From Gaussian Data to a Covariance Objective
In Gaussian factor analysis, the observations are modeled through a covariance matrix assembled from loadings and specific variances. Maximum likelihood therefore searches over parameters, but the data enter the covariance part of the objective through a compact summary: the sample covariance matrix. Deriving this reduction makes the optimization target precise and clarifies which constraints the factor model imposes.
Assume \(X_1,\ldots,X_n\) are independent observations from a \(p\)-variate normal distribution with mean vector \(\mu\) and covariance matrix \(\Sigma\). For a common-factor model with \(q\) factors, take the factor covariance to be the identity matrix and assume the specific errors are uncorrelated with one another and with the factors. The implied covariance is \(\Sigma=\Lambda\Lambda^{\mathsf T}+\Psi\), where \(\Lambda\) is a \(p\times q\) loading matrix and \(\Psi\) is diagonal.
The strict positivity condition on each uniqueness \(\psi_j\) guarantees that \(\Sigma\) is positive definite, even when the loading matrix does not have full column rank. Indeed, for every nonzero vector \(a\), \(a^{\mathsf T}\Sigma a=\|\Lambda^{\mathsf T}a\|^2+\sum_j\psi_j a_j^2>0\). Thus its determinant and inverse, both needed in the normal likelihood, are well defined.
Profiling Out the Mean
For fixed positive-definite \(\Sigma\), the Gaussian log-likelihood for the full sample is
The mean can be maximized out before optimizing the factor parameters. The following derivation identifies its maximizing value and shows exactly where \(S\) enters.
To prove the result, write \(X_i-\mu=(X_i-\bar X)+(\bar X-\mu)\) and expand the quadratic sum. The cross terms add to zero because \(\sum_i(X_i-\bar X)=0\). Therefore
The second term is nonnegative and equals zero exactly when \(\mu=\bar X\), since \(\Sigma^{-1}\) is positive definite. For the first term, use \(\operatorname{tr}(uv^{\mathsf T}A)=v^{\mathsf T}Au\) to obtain
Substitution into the log-likelihood proves both claims. The result does not require \(S\) to be invertible: the trace expression remains defined because \(\Sigma\) is positive definite. \(\square\)
Worked Example: Profiling the Mean in Two Dimensions
Suppose the three observations are \(X_1=(0,0)^{\mathsf T}\), \(X_2=(2,1)^{\mathsf T}\), and \(X_3=(4,2)^{\mathsf T}\). Evaluate the profiled log-likelihood at \(\Sigma=\operatorname{diag}(4,1)\).
The sample mean is \(\bar X=(2,1)^{\mathsf T}\), so the centered observations are \((-2,-1)^{\mathsf T},(0,0)^{\mathsf T},(2,1)^{\mathsf T}\). Consequently,
Here \(|\Sigma|=4\) and \(\Sigma^{-1}=\operatorname{diag}(1/4,1)\). The trace is \(\operatorname{tr}(S\Sigma^{-1})=(8/3)(1/4)+2/3=4/3\). Although \(S\) is singular, this calculation is valid. With \(p=2\) and \(n=3\), the profile log-likelihood is
The sample covariance summarizes the centered observations for this calculation; no inverse of \(S\) is needed.
The Constrained Factor-Analysis Optimization
Since the leading term of \(\ell_{\mathrm{prof}}\) is constant in the parameters, maximizing the profile likelihood is equivalent to minimizing
over \(\Lambda\in\mathbb R^{p\times q}\) and diagonal \(\Psi\) with strictly positive entries. This is the concentrated Gaussian likelihood objective. Its two terms have distinct roles: the log determinant penalizes covariance volume, while the trace term measures how the model covariance weights the observed sample covariance. Neither term alone is the likelihood criterion; both arise from the normal density.
For a proof, let \(A=S^{-1/2}\Sigma S^{-1/2}\), which is positive definite, and let \(\lambda_1,\ldots,\lambda_p\) be its positive eigenvalues. Determinant multiplication and cyclic invariance of trace give
For every \(\lambda>0\), \(h(\lambda)=\log\lambda+1/\lambda-1\) is nonnegative. In fact, \(h'(\lambda)=(\lambda-1)/\lambda^2\), so it decreases up to \(1\), increases after \(1\), and has minimum \(h(1)=0\). Summing this inequality over the eigenvalues proves the stated bound. Equality requires every \(\lambda_j=1\), which means \(A=I\) and hence \(\Sigma=S\). Conversely, \(\Sigma=S\) gives equality directly. \(\square\)
This result supplies a useful benchmark, not an automatic solution to factor analysis. The unconstrained covariance optimum is \(S\) when \(S\) is positive definite. The factor model attains that benchmark only if \(S\) can be represented as \(\Lambda\Lambda^{\mathsf T}+\Psi\) with the chosen factor count and positive diagonal uniquenesses. Otherwise the maximum-likelihood fit is the best covariance within the restricted model, not an arbitrary covariance matrix.
