Weak-IV Tests with Multiple Endogenous Regressors

Why Joint Inference Needs Its Own Procedure

When a linear instrumental-variables model has several endogenous regressors, the parameter of interest is a vector. A claim about that vector may concern all coefficients at once, such as whether a treatment and a second endogenous exposure have specified effects. Testing each coefficient separately does not answer that joint question, and weak instruments make ordinary 2SLS-based Wald inference unreliable.

The Anderson–Rubin (AR) approach tests a proposed coefficient vector by asking whether the outcome left after subtracting that vector’s fitted endogenous-regressor contribution is still related to the instruments. Under the null, that residualized outcome should have no instrument association. Crucially, the test does not divide by estimated first-stage effects, so it does not require those effects to be strong.

Model, Null, and Assumptions

Let \(Y\) be an \(n\)-vector of outcomes, \(D\) an \(n\times K\) matrix of endogenous regressors, \(Z\) an \(n\times L\) matrix of excluded instruments, and \(W\) a matrix of included exogenous controls (including an intercept if appropriate). Consider the linear structural model

Y = D\beta + W\gamma + u,

where \(\beta\) is the \(K\)-vector of coefficients to be tested. The joint null is \(H_0:\beta=\beta_0\), for a specified vector \(\beta_0\). The instruments must satisfy the model’s exclusion and exogeneity conditions: after accounting for \(W\), they are excluded from the structural equation and have zero conditional association with the structural error. The sampling conditions must also justify the chosen covariance estimator; for example, independent observations with finite moments for a heteroskedasticity-robust large-sample test, or a suitable cluster structure with a cluster-robust covariance estimator.

The AR test does not require a strong first stage or full column rank of the first-stage coefficient matrix. It does require enough variation in the instruments for the covariance matrix of the tested instrument associations to be estimable and nonsingular. Instrument validity is an assumption, not something established by the AR test.

Definition: For a proposed \(\beta_0\), form the transformed outcome \(Y_0=Y-D\beta_0\). Regress \(Y_0\) on \(W\) and \(Z\), and test the joint null that all \(L\) coefficients on the excluded instruments \(Z\) are zero. This is the Anderson–Rubin test of \(H_0:\beta=\beta_0\).

Under \(H_0\), the model implies \(Y_0=W\gamma+u\). Instrument exogeneity therefore implies that the excluded instruments have zero population coefficients in this regression. For a heteroskedasticity-robust implementation, let \(\hat\delta(\beta_0)\) be the vector of estimated coefficients on \(Z\), and let \(\widehat V(\beta_0)\) be its robust covariance matrix. The Wald form of the AR statistic is

AR(\beta_0)=\hat\delta(\beta_0)^\mathsf{T} \widehat V(\beta_0)^{-1}\hat\delta(\beta_0).

Under the null and the stated large-sample conditions, this statistic is compared with a chi-square distribution with \(L\) degrees of freedom. With independent, normally distributed, homoskedastic errors, the classical regression \(F\)-test of the \(L\) excluded-instrument coefficients gives a finite-sample version, with numerator degrees of freedom \(L\) and denominator degrees of freedom equal to the residual degrees of freedom in that regression. The covariance method and reference distribution must match the sampling assumptions.

Worked Joint Test

Worked Example: Testing a Candidate Pair of Coefficients

Suppose a hypothetical study has two endogenous regressors and two excluded instruments. At a proposed value \(\beta_0\), the robust regression of \(Y-D\beta_0\) on the controls and instruments gives instrument-coefficient estimates \((0.12,-0.08)^\mathsf{T}\). Suppose their estimated covariance matrix is diagonal, with standard errors \(0.05\) and \(0.04\). These are hypothetical regression results, and the stated covariance matrix is assumed valid for the sampling design.

The joint null sets both instrument coefficients to zero. With the given diagonal covariance, the AR statistic is

AR(\beta_0) =\left(\frac{0.12}{0.05}\right)^2 +\left(\frac{-0.08}{0.04}\right)^2 =2.4^2+(-2)^2 =5.76+4.00 =9.76.

There are \(L=2\) instruments, so the large-sample reference distribution is \(\chi^2_2\). Its upper-tail p-value is \(P(\chi^2_2\ge 9.76)=\exp(-9.76/2)\approx0.0076\). At the 5% significance level, reject this candidate coefficient vector: under the maintained instrument-validity and sampling assumptions, the transformed outcome shows evidence of an association with the instruments. This conclusion is about the specified pair \(\beta_0\); it is not a test that either coefficient alone is zero.

Inverting the Test: A Joint Confidence Region

A confidence region is obtained by testing candidate vectors and retaining those not rejected. For a test of size \(\alpha\), define

\mathcal C_{1-\alpha} =\{\beta_0: AR(\beta_0)\le c_{1-\alpha}\},

where \(c_{1-\alpha}\) is the appropriate upper-tail critical value (for example, the \(1-\alpha\) quantile of \(\chi^2_L\) for the robust large-sample test). Under the maintained assumptions, this inversion has coverage at least approximately \(1-\alpha\): the true coefficient vector is retained with probability at least \(1-\alpha\), up to the test’s size approximation. The set may be bounded, unbounded, disconnected, or empty in a finite sample. An unbounded set is meaningful evidence that the data do not sharply locate the coefficients; it is not a computational failure.

