Hausman Comparisons Under Competing Covariance Structures

Why the Covariance Structure Matters

A Hausman comparison asks whether two estimators of the same structural parameter differ by more than sampling variation would explain. In an instrumental-variables setting, the usual comparison is between an estimator that is efficient if the regressors are exogenous and an instrumental-variables estimator that remains consistent when some regressors are endogenous, provided the instruments are valid. The comparison is informative only when the covariance used for the difference matches the assumptions that justify the test.

This qualification matters under heteroskedasticity. A familiar formula subtracts the covariance matrix of the efficient estimator from that of the less efficient estimator. That subtraction is justified under a particular efficiency relationship; it is not a general rule for combining two covariance estimates. Under heteroskedasticity, ordinary least squares (OLS) is generally not efficient, and subtracting two heteroskedasticity-robust covariance matrices does not, by itself, estimate the variance of the difference.

Setup and the Conventional Hausman Statistic

Let the structural equation be $y=X\beta+W\gamma+u$, where $X$ contains included exogenous regressors and $W$ contains one or more regressors whose exogeneity is in question. Let $Z$ denote excluded instruments for $W$. Write $\theta=(\beta',\gamma')'$ for the vector of structural coefficients. The comparison uses an OLS estimate $\hat{\theta}_{OLS}$ and an instrumental-variables estimate $\hat{\theta}_{IV}$ of this same vector.

The null hypothesis is that the regressors in $W$ are exogenous, so that OLS and instrumental variables estimate the same structural parameter consistently. The alternative is that at least one regressor in $W$ is endogenous, in which case OLS is generally inconsistent while instrumental variables can remain consistent. This interpretation requires a correctly specified structural equation and instruments that are relevant and satisfy the exclusion and exogeneity conditions under both hypotheses. If those instrument conditions fail, the comparison does not isolate endogeneity.

Definition: Under the conventional covariance-difference Hausman test, $V_{IV}$ and $V_{OLS}$ denote covariance matrices for the two estimators under the null. If OLS is efficient under the null and the instrumental-variables estimator remains consistent under the alternative, the statistic is $$ H=(\hat{\theta}_{IV}-\hat{\theta}_{OLS})'(V_{IV}-V_{OLS})^{-1}(\hat{\theta}_{IV}-\hat{\theta}_{OLS}). $$ Under the null and regularity conditions, $H$ has an asymptotic chi-square distribution with degrees of freedom equal to the rank of the tested coefficient difference.

In the usual linear model, OLS is efficient under the null when the structural disturbance is conditionally homoskedastic and the standard linear-model conditions hold. The covariance matrices in the formula must correspond to the same coefficients and parameterization. If only a subset of coefficients is being tested, the difference vector and covariance matrices must be restricted to that subset.

Why the Covariance Difference Appears

The subtraction in the conventional statistic is not an arbitrary way to combine standard errors. It follows from the efficiency of OLS under the null. Let $d=\hat{\theta}_{IV}-\hat{\theta}_{OLS}$. Under that null, efficiency implies that the difference $d$ is asymptotically uncorrelated with $\hat{\theta}_{OLS}$. Equivalently, the covariance between the two estimates is $V_{OLS}$. Expanding the variance of the difference then gives the covariance-difference formula.

\begin{aligned} \operatorname{Cov}(d,\hat{\theta}_{OLS})&=0,\\ \operatorname{Cov}(\hat{\theta}_{IV},\hat{\theta}_{OLS})&=V_{OLS},\\ \operatorname{Var}(d) &=V_{IV}+V_{OLS} -\operatorname{Cov}(\hat{\theta}_{IV},\hat{\theta}_{OLS}) -\operatorname{Cov}(\hat{\theta}_{OLS},\hat{\theta}_{IV})\\ &=V_{IV}+V_{OLS}-V_{OLS}-V_{OLS}\\ &=V_{IV}-V_{OLS}. \end{aligned}

Thus the conventional matrix is the variance of the estimator difference because of a specific covariance identity. Under the null, the statistic standardizes that difference by its estimated variance. If the matrix is nonsingular and the null holds, the standardized quadratic form has the stated asymptotic chi-square reference distribution.

