Young et al. [13] | Kent et al. [14] |
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prob(“no problems”) = ϑ_{1} = 0.5 – tan^{− 1} (−α_{1} + β’X)/ πprob.(“some problems”) = ϑ_{2} = 0.5 – tan^{− 1} (−α_{2} + β’X)/ π- ϑ_{1}prob.(“extreme problems”) = ϑ_{3} = 1 – ϑ_{1} – ϑ_{2}Where α_{i} (i_{=} 1; 2) were the constant terms in the linearpredictor for “no problems” and “some problems,” respectively, and β’X in the linear predictor related thefunctioning levels of an EQ-5D-3L dimension with the PDQ-39 subscale scores. Based on individual pattern X, the predicted functioning level can be obtained by assigning the category with the largest estimated probability (that is, the maximum of ϑ_{1,}ϑ_{2, and}ϑ_{3}) | Dimension i = β(Dimension, i)*x Where i = 1,2,3. Dimension i corresponds to getting a response to an EQ-5D-3L question (i.e. Mobility 2 indicates the response “some problems” in the Mobility dimension). β is the vector of the regression coefficients. x is the matrix of the PDQ-39 scores of the subscales. \( P\ \left( Dimension\ i\right)=\frac{e^{Dimension\ i}}{e^{Dimension\ 1}+{e}^{Domension\ 2}+{e}^{Dimension\ 3}} \) Where i = 1,2,3, and Dimension 1 = 0 (The pivot outcome); e ^{x} is the natural exponential function |