For inference on one component, say \(\beta_1\), project the joint region onto that coordinate. This means retaining every value of \(\beta_1\) for which at least one choice of the other coefficients belongs to the region:

\mathcal C^{\mathrm{proj}}_{1-\alpha} =\{\beta_1:\text{ there exists }\beta_{-1} \text{ such that }(\beta_1,\beta_{-1})\in\mathcal C_{1-\alpha}\}.

Projection allows the other coefficients to vary over all values compatible with the joint region. It is not the same operation as fixing them at estimated values or at the center of the region.

Worked Example: Projecting a Circular Region

Suppose a hypothetical two-coefficient AR inversion produces the region

(\beta_1-1)^2+(\beta_2-2)^2\le 5.991.

This form could arise from a two-instrument large-sample test using the 95th percentile \(5.991\) of \(\chi^2_2\), with the displayed quadratic statistic. To find the projection onto \(\beta_1\), ask whether some \(\beta_2\) makes the inequality hold. For any fixed \(\beta_1\), the left side is minimized at \(\beta_2=2\), where it equals \((\beta_1-1)^2\). Thus a value is in the projection exactly when

(\beta_1-1)^2\le5.991 \quad\Longleftrightarrow\quad 1-\sqrt{5.991}\le\beta_1\le1+\sqrt{5.991}.

Since \(\sqrt{5.991}\approx2.448\), the projected confidence set is approximately \([-1.448,\,3.448]\). The projection explicitly allows the other coefficient to vary. In this circular example, the projected endpoints happen to coincide with those obtained by fixing \(\beta_2=2\), its center; that need not hold in general.

Worked Example: A Correlated Region Changes the Projection

Now suppose a hypothetical joint AR region is centered at \((1,2)\) and has the following quadratic form, with the same 95% cutoff:

\frac{(\beta_1-1)^2-2(\beta_1-1)(\beta_2-2)+4(\beta_2-2)^2}{3} \le5.991.

Set \(x_1=\beta_1-1\) and \(x_2=\beta_2-2\). Completing the square gives

\frac{x_1^2-2x_1x_2+4x_2^2}{3} =\frac{4}{3}\left(x_2-\frac{x_1}{4}\right)^2+\frac{x_1^2}{4}.

For each \(x_1\), the expression is minimized at \(x_2=x_1/4\). Therefore a value of \(\beta_1\) is in the projection if and only if \(x_1^2/4\le5.991\). The projected interval is

1-\sqrt{4(5.991)}\le\beta_1\le1+\sqrt{4(5.991)},

or approximately \([-3.895,\,5.895]\). If instead \(\beta_2\) is fixed at its center, then \(x_2=0\), and the condition becomes \(x_1^2/3\le5.991\). This gives the narrower interval \([1-\sqrt{3(5.991)},\,1+\sqrt{3(5.991)}]\), approximately \([-3.240,\,5.240]\). The difference occurs because the projection can use noncentral values of \(\beta_2\) to remain inside the tilted region.

Joint Versus Separate Coverage

A \(1-\alpha\) interval for each coefficient gives a marginal coverage statement for each coefficient, not automatically a \(1-\alpha\) statement that all intervals cover their coefficients simultaneously. For two independent procedures with 95% coverage each, the probability that both cover is \(0.95^2=0.9025\), not 0.95. Without independence, the joint coverage depends on their dependence, but separate 95% marginal guarantees still do not ensure 95% simultaneous coverage.

A joint confidence region addresses the vector as a whole. Projecting the same \(1-\alpha\) joint region onto each coordinate produces coordinate sets that jointly contain the true vector’s coordinates whenever the region contains the true vector. Consequently, their simultaneous coverage is at least the joint region’s coverage, though the projected sets can be wider than separate marginal intervals. The price of this protection is appropriate: a statement about several parameters at once must account for uncertainty in the vector, not just uncertainty in each component considered separately.

Common Mistakes and Practical Interpretation

Using a conventional 2SLS Wald test because it is familiar. Under weak identification, its usual approximation can fail. The AR test instead tests instrument association after subtracting a candidate coefficient vector.
Testing only one instrument coefficient at a time. The joint null concerns all \(L\) excluded-instrument coefficients. The AR statistic combines their estimated associations and covariance.
Confusing failure to reject with confirmation. A candidate vector retained in the confidence region has not been ruled out at the chosen level; it has not been proven correct. Weak or uninformative instruments can leave a very large set.
Fixing nuisance coefficients when the goal is projection. Projection asks whether any values of the other coefficients make the joint test acceptable. Fixing them answers a different, conditional question.
Treating separate intervals as a joint guarantee. Marginal 95% coverage for each coefficient does not generally imply 95% simultaneous coverage. Use a joint procedure when the scientific claim concerns the coefficient vector or simultaneous coverage.

Key takeaway: The joint Anderson–Rubin test evaluates a candidate vector through the instrument association of \(Y-D\beta_0\). Inverting it gives a confidence region that remains useful without strong first stages; projection carries that joint uncertainty into inference on individual coefficients.

Check Your Understanding

Answer each question using the definitions and assumptions above.

In the AR regression for a candidate \(\beta_0\), what joint null is tested for the excluded instruments?
Why does the AR procedure avoid the particular weak-identification problem associated with dividing by estimated first-stage effects?
For the circular region \((\beta_1-1)^2+(\beta_2-2)^2\le5.991\), derive the projection onto \(\beta_1\).
In a tilted joint region, why can the projected interval differ from the interval obtained by fixing \(\beta_2\) at its center?
If two separate intervals each have 95% coverage and their coverage events are independent, what is the probability that both cover?