Erratum to: Multivariate normal distribution approaches for dependently truncated data

Erratum to: Stat Papers (2012) 53:133–149 DOI 10.1007/s00362-010-0321-x

As per the request of Dr. Takeshi Emura, his address has been updated in the affiliation section.

We provide corrections for Emura and Konno (2012). We also numerically verify the corrected formulae. Appendix gives a real data used for numerical analysis.

Correction in the score function [p. 138, definition of ${\mathbf{U}}_{i}^{*} ({\varvec{\uptheta }})$]

For ${{\varvec{\uptheta }'}}=(\mu _L ,\;\mu _X ,\;\sigma _L^2 ,\;\sigma _X^2)$, the corrected score function is

$$\begin{aligned} \frac{\partial l({\varvec{\uptheta }})}{\partial {\varvec{\uptheta }}}=-n\frac{\dot{c}^*({\varvec{\uptheta }})}{c^*({\varvec{\uptheta }})}+\sum \limits _i {{\mathbf{U}}_i^*({\varvec{\uptheta }})}, \end{aligned}$$

(1)

where

$$\begin{aligned} c^*({\varvec{\uptheta }})=\Phi \left( {\frac{\mu _X -\mu _L }{\sqrt{\sigma _X^2 +\sigma _L^2 } }}\right) , \quad {\mathbf{U}}_i^*({\varvec{\uptheta }})=\left[ \begin{array}{c} {(L_i -\mu _L )/\sigma _L^2 } \\ {(X_i -\mu _X )/\sigma _X^2 }\\ {-/(2\sigma _L^2 )+(L_i -\mu _L )^2/(2\sigma _L^4 )} \\ {-/(2\sigma _X^2 )+(X_i -\mu _X )^2/(2\sigma _X^4 )} \\ \end{array}\right] . \end{aligned}$$

(2)

The error occurred in the third and fourth components of ${\mathbf{U}}_{i}^{*} ({\varvec{\uptheta }})$.

To confirm that Eqs. (1) and (2) are correct, we focus on the third component of Eq. (1):

$$\begin{aligned} \frac{\partial l({\varvec{\uptheta }})}{\partial (\sigma _L^2 )}=-n\frac{1}{c^*({\varvec{\uptheta }})}\frac{\partial c^*({\varvec{\uptheta }})}{\partial (\sigma _L^2 )}+\sum \limits _i{\left\{ {-\frac{1}{2\sigma _L^2 }+\frac{(L_i -\mu _L )^2}{2\sigma _L^4 }} \right\} }. \end{aligned}$$

(3)

We compute the score functions using a real data from Appendix. Then, compare Eq. (3) with the numerical derivative

$$\begin{aligned} \{l(\mu _L ,\mu _X ,\sigma _L^2 +h,\sigma _X^2 ,0)-l(\mu _L ,\mu _X ,\sigma _L^2 ,\sigma _X^2 ,0)\}/h, \end{aligned}$$

where $h=10^{-7}$. The results are given in Table 1. We see that there is virtually no difference between the corrected formula and the numerical derivative. On the other hand, the values of the formula of Emura and Konno (2012) are remarkably different from those of the numerical derivative.

Table 1 Calculations of the score function using the three methods

Correction in the function ${\dot{w}}(c)$ [p. 139]

Emura and Konno (2012) considered a function $w(\cdot ):\,\;(0,1)\rightarrow [0,1]$, defined as

$$\begin{aligned} w(c)=\frac{{\varvec{\Phi }}^{-1}(c)\phi \{{\varvec{\Phi }}^{-1}(c)\}}{c}+\frac{\phi \{{\varvec{\Phi }} ^{-1}(c)\}^2}{c^2}. \end{aligned}$$

They showed that $w$ is strictly decreasing, reflecting the decreasing loss of information at inclusion probability $c$. However, they do not give the formula of $\dot{w}(c)=dw(c)/dc$, and their claim $\dot{w}(1/2)= {\sqrt{{2/\pi }}} (1-4/\pi )$ is incorrect.

Here we provide an explicit derivative given by

$$\begin{aligned} {\dot{w}}(c)=\frac{\{\;1-{\varvec{\Phi }}^{{-1}}(c)^2\;\}c-{\varvec{\Phi }} ^{{-1}}(c)\phi \{{\varvec{\Phi }}^{{-1}}(c)\}}{c^2}-\frac{2\phi \{{\varvec{\Phi }}^{{-1}}(c)\}}{c^3}[c{\varvec{\Phi }} ^{{-1}}(c)+\phi \{{\varvec{\Phi }}^{{-1}}(c)\}] \end{aligned}$$

(4)

With this formula, one has

$$\begin{aligned} \dot{w}(1/2)=\frac{1}{1/2}-\frac{1}{2\pi (1/2)^4}=2-\frac{8}{\pi }\cong -0.5464791. \end{aligned}$$

(5)

We have confirmed the correctness of Eqs. (4) and (5) in Table 2.

Table 2 Numerical calculations of the functions $w(c)$ and ${\dot{w}}(c)$

Appendix: Data on scores from examination and homework (Chung 2013)

The example is due to Chung (2013). Table 3 records two evaluation items: $X^{O}=$  score from examination and $Y^O=$  score from homework are recorded for 10 students. Define a weighted mean $Z^O=0.75X^O+0.25Y^O$. Students receive a “fair” if $Z^{O}\ge 60$ and a “fail” if $Z^{O}<60$, respectively. Suppose that a teacher can obtain samples for those students with $Z^{O}\ge 60$. Then, $Z^{O}\!\ge 60$ is equal to $X^{O}\ge L^O$, where $L^{O}=80-Y^{O}/3$. Accordingly, the observed data is $(L_j , X_j )$, subject to $L_j \le X_j$, for $j=1,2,\ldots , 7$ in which 7 out of 10 samples are included.

Table 3 Test score data for two items $X^{O}=$ score from examination, and $Y^{O}=$ score from homework for 10 students