Deriving the Anderson-Rubin Test

Why Test a Transformed Outcome?

Instrumental-variables estimates are often expressed as ratios: a reduced-form effect of the instrument on the outcome divided by a first-stage effect of the instrument on treatment. When the first-stage effect is weak, sampling variation in the denominator can make ordinary Wald inference unreliable. The Anderson–Rubin (AR) test takes a different route: for each candidate treatment effect, it tests a restriction on reduced-form coefficients without dividing by an estimated first-stage coefficient.

This tutorial derives that restriction and the classical test for it. The focus is a single endogenous treatment and \(q\) instruments, with no included covariates. The derivation shows why the AR procedure can test a candidate effect even when the first stage is weak, and also why its tested restriction does not identify the effect if the instruments have no first-stage relevance.

Model and the Restriction Implied by a Candidate Effect

Let \(Y\) be an outcome, \(D\) a scalar treatment, and \(Z=(Z_1,\ldots,Z_q)'\) a vector of instruments. Consider the structural equation and first-stage conditional mean

Y=\alpha+\beta D+u,\qquad E[D\mid Z]=\delta+Z'\pi.

Here \(\beta\) is the structural treatment effect, \(\pi\) is the vector of first-stage slopes, and \(u\) is the structural error. Instrument validity requires that the instruments be excluded from the structural outcome equation and satisfy \(E[u\mid Z]=0\). The reduced-form conditional mean for \(Y\) is then

E[Y\mid Z]=\alpha+\beta\delta+Z'(\beta\pi).

Write this reduced form as \(E[Y\mid Z]=\alpha_Y+Z'\gamma\). It follows that \(\gamma=\beta\pi\). Thus, if a candidate value \(\beta_0\) is the true effect, the reduced-form slopes must satisfy \(\gamma-\beta_0\pi=0\). This is a restriction implied by the structural null \(H_0:\beta=\beta_0\). It is equivalent to that null only when \(\pi\ne 0\): if \(\pi=0\), the restriction holds for every \(\beta_0\), so it cannot distinguish candidate effects.

Definition: For a candidate effect \(\beta_0\), the classical Anderson–Rubin test tests the joint restriction that the instrument slopes in the regression of \(Y-\beta_0D\) on an intercept and \(Z\) are zero. Equivalently, it tests the reduced-form restriction \(\gamma-\beta_0\pi=0\), which is implied by \(H_0:\beta=\beta_0\). The restriction is equivalent to the structural null only if the instruments identify the effect, meaning \(\pi\ne 0\) in this single-treatment setup.

Why the Regression Tests the Restriction

For any fixed candidate \(\beta_0\), define the transformed outcome \(Y-\beta_0D\). By subtracting the first-stage conditional mean multiplied by \(\beta_0\) from the reduced-form conditional mean for \(Y\), we obtain

E[Y-\beta_0D\mid Z] =(\alpha_Y-\beta_0\delta)+Z'(\gamma-\beta_0\pi).

The coefficient vector on \(Z\) in this regression is therefore \(\gamma-\beta_0\pi\). Testing whether all those coefficients equal zero directly tests the reduced-form restriction. Under the structural null, \(Y-\beta_0D=\alpha+u\): the treatment term cancels, leaving no instrument slope when \(E[u\mid Z]=0\). This is the key justification for the test.

For the usual finite-sample F reference distribution, suppose observations are independent, the instrument design matrix has full column rank, and the transformed regression error under the null is normally distributed with a common variance, conditional on the instruments. The regression includes an intercept and all \(q\) instruments. Under these conditions, the ordinary partial F test of the \(q\) instrument slopes has an \(F_{q,n-q-1}\) distribution under the null. Without normality, the same classical test may be justified approximately in sufficiently large samples under suitable regularity conditions; heteroskedasticity or dependence can invalidate its usual standard-error calculation and reference distribution.

The test does not estimate \(\beta\) by taking a ratio, nor does it need the estimated first-stage slope to be bounded away from zero to calculate a statistic. Its validity as a test of the structural null still depends on instrument validity and the stated inference conditions. A weak first stage can make the test have little power, but it does not create the unstable denominator that causes conventional Wald inference to fail.

