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Multivariate Dependence Testing · Tutorial 449 of 975

Degenerate Null Calibration for Dependence Statistics

Learn how degeneracy shapes the null distribution of distance and kernel dependence statistics, and how to choose between permutation and asymptotic calibration.

Advanced 12 min read

What You'll Learn

  • Identify degeneracy under the independence null and explain its effect on the usual normal approximation.
  • Derive the weighted chi-squared limit for biased HSIC with bounded kernels.
  • Distinguish the asymptotic limits of degenerate V-statistics and U-statistics.
  • Explain when permutation calibration is exact and when its asymptotic distribution matches the null limit.
  • Recognize how estimating nuisance parameters can change a statistic’s limiting law.
  • Avoid treating a weighted chi-squared limit as an ordinary chi-squared distribution.

Why the Null Distribution Is Not Usually Normal

Many statistics for testing independence are built from a kernel or distance that measures joint dependence. Under independence, the population dependence measure is zero. More importantly, the first-order contribution to its sampling fluctuation also vanishes: the relevant kernel is degenerate. That cancellation changes the scale and the form of the null limit. A standard normal approximation that would apply to a nondegenerate average can therefore give the wrong calibration.

This tutorial focuses on the resulting calibration problem. The two main approaches are a weighted chi-squared approximation, obtained from the spectrum of the degenerate kernel, and a permutation distribution, obtained by breaking the observed pairings while preserving the two marginal samples. These methods are related, but they are not interchangeable in every finite sample or every nuisance-parameter setting.

Definition: A symmetric, square-integrable kernel \(h(Z,Z')\), where \(Z,Z'\) are independent with the same distribution, is degenerate if \(E[h(Z,Z')\mid Z]=0\) almost surely. Equivalently, its conditional expectation given either argument is zero. Degeneracy makes the first-order projection of the associated statistic vanish.

For independence testing, \(Z=(X,Y)\), and the null hypothesis is that \(X\) and \(Y\) are independent. Centered distance and kernel constructions lead to degenerate kernels under this null. The statistic is then driven by second-order fluctuation rather than by an ordinary average with a nonzero first-order variance.

Weighted Chi-Squared Limits

A useful general result describes a degenerate, order-two U-statistic. Suppose \(Z_1,\ldots,Z_n\) are independent and identically distributed, and \(h\) is symmetric, square-integrable, and degenerate. Let the integral operator associated with \(h\) have eigenvalues \(\lambda_j\), counted with multiplicity, and an orthonormal set of eigenfunctions \(e_j\), so that \(h(z,z')=\sum_j\lambda_j e_j(z)e_j(z')\) in squared-mean sense. Then, under the usual spectral conditions for this expansion,

$$ n\,\frac{1}{\binom{n}{2}}\sum_{1\le i<j\le n}h(Z_i,Z_j) \ \xrightarrow{d}\ \sum_j \lambda_j(Z_j^2-1), $$

where the \(Z_j\) on the right are independent standard normal random variables. The limit is generally not normal. Its weights can be positive or negative, and the subtraction of 1 reflects the off-diagonal U-statistic form.

Theorem: For a bounded, positive semidefinite kernel \(k\) and an independent sample \(X_1,\ldots,X_n\), the associated centered-kernel V-statistic has a weighted chi-squared limit at scale \(n\). If \(\tilde k\) is the population-centered kernel and its integral operator has eigenvalues \(\alpha_r\ge 0\), then $$ n\,\frac{1}{n^2}\sum_{i,j=1}^n \tilde k(X_i,X_j) \ \xrightarrow{d}\ \sum_r \alpha_r Z_r^2. $$ This is a nonnegative weighted sum, unlike the centered limit for a canonical U-statistic.

For HSIC, there are two independent samples under the null: \(X_1,\ldots,X_n\) and \(Y_1,\ldots,Y_n\). Let \(k\) and \(l\) be bounded positive semidefinite kernels, and let their population-centered integral operators have eigenvalues \(\alpha_r\ge0\) and \(\beta_s\ge0\), respectively. The biased HSIC estimator uses empirically centered Gram matrices. Its null limit is

$$ n\,\widehat{\operatorname{HSIC}}_{\mathrm{biased}} \ \xrightarrow{d}\ \sum_{r,s}\alpha_r\beta_s Z_{rs}^2, $$

where the \(Z_{rs}\) are independent standard normal variables. Boundedness ensures the feature maps have finite second moments; the spectral sums are well-defined, and the result follows from the Hilbert-space central limit theorem. The product weights arise because the limiting covariance operator for the tensor-product features has eigenvalues \(\alpha_r\beta_s\).