Worked Example: A Conventional Covariance-Difference Test

Suppose a hypothetical study estimates a two-coefficient structural equation by OLS and instrumental variables. Assume the null model is correctly specified, the instruments are relevant and valid, and the disturbance is homoskedastic, so OLS is efficient under the null. Suppose the estimates are $\hat{\theta}_{IV}=(1.12,\,0.42)'$ and $\hat{\theta}_{OLS}=(1.00,\,0.50)'$, and their estimated covariance matrices are $V_{IV}=\begin{pmatrix}0.09&0.018\\0.018&0.05\end{pmatrix}$ and $V_{OLS}=\begin{pmatrix}0.04&0.010\\0.010&0.03\end{pmatrix}$.

The difference in estimates is $d=(0.12,\,-0.08)'$. The estimated covariance of this difference is \[ V_{IV}-V_{OLS} =\begin{pmatrix}0.05&0.008\\0.008&0.02\end{pmatrix}. \] Its determinant is $0.05(0.02)-0.008^2=0.000936>0$, and its diagonal entries are positive, so it is positive definite.

For a symmetric two-by-two matrix, the inverse is its adjugate divided by its determinant. Therefore, \[ H=d'(V_{IV}-V_{OLS})^{-1}d =\frac{0.02(0.12)^2+0.05(-0.08)^2-2(0.008)(0.12)(-0.08)}{0.000936} =\frac{0.0007616}{0.000936} =0.8137. \]

With two tested coefficients, the reference distribution is chi-square with 2 degrees of freedom. Its upper-tail p-value is $P(\chi^2_2\geq0.8137)=\exp(-0.8137/2)\approx0.6657$. At a 5% significance level, this result does not provide evidence against exogeneity. It does not establish that the regressors are exogenous; it indicates that this comparison does not detect a statistically significant difference.

Why Robust Covariance Subtraction Can Fail

Under heteroskedasticity, OLS can remain consistent under the exogeneity null, but it is generally no longer efficient. Consequently, the covariance identity in the derivation above need not hold. In general, the variance of the difference is

\operatorname{Var}(\hat{\theta}_{IV}-\hat{\theta}_{OLS}) =V_{IV}+V_{OLS} -C_{IV,OLS}-C_{IV,OLS}',

where $C_{IV,OLS}=\operatorname{Cov}(\hat{\theta}_{IV},\hat{\theta}_{OLS})$. The two estimators are calculated from the same observations, so their covariance cannot generally be assumed to vanish. Robust covariance estimates for each estimator separately do not supply this cross-covariance, nor do they restore the efficiency identity. Subtracting the two robust covariance matrices alone therefore need not give the variance of the difference.

A further warning sign is a difference matrix that is not positive semidefinite. A genuine covariance matrix cannot have a negative variance in any direction. A finite-sample estimate of a theoretically justified covariance matrix can have numerical problems, but a materially indefinite subtraction is not repaired merely by inverting it or reporting the resulting statistic. It signals that the assumed covariance-difference interpretation may not apply.

Worked Example: An Indefinite Robust Covariance Difference

Consider a second hypothetical data set with heteroskedastic disturbances. Suppose the reported robust covariance estimates for the same two coefficients are $V_{IV}^{rob}=\begin{pmatrix}0.06&0.014\\0.014&0.03\end{pmatrix}$ and $V_{OLS}^{rob}=\begin{pmatrix}0.04&0.010\\0.010&0.035\end{pmatrix}$. Assume for this illustration that the instruments are relevant and valid and that the structural equation is correctly specified. Heteroskedasticity means, however, that OLS efficiency under the null is not available.

The direct subtraction is \[ V_{IV}^{rob}-V_{OLS}^{rob} =\begin{pmatrix}0.02&0.004\\0.004&-0.005\end{pmatrix}. \] Its determinant is $0.02(-0.005)-0.004^2=-0.000116<0$. The matrix is indefinite, so it cannot be used as a covariance matrix for the estimator difference. The problem is not simply that one coefficient has an inconvenient standard error: the subtraction lacks the required covariance justification, and the actual variance also depends on the cross-covariance of the two estimates. A conventional Hausman statistic formed by inverting this matrix is not a valid test.

A Heteroskedasticity-Robust Auxiliary-Regression Test

An alternative is to express the endogeneity null as a restriction in an auxiliary regression. Regress each suspect endogenous regressor in $W$ on the included exogenous regressors $X$ and excluded instruments $Z$, using the same sample as the structural equation. Let $\hat{V}$ be the matrix of first-stage residuals. Then estimate the augmented equation

y=X\beta+W\gamma+\hat{V}\delta+e.

