Why Robust Anderson–Rubin Inference Is Needed
A conventional instrumental-variable Wald test relies on estimating the structural coefficient precisely enough to use its estimated standard error. When instruments are weak, that approximation can fail. The Anderson–Rubin (AR) approach instead tests a proposed coefficient value by asking whether the corresponding null-imposed outcome is associated with the instruments. This avoids dividing by an estimated first-stage coefficient.
With heteroskedastic errors, the AR regression still gives a direct test, but its covariance estimate must allow the error variance to differ across observations. The resulting robust Wald statistic is generally calibrated by a large-sample chi-square distribution. It is not, in general, the exact finite-sample F test available under Gaussian homoskedastic errors.
Null-Imposed Regression and Robust Statistic
Consider independent, identically distributed observations from the structural model \(y_i=x_i\beta+w_i'\gamma+u_i\), where \(x_i\) is an endogenous regressor, \(w_i\) is a vector of included exogenous controls, and \(z_i\) is a \(q\)-vector of excluded instruments. To test \(H_0:\beta=\beta_0\), form \(v_i(\beta_0)=y_i-x_i\beta_0\) and regress it on \(w_i\) and \(z_i\). Under the null, the coefficient on \(z_i\) should be zero.
Write \(a_i=(w_i',z_i')'\), let \(\hat b(\beta_0)\) be the ordinary least-squares coefficient vector in that regression, and let \(R\) select its \(q\) instrument coefficients. The HC0 sandwich estimate and AR statistic are
where \(A\) is the matrix whose rows are \(a_i'\), and \(\hat e_i\) are residuals from the null-imposed regression. HC1 or other standard consistent heteroskedasticity-robust corrections can be used instead; with fixed numbers of regressors, their asymptotic limits agree with HC0. The test rejects at asymptotic level \(\alpha\) when \(AR_{\mathrm{rob}}(\beta_0)\) exceeds the \(1-\alpha\) quantile of \(\chi^2_q\).
Worked Example: A One-Instrument Robust AR Statistic
Suppose the null-imposed outcome \(v_i(\beta_0)\) is regressed on one instrument \(z_i\), with no intercept or other controls. The four observed pairs are \((z_i,v_i)=(1,2),(1,0),(-1,1),(-1,-1)\). Compute the HC0 AR statistic.
The OLS instrument coefficient is \( \hat b=\sum_i z_i v_i/\sum_i z_i^2=(2+0-1+1)/4=0.5\). The fitted values are \(0.5,0.5,-0.5,-0.5\), so the residuals are \(1.5,-0.5,1.5,-0.5\). The robust sandwich variance of this coefficient is
Therefore \(AR_{\mathrm{rob}}=0.5^2/(5/16)=0.8\). This calculation illustrates how the robust statistic uses the observation-specific squared residuals. With only four observations, its asymptotic chi-square calibration should not be treated as a reliable finite-sample approximation.
Conditions for Asymptotic Validity
The relevant restriction is a statement about the instrument coefficients in a population linear projection. A useful sufficient set of conditions is the following: observations are iid; \(E[u_i\mid w_i,z_i]=0\); \(E[\|a_i\|^4]<\infty\) and \(E[u_i^4]<\infty\); \(E[a_i a_i']\) is finite and positive definite; and the covariance of the instrument score after accounting for controls is positive definite. The conditional mean condition implies that the null-imposed outcome has zero population linear-projection coefficients on the excluded instruments under \(H_0\). It allows the conditional variance of \(u_i\) to depend on \(w_i,z_i\).
For the last condition, define the population linear projection of \(z_i\) on \(w_i\) and its residual \(\tilde z_i\). Require \(Q=E[\tilde z_i\tilde z_i']\) and \(\Omega=E[u_i^2\tilde z_i\tilde z_i']\) to be positive definite. These conditions ensure that the instrument coefficients are identified in the projection and that their robust asymptotic covariance can be inverted. They do not require \(z_i\) to predict \(x_i\) strongly.
Proof. Under the null, the null-imposed outcome is \(w_i'\gamma+u_i\). By the conditional mean assumption, the population regression of this outcome on \(w_i,z_i\) has coefficient \(\gamma\) on \(w_i\) and zero on \(z_i\). Partialling out the linear projection on \(w_i\), the instrument-coefficient estimation error is driven by the score \(\tilde z_i u_i\). The iid central limit theorem applies because this score has finite second moments:
The sample projection residuals for \(z_i\) converge to \(\tilde z_i\), and the fitted null-regression residuals converge to \(u_i\) in the sample averages defining the sandwich “meat.” The moment conditions and the law of large numbers therefore give consistency of the robust covariance estimate for the asymptotic covariance of \(\hat b_z\), scaled by \(1/n\). Inverting this consistent covariance and applying the continuous mapping theorem to the Wald quadratic form gives the sum of squares of \(q\) independent standard normal variables in the limit, which is \(\chi^2_q\). \(\square\)
Worked Example: A Two-Instrument Robust Quadratic Form
Suppose a fitted null regression gives \(\sqrt n\,\hat b_z=(0.12,-0.08)'\), and the estimated covariance for this scaled coefficient vector is \(\hat\Sigma=\begin{pmatrix}4&1\\1&2\end{pmatrix}\). For \(n=100\), calculate the robust Wald statistic and compare it with the 5% asymptotic critical value for two restrictions.
