A Perfect Match Between a Model and a Mode

Abstract

When the partial least squares estimation methods, the “modes,” are applied to the standard latent factor model against which methods are designed and calibrated in PLS, they will not yield consistent estimators without adjustments. We specify a different model in terms of observables only, that satisfies the same rank constraints as the latent variable model, and show that now mode B is perfectly suitable without the need for corrections. The model explicitly uses composites, linear combinations of observables, instead of latent factors. The composites may satisfy identifiable linear structural equations, which need not be regression equations, estimable via 2SLS or 3SLS. Each time practitioners contemplate the use of PLS’ basic design model the composites model is a viable alternative. The chapter is conceptual mainly, but a small Monte Carlo study exemplifies the feasibility of the new approach.

This chapter “continues” a sometimes rather spirited discussion with Wold, that started in 1977, at the Wharton School in Philadelphia, via my PhD thesis, Dijkstra (1981), and a paper Dijkstra (1983). There was a long silence, until about 2008, when Peter M. Bentler (UCLA) rekindled my interest in PLS, one of the many things for which I owe him my gratitude. Crucial also is the collaboration with Joerg Henseler (Twente), that led to a number of papers on PLS and on ways to get consistency without the need to increase the number of indicators, PLSc, as well as to a software program ADANCO for composites. I am very much in his debt too. The present chapter expands on Dijkstra (2010) by avoiding unobservables much as possible while still adhering to Wold’s fundamental principle of soft modeling.

Working paper version of a chapter in “Recent Developments in PLS SEM,” H. Latan and R. Noonan (eds.) to be published in 2017/2018.

Appendix

Here we will prove that $\boldsymbol{\Sigma }$ is positive definite when and only when the correlation matrix of the composites, R _c , is positive definite. The “only when”-part is trivial. The proof that {R _c is p.d.} implies {$\boldsymbol{\Sigma }$ is p.d.} is a bit more involved. It is helpful to note for that purpose that we may assume that each $\boldsymbol{\Sigma }_{ii}$ is a unit matrix (pre-multiply and post-multiply by a block-diagonal matrix with $\boldsymbol{\Sigma }_{ii}^{-\frac{1} {2} }$ on the diagonal, and redefine w _i such that $\mathbf{w}_{i}^{\intercal }\mathbf{w}_{i} = 1$ for each i). So if we want to know whether the eigenvalues of $\boldsymbol{\Sigma }$ are positive it suffices to study the eigenvalue problem $\widetilde{\boldsymbol{\Sigma }}\mathbf{x =}\gamma \mathbf{x}$:

$$\displaystyle{ \left [\begin{array}{*{10}c} \mathbf{I}_{p_{1}} & r_{12}\mathbf{w}_{1}\mathbf{w}_{2}^{\intercal }&r_{13}\mathbf{w}_{1}\mathbf{w}_{3}^{\intercal }& \cdot & r_{1N}\mathbf{w}_{1}\mathbf{w}_{N}^{\intercal } \\ & \mathbf{I}_{p_{2}} & r_{23}\mathbf{w}_{2}\mathbf{w}_{3}^{\intercal }& \cdot & r_{2N}\mathbf{w}_{2}\mathbf{w}_{N}^{\intercal } \\ & & \cdot & \cdot & \cdot \\ & & &\mathbf{I} _{ p_{N-1}} & r_{N-1,N}\mathbf{w}_{N-1}\mathbf{w}_{N}^{\intercal } \\ & & & & \mathbf{I}_{p_{N}} \end{array} \right ]\left [\begin{array}{*{10}c} \mathbf{x}_{1} \\ \mathbf{x}_{2} \\ \cdot \\ \mathbf{x} _{ N-1} \\ \mathbf{x}_{N} \end{array} \right ] =\gamma \left [\begin{array}{*{10}c} \mathbf{x}_{1} \\ \mathbf{x}_{2} \\ \cdot \\ \mathbf{x} _{ N-1} \\ \mathbf{x}_{N} \end{array} \right ] }$$

(4.59)

with obvious implied definitions. Observe that every nonzero solution of

$$\displaystyle{ \left [\begin{array}{*{10}c} \mathbf{w}_{1}^{\intercal }& \mathbf{0} & \cdot & \cdot & \mathbf{0} \\ \mathbf{0} &\mathbf{w}_{2}^{\intercal }&\mathbf{0}& \cdot & \mathbf{0}\\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & \cdot &\mathbf{0}&\mathbf{w}_{N-1}^{\intercal }& \mathbf{0} \\ \mathbf{0} & \cdot &\cdot & \mathbf{0} &\mathbf{w}_{N}^{\intercal } \end{array} \right ]\left [\begin{array}{*{10}c} \mathbf{x}_{1} \\ \mathbf{x}_{2} \\ \cdot \\ \mathbf{x} _{ N-1} \\ \mathbf{x}_{N} \end{array} \right ] = \mathbf{0} }$$

(4.60)

corresponds with γ = 1, and there are $\mathop{\sum }_{i=1}^{N}p_{i} - N$ linearly independent solutions. The multiplicity of the root γ = 1 is therefore $\mathop{\sum }_{i=1}^{N}p_{i} - N$ and we need to find N more roots. By assumption R _c has N positive roots. Let u be an eigenvector with eigenvalue μ, so R _c u = μ ⋅u. We have

$$\displaystyle{ \widetilde{\boldsymbol{\Sigma }}\left [\begin{array}{*{10}c} u_{1}\mathbf{w}_{1} \\ u_{2}\mathbf{w}_{2} \\ \cdot \\ u_{ N}\mathbf{w}_{N} \end{array} \right ] = \left [\begin{array}{*{10}c} \left (u_{1} + r_{12}u_{2} + \cdot + r_{1N}u_{N}\right )\mathbf{w}_{1} \\ \left (r_{21}u_{1} + u_{2} + \cdot + r_{2N}u_{N}\right )\mathbf{w}_{2} \\ \cdot \\ \left (r_{N1}u_{1} + r_{N2}u_{2} + \cdot + u_{N}\right )\mathbf{w}_{N} \end{array} \right ] =\mu \left [\begin{array}{*{10}c} u_{1}\mathbf{w}_{1} \\ u_{2}\mathbf{w}_{2} \\ \cdot \\ u_{ N}\mathbf{w}_{N} \end{array} \right ] }$$

(4.61)

In other words, the remaining eigenvalues are those of R _c , and so all eigenvalues of $\widetilde{\boldsymbol{\Sigma }}$ are positive. Therefore $\boldsymbol{\Sigma }$ is p.d., as claimed.

Note for the determinant of $\boldsymbol{\Sigma }$ that

$$\displaystyle{ \text{det}\left (\boldsymbol{\Sigma }\right ) = \text{det}\left (\mathbf{R}_{c}\right ) \times \text{det}\left (\boldsymbol{\Sigma }_{11}\right ) \times \text{det}\left (\boldsymbol{\Sigma }_{22}\right ) \times \text{det}\left (\boldsymbol{\Sigma }_{33}\right ) \times \ldots \times \text{det}\left (\boldsymbol{\Sigma }_{NN}\right ) }$$

(4.62)

and so the Kullback–Leibler’ divergence between the Gaussian density for block-independence and the Gaussian density for the composites model is $-\frac{1} {2}$ log(det$\left (\mathbf{R}_{c}\right )$). It is well known that 0 ≤ det$\left (\mathbf{R}_{c}\right ) \leq 1$, with 0 in case of a perfect linear relationship between the composites, so Kullback–Leibler divergence is infinitely large, and 1 in case of zero correlations between all composites, with zero divergence.