Worked Example: Reading the HSIC Limit Weights

Suppose the centered kernel operator for \(X\) has eigenvalues \(0.6\) and \(0.2\), and the centered kernel operator for \(Y\) has eigenvalues \(0.5\) and \(0.1\). Assume all other eigenvalues are zero. Under independence, the limiting distribution of \(n\widehat{\operatorname{HSIC}}_{\mathrm{biased}}\) is

$$ 0.30Z_{11}^2+0.06Z_{12}^2+0.10Z_{21}^2+0.02Z_{22}^2. $$

Each weight is a product: \(0.6(0.5)=0.30\), \(0.6(0.1)=0.06\), \(0.2(0.5)=0.10\), and \(0.2(0.1)=0.02\). The mean of this limiting variable is the sum of the weights, \(0.30+0.06+0.10+0.02=0.48\), since \(E[Z_{rs}^2]=1\). This is a weighted chi-squared distribution, not a single \(\chi^2\) distribution with an unspecified degrees of freedom.

Why the Spectral Limit Holds

For bounded kernels, represent the centered kernels by feature maps \(\tilde\phi(X)\) and \(\tilde\psi(Y)\) in their respective Hilbert spaces. The population-centered version of the biased HSIC statistic is the squared norm of the empirical cross-feature average:

$$ \widehat{\operatorname{HSIC}}_{\mathrm{pop}} = \left\| \frac{1}{n}\sum_{i=1}^n \tilde\phi(X_i)\otimes\tilde\psi(Y_i) \right\|^2. $$

Under independence, each tensor feature has mean zero. Since the feature maps are bounded, these random elements have finite second moments, so the Hilbert-space central limit theorem gives convergence of their normalized sum to a centered Gaussian element. Empirical centering rather than population centering does not affect this limit: the difference between the two normalized sums is \(\sqrt n\,\bar\phi\otimes\bar\psi\), which is \(O_p(n^{-1/2})\) under independence and hence converges to zero in probability.

In eigenfunction coordinates, the covariance operator of the Gaussian limit has eigenvalues \(\alpha_r\beta_s\). The squared norm of a centered Gaussian element is the sum of these eigenvalues times independent squared standard normals. This proves the stated weighted chi-squared limit. Boundedness also gives a finite trace for the covariance operator, so the sum of the nonnegative terms is finite almost surely. \(\square\)

The U-statistic and V-statistic limits should not be conflated. A U-statistic excludes diagonal terms, and for a canonical order-two kernel its limit contains terms of the form \(\lambda_j(Z_j^2-1)\). A biased V-statistic includes diagonal terms; for nonnegative kernel eigenvalues its limit instead has the nonnegative form \(\sum_j\lambda_j Z_j^2\). Distance covariance and HSIC have biased and unbiased estimators, so calibration must match the estimator actually used.

Worked Example: A Rank-One Independence Statistic

Let \(X_i\) and \(Y_i\) be independent Rademacher variables: each takes values \(-1\) and \(1\) with equal probability. Use the rank-one kernels \(k(x,x')=xx'\) and \(l(y,y')=yy'\). Their feature maps have mean zero, and each centered kernel operator has its single nonzero eigenvalue equal to 1.

The biased HSIC statistic in this example is

$$ \widehat{\operatorname{HSIC}}_{\mathrm{biased}} = \left(\frac{1}{n}\sum_{i=1}^n X_iY_i\right)^2. $$

Because \(X_iY_i\) are independent Rademacher variables with mean zero and variance one, the central limit theorem gives \(n^{-1/2}\sum_i X_iY_i\xrightarrow{d}N(0,1)\). Squaring yields \(n\widehat{\operatorname{HSIC}}_{\mathrm{biased}}\xrightarrow{d}\chi_1^2\), which is the one-weight special case of the spectral result. Here a single chi-squared distribution is appropriate because there is only one nonzero product eigenvalue.

Permutation Calibration and Its Scope

A permutation test for independence holds the observed \(X\) values fixed and randomly reassigns the observed \(Y\) values to them. Under the null of independence, and for independent identically distributed pairs, the pairings are exchangeable conditional on the two marginal samples. Consequently, a test that compares the observed statistic with its permutation distribution has exact finite-sample level when all permutations are used and ties are handled conservatively (or randomized). With a random subset of permutations, the Monte Carlo procedure must also account for the observed arrangement to retain valid testing.

