Why an Exact Null Law Is Useful
Bartlett’s chi-square approximation provides a convenient way to test whether all population canonical correlations vanish. When the sample is not large, however, it is useful to know the finite-sample null distribution rather than rely on an approximation. In the classical Gaussian setting, Wilks’ lambda has an exact distribution that can be expressed as a product of independent beta variables.
The parameters in that product depend on the number of observations and on the dimensions of both variable sets. The result is conditional on a specific model: independent observations from a joint multivariate normal distribution, a global null of zero cross-covariance, and enough observations for the sample covariance matrices to be nonsingular.
Under joint Gaussianity, zero cross-covariance implies independence between the two variable vectors. This independence is what makes the exact null derivation possible. It is not enough that two sets merely have zero cross-covariance in an arbitrary, non-Gaussian population.
The Exact Finite-Sample Law
Write \(d=\max(p,q)\) for the larger set dimension and \(s=\min(p,q)\) for the smaller. Let \(n\) observations be independent and identically distributed, and suppose \(n>d+s\). This sample-size condition ensures that the centered sample covariance matrix for the combined variables is positive definite with probability one under the model. Define \(m=n-1\), the degrees of freedom remaining after estimating the means.
The ordering in the formula is chosen so that the first beta shape parameter decreases as \(i\) increases. The last such parameter is \((m-d-s+1)/2\), which is positive under \(n>d+s\). The formula is symmetric in the two variable-set dimensions because it uses only their maximum \(d\) and minimum \(s\).
Why the Product Representation Holds
A regression projection reduces the determinant in Wilks’ lambda to a ratio of determinants of independent Wishart matrices. To use that reduction, take the larger variable set as the predictors and the smaller set as the responses. Let the centered data matrix for the predictors have \(d\) columns and the centered response matrix have \(s\) columns. Under the null, these matrices are independent.
Let \(P_0\) be the projection onto the constant vector and \(P_X\) the projection onto the centered predictor columns. With probability one, the predictor columns have rank \(d\). The centered observation space has dimension \(m\), and the part of that space orthogonal to both the constant vector and the predictors has dimension \(m-d\). Define $$ H=Y^\mathsf{T}P_XY,\qquad E=Y^\mathsf{T}(I-P_0-P_X)Y,\qquad T=Y^\mathsf{T}(I-P_0)Y=H+E, $$ where \(Y\) is the uncentered response data matrix. The projections remove the mean, so \(T\) is the total centered response sum-of-products matrix.
Conditional on the predictors, choose an orthonormal basis for the centered observation space whose first \(d\) directions span the predictor columns. Under the null, the response coordinates in this basis are independent Gaussian vectors with covariance \(\Sigma_Y\). Consequently, \(H\) and \(E\) are independent Wishart matrices with scale \(\Sigma_Y\) and degrees of freedom \(d\) and \(m-d\), respectively. These conditional distributions do not depend on the particular predictor matrix.
The determinant identity for canonical correlations gives \(\Lambda=|E|/|T|\). For example, the determinant of the residual response covariance divided by the determinant of the total response covariance is the product of \(1-\hat\rho_j^2\). Multiplying the response coordinates by a nonsingular whitening transformation changes both determinants by the same factor, so the ratio is unaffected. We can therefore take \(\Sigma_Y=I\) in deriving its distribution.
For a nonnegative integer \(r\), integrating \(|B|^r\) against this density changes \(a\) to \(a+r\) in the matrix beta normalizing integral. Thus $$ E[\Lambda^r]= \frac{\Gamma_s(a+r)\Gamma_s(a+b)} {\Gamma_s(a)\Gamma_s(a+b+r)} = \prod_{i=1}^{s} \frac{\Gamma(a-(i-1)/2+r)\Gamma(a+b-(i-1)/2)} {\Gamma(a-(i-1)/2)\Gamma(a+b-(i-1)/2+r)}. $$ Each factor is the \(r\)th moment of a \(\operatorname{Beta}(a-(i-1)/2,b)\) random variable. The product of independent variables with those beta distributions therefore has the same moments as \(\Lambda\). Both distributions are supported on \([0,1]\), where moments determine the distribution: polynomials are dense among continuous functions on this interval, so equality of all moments implies equality of the probability measures. Substituting \(a=(m-d)/2\) and \(b=d/2\) proves the theorem. \(\square\)
Worked Examples
Worked Example: One Variable Versus Two
Suppose one set has dimension \(1\), the other has dimension \(2\), and \(n=10\) observations are collected. Find the exact null distribution of Wilks’ lambda.
