在研究有關產品可靠度方面的問題時，通常需進行壽命試驗，而在試驗進行時，通常希望能預測產品未來壽命時間，作為決定是否要變更生產計畫或改採行其他決策的參考。但實驗當中因時間、成本或其他限制的考量而往往無法取得完整的樣本資料。本文假設產品壽命時間服從雙參數柏拉圖分配，利用貝氏方法在逐步型II設限下對未被觀察到的產品壽命進行區間預測，做為評估及改善產品可靠度的依據。

    本文亦舉一些數值例子來示範如何建立未來觀測值之貝氏預測區間，模擬研究結果顯示貝氏預測區間皆有達到名目的信心水準。對於雙樣本的問題，本文也有提出其觀測值的貝氏預測區間，模擬結果也顯示有達到名目信心水準。

For reliability analysis, the lifetime test is usually needed. Experimenters usually want to predict the lifetime of unobserved sample so that they can determine whether the production schedule is needed to be redesigned. In many life testing experiments, we may not be able to obtain a complete sample due to time limitation, budget of or other restrictions. This paper is proposing a Bayesian predictive interval of future observation for the Pareto distribution based on a progressively Type II censored sample.

  Some numerical examples are given to demonstrate all predictive intervals. A simulation study is done and the results show that the coverage percentage of proposed predictive interval can reach the nominal confidence coefficient. The two-sample problem is also discussed in this paper. At last, some examples are given to illustrate the predictive intervals.

目錄
第一章  緒論 1
1.1 前言 1
1.2 研究動機與目的 2
1.3 本文架構 5
第二章  文獻探討 6
2.1 逐步型II設限之相關文獻的探討 6
2.2 推論方法之相關文獻的探討 7
2.3 分配介紹 8
第三章 母體分配為柏拉圖分配的未來順序觀測值之貝氏預測區間 10
3.1 尺度參數已知下，對未來順序觀測值之貝氏預測區間之建構 12
3.2 尺度參數及形狀參數未知下，對未來順序觀測值之貝氏預測區間之建構 18
3.3 在一般化無資訊事前分配下的未來觀測值之貝氏預測區間之建構 24
第四章 數值實例示範與模擬比較 28
4.1 數值實例 28
4.1.1 尺度參數已知下的未來觀測值之預測區間 28
4.1.2 尺度及形狀參數皆未知下的未來觀測值之預測區間 34 
4.2 統計模擬 40
第五章  雙樣本情形之貝氏預測區間 48
5.1 雙樣本情形之貝氏預測區間建構 48
5.2 數值實例 51
5.3 統計模擬 56
第六章  結論 62
參考文獻 64

