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The measure associated to the free Poisson law is given by

The measure associated to the free Poisson law is given by

 

 

====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>

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

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

给出了自由泊松定律的 r- 变换

以下给出自由泊松定律的 r- 变换

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)是由

柯西变换(即 Stieltjes 变换的负变换)由以下给出

The S-transform is given by

The S-transform is given by

给出了 s 变换的一般形式，并给出了 s 变换的一般形式

以下给出了 s 变换的一般形式

== Statistical Inference ==

== Statistical Inference 统计学推论==

{{See also|Poisson regression}}

{{See also|Poisson regression}}

 

=== Parameter estimation参数估计 ===

=== Parameter estimation ===

Given a sample of ''n'' measured values <math> k_i \in \{0,1,...\}</math>, for ''i''&nbsp;=&nbsp;1,&nbsp;...,&nbsp;''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 <ref>{{cite web |last=Paszek|first=Ewa |title=Maximum Likelihood Estimation – Examples |url = http://cnx.org/content/m13500/latest/?collection=col10343/latest}}</ref>

Given a sample of ''n'' measured values <math> k_i \in \{0,1,...\}</math>, for ''i''&nbsp;=&nbsp;1,&nbsp;...,&nbsp;''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 <ref>{{cite web |last=Paszek|first=Ewa |title=Maximum Likelihood Estimation – Examples |url = http://cnx.org/content/m13500/latest/?collection=col10343/latest}}</ref>

Given a sample of n measured values <math> k_i \in \{0,1,...\}</math>, for i&nbsp;=&nbsp;1,&nbsp;...,&nbsp;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  

Given a sample of n measured values <math> k_i \in \{0,1,...\}</math>, for i&nbsp;=&nbsp;1,&nbsp;...,&nbsp;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  

给定一个 n 个测量值的样本{0,1，... } </math > ，对于 i = 1，... ，n，我们希望估计取样的泊松总体参数的值。最大似然估计是

给定一个 n 个测量值的样本{0,1，... } </math > ，对于 i = 1，... ，n，我们希望估计取样的泊松总体参数的值。'''<font color="#ff8000"> 最大似然估计The maximum likelihood estimate</font>'''是

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.

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.

因为每个观测值都有期望值，所以样本的意义也是如此。因此，最大似然估计是。由于其方差达到了 CRLB 下界，因此它也是一个有效的估计量。因此它是最小方差无偏的。也可以证明和(因此样本平均值是和的单射)是一个完整充分的统计量。

因为每个观测值都有期望值，所以样本的意义也是如此。因此，'''<font color="#ff8000"> 最大似然估计The maximum likelihood estimate</font>'''是。由于其方差达到了 CRLB 下界，因此它也是一个有效的估计量。它是最小方差无偏的。也可以证明和(因此样本平均值是和的单射)是一个完整充分的统计量。

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 <math>\mathbf{x}</math> (called <math>h(\mathbf{x})</math>) and one that depends on the parameter <math>\lambda</math> and the sample <math>\mathbf{x}</math> only through the function <math>T(\mathbf{x})</math>. Then <math>T(\mathbf{x})</math> is a sufficient statistic for <math>\lambda</math>.

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 <math>\mathbf{x}</math> (called <math>h(\mathbf{x})</math>) and one that depends on the parameter <math>\lambda</math> and the sample <math>\mathbf{x}</math> only through the function <math>T(\mathbf{x})</math>. Then <math>T(\mathbf{x})</math> is a sufficient statistic for <math>\lambda</math>.

为了证明充分性，我们可以用因子分解定理。考虑将样本的关节概率质量函数分成两部分: 一部分仅依赖于样本 < math > mathbf { x } </math > (称为 < math > h (mathbf { x }) </math >) ，另一部分依赖于参数 < math > lambda </math > 和样本 < math > mathbf { x } </math > 只通过函数 math < t (mathbf { x }) </math > 。那么 < math > t (mathbf { x }) </math > 就是 < math > lambda </math > 的一个充分的统计量。

为了证明充分性，我们可以用'''<font color="#ff8000"> 因子分解定理Factorization theorem</font>'''。考虑将'''<font color="#ff8000"> 联合泊松分布Joint Poisson distribution</font>'''的'''<font color="#ff8000"> 概率质量函数Probability mass function</font>'''分成两部分: 一部分仅依赖于样本 < math > mathbf { x } </math > (称为 < math > h (mathbf { x }) </math >) ，另一部分依赖于参数 < math > lambda </math > 和样本 < math > mathbf { x } </math > 只通过函数 math < t (mathbf { x }) </math > 。那么 < math > t (mathbf { x }) </math > 就是 < math > lambda </math > 的一个充分的统计量。

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

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

第一个学期，< math > h (mathbf { x }) </math > ，仅依赖于 < math > mathbf { x } </math > 。第二个术语，< math > g (t (mathbf { x }) | lambda) </math > ，仅通过 < math > t (mathbf { x }) = sum { i = 1} ^ nx _ i </math > 取决于样本。因此，< math > t (mathbf { x }) </math > 就足够了。

第一个术语，< math > h (mathbf { x }) </math > ，仅依赖于 < math > mathbf { x } </math > 。第二个术语，< math > g (t (mathbf { x }) | lambda) </math > ，仅通过 < math > t (mathbf { x }) = sum { i = 1} ^ nx _ i </math > 取决于样本。因此，< math > t (mathbf { x }) </math > 就足够了。

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:

为了找到泊松种群概率密度函数最大的参数，我们可以使用似然函数的对数:

为了找到泊松族群概率密度函数最大的参数λ ，我们可以使用'''<font color="#ff8000"> 似然函数Likelihood function</font>'''的对数:

Solving for λ gives a stationary point.

Solving for λ gives a stationary point.