## 2 Estimation of \(R_{s,k}\) by Maximum Likelihood
With the obtrusion \(\hat{\beta}\), \(\hat{\alpha}_{1}\) is obtained from Equation (10) and \(\hat{\alpha}_{2}\) from Equation (11), and the maximum likelihood estimate of \(R_{s,k}\) is as follows:  
\[\hat{R}_{s,k}=\hat{\
u}\sum\limits_{i=s}^{k}\frac{k!}{(k-i)!}\left(\prod\limits _{j=0}^{i}\left(k+\hat{\
u}-j\right)\right)^{-1},\text{ where }\hat{\
u}=\frac{\hat{ \alpha}_{2}}{\hat{\alpha}_{1}}. \tag{14}\]  
The variances of the maximum likelihood estimators are the following:  
\[V(\hat{\alpha}_{1})=\left(E\left(-\frac{\partial^{2}\ln L}{\partial\alpha_{1} ^{2}}\right)\right)^{-1}=\frac{\alpha_{1}^{2}}{n}\text{ and }V(\hat{\alpha}_{2})=\left(E\left(-\frac{ \partial^{2}\ln L}{\partial\alpha_{2}^{2}}\right)\right)^{-1}=\frac{\alpha_{2} ^{2}}{m}. \tag{15}\]  
The asymptotic variance of an estimate of \(R_{s,k}\), which is a function of two independent statistics \(\alpha_{1}\) and \(\alpha_{2}\), is given by [PERSON] (1961):  
\[V\left(\hat{R}_{s,k}\right)\cong V(\hat{\alpha}_{1})\left(\frac{\partial R_{ s,k}}{\partial\alpha_{1}}\right)^{2}+V(\hat{\alpha}_{2})\left(\frac{ \partial R_{s,k}}{\partial\alpha_{2}}\right)^{2}. \tag{16}\]  
Equation (16) yields the asymptotic variance of \(\hat{R}_{s,k}\):  
\[\frac{\partial\hat{R}_{1,3}}{\partial\alpha_{1}}=\frac{3\hat{\
u}}{\hat{ \alpha}_{1}(3+\hat{\
u})^{2}}\text{ and }\frac{\partial\hat{R}_{1,3}}{\partial\alpha_{2}}=\frac{-3}{\hat{\alpha}_{1}(3 +\hat{\
u})^{2}}. \tag{17}\]  
The derivatives of \(R_{s,k}\) for (\(s\), \(k\)) (\(1\), \(3\)) and (\(s\), \(k\)) (\(2\), \(4\)) are obtained separately:  
\[\frac{\partial\hat{R}_{2,4}}{\partial\alpha_{1}}=\frac{12(7+2\hat{\
u})}{\hat{ \alpha}_{1}\left((3+\hat{\
u})(4+\hat{\
u})\right)^{2}}\text{ and }\frac{\partial\hat{R}_{2,4}}{\partial\alpha_{2}}=\frac{-12(7+2\hat{\
u})}{ \hat{\alpha}_{1}\left((3+\hat{\
u})(4+\hat{\
u})\right)^{2}}. \tag{18}\]  
Thus,  
\[V(\hat{R}_{1,3})=\frac{9\hat{\
u}^{2}}{\left(3+\hat{\
u}\right)^{4}}\Big{(} \frac{1}{n}+\frac{1}{m}\Big{)}, \tag{19}\]  
\[V(\hat{R}_{2,4})=\frac{144\hat{\
u}^{2}\left(2\hat{\
u}+7\right)^{2}}{\left((3 +\hat{\
u})(4+\hat{\
u})\right)^{4}}\Big{(}\frac{1}{n}+\frac{1}{m}\Big{)}. \tag{20}\]Under the usual regularity conditions (Rao, [28]):  
as \(n\rightarrow\infty,m\rightarrow\infty,\frac{\hat{R}_{s,k}-R_{s,k}}{V\left(R_{s,k} \right)}\overset{d}{\rightarrow}N(0,1)\). The asymptotic \(100(1-\alpha)\%\) confidence interval for \(R_{s,k}\) is as follows:  
\[\left[\hat{R}_{s,k}\mp Z_{(1-\alpha/2)}\left(V\left(\hat{R}_{s,k}\right) \right)^{1/2}\right]. \tag{21}\]  
The asymptotic \(100(1-\alpha)\%\) confidence interval for \(R_{1,3}\) is as follows:  
\[\left[\hat{R}_{1,3}\mp Z_{(1-\alpha/2)}\frac{3\hat{v}}{(3+\hat{v})^{2}}\Big{(} \frac{1}{n}+\frac{1}{m}\Big{)}^{1/2}\right], \tag{22}\]  
where \(\hat{v}=\hat{\alpha}_{2}/\hat{\alpha}_{1}\). The asymptotic \(100(1-\alpha)\%\) confidence interval for \(R_{2,4}\) is as follows:  
\[\left[\hat{R}_{2,4}\mp Z_{(1-\alpha/2)}\frac{12\hat{v}(2\hat{v}+7)}{\left((3+ \hat{v})(4+\hat{v})\right)^{2}}\Big{(}\frac{1}{n}+\frac{1}{m}\Big{)}^{1/2} \right], \tag{23}\]  
where \(\hat{v}=\hat{\alpha}_{2}/\hat{\alpha}_{1}\) and \(Z_{(1-\alpha/2)}\) is the \(\left(1-\alpha/2\right)^{\text{th}}\) percentile of the standard normal distribution. The following Algorithm 1 is used for the estimation of \(R_{s,k}\)  
```
Step 1: For given values of \(\beta\), \(\alpha_{1}\), \(\alpha_{2}\) using Monte Carlo simulation, generate 10,000 samples such that X-EMED \((\alpha_{1},\beta)\) and Y-EMED \((\alpha_{2},\lambda)\). Step 2: Determine \(R_{1,3}\) and \(R_{24}\) using given parametric values. Step 3: Find maximum likelihood estimate for \(\alpha_{1}\) and \(\alpha_{2}\) via simulation. Step 4: Estimate multicomponent stress-strength reliability estimate \(R_{1,3}\) and \(R_{24}\) using maximum likelihood function for \(\alpha_{1}\) and \(\alpha_{2}\). Step 5: Compute average bias, average mean squared errors, average length of the confidence interval, and average coverage probability (ACP)of \(R_{1,3}\) and \(R_{24}\).