## 3 Interval Estimation  
### Asymptotic Confidence Interval  
Similar to the work of [PERSON] et al. [17], we construct an ACI using the asymptotic normality of the MLE. Let \(\hat{\eta}=(\hat{\theta_{1}},\ \hat{\theta_{2}},\ \hat{\lambda})\) be the MLEs of \(\eta=(\theta_{1},\ \hat{\theta_{2}},\ \lambda)\). The observed Fisher information matrix is given by \(I(\hat{\eta})=[I_{ij}]=[-\frac{\hat{\eta}^{2}l}{\hat{\eta}_{i}\hat{\eta}_{j}}] _{\hat{\eta}=\hat{\eta}},\ i,\ j=1,\ 2,\ 3\), where  
\[\begin{split} I_{11}=&\frac{nk}{\theta_{1}^{2}},\ I_{22}= \frac{n}{\theta_{2}{}^{2}},\ I_{12}=I_{12}=0,\\ I_{13}=& I_{31}\!=\!\sum_{i=1}^{n}\sum_{j=1}^{k}\left(1+S_{j} \right)e^{x_{ij}{}^{\lambda}}x_{ij}{}^{\lambda}\ln x_{ij},\ I_{23}=I_{32}=\sum_{i =1}^{n}\left(1+T_{i}\right)e^{y_{i}{}^{\lambda}}y_{i}{}^{\lambda}\ln y_{i}, \\ I_{33}=&\frac{n(k+1)}{\lambda^{2}}-\sum_{i=1}^{n}\sum_{j=1}^{k} \left[x_{ij}{}^{\lambda}(\ln x_{ij})^{2}-\theta_{1}(1+S_{j})e^{x_{ij}{}^{ \lambda}}x_{ij}{}^{\lambda}(\ln x_{ij})^{2}(x_{ij}{}^{\lambda}+1)\right]\\ &-\sum_{i=1}^{n}\left[y_{i}{}^{\lambda}(\ln y_{i})^{2}-\theta_{2}(1+T_{i })e^{y_{i}{}^{\lambda}}y_{i}{}^{\lambda}(\ln y_{i})^{2}(y_{i}{}^{\lambda}+1) \right].\end{split}\]  
Using the delta method (see [PERSON] and [PERSON] [18]), the variance of \(\hat{R}_{s,k}\) is derived as follows:  
\[Var(\hat{R}_{s,k})=(\frac{\partial R_{s,k}}{\partial\theta_{1}})^{2}[I^{-1}]_{11} +(\frac{\partial R_{s,k}}{\partial\theta_{2}})^{2}[I^{-1}]_{22}+2(\frac{ \partial R_{s,k}}{\partial\theta_{1}})(\frac{\partial R_{s,k}}{\partial\theta_{ 2}})[I^{-1}]_{12},\]where  
\[\frac{\partial R_{s,k}}{\partial\theta_{1}} =\sum_{m=s}^{k}\sum_{n=0}^{k-m}\left(\begin{array}{c}k\\ m\end{array}\right)\left(\begin{array}{c}k-m\\ n\end{array}\right)(-1)^{n+1}\frac{(m+n)\theta_{2}}{\left[(m+n)\theta_{1}+ \theta_{2}\right]^{2}},\] \[\frac{\partial R_{s,k}}{\partial\theta_{2}} =\sum_{m=s}^{k}\sum_{n=0}^{k-m}\left(\begin{array}{c}k\\ m\end{array}\right)\left(\begin{array}{c}k-m\\ n\end{array}\right)(-1)^{n}\frac{(m+n)\theta_{1}}{\left[(m+n)\theta_{1}+ \theta_{2}\right]^{2}}.\]  
Here, the parameters \((\theta_{1},\ \theta_{2},\ \lambda)\) are computed as the MLEs \((\hat{\theta}_{1},\ \hat{\theta}_{2},\ \hat{\lambda})\). Therefore, the \(100(1-\alpha)\%\) ACI for \(R_{s,k}\) is  
\[\left(\hat{R}_{s,k}-z_{\alpha/2}\sqrt{\text{Var}\big{(}\hat{R}_{s,k}\big{)}}, \ \hat{R}_{s,k}+z_{\alpha/2}\sqrt{\text{Var}\big{(}\hat{R}_{s,k}\big{)}}\right). \tag{18}\]  
where \(z_{\alpha}\) is \(100(1-\alpha)th\) percentile of \(N(0,1)\).  
A negative lower bound might be generated by the ACI proposed above but \(0<R_{s,k}<1\). To avoid this problem, the logit transformation can be used to provide a more accurate confidence interval, as proposed by [PERSON] and [PERSON] [19]. Let \(\xi=\ln(\hat{R}_{s,k}/(1-\hat{R}_{s,k}))\) be the MLE of \(\xi=\ln(R_{s,k}/(1-R_{s,k}))\). Then, the \(100(1-\alpha)\%\) ACI of \(\xi\) is \((\xi_{L},\ \xi_{U})\), where  
\[\xi_{L}=\ln(\frac{\hat{R}_{s,k}}{1-\hat{R}_{s,k}})-z_{\alpha/2}\frac{\sqrt{ Var(\hat{R}_{s,k})}}{\hat{R}_{s,k}(1-\hat{R}_{s,k})},\]  
and  
\[\xi_{U}=\ln(\frac{\hat{R}_{s,k}}{1-\hat{R}_{s,k}})+z_{\alpha/2}\frac{\sqrt{ Var(\hat{R}_{s,k})}}{\hat{R}_{s,k}(1-\hat{R}_{s,k})}.\]  
Then, the ACI of \(R_{s,k}\) based on the logit scale is  
\[(\frac{\exp(\xi_{L})}{1+\exp(\xi_{L})},\ \frac{\exp(\xi_{U})}{1+\exp(\xi_{U})}). \tag{19}\]