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Two Gaussian latent factors admit a nontrivial orthogonal rotation

Example

Let $p = 3$ , $q = 2$ ,

$A = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{array}), D = I_{3}, R = \frac{1}{\sqrt{2}} (\begin{array}{cc} 1 & - 1 \\ 1 & 1 \end{array}) .$

Let $S$ have density $(2 π)^{- 1} \exp (- ∥ s ∥^{2} / 2)$ on $R^{2}$ , and let $ε$ be independent of $S$ with density $(2 π)^{- 3 / 2} \exp (- ∥ e ∥^{2} / 2)$ on $R^{3}$ . With $X = A S + ε$ , the two parameterizations $(A, S)$ and $(A R^{T}, R S)$ have the same covariance

$Cov (X) = (\begin{array}{ccc} 2 & 0 & 1 \\ 0 & 2 & 1 \\ 1 & 1 & 3 \end{array})$

and the same law of $X$ . The matrix $R$ is not a signed permutation, so the Gaussian latent orientation is not uniquely determined.

Facts & Assumptions

Given: The matrices and independent Gaussian random vectors in the example statement.

[F1]

Orthogonal rotation preserves the covariance factor model: replacing $A, S$ by $A R^{T}, R S$ for orthogonal $R$ preserves the covariance factor model and its covariance matrix.

[F2]

Covariance factor model with diagonal residual covariance: the factor model requires centered factors with covariance identity, diagonal positive residual covariance, and zero factor-residual cross-covariance.

Verification

1.1givenalgebra

The displayed densities factor as products of the one-dimensional standard Gaussian density $g (x) = (2 π)^{- 1 / 2} e^{- x^{2} / 2}$ , so the coordinates of $S$ and $ε$ are independent. Symmetry gives $E Z = 0$ for a coordinate $Z$ , and $(x e^{- x^{2} / 2})^{'} = e^{- x^{2} / 2} - x^{2} e^{- x^{2} / 2}$ with vanishing boundary terms gives $\int_{R} x^{2} g (x) d x = \int_{R} g (x) d x = 1$ . Thus $E S = E ε = 0$ , $Cov (S) = I_{2}$ , and $Cov (ε) = I_{3}$ . The given independence of $S$ and $ε$ makes their cross-covariance zero.

2.1F1givenstep 1.1algebra

Direct multiplication gives $R^{T} R = I_{2}$ and $(A R^{T}) (R S) = A S$ , so the transformed representation yields the same $X$ for every outcome. By [F1], its covariance is unchanged; here $A A^{T} + I_{3} = (\begin{array}{ccc} 2 & 0 & 1 \\ 0 & 2 & 1 \\ 1 & 1 & 3 \end{array})$ .

3.1F2step 1.1step 2.1algebra∎

The density of $R S$ is $(2 π)^{- 1} \exp (- ∥ R^{T} s ∥^{2} / 2) ∣ \det R ∣ = (2 π)^{- 1} \exp (- ∥ s ∥^{2} / 2)$ because $R$ is orthogonal and $∣ \det R ∣ = 1$ . Thus $R S$ again has independent standard Gaussian coordinates; it remains independent of $ε$ , and [F2] together with step 1.1 shows both parameterizations satisfy the model. Since $R$ has four nonzero entries, it is not a signed permutation.

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