## 3 Maximum Likelihood Estimator of \(R\)  
In this section, we consider the problem of estimating \(R=P(Y<X)\) under the assumption that \(X\sim IW(\alpha,\theta_{1})\) and \(Y\sim IW(\alpha,\theta_{2})\). Then, it can be easily calculated that  
\[R=P(Y<X)=\frac{\theta_{2}}{\theta_{1}+\theta_{2}}. \tag{5}\]  
To computer the MLE of \(R\), first we obtain the MLEs of \(\theta_{1}\) and \(\theta_{2}\). Suppose \(X_{1},X_{2},...,X_{n}\) is a random sample from \(IW(\alpha,\theta_{1})\) and \(Y_{1},Y_{2},...,Y_{m}\) is a random sample from \(IW(\alpha,\theta_{2})\). The joint likelihood function is:  
\[l(\alpha,\theta_{1},\theta_{2})=\frac{\alpha^{n+m}}{\theta_{1}^{n}\theta_{2} ^{m}}(\prod_{i=1}^{n}x_{i}^{-\alpha-1})(\prod_{j=1}^{m}y_{j}^{-\alpha-1})e^{ -(\sum_{i=1}^{n}\frac{x_{i}^{-\alpha}}{\theta_{1}}+\sum_{j=1}^{n}\frac{y_{j}^ {-\alpha}}{\theta_{2}})}.\]Then, the log-likelihood function is  
\[\begin{split} L(\alpha,\theta_{1},\theta_{2})&=(m+n) \ln\alpha-n\ln\theta_{1}-m\ln\theta_{2}-(\alpha+1)\left[\sum_{i=1}^{n}\ln x_{i} +\sum_{j=1}^{m}\ln y_{j}\right]\\ &-\frac{1}{\theta_{1}}\sum_{i=1}^{n}x_{i}^{-\alpha}-\frac{1}{ \theta_{2}}\sum_{j=1}^{m}y_{j}^{-\alpha}.\end{split} \tag{6}\]  
\(\hat{\alpha}\), \(\hat{\theta_{1}}\) and \(\hat{\theta_{2}}\), the MLEs of the parameters \(\alpha\), \(\theta_{1}\) and \(\theta_{2}\), can be numerically obtained by solving the following equations:  
\[\frac{\partial L}{\partial\alpha} =\frac{m+n}{\alpha}-\sum_{i=1}^{n}\ln x_{i}-\sum_{j=1}^{m}\ln y_{j }+\frac{1}{\theta_{1}}\sum_{i=1}^{n}x_{i}^{-\alpha}\ln x_{i}+\frac{1}{\theta_ {2}}\sum_{j=1}^{m}y_{j}^{-\alpha}\ln y_{j}=0, \tag{7}\] \[\frac{\partial L}{\partial\theta_{1}} =-\frac{n}{\theta_{1}}+\frac{1}{\theta_{2}^{2}}\sum_{i=1}^{n}x_{i }^{-\alpha}=0,\] (8) \[\frac{\partial L}{\partial\theta_{2}} =-\frac{m}{\theta_{2}}+\frac{1}{\theta_{2}^{2}}\sum_{j=1}^{m}y_{j }^{-\alpha}=0. \tag{9}\]  
From (8) and (9), we obtain  
\[\hat{\theta_{1}}(\alpha)=\frac{1}{n}\sum_{i=1}^{n}x_{i}^{-\alpha}\quad\text{ and}\quad\hat{\theta_{2}}(\alpha)=\frac{1}{m}\sum_{j=1}^{m}y_{j}^{-\alpha}. \tag{10}\]  
Putting the expressions of \(\hat{\theta_{1}}(\alpha)\) and \(\hat{\theta_{2}}(\alpha)\) into (7), we obtain  
\[\frac{m+n}{\alpha}-\frac{1}{\alpha}\left[\sum_{i=1}^{n}\ln x_{i}^{ \alpha}+\sum_{j=1}^{m}\ln x_{j}^{\alpha}\right]+\frac{\sum_{i=1}^{n}x_{i}^{- \alpha}\ln x_{i}}{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{-\alpha}}+\frac{\sum_{j=1}^ {m}y_{j}^{-\alpha}\ln y_{j}}{\frac{1}{m}\sum_{j=1}^{m}y_{j}^{-\alpha}} = 0. \tag{11}\]  
Therefore, \(\hat{\alpha}\) can be obtained as a fixed point solution of the non-linear equation of the form  
\[h(\alpha) = \alpha, \tag{12}\]  
where  
\[h(\alpha) = \frac{-m-n+\sum_{i=1}^{n}\ln x_{i}^{\alpha}+\sum_{j=1}^{m}\ln y_ {j}^{\alpha}}{\frac{\sum_{i=1}^{n}x_{i}^{-\alpha}\ln x_{j}}{\frac{1}{n}\sum_ {i=1}^{n}x_{i}^{-\alpha}}+\frac{\sum_{j=1}^{n}y_{j}^{-\alpha}\ln y_{j}}{\frac {1}{n}\sum_{j=1}^{n}y_{j}^{-\alpha}}}. \tag{13}\]  
Using a simple iterative procedure \(h(\alpha_{(j)})=\alpha_{(j+1)}\), where \(\alpha_{(j)}\) is the j-th iterate of \(\hat{\alpha}\), we stop the iterative procedure when \(|\alpha_{(j)}-\alpha_{(j+1)}|\) is adequately small less than a specified level. Once we obtain \(\hat{\alpha}\), then \(\hat{\theta_{1}}\) and \(\hat{\theta_{2}}\) can be calculated from (10). Therefore, we obtain the MLE of \(R=P(Y<X)\) as  
\[\hat{R} = \frac{\frac{1}{m}\sum_{j=1}^{m}y_{j}^{-\hat{\alpha}}}{\frac{1}{n} \sum_{i=1}^{n}x_{i}^{-\hat{\alpha}}+\frac{1}{m}\sum_{j=1}^{m}y_{j}^{-\hat{ \alpha}}}. \tag{14}\]