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Heteroskedasticity-Robust Inference · Tutorial 115 of 952

Breusch-Pagan and Koenker Score Diagnostics

Learn how squared-residual auxiliary regressions produce Breusch–Pagan and Koenker diagnostics, and what their different distributional assumptions mean for inference.

Intermediate 11 min read

What You'll Learn

  • State the variance-model null hypothesis and choose the covariates used in the test.
  • Derive the Breusch–Pagan score statistic from the normal likelihood.
  • Set up the auxiliary regression of squared OLS residuals.
  • Compute and interpret chi-square diagnostics with the appropriate degrees of freedom.
  • Distinguish the classical normal-theory test from Koenker’s studentized variant.
  • Recognize what rejection and non-rejection do—and do not—establish.

Why Test for Variance Dependence?

A residual plot can suggest that error variability changes with the predictors, but a visual pattern does not quantify how much evidence there is against constant conditional variance. The Breusch–Pagan and Koenker tests turn that diagnostic question into a joint test: do the squared residuals systematically vary with a specified set of covariates? Their shared auxiliary-regression structure makes them convenient, while their different scaling reflects different assumptions about the error distribution.

These are diagnostics for the variance specification, not tests of whether the conditional mean is correctly specified. Let the mean model be \(y_i=x_i'\beta+u_i\), estimated by OLS, and let \(\hat u_i\) be its residuals. The covariate vector \(z_i\) contains the variables whose relationship with the error variance is being tested. It may, but need not, contain all the regressors in the mean model.

Definition: A variance-model score diagnostic tests whether the conditional variance depends on specified covariates. For example, the model \(\operatorname{Var}(u_i\mid x_i,z_i)=\sigma^2\exp(z_i'\gamma)\) has the null hypothesis \(H_0:\gamma=0\), against the alternative that at least one component of \(\gamma\) is nonzero. Under the null, the conditional variance is constant at \(\sigma^2\).

The exponential form ensures a positive variance and provides a convenient parametric model for deriving the score test. The null is the substantive target: the test asks whether the chosen variance covariates provide evidence of departures from constant variance. It does not test every possible form of heteroskedasticity.

The Auxiliary Regression

First estimate the conditional mean by OLS and calculate \(\hat u_i^2\). Then regress these squared residuals on an intercept and the chosen variance covariates:

$$ \hat u_i^2 = a + z_i'\delta + v_i. $$

Let \(R^2_{\text{aux}}\) denote the coefficient of determination from this auxiliary regression, and let \(q\) be the number of nonconstant covariates in \(z_i\). Under the null and suitable regularity conditions, the Koenker statistic is \(nR^2_{\text{aux}}\), which is compared with a chi-square distribution with \(q\) degrees of freedom. The conventional Breusch–Pagan score test uses normal-theory scaling; its familiar large-sample form is also \(nR^2_{\text{aux}}\). A finite-sample score expression makes the difference in scaling explicit, as derived below.

The intercept is included to allow the average squared residual to be nonzero. It is not counted among the \(q\) tested coefficients. The test is joint: the null is that all \(q\) slopes are zero, not that each slope is assessed by a separate test.

1
Fit the mean model.
Obtain OLS residuals \(\hat u_i\) from the specified conditional-mean regression.
2
Choose variance covariates.
Specify \(z_i\) from the variance question of interest; do not choose them solely because they produce a small p-value.
3
Regress squared residuals.
Fit \(\hat u_i^2\) on an intercept and the \(q\) covariates, and record \(R^2_{\text{aux}}\).
4
Calculate and compare.
For the Koenker form, calculate \(nR^2_{\text{aux}}\) and compare it with \(\chi^2_q\); for a classical Breusch–Pagan score calculation, use its normal-theory scaling.

