## E-Bayesian Estimation and Its E-MSE of Poisson Distribution Parameter Under Different Loss Functions

Han Ming,

 基金资助: 浙江省自然科学基金.  LY18A010026

 Fund supported: the Natural Science Foundation of Zhejiang Province.  LY18A010026

Abstract

In order to measure the error of E-Bayesian estimation, this paper the definition of E-MSE(expected mean square error) is introduced based on the definition of E-Bayesian estimation. For parameter of Poisson distribution, under different loss functions (including:squared error loss, K-loss and weighted squared error loss), the formulas of E-Bayesian estimation and formulas of E-MSE are given respectively. Monte Carlo simulations are performed to compare the performances of the proposed methods of estimation and a real data set have been analysed for illustrative purposes, results are compared on the basis of E-MSE, the results show that the proposed method is feasible and convenient for application.

Keywords： Poisson distribution ; E-Bayesian estimation ; E-MSE ; Loss function ; Monte Carlo method

Han Ming. E-Bayesian Estimation and Its E-MSE of Poisson Distribution Parameter Under Different Loss Functions. Acta Mathematica Scientia[J], 2019, 39(3): 664-673 doi:

## 2 E-Bayes估计及其E-MSE的定义

$P\{ X=x_{i}\}=\frac{e^{-\lambda}\lambda ^{x_{i}}}{x_{i}!}, \ \lambda>0, x_{i}=0, 1, 2, \cdots.$

$x=(x_{1}, x_{2}, \cdots, x_{n})$为来自Poisson分布(2.1)的样本观察值,则样本的似然函数为

$L(x|\lambda)=\prod\limits_{i=1}^n \frac{e^{-\lambda}\lambda ^{x_{i}}}{x_{i}!}=\frac{e^{-n\lambda}\lambda^{T}}{\prod\limits_{i=1}^n x_{i}!},$

$\pi(\lambda|a, b)=\frac{b^{a}\lambda^{a-1}\exp(-b\lambda)}{\Gamma(a)}, \ \lambda>0,$

$\widehat{\lambda}_{EB}$的E-MSE (expected mean square error).其中$\iint_{D} MSE[\widehat{\lambda}_{B}(a, b)]\pi(a, b)\mbox{d}a\mbox{d}b$是存在的, $D$为超参数$a$$b取值的集合, \pi(a, b)$$a$$b在集合D上的密度函数, MSE[\widehat \lambda_{B}(a, b)]$$\widehat{\lambda}_{B}(a, b)$的MSE (用超参数$a$$b表示). 从定义2.2可以看出, \widehat \lambda_{EB}的E-MSE \lambda的Bayes估计的MSE (MSE[\widehat{\lambda}_{B}(a, b)])对超参数的数学期望. ## 3 不同损失函数下参数的Bayes估计 损失函数在贝叶斯统计推断中是非常重要的,最常用的是平方损失函数下的贝叶斯估计.根据文献[1],有如下引理3.1. 引理3.1 设x=(x_{1}, x_{2}, \cdots, x_{n})为来自某总体的样本观察值, \theta是该总体的未知参数,在不同损失函数下,有以下结论: (ⅰ)在平方损失函数L_{1}(\theta, \delta)=(\theta-\delta)^{2}下, \theta的Bayes估计为\widehat{\theta}_{B1}(x)=E(\theta|x); (ⅱ)在K -损失函数L_{2}(\theta, \delta)=\Big(\sqrt{\frac{\theta}{\delta}}- \sqrt{\frac{\delta}{\theta}}\Big)^{2}下, \theta的Bayes估计为\widehat{\theta}_{B2}(x)=\sqrt{\frac{E(\theta|x)}{E(\theta^{-1}|x)}}; (ⅲ)在加权平方损失函数L_{3}(\theta, \delta)=\theta^{-1}(\theta-\delta)^{2}下, \theta的Bayes估计为\widehat{\theta}_{B3}(x)=\left[E(\theta^{-1}|x)\right]^{-1}. 这里\delta是参数\theta的一个估计. ## 4 不同损失函数下\lambda的E-Bayes估计 定理4.1 设x_{1}, x_{2}, \cdots, x_{n}为来自Poisson分布(2.1)式的样本观察值,似然函数由(2.2)式给出,若\lambda的先验分布为Gamma分布,其密度函数由(2.3)式给出,超参数a$$b$的先验密度函数由

$\pi(a, b)=\frac{1}{c}, \ 0<a<1, 0<b<c$

(ⅰ)在平方损失函数下, $\lambda$的Bayes估计为$\widehat{\lambda}_{B1}(a, b)= \frac{T+a}{n+b}$, $\lambda$的E-Bayes估计为

(ⅱ)在K-损失函数下, $\lambda$的Bayes估计为$\widehat{\lambda}_{B2}(a, b)=\frac{\sqrt{(T+a-1)(T+a)}}{n+b}$, $\lambda$的E-Bayes估计为

(ⅲ)在加权平方损失函数下, $\lambda$的Bayes估计为$\widehat{\lambda}_{B3}(a, b)=\frac{T+a-1}{n+b}$; $\lambda$的E-Bayes估计为

(ⅰ)的证明已在文献[14]中给出了,这里从略.