The determinant of \(\hat\Sigma\) is \(4(2)-1(1)=7\), so \(\hat\Sigma^{-1}=\frac{1}{7}\begin{pmatrix}2&-1\\-1&4\end{pmatrix}\). Multiplication gives \(\hat\Sigma^{-1}(0.12,-0.08)'=(0.32/7,-0.44/7)'\). Hence
Because the supplied covariance is for the scaled coefficient vector, this is the Wald statistic directly; equivalently it uses \(n\hat b_z'\widehat{\operatorname{Var}}(\hat b_z)^{-1}\hat b_z\). The 5% critical value is approximately \(5.991\) for \(\chi^2_2\). Since \(0.0105<5.991\), the null is not rejected by this asymptotic test.
Weak Instruments and the Role of Calibration
The AR construction tests the implication of the proposed structural value for the reduced-form relationship between the null-imposed outcome and the instruments. It does not estimate \(\beta\) by dividing by a first-stage coefficient. Consequently, if first-stage relevance becomes weak while the instrument-score conditions remain valid and nondegenerate, that weakness alone does not invalidate the null statistic’s asymptotic argument. Weak instruments can nevertheless make the test have low power: the null-imposed outcome may be difficult to distinguish from the null when instruments provide little information about \(x_i\).
The qualification matters. The theorem is a fixed-\(q\), iid, large-sample result with a consistently estimable, nonsingular score covariance. It does not claim exact size in a small sample, nor does it guarantee validity if the instrument moments are invalid, the robust covariance is singular, or the assumptions needed for the central limit theorem fail.
Worked Example: Exact F Calibration Is Not Robust Chi-Square Calibration
Suppose a classical homoskedastic AR analysis with one instrument reports an F statistic of \(4.2\) with 18 denominator degrees of freedom. Under the Gaussian homoskedastic model, the exact 5% critical value is approximately \(F_{1,18}=4.414\), so \(4.2<4.414\) and the exact test does not reject. The large-sample robust reference for one restriction is \(\chi^2_1\), whose 5% critical value is approximately \(3.841\); numerically, \(4.2>3.841\).
These comparisons show why the reference distribution must be stated, not assumed. The robust AR procedure uses a sandwich Wald statistic and asymptotic chi-square calibration. The exact homoskedastic procedure uses the classical F statistic and its finite-sample F distribution. They are distinct procedures; one must not attach the exact F claim to the heteroskedasticity-robust statistic.
Why the Homoskedastic F Test Can Be Exact
For contrast, suppose the regression design is fixed and the null-imposed regression errors are independent \(N(0,\sigma^2)\), with common variance, and the usual regression rank conditions hold. Under the null, the classical partial F statistic for \(q\) restrictions is the ratio of two independent scaled chi-square quantities. The numerator is the reduction in residual sum of squares attributable to the restrictions, divided by \(q\sigma^2\), and has a \(\chi^2_q/q\) distribution. The unrestricted residual sum of squares divided by \(\sigma^2\) has a \(\chi^2_{n-k}\) distribution, where \(k\) is the number of fitted coefficients. Independence follows from orthogonality of the corresponding Gaussian projections. Their ratio is therefore \(F_{q,n-k}\), exactly in finite samples. With heteroskedastic errors, this common-variance chi-square argument does not apply; robust Wald inference instead relies on the asymptotic result above.
Common Pitfall
- Wrong reasoning: “The Anderson–Rubin test is robust to weak instruments, so the usual finite-sample F calibration is automatically valid.” Weak-instrument robustness concerns the test’s construction and its behavior under weak relevance; it does not make a heteroskedasticity-robust Wald statistic exactly F-distributed.
- Correction: State the covariance estimator and reference distribution together. For heteroskedasticity-robust AR inference with fixed \(q\), use the sandwich Wald statistic and its asymptotic \(\chi^2_q\) calibration under the stated moment conditions. Claim an exact F distribution only under the classical Gaussian homoskedastic conditions.
Check Your Understanding
Use these questions to check whether you can distinguish the test construction from its calibration.
- Why does the null-imposed regression test instrument coefficients instead of estimating the structural coefficient by an IV ratio?
- What covariance estimator enters the robust AR Wald statistic, and what is its asymptotic reference distribution for \(q\) restrictions?
- Name two assumptions needed for the asymptotic validity result and explain the role of positive definiteness of the instrument-score covariance.
- Why does weak first-stage relevance not, by itself, invalidate the AR test’s null moment argument?
- What additional assumptions allow an exact finite-sample F calibration, and why should it not be assigned automatically to the robust statistic?