Here \(d=2\), \(s=1\), and \(m=n-1=9\). There is one beta factor, with parameters $$ \frac{m-d-1+1}{2}=\frac{9-2}{2}=\frac{7}{2}, \qquad \frac{d}{2}=1. $$ Thus \(\Lambda\sim\operatorname{Beta}(7/2,1)\) under the global null. Its mean is \( (7/2)/(7/2+1)=7/9\). Since there is only one canonical correlation, \(\Lambda=1-\hat\rho_1^2\); accordingly, \(E[\hat\rho_1^2]=1-7/9=2/9\). This calculation concerns the exact null expectation, not a test decision or a large-sample approximation.
Worked Example: A Two-Dimensional Smaller Set
Suppose the variable sets have dimensions \(2\) and \(3\), with \(n=14\). Find the two beta factors and the expected value of Wilks’ lambda under the null.
The dimensions give \(d=3\) and \(s=2\), while \(m=13\). The sample-size condition holds because \(14>3+2\). For \(i=1\), the first beta parameter is \((13-3-1+1)/2=5\); for \(i=2\), it is \((13-3-2+1)/2=9/2\). The common second parameter is \(d/2=3/2\). Therefore $$ \Lambda\ \overset{d}{=}\ B_1B_2,\qquad B_1\sim\operatorname{Beta}(5,3/2),\quad B_2\sim\operatorname{Beta}(9/2,3/2), $$ with \(B_1\) and \(B_2\) independent. A beta variable with parameters \(a,b\) has mean \(a/(a+b)\), so $$ E[\Lambda]=E[B_1]E[B_2] =\frac{5}{5+3/2}\frac{9/2}{9/2+3/2} =\frac{10}{13}\cdot\frac{3}{4} =\frac{15}{26}. $$ Independence justifies multiplying the expectations.
Worked Example: Checking How the Dimensions Enter
Suppose \(p=3\), \(q=2\), and \(n=12\). Determine the exact null law and its mean.
Now \(d=3\), \(s=2\), and \(m=11\). Since \(12>3+2\), the required sample-size condition holds. The beta parameters are $$ B_1\sim\operatorname{Beta}\left(\frac{11-3-1+1}{2},\frac{3}{2}\right) =\operatorname{Beta}(4,3/2), \qquad B_2\sim\operatorname{Beta}\left(\frac{11-3-2+1}{2},\frac{3}{2}\right) =\operatorname{Beta}(7/2,3/2). $$ The factors are independent, giving $$ E[\Lambda]=\frac{4}{4+3/2}\frac{7/2}{7/2+3/2} =\frac{8}{11}\cdot\frac{7}{10} =\frac{28}{55}. $$ If the set dimensions were listed in the opposite order, \(d\) and \(s\) would be unchanged, so the exact law would be unchanged as well. The test concerns association between the two sets, not which set is labelled first.
Using the Law and Avoiding a Common Mistake
The beta-product representation specifies a finite-sample null law, including the dependence of its parameters on sample size and both set dimensions. It does not say that the observed canonical roots are independent beta variables. The theorem describes the distribution of their product through independent auxiliary factors; those factors need not correspond one-for-one to the observed roots.
The sample-size condition matters as well. If \(n\le p+q\), the centered joint sample covariance matrix cannot be positive definite in the usual classical setting. The nonsingular determinant ratio and the exact law derived here then do not apply as stated. A computational convention that returns zero roots or a numerical value of lambda in a rank-deficient sample does not restore the classical assumptions.
For large samples, Bartlett’s chi-square approximation can be a useful alternative. The exact law is valuable for understanding what that approximation is approximating, and for finite-sample calculations when its assumptions hold. Neither law is automatically appropriate if observations are not independent, the joint distribution is not Gaussian, or the global null is false.
- Wrong reasoning: “There are \(s\) canonical roots, so each \(1-\hat\rho_j^2\) is one of the independent beta factors.”
- Correction: The exact theorem gives a product distribution for Wilks’ lambda. The independent beta variables are a representation of that product’s law, not a claim about the separate canonical roots.
Check Your Understanding
Use the exact null law and its assumptions to answer these questions.
- For dimensions \(p=1\), \(q=3\), and sample size \(n=12\), identify \(d\) and \(s\), and give the beta parameters for Wilks’ lambda.
- Why does the derivation use the larger variable set as predictors and the smaller one as responses?
- For dimensions \(p=2\), \(q=2\), and \(n=8\), calculate the two beta parameter pairs.
- What does independence in the beta-product representation establish, and what does it not establish about the canonical roots?
- Name two assumptions that are essential to the exact null law derived here.