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第291行： − + + + − ===Genera通常l===+ − --[[用户:fairywang|fairywang]]([[用户讨论:fairywang|讨论]]) 【审校】“Genera通常l”改为“General通常” − * If <math>X_1 \sim \mathrm{Pois}(\lambda_1)\,</math> and <math>X_2 \sim \mathrm{Pois}(\lambda_2)\,</math> are independent, then the difference <math> Y = X_1 - X_2</math> follows a [[Skellam distribution]]. − * If <math>X_1 \sim \mathrm{Pois}(\lambda_1)\,</math> and <math>X_2 \sim \mathrm{Pois}(\lambda_2)\,</math> are independent, then the distribution of <math>X_1</math> conditional on <math>X_1+X_2</math> is a [[binomial distribution]]. − + − Specifically, if <math>X_1+X_2=k</math>, then <math>\!X_1\sim \mathrm{Binom}(k, \lambda_1/(\lambda_1+\lambda_2))</math>.+ − 具体来说，如果 < math > x _ 1 + x _ 2 = k </math > ，那么 < math > ! x _ 1 sim mathrm { Binom }(k，lambda _ 1/(lambda _ 1 + lambda _ 2)) </math > 。 − − :More generally, if ''X''<sub>1</sub>, ''X''<sub>2</sub>,..., ''X''<sub>''n''</sub> are independent Poisson random variables with parameters ''λ''<sub>1</sub>, ''λ''<sub>2</sub>,..., ''λ''<sub>''n''</sub> then − − More generally, if X<sub>1</sub>, X<sub>2</sub>,..., X<sub>n</sub> are independent Poisson random variables with parameters λ<sub>1</sub>, λ<sub>2</sub>,..., λ<sub>n</sub> then − − 更一般地说，如果 x < sub > 1 </sub > ，x < sub > 2 </sub > ，... ，x < sub > n </sub > 是独立的随机变量，参数 < sub > 1 </sub > ，< sub > 2 </sub > ，... ，< sub > n </sub > 然后 − − :: given <math>\sum_{j=1}^n X_j=k,</math> <math>X_i \sim \mathrm{Binom}\left(k, \frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right)</math>. In fact, <math>\{X_i\} \sim \mathrm{Multinom}\left(k, \left\{\frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right\}\right)</math>. − − given <math>\sum_{j=1}^n X_j=k,</math> <math>X_i \sim \mathrm{Binom}\left(k, \frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right)</math>. In fact, <math>\{X_i\} \sim \mathrm{Multinom}\left(k, \left\{\frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right\}\right)</math>. − − 给定的数学公式 sum { j = 1} ^ n x _ j = k，</math > x _ i sim mathrm { Binom }左(k，frac { lambda _ i }{ sum { j = 1} ^ n lambda _ j }右) </math > 。事实上，[ math ]{ xi } sim mathrm { mothom } left (k，left { frac { lambda _ i }{ sum { j = 1} ^ n lambda _ j } right }}) </math > 。 第323行：
第309行： − <math>F_\mathrm{Binomial}(k;n, p) \approx F_\mathrm{Poisson}(k;\lambda=np)\,</math> − (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.{{r|Zhang2013|Zhang2016}} The discrete compound Poisson distribution can be deduced from the limiting distribution of univariate multinomial distribution. It is also a [[compound Poisson distribution#Special cases|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]] 。{{r|Zhang2013|Zhang2016}}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 <math>\sqrt{\lambda}</math>) 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). − <math>F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)\,</math> − − (x; λ) approx f _ mathrm { normal }(x; mu = lambda，sigma ^ 2 = lambda) ，</math > − + − − <math>Y = 2 \sqrt{X} \approx \mathcal{N}(2\sqrt{\lambda};1)</math>, − − (2 sqrt { lambda } ; 1) </math > , − and − − 及 − + − <math>Y = \sqrt{X} \approx \mathcal{N}(\sqrt{\lambda};1/4)</math>. − + − :Under this transformation, the convergence to normality (as <math>\lambda</math> increases) is far faster than the untransformed variable.{{Citation needed|date=May 2012}} Other, slightly more complicated, variance stabilizing transformations are available,{{r|Johnson2005|p=168}} one of which is [[Anscombe transform]].{{r|Anscombe1948}} See [[Data transformation (statistics)]] for more general uses of transformations. − − Under this transformation, the convergence to normality (as <math>\lambda</math> 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. − − 在这种转换下，收敛到正态的速度(如 < math > lambda </math > 增加)远远快于未转换的变量。还有一些稍微复杂一些的稳定方差的变换，其中之一就是安斯科姆变换。有关转换的更多一般用途，请参见数据转换(统计信息)。 − * 第373行：