Worked Example: Comparing Two Admissible Factor Covariances
Take \(p=2\), \(q=1\), and sample covariance \(S=\begin{pmatrix}1&0.4\\0.4&1\end{pmatrix}\). First choose \(\Lambda=(0.6,0.8)^{\mathsf T}\) and \(\Psi=\operatorname{diag}(0.64,0.36)\). Both uniquenesses are positive, and
Its determinant is \(1-0.48^2=0.7696\), and its inverse is \(\frac{1}{0.7696}\begin{pmatrix}1&-0.48\\-0.48&1\end{pmatrix}\). Therefore,
For comparison, \(\Lambda=(0,0)^{\mathsf T}\) and \(\Psi=I\) give \(\Sigma=I\), with \(F=\log|I|+\operatorname{tr}(S)=2\). Since a smaller \(F\) means a larger profile likelihood, the first admissible covariance fits better than this second choice.
Worked Example: When a One-Factor Model Reaches the Covariance Benchmark
Consider the positive-definite sample covariance \(S=\begin{pmatrix}1&0.20&0.15\\0.20&1&0.12\\0.15&0.12&1\end{pmatrix}\). Set \(\Lambda=(0.5,0.4,0.3)^{\mathsf T}\) and \(\Psi=\operatorname{diag}(0.75,0.84,0.91)\). The uniquenesses are all positive. Multiplication gives diagonal entries \(0.25+0.75=1\), \(0.16+0.84=1\), and \(0.09+0.91=1\), and off-diagonal entries \(0.5(0.4)=0.20\), \(0.5(0.3)=0.15\), and \(0.4(0.3)=0.12\). Thus \(\Sigma=S\).
The determinant is \[ |S|=1+2(0.20)(0.15)(0.12)-(0.20)^2-(0.15)^2-(0.12)^2 =0.9303. \] This matrix is positive definite: it is symmetric, has positive diagonal entries, and is strictly diagonally dominant. The covariance benchmark therefore applies. The fitted objective is
Because this one-factor covariance equals \(S\), no other positive-definite covariance can have a smaller value of the objective. This example illustrates why representability matters: a restricted model can nevertheless attain the unconstrained benchmark when the observed covariance has the required factor form.
What the Concentrated Likelihood Does—and Does Not—Say
For fixed data and a fixed factor count, the profiled likelihood depends on \(\Lambda\) and \(\Psi\) only through their implied covariance matrix. Thus two parameter values that produce the same \(\Sigma\) have the same likelihood. The likelihood itself does not distinguish parameter representations that leave the covariance unchanged; constraints used to select or identify a representation are separate from the covariance objective.
The Gaussian assumption is essential to this reduction: under a multivariate normal model, the covariance parameters enter the likelihood through the log determinant and the quadratic form summarized by \(S\). This should not be generalized to an arbitrary non-Gaussian model merely because its covariance is specified by the same factor decomposition.
Common Mistake
A tempting but incorrect argument is: “The sample covariance is the maximum-likelihood covariance, so the factor-analysis estimate is \(S\).” The first part is true only for an unrestricted positive-definite covariance model when \(S\) is positive definite. In factor analysis, the covariance must also have the specified loading-plus-diagonal form. If \(S\) does not have that form for the chosen \(q\), it is not an admissible factor-model covariance. The correction is to minimize \(F(\Lambda,\Psi)\) over the stated constraints; the saturated benchmark tells us the best possible value, not that the restricted model must attain it.
Check Your Understanding
Use the likelihood reduction and constraints developed above to answer these questions.
- Why does maximizing the Gaussian likelihood over the mean give \(\hat\mu=\bar X\) for every fixed positive-definite covariance?
- Write the concentrated covariance objective for a \(p\)-variable, \(q\)-factor model, including the restrictions on \(\Psi\).
- Why is the trace term \(\operatorname{tr}(S\Sigma^{-1})\) defined even if the sample covariance is singular?
- Under what conditions does the unrestricted covariance benchmark equal \(\log|S|+p\), and which covariance attains it?
- Explain why matching the saturated covariance is not guaranteed for a constrained factor model.