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Data fusion using factor analysis and low-rank matrix completion

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Data fusion involves the integration of multiple related datasets. The statistical file-matching problem is a canonical data fusion problem in multivariate analysis, where the objective is to characterise the joint distribution of a set of variables when only strict subsets of marginal distributions have been observed. Estimation of the covariance matrix of the full set of variables is challenging given the missing-data pattern. Factor analysis models use lower-dimensional latent variables in the data-generating process, and this introduces low-rank components in the complete-data matrix and the population covariance matrix. The low-rank structure of the factor analysis model can be exploited to estimate the full covariance matrix from incomplete data via low-rank matrix completion. We prove the identifiability of the factor analysis model in the statistical file-matching problem under conditions on the number of factors and the number of shared variables over the observed marginal subsets. Additionally, we provide an EM algorithm for parameter estimation. On several real datasets, the factor model gives smaller reconstruction errors in file-matching problems than the common approaches for low-rank matrix completion.

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We would like to thank the reviewers for thoughtful suggestions that have helped to shape and clarify the manuscript.

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Correspondence to Daniel Ahfock.

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This research was partially funded by the Australian Government through the Australian Research Council (Project Number DP180101192)



1.1 Proof of Lemma 1

Due to the rotational invariance of the factor model, we have that

$$\begin{aligned}&\begin{bmatrix} \varvec{\varLambda }_{X}^{A} \\ \varvec{\varLambda }_{Y}^{A} \\ \varvec{\varLambda }_{Z}^{A} \end{bmatrix} = \begin{bmatrix} \varvec{\varLambda }_{X} \\ \varvec{\varLambda }_{Y} \\ \varvec{\varLambda }_{Z} \end{bmatrix}\varvec{R}_{1}, \quad \begin{bmatrix} \varvec{\varLambda }_{X}^{B} \\ \varvec{\varLambda }_{Y}^{B} \\ \varvec{\varLambda }_{Z}^{B} \end{bmatrix} = \begin{bmatrix} \varvec{\varLambda }_{X} \\ \varvec{\varLambda }_{Y} \\ \varvec{\varLambda }_{Z} \end{bmatrix}\varvec{R}_{2},\nonumber \\&\begin{bmatrix} \varvec{\varLambda }_{X}^{B} \\ \varvec{\varLambda }_{Y}^{B} \\ \varvec{\varLambda }_{Z}^{B} \end{bmatrix} = \begin{bmatrix} \varvec{\varLambda }_{X}^{A} \\ \varvec{\varLambda }_{Y}^{A} \\ \varvec{\varLambda }_{Z}^{A} \end{bmatrix}\varvec{R}_{3} , \end{aligned}$$

for orthogonal matrices \(\varvec{R}_{1}, \varvec{R}_{2}\), and \(\varvec{R}_{3}\). The alignment of \(\varvec{\varLambda }_{X}^{A}\) and \(\varvec{\varLambda }_{X}^{B}\) is an orthogonal Procrustes problem. Let \(\varvec{R}\) be the solution to the optimisation problem

$$\begin{aligned} \varvec{R}&= {{\,\mathrm{argmin}\,}}\ \Vert \varvec{\varLambda }_{X}^{A} - \varvec{\varLambda }_{X}^{B}\varvec{R} \Vert _{F}, \quad \text { subject to } \varvec{R}^{\mathsf {T}}\varvec{R} = \varvec{I}. \end{aligned}$$

