Tutorials › Multivariate Statistics (Graduate) › Cross-Covariance Tests Under Gaussian Sampling

Multivariate Dependence Testing · Tutorial 432 of 975

Cross-Covariance Tests Under Gaussian Sampling

Construct and interpret a global test of zero cross-covariance for paired multivariate Gaussian observations, and understand what the test can and cannot conclude.

Advanced 11 min read

What You'll Learn

  • Define the zero cross-covariance null for paired multivariate observations
  • Derive the Gaussian likelihood-ratio statistic from the fitted covariance matrices
  • Express the determinant ratio through sample canonical correlations
  • Use the fixed-dimension chi-square approximation with the correct degrees of freedom
  • Interpret the null as independence under joint Gaussian sampling
  • Recognize dimension, sample-size, and rank limitations

Why Test a Cross-Covariance Matrix?

When each subject contributes two multivariate measurements, the question of association concerns more than one pair of variables. Let \(X_i\) be a \(p\)-dimensional vector and \(Y_i\) a \(q\)-dimensional vector for subject \(i\), with \(i=1,\ldots,n\). Their cross-covariance matrix contains \(pq\) covariances, one for each pairing of a component of \(X_i\) with a component of \(Y_i\). A global test asks whether all of those covariances are zero at once.

Assume throughout that the paired observations \((X_i,Y_i)\) are independent and identically distributed from a joint multivariate normal distribution, with finite-dimensional vectors and a positive-definite covariance matrix. The test is useful because it combines evidence across the whole matrix. It also requires care: the number of restrictions grows with both dimensions, and a zero cross-covariance test is not generally an omnibus test of dependence outside the Gaussian model.

Definition: Write the joint mean and covariance as \[ E\begin{bmatrix}X_i\\Y_i\end{bmatrix} = \begin{bmatrix}\mu_X\\\mu_Y\end{bmatrix}, \qquad \operatorname{Cov}\begin{pmatrix}X_i\\Y_i\end{pmatrix} = \begin{bmatrix} \Sigma_{XX} & \Sigma_{XY}\\ \Sigma_{YX} & \Sigma_{YY} \end{bmatrix}, \] where \(\Sigma_{XY}=\operatorname{Cov}(X_i,Y_i)\) is a \(p\times q\) matrix and \(\Sigma_{YX}=\Sigma_{XY}^{\mathsf T}\). The null hypothesis is \(H_0:\Sigma_{XY}=0\), meaning that all \(pq\) cross-covariances are zero. The alternative is \(H_1:\Sigma_{XY}\ne 0\).

Under the joint Gaussian model, a block-diagonal covariance matrix makes the joint density factor into the two marginal densities. Thus, in this model, \(H_0\) is also the null of independence between \(X_i\) and \(Y_i\). Outside joint Gaussian sampling, zero cross-covariance need not imply independence; uncorrelatedness and independence are distinct concepts.

The Gaussian Likelihood-Ratio Test

Let \(Z_i=(X_i^{\mathsf T},Y_i^{\mathsf T})^{\mathsf T}\), let \(\bar Z\) be its sample mean, and define the centered sample covariance using divisor \(n\): \[ S=\frac{1}{n}\sum_{i=1}^n(Z_i-\bar Z)(Z_i-\bar Z)^{\mathsf T} = \begin{bmatrix} S_{XX}&S_{XY}\\ S_{YX}&S_{YY} \end{bmatrix}. \] The divisor convention does not affect the determinant ratio below, since the common scaling cancels.

For a nonsingular \(S\), the unrestricted Gaussian maximum-likelihood estimate of the covariance is \(S\). Under \(H_0\), the covariance is constrained to be block diagonal; its maximum-likelihood estimate has diagonal blocks \(S_{XX}\) and \(S_{YY}\). Comparing the maximized likelihoods gives Wilks' likelihood-ratio statistic.

Theorem: If \(n>p+q\), so the centered sample covariance can be nonsingular, the likelihood-ratio statistic for \(H_0:\Sigma_{XY}=0\) is \[ \Lambda=\left(\frac{|S|}{|S_{XX}||S_{YY}|}\right)^{n/2}, \qquad T=-2\log\Lambda =n\log\left(\frac{|S_{XX}||S_{YY}|}{|S|}\right). \] Small values of \(\Lambda\), equivalently large values of \(T\), count against \(H_0\).

