Testing Linear Combinations of 2SLS Coefficients

Why Test a Linear Combination?

A hypothesis about 2SLS coefficients need not concern one coefficient at a time. A researcher may want to test whether two effects are equal, whether a weighted sum of effects is zero, or whether several coefficients jointly satisfy specified restrictions. These questions require the uncertainty of the relevant combination, not just the separate standard errors of its coefficients.

The key calculation is a covariance transformation. If the 2SLS estimator has covariance matrix \(V\), then the variance of a linear combination is obtained by placing the combination’s coefficients on both sides of \(V\). This automatically accounts for covariance between coefficient estimates.

Definition: Let \(\hat{\beta}\) be a \(k\)-component 2SLS coefficient vector, with estimated covariance matrix \(\widehat{V}\). A scalar linear contrast is \(c'\beta\), where \(c\) is a specified \(k\)-component vector. Its estimated value is \(c'\hat{\beta}\), and its estimated variance is \(c'\widehat{V}c\). A collection of \(q\) linear restrictions is written \(R\beta=r\), where \(R\) is a \(q\times k\) matrix and \(r\) is a specified \(q\)-component vector.

Assumptions and the Wald Test

The usual large-sample Wald test relies on the 2SLS estimator being consistent and asymptotically normal under the model’s instrument validity and identification conditions. The covariance estimator \(\widehat{V}\) must be consistent for the covariance of \(\hat{\beta}\) under the sampling structure. Use the covariance estimator appropriate to the setting, such as a homoskedastic, heteroskedasticity-robust, cluster-robust, or HAC estimator. The null restrictions must also be estimable: for a joint test, \(R\) must have full row rank and \(R\widehat{V}R'\) must be invertible.

For a scalar contrast, test \(H_0:c'\beta=q_0\) against \(H_a:c'\beta\ne q_0\). Its estimated standard error is the square root of \(c'\widehat{V}c\). The asymptotic Wald statistic is the difference between the estimated contrast and its null value, divided by that standard error. Under \(H_0\), it is approximately standard normal in large samples.

SE(c'\hat{\beta})=\sqrt{c'\widehat{V}c}, \qquad z=\frac{c'\hat{\beta}-q_0}{\sqrt{c'\widehat{V}c}}.

For \(q\) joint restrictions, the Wald statistic measures the discrepancy between \(R\hat{\beta}\) and \(r\), scaled by their estimated covariance matrix. Under the null and the stated large-sample conditions, it has an approximate chi-square distribution with \(q\) degrees of freedom.

W=(R\hat{\beta}-r)'(R\widehat{V}R')^{-1}(R\hat{\beta}-r) \ \overset{approx}{\sim}\ \chi^2_q.

The restriction vector and covariance matrix must be transformed together. In particular, the covariance of \(R\hat{\beta}\) is \(R\widehat{V}R'\); it is not generally obtained by retaining only the diagonal elements of \(\widehat{V}\). For one restriction, the joint Wald statistic is the square of the scalar \(z\) statistic, so \(W=z^2\) and its reference distribution is approximately \(\chi^2_1\).

Why the Covariance Transformation Works

The justification follows directly from the distribution of the coefficient estimator. Suppose, under the null conditions and regularity assumptions, \(\hat{\beta}\) is approximately multivariate normal with mean \(\beta\) and covariance matrix \(V\). A fixed linear transformation of a multivariate normal vector is also multivariate normal. Therefore \(R\hat{\beta}\) has approximate mean \(R\beta\) and covariance \(RVR'\). Under \(H_0:R\beta=r\), the transformed difference \(R\hat{\beta}-r\) is centered at zero.

If \(RVR'\) is nonsingular, standardizing this \(q\)-component difference by its covariance produces a squared length with an approximate chi-square distribution on \(q\) degrees of freedom. Replacing the unknown \(V\) with a consistent \(\widehat{V}\) gives the feasible Wald statistic above. For a single contrast, the same argument gives variance \(c'Vc\), since the variance of a fixed weighted sum is the weights multiplied on both sides of the covariance matrix.

\operatorname{Var}(c'\hat{\beta}) =c'\operatorname{Var}(\hat{\beta})c =c'Vc.

Worked Scalar Contrasts

Worked Example: Testing a Sum of Two 2SLS Coefficients

Suppose a hypothetical 2SLS analysis estimates two coefficients for treatment components measured in the same units. The outcome is a symptom score in points. The estimates and estimated covariance matrix are

\hat{\beta}= \begin{pmatrix}2.4\\-0.8\end{pmatrix}, \qquad \widehat{V}= \begin{pmatrix}0.36&0.12\\0.12&0.25\end{pmatrix}.