Deriving the Breusch–Pagan Score

The classical derivation assumes independent normal errors conditional on the covariates, with variance \(\sigma^2\exp(z_i'\gamma)\). Under \(H_0:\gamma=0\), the score for \(\gamma\), if the errors and \(\sigma^2\) were known, is

$$ S_\gamma=\frac{1}{2}\sum_{i=1}^n z_i\left(\frac{u_i^2}{\sigma^2}-1\right). $$

Because the intercept in \(z_i\) would describe the overall scale \(\sigma^2\), rather than a change in variance with covariates, the scale is a nuisance parameter. Removing that nuisance direction centers the tested covariates. Write \(Z_c\) for the matrix whose rows are \(z_i-\bar z\), where \(\bar z=n^{-1}\sum_i z_i\). Under normality, the efficient information for the centered variance slopes is \(\tfrac12 Z_c'Z_c\). Thus the score quadratic form is

$$ LM_{\mathrm{BP}} = \frac{1}{2}r'P_{Z_c}r, \qquad r_i=\frac{u_i^2}{\sigma^2}-1, \qquad P_{Z_c}=Z_c(Z_c'Z_c)^{-1}Z_c'. $$

This expression shows why the test is related to an auxiliary regression: \(r'P_{Z_c}r\) is the explained sum of squares when the centered, scaled squared errors are projected onto the centered variance covariates. In practice, the errors and \(\sigma^2\) are unknown, so OLS residuals and a residual-based scale estimate are used. Under the null, normality gives \(\operatorname{Var}(u_i^2/\sigma^2)=2\). Consequently the total variation in scaled squared errors is approximately \(2n\), and the score statistic is asymptotically equivalent to \(nR^2_{\text{aux}}\). With \(q\) tested slopes, its null reference distribution is \(\chi^2_q\).

Theorem: Under the constant-variance null, a correctly specified conditional-mean model, independent observations, full-rank tested covariates, and standard large-sample regularity conditions, the Breusch–Pagan score statistic has an asymptotic \(\chi^2_q\) distribution when the conditional errors are normal. Its usual auxiliary-regression form is \(nR^2_{\text{aux}}\), interpreted as a large-sample score statistic.

Koenker’s Studentized Variant

The normal-theory score derivation uses the fact that a normal error has fourth moment \(3\sigma^4\), so the variance of its square is \(2\sigma^4\). If errors are not normal, the variation of squared errors need not have that value. Koenker’s studentized version adjusts the scaling using the observed variation in squared residuals. In its standard auxiliary-regression form, the statistic is \(nR^2_{\text{aux}}\), with a \(\chi^2_q\) reference distribution under the null.

The studentization makes this version less sensitive to nonnormality than the classical normal-theory score test. It does not remove the need for assumptions: the usual large-sample justification relies on independent observations, a suitable conditional-mean model, constant conditional variance under the null, and finite moments sufficient for the squared residuals and their variation to behave regularly. In particular, a heavy-tailed distribution with no finite fourth moment can undermine the usual chi-square approximation.

Both tests use the same chosen variance covariates and both are asymptotic diagnostics. Their statistics need not be identical in a finite sample because the classical BP score uses normal-theory scaling, whereas Koenker’s version studentizes using the empirical variation in squared residuals. The choice between them should reflect the error-distribution assumptions that are credible for the application, not a search for whichever p-value is smaller.

Worked Examples

Worked Example: A Joint Test With Two Variance Covariates

Suppose a hypothetical OLS regression of weekly clinic workload on its mean-model covariates uses \(n=40\) independent observations. A researcher tests whether two specified covariates, patient volume and staffing level, are related to the error variance. The squared-residual auxiliary regression, including an intercept, has \(R^2_{\text{aux}}=0.180\). Use the Koenker form at significance level \(\alpha=0.05\).

The null is that both variance slopes are zero, \(H_0:\delta_1=\delta_2=0\); the alternative is that at least one is nonzero. There are \(q=2\) tested slopes. The assumptions for the usual large-sample diagnostic are independent observations, an adequate conditional-mean model, and a constant conditional variance under the null, with suitable finite moments. Here we treat the specified observations and model as satisfying those working conditions.

Statistic. The Koenker statistic is

$$ nR^2_{\text{aux}}=40(0.180)=7.20. $$

Under \(H_0\), compare this with \(\chi^2_2\). For two degrees of freedom the upper-tail probability is \(\exp(-7.20/2)=0.0273\). Since \(0.0273<0.05\), reject \(H_0\) at the 5% level. These data provide evidence that the squared residuals are related to at least one of the two specified covariates. The result does not identify which covariate matters or establish a causal effect on variance.

