Erratum to: Linking Item Response Model Parameters

Erratum to: Psychometrika (2016) 81(3):650–673 DOI 10.1007/s11336-015-9469-6

The following argument should have been added to the proof of Theorem 3 to show that the linking function ${\xi }^{\ast }=\phi \left(\xi \right)$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\xi }^{{*}}= \varphi (\varvec{\xi })$$\end{document} has to be separable in the components of $\xi$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\xi }$$\end{document} : as the linking problem is symmetric in ${\xi }^{\ast }$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\xi }^{{*}}$$\end{document} and $\xi$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\xi }$$\end{document} , $\phi$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi $$\end{document} has to be bijective (i.e., has an inverse that returns the same unique $\xi$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\xi }$$\end{document} from which the linking departs). In addition, to allow for the fact that the two calibrations may yield the same value for some of the parameters, $\phi$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi $$\end{document} should always be able to return ${\xi }_j^{\ast }={\xi }_j,j=1,\dots ,d$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\xi _{j}^{{*}}=\xi _{j}, j=1,\ldots ,d$$\end{document} , for all values of $\xi$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\xi }$$\end{document} . The separable form of $\phi \left(\xi \right)$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi (\varvec{\xi })$$\end{document} in (31) does have both properties: each of its component functions is monotone and thus has an inverse, while the identity function is a special case of a monotone function. Now, if $\phi \left(\xi \right)$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi (\varvec{\xi })$$\end{document} would not be separable in its components, it would hold that ${\xi }_j^{\ast }={\phi }_j({\xi }_1,\dots ,{\xi }_d\})$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\xi _{j}^{{*}}=\varphi _{j}(\xi _{1},\ldots ,\xi _{d}\})$$\end{document} for some $j=1,\dots ,d$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$j=1,\ldots ,d$$\end{document} . However, ${\xi }_j^{\ast }={\phi }_j({\xi }_1,\dots ,{\xi }_d)$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\xi _{j}^{{*}}=\varphi _{j}(\xi _{1},\ldots ,\xi _{d})$$\end{document} is only able to always return ${\phi }_j$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi _{j}$$\end{document} ( ${\xi }_1,\dots ,{\xi }_j,\dots ,{\xi }_d)={\xi }_j$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\xi _{1},\ldots ,\xi _{j},\ldots ,\xi _{d})=\xi _{j}$$\end{document} when it is independent of ( ${\xi }_1,\dots ,{\xi }_{j-1},{\xi }_{j+1},\dots ,{\xi }_d)$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\xi _{1},\ldots ,\xi _{j-1},\xi _{j+1},\ldots ,\xi _{d})$$\end{document} , that is, does not vary as a function of any of the other parameters. It follows that ${\xi }^{\ast }=\phi \left(\xi \right)$ \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\xi }^{{*}}=\varphi (\varvec{\xi })$$\end{document} has to be separable in its components.