#### 期刊菜单

Comparison for Quantiles of Two-Parameter Exponential Distributions
DOI: 10.12677/SA.2019.82027, PDF, HTML, XML, 下载: 582  浏览: 1,198

Abstract: This paper considers the testing problem for comparing quantiles of the two-parameter exponential distribution. We provide two approaches, the parameters bootstrap and Fiducial method. Simulation studies were conducted to compare two methods. The above methods were also applied in an example.

1. 引言

1989年Tsui和Weerahandi [5] 提出了广义推断的概念。2007年Li，Xu和Li [6] 给出了广义P值的Fiducial推断。1979年Efron [7] 提出的Bootstrap方法应用十分广泛，对于复杂分布的参数和复杂估计可直接导出标准误和区间估计。近年来参数Bootstrap方法也得到了广泛的应用和改进 [8] 。本文将分别使用Fiducial推断和参数Bootstrap似然比方法，从假设检验的角度对两个双参数指数分布总体的分位数进行比较。

2. 问题表述

$p\left(x;\mu ,\sigma \right)=\left\{\begin{array}{cc}\begin{array}{l}\frac{1}{\sigma }\mathrm{exp}\left\{-\frac{x-\mu }{\sigma }\right\}\\ 0\end{array}& \begin{array}{l}x>\mu \\ 其他\end{array}\end{array}$

$\left({X}_{i,1},{X}_{i,2},\cdots ,{X}_{i,{n}_{i}}\right)$ 为来自上述分布的容量为 $n$ 的样本，设观测到的数据为定数截断数据为 ${X}_{\left(i1\right)}<{X}_{\left(i2\right)}<\cdots <{X}_{\left(i{r}_{i}\right)}$$i=1,2$。分布 Exp (μI, σi) 的 $p$ 分位数表示为 ${Q}_{{}_{i,p}}={\mu }_{i}-{\sigma }_{i}\mathrm{log}\left(1-p\right)$，从而两个总体分位数的差可写为 ${Q}_{d}={Q}_{2,p}-{Q}_{1,p}$。易知

${W}_{i}={X}_{i,1}$${V}_{i}={\sum }_{j=1}^{{r}_{i}}{X}_{i,j}+\left({n}_{i}-{r}_{i}\right){X}_{i,{r}_{i}}-{n}_{i}{X}_{i,1}$ (1)

${\stackrel{^}{\mu }}_{i}={W}_{i}-\frac{1}{{n}_{i}\left({r}_{i}-1\right)}{V}_{i}$${\stackrel{^}{\sigma }}_{i}=\frac{1}{{r}_{i}-1}{V}_{i}$ (2)

${Q}_{d}$ 的估计为

${\stackrel{^}{Q}}_{d}=\left[{\stackrel{^}{\mu }}_{2}-{\stackrel{^}{\sigma }}_{2}\mathrm{log}\left(1-p\right)\right]-\left[{\stackrel{^}{\mu }}_{1}-{\stackrel{^}{\sigma }}_{1}\mathrm{log}\left(1-p\right)\right]$ (3)

${H}_{0}:{Q}_{d}\le 0\text{ }\text{VS}\text{ }{H}_{1}:{Q}_{d}>0$ (4)

3. 参数Bootstrap方法

Bootstrap方法由Efron [7] 首先提出，其原理是从估计模型中进行重复抽样，常采用的为非参数Bootstrap方法。本文采用参数Bootstrap，用参数的估计取代原参数得到参数模型。

${W}_{i}={E}_{\left(i,1\right)}$${V}_{bi}={\sum }_{j=1}^{{r}_{i}}{E}_{\left(i,j\right)}+\left({n}_{i}-{r}_{i}\right){E}_{\left(i,{r}_{i}\right)}-{n}_{i}{E}_{\left(i,1\right)}$ (5)

${\stackrel{^}{Q}}_{b}=\left[{\stackrel{^}{\mu }}_{b2}-{\stackrel{^}{\sigma }}_{b2}\mathrm{log}\mathrm{log}\left(1-p\right)\right]-\left[{\stackrel{^}{\mu }}_{b1}-{\stackrel{^}{\sigma }}_{b1}\mathrm{log}\left(1-p\right)\right]$ (6)

${\stackrel{^}{\mu }}_{bi}={W}_{bi}-\frac{1}{{n}_{i}\left({r}_{i}-1\right)}{V}_{bi}$${\stackrel{^}{\sigma }}_{bi}=\frac{1}{{r}_{i}-1}{V}_{bi}$$i=1,2$ (7)