Assuming that \(\varvec{\varLambda }_{X}^{A}\) and \(\varvec{\varLambda }_{X}^{B}\) are of full column rank, Schönemann (1966) showed that there is a unique solution to (18). As \(\text {rank}(\varvec{\varLambda }_{X}^{A})=\text {rank}(\varvec{\varLambda }_{X}^{B})=\text {rank}(\varvec{\varLambda }_{X})\), both \(\varvec{\varLambda }_{X}^{A}\) and \(\varvec{\varLambda }_{X}^{B}\) are of rank q under Assumption 1. Define \(\varvec{M} = (\varvec{\varLambda }_{X}^{B})^{\mathsf {T}}{\varvec{\varLambda }_{X}^{A}}\) and let the singular value decomposition of \(\varvec{M}\) be given by \(\varvec{M} =\varvec{W}\varvec{D}\varvec{Q}^{\mathsf {T}} \). Then using the result from Schönemann (1966), the unique solution to (18) is given by \(\varvec{R} = \varvec{W}\varvec{Q}^{\mathsf {T}}\). The uniqueness of the solution implies that \(\varvec{R}=\varvec{R}_{3}^{\mathsf {T}}\) as \(\varLambda _{X}^{B}\varvec{R}_{3}\varvec{R}_{3}^{\mathsf {T}} = \varvec{\varLambda }_{X}^{A}\) from (17). Then \(\varvec{\varLambda }_{Z}^{B}\varvec{R} = \varvec{\varLambda }_{Z}^{B}\varvec{R}_{3}^{\mathsf {T}} =\varvec{\varLambda }_{Z}^{A}\varvec{R}_{3}\varvec{R}_{3}^{\mathsf {T}} = \varvec{\varLambda }_{Z}^{A}\) again using (17). Finally, \(\varvec{\varLambda }_{Y}^{A}(\varvec{\varLambda }_{Z}^{B}\varvec{R})^{\mathsf {T}} = \varvec{\varLambda }_{Y}^{A}(\varvec{\varLambda }_{Z}^{A})^{\mathsf {T}} = \varvec{\varLambda }_{Y}\varvec{R}_{1}\varvec{R}_{1}^{\mathsf {T}}\varvec{\varLambda }_{Z}^{\mathsf {T}} = \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Z}^{\mathsf {T}}\).

1.2 Proof of Theorem 1

Using Theorem 5.1 in Anderson and Rubin (1956), Assumption 2 guarantees that if

$$\begin{aligned}&\begin{pmatrix} \varvec{\varLambda }_{X}\varvec{\varLambda }_{X}^{\mathsf {T}} + \varvec{\varPsi }_{X} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Y}^{\mathsf {T}} \\ \varvec{\varLambda }_{Y}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Y}^{\mathsf {T}} + \varvec{\varPsi }_{Y} \end{pmatrix} \\&\quad = \begin{pmatrix} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} + \varvec{\varPsi }_{X}^{*} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} + \varvec{\varPsi }_{Y}^{*} \end{pmatrix}, \\&\begin{pmatrix} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Z}^{\mathsf {T}} + \varvec{\varPsi }_{X} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Z}^{\mathsf {T}} \\ \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Z}^{\mathsf {T}} &{} \varvec{\varLambda }_{Z}\varvec{\varLambda }_{Z}^{\mathsf {T}} + \varvec{\varPsi }_{Z} \end{pmatrix} \\&\quad = \begin{pmatrix} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} + \varvec{\varPsi }_{X}^{*} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} + \varvec{\varPsi }_{Z}^{*} \end{pmatrix}. \end{aligned}$$

then the uniquenesses are equal, \(\varvec{\varPsi }_{X}=\varvec{\varPsi }_{X}^{*}\), \(\varvec{\varPsi }_{Y}=\varvec{\varPsi }_{Y}^{*}\), and \(\varvec{\varPsi }_{Z}=\varvec{\varPsi }_{Z}^{*}\), implying

$$\begin{aligned} \begin{pmatrix} \varvec{\varLambda }_{X}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Y}^{\mathsf {T}} \\ \varvec{\varLambda }_{Y}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Y}^{\mathsf {T}} \end{pmatrix}&= \begin{pmatrix} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} \end{pmatrix}, \end{aligned}$$
$$\begin{aligned} \begin{pmatrix} \varvec{\varLambda }_{X}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Z}^{\mathsf {T}} \\ \varvec{\varLambda }_{Z}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{Z}\varvec{\varLambda }_{Z}^{\mathsf {T}} \end{pmatrix}&= \begin{pmatrix} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \end{pmatrix}. \end{aligned}$$