第338行： − <math>F_\text{Poisson}(k;\lambda) = 1-F_{\chi^2}(2\lambda;2(k+1)) \quad\quad \text{ integer } k,</math> − 1-F _ { chi ^ 2}(2 lambda; 2(k + 1)) quad text { integer } k，</math > + − − − − − 及+ − ::<math>\Pr(X=k)=F_{\chi^2}(2\lambda;2(k+1)) -F_{\chi^2}(2\lambda;2k) . − <math>\Pr(X=k)=F_{\chi^2}(2\lambda;2(k+1)) -F_{\chi^2}(2\lambda;2k) . − + + − </math> − − </math> − − 数学 − − − − ===''' Poisson Approximation 泊松近似'''=== − − − − Assume <math>X_1\sim\operatorname{Pois}(\lambda_1), X_2\sim\operatorname{Pois}(\lambda_2), \dots, X_n\sim\operatorname{Pois}(\lambda_n)</math> where <math>\lambda_1 + \lambda_2 + \dots + \lambda_n=1</math>, then<ref>{{Cite web | url=https://newonlinecourses.science.psu.edu/stat504/node/48/ | title=1.7.7 – Relationship between the Multinomial and Poisson | STAT 504}}</ref> <math>(X_1, X_2, \dots, X_n)</math> is [[Multinomial distribution|multinomially distributed]] − − Assume <math>X_1\sim\operatorname{Pois}(\lambda_1), X_2\sim\operatorname{Pois}(\lambda_2), \dots, X_n\sim\operatorname{Pois}(\lambda_n)</math> where <math>\lambda_1 + \lambda_2 + \dots + \lambda_n=1</math>, then <math>(X_1, X_2, \dots, X_n)</math> is multinomially distributed − − 假设 x _ 1 sim 操作者名为{ Pois }(lambda _ 1) ，x _ 2 sim 操作者名为{ Pois }(lambda _ 2) ，点，xn sim 操作者名为{ Pois }(lambda _ n) </math > 其中 < math > lambda _ 1 + lambda _ 2 + dots + lambda _ n = 1 </math 多项式 > ，那么 < math > (x _ 1，x _ 2，dots，x _ n) </math > 是统一分布的 − <math>(X_1, X_2, \dots, X_n) \sim \operatorname{Mult}(N, \lambda_1, \lambda_2, \dots, \lambda_n)</math> conditioned on <math>N = X_1 + X_2 + \dots X_n</math>. − − (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{{r|Mitzenmacher2005|p=101-102}}, among other things, that for any nonnegative function <math>f(x_1,x_2,\dots,x_n)</math>, − − This means, among other things, that for any nonnegative function <math>f(x_1,x_2,\dots,x_n)</math>, − − 这意味着，对于任何非负函数 f (x _ 1，x _ 2，dots，x _ n) </math > , − if <math>(Y_1, Y_2, \dots, Y_n)\sim\operatorname{Mult}(m, \mathbf{p})</math> is multinomially distributed, then+ − if <math>(Y_1, Y_2, \dots, Y_n)\sim\operatorname{Mult}(m, \mathbf{p})</math> is multinomially distributed, then − + − − <math> − − 《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)] − − 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)] − </math> − 数学+ − where <math>(X_1, X_2, \dots, X_n)\sim\operatorname{Pois}(\mathbf{p})</math>. − where <math>(X_1, X_2, \dots, X_n)\sim\operatorname{Pois}(\mathbf{p})</math>.+ − 其中 < math > (x _ 1，x _ 2，dots，x _ n) sim 操作员名称{ Pois }(mathbf { p }) </math > 。 − − − − The factor of <math>e\sqrt{m}</math> can be removed if <math>f</math> is further assumed to be monotonically increasing or decreasing. − − The factor of <math>e\sqrt{m}</math> can be removed if <math>f</math> is further assumed to be monotonically increasing or decreasing. − − 如果进一步假定 < math > f </math > 是单调递增或递减的，则可以去掉 < math > e sqrt { m } </math > 的因子。 − − − − === ''' 二元泊松分布Bivariate Poisson distribution'''=== − − This distribution has been extended to the [[bivariate]] case.{{r|Loukas1986}} The [[generating function]] for this distribution is − − This distribution has been extended to the bivariate case. The generating function for this distribution is + − − <math> g( u, v ) = \exp[ ( \theta_1 - \theta_{12} )( u - 1 ) + ( \theta_2 - \theta_{12} )(v - 1) + \theta_{12} ( uv - 1 ) ] </math> − − < math > g (u，v) = exp [(theta _ 1-theta _ {12})(u-1) + (theta _ 2-theta _ {12})(v-1) + theta _ {12}(uv-1)] </math > − − − − with − − with 第485行：
