Bartlett's Chi-Square Approximation

Why Correct the Log-Wilks Statistic?

Wilks’ lambda summarizes a set of canonical roots in one number. Under a null hypothesis of zero canonical correlations, its logarithm is related to a likelihood-ratio statistic. A large-sample likelihood-ratio result suggests a chi-square reference distribution, but the raw statistic can be noticeably miscalibrated at finite sample sizes. Bartlett’s correction adjusts its scale to improve that approximation.

The correction is useful both for testing all canonical correlations and for testing whether any correlations remain after the first few roots. It is an approximation under specified sampling conditions, not a claim that the finite-sample distribution is exactly chi-square.

Definition: Let $p$ and $q$ be the numbers of variables in the two sets, and let $m=\min(p,q)$. Write the ordered sample canonical correlations as $\hat\rho_1\ge \cdots \ge \hat\rho_m\ge 0$. For a test beginning at root $k+1$, where $0\le k<m$, Wilks’ lambda is $$ \Lambda_k=\prod_{i=k+1}^{m}(1-\hat\rho_i^2). $$ The global test is the case $k=0$. The sequential null hypothesis is that the population canonical correlations from $\rho_{k+1}$ through $\rho_m$ are all zero.

For the global test, the null says that the two variable sets are uncorrelated. For a sequential test, the first $k$ canonical roots are not part of the tested set; the question is whether the remaining roots contribute any association. The degrees of freedom therefore depend on both the number of roots being tested and the original dimensions $p$ and $q$.

Deriving the Global Bartlett Correction

The global derivation shows where the correction comes from. Assume $n$ independent observations from a joint multivariate normal distribution, with nonsingular within-set covariance matrices. Under the global null, the cross-covariance between the two sets is zero. Let $W$ be the centered sum-of-products matrix for all $p+q$ variables, and let $W_{XX}$ and $W_{YY}$ be its diagonal blocks. In this setting, $$ \Lambda_0=\frac{|W|}{|W_{XX}|\,|W_{YY}|}. $$ The centered matrix $W$ has a Wishart distribution with $\nu=n-1$ degrees of freedom under the null.

A log-determinant identity for a Wishart matrix gives a way to calculate the mean of $-\log\Lambda_0$. Specifically, if $A$ is a $d$-dimensional Wishart matrix with $\nu$ degrees of freedom and identity scale, then $$ E[\log|A|]=d\log 2+\sum_{j=1}^{d}\psi\!\left(\frac{\nu+1-j}{2}\right), $$ where $\psi$ is the digamma function. This identity follows from Bartlett decomposition: the determinant is a product of independent chi-square variables with degrees of freedom $\nu,\nu-1,\ldots,\nu-d+1$, and a chi-square variable with $a$ degrees of freedom has expected log equal to $\psi(a/2)+\log 2$.

Applying the identity to the two marginal blocks and the full matrix cancels the $d\log 2$ terms. For fixed $p,q$ and large $\nu$, the digamma expansion is $$ \psi\!\left(\frac{\nu+1-j}{2}\right) =\log\!\left(\frac{\nu}{2}\right)-\frac{j}{\nu} +\frac{1/6-j^2/2}{\nu^2}+O(\nu^{-3}). $$ The logarithmic terms cancel in the determinant ratio. The terms of order $\nu^{-1}$ contribute $pq/\nu$. At order $\nu^{-2}$, the constant parts cancel and the squared-index terms contribute $pq(p+q+1)/(2\nu^2)$. Thus $$ E[-\log\Lambda_0] =\frac{pq}{\nu}\left\{1+\frac{p+q+1}{2\nu}+O(\nu^{-2})\right\}. $$ This expansion explains the scale correction: multiplying $-\log\Lambda_0$ by $\nu-(p+q+1)/2$ makes its mean agree with the chi-square degrees of freedom $pq$ to this order. Since $\nu=n-1$, the factor is $n-(p+q+3)/2$.

Theorem: Under independent multivariate normal sampling, the Bartlett chi-square approximation for testing the global null of zero canonical correlations is $$ T_0=-\left[n-\frac{p+q+3}{2}\right]\log\Lambda_0 \ \dot{\sim}\ \chi^2_{pq}. $$ Here $\dot{\sim}$ denotes an approximation, not equality in distribution. The log-determinant expansion derives the first-order scale adjustment; it does not make the chi-square law exact at finite $n$.

Sequential Tests and Degrees of Freedom

For a test of roots $k+1,\ldots,m$, the number of remaining dimensions in the two sets is $p-k$ and $q-k$. Their product gives the chi-square degrees of freedom. The Bartlett factor for this sequential likelihood-ratio statistic includes the $k$ roots already set aside. The resulting statistic is $$ T_k=-\left[n-k-\frac{p+q+3}{2}\right]\log\Lambda_k, \qquad T_k\ \dot{\sim}\ \chi^2_{(p-k)(q-k)}. $$ This is the standard first-order Bartlett approximation for the sequential test. It uses the original $p,q$ in the correction factor, while the degrees of freedom use the remaining dimensions. In the global case $k=0$, it reduces to the result above.