(ⅱ)设$x_{1}, x_{2}, \cdots, x_{n}$为来自Poisson分布(2.1)式的样本观察值,似然函数由(2.2)式给出,若$\lambda$的先验分布为Gamma分布,其密度函数由(2.3)式给出,根据Bayes定理,则$\lambda$的后验密度函数为

## 6 模拟算例

 $n$ 20 40 60 80 100 $\widehat\lambda_{EB1}$ 0.5074 0.5057 0.5049 0.5022 0.5015 $\widehat\lambda_{EB2}$ 0.4873 0.4932 0.4966 0.496 0.4964 $\widehat\lambda_{EB3}$ 0.4636 0.481 0.4884 0.4898 0.4915 E-MSE$(\widehat\lambda_{EB1})$ 0.025 0.0125 0.0083 0.0062 0.0048 E-MSE$(\widehat\lambda_{EB2})$ 0.0256 0.0126 0.0084 0.0063 0.005 E-MSE$(\widehat\lambda_{EB3})$ 0.0274 0.0131 0.0086 0.0064 0.0051

 $n$ 20 40 60 80 100 $\widehat\lambda_{EB1}$ 0.9977 1.0003 0.9988 0.9997 0.9999 $\widehat\lambda_{EB2}$ 0.973 0.9879 0.9905 0.9936 0.9947 $\widehat\lambda_{EB3}$ 0.9489 0.9756 0.9823 0.9874 0.9898 E-MSE$(\widehat\lambda_{EB1})$ 0.0487 0.0247 0.0165 0.0124 0.0099 E-MSE$(\widehat\lambda_{EB2})$ 0.0493 0.0249 0.0166 0.0125 0.01 E-MSE$(\widehat\lambda_{EB3})$ 0.0511 0.0253 0.0168 0.0126 0.0101

 $n$ 20 40 60 80 100 $\widehat\lambda_{EB1}$ 1.9735 1.9882 1.9929 1.9936 1.9941 $\widehat\lambda_{EB2}$ 1.9489 1.9758 1.9846 1.9873 1.9891 $\widehat\lambda_{EB3}$ 1.9247 1.9635 1.9764 1.9811 1.9841 E-MSE$(\widehat\lambda_{EB1})$ 0.0963 0.0491 0.0329 0.0247 0.0197 E-MSE$(\widehat\lambda_{EB2})$ 0.0969 0.0492 0.033 0.0248 0.0198 E-MSE$(\widehat\lambda_{EB3})$ 0.0987 0.0497 0.0332 0.0249 0.0199

 $n$ 20 40 60 80 100 $\widehat\lambda_{EB1}$ 3.9248 3.9617 3.972 3.9787 3.9885 $\widehat\lambda_{EB2}$ 3.9003 3.9493 3.9638 3.9725 3.9886 $\widehat\lambda_{EB3}$ 3.876 3.937 3.9555 3.9663 3.9887 E-MSE$(\widehat\lambda_{EB1})$ 0.1915 0.0978 0.0656 0.0494 0.0395 E-MSE$(\widehat\lambda_{EB2})$ 0.1921 0.098 0.0657 0.0495 0.0397 E-MSE$(\widehat\lambda_{EB3})$ 0.1939 0.0984 0.0659 0.0496 0.0398

## 7 应用实例

 $k$ 0 1 2 3 4 5 6 7 8 9 10 11 总数 $n_{k}$ 64 171 239 220 155 83 46 20 6 3 0 1 1008

 $c$ 1 3 5 7 9 极差 $\widehat{\lambda}_{EB1}$ 2.82251 2.81971 2.81693 2.81415 2.81138 0.0111324 $\widehat{\lambda}_{EB2}$ 2.82201 2.81922 2.81643 2.81365 2.81088 0.0111304 $\widehat{\lambda}_{EB3}$ 2.82152 2.81872 2.81594 2.81316 2.81039 0.0111284 E-MSE$(\widehat{\lambda}_{EB1})$ 0.0027987 0.0027932 0.0027877 0.0027822 0.0027767 2.20e-005 E-MSE$(\widehat{\lambda}_{EB2})$ 0.002799 0.0027934 0.0027879 0.0027824 0.0027769 2.21e-005 E-MSE$(\widehat{\lambda}_{EB3})$ 0.0027997 0.0027942 0.0027886 0.0027832 0.0027777 2.20e-005

## 8 结论

(1)不同的损失函数对$\widehat\lambda_{EBi}\ (i=1, 2, 3)$和E-MSE$(\widehat{\lambda}_{EBi})\ (i=1, 2, 3)$有一些的影响,并且随着$n$(样本容量)的增加其影响越来越小.

(2)对E-MSE$(\widehat{\lambda}_{EBi})$有如下顺序关系: E-MSE$(\widehat\lambda_{EB1})<$E-MSE$(\widehat\lambda_{EB2})<$E-MSE$(\widehat\lambda_{EB3})$.

(3)如果以E-MSE作为评价标准,在E-MSE越小越优的意义下, $\widehat \lambda_{EB1}$ "优于"$\widehat \lambda_{EB2}$, $\widehat \lambda_{EB2}$ "优于"$\widehat \lambda_{EB3}$.

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