Bartlett Correction for Factor-Model Fit Tests

Why Correct the Fit Statistic?

For a Gaussian factor model, the likelihood-ratio statistic for overall fit has an asymptotic chi-square distribution when the model is correct and its parameters are regular. In finite samples, however, its null distribution can be shifted relative to that limiting distribution. The Bartlett correction rescales the statistic to remove the leading finite-sample distortion in the regular Gaussian setting.

The adjustment is a scale correction: it changes the observed statistic, not the model being tested or the reference degrees of freedom. Its derivation uses a second-order likelihood expansion, so it is an approximation rather than an exact finite-sample result.

Likelihood-Ratio Statistic and Degrees of Freedom

Assume $X_1,\ldots,X_n$ are independent observations from a $p$-variate normal distribution with mean $\mu$ and covariance matrix $\Sigma=\Lambda\Lambda^\mathsf{T}+\Psi$. Here $\Lambda$ is a $p\times q$ loading matrix, $q$ is fixed, and $\Psi$ is diagonal with strictly positive diagonal entries. Let $S$ be the centered sample covariance matrix formed with divisor $\nu=n-1$. The Gaussian likelihood-ratio discrepancy at a candidate covariance matrix is

F(S,\Sigma) =\log|\Sigma|+\operatorname{tr}(S\Sigma^{-1}) -\log|S|-p.

Let $\hat\Sigma$ be the covariance matrix at the fitted factor model, with the likelihood maximized over its parameters. The uncorrected likelihood-ratio statistic is $T=\nu F(S,\hat\Sigma)$. Its asymptotic reference distribution has degrees of freedom equal to the number of covariance moments minus the number of locally identifiable covariance parameters.

Definition: For a regular $q$-factor Gaussian model with $p$ observed variables, the likelihood-ratio goodness-of-fit statistic is $T=\nu F(S,\hat\Sigma)$, where $\nu=n-1$. Its asymptotic chi-square degrees of freedom are $$ d=\frac{(p-q)^2-p-q}{2}. $$

There are $p(p+1)/2$ distinct entries in a symmetric covariance matrix. The factor representation has $pq$ loading entries and $p$ uniquenesses. Orthogonal changes of factor coordinates do not change $\Lambda\Lambda^\mathsf{T}$, so the loading representation has $q(q-1)/2$ redundant directions. Under generic local identification, the model covariance dimension is therefore $pq-q(q-1)/2+p$. Subtracting this from the unrestricted covariance dimension gives

\begin{aligned} d &=\frac{p(p+1)}{2} -\left(pq-\frac{q(q-1)}{2}+p\right)\\ &=\frac{p^2-p-2pq+q^2-q}{2} =\frac{(p-q)^2-p-q}{2}. \end{aligned}

This parameter count assumes a regular, locally identified covariance model at an interior point. It is not a license to apply the chi-square calibration at boundary or nonidentified fits.

The Bartlett Correction

Write $\nu=n-1$, and define the factor-specific second-order constant $a=(2p+5)/6+2q/3$. In the regular Gaussian factor-model setting, the second-order null expansion of the likelihood-ratio statistic has leading scale inflation $1+a/\nu$. Equivalently, its mean has expansion $E[T]=d(1+a/\nu)+O(\nu^{-2})$, with the corresponding first-order distortion in its null distribution. This expansion is obtained by expanding the profiled Gaussian likelihood about the true covariance and using the Wishart fluctuations of $S$; it is a large-sample expansion with $p$ and $q$ fixed.

Theorem: Suppose the Gaussian factor model is correctly specified, $p$ and $q$ are fixed as $n$ grows, the true covariance is an interior point with positive uniquenesses, and the covariance model is locally identifiable after accounting for rotations. Under the regular likelihood expansion, the Bartlett-corrected statistic is $$ T_{\mathrm B} =\left(1-\frac{a}{\nu}\right)T =\left(1-\frac{(2p+5)/6+2q/3}{n-1}\right)T. $$ It is calibrated against the same asymptotic $\chi^2_d$ distribution as $T$. The scaling removes the leading $O(\nu^{-1})$ null distortion; it does not make the test exact.

The correction follows directly from the expansion. If $T$ has leading scale inflation $1+a/\nu$, multiply by the reciprocal scale, expanded to first order:

\frac{1}{1+a/\nu} =1-\frac{a}{\nu}+O(\nu^{-2}).

Thus the corrected statistic is $(1-a/\nu)T$ to the order retained. Its mean, for example, satisfies

E[T_{\mathrm B}] =\left(1-\frac{a}{\nu}\right) d\left(1+\frac{a}{\nu}\right)+O(\nu^{-2}) =d+O(\nu^{-2}),

because the product of the two first-order factors is $1-a^2/\nu^2$. The same second-order likelihood expansion motivates using this scale adjustment for the null calibration, not just for the mean. The formula requires a positive correction multiplier for its usual interpretation; even when positive, it remains an approximation.

