| Step 1: Sample \documentclass[12pt]{minimal}
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\begin{document}$$Y$$\end{document}Y and \documentclass[12pt]{minimal}
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\begin{document}$$X$$\end{document}X
| Sample 5000 values of \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u}$$\end{document}Yu from \documentclass[12pt]{minimal}
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\begin{document}$${\text{Beta}}(6.52, 5.55)$$\end{document}Beta(6.52,5.55) and 5000 values of \documentclass[12pt]{minimal}
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\begin{document}$$X_{u}$$\end{document}Xu from \documentclass[12pt]{minimal}
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\begin{document}$${\rm Beta}(8.49,3.64)$$\end{document}Beta(8.49,3.64), where the Beta distribution parameters were calculated using Eqs. (5) and (6). | Sample 5000 values of \documentclass[12pt]{minimal}
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\begin{document}$$Y_{c}$$\end{document}Yc from \documentclass[12pt]{minimal}
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\begin{document}$${\text{Gamma}}(1000,0.100)$$\end{document}Gamma(1000,0.100) and 5000 values of \documentclass[12pt]{minimal}
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\begin{document}$$X_{c}$$\end{document}Xc from \documentclass[12pt]{minimal}
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\begin{document}$${\text{Gamma}}(806.67, 0.136)$$\end{document}Gamma(806.67,0.136), where the Gamma distribution parameters were calculated using Eqs. (3) and (4). |
| Step 2: Transform sampled \documentclass[12pt]{minimal}
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\begin{document}$$Y$$\end{document}Y and \documentclass[12pt]{minimal}
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\begin{document}$$X$$\end{document}X from step 1 to unbounded range | Transform sampled \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u}$$\end{document}Yu and \documentclass[12pt]{minimal}
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\begin{document}$$X_{u}$$\end{document}Xu from step 1 to unbounded range using logit function: \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u}^{\prime } = {\text{logit}}(Y_{u} )$$\end{document}Yu′=logit(Yu) and \documentclass[12pt]{minimal}
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\begin{document}$$X_{u}^{\prime } = {\text{logit}}(X_{u} )$$\end{document}Xu′=logit(Xu).The mean and variance for the transformed variables are \documentclass[12pt]{minimal}
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\begin{document}$$\mu_{{Y_{u} }}^{\prime} = 0.928$$\end{document}μYu′=0.928, \documentclass[12pt]{minimal}
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\begin{document}$$\sigma_{{Y_{u} }}^{\prime 2} = 0.436$$\end{document}σYu′2=0.436, \documentclass[12pt]{minimal}
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\begin{document}$$\mu_{{X_{u} }}^{\prime } = 0.186$$\end{document}μXu′=0.186, and \documentclass[12pt]{minimal}
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\begin{document}$$\sigma_{{X_{u} }}^{\prime 2} = 0.364$$\end{document}σXu′2=0.364. | Transform sampled \documentclass[12pt]{minimal}
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\begin{document}$$Y_{c}$$\end{document}Yc and \documentclass[12pt]{minimal}
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\begin{document}$$X_{c}$$\end{document}Xc from step 1 to unbounded range using log function: \documentclass[12pt]{minimal}
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\begin{document}$$Y_{c}^{\prime } = \log (Y_{c} )$$\end{document}Yc′=log(Yc) and \documentclass[12pt]{minimal}
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\begin{document}$$X_{c}^{\prime } = \log (X_{c} )$$\end{document}Xc′=log(Xc).The mean and variance for the transformed variables are \documentclass[12pt]{minimal}
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\begin{document}$$\mu_{{Y_{c} }}^{\prime } = 4.60$$\end{document}μYc′=4.60, \documentclass[12pt]{minimal}
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\begin{document}$$\sigma_{{Y_{c} }}^{\prime 2} = 0.00101$$\end{document}σYc′2=0.00101, \documentclass[12pt]{minimal}
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\begin{document}$$\mu_{{X_{c} }}^{\prime } = 4.70$$\end{document}μXc′=4.70, and \documentclass[12pt]{minimal}
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\begin{document}$$\sigma_{{X_{c} }}^{\prime 2} = 0.00124$$\end{document}σXc′2=0.00124. |
| Step 3: Sample the difference between transformed \documentclass[12pt]{minimal}
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\begin{document}$$Y$$\end{document}Y and \documentclass[12pt]{minimal}