第379行： − <math> \theta_1, \theta_2 > \theta_{ 12 } > 0 \, </math> − − 1，theta 2，theta {12} > 0，</math > − − − − The marginal distributions are Poisson(''θ''<sub>1</sub>) and Poisson(''θ''<sub>2</sub>) and the correlation coefficient is limited to the range − The marginal distributions are Poisson(θ<sub>1</sub>) and Poisson(θ<sub>2</sub>) and the correlation coefficient is limited to the range+ − − 边缘分布为 Poisson (< sub > 1 </sub >)和 Poisson (< sub > 2 </sub >) ，相关系数仅限于一定范围 − <math> 0 \le \rho \le \min\left\{ \frac{ \theta_1 }{ \theta_2 }, \frac{ \theta_2 }{ \theta_1 } \right\}</math> − [数学][数学][数学][数学]+ − − A simple way to generate a bivariate Poisson distribution <math>X_1,X_2</math> is to take three independent Poisson distributions <math>Y_1,Y_2,Y_3</math> with means <math>\lambda_1,\lambda_2,\lambda_3</math> and then set <math>X_1 = Y_1 + Y_3,X_2 = Y_2 + Y_3</math>. The probability function of the bivariate Poisson distribution is − − A simple way to generate a bivariate Poisson distribution <math>X_1,X_2</math> is to take three independent Poisson distributions <math>Y_1,Y_2,Y_3</math> with means <math>\lambda_1,\lambda_2,\lambda_3</math> and then set <math>X_1 = Y_1 + Y_3,X_2 = Y_2 + Y_3</math>. The probability function of the bivariate Poisson distribution is − − 一个简单的方法来产生一个二变量的泊松分布分布: 取3个独立的 Poisson 分布 < math > y _ 1，y _ 2，y _ 3 </math > ，用 < math > lambda _ 1，lambda _ 2，lambda _ 3 </math > 然后设置 < math > x _ 1 = y _ 1 + y _ 3，x _ 2 = y _ 2 + y _ 3 </math > 。二元概率密度函数变量的泊松分布变量是 − --[[用户:fairywang|fairywang]]([[用户讨论:fairywang|讨论]]) 【审校】“泊松分布分布”改为“泊松分布”，“泊松分布变量”改为“泊松分布变量” − − <math> − − 《数学》 − − \begin{align} − − − 开始{ align } − − & \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! }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} − − 结束{ align } − − </math> − − 数学 − − − − ===''' 自由泊松分布Free Poisson distribution'''=== − − The free Poisson distribution<ref>Free Random Variables by D. Voiculescu, K. Dykema, A. Nica, CRM Monograph Series, American Mathematical Society, Providence RI, 1992</ref> with jump size <math>\alpha</math> and rate <math>\lambda</math> arises in [[free probability]] theory as the limit of repeated [[free convolution]] − − The free Poisson distribution with jump size <math>\alpha</math> and rate <math>\lambda</math> arises in free probability theory as the limit of repeated free convolution − − 带有跳跃大小 < math > alpha </math > 和速率 < math > lambda </math > 的''' 自由泊松分布Free Poisson distribution'''作为重复自由卷积的极限在自由概率论中出现 + + − − <math> − − 《数学》 − − \left( \left(1-\frac{\lambda}{N}\right)\delta_0 + \frac{\lambda}{N}\delta_\alpha\right)^{\boxplus N}</math> − − 左(左(1-frac { lambda }{ n }右) delta _ 0 + frac { lambda }{ n } delta _ alpha 右) ^ { boxplus n } </math > − − − − as N → ∞. − − as N → ∞. − − − − In other words, let <math>X_N</math> be random variables so that <math>X_N</math> has value <math>\alpha</math> with probability <math>\frac{\lambda}{N}</math> and value 0 with the remaining probability. Assume also that the family <math>X_1,X_2,\ldots</math> are [[free independence|freely independent]]. Then the limit as <math>N\to\infty</math> of the law of <math>X_1+\cdots +X_N</math> − − In other words, let <math>X_N</math> be random variables so that <math>X_N</math> has value <math>\alpha</math> with probability <math>\frac{\lambda}{N}</math> and value 0 with the remaining probability. Assume also that the family <math>X_1,X_2,\ldots</math> are freely independent. Then the limit as <math>N\to\infty</math> of the law of <math>X_1+\cdots +X_N</math> − − 换句话说，让 x _ n </math > 是随机变量，因此 x _ n </math > 具有值 < math > alpha </math > 具有概率 < math > frac { lambda }{ n } </math > ，值0具有剩余的概率。