To verify the likelihood calculation, the Gaussian log likelihood for a covariance matrix \(\Sigma\) and mean \(\mu\), apart from a constant, is \[ \ell(\mu,\Sigma)=-\frac{n}{2}\log|\Sigma|-\frac{1}{2}\sum_{i=1}^n(Z_i-\mu)^{\mathsf T}\Sigma^{-1}(Z_i-\mu). \] At the maximum over \(\mu\), \(\hat\mu=\bar Z\). For any positive-definite covariance estimate, the quadratic term then equals \(n\,\operatorname{tr}(\Sigma^{-1}S)\). Maximizing over unrestricted \(\Sigma\) gives \(\hat\Sigma=S\); maximizing over block-diagonal \(\Sigma\) gives \(\hat\Sigma_0=\operatorname{diag}(S_{XX},S_{YY})\). At either maximum, the trace term is \(n(p+q)\), so it cancels in the log-likelihood difference. The remaining determinant terms yield \[ \frac{L(\hat\Sigma_0)}{L(\hat\Sigma)} =\left(\frac{|S|}{|S_{XX}||S_{YY}|}\right)^{n/2}, \] as stated. \(\square\)

Canonical Correlations and the Determinant Ratio

The determinant ratio has a useful interpretation in terms of sample canonical correlations. These are the singular values \(r_1,\ldots,r_s\), where \(s=\min(p,q)\), of the standardized cross-covariance matrix \(S_{XX}^{-1/2}S_{XY}S_{YY}^{-1/2}\). They summarize the strongest linear associations between combinations of the components of \(X\) and combinations of the components of \(Y\).

Result: If \(S_{XX}\) and \(S_{YY}\) are positive definite, then \[ \frac{|S|}{|S_{XX}||S_{YY}|} =\prod_{j=1}^{s}(1-r_j^2). \] In particular, the likelihood-ratio statistic can be written as \[ T=-n\sum_{j=1}^{s}\log(1-r_j^2). \]

Here is a complete determinant argument, including what happens when the two vector dimensions differ. The block determinant identity gives \[ |S|=|S_{XX}|\left|S_{YY}-S_{YX}S_{XX}^{-1}S_{XY}\right|. \] Factoring \(S_{YY}^{1/2}\) from the second determinant on both sides gives \[ \frac{|S|}{|S_{XX}||S_{YY}|} = \left|I_q-S_{YY}^{-1/2}S_{YX}S_{XX}^{-1}S_{XY}S_{YY}^{-1/2}\right|. \] Let \(B=S_{XX}^{-1/2}S_{XY}S_{YY}^{-1/2}\), a \(p\times q\) matrix. The matrix inside this last determinant is \(I_q-B^{\mathsf T}B\). The eigenvalues of \(B^{\mathsf T}B\) are \(r_1^2,\ldots,r_s^2\), together with \(q-s\) zeros if \(q>p\). Therefore the eigenvalues of \(I_q-B^{\mathsf T}B\) are \(1-r_j^2\) for \(j=1,\ldots,s\), and, when \(q>s\), an additional \(q-s\) eigenvalues equal to \(1\). The additional eigenvalues contribute a factor of \(1\) to the determinant. Hence the determinant is exactly \(\prod_{j=1}^{s}(1-r_j^2)\), not a product over nonzero eigenvalues of the factored matrix. \(\square\)

Each canonical correlation lies between zero and one when the sample covariance is positive definite. A large value makes \(1-r_j^2\) small and therefore reduces \(\Lambda\), increasing evidence against the null. The product shows that evidence can accumulate across several moderate canonical correlations; the test is not determined by just the largest one.

Worked Example: A Test With One Variable in Each Vector

Suppose \(p=q=1\), \(n=40\), and the sample covariance matrix is \[ S=\begin{bmatrix}4&3\\3&9\end{bmatrix}. \] Its determinant is \(4(9)-3^2=27>0\), and the sample correlation is \(r=3/(\sqrt4\sqrt9)=0.5\).

The determinant ratio is \[ \Lambda^{2/n}=\frac{|S|}{|S_{XX}||S_{YY}|} =\frac{27}{4(9)}=0.75=1-r^2. \] Thus \(T=-40\log(0.75)=11.507\). The null has \(pq=1\) restriction, so the large-sample reference distribution is \(\chi^2_1\). Its upper-tail probability at 11.507 is approximately \(0.00069\). On that approximation, the data provide strong evidence against zero covariance.

Worked Example: One Vector Has Two Components

Take \(p=2\), \(q=1\), and suppose the sample covariance blocks are \[ S_{XX}=I_2,\qquad S_{YY}=1,\qquad S_{XY}=\begin{bmatrix}0.3\\0.4\end{bmatrix}. \] The full sample covariance is positive definite: its determinant is \(1-(0.3^2+0.4^2)=0.75>0\), and \(S_{XX}\) is positive definite.

There is one sample canonical correlation, equal to the length of \(S_{XY}\) here: \[ r_1=\sqrt{0.3^2+0.4^2}=0.5. \] Consequently, the determinant ratio is \(1-r_1^2=0.75\). If \(n=50\), then \(T=-50\log(0.75)=14.384\). The null has \(pq=2\) restrictions, not one, because both entries of the cross-covariance vector must be zero. The \(\chi^2_2\) upper-tail probability is \(\exp(-14.384/2)\), approximately \(0.00075\).

