Why Wald Tests Are Useful in 2SLS
A 2SLS analysis often asks whether one estimated coefficient differs from a specified value, or whether a group of coefficients is jointly consistent with a set of specified values. A Wald test answers these questions using the 2SLS estimate and an estimated covariance matrix. The central calculation is straightforward, but the conclusion depends on using a covariance matrix appropriate to the data and on the large-sample approximation being adequate.
Write the structural coefficient vector as \(\beta\), its 2SLS estimate as \(\hat{\beta}\), and the estimated covariance matrix of that estimate as \(\widehat{\operatorname{Var}}(\hat{\beta})\). This tutorial treats inference conditional on a correctly specified instrumental-variables model, valid instruments, identification of the coefficients being tested, and regularity conditions supporting the relevant asymptotic approximation. A Wald test does not establish those conditions.
A Test of One Coefficient
For coefficient \(\beta_j\), consider \(H_0:\beta_j=\beta_{j0}\) against a two-sided alternative \(H_a:\beta_j\ne\beta_{j0}\). Let \(SE(\hat{\beta}_j)\) be the square root of the \(j\)th diagonal element of the covariance matrix used for inference. The single-coefficient Wald statistic is
Under the null and the stated large-sample conditions, \(z\) is approximately standard normal and \(W\) is approximately chi-square with one degree of freedom. The two-sided p-value can be calculated as \(2P(Z\geq |z|)\) for \(Z\sim N(0,1)\), or equivalently as the upper-tail probability for a chi-square random variable with one degree of freedom. At significance level \(\alpha\), reject \(H_0\) when the p-value is less than \(\alpha\).
The coefficient’s estimated variance must come from the covariance estimator chosen for the error structure: for example, homoskedastic, heteroskedasticity-robust, cluster-robust, or HAC. The point estimate may be the same across these choices while its standard error, test statistic, and p-value differ. Use the 2SLS covariance estimate, not the ordinary least-squares second-stage standard error from a regression of the outcome on fitted endogenous regressors.
Worked Example: Testing One 2SLS Coefficient
Suppose a hypothetical 2SLS analysis estimates an effect of \(\hat{\beta}_1=0.42\) outcome units per one-unit increase in an endogenous regressor. Its appropriate heteroskedasticity-robust standard error is \(0.18\) outcome units. Test \(H_0:\beta_1=0\) against \(H_a:\beta_1\ne0\) at \(\alpha=0.05\).
Assume for this example that the instrument is valid, the coefficient is identified sufficiently well for the large-sample approximation, and the robust covariance estimate is consistent for the sampling variance. Then
The two-sided standard-normal p-value is approximately \(0.0196\). Since \(0.0196<0.05\), reject \(H_0\). The data provide evidence against a zero coefficient under the model and inference assumptions; they do not show that the instrument is valid or that the estimate is causal without those assumptions. The coefficient is expressed in outcome units per unit of the regressor.
Joint Wald Tests
A joint test asks whether several restrictions hold simultaneously. Suppose the restrictions select \(q\) coefficients from a \(k\)-coefficient vector and set them to specified values. Let \(R\) be the \(q\)-by-\(k\) matrix that selects those coefficients, and let \(r\) be the \(q\)-vector of null values. The hypotheses are \(H_0:R\beta=r\) and \(H_a:R\beta\ne r\), where the alternative means that at least one restriction fails. The rows of \(R\) must be linearly independent so that the test contains \(q\) distinct restrictions.
The estimated restrictions are \(R\hat{\beta}\). Their estimated covariance matrix is \(R\widehat{\operatorname{Var}}(\hat{\beta})R'\). The joint Wald statistic is
Under \(H_0\), if the covariance estimate is consistent, the restrictions are identified, and the relevant asymptotic approximation holds, \(W\) has an approximate chi-square distribution with \(q\) degrees of freedom. Reject at level \(\alpha\) if \(W\) exceeds the \(1-\alpha\) quantile of that distribution, or equivalently if its upper-tail p-value is less than \(\alpha\). The full covariance matrix matters: when tested coefficient estimates are correlated, separate variances alone do not determine the joint statistic.
Why the Joint Reference Distribution Applies
The justification follows from the asymptotic distribution of the 2SLS estimate. Under the null, the restriction error \(R\hat{\beta}-r\) is centered around zero, and its asymptotic covariance is estimated by \(R\widehat{\operatorname{Var}}(\hat{\beta})R'\). Standardizing the restriction error therefore gives an approximately multivariate standard normal vector. Its squared length is the Wald statistic.
For justification, if \(A\) is the covariance matrix of a mean-zero multivariate normal vector \(U\), then \(A^{-1/2}U\) has covariance equal to the identity matrix. Its \(q\) components are standard normal, so the sum of their squares has a chi-square distribution with \(q\) degrees of freedom. Substituting the estimated covariance for \(A\) yields the large-sample Wald approximation. This argument explains why covariance estimation and adequate approximation matter; it does not make the test exact in a finite sample.
Worked Example: Testing Two Coefficients Jointly
In a hypothetical model, two selected 2SLS coefficients have estimates \(\hat{\beta}_1=0.30\) and \(\hat{\beta}_2=-0.20\). Their estimated covariance matrix, in squared coefficient units, is
Test \(H_0:\beta_1=0,\ \beta_2=0\) against the alternative that at least one coefficient is nonzero. For these restrictions \(R\) is the \(2\)-by-\(2\) identity matrix and \(r=(0,0)'\). The determinant of the covariance matrix is \(0.04(0.04)-(0.01)^2=0.0015\), so its inverse is
Thus the statistic is
There are \(q=2\) restrictions. For a chi-square distribution with two degrees of freedom, the upper-tail probability at \(4.2667\) is \(\exp(-4.2667/2)\approx0.1184\). At the 5% level, do not reject the joint null. This is not the same as finding that both coefficients are zero: the test does not provide sufficient evidence against the joint null at this significance level. The calculation assumes the given covariance estimate is appropriate and the large-sample approximation is adequate.