同时假设家庭成员 x1，x2，ldots </math > 是自由独立的。然后将极限值设为 < math > n，以确定 < math > x1 + cdots + xn </math > 的规律 − --[[用户:fairywang|fairywang]]([[用户讨论: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 <math>\lambda,\alpha</math>. − − is given by the Free Poisson law with parameters <math>\lambda,\alpha</math>. − − 是由带参数的自由泊松定律给出的。 + − 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''' 泊松分布Poisson distribution.'''的一种方法。 − --[[用户:fairywang|fairywang]]([[用户讨论:fairywang|讨论]]) 【审校】“泊松分布 ”改为“泊松分布 ”+ − − The measure associated to the free Poisson law is given by<ref>James A. Mingo, Roland Speicher: Free Probability and Random Matrices. Fields Institute Monographs, Vol. 35, Springer, New York, 2017.</ref> − − 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 − − <math>\mu=\begin{cases} (1-\lambda) \delta_0 + \lambda \nu,& \text{if } 0\leq \lambda \leq 1 \\ − − < math > mu = begin { cases }(1-lambda) delta _ 0 + lambda nu，& text { if }0 leq lambda leq 1 − − \nu, & \text{if }\lambda >1, − − 1，& text { if } lambda > 1, − − \end{cases} − − 结束{ cases } − − </math> − − 数学 − − − − where − − where − − − <math>\nu = \frac{1}{2\pi\alpha t}\sqrt{4\lambda \alpha^2 - ( t - \alpha (1+\lambda))^2} \, dt</math> − − 4 lambda alpha ^ 2-(t-alpha (1 + lambda)) ^ 2} ，dt </math > − − − − and has support <math>[\alpha (1-\sqrt{\lambda})^2,\alpha (1+\sqrt{\lambda})^2]</math>. − − and has support <math>[\alpha (1-\sqrt{\lambda})^2,\alpha (1+\sqrt{\lambda})^2]</math>. − − 并且支持 < math > [ alpha (1-sqrt { lambda }) ^ 2，alpha (1 + sqrt { lambda }) ^ 2] </math > 。 − − − − This law also arises in [[random matrix]] theory as the [[Marchenko–Pastur law]]. Its [[free cumulants]] are equal to <math>\kappa_n=\lambda\alpha^n</math>. − − This law also arises in random matrix theory as the Marchenko–Pastur law. Its free cumulants are equal to <math>\kappa_n=\lambda\alpha^n</math>. − − 这个定律也出现在随机矩阵理论中，称为马尔琴科-帕斯图定律。它的自由累积量等于 < math > kappa _ n = lambda alpha ^ n </math > 。 − − − − ====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 − 我们给出了自由泊松定律的一些重要变换的值。在 a. Nica 和 r. Speicher 合著的《自由概率组合学讲座》一书中+ + − The [[R-transform]] of the free Poisson law is given by+ − − The R-transform of the free Poisson law is given by − − 以下给出自由泊松定律的 r- 变换 + − − <math>R(z)=\frac{\lambda \alpha}{1-\alpha z}. </math> − − 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 − − 柯西变换(即 Stieltjes 变换的负变换)由以下给出 + − − <math> − − 《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 } − − </math> − − 数学+ − − − − The [[S-transform]] is given by − − The S-transform is given by − − 以下给出了 s 变换的一般形式 − − − − <math> − − 《数学》 − − S(z) = \frac{1}{z+\lambda} − − − (z) = frac {1}{ z + lambda } − − </math> − − − 数学 − − − − in the case that <math>\alpha=1</math>. − − in the case that <math>\alpha=1</math>. − +
→' 相关分布'Related distributions
<br>
<br>
== ''' 相关分布'Related distributions''==
==相关分布==
===通常===
* 如果<math>X_1 \sim \mathrm{Pois}(\lambda_1)\,</math> 且<math>X_2 \sim \mathrm{Pois}(\lambda_2)\,</math>独立, 则差值<math> Y = X_1 - X_2</math> 遵循[[Skellam分布]].
