Anderson-Rubin Inference for Several Endogenous Regressors

Why Test a Coefficient Vector Directly?

With several endogenous regressors, a proposed structural coefficient is a vector, and the inferential question may concern the whole vector at once. Conventional tests based on 2SLS estimates can behave poorly when instruments are weak. The Anderson–Rubin (AR) construction takes a different route: for each proposed coefficient vector, subtract its contribution from the outcome and test whether the instruments explain what remains.

The resulting test does not require the first stages to be strong. Under a fixed-design Gaussian model with homoskedastic errors, its reference distribution is exact. Inverting the test gives a joint confidence region, which can then be projected onto any linear combination or set of linear restrictions. The projection may be conservative, but it does not rely on pretending that a weakly identified coefficient estimate has a reliable normal approximation.

Model and Joint AR Test

Consider the structural equation

where \(y\) is an \(n\)-vector, \(W\) contains included exogenous regressors, \(X\) contains \(k\) endogenous regressors, and \(\beta\in\mathbb{R}^k\) is the coefficient vector of interest. The excluded instruments are the columns of \(Z\). Let \(q=\operatorname{rank}(M_WZ)\), where \(M_W=I-P_W\) is the residual-maker for \(W\), and let \(A=[W,Z]\). Assume \(W\) has full column rank and \(q>0\). The test will also require \(\nu=n-\operatorname{rank}(A)>0\).

For a candidate vector \(b\), form the adjusted outcome \(y-Xb\). Define \(P_B\) as the orthogonal projection onto the column space of a matrix \(B\), and \(M_B=I-P_B\). The two residual sums of squares are

Here \(S_0(b)-S_1(b)\) is the reduction in residual sum of squares from adding the excluded instruments to a regression that already includes \(W\). The AR test of the joint null \(H_0:\beta=b\) rejects for a sufficiently large value of

This is the partial \(F\)-test for zero coefficients on \(Z\) in the regression of \(y-Xb\) on \([W,Z]\). It tests the candidate vector jointly; it is not a collection of separate one-coefficient tests.

Why the Reference Distribution Is Exact

The relevant assumptions are that, conditional on the fixed design matrices \(W\) and \(Z\), the structural errors satisfy \(u\sim N(0,\sigma^2 I_n)\) for some \(\sigma^2>0\), and the null model holds at the tested value \(b\): \(y=W\gamma+Xb+u\). No condition that instruments strongly predict \(X\) is needed. In particular, the argument does not condition on \(X\), which may be endogenous.

Proof. Under the null, \(y-Xb=W\gamma+u\). Since \(M_WW=0\) and \(M_AW=0\),

The column spaces of \(W\) and \(A\) are nested, so \(P_A-P_W\) is an orthogonal projection. Its rank is \(\operatorname{rank}(A)-\operatorname{rank}(W)=q\). The projection \(M_A\) also is orthogonal and has rank \(\nu\). Their ranges are orthogonal: the first is contained in the column space of \(A\), while the second is orthogonal to that space. For a spherical Gaussian vector, orthogonal projections onto orthogonal subspaces give independent squared lengths. Therefore

and the two variables are independent. Their ratio after division by their respective degrees of freedom has the \(F_{q,\nu}\) distribution. Since \(\nu>0\), \(M_Au\) is a nondegenerate Gaussian vector on a \(\nu\)-dimensional subspace, so it is nonzero with probability one. Thus \(S_1(b)>0\) almost surely, completing the proof. \(\square\)

Inverting the Test: A Joint Confidence Region

Inverting the level-\(\alpha\) test means retaining every coefficient vector not rejected. It is safest to define the region by an inequality between sums of squares, rather than by a ratio that may be undefined at some candidate vectors:

Where \(S_1(b)>0\), this inequality is equivalent to \(F_{\mathrm{AR}}(b)\leq F_{q,\nu;1-\alpha}\). If both sums of squares vanish, the inequality includes that point; the ratio there would be \(0/0\) and is undefined. This convention makes the region well-defined for every \(b\). At the true \(\beta\), \(S_1(\beta)>0\) almost surely under the theorem’s assumptions, so this convention does not change its coverage.

Proof. At the true \(\beta\), the denominator \(S_1(\beta)\) is positive almost surely, and the theorem gives \(F_{\mathrm{AR}}(\beta)\sim F_{q,\nu}\), conditional on \(W,Z\). Consequently,

The equality uses the continuous \(F\) distribution. No approximation to the distribution of an estimator is involved. \(\square\)

A Multivariate Confidence-Region Calculation

The next example shows how the inequality can produce a region in two dimensions. Continue to use \(q=2\), \(\nu=10\), and the 5% critical value \(4.103\). Suppose the residualized data have the following sums of squares for every \(b=(b_1,b_2)'\):

These expressions are compatible with a two-dimensional instrument projection whose adjusted-outcome coordinates are \((2-2b_1,\,1-2b_2)\), and an orthogonal residual component with squared length \(10+b_1^2+b_2^2\). In particular, \(S_1(b)>0\) everywhere here.

Using the Region for Linear Restrictions

A joint AR test directly tests a fully specified vector \(b\). For a restriction \(R\beta=r\), where \(R\) is an \(s\times k\) matrix of rank \(s\), there may be nuisance directions: many coefficient vectors can satisfy the restriction. A simple valid procedure is to project the joint confidence region:

The resulting set is a confidence set for \(R\beta\). To test \(H_0:R\beta=r\), fail to reject if there is at least one \(b\in\mathcal{C}_{1-\alpha}\) with \(Rb=r\); reject otherwise. This is a valid level-\(\alpha\) test: if the null is true and the test rejects, then the true \(\beta\) cannot be in \(\mathcal{C}_{1-\alpha}\). Hence the rejection probability is no greater than \(P(\beta\notin\mathcal{C}_{1-\alpha})=\alpha\). The projected test can be conservative, because it accepts whenever any vector satisfying the restriction remains in the region.

Scope and Common Pitfalls

The exact \(F\) result depends on the stated Gaussian, homoskedastic error model and the instrument rank and residual-degree-of-freedom conditions. It is not the heteroskedasticity-robust AR procedure. The AR construction is useful under weak identification precisely because its null distribution does not require a strong first stage; that robustness to instrument weakness does not remove the need for appropriate error assumptions.

• Using a Wald test because it is familiar. A 2SLS Wald approximation can be unreliable under weak instruments. The AR test instead evaluates the instrument contribution to \(y-Xb\) at each proposed \(b\).
• Using the wrong numerator degrees of freedom. The numerator degrees of freedom are \(q=\operatorname{rank}(M_WZ)\), not necessarily the raw number of instrument columns if some are redundant after accounting for \(W\).
• Dividing when the denominator is zero. The ratio is undefined when \(S_1(b)=0\). Define the inverted region by its sum-of-squares inequality; it agrees with the ratio rule wherever the denominator is positive.
• Calling the projection an exact restriction test. Projection gives a valid, possibly conservative test. It does not generally have the same rejection rule or size as a separately derived optimal test of \(R\beta=r\).