(a) 通过(2)式分别计算 ${\stackrel{^}{\mu }}_{i}$${\stackrel{^}{\sigma }}_{i}$

(b) 生成对应的样本容量为 ${r}_{i}$ 的Bootstrap样本并进行排序，

(c) 通过(5) (6) (7)式计算的实现值，

(d) 重复步骤(a)~(c) $N$ 次( $N$ 一般大于1000)，

(e) 给出p值 $p=2\mathrm{min}\left\{\frac{p#\left\{{\stackrel{^}{Q}}_{b}\le 0\right\}}{N},\frac{p#\left\{{\stackrel{^}{Q}}_{b}>0\right\}}{N}\right\}$

4. Fiducial方法

Fiducial推断由Fisher首先提出并研究，这类方法不需要先验分布，而是根据数据给出与参数的后验分布有相同作用的Fiducial分布．其后有不少统计学者对之进行讨论。Hannig等也讨论了利用Fiducial推断方法构造广义区间估计。Li等 [6] 给出了利用Fiducial推断进行假设检验的方法。本节利用Fiducial推断研究分位数差的假设检验。

$2{n}_{i}\frac{{W}_{i}-{\mu }_{i}}{{\sigma }_{i}}~{\chi }^{2}\left(2\right)$$\frac{2{V}_{i}}{{\sigma }_{i}}~{\chi }^{2}\left(2{r}_{i}-2\right)$$i=1,2$

${E}_{i,1}~{\chi }^{2}\left(2\right)$${E}_{i,2}~{\chi }^{2}\left(2{r}_{i}-2\right)$ 独立，有

$\left({W}_{i},{V}_{i}\right)=\left(\frac{{\sigma }_{i}{E}_{i,1}}{2{n}_{i}}+{\mu }_{i},\frac{{\sigma }_{i}{E}_{i,2}}{2}\right)$

$\left({\mu }_{i},{\sigma }_{i}\right)=\left({w}_{i}-\frac{{v}_{i}{e}_{1}}{{n}_{i}{e}_{2}},\frac{2{v}_{i}}{{e}_{2}}\right)$

${T}_{{\mu }_{i}}={w}_{i}-\frac{{v}_{i}{E}_{1}}{{n}_{i}{E}_{2}},{T}_{{\sigma }_{i}}=\frac{2{v}_{i}}{{E}_{2}}$

${T}_{{Q}_{d}}=\left[{T}_{{\mu }_{2}}-{T}_{{\sigma }_{2}}\mathrm{log}\left(1-p\right)\right]-\left[{T}_{{\mu }_{1}}-{T}_{{\sigma }_{1}}\mathrm{log}\left(1-p\right)\right]$

$p=2\mathrm{min}\left\{p\left({T}_{{Q}_{d}}<0\right),p\left({T}_{{Q}_{d}}>0\right)\right\}$

1) 对给定数据，选取一个较大的模拟样本数N，

2) 对于 $j=1,\cdots ,N$，通过 ${E}_{i}$ 的分布 $Exp\left({\mu }_{i},{\sigma }_{i}\right)$ 随机生成 ${e}_{j}$

3) 求方程 $\left({w}_{i},{v}_{i}\right)=\left(\frac{{\sigma }_{i}{e}_{i,1}}{2{n}_{i}}+{\mu }_{i},\frac{{\sigma }_{i}{e}_{i,2}}{2}\right)$ 的解并计算 ${T}_{{Q}_{d}}$ 的实现值 ${T}_{i}$${T}_{{Q}_{d}}=\left[{T}_{{\mu }_{2}}-{T}_{{\sigma }_{2}}\mathrm{log}\left(1-p\right)\right]-\left[{T}_{{\mu }_{1}}-{T}_{{\sigma }_{1}}\mathrm{log}\left(1-p\right)\right]$

4) 计算p值 $p=2\mathrm{min}\left\{\frac{p#\left\{{T}_{i}<0\right\}}{N},\frac{p#\left\{{T}_{i}>0\right\}}{N}\right\}$

5. 数据仿真模拟

Table 1. Power of two methods (n = 10, r = 8)

Table 2. Power of two methods (n = 50, r = 40)

6. 实例分析

Table 3. Lifetime data of tyre

7. 结论

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