The coefficient vector $\delta$ describes the association between the first-stage residual components and the structural outcome after controlling for $X$ and $W$. Under the exogeneity null and the linear control-function representation, the restriction is $H_0:\delta=0$. Evidence against that restriction is evidence of endogeneity, conditional on the validity of the instruments and the model assumptions. With one suspect regressor, this is a single-coefficient test; with several, the joint null sets all corresponding residual coefficients to zero.

To allow heteroskedasticity, test the restriction using a heteroskedasticity-robust Wald covariance matrix from the augmented regression. For $q$ restrictions, the Wald statistic has an asymptotic chi-square reference distribution with $q$ degrees of freedom under the null and standard regularity conditions. This auxiliary-regression approach does not require OLS to be efficient and does not rely on subtracting the two estimators’ robust covariance matrices. The first-stage regressions must be appropriate linear projections, the instruments must satisfy the stated conditions, and the robust inference is asymptotic rather than an exact small-sample result.

Worked Example: A Robust Residual-Inclusion Wald Test

Suppose a hypothetical sample of 240 workers is used to study log hourly earnings, measured in log points, as a function of years of schooling and other exogenous controls. Schooling is treated as potentially endogenous. Assume a relevant excluded instrument is available, the instrument is exogenous and excluded from the earnings equation, and the linear structural specification is appropriate. The first stage regresses schooling on the controls and instrument; its residual is denoted $\hat{v}$.

The augmented earnings regression includes schooling, the same exogenous controls, and $\hat{v}$. Suppose the estimated coefficient on $\hat{v}$ is $0.30$ log points per residual year of schooling, with heteroskedasticity-robust standard error $0.12$ log points per residual year. The null is $H_0:\delta=0$. The robust Wald statistic for this one restriction is \[ W=\left(\frac{0.30}{0.12}\right)^2=(2.500)^2=6.250. \]

Under $H_0$, $W$ is compared with a chi-square distribution with 1 degree of freedom. The p-value is approximately $0.0124$. At the 5% level, reject the null of exogeneity: the residual coefficient provides evidence that schooling is endogenous under the maintained model and instrument assumptions. The result is not proof that the instrument is valid; the test relies on that condition rather than establishing it.

Choosing and Reporting the Comparison

The two approaches answer a related question but rest on different covariance calculations. Use the conventional covariance-difference statistic when the efficiency condition under the null is credible and the resulting difference matrix is estimable and well behaved. When heteroskedasticity makes OLS inefficient, do not assume that replacing each covariance estimate with a robust one makes their difference valid. A robust auxiliary-regression Wald test offers a direct alternative under its own linear-model and instrument assumptions.

A careful report names the estimator comparison or auxiliary regression, states the null, identifies the covariance estimator, gives the statistic, degrees of freedom, and p-value, and describes the conclusion in context. The test is about the exogeneity restriction. It does not by itself establish instrument validity, and failure to reject is not proof that endogeneity is absent.

Common Mistake

Wrong reasoning: “Both estimators have heteroskedasticity-robust standard errors, so subtracting their robust covariance matrices gives a robust Hausman test.”
Correction: Separate robust covariance estimates do not include the cross-covariance between estimators, and under heteroskedasticity OLS generally lacks the efficiency property that makes the conventional subtraction equal the variance of the difference. Use a test with a justified covariance structure, such as the robust auxiliary-regression Wald test, or estimate the full joint covariance of the estimator difference using a method appropriate to the model.

Key takeaway: The covariance-difference Hausman statistic is justified by OLS efficiency under the null, not by the fact that two estimators are available. Under heteroskedasticity, robust standard errors for each estimator do not validate a simple subtraction; a robust residual-inclusion Wald test can test the endogeneity restriction without that efficiency assumption.

Check Your Understanding

Answer each question by identifying the relevant covariance or maintained assumption.

What efficiency property makes the conventional covariance difference equal the variance of the estimator difference?
Why can the difference of two heteroskedasticity-robust covariance matrices be indefinite?
In the auxiliary regression, what restriction is tested when there is one potentially endogenous regressor?
A robust auxiliary-regression test rejects exogeneity. What instrument assumptions are still needed to interpret this as evidence of endogeneity?
Why is failure to reject the endogeneity null not proof that the regressor is exogenous?