Factor analysis for paired ranked data with application on parent–child value orientation preference data

Abstract

Ranking data appear in everyday life and arise in many fields of study such as marketing, psychology and politics. Very often, the key objective of analyzing and modeling ranking data is to identify underlying factors that affect the individuals’ choice behavior. Factor analysis for ranking data is one of the most widely used methods to tackle the aforementioned problem. Recently, Yu et al. [J R Stat Soc Ser A (Statistics in Society) 168:583–597, 2005] have developed factor models for ranked data in which each individual is asked to rank a set of items. However, paired ranked data may arise when the same set of items are ranked by a pair of judges such as a couple in a family. This paper extended the factor model to accommodate such paired ranked data. The Monte Carlo expectation-maximization algorithm was used for parameter estimation, at which the E-step is implemented via the Gibbs Sampler. For model assessment and selection, a tailor-made method called the bootstrap predictive checks approach was proposed. Simulation studies were conducted to illustrate the proposed estimation and model selection method. The proposed method was applied to analyze a parent–child partially ranked data collected from a value priorities survey carried out in the United States.

Appendix

The complete-data log-likelihood function is:

$$\begin{aligned}&= \! - \frac{n}{2} \; \left\{ {\sum \limits _{c=1}^{2}} \log \left| {\varvec{\varPsi }} \right| + \log \left| {\varvec{\varSigma }}_{f} \right| \right\} \\&\!- \frac{1}{2}{\sum \limits _{i=1}^{n}} \!\left\{ {\sum \limits _{c=1}^{2}} tr \!\left[ {\varvec{\varPsi }}^{-1} ({\varvec{U}}_{ic} \!\!-\!\! {\varvec{\mu }}_{c} \!\!-\!\! {\varvec{\varLambda }}^{T} {\varvec{f}}_{ic}) ({\varvec{U}}_{ic} \!-\! {\varvec{\mu }}_{c} \!-\! {\varvec{\varLambda }}^{T} {\varvec{f}}_{ic})^{T} \right] \!+\! tr \left[ {\varvec{\varSigma }}_{f}^{-1} {\varvec{f}}_{i} {\varvec{f}}_{i}^{T} \right] \right\} , \end{aligned}$$

where

$$\begin{aligned} {\varvec{\varSigma }}_{f}= \begin{pmatrix} {\varvec{I}}_d\quad&\quad {\varvec{\rho }}_{12}\\ {\varvec{\rho }}_{12}&\quad {\varvec{I}}_d\\ \end{pmatrix}, \end{aligned}$$

assuming that ${\varvec{\rho }}_{12}$ is a $d \times d$ diagonal matrix with ($\ell ,\ell$)th element equals ${\rho }_{12,\ell }$. Then, $\log |{\varvec{\varSigma }}_{f}|$ = $\log \prod _{\ell =1}^d (1-\rho ^2_{12,\ell })$ = $\sum _{\ell =1}^d \log (1-\rho ^2_{12,\ell })$.

Also,³

$$\begin{aligned} {\varvec{\varSigma }}_{f}^{-1} = \begin{pmatrix} {\varvec{I}}_d+{\varvec{\rho }}_{12}({\varvec{I}}_d-{\varvec{\rho }}^2)^{-1}{\varvec{\rho }}_{12}&\quad -{\varvec{\rho }}_{12}({\varvec{I}}_d-{\varvec{\rho }}_{12}^2)^{-1} \\ -({\varvec{I}}_d-{\varvec{\rho }}_{12}^2)^{-1}{\varvec{\rho }}_{12}&({\varvec{I}}_d-{\varvec{\rho }}_{12}^2)^{-1} \\ \end{pmatrix}, \end{aligned}$$

therefore,

$$\begin{aligned} \sum _{i=1}^n tr \left[ {\varvec{\varSigma }}_{f}^{-1} {\varvec{f}}_{i} {\varvec{f}}_{i}^{T} \right]&= tr\left[({\varvec{I}}_d+{\varvec{\rho }}_{12}({\varvec{I}}_d-{\varvec{\rho }}^2)^{-1}{\varvec{\rho }}_{12})\sum _{i=1}^n {\varvec{f}}_{i1}{\varvec{f}}_{i1}^T \right]\\&-tr\left[{\varvec{\rho }}_{12}({\varvec{I}}_d-{\varvec{\rho }}_{12}^2)^{-1} \sum _{i=1}^n{\varvec{f}}_{i2}{\varvec{f}}_{i1}^T\right]\\&-tr\left[({\varvec{I}}_d-{\varvec{\rho }}_{12}^2)^{-1}{\varvec{\rho }}_{12} \sum _{i=1}^n{\varvec{f}}_{i1}{\varvec{f}}_{i2}^T\right]\\&+tr\left[({\varvec{I}}_d-{\varvec{\rho }}_{12}^2)^{-1} \sum _{i=1}^n{\varvec{f}}_{i2}{\varvec{f}}_{i2}^T\right]. \end{aligned}$$

Ignoring the terms independent of ${\rho }_{12,\ell }$, this becomes $(1+\frac{{\rho }_{12,\ell }^2}{1-{\rho }_{12,\ell }^2})\sum _{i=1}^n f_{i1\ell }^2$ - $(\frac{2{\rho }_{12,\ell }}{1-{\rho }_{12,\ell }^2})\sum _{i=1}^n f_{i1\ell }f_{i2\ell }$ + $(\frac{1}{1-{\rho }_{12,\ell }^2})\sum _{i=1}^n f_{i2\ell }^2$.

Hence, the derivative of complete log-likelihood with respect to ${\rho }_{12,\ell }$ equals

$$\begin{aligned}&\frac{n{\rho }_{12,\ell }}{1-{\rho }_{12,\ell }^2} - \frac{1}{2}\left(\frac{(1-{\rho }_{12,\ell }^2)\times 2{\rho }_{12,\ell }+{\rho }_{12,\ell }^2\times 2{\rho }_{12,\ell }}{(1-{\rho }_{12,\ell }^2)^2}\right.\sum _{i=1}^n f_{i1\ell }^2\\&\left.\quad \quad \quad \quad \quad -\frac{2\left[(1-{\rho }_{12,\ell }^2)+2{\rho }_{12,\ell }^2\right]}{(1-{\rho }_{12,\ell }^2)^2} \sum _{i=1}^n f_{i1\ell }f_{i2\ell } + \frac{2{\rho }_{12,\ell }}{(1-{\rho }_{12,\ell }^2)^2} \sum _{i=1}^n f_{i2\ell }^2 \right) \\&\quad = -\frac{1}{(1-{\rho }_{12,\ell }^2)^2}\left[n \rho _{12\ell }^{3} - \left( \sum _{i=1}^{n}{f_{i1\ell } f_{i2\ell }} \right) \rho _{12\ell }^{2} + \left( \sum _{i=1}^{n}{\left[ f_{i1\ell }^{2} + f_{i2\ell }^{2} \right]} - n \right) \rho _{12\ell }\right.\nonumber \\&\qquad \qquad \qquad \qquad \qquad \left.-\sum _{i=1}^{n}{f_{i1\ell } f_{i2\ell }} \right]. \end{aligned}$$

Note that this is a cubic function and is decreasing on $\rho _{12\ell }$. Therefore, if there is only one real solution of $\rho _{12\ell }$, the solution will maximize the complete log-likelihood function.