Worked Example: Deriving the Candidate-Effect Restriction

Suppose there are two instruments, and the hypothetical population first-stage slopes are \(\pi=(0.12,0.04)'\), while the outcome reduced-form slopes are \(\gamma=(0.18,0.06)'\). Find the slope restriction for \(\beta_0=1.5\), and compare it with the restriction for \(\beta_0=2\).

For \(\beta_0=1.5\),

\gamma-\beta_0\pi = \begin{pmatrix}0.18\\0.06\end{pmatrix} - 1.5\begin{pmatrix}0.12\\0.04\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}.

Thus, the reduced-form slope restriction implied by the candidate \(1.5\) holds for these coefficients. For \(\beta_0=2\),

\gamma-2\pi = \begin{pmatrix}0.18\\0.06\end{pmatrix} - \begin{pmatrix}0.24\\0.08\end{pmatrix} = \begin{pmatrix}-0.06\\-0.02\end{pmatrix}.

That candidate implies nonzero slopes in the transformed-outcome regression. These coefficient values alone do not establish whether either restriction should be rejected: inference also requires the sampling uncertainty and the joint F test.

Computing the Classical AR F Statistic

Let \(RSS_U\) be the residual sum of squares from the unrestricted regression of \(Y-\beta_0D\) on an intercept and all \(q\) instruments. Let \(RSS_R\) be the residual sum of squares from the restricted regression containing only an intercept. The classical statistic is

F_{\mathrm{AR}}= \frac{(RSS_R-RSS_U)/q}{RSS_U/(n-q-1)}.

The numerator measures the reduction in residual variation from adding the \(q\) instrument slopes, per restriction. The denominator is the unrestricted regression’s residual variance estimate. Under the classical assumptions above, compare this statistic with an \(F\) distribution having \(q\) and \(n-q-1\) degrees of freedom. A small p-value is evidence against the tested restriction, and therefore against the candidate structural effect when relevance holds. It is not the probability that the null is true.

Worked Example: Testing Two Instrument Slopes

Suppose a hypothetical sample has \(n=80\) observations and two instruments. In the regression of \(Y-\beta_0D\) on an intercept and both instruments, the unrestricted \(R^2\) is \(0.0500\). Calculate the classical AR statistic and interpret it at the 5% level.

With an intercept, \(q=2\) slopes, and \(n=80\), the unrestricted residual degrees of freedom are \(80-2-1=77\). The partial F statistic from \(R^2\) is

F_{\mathrm{AR}} = \frac{R^2/q}{(1-R^2)/(n-q-1)} = \frac{0.0500/2}{0.9500/77} = \frac{0.0250}{0.0123377} \approx 2.026.

The reference distribution is \(F_{2,77}\), and the upper-tail p-value is approximately \(0.139\). Since \(0.139>0.05\), this test does not reject the zero-slope restriction at the 5% level. If the instruments are valid and relevant and the classical assumptions hold, the result does not provide sufficient evidence against the candidate \(\beta_0\). It does not prove that the candidate is correct.

Identification, Interpretation, and Failure Modes

The distinction between testing an implied restriction and identifying an effect is essential. If \(\pi=0\), then validity implies \(\gamma=\beta\pi=0\), regardless of \(\beta\). Consequently, \(\gamma-\beta_0\pi=0\) for every candidate. In a finite sample, the AR test remains computable, but its results cannot distinguish candidate effects when the population first stage is exactly zero. With a very weak first stage, tests can also have low power, so many different candidate values may fail to be rejected.

Worked Example: Zero First-Stage Slopes

Suppose a hypothetical population has \(\pi=(0,0)'\) and \(\gamma=(0,0)'\). Evaluate the reduced-form restriction for \(\beta_0=1\) and for \(\beta_0=5\).

For \(\beta_0=1\), \(\gamma-\beta_0\pi=(0,0)'-1(0,0)'=(0,0)'\). For \(\beta_0=5\), \(\gamma-\beta_0\pi=(0,0)'-5(0,0)'=(0,0)'\) as well. Both candidates satisfy the restriction. The data-generating reduced forms in this example contain no instrument slopes for either treatment or outