量函数的'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''。
==== Examples of probability for Poisson distributions ====
==== Examples of probability for Poisson distributions ====
==='''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''概率的例子===
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.
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.
在某一条河流上，洪水平均每100年发生泛滥一次。计算在100年间洪水泛滥次数= 0,1,2,3,4,5，或6次的概率，假设（其分布）适用泊松模型。
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
因为平均事件率是每100年发一次洪水，λ = 1
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.
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.
乌加特和他的同事们报告说，世界杯足球赛的平均进球数约为2.5个，泊松模型是适当的。
乌加特和他的同事们报告说，世界杯足球赛的平均进球数约为2.5个，适用泊松模型。
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.
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.
因为平均每场比赛有2.5个进球，= 2.5个。
因为平均每场比赛有2.5个进球，λ = 2.5个。
====Once in an interval events: The special case of ''λ'' = 1 and ''k'' = 0 ====
====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 {{mvar|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 {{mvar|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?
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?
假设天文学家估计，大型陨石(超过一定大小)平均每100年撞击地球一次(= 每100年撞击一次) ，而且陨石撞击的次数紧随泊松分佈之后。在接下来的100年里，陨石击中0的概率是多少？
假设天文学家估计，大型陨石(超过一定大小)平均每100年撞击地球一次(= 每100年撞击一次) ，而且陨石撞击的次数紧随'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''之后。在接下来的100年里，被陨石击中k=0的概率是多少？
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.
根据这些假设，未来100年内没有大陨石撞击地球的概率大约为0.37。剩下的1-0.37 = 0.63是未来100年内1,2,3或更多大型陨石撞击的概率。
根据这些假设，未来100年内没有大陨石撞击地球的概率大约为0.37。剩下的1-0.37 = 0.63是未来100年内被1,2,3或更多大型陨石撞击的概率。
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 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.
在上面的一个例子中，洪水每100年发生泛滥一次(''λ'' = 1)。根据同样的计算，100年内不会有洪水泛滥的概率大约是0.37。
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.
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.
一般来说，如果一个事件平均每个时间间隔发生一次(''λ'' = 1) ，并且事件遵循'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''，那么p (下一个间隔中正好有一个事件) = 0.37，如洪水泛滥的表所示。
=== Examples that violate the Poisson assumptions ===
=== 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).
每分钟抵达学生会的学生人数可能不会遵循一个'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''，因为这个比率不是恒定的(上课时间的低比率，课间时的高比率) ，而且每个学生的到达也不是独立的(学生往往是成群结队来的)。
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.
一次大的强震会增加发生类似震级余震的可能性，那么一个国家每年发生5级地震的次数可能不会服从'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''。
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.
至少有一个事件确定发生的例子不是 Poission 分布式的，但也许可以使用零截断'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''进行建模。
== Properties ==
== Properties ==
==
If <math>X_i \sim \operatorname{Pois}(\lambda_i)</math> for <math>i=1,\dotsc,n</math> are independent, then <math>\sum_{i=1}^n X_i \sim \operatorname{Pois}\left(\sum_{i=1}^n \lambda_i\right)</math>. 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.
If <math>X_i \sim \operatorname{Pois}(\lambda_i)</math> for <math>i=1,\dotsc,n</math> are independent, then <math>\sum_{i=1}^n X_i \sim \operatorname{Pois}\left(\sum_{i=1}^n \lambda_i\right)</math>. 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.
如果对于 < math > i = 1，dotsc，n </math > 是独立的，那么 < math > sum { i = 1} ^ n xi sim 操作者名{ Pois }左(sum { i = 1} ^ n lambda _ i 右) </math > 。一个逆定理是雷科夫定理，它说如果两个独立的随机变量之和是泊松分布的，那么这两个独立的随机变量之和也是泊松分布的。
如果对于 < math > i = 1，dotsc，n </math > 是独立的，那么 < math > sum { i = 1} ^ n xi sim 操作者名{ Pois }左(sum { i = 1} ^ n lambda _ i 右) </math > 。一个逆定理是雷科夫定理，它说如果两个独立的随机变量之和是'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''的，那么这两个独立的随机变量之和也是'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''的。
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.
这个定义类似于从(经典)泊松过程获得经典泊松分佈的一种方法。
这个定义类似于从(经典)泊松过程获得经典'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''的一种方法。
where
where
其中
====Some transforms of this law====
====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<ref>Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher, pp. 203–204, Cambridge Univ. Press 2006</ref>
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<ref>Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher, pp. 203–204, Cambridge Univ. Press 2006</ref>
which is the negative of n times the reciprocal of the average of the k<sub>i</sub>. 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 k<sub>i</sub>. This expression is negative when the average is positive. If this is satisfied, then the stationary point maximizes the probability function.
它是 n 乘以 k < sub > i </sub > 平均值的倒数。当平均数为正时，这个表达式是负的。如果这一点得到了满足，那么驻点最大化了概率密度函数。
它是 n 乘以 k < sub > i </sub > 平均值的倒数。当平均数为正时，这个表达式是负的。如果这一点得到了满足，那么'''<font color="#ff8000"> 驻点The stationary point</font>'''最大化了概率密度函数。
For this equality to hold, <math>g(t)</math> must be 0. This follows from the fact that none of the other terms will be 0 for all <math>t</math> in the sum and for all possible values of <math>\lambda</math>. Hence, <math> E(g(T)) = 0</math> for all <math>\lambda</math> implies that <math>P_\lambda(g(T) = 0) = 1</math>, and the statistic has been shown to be complete.
For this equality to hold, <math>g(t)</math> must be 0. This follows from the fact that none of the other terms will be 0 for all <math>t</math> in the sum and for all possible values of <math>\lambda</math>. Hence, <math> E(g(T)) = 0</math> for all <math>\lambda</math> implies that <math>P_\lambda(g(T) = 0) = 1</math>, and the statistic has been shown to be complete.
要保证这个等式成立，< math > g (t) </math > 必须为0。这源于这样一个事实: 对于所有 < math > t </math > 的和和以及 < math > > lambda </math > 的所有可能值，其他项都不会为0。因此，e (g (t)) = 0 </math > > lambda </math > 意味着 < math > p _ lambda (g (t) = 0 = 1 </math > ，统计已被证明是完整的。
=== Confidence interval ===
=== Confidence interval ===
==='''<font color="#ff8000">置信区间Confidence interval </font>'''===
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 distribution]]s. 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 {{math|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 distribution]]s. 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 {{math|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
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
'''<font color="#ff8000">置信区间Confidence interval </font>'''的平均'''<font color="#ff8000"> 泊松分布Poisson distribution</font>'''可以用泊松分布和卡方分布的累积分布函数之间的关系来表示。卡方分布本身与伽玛分布密切相关，这导致了另一种表达方式。给定一个来自平均泊松分佈的观测值 k，一个带有置信水平的置信区间是