Using Lemma 1, \(\varLambda _{Y}\varLambda _{Z}^{\mathsf {T}}\) can be uniquely recovered given the matrices on the left-hand side of (19) and (20). Likewise, \(\varLambda _{Y}^{*}\varLambda _{Z}^{*\mathsf {T}}\) can be uniquely recovered given the matrices on the right hand side of (19) and (20). It remains to show that \(\varLambda _{Y}\varLambda _{Z}^{\mathsf {T}} = \varLambda _{Y}^{*}\varLambda _{Z}^{*\mathsf {T}}\). To do so, define the eigendecompositions

$$\begin{aligned} \varvec{V}_{A}\varvec{D}_{A}\varvec{V}_{A}^{\mathsf {T}}&= \begin{pmatrix} \varvec{\varLambda }_{X}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Y}^{\mathsf {T}} \\ \varvec{\varLambda }_{Y}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Y}^{\mathsf {T}} \end{pmatrix}=\begin{pmatrix} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} \end{pmatrix} , \\ \varvec{V}_{B}\varvec{D}_{B}\varvec{V}_{B}^{\mathsf {T}}&= \begin{pmatrix} \varvec{\varLambda }_{X}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Z}^{\mathsf {T}} \\ \varvec{\varLambda }_{Z}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{Z}\varvec{\varLambda }_{Z}^{\mathsf {T}} \end{pmatrix}= \begin{pmatrix} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \end{pmatrix}, \end{aligned}$$

and the rotated and scaled eigenvectors

$$\begin{aligned} \begin{pmatrix}\varvec{\varvec{\varGamma }}_{X}^{A} \\ \varvec{\varvec{\varGamma }}_{Y}^{A} \end{pmatrix}&= \varvec{V}_{A}\varvec{D}_{A}^{1/2}, \quad \begin{pmatrix} \varvec{\varvec{\varGamma }}_{X}^{B} \\ \varvec{\varvec{\varGamma }}_{Z}^{B}\end{pmatrix} = \varvec{V}_{B}\varvec{D}_{B}^{1/2}. \end{aligned}$$

Using Assumption 1 and Lemma 1, the equality

$$\begin{aligned} \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Z}^{\mathsf {T}}&= \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}}= \varvec{\varGamma }_{Y}^{A}\varvec{\varGamma }_{Z}^{B}\varvec{W}\varvec{Q}^{\mathsf {T}}, \end{aligned}$$

must hold, where \(\varvec{W}\) and \(\varvec{Q}\) are the left and right singular vectors of the matrix \(\varvec{M} = (\varvec{\varGamma }_{X}^{B})^{\mathsf {T}}{\varvec{\varGamma }_{X}^{A}} = \varvec{W}\varvec{D}\varvec{Q}^{\mathsf {T}}\). Combining the equalities in (19), (20) and (21) gives the main result

$$\begin{aligned} \begin{pmatrix} \varvec{\varLambda }_{X}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Y}^{\mathsf {T}} &{} \varvec{\varLambda }_{X}\varvec{\varLambda }_{Z}^{\mathsf {T}} \\ \varvec{\varLambda }_{Y}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Y}^{\mathsf {T}} &{} \varvec{\varLambda }_{Y}\varvec{\varLambda }_{Z}^{\mathsf {T}} \\ \varvec{\varLambda }_{Z}\varvec{\varLambda }_{X}^{\mathsf {T}} &{} \varvec{\varLambda }_{Z}\varvec{\varLambda }_{Y}^{\mathsf {T}} &{} \varvec{\varLambda }_{Z}\varvec{\varLambda }_{Z}^{\mathsf {T}} \end{pmatrix}&= \begin{pmatrix} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} &{} \varvec{\varLambda }_{X}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Y}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \\ \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{X}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{Y}^{*\mathsf {T}} &{} \varvec{\varLambda }_{Z}^{*}\varvec{\varLambda }_{Z}^{*\mathsf {T}} \end{pmatrix}. \end{aligned}$$

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Ahfock, D., Pyne, S. & McLachlan, G.J. Data fusion using factor analysis and low-rank matrix completion. Stat Comput 31, 58 (2021).

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