Worked Calculations

Worked Example: Corrected and Uncorrected Fit Tests

A centered sample has $n=101$ observations on $p=6$ variables. A one-factor Gaussian model is fitted, and suppose the maximized likelihood gives $F(S,\hat\Sigma)=0.1187513$. Calculate both statistics and their approximate upper-tail probabilities.

Here $\nu=n-1=100$, $q=1$, and

d=\frac{(6-1)^2-6-1}{2} =\frac{25-7}{2}=9.

The correction constant and multiplier are

a=\frac{2(6)+5}{6}+\frac{2(1)}{3} =\frac{17}{6}+\frac{4}{6} =\frac{21}{6}=3.5, \qquad 1-\frac{a}{\nu}=1-\frac{3.5}{100}=0.965.

The uncorrected statistic is $T=100(0.1187513)=11.87513$. Applying the correction gives $T_{\mathrm B}=0.965(11.87513)=11.46$, to two decimal places. Using a $\chi^2_9$ reference distribution, the uncorrected upper-tail probability is approximately $0.2201$, or $0.22$ to two decimal places. For the corrected statistic, $P(\chi^2_9\ge 11.46)\approx0.2455$, or $0.25$ to two decimal places. The correction lowers the statistic and therefore raises its upper-tail probability; both calculations use 9 degrees of freedom.

Worked Example: A Different Number of Variables and Factors

For $p=8$, $q=2$, and a sample of $n=61$, suppose the fitted discrepancy is $F(S,\hat\Sigma)=0.30$. Find the degrees of freedom and corrected statistic.

There are $\nu=60$ covariance degrees of freedom in the sample covariance convention used here. The test degrees of freedom are

d=\frac{(8-2)^2-8-2}{2} =\frac{36-10}{2}=13.

The correction constant is $a=(2(8)+5)/6+2(2)/3=21/6+4/3=7/2+4/3=29/6$. Thus the multiplier is

1-\frac{a}{\nu} =1-\frac{29}{360} =\frac{331}{360} \approx0.91944.

The uncorrected statistic is $T=60(0.30)=18$. Its corrected value is $T_{\mathrm B}=(331/360)(18)=16.55$. The two statistics are compared with $\chi^2_{13}$, not with different reference degrees of freedom. This example shows that the adjustment depends on $p$, $q$, and sample size, while the fitted discrepancy supplies the observed statistic.

Worked Example: Detecting an Unusable Multiplier

Consider $p=10$, $q=2$, and $n=6$. Before applying a Bartlett correction, check its multiplier.

Here $\nu=5$, and $a=(2(10)+5)/6+2(2)/3=25/6+8/6=33/6=5.5$. The proposed multiplier is

1-\frac{a}{\nu}=1-\frac{5.5}{5}=-0.1.

Multiplying a nonnegative likelihood-ratio statistic by $-0.1$ would produce a negative value, which cannot serve as an ordinary chi-square test statistic. The formula therefore gives no usable corrected test in this setting. The degrees of freedom count is $d=((10-2)^2-10-2)/2=26$, but that count does not rescue the invalid scale factor. The sample is too small for this correction to provide a meaningful calibration.

Interpretation and Common Pitfalls

The correction matters because a likelihood-ratio statistic can be systematically inflated in finite samples even when its first-order large-sample limit is chi-square. Rescaling can improve the approximation to the null distribution when the Gaussian model and regularity conditions are credible. It does not repair a poor factor model, a boundary estimate, a failure of local identification, or an inappropriate distributional assumption.

Changing the degrees of freedom: The correction rescales $T$; it does not alter $d$. Use the same $\chi^2_d$ reference distribution for the corrected statistic.
Reversing the multiplier: The usual correction is $1-a/\nu$, which is below one when $a>0$. It typically reduces the statistic; it is not $1+a/\nu$.
Treating the result as exact: Bartlett correction removes a leading-order distortion under regularity conditions. It is not an exact finite-sample distributional identity.
Ignoring the model conditions: Positive uniquenesses, an interior fit, and local identifiability matter. A correction factor cannot make a boundary or nonregular likelihood-ratio test regular.
Using a nonpositive factor: If $1-a/\nu\leq0$, the corrected statistic is not a sensible nonnegative test statistic. Do not report it as an ordinary chi-square calibration.

In practice, report the uncorrected statistic, the corrected statistic, the degrees of freedom, and which statistic supplied the reported $p$-value. That makes the finite-sample adjustment transparent and helps readers distinguish it from changes to the model or the test’s reference distribution.

Check Your Understanding

Answer these questions using the definitions and assumptions above.

For $p=7$ variables and $q=1$ factor, what is the asymptotic degrees of freedom?
What does the Bartlett correction change, and what does it leave unchanged?
For $p=6$, $q=1$, and $n=101$, calculate the correction multiplier.
Why is it important that the true uniquenesses be positive and the fitted model locally identifiable?
What should you conclude if the proposed Bartlett multiplier is negative?