Test whether the sum of the two coefficients is zero: \(H_0:\beta_1+\beta_2=0\) against a two-sided alternative. Here \(c=(1,1)'\), so the estimated sum is \(c'\hat{\beta}=2.4-0.8=1.6\) points. Transform the entire covariance matrix:

c'\widehat{V}c =(1,1) \begin{pmatrix}0.36&0.12\\0.12&0.25\end{pmatrix} \begin{pmatrix}1\\1\end{pmatrix} =0.36+0.12+0.12+0.25=0.85.

Thus \(SE=\sqrt{0.85}=0.9220\) points, and \(z=1.6/0.9220=1.735\). The two-sided large-sample p-value is approximately \(0.0828\). At \(\alpha=0.05\), the result does not provide sufficient evidence against the null that the sum is zero. This is not evidence that the null is true; it says the observed contrast is not sufficiently far from zero relative to its estimated uncertainty under this test.

The covariance terms add here because both coefficients enter with positive weights. Omitting them would give a variance of \(0.36+0.25=0.61\), which is not the variance of the estimated sum.

Worked Example: Testing Equality of Two Coefficients

Use the same hypothetical estimates and covariance matrix to test whether the two coefficients are equal. The null is \(H_0:\beta_1-\beta_2=0\), so choose \(c=(1,-1)'\). The estimated difference is \(2.4-(-0.8)=3.2\) points.

c'\widehat{V}c =(1,-1) \begin{pmatrix}0.36&0.12\\0.12&0.25\end{pmatrix} \begin{pmatrix}1\\-1\end{pmatrix} =0.36-0.12-0.12+0.25=0.37.

The estimated standard error is \(\sqrt{0.37}=0.6083\) points, giving \(z=3.2/0.6083=5.260\). The two-sided standard-normal p-value is approximately \(1.44\times10^{-7}\). At the 5% significance level, reject equality of the coefficients. The contrast’s uncertainty is smaller than it would be if the estimates were uncorrelated, because the positive covariance reduces the variance of their difference.

Joint Restrictions

A joint hypothesis is useful when the scientific or policy question concerns several restrictions at once. For example, “both coefficients are zero” is a two-restriction hypothesis, not two separate scalar questions. The joint test uses the covariance between the restricted coefficient estimates, so it can differ from a summary of separate tests.

Worked Example: Testing Two Coefficients Jointly

Consider a hypothetical 2SLS model with two coefficients, estimated as \(\hat{\beta}=(1,2)'\), and estimated covariance matrix

\widehat{V}= \begin{pmatrix}0.25&0.10\\0.10&0.36\end{pmatrix}.

Test \(H_0:\beta_1=0\) and \(\beta_2=0\) jointly. Set \(R=I_2\), \(r=(0,0)'\), and \(q=2\). The covariance matrix of the restricted estimates is \(\widehat{V}\). Its determinant is \(0.25(0.36)-0.10^2=0.08\), so it is invertible, with

\widehat{V}^{-1} =\frac{1}{0.08} \begin{pmatrix}0.36&-0.10\\-0.10&0.25\end{pmatrix} = \begin{pmatrix}4.5&-1.25\\-1.25&3.125\end{pmatrix}.

The Wald statistic is

W= \begin{pmatrix}1&2\end{pmatrix} \begin{pmatrix}4.5&-1.25\\-1.25&3.125\end{pmatrix} \begin{pmatrix}1\\2\end{pmatrix} =(1,2)\begin{pmatrix}2\\5\end{pmatrix} =12.

Under the null, compare \(W=12\) with a chi-square distribution with 2 degrees of freedom. The p-value is approximately \(0.00248\). At \(\alpha=0.05\), reject the joint null: the data provide evidence that at least one of the two population coefficients is nonzero. This test alone does not identify which restriction is responsible for the rejection.

Implementation, Diagnostics, and Limitations

A reliable calculation follows a short sequence. Write the null in matrix form, obtain the appropriate covariance matrix for the 2SLS estimator, transform that matrix, and then use the reference distribution for the number of independent restrictions.

Specify the target.
Write the scalar contrast \(c'\beta=q_0\), or the joint restrictions \(R\beta=r\). Check that the rows of \(R\) represent distinct restrictions.

Use the matching covariance estimator.
Use \(\widehat{V}\) for the 2SLS estimator, with a variance estimator suited to the design and dependence structure. Do not substitute the ordinary second-stage OLS covariance matrix.

