机械设计制造及其自动化中英文翻译--逐次定数截尾数据下对一个简单的步进应力模型的估计(编辑修改稿)

机械设计制造及其自动化中英文翻译--逐次定数截尾数据下对一个简单的步进应力模型的估计(编辑修改稿)内容摘要：

ote that, regardless of the values of 2 , the loglikelihood function is an increasingfunction of 1 . Hence, no MLE for 0 and 1 exists. The MLE for 2 is   ^211 1mi i irXm   . 4. Interval Estimations In this section, an exact  100 1 % confidence interval for 1 and an at least  100 1 % confidence interval for 0 are constructed. Consider the following transformation: 6       112 1 2 11 1 1, 1 ,... ... ... 1 .m m m mZ nYZ n r Y YZ n r r m Y Y   (10) Balakrishnan and Aggarwala (Ref. 2, p. 18) showed that the generalized spacings 12, , , mZ Z Z , as defined in (10), are independent and identically distributed as the standard exponential distribution. Hence, by the Theorem in Lawless,7 1122V Z nY has a chisquare distribution with 2 degrees of freedom and  212 2 1mmi i i iiiU Z r Y n Y    has a chisquare distribution with2m2 degrees of freedom. We can also show that U and V are independent. In the following discussion, let  12,F  be the upper a percentage point of the F distribution with 1 and 2 degrees of freedom and let 2 be the upper a percentage point of the chisquare distribution with  degrees of freedom. . Confidence Intervals for 1 Consider the case that 10 nm.Let 12WWD,where      1111 1n iiiW r X n X     , and   112 1 1niiinW r X   . It is easy to show that  111mi i iD r Y nY    and, hence 2DU has a chisquare distribution with 2m2 degrees of freedom. It is also easy to see that 112nX V  has a chisquare distribution with 2 degrees of freedom, and 1X and D are independent. Now, we are going to derive a confidence interval for 1 . Consider the pivotal quantity    1 2 ,2 211 mn m X FD  . Let  1 1 2,2 22 mAF and  2 2,2 22 mAF .For 01,we have       1 21112111112 2 2 1111111 vvn m XP A ADn m X n m XP W e WW A W A                  Hence, if   1121 0n m X WA ,a  100 1 % confidence interval for 1 is  12,ll , where 7   22 1 1 11 l o g , 1 , 21 ii iWAliv v n m X W A  . (11) If   1121 0n m X WA ， a  100 1 % confidence interval for 1 is  1,l  . Note that the previous confidence interval for 1 is valid under the condition 10 nm However, in practice, 1n can be 0 or m. Thus, we have the following two remarks. Remark 2. When 1nm ,consider the pivotal quantity    1 2 ,2 201 mn m X FW  . where  0 1 11mi i iW r X nX   .Then the  100 1 % confidence interval for 1 is  , ,for   11201n m XAAW， an empty set  elsewhere. Remark 3. when 1 0n and 2nm ,consider the pivotal quantity   2122 , 2 201mn m XFW  . Let  02 1211 l o g , 3 , 41ii WA Xliv v n m    . Then, the  100 1 % confidence interval for 1 is  34,ll ,for    01 1 01WA Xnm    , 4,l elsewhere. . Confidence Interval for 0 Suppose that 10 nm .We have 112nX is distributed as 22 , 121222WWD  is distributed as  222m , and D and 1X are independent. Let  21 11 2212B   ,  22 11 2212B   ,  23 11 22212mB    and  24 11 22212mB   . For 01, we have 1 1 21 2 3 41 1 21 1 1 2 2 1 11 3 42 1 1 2 11 1 1,n X W WP B B B Bn X n X W B W W BP B BB B n X n X                   Hence, if 123 1 0WBB nX, an at least  100 1 % confidence interval for 0 is  56,ll ,where 8 125 2 1122 1 2311 l o g l o gn X Wl v vWBv v B BnX   , (12) and 126 2 1112 1 1411 l o g l o gn X Wl v vWBv v B BnX   . (13) If 123 1 0WBB nX, an at least  100 1 % confidence interval for 0 is  6,l . Note that the above discussion is under the condition 10 nm. However, inpractice, 1n can be 0 or m. Therefore, we have the following two remarks. Remark4. When 1nm , we have 011 2 3 4011 1 24 0 31, WnXB B B BWB nX BPB W B         Hence, an at least  100 1 % confidence interval for 0 is  , , for 1 1 24 0 3B nX BB W B, an empty set  elsewhere. Remark5. When 1 0n and 2nm , we have    011 2 3 42 1 24 1 3 1 001220 1 0 4 31,WXP B n B B BB X B X WWBBPn W n W B B                 Thus, an at least  100 1 % confidence interval for 0 is  78,ll , for  4110 0BXBnW, where  007 2 12 1 0 2 3 1 31 l o g l o gn W Wl v vv v W B n B X B      ， and  008 2 12 1 0 1 4 1 41 l o g l o gn W Wl v vv v W B n B X B      . If  4110 0BXBnW, the confidence interval is  7,l  . 5. An Illustrative Example 9 To illustrate the use of the methods given in this paper, Table 1 presents the simulated data from a simple stepstress ALT model with progressive type II censoring. These data were simulated by generating a progressively type II censored sample from the standard exponential distribution using the algorithm presented in Balakrishnan and Aggarwala (Ref. 2, p. 32), and then the transformation (9) is used to get the sample from model (2). We choose 5n , 30m , 0   , 1   , 1  , 2  ,  , and censoring scheme  1 2 30, ,...,r r r listed in Table 1. Table1. Simulated failure time data. stress 1v i 1 2 3 4 5 6 7 8 ix ir 2 0 0 2 0 0 0 2 stress。