:具体来说，如果<math>X_1+X_2=k</math> ，那么<math>\!X_1\sim \mathrm{Binom}(k, \lambda_1/(\lambda_1+\lambda_2))</math>。
:Specifically, if <math>X_1+X_2=k</math>, then <math>\!X_1\sim \mathrm{Binom}(k, \lambda_1/(\lambda_1+\lambda_2))</math>.
:更一般地说，如果''X''<sub>1</sub>, ''X''<sub>2</sub>,..., ''X''<sub>''n''</sub> 是独立的随机变量，参数λ<sub>1</sub>, λ<sub>2</sub>,..., λ<sub>n</sub>然后
:: 给定 <math>\sum_{j=1}^n X_j=k,</math> <math>X_i \sim \mathrm{Binom}\left(k, \frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right)</math>. 事实上， <math>\{X_i\} \sim \mathrm{Multinom}\left(k, \left\{\frac{\lambda_i}{\sum_{j=1}^n\lambda_j}\right\}\right)</math>.
* If <math>X \sim \mathrm{Pois}(\lambda)\,</math> and the distribution of <math>Y</math>, conditional on ''X'' = ''k'', is a [[binomial distribution]], <math>Y \mid (X = k) \sim \mathrm{Binom}(k, p)</math>, then the distribution of Y follows a Poisson distribution <math>Y \sim \mathrm{Pois}(\lambda \cdot p)\,</math>. In fact, if <math>\{Y_i\} </math>, conditional on X = k, follows a multinomial distribution, <math>\{Y_i\} \mid (X = k) \sim \mathrm{Multinom}\left(k, p_i\right)</math>, then each <math>Y_i</math> follows an independent Poisson distribution <math>Y_i \sim \mathrm{Pois}(\lambda \cdot p_i), \rho(Y_i, Y_j) = 0</math>.
* If <math>X \sim \mathrm{Pois}(\lambda)\,</math> and the distribution of <math>Y</math>, conditional on ''X'' = ''k'', is a [[binomial distribution]], <math>Y \mid (X = k) \sim \mathrm{Binom}(k, p)</math>, then the distribution of Y follows a Poisson distribution <math>Y \sim \mathrm{Pois}(\lambda \cdot p)\,</math>. In fact, if <math>\{Y_i\} </math>, conditional on X = k, follows a multinomial distribution, <math>\{Y_i\} \mid (X = k) \sim \mathrm{Multinom}\left(k, p_i\right)</math>, then each <math>Y_i</math> follows an independent Poisson distribution <math>Y_i \sim \mathrm{Pois}(\lambda \cdot p_i), \rho(Y_i, Y_j) = 0</math>.
:: <math>F_\mathrm{Binomial}(k;n, p) \approx F_\mathrm{Poisson}(k;\lambda=np)\,</math>
:: <math>F_\mathrm{Binomial}(k;n, p) \approx F_\mathrm{Poisson}(k;\lambda=np)\,</math>
*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 。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的一个特例。
*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 <math>\sqrt{\lambda}</math>），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)。
*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 <math>\sqrt{\lambda}</math>），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)。
:: <math>F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)\,</math>
:: <math>F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)\,</math>
* [[Variance-stabilizing transformation]]: If <math>X \sim \mathrm{Pois}(\lambda)\,</math>, then
* [[Variance-stabilizing transformation]]: If <math>X \sim \mathrm{Pois}(\lambda)\,</math>, then
::<math>Y = 2 \sqrt{X} \approx \mathcal{N}(2\sqrt{\lambda};1)</math>,{{r|Johnson2005|p=168}}
::<math>Y = 2 \sqrt{X} \approx \mathcal{N}(2\sqrt{\lambda};1)</math>,
:and
:and
::<math>Y = \sqrt{X} \approx \mathcal{N}(\sqrt{\lambda};1/4)</math>.{{r|McCullagh1989|p=196}}
::<math>Y = \sqrt{X} \approx \mathcal{N}(\sqrt{\lambda};1/4)</math>.
(sqrt { lambda } ; 1/4) </math > .
