1 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 \(\varvec{\xi }^{{*}}= \varphi (\varvec{\xi })\) has to be separable in the components of \(\varvec{\xi }\): as the linking problem is symmetric in \(\varvec{\xi }^{{*}}\) and \(\varvec{\xi }\), \(\varphi \) has to be bijective (i.e., has an inverse that returns the same unique \(\varvec{\xi }\) 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, \(\varphi \) should always be able to return \(\xi _{j}^{{*}}=\xi _{j}, j=1,\ldots ,d\), for all values of \(\varvec{\xi }\). The separable form of \(\varphi (\varvec{\xi })\) 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 \(\varphi (\varvec{\xi })\) would not be separable in its components, it would hold that \(\xi _{j}^{{*}}=\varphi _{j}(\xi _{1},\ldots ,\xi _{d}\})\) for some \(j=1,\ldots ,d\). However, \(\xi _{j}^{{*}}=\varphi _{j}(\xi _{1},\ldots ,\xi _{d})\) is only able to always return \(\varphi _{j}\) (\(\xi _{1},\ldots ,\xi _{j},\ldots ,\xi _{d})=\xi _{j}\) when it is independent of (\(\xi _{1},\ldots ,\xi _{j-1},\xi _{j+1},\ldots ,\xi _{d})\), that is, does not vary as a function of any of the other parameters. It follows that \(\varvec{\xi }^{{*}}=\varphi (\varvec{\xi })\) has to be separable in its components.