White’s Test With Auxiliary Regressions

Why Regress Squared Residuals?

Heteroskedasticity is a feature of the conditional variance: it occurs when the variance of the regression error changes with the regressors. A residual-versus-fitted plot can suggest such a pattern, but White’s test turns a broader search for variance dependence into a formal auxiliary regression. Its key idea is to ask whether the squared residuals are systematically related to functions of the regressors.

The test is useful because it does not require the analyst to specify one particular variance pattern, such as variance increasing linearly with a single regressor. Its flexibility has a cost: including many squares and interactions can use many degrees of freedom and make the test difficult to use effectively in a modest sample.

The null hypothesis is conditional homoskedasticity, \(E(u_i^2\mid X)=\sigma^2\), where \(X\) denotes the regressors in the original model. The alternative, as assessed by a particular auxiliary regression, is that the conditional variance is associated with at least one of the included functions. The test cannot detect a pattern that its auxiliary regressors fail to represent.

Two Auxiliary-Regression Choices

The compact version regresses squared residuals on the fitted value and its square:

If these two nonconstant terms are linearly independent in the sample, the joint null is \(H_0:a_1=a_2=0\), so the test has \(q=2\) restrictions. This version checks for a variance pattern that can be represented as a quadratic function of the fitted value. It is compact, but it is not a general test of every possible nonlinear relationship between variance and the original regressors.

The expanded version uses the original regressors, their squares, and their pairwise cross-products. For example, with two nonconstant regressors \(x_{i1}\) and \(x_{i2}\), the auxiliary regression is

The null is that all five slopes are zero. With \(k\) nonconstant regressors, the nominal number of nonconstant terms is

This count includes \(k\) linear terms, \(k\) squares, and \(k(k-1)/2\) distinct pairwise interactions. It is only a nominal count: duplicated or linearly dependent columns reduce the effective number of restrictions. For example, the square of a binary indicator equals that indicator, so it does not add an independent term.

Why the Chi-Square Calibration Applies

The standard calibration is an asymptotic score-test result. Under the null, the included variance-related slopes are zero. The score measures how much the likelihood would improve by allowing those slopes to depart from zero; after accounting for the nuisance variance parameter, its quadratic form has an asymptotic chi-square distribution with one degree of freedom per independent restriction. The auxiliary regression provides a convenient form of this test: its explained variation is summarized by \(R^2\), giving the conventional statistic \(nR^2\).

For the classical derivation, assume independent observations, a correctly specified conditional mean, conditional normal errors under the null, and regularity conditions such as a fixed number of linearly independent auxiliary terms as \(n\) grows. Under conditional normality, the variance of \(u_i^2\) at the null is \(2\sigma^4\), which gives the standard score-test scaling represented by \(nR^2\). The chi-square approximation is asymptotic, not an exact finite-sample result.

The degrees of freedom are the number \(q\) of independent nonconstant auxiliary regressors being tested, not the residual degrees of freedom in the auxiliary regression. If the expanded model has redundant columns, use the effective rank of its nonconstant terms. A full-rank regression also requires enough observations to estimate the intercept and all auxiliary coefficients.

The chi-square reference distribution is not a claim that the finite-sample statistic is exactly chi-square. Nor does conditional homoskedasticity alone guarantee that every unadjusted score statistic has the classical calibration under arbitrary nonnormal errors. The conventional \(nR^2\) calibration uses the stated conditions; when those conditions are doubtful, the analyst should consider an appropriate studentized diagnostic rather than assume the reference distribution remains accurate.

Degrees-of-Freedom Costs and Practical Choices

An expanded White regression can become large quickly. If the original model has \(k=8\) nonconstant regressors, the full expansion contains \(q=8(11)/2=44\) nonconstant terms, before the auxiliary intercept. With \(n=60\), this requires estimating 45 auxiliary coefficients and leaves only \(60-45=15\) residual degrees of freedom, provided the design has full rank. The chi-square calibration still uses \(q=44\), not 15. The small residual degrees of freedom also signal that the auxiliary regression may be unstable and the test may have low power against many alternatives.

A fitted-value version uses only two restrictions when both terms vary independently, so it avoids much of this degrees-of-freedom cost. But its alternative is narrower: it detects patterns represented by the fitted value and its square, not every variance relationship captured by an expanded regression. The choice should reflect the sample size, the number and nature of regressors, and the kind of variance dependence the analyst wants to detect. With categorical regressors, polynomial terms may be redundant; count the independent columns actually available rather than blindly applying the nominal formula.

Common Mistake and Interpretation

A frequent mistake is to treat the auxiliary regression as though it were an ordinary regression whose individual coefficient \(t\)-tests answer the question, or to compare \(nR^2\) with a chi-square distribution having the auxiliary residual degrees of freedom. White’s test is a joint test: the statistic is \(nR^2\), and its reference degrees of freedom equal the number of independent nonconstant auxiliary regressors under test. Another mistake is to conclude that a rejection proves heteroskedasticity of a particular form. It indicates evidence against the null under the test’s assumptions and chosen expansion; it does not by itself isolate a cause. Conversely, failure to reject is not proof that the variance is constant, especially when the expansion is large relative to the sample.