:在这种转换下，收敛到正态的速度(如<math>\lambda</math>增加)远远快于未转换的变量。还有一些稍微复杂一些的稳定方差的变换，其中之一就是安斯科姆变换。有关转换的更多一般用途，请参见数据转换(统计信息)。
* 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 distribution|exponential]] random variables having mean 1/''λ''.{{r|Ross2010|p=317–319}}
* 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 distribution|exponential]] random variables having mean 1/''λ''.{{r|Ross2010|p=317–319}}
* The [[cumulative distribution function]]s of the Poisson and [[chi-squared distribution]]s are related in the following ways:{{r|Johnson2005|p=167}}
* The [[cumulative distribution function]]s of the Poisson and [[chi-squared distribution]]s are related in the following ways:{{r|Johnson2005|p=167}}
::<math>F_\text{Poisson}(k;\lambda) = 1-F_{\chi^2}(2\lambda;2(k+1)) \quad\quad \text{ integer } k,</math>
::<math>F_\text{Poisson}(k;\lambda) = 1-F_{\chi^2}(2\lambda;2(k+1)) \quad\quad \text{ integer } k,</math>
:and
:and{{r|Johnson2005|p=158}}
and
::<math>\Pr(X=k)=F_{\chi^2}(2\lambda;2(k+1)) -F_{\chi^2}(2\lambda;2k) .</math>
< math > Pr (x = k) = f { chi ^ 2}(2 lambda; 2(k + 1))-f { chi ^ 2}(2 lambda; 2k).
===泊松近似===
假设<math>X_1\sim\operatorname{Pois}(\lambda_1), X_2\sim\operatorname{Pois}(\lambda_2), \dots, X_n\sim\operatorname{Pois}(\lambda_n)</math>其中 <math>\lambda_1 + \lambda_2 + \dots + \lambda_n=1</math>，那么<ref>{{Cite web | url=https://newonlinecourses.science.psu.edu/stat504/node/48/ | title=1.7.7 – Relationship between the Multinomial and Poisson | STAT 504}}</ref> <math>(X_1, X_2, \dots, X_n)</math>是统一分布的。
<math>(X_1, X_2, \dots, X_n) \sim \operatorname{Mult}(N, \lambda_1, \lambda_2, \dots, \lambda_n)</math> conditioned on <math>N = X_1 + X_2 + \dots X_n</math>.
<math>(X_1, X_2, \dots, X_n) \sim \operatorname{Mult}(N, \lambda_1, \lambda_2, \dots, \lambda_n)</math> conditioned on <math>N = X_1 + X_2 + \dots X_n</math>.
这意味着，对于任何非负函数<math>f(x_1,x_2,\dots,x_n)</math> ,
如果[ math ](y _ 1，y _ 2，dots，y _ n) sim 操作符名称{ Mult }(m，mathbf { p }) </math > 是多项式分布，则
如果 <math>(Y_1, Y_2, \dots, Y_n)\sim\operatorname{Mult}(m, \mathbf{p})</math> 是多项式分布，则
:<math>
:<math>
\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)]
</math>
</math>
其中 <math>(X_1, X_2, \dots, X_n)\sim\operatorname{Pois}(\mathbf{p})</math>。
如果进一步假定<math>f</math>是单调递增或递减的，则可以去掉<math>e\sqrt{m}</math> 的因子。
=== 二元泊松分布===
这种分布已经扩展到二元情况。这个分布的母函数是
这种分布已经扩展到二元情况。这个分布的母函数是
: <math> g( u, v ) = \exp[ ( \theta_1 - \theta_{12} )( u - 1 ) + ( \theta_2 - \theta_{12} )(v - 1) + \theta_{12} ( uv - 1 ) ] </math>
: <math> g( u, v ) = \exp[ ( \theta_1 - \theta_{12} )( u - 1 ) + ( \theta_2 - \theta_{12} )(v - 1) + \theta_{12} ( uv - 1 ) ] </math>
与
与
: <math> \theta_1, \theta_2 > \theta_{ 12 } > 0 \, </math>
: <math> \theta_1, \theta_2 > \theta_{ 12 } > 0 \, </math>
边缘分布为Poisson(''θ''<sub>1</sub>)和Poisson(''θ''<sub>2</sub>)，相关系数仅限于一定范围