Transform and test.
Calculate \(c'\widehat{V}c\) for one contrast or \(R\widehat{V}R'\) for several restrictions; then calculate the statistic and its large-sample p-value.

Several checks help identify mistakes or numerical problems. The estimated variance of a scalar contrast should be nonnegative, and a covariance matrix used in a joint test should be symmetric and positive definite on the tested restrictions. If \(R\widehat{V}R'\) is singular, the stated joint statistic cannot be calculated using an ordinary inverse; the restrictions may be redundant, or the data may not separately estimate them. Check the restriction matrix and model identification rather than silently dropping a restriction.

Conventional Wald inference is also a large-sample approximation. A valid, consistent covariance estimator does not by itself guarantee that the approximation is accurate in a small sample or when instruments provide weak identification. In such settings, a Wald statistic can have poor size, meaning its actual Type I error rate may differ from the nominal \(\alpha\). Interpret a conventional Wald p-value with that limitation in mind.

Testing a linear combination differs from testing each coefficient separately. Separate tests answer whether each individual coefficient equals a specified value; they do not directly test whether a sum, difference, or other contrast equals a value. Reparameterizing the model so that the desired contrast is a coefficient can express the same hypothesis, but it does not remove the need to account for the covariance structure. The contrast calculation makes that dependence explicit.

Common Mistakes

Ignoring covariance. Adding the separate variances is correct only when the relevant covariance terms are zero. For a contrast, calculate \(c'\widehat{V}c\).
Using the wrong signs. A difference \(\beta_1-\beta_2\) uses \(c=(1,-1)'\), which makes the covariance contribution negative when the estimated covariance is positive.
Testing coefficients separately instead of testing the stated restriction. Individual results do not substitute for a test of a sum, equality, or joint null.
Using an unsuitable covariance matrix. The transformation must start from a covariance estimator for 2SLS that reflects the relevant heteroskedasticity or dependence structure.
Overstating a joint rejection. Rejecting that all restrictions hold does not establish that every individual restriction is false.

Key takeaway: Linear restrictions on 2SLS coefficients are tested by transforming the full estimated covariance matrix. For one contrast use \(c'\widehat{V}c\); for several restrictions use \(R\widehat{V}R'\), with a chi-square reference distribution whose degrees of freedom equal the number of independent restrictions.

Check Your Understanding

For each problem, identify the contrast or restrictions before calculating. Complete solutions follow the questions.

Suppose \(c=(2,1)'\), \(\hat{\beta}=(1,3)'\), and \(\widehat{V}=\begin{pmatrix}0.25&0.05\\0.05&0.16\end{pmatrix}\). What are the estimated contrast and its standard error?
With \(\hat{\beta}=(2,1)'\) and \(\widehat{V}=\begin{pmatrix}0.49&0\\0&0.25\end{pmatrix}\), test \(H_0:\beta_1-\beta_2=0\) using a two-sided large-sample test. Give the statistic and conclusion at \(\alpha=0.05\).
For a joint test \(R\beta=r\), \(R\) has three rows but rank two. How many independent restrictions are represented, and what issue arises if the test is treated as having three independent restrictions?
Why is the variance of \(\hat{\beta}_1+\hat{\beta}_2\) not generally the sum of the two estimated variances?

Solutions. 1. The estimate is \(2(1)+1(3)=5\). Its variance is \(c'\widehat{V}c=4(0.25)+2(0.05)+2(0.05)+0.16=1.36\), so the standard error is \(\sqrt{1.36}=1.166\). 2. Use \(c=(1,-1)'\). The estimated difference is \(2-1=1\), its variance is \(0.49+0.25=0.74\), and its standard error is \(\sqrt{0.74}=0.8602\). Thus \(z=1/0.8602=1.163\), with a two-sided p-value of approximately \(0.245\). Do not reject the null at \(\alpha=0.05\). 3. The rows represent only two independent restrictions. Treating them as three independent restrictions gives the wrong degrees of freedom and an invalid ordinary inverse if the transformed covariance is singular. Identify the redundancy or formulate an equivalent full-rank restriction matrix. 4. The variance includes twice the covariance: \(\operatorname{Var}(\hat{\beta}_1+\hat{\beta}_2)=\operatorname{Var}(\hat{\beta}_1)+\operatorname{Var}(\hat{\beta}_2)+2\operatorname{Cov}(\hat{\beta}_1,\hat{\beta}_2)\). It equals the sum of variances only if that covariance is zero.