## 3 Classical Inferences
### Maximum Likelihood Method
 \tag{4}\]  
then upon computing the first partial derivatives of \(l(\alpha,\lambda,\beta)\) concerning the three parameters and making them equal to 0, one arrives at the set of equations below:\[\frac{\partial l}{\partial\alpha}= (m_{1}+m_{2}-r_{1}-r_{2})\frac{1}{\alpha} \tag{5}\] \[-\frac{r_{1}\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^ {\alpha}\ln\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}}{1-\Big{(}1- e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^{\alpha}}\] \[-\frac{\beta r_{2}\Big{(}1-e^{-\lambda/x_{2_{\mathcal{V}}_{2}+1}} \Big{)}^{\beta\alpha}\ln\Big{(}1-e^{-\lambda/x_{2_{\mathcal{V}}_{2}+1}}\Big{)} }{1-\Big{(}1-e^{-\lambda/x_{2_{\mathcal{V}}_{2}+1}}\Big{)}^{\beta\alpha}}\] \[+\sum_{i=r_{1}+1}^{m_{1}}\ln\Big{(}1-e^{-\lambda/x_{1i}}\Big{)}+ \beta\sum_{i=r_{2}+1}^{m_{2}}\ln\Big{(}1-e^{-\lambda/x_{2i}}\Big{)}\] \[+\sum_{i=r_{1}+1}^{m_{1}}R_{1i}\ln\Big{(}1-e^{-\lambda/x_{1i}} \Big{)}+\beta\sum_{i=r_{2}+1}^{m_{2}}R_{2i}\ln\Big{(}1-e^{-\lambda/x_{2i}} \Big{)}=0,\]  
\[\frac{\partial l}{\partial\lambda}= (m_{1}+m_{2}-r_{1}-r_{2})\frac{1}{\lambda}-\sum_{i=r_{1}+1}^{m_{ 1}}\frac{1}{x_{1i}}-\sum_{i=r_{2}+1}^{m_{2}}\frac{1}{x_{2i}} \tag{6}\] \[-\frac{\alpha r_{1}\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1 }}\Big{)}^{\alpha-1}e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}}{\Big{[}1-\Big{(}1 -e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^{\alpha}\Big{]}x_{1_{\mathcal{ V}}_{1}+1}}-\frac{\beta\alpha r_{2}\Big{(}1-e^{-\lambda/x_{2_{\mathcal{V}}_{2}+1}} \Big{)}^{\beta\alpha-1}e^{-\lambda/x_{2_{\mathcal{V}}_{2}+1}}}{\Big{[}1-\Big{(} 1-e^{-\lambda/x_{2_{\mathcal{V}}_{2}+1}}\Big{)}^{\beta\alpha}\Big{]}x_{2_{ \mathcal{V}}_{2}+1}}\] \[+(\alpha-1)\sum_{i=r_{1}+1}^{m_{1}}\frac{e^{-\lambda/x_{1i}}}{ \big{(}1-e^{-\lambda/x_{1i}}\big{)}x_{1i}}+(\beta\alpha-1)\sum_{i=r_{2}+1}^{m _{2}}\frac{e^{-\lambda/x_{2i}}}{\big{(}1-e^{-\lambda/x_{2i}}\big{)}x_{2i}}\] \[+\alpha\sum_{i=r_{1}+1}^{m_{1}}R_{1i}\frac{e^{-\lambda/x_{1i}}}{ \big{(}1-e^{-\lambda/x_{1i}}\big{)}x_{1i}}+\beta\alpha\sum_{i=r_{2}+1}^{m_{2} }R_{2i}\frac{e^{-\lambda/x_{2i}}}{\big{(}1-e^{-\lambda/x_{2i}}\big{)}x_{2i}}=0\]  
and  
\[\frac{\partial l}{\partial\beta}= (m_{2}-r_{2})\frac{1}{\beta}-\frac{\alpha r_{2}\Big{(}1-e^{- \lambda/x_{2_{\mathcal{V}}_{2}+1}}\Big{)}^{\beta\alpha}\ln\Big{(}1-e^{- \lambda/x_{2_{\mathcal{V}}_{2}+1}}\Big{)}}{1-\Big{(}1-e^{-\lambda/x_{2_{ \mathcal{V}}_{2}+1}}\Big{)}^{\beta\alpha}} \tag{7}\] \[-\frac{r_{1}\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^ {\alpha}\ln\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}}{1-\Big{(}1- e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^{\alpha}}\]0  
The MLEs of \(\alpha\), \(\lambda\) and \(\beta\) symbolized by \(\hat{\alpha}\), \(\hat{\lambda}\) and \(\hat{\beta}\), are then derived from Equations (5)-(7). Due to the challenge of obtaining the precise analytical solution, we utilize the function 'optim' in R to derive the MLEs of the three parameters. In addition, one can obtain the MLE of the mean of the GIED straightforwardly due to the invariance property of MLE. According to [3], the mean \(\mu\) of GIED is  
\[-\frac{r_{1}\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^ {\alpha}\ln\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}}{1-\Big{(}1- e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^{\alpha}}\]1  
so the MLE of the mean \(\mu\) is  
\[-\frac{r_{1}\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^ {\alpha}\ln\Big{(}1-e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}}{1-\Big{(}1- e^{-\lambda/x_{1_{\mathcal{V}}_{1}+1}}\Big{)}^{\alpha}}\]2