Why Robust First-Stage Diagnostics Are Needed
A first-stage statistic asks whether excluded instruments predict an endogenous regressor after accounting for included exogenous controls. Under homoskedasticity, the familiar first-stage F statistic summarizes a joint test of zero coefficients on the excluded instruments. With heteroskedastic errors, the usual covariance formula for that test is generally wrong: the variance of the coefficient estimates depends on the pattern of error variances across observations. A heteroskedasticity-robust covariance estimate changes the test statistic accordingly.
This adjustment matters, but it has a precise scope. A robust Wald statistic tests the null that the first-stage coefficients are zero and provides a relevance diagnostic. Its large-sample null reference distribution does not, by itself, ensure that conventional two-stage least-squares inference is reliable when instruments are weak. The weak-instrument results and Stock-Yogo thresholds from the earlier material address different calibration questions and should not be transferred automatically to a heteroskedasticity-robust statistic.
Partialling Out Controls and Defining the Test
Let \(x_i\) be one endogenous regressor, \(z_i\) a \(q\)-vector of excluded instruments, and \(w_i\) a \(p\)-vector of included exogenous controls. Assume the first-stage linear projection is \(x_i=w_i'\gamma+z_i'\pi+v_i\), where \(\pi\) is the \(q\)-vector of coefficients whose relevance is being tested. The null hypothesis is \(H_0:\pi=0\). The intercept, if included, is part of \(w_i\).
Write \(W\) and \(Z\) for the sample matrices with rows \(w_i'\) and \(z_i'\). Define the projection residuals of the instruments on the controls by \(\hat r_i=z_i-\hat\Pi'w_i\), where \(\hat\Pi=(W'W)^{-1}W'Z\), assuming \(W'W\) is invertible. Thus, the vectors \(\hat r_i\) are the rows of the residualized-instrument matrix \(M_WZ\), with \(M_W=I-W(W'W)^{-1}W'\). Define \(\hat Q=n^{-1}\sum_i\hat r_i\hat r_i'\).
This is the HC0 sandwich version: there is no degrees-of-freedom correction multiplying the “meat” \(\hat\Omega\). Other HC versions and finite-sample scaling conventions exist, so a reported statistic should be read with its covariance and scaling convention in view. The essential distinction is between the unrestricted residuals \(\hat v_i\), which estimate the error variance in the sandwich, and the errors used to derive the sampling behavior of \(\hat\pi\).
Why the Robust Wald Statistic Has a Chi-Square Null Limit
The following result gives the large-sample justification. It is an asymptotic result, not an assertion that the statistic has an exact finite-sample F distribution.
Here heteroskedasticity is allowed: no requirement says that \(E[v_i^2\mid w_i,z_i]\) is constant. Positive definiteness of \(\Omega\) ensures that the limiting covariance is invertible. The stated moment and rank conditions rule out degenerate designs and support the approximations below.
Proof. Let \(\hat r_i=z_i-\hat\Pi'w_i\). The partial-regression identity gives the exact coefficient expansion
The \(v_i\) here are the reduced-form errors in \(x_i=w_i'\gamma+v_i\) under the null, not the residuals from the unrestricted first-stage regression. To obtain the limit, note that \(\hat\Pi-\Pi=O_p(n^{-1/2})\), while \(n^{-1/2}\sum_iw_iv_i=O_p(1)\) because \(E[w_iv_i]=0\). Therefore,
Also, \(\hat Q\xrightarrow{p}Q\). The multivariate central limit theorem gives \(n^{-1/2}\sum_i r_iv_i\xrightarrow{d}N(0,\Omega)\), so Slutsky’s theorem yields
It remains to justify the estimated covariance. The unrestricted OLS residuals satisfy \(\hat v_i=v_i-w_i'(\hat\gamma-\gamma)-z_i'\hat\pi\) under the null. OLS consistency and the moment conditions imply that replacing \(v_i^2\) by \(\hat v_i^2\) in the sample average changes it by \(o_p(1)\); this follows by expanding the square and applying the law of large numbers and Cauchy-Schwarz to the terms involving coefficient errors. Replacing \(r_i\) by \(\hat r_i\) likewise changes the average by \(o_p(1)\), using \(\hat\Pi-\Pi=o_p(1)\) and the finite moments. Consequently \(\hat\Omega\xrightarrow{p}\Omega\), and \(\hat Q\xrightarrow{p}Q\).
The estimated covariance of \(\sqrt n\,\hat\pi\) is \(\hat Q^{-1}\hat\Omega\hat Q^{-1}\), which converges in probability to \(Q^{-1}\Omega Q^{-1}\). The Wald statistic is the quadratic form of the asymptotically standard normal vector obtained by studentizing \(\sqrt n\,\hat\pi\). Its limit is therefore \(\chi_q^2\). Dividing by \(q\) gives the stated limit for \(F_{\mathrm{rob}}\). \(\square\)
Worked Calculations
Worked Example: A Two-Instrument Robust Wald Test
Suppose a first-stage calculation after partialling out controls gives \(n=200\), \(\hat Q=\begin{pmatrix}4&0\\0&1\end{pmatrix}\), \(\hat\Omega=\begin{pmatrix}16&0\\0&4\end{pmatrix}\), and \(\hat\pi=(0.10,0.20)'\). Compute the robust Wald statistic and its F scaling.