Worked Example: An Omnibus Test of Three Coefficients
Suppose a hypothetical 2SLS model reports three selected coefficient estimates \(0.40\), \(-0.30\), and \(0.20\), with an appropriate estimated covariance matrix
Here \(S\) denotes the selected coefficients. Test whether all three are zero, using \(q=3\) restrictions. This example assumes the selected coefficients are identified, the covariance estimate is consistent for the stated sampling design, and a large-sample reference distribution is reasonable. Because the covariance matrix is diagonal, the Wald statistic is the sum of the three squared estimate-to-standard-error ratios:
The upper-tail probability for \(\chi^2_3\) at 12 is approximately \(0.00738\). At \(\alpha=0.05\), reject the null that all three coefficients are zero. This omnibus conclusion says at least one of the three restrictions is inconsistent with the data at the chosen level; the joint test alone does not identify which coefficient is responsible.
Choosing the Covariance Estimate and Checking the Test
The Wald formula does not determine which covariance estimate is appropriate. That choice follows the sampling and error structure, as in the earlier material on 2SLS inference. Homoskedastic standard errors require a homoskedastic error model; heteroskedasticity-robust standard errors allow nonconstant error variance; cluster-robust standard errors allow dependence within the specified clusters; and HAC standard errors address suitable forms of serial dependence. In every case, use the covariance estimator for the 2SLS estimator and apply the same choice throughout the test.
Before interpreting a result, check that the tested restrictions are genuinely distinct, the restricted covariance matrix is invertible, and the chosen estimator matches the data structure. If the matrix is singular or nearly singular, the restrictions may be redundant or the estimated covariance may not support the calculation reliably. Check that the software is testing the intended null and reports the intended covariance type, degrees of freedom, and reference distribution.
- Weak identification: Conventional Wald tests can have poor size and misleading p-values when instruments provide little information about the endogenous regressors. A robust covariance estimator does not, by itself, repair weak-identification problems.
- Few independent clusters: A cluster-robust covariance estimate and a large-sample chi-square reference may be unreliable when the number of independent clusters is small. Finite-sample corrections and an appropriate reference distribution may be needed, but no correction guarantees accurate inference in every design.
- Wrong dependence assumptions: Treating correlated observations as independent can understate uncertainty. Conversely, choosing a covariance estimator that does not match the design does not make a Wald test valid.
- Multiple separate tests: Testing each coefficient separately is not a substitute for a joint test. The separate tests do not account for the covariance among estimates and answer different questions.
Large-Sample Tests and Alternatives
For one coefficient, a two-sided Wald test is equivalent to comparing the absolute \(z\) statistic with a standard-normal critical value. For \(q\) restrictions, the large-sample reference is \(\chi^2_q\). In finite samples, some software reports an \(F\)-type statistic or uses a finite-sample reference approximation. Such a convention should be identified explicitly: it is not a universal exact \(F\) result for 2SLS, and a variance correction alone does not make the Wald distribution exact.
Wald tests are convenient because they use an unrestricted estimate and its covariance matrix. Score-type tests instead evaluate departures from the null using estimation under the restrictions. Likelihood-ratio tests compare likelihoods and require an appropriate likelihood framework; they are not a generic replacement for Wald tests in every IV analysis. The choice among available procedures depends on the model and assumptions, while weak identification requires particular caution with conventional Wald inference.
A common mistake is to say, “The first coefficient is individually insignificant, so the coefficients are jointly insignificant.” This does not follow: a joint test uses the entire vector of estimates and their covariance, and can reach a different conclusion. Another mistake is to use the ordinary second-stage OLS standard error because the second-stage regression output displays one. That standard error generally fails to account for estimation of the fitted endogenous regressor; the appropriate 2SLS covariance estimate is required.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A coefficient estimate is \(0.24\), its appropriate standard error is \(0.10\), and the null value is zero. Calculate the two-sided Wald statistic and approximate p-value. What is the decision at the 5% level?
- A joint test has \(q=2\) restrictions and \(W=6.00\). The chi-square upper-tail probability is approximately \(0.0498\). State the decision at the 5% level and interpret it.
- Why can two separate coefficient tests give a different conclusion from a joint Wald test?
- What should you check if the covariance matrix of the estimated restrictions cannot be inverted?
Solutions. (1) \(z=0.24/0.10=2.40\), so \(W=2.40^2=5.76\). The two-sided p-value is approximately \(0.0164\); reject the zero-coefficient null at 5%, conditional on the model and asymptotic assumptions. (2) Since \(0.0498<0.05\), reject the joint null at the 5% level; the result is evidence that at least one of the two restrictions fails, not proof of which one. (3) The joint statistic uses the covariance between estimates as well as their variances, and addresses a different, simultaneous null. Separate tests do not incorporate that covariance into a single joint decision. (4) Check whether restrictions are redundant or linearly dependent, and whether the covariance estimate is singular because of the design or unreliable estimation. Do not proceed by simply inverting a singular matrix; the test setup or estimation may need correction.