Reference Distribution and Dimension

The null imposes \(pq\) scalar restrictions: every entry of a \(p\times q\) matrix must vanish. Under the usual regularity conditions for Wilks' theorem—independent observations from the stated Gaussian model, a true parameter in the interior of the positive-definite covariance parameter space, fixed \(p\) and \(q\), and sample size increasing—the likelihood-ratio statistic satisfies \[ T\xrightarrow{d}\chi^2_{pq}\qquad\text{under }H_0. \] The degrees of freedom are the number of cross-covariance parameters excluded by the null. This is a large-sample approximation, not a claim that \(T\) has this chi-square distribution exactly for every finite sample.

For a level-\(\alpha\) test, reject when \(T\) exceeds the \(1-\alpha\) quantile of \(\chi^2_{pq}\), or equivalently report the upper-tail probability \(P(\chi^2_{pq}\ge T)\) as an approximate \(p\)-value. The statistic is nonnegative because the unrestricted covariance fit cannot have a lower maximized likelihood than the constrained fit.

Worked Example: Several Canonical Directions

Suppose \(p=3\), \(q=2\), \(n=80\), and the two sample canonical correlations are \(0.6\) and \(0.2\). The determinant ratio is \[ \Lambda^{2/n}=(1-0.6^2)(1-0.2^2)=(0.64)(0.96)=0.6144. \] The likelihood-ratio statistic is \[ T=-80\log(0.6144)\approx 38.97. \] Here \(pq=6\), so the approximate null reference is \(\chi^2_6\). Its 95th percentile is about \(12.592\); since \(38.97>12.592\), the test rejects at the 5% level. Both canonical directions contribute to the statistic: their separate factors are \(0.64\) and \(0.96\), whose product is smaller than either factor alone.

Dimension affects both interpretation and calibration. Increasing \(p\) or \(q\) increases the number of cross-covariances being tested and therefore the chi-square degrees of freedom. Even when all individual sample associations look modest, several directions can combine to give a small determinant ratio. Conversely, a larger degrees-of-freedom reference distribution changes the rejection threshold. Report the dimensions, sample size, statistic, degrees of freedom, and whether the result uses an asymptotic approximation.

The fixed-dimension approximation should not be treated as reliable automatically when dimensions are large relative to sample size. Moreover, the likelihood-ratio formula using \(|S|\) requires a nonsingular full sample covariance; for continuous Gaussian data, \(n>p+q\) is a convenient sufficient condition. If the covariance estimate is singular, the ordinary determinant-based likelihood-ratio test is not available in this form. Specialized high-dimensional methods or a justified dimension-reduction strategy may be needed.

Common Mistake: Treating This as a Collection of Separate Tests

A tempting argument is: “There are \(pq\) cross-covariances, so test each one separately and declare dependence if any individual \(p\)-value is small.” That is not the global likelihood-ratio test. Separate tests do not use the determinant structure, and their combined false-positive rate is not generally the nominal level unless a suitable multiple-testing procedure is used. The correction is to recognize \(H_0:\Sigma_{XY}=0\) as one matrix hypothesis with \(pq\) restrictions and use the global statistic with its corresponding reference distribution.

A second interpretive caution is important: rejecting \(H_0\) establishes evidence of nonzero cross-covariance under the model, not the direction of causation. Failing to reject does not establish independence in arbitrary distributions. The independence interpretation of this null is specific to joint Gaussian sampling.

Key takeaway: For paired Gaussian vectors, the likelihood-ratio test of zero cross-covariance uses \(T=-n\log\{|S|/(|S_{XX}||S_{YY}|)\}\), equivalently the sum of \(-n\log(1-r_j^2)\) over sample canonical correlations. Its fixed-dimension null approximation is \(\chi^2_{pq}\); both the determinant and the degrees of freedom reflect the two dimensions.

Check Your Understanding

Work through these questions using the definitions and test construction above.

  1. For \(p=4\) and \(q=3\), how many scalar restrictions are imposed by the zero cross-covariance null, and what is the asymptotic reference distribution?
  2. If the sample canonical correlations are \(0.4\) and \(0.1\), write the determinant ratio as a product and explain which canonical direction contributes more to the statistic.
  3. Why does a zero cross-covariance matrix have an independence interpretation under joint Gaussian sampling but not in general?
  4. What condition on the sample covariance is needed for the ordinary determinant-based likelihood-ratio formula, and why can this be a concern when dimensions are large?
  5. Explain why testing each entry of \(S_{XY}\) separately is not the same procedure as the global likelihood-ratio test.
Clinical Biostats