Worked Example: A Small Auxiliary R-Squared Is Not Automatically Evidence

In a second hypothetical regression with \(n=60\), an analyst tests whether two preselected covariates are associated with variance. The auxiliary regression of squared OLS residuals on an intercept and those two variables has \(R^2_{\text{aux}}=0.045\). Assume the same independence, mean-model, null-variance, and moment conditions needed for the Koenker approximation.

The joint statistic is

$$ nR^2_{\text{aux}}=60(0.045)=2.70. $$

With \(q=2\), the upper-tail probability is \(\exp(-2.70/2)=0.259\). This exceeds a 5% significance level, so the test does not reject the null. The appropriate conclusion is that this test does not provide strong evidence of variance dependence on these two covariates. It is not proof that the conditional variance is constant: the test may have limited power, and it only addresses the specified covariates and alternatives.

Worked Example: How Non-Normality Can Change the Scaling

Consider a hypothetical sample with \(n=50\) and one tested variance covariate. In the auxiliary regression, scale the squared residuals by their sample mean, so their mean is one. Suppose their centered total sum of squares is \(80\), and the covariate explains \(7\%\) of that variation. Thus \(R^2_{\text{aux}}=0.070\), and the explained sum of squares is \(0.070(80)=5.60\). For illustration, suppose the variation in these scaled squared residuals is lower than the normal-theory value, with total sum of squares per observation \(80/50=1.60\), rather than approximately \(2\).

The normal-theory score expression is one-half the explained sum of squares, giving

$$ LM_{\mathrm{BP}}=\frac{1}{2}(5.60)=2.80. $$

Its \(\chi^2_1\) upper-tail probability is approximately \(0.0943\), so it does not reject at the 5% level. The Koenker statistic is

$$ nR^2_{\text{aux}}=50(0.070)=3.50, $$

whose \(\chi^2_1\) upper-tail probability is approximately \(0.0614\); this also does not reject at 5%, though the evidence is closer to that threshold. The difference illustrates the role of scaling: the classical score uses its normal-theory squared-error variation, while the studentized form reflects the observed variation. The example does not establish that either approximation is reliable for every nonnormal distribution; the necessary independence and moment conditions still matter.

Interpretation, Limits, and a Common Pitfall

A small p-value indicates that the observed auxiliary-regression association would be unusual under the constant-variance null and the assumptions used for the reference distribution. It is evidence against constant variance in the direction represented by the chosen \(z_i\). A large p-value is a failure to detect that pattern, not confirmation of homoskedasticity. The tests also do not measure the practical size of heteroskedasticity; inspect the residual scale and consider the substantive importance of any detected pattern.

A common mistake is to regress squared residuals on every available predictor, see a significant result, and conclude that the variance model is wrong in every possible way. The test only addresses its chosen covariate set and its asymptotic approximation. Another mistake is to regard the Koenker version as assumption-free because it is studentized. It relaxes the normality requirement, but still relies on independent observations and appropriate moment and model conditions.

Finally, a rejection does not invalidate OLS coefficient estimates by itself. Under the appropriate conditional-mean assumptions, OLS can still estimate the mean-model coefficients, while conventional variance formulas may be unsuitable under heteroskedasticity. Robust covariance estimation is a separate inferential response; the diagnostic tests do not substitute for deciding how uncertainty should be estimated.

Key takeaway: Regress squared OLS residuals on a prespecified set of variance covariates. The Breusch–Pagan score test uses normal-theory scaling; Koenker studentizes the diagnostic to reduce sensitivity to nonnormality. In either case, specify the null, count only the tested slopes for the degrees of freedom, and interpret the result relative to its assumptions.

Check Your Understanding

Answer each question using the definitions and procedures above.

  1. In the auxiliary regression of squared residuals on an intercept and three variance covariates, what is the null hypothesis and how many degrees of freedom are used?
  2. A Koenker test has \(n=80\) and \(R^2_{\text{aux}}=0.025\). Calculate its statistic.
  3. Why does the classical Breusch–Pagan score test rely more directly on normality than Koenker’s studentized variant?
  4. What does failure to reject a constant-variance null allow you to conclude, and what does it not establish?
  5. Why is a Koenker test not assumption-free even though it is less sensitive to nonnormality?
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