First, \(\hat Q^{-1}=\operatorname{diag}(1/4,1)\). Thus the estimated covariance of \(\sqrt n\,\hat\pi\) is \(\hat Q^{-1}\hat\Omega\hat Q^{-1}=\operatorname{diag}(1,4)\), whose inverse is \(\operatorname{diag}(1,1/4)\). Hence
There are \(q=2\) excluded instruments, so \(F_{\mathrm{rob}}=4/2=2\). Under the null and the theorem’s conditions, \(W_{\mathrm{rob}}\) is compared asymptotically with \(\chi_2^2\), or equivalently \(F_{\mathrm{rob}}\) with \(\chi_2^2/2\). It is not thereby an exact \(F_{2,\nu}\) statistic.
Worked Example: Robust and Homoskedastic Tests Can Disagree
Suppose \(q=2\), \(n=100\), \(\hat Q=I_2\), and \(\hat\pi=(0.30,0.20)'\). Let the robust meat estimate be \(\hat\Omega=\operatorname{diag}(9,1)\). Compute the robust statistic, then compare it with the homoskedastic Wald calculation that would use \(\hat\Omega=I_2\).
For the robust calculation, \(\hat Q\) is the identity, so \(\hat Q\hat\Omega^{-1}\hat Q=\operatorname{diag}(1/9,1)\). Therefore
If one instead imposed the homoskedastic covariance with \(\hat\Omega=I_2\), the calculation would be \(W=100(0.30^2+0.20^2)=100(0.09+0.04)=13\), or \(F=13/2=6.5\). The discrepancy arises because the robust estimate assigns a larger variance to the first coefficient, reducing the evidence it contributes against the null. Neither number is a universal measure independent of the assumed covariance structure.
Worked Example: One Instrument and a Robust HC0 Statistic
Consider a first stage with an intercept and one excluded instrument. After removing the intercept, suppose the residualized instrument values are \((-1,-1,1,1)\), and the unrestricted first-stage residuals are \((1,-1,1,-1)\). The estimated coefficient is \(\hat\pi=0.5\). Calculate the HC0 robust Wald statistic.
The sample residualized-instrument second moment is \(\hat Q=(1/4)\sum_i\hat r_i^2=1\). The HC0 meat is \(\hat\Omega=(1/4)\sum_i\hat r_i^2\hat v_i^2=1\), since every squared value in the sum is 1. Thus the estimated variance of \(\hat\pi\) is \(\hat Q^{-1}\hat\Omega\hat Q^{-1}/n=1/4\). The one-degree-of-freedom Wald statistic is
With one excluded instrument, \(q=1\), so the F-scaled statistic is also 1. This small numerical illustration checks the sandwich arithmetic; its sample size is far too small to justify relying on the asymptotic reference distribution as an exact finite-sample calibration.
Kleibergen-Paap-Style Interpretation
For a single endogenous regressor, a heteroskedasticity-robust joint Wald test of the excluded-instrument coefficients is the relevant robust rank/relevance test. A common reported form is called a Kleibergen-Paap rk Wald F statistic. With the convention used here, its F-scaled form is \(W_{\mathrm{rob}}/q\). Software can apply different finite-sample or normalization conventions, so the statistic’s label alone does not specify every numerical detail. The null being tested remains that the excluded instruments have zero first-stage coefficients.
This robustification replaces the homoskedastic covariance estimate with a sandwich estimate; it does not turn the diagnostic into a general weak-IV inference procedure. In particular, the chi-square null approximation describes behavior under the zero-relevance null with regular fixed-rank conditions. It does not establish that conventional 2SLS Wald confidence intervals have reliable coverage when the first stage is weak. Nor should a heteroskedasticity-robust statistic be compared mechanically with Stock-Yogo thresholds derived for a different covariance model and calibration objective.
What the Diagnostic Does—and Does Not—Establish
A small robust Wald statistic indicates that the data provide limited evidence against zero first-stage relevance under the stated robust test. A large value is evidence against that null. Neither conclusion certifies the exclusion restriction, proves that the instruments are exogenous, or establishes that 2SLS behaves well in the sample. Instrument validity is a separate modeling requirement, and weak-IV inference requires methods whose guarantees address weak identification rather than merely using a robust standard error.
A common mistake is to take a robust statistic, compare it to a conventional first-stage F threshold, and conclude that passing the threshold guarantees valid 2SLS inference. This reasoning mixes a heteroskedasticity-robust null test with a threshold calibrated under different assumptions and treats a diagnostic as an inference guarantee. The correction is to identify the covariance estimator and reference distribution, interpret the statistic as evidence about first-stage relevance, and separately assess whether the desired structural inference is valid under possible weak instruments.
Check Your Understanding
Use the definitions and results above to answer these questions.
- Why are unrestricted first-stage residuals appropriate for estimating the HC0 meat but not for the coefficient’s score expansion?
- For \(q=3\) instruments and a robust Wald statistic of 12, what is the F-scaled statistic, and what is its asymptotic null reference distribution?
- Which nonsingularity condition is needed for \(\hat Q^{-1}\), and what does it mean for the residualized instruments?
- Why can heteroskedasticity make the ordinary homoskedastic first-stage F statistic misleading?
- Why does rejection of zero first-stage coefficients not by itself establish that conventional 2SLS inference is reliable?