For standard distance and kernel statistics under regularity conditions, the conditional permutation distribution of the scaled statistic converges in probability to the same weighted chi-squared null law as the asymptotic theory. This makes permutation calibration a practical alternative when the relevant eigenvalues are difficult to estimate. It also adapts to the observed marginal samples instead of requiring a separate estimate of the full spectrum.

Worked Example: Exact Permutation Calibration at Small Sample Size

Take \(n=4\), with observed signs \(X=(1,1,-1,-1)\) and \(Y=(1,1,-1,-1)\). For the rank-one statistic above, define \(T=n\widehat{\operatorname{HSIC}}_{\mathrm{biased}}=(\sum_i X_iY_i)^2/n\). The observed pairing gives \(\sum_iX_iY_i=4\), so \(T=16/4=4\).

Under the 24 equally likely permutations of the four \(Y\) values, there are six distinct placements of the two positive signs. If both positive \(Y\) values land on positive \(X\) values, or both land on negative \(X\) values, the sum is \(4\) or \(-4\); each case accounts for \(2!2!=4\) permutations. In the other 16 permutations, the sum is zero. Thus \(T=4\) for 8 of the 24 permutations and \(T=0\) for the other 16. The exact upper-tail permutation \(p\)-value is \(8/24=1/3\). The asymptotic \(\chi_1^2\) tail probability at 4 is about \(0.046\), illustrating why the asymptotic approximation can be poor at this sample size.

Nuisance Estimation Can Change the Limit

A weighted chi-squared reference law depends on the kernel and on the distribution under the null. If the statistic also depends on an estimated nuisance parameter, substituting its estimate can change the first-order fluctuation. The resulting limit may involve a projected Gaussian process, a modified covariance operator, or additional terms from the estimator. It is therefore not generally valid to calculate the fixed-parameter eigenvalues and use them unchanged after nuisance estimation.

A simple calculation shows the mechanism. Suppose \(W_1,\ldots,W_n\) are independent \(N(\theta,1)\), and consider the squared, scaled residual-mean statistic. If \(\theta\) is known, it is

$$ n(\bar W-\theta)^2 = \left(\sqrt n(\bar W-\theta)\right)^2 \xrightarrow{d}\chi_1^2. $$

If instead \(\theta\) is estimated by \(\hat\theta=\bar W\), the same residual statistic is exactly \(n(\bar W-\hat\theta)^2=0\) for every sample. Estimation has removed the fluctuating direction entirely. This is not itself an independence test; it isolates the algebra behind the issue. In a dependence test with fitted margins, copula parameters, or other nuisance quantities, the effect is usually less extreme but must still be included in the calibration.

A parametric bootstrap can incorporate nuisance estimation by fitting the null model, simulating from that fitted model, and repeating the full fitting and testing procedure for each simulated sample. A permutation test is exact only when the relevant exchangeability is preserved. If nuisance fitting or design restrictions affect that exchangeability, the fitting and permutation scheme must respect those restrictions; otherwise, exactness does not follow merely because the statistic is permuted.

Common Calibration Pitfall

  • Wrong reasoning: “The null statistic is nonnegative, so its limit must be an ordinary chi-squared variable.”
  • Correction: Nonnegativity only rules out negative values. The limit can be a weighted sum of many independent squared normals, with weights determined by the kernel spectrum. It is a single scaled chi-squared variable only in special cases, such as a single nonzero weight.
  • Also check: Whether the statistic is a biased V-statistic or an unbiased U-statistic, whether nuisance parameters were estimated, and whether the permutation scheme preserves the required exchangeability.
Key takeaway: Degeneracy under independence leads to second-order, generally nonnormal limits. Spectral weighted chi-squared approximations and permutation calibration can agree asymptotically, but finite-sample accuracy, estimator choice, and nuisance estimation determine which calibration is appropriate.

Check Your Understanding

Use the definitions and results above to check your understanding.

  1. What property of a kernel makes its first-order projection vanish under the null?
  2. Why does the biased HSIC limit contain nonnegative weights multiplying \(Z_{rs}^2\)?
  3. How does the limiting form for a canonical order-two U-statistic differ from the V-statistic form?
  4. Under what condition does permutation calibration have exact finite-sample level for an independence test?
  5. Why can a nuisance estimate change the weighted chi-squared reference distribution?
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