表格目錄
表 4.1：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = =20，m = 15且設限計畫
為R = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,5) 下之Y(s) , s = 1,2,...,5的95%預測區間及其長度 30
表 4.2：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 15且設限計畫
為R = (1,1,0,0,0,0,0,0,0,0,0,0,0,0,3) 下之Y(s) , s = 1,2,3的95%預測區間及其長度 31
表 4.3：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 15且設限計畫
為R = (0,0,0,0,0,0,0,2,0,0,0,0,0,0,3) 下之Y(s) , s = 1,2,3的95%預測區間及其長度 32
表 4.4：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 15且設限計畫
為R = (0,0,0,0,0,0,0,0,0,0,0,0,1,1,3) 下之Y(s) , s = 1,2,3的95%預測區間及其長度 32
表 4.5：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 10且設限計畫
為R = (4,1,0,0,0,0,0,0,0,5) 下之Y(s) , s = 1,2,...,5的95%預測區間及其長度 33
表 4.6：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m =10且設限計畫
為R = (0,0,0,1,2,2,0,0,0,5) 下之Y(s) , s = 1,2,...,5的95%預測區間及其長度 33
表 4.7：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m =15且設限計畫
為R = (0,0,0,0,0,0,0,1,4,5) 下之Y(s) , s = 1,2,...,5的95%預測區間及其長度 34
表 4.8：雙參數皆未知，當n = 20，m = 15時pi1(a|k)~Gamma(4,0.5)且pi2(k)~Gamma(2.3,2)下，設限計畫為R =(0,0,0,0,0,0,0,0,0,0,0,0,0,0,5) 時，Y(s) , s = 1,2,...,5的95%預測區間及長度 36
表 4.9：雙參數皆未知，當n = 20，m = 15時pi1(a|k)~Gamma(4,0.5)且pi2(k)~Gamma(2.3,2)下，，設限計畫為R = (1,1,0,0,0,0,0,0,0,0,0,0,0,0,3) 時，Y(s) , s = 1,2,3的95%預測區間及長度 37
表 4.10：雙參數皆未知，當n = 20，m = 15時pi1(a|k)~Gamma(4,0.5)且pi2(k)~Gamma(2.3,2)下，，設限計畫為R = (0,0,0,0,0,0,0,2,0,0,0,0,0,0,3) 時，Y(s) , s = 1,2,3的95%預測區間及長度 37
表 4.11：雙參數皆未知，當n = 20，m = 15時pi1(a|k)~Gamma(4,0.5)且pi2(k)~Gamma(2.3,2)下，設限計畫為R = (0,0,0,0,0,0,0,0,0,0,0,0,1,1,3) 時，Y(s) , s = 1,2,3的95%預測區間及長度 38
表 4.12：雙參數皆未知，當n = 20，m = 15時pi1(a|k)~Gamma(4,0.5)且pi2(k)~Gamma(2.3,2)下，設限計畫為R = (4,1,0,0,0,0,0,0,0,5)時，Y(s) , s = 1,2,...,5的95%預測區間及長度 38
表 4.13：雙參數皆未知，當n = 20，m = 15時pi1(a|k)~Gamma(4,0.5)且pi2(k)~Gamma(2.3,2)下，設限計畫為R = (0,0,0,1,2,2,0,0,0,5)時，Y(s) , s = 1,2,...,5的95%預測區間及長度 39
表 4.14：雙參數皆未知，當n = 20，m = 15時pi1(a|k)~Gamma(4,0.5)且pi2(k)~Gamma(2.3,2)下，設限計畫為R = (0,0,0,0,0,0,0,1,4,5)時，Y(s) , s = 1,2,...,5的95%預測區間及長度 40
表 4.15：單參數c = 0.5已知下，當γ = 0.5時，(n,m) =(20,10) ，Y(s)的90%及95%平均預測區間長度及覆蓋率 43
表 4.16：單參數c = 0.5已知下，當γ = 0.5時，(n,m) = (20,15) ，Y(s)的90%及95%平均預測區間長度及覆蓋率 44
表 4.17：單參數c = 0.5已知下，當γ = 0.5時，(n,m) = (30,20) ，Y(s)的90%及95%平均預測區間長度及覆蓋率 45
表 4.18：單參數c = 0.5已知下，當γ = 0.5時，(n,m) = (30,25) ，Y(s)的90%及95%平均預測區間長度及覆蓋率 46
表 5.1：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 15且設限計畫為R = (0,0,0,0,0,0,0,0,0,0,0,0,0,0,5) 下之z(b) ,b =1,2,...,5的95%預測區間及其長度 51
表 5.2：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 15且設限計畫為R = (1,1,0,0,0,0,0,0,0,0,0,0,0,0,3) 下之z(b) ,b =1,2,...,5的95%預測區間及其長度 52
表 5.3：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 15且設限計畫為R = (0,0,0,0,0,0,0,2,0,0,0,0,0,0,3) 下之z(b) ,b =1,2,...,5的95%預測區間及其長度 53
表 5.4：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 15且設限計畫為R = (0,0,0,0,0,0,0,0,0,0,0,0,1,1,3) 下之z(b) ,b =1,2,...,5的95%預測區間及其長度 53
表 5.5：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 10且設限計畫為R = (4,1,0,0,0,0,0,0,0,5) 下之z(b) ,b =1,2,...,5的95%預測區間及其長度 54
表 5.6：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 10且設限計畫為R = (0,0,0,1,2,2,0,0,0,5) 下之z(b) ,b =1,2,...,5的95%預測區間及其長度 55
表 5.7：單參數c = 0.5 已知下， a 服從Gamma (4,0.5) ， n = 20，m = 10且設限計畫為R = (0,0,0,0,0,0,0,1,4,5) 下之z(b) ,b =1,2,...,5的95%預測區間及其長度 55
表 5.8：單參數c = 0.5已知下，當γ = 0.5時，(n,m) = (30,25)，z(b) ,b =1,2,...,5的90%及95%平均預測區間長度及覆蓋率 57
表 5.9：單參數c = 0.5已知下，當γ = 0.5時，(n,m) = (30,20)，z(b) ,b =1,2,...,5的90%及95%平均預測區間長度及覆蓋率 58
表 5.10：單參數c = 0.5已知下，當γ = 0.5時，(n,m) =(20,15)，z(b) ,b =1,2,...,5的90%及95%平均預測區間長度及覆蓋率 59
表 5.11：單參數c = 0.5已知下，當γ = 0.5時，(n,m) =(20,10)，z(b) ,b =1,2,...,5的90%及95%平均預測區間長度及覆蓋率 60