: <math> 0 \le \rho \le \min\left\{ \frac{ \theta_1 }{ \theta_2 }, \frac{ \theta_2 }{ \theta_1 } \right\}</math>
: <math> 0 \le \rho \le \min\left\{ \frac{ \theta_1 }{ \theta_2 }, \frac{ \theta_2 }{ \theta_1 } \right\}</math>
一个简单的方法来产生一个二变量的泊松分布<math>X_1,X_2</math>: 取3个独立的 Poisson 分布 <math>Y_1,Y_2,Y_3</math> ，用<math>\lambda_1,\lambda_2,\lambda_3</math>然后设置 <math>X_1 = Y_1 + Y_3,X_2 = Y_2 + Y_3</math>。二元概率密度函数变量的泊松分布变量是
: <math>
: <math>
\begin{align}
\begin{align}
& \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
\end{align}
\end{align}
</math>
</math>
===自由泊松分布===
带有跳跃大小<math>\alpha</math>和速率<math>\lambda</math>的自由泊松分布<ref>Free Random Variables by D. Voiculescu, K. Dykema, A. Nica, CRM Monograph Series, American Mathematical Society, Providence RI, 1992</ref>作为重复自由卷积的极限在自由概率论中出现
: <math>
: <math>
\left( \left(1-\frac{\lambda}{N}\right)\delta_0 + \frac{\lambda}{N}\delta_\alpha\right)^{\boxplus N}</math>
\left( \left(1-\frac{\lambda}{N}\right)\delta_0 + \frac{\lambda}{N}\delta_\alpha\right)^{\boxplus N}</math>
as ''N'' → ∞.
as ''N'' → ∞.
换句话说，让<math>X_N</math>是随机变量，因此<math>X_N</math>具有值<math>\alpha</math>具有概率<math>\frac{\lambda}{N}</math>，值0具有剩余的概率。同时假设集合<math>X_1,X_2,\ldots</math>是自由独立的。然后将极限值设为<math>N\to\infty</math>以确定<math>X_1+\cdots +X_N</math>的规律是由带参数<math>\lambda,\alpha</math>的自由泊松定律给出的。
这个定义类似于从(经典)泊松过程获得经典泊松分布的一种方法。
与自由泊松定律相关的测度由以下给出<ref>James A. Mingo, Roland Speicher: Free Probability and Random Matrices. Fields Institute Monographs, Vol. 35, Springer, New York, 2017.</ref>
:<math>\mu=\begin{cases} (1-\lambda) \delta_0 + \lambda \nu,& \text{if } 0\leq \lambda \leq 1 \\
:<math>\mu=\begin{cases} (1-\lambda) \delta_0 + \lambda \nu,& \text{if } 0\leq \lambda \leq 1 \\
\nu, & \text{if }\lambda >1,
\nu, & \text{if }\lambda >1,
\end{cases}
\end{cases}
</math>
</math>
其中
其中
: <math>\nu = \frac{1}{2\pi\alpha t}\sqrt{4\lambda \alpha^2 - ( t - \alpha (1+\lambda))^2} \, dt</math>
: <math>\nu = \frac{1}{2\pi\alpha t}\sqrt{4\lambda \alpha^2 - ( t - \alpha (1+\lambda))^2} \, dt</math>
并且支持<math>[\alpha (1-\sqrt{\lambda})^2,\alpha (1+\sqrt{\lambda})^2]</math>。
这个定律也出现在随机矩阵理论中，称为'''马尔琴科-帕斯图定律 Marchenko–Pastur law'''。它的自由累积量等于<math>\kappa_n=\lambda\alpha^n</math>。
====这一定律的一些变换====
我们给出了自由泊松定律的一些重要变换的值。在 a. Nica 和 r. Speicher 合著的《自由概率组合学讲座 Lectures on the Combinatorics of Free Probability》一书中<ref>Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher, pp. 203–204, Cambridge Univ. Press 2006</ref>
以下给出自由泊松定律的R-变换
: <math>R(z)=\frac{\lambda \alpha}{1-\alpha z}. </math>
: <math>R(z)=\frac{\lambda \alpha}{1-\alpha z}. </math>
柯西变换 Cauchy transform(即 Stieltjes变换的负变换)由以下给出
: <math>
: <math>
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}
</math>
</math>
以下给出了S变换的一般形式
: <math>
: <math>
S(z) = \frac{1}{z+\lambda}
S(z) = \frac{1}{z+\lambda}
</math>
</math>
在这种情况下。
在这种情况下。
<br>
==统计学推论==
==统计学推论==