## 5 Confidence Intervals of \(R\)
### Asymptotic Confidence Interval of \(R\)  
In this subsection, we derive the asymptotic distribution of the MLE \(\hat{\theta}=(\hat{\alpha},\hat{\theta_{1}},\hat{\theta_{2}})\) and \(\hat{R}\). Based on the asymptotic distribution of \(\hat{R}\), the corresponding asymptotic confidence interval of \(R\) is obtained. We denote the exact Fisher information matrix of \(\theta=(\alpha,\theta_{1},\theta_{2})\) as \(J(\theta)=-\mathbb{E}(I;\theta)\), where \(I=[I_{ij}]_{i,j=1,2,3}\), \(I_{ij}=\partial^{2}L/\partial\theta_{i}\partial\theta_{j}\) and \(L\) is given in (6):  
\[I(\theta)=\left( \begin{array}{ccc}\frac{\partial^{2}L}{ \partial\alpha^{2}}&\frac{\partial^{2}L}{\partial\alpha\partial\theta_{1}}& \frac{\partial^{2}L}{\partial\alpha\partial\theta_{2}}\\ \frac{\partial^{2}L}{\partial\theta_{1}\partial\alpha}&\frac{\partial^{2}L}{ \partial\theta_{1}^{2}}&\frac{\partial^{2}L}{\partial\theta_{1}\partial\theta_ {2}}\\ \frac{\partial^{2}L}{\partial\theta_{2}\partial\alpha}&\frac{\partial^{2}L}{ \partial\theta_{2}\partial\theta_{1}}&\frac{\partial^{2}L}{\partial\theta_{ 2}^{2}}\end{array} \right)=\left( \begin{array}{ccc}I_{11}&I_{12}&I_{13}\\ I_{21}&I_{22}&I_{23}\\ I_{31}&I_{32}&I_{33}\end{array} \right).\]  
It is easy to see that  
\[I_{11}=-\frac{m+n}{\alpha^{2}}-\frac{1}{\theta_{1}}\sum_{i=1}^{ n}(\ln x_{i})^{2}x_{i}^{-\alpha}-\frac{1}{\theta_{2}}\sum_{j=1}^{m}(\ln y_{j})^{2}y _{j}^{-\alpha},\] \[I_{12}=I_{21}=-\frac{1}{\theta_{1}^{2}}\sum_{i=1}^{n}x_{i}^{- \alpha}\ln x_{i},\] \[I_{13}=I_{31}=-\frac{1}{\theta_{2}^{2}}\sum_{j=1}^{m}y_{j}^{- \alpha}\ln y_{j},\] \[I_{22}=\frac{n}{\theta_{1}^{2}}-\frac{2}{\theta_{1}^{3}}\sum_{i =1}^{n}x_{i}^{-\alpha},\] \[I_{33}=\frac{m}{\theta_{2}^{2}}-\frac{2}{\theta_{2}^{3}}\sum_{j =1}^{m}y_{j}^{-\alpha},\] \[I_{23}=I_{32}=0.\]Moreover,  
\[J_{11} =-\mathbb{E}(\frac{\partial^{2}L}{\partial\alpha^{2}})=\frac{1}{ \alpha^{2}}\left[(m+n)(1+\gamma^{\prime\prime}(2))+n(\ln\theta_{1})^{2}+m(\ln \theta_{2})^{2}+2\gamma^{\prime}(2)(n\ln\theta_{1}+m\ln\theta_{2})\right],\] \[J_{12} =J_{21}=\frac{1}{\theta_{1}^{2}}\sum_{i=1}^{n}\mathbb{E}(x_{i}^{- \alpha}\ln x_{i})=-\frac{n}{\theta_{1}\alpha}\left[\ln\theta_{1}+\gamma^{ \prime}(2)\right],\] \[J_{13} =J_{31}=\frac{1}{\theta_{2}^{2}}\sum_{j=1}^{m}\mathbb{E}(y_{j}^{ -\alpha}\ln y_{j})=-\frac{m}{\theta_{2}\alpha}\left[\ln\theta_{2}+\gamma^{ \prime}(2)\right],\] \[J_{22} =\frac{n}{\theta_{1}^{2}},\] \[I_{12}=I_{21}=-\frac{1}{\theta_{1}^{2}}\sum_{i=1}^{n}x_{i}^{- \alpha}\ln x_{i},\]0 \[I_{12}=I_{21}=-\frac{1}{\theta_{1}^{2}}\sum_{i=1}^{n}x_{i}^{- \alpha}\ln x_{i},\]1  
where \(\gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}dx\) is the Gamma function. **Theorem 1**. : _As \(n\to\infty\) and \(m\to\infty\) and \(\frac{n}{m}\to p\), then_  
\[I_{12}=I_{21}=-\frac{1}{\theta_{1}^{2}}\sum_{i=1}^{n}x_{i}^{- \alpha}\ln x_{i},\]2  
_where_  
\[A(\alpha,\theta_{1},\theta_{2})=\left( \begin{array}{ccc}a_{11}&a _{12}&a_{13}\\ a_{21}&a_{22}&0\\ a_{31}&0&a_{33}\end{array} \right),\]  
_and_  
\[a_{11} =\frac{1}{\alpha^{2}}\left[(1+p)(1+\gamma^{\prime\prime}(2))+p( \ln\theta_{1})^{2}+(\ln\theta_{2})^{2}+2\gamma^{\prime}(2)(p\ln\theta_{1}+ \ln\theta_{2})\right]=\lim_{n,m\to\infty}\frac{J_{11}}{m},\] \[a_{12} =a_{21}=-\frac{\sqrt{p}}{\theta_{1}\alpha}\left[ln\theta_{1}+ \gamma^{\prime}(2)\right]=\lim_{n,m\to\infty}\frac{\sqrt{p}}{n}J_{12},\] \[a_{13} =a_{31}=-\frac{1}{\theta_{2}\alpha}\left[ln\theta_{2}+\gamma^{ \prime}(2)\right]=\lim_{n,m\to\infty}\frac{1}{m}I_{13},\] \[a_{22} =\frac{1}{\theta_{1}^{2}}=\lim_{n,m\to\infty}\frac{1}{n}J_{22},\] \[a_{33} =\frac{1}{\theta_{2}^{2}}=\lim_{n,m\to\infty}\frac{1}{m}I_{33},\] \[a_{23} =a_{32}=0.\]  
Proof. : We can use the asymptotic properties of MLEs to prove it. **Theorem 2**.