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\begin{document}$$X$$\end{document}X
| Since \documentclass[12pt]{minimal}
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\begin{document}$$\sigma_{{Y_{u} }}^{\prime 2} > \sigma_{{X_{u} }}^{\prime 2}$$\end{document}σYu′2>σXu′2, we define \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u}^{\prime } = X_{u}^{\prime } + \Delta_{u}^{\prime }$$\end{document}Yu′=Xu′+Δu′, where the mean and variance of \documentclass[12pt]{minimal}
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\begin{document}$$\Delta_{u}^{\prime }$$\end{document}Δu′ is 0.742 and 0.072, respectively.Sample \documentclass[12pt]{minimal}
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\begin{document}$$\Delta_{u}^{\prime }$$\end{document}Δu′ from \documentclass[12pt]{minimal}
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\begin{document}$${\rm Gamma}(7.64, 0.10)$$\end{document}Gamma(7.64,0.10), where the Gamma distribution parameters were calculated using Eqs. (3) and (4). Compute sampled values of \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u}^{\prime}$$\end{document}Yu′ by adding sampled \documentclass[12pt]{minimal}
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\begin{document}$${\Delta }_{u}^{\prime}$$\end{document}Δu′ and sampled \documentclass[12pt]{minimal}
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\begin{document}$$X_{u}^{\prime}$$\end{document}Xu′ from step 2. | Since \documentclass[12pt]{minimal}
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\begin{document}$$\sigma_{{Y_{c} }}^{\prime 2} < \sigma_{{X_{c} }}^{\prime 2}$$\end{document}σYc′2<σXc′2, we define \documentclass[12pt]{minimal}
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\begin{document}$$X_{c}^{\prime} = Y_{c}^{\prime} + {\Delta }_{c}^{\prime}$$\end{document}Xc′=Yc′+Δc′, where the mean and variance of \documentclass[12pt]{minimal}
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\begin{document}$$\Delta_{c}^{\prime }$$\end{document}Δc′ is 0.10 and 0.0002, respectively.Sample \documentclass[12pt]{minimal}
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\begin{document}$$\Delta_{c}^{\prime }$$\end{document}Δc′ from \documentclass[12pt]{minimal}
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\begin{document}$${\rm Gamma}(41.07, 0.002)$$\end{document}Gamma(41.07,0.002), where the Gamma distribution parameters were calculated using Eqs. (3) and (4). Compute sampled values of \documentclass[12pt]{minimal}
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\begin{document}$$X_{c}^{\prime}$$\end{document}Xc′ by subtracting sampled \documentclass[12pt]{minimal}
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\begin{document}$${\Delta }_{c}^{\prime}$$\end{document}Δc′ from sampled \documentclass[12pt]{minimal}
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\begin{document}$$Y_{c}^{\prime}$$\end{document}Yc′ from step 2. |
| Step 4: Back transform | Back transform sampled \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u}^{\prime}$$\end{document}Yu′ and \documentclass[12pt]{minimal}
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\begin{document}$$X_{u}^{\prime}$$\end{document}Xu′ to \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u}$$\end{document}Yu and \documentclass[12pt]{minimal}
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\begin{document}$$X_{u}$$\end{document}Xu using \documentclass[12pt]{minimal}
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\begin{document}$$Y_{u} = \frac{{e^{{Y_{u}^{\prime} }} }}{{1 + e^{{Y_{u}^{\prime} }} }}$$\end{document}Yu=eYu′1+eYu′ and \documentclass[12pt]{minimal}
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\begin{document}$$X_{u} = \frac{{e^{{X_{u}^{\prime} }} }}{{1 + e^{{X_{u}^{\prime} }} }}$$\end{document}Xu=eXu′1+eXu′. | Back transform sampled \documentclass[12pt]{minimal}
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\begin{document}$$X_{c}^{\prime}$$\end{document}Xc′ and \documentclass[12pt]{minimal}
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\begin{document}$$Y_{c}^{\prime}$$\end{document}Yc′ to \documentclass[12pt]{minimal}
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\begin{document}$$X_{c}$$\end{document}Xc and \documentclass[12pt]{minimal}
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\begin{document}$$Y_{c}$$\end{document}Yc using \documentclass[12pt]{minimal}
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\begin{document}$$X_{c} = e^{{X_{c}^{\prime} }}$$\end{document}Xc=eXc′ and \documentclass[12pt]{minimal}
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\begin{document}$$Y_{c} = e^{{Y_{c}^{\prime} }}$$\end{document}Yc=eYc′. |