圖例目錄
圖 1.1 ： 逐步型II 右設限流程圖 4
圖 4.1 ： 雙參數柏拉圖分配的機率密度函數 41

[1] Aggarwala, R. and Balakrishnan, N. (1998), Some properties of progressive censored order statistics from the pareto distribution with applications to inference, J:Statist:P lann:Inference, 70, 35-49.

[2] Aitchison, J. and Dunsmore, I. R.(1975), Statistical Prediction Analysis, Cambridge University Press, Cambridge.

[3] Ali Mousa, M. A. M. (2003), Bayesian Predcition based on Pareto Dounly Censored data, Statistics, 37(1), 65-72

[4] Arnold, B. C. and Press, S. J. (1989), Bayesian estimation and prediction for Pareto data, J: Amer: Statistical Assoc:, 84, 1079-1084.

[5] Balakrishnan, N. and Aggarwala, R. (2000), Progressive Censoring Theory; Methods; and Applications, Birkh user. Boston.

[6] Cohen, A. C. (1963), Progressively censored samples in the life testig, Technometrics, 5, 327-339.

[7] Cohen, A. C. (1976), Progressively censored sampling in the three parameter log-normal distribution, Technometrics, 17, 347-351.

[8] Cohen, A. C. and Norgaard, N.J. (1977), Progressively censored sampling in the three parameter gamma distribution, Technometrics, 19, 333-340.

[9] David, H. A. (1981), Order Statistics, 2nd ed, John Wiley and Sons, Inc., New York.

[10] Lawless, J. F. (1971), A prediction problem concerning samples from the exponential distribution with application in life testing, Technometrics, 13, 725-730.

[11] Like·s, J. (1974), Prediction of s-th ordered observation for the two-parameter exponential distribution, Technometrics, 16, 241-244.

[12] Lingappaiah, G. S. (1981), Sequential life testing with spacings, exponential model, IEEE Transactions on Reliability, R-30(4), 370-374.

[13] Maritz, J. L. and Lwin, T. (1989), Empirical Bayes Methods, 2nd ed, Chapman and Hall.

[14] Martz, H. F. and Waller, R. A. (1982), Bayesian Reliability Analysis, Wiley, New York.

[15] McNolty, F., Doyle, J. and Hansen, E. (1980), Properties of the mixed exponen- tial failure process, Technometrics, 22, 555-565.

[16] Nigm, A. M., Al-Hussaini, E. K. and Jaheen, Z. F. (2003), Bayesian one-sample prediction of future observations under Pareto distribution, Statistics, 37, 527-536.

[17] Ouyang, L. Y. and Wu, S. J. (1994), Prediction Inter-
vals for an Ordered Observation from a Pareto Distribution,
IEEE Transactions on Reliability, 43, 264-269

[18] Waller, R. A. and Waterman, M, S. (1978), Percentiles for the Gamma Distribution, SIAM Review, 20(1), 186-187