The adjustment can be read as a finite-sample refinement of the likelihood-ratio chi-square approximation: the sequential likelihood expansion accounts for the $k$ leading roots, and the Bartlett scale factor corrects the leading finite-sample discrepancy. The correction is most credible when the sample is large relative to the dimensions. If its multiplier is nonpositive, the approximation is not usable; that is a warning that the sample is too small for this formula, not a reason to replace the multiplier by an arbitrary positive value.

Worked Example: Testing All Roots

Suppose $p=3$, $q=2$, and $n=80$. The sample canonical correlations are $\hat\rho_1=0.60$ and $\hat\rho_2=0.40$. Test the global null at the 5% level using Bartlett’s approximation.

First calculate Wilks’ lambda: $$ \Lambda_0=(1-0.60^2)(1-0.40^2) =(0.64)(0.84)=0.5376. $$ The correction factor is $$ 80-\frac{3+2+3}{2}=80-4=76. $$ Therefore $$ T_0=-76\log(0.5376)\approx -76(-0.6206)=47.17. $$ The degrees of freedom are $pq=3(2)=6$. The 5% upper-tail critical value for $\chi^2_6$ is approximately $12.59$. Since $47.17>12.59$, reject the global null. This test indicates evidence that at least one population canonical correlation is nonzero; it does not, by itself, identify which root is responsible.

Worked Example: Testing the Roots After the First

Let $p=4$, $q=3$, and $n=60$. Suppose the sample roots are $0.70, 0.50, 0.30$. Test whether any association remains after the first root, so $k=1$.

The first root is omitted from the Wilks product: $$ \Lambda_1=(1-0.50^2)(1-0.30^2) =(0.75)(0.91)=0.6825. $$ The sequential correction factor is $$ 60-1-\frac{4+3+3}{2}=59-5=54. $$ Hence $$ T_1=-54\log(0.6825)\approx -54(-0.3820)=20.63. $$ The degrees of freedom are $$ (p-k)(q-k)=(4-1)(3-1)=6. $$ Because $20.63$ exceeds the 5% $\chi^2_6$ critical value $12.59$, reject the null that the remaining roots are zero. The conclusion concerns roots 2 and 3 together, not root 2 considered in isolation.

Worked Example: A Sequential Test That Does Not Reject

Take $p=2$, $q=5$, $n=40$, and test the roots after the first, $k=1$. Suppose the remaining sample root is $0.30$. Here $m=2$, so the tested set contains just that one root.

The sequential Wilks product and correction factor are $$ \Lambda_1=1-0.30^2=0.91, \qquad 40-1-\frac{2+5+3}{2}=34. $$ The statistic is $$ T_1=-34\log(0.91)\approx -34(-0.09431)=3.21. $$ There are $(2-1)(5-1)=4$ degrees of freedom. The 5% upper-tail critical value for $\chi^2_4$ is approximately $9.49$. Since $3.21<9.49$, do not reject the null for the remaining root at this level. This is a failure to find sufficient evidence against that null, not proof that the population root is exactly zero.

Assumptions, Interpretation, and a Common Pitfall

The stated approximation assumes independent observations from a joint multivariate normal population and a sample covariance structure of adequate rank. In particular, the covariance blocks must be estimable and the correction multiplier must be positive. The approximation uses the sample canonical correlations through $\Lambda_k$, but it is a statement about the sampling distribution of the test statistic under the specified null. For small samples or many variables relative to $n$, the chi-square calibration may be poor.

A common mistake is to use $pq$ degrees of freedom for every test. Those are the global-null degrees of freedom only. When testing after $k$ roots, the correct degrees of freedom are $(p-k)(q-k)$. A second mistake is to use $n-(p+q+3)/2$ unchanged for every sequential test; the sequential factor subtracts $k$ as well. For example, when $p=4,q=3,k=1$, the degrees of freedom are $6$, and the multiplier is $n-6$, not $n-5$.

Key takeaway: For a test beginning at root $k+1$, compute $\Lambda_k$ from the remaining roots, use multiplier $n-k-(p+q+3)/2$, and compare the resulting statistic with $\chi^2_{(p-k)(q-k)}$. Treat this as a normal-theory approximation, not an exact finite-sample law.

Check Your Understanding

Use the definitions and formulas above to answer the following.

For $p=5,q=3,n=100$, what are the degrees of freedom for the global test and for the test after the first two roots?
Why does the global Bartlett factor differ from $n$, even though the leading likelihood-ratio approximation is chi-square?
For $p=4,q=2,n=50,k=1$, write the sequential multiplier and degrees of freedom.
What does rejection of a sequential null tell you, and what does it not tell you about individual roots?
Why is a nonpositive Bartlett multiplier a warning rather than a quantity to modify by hand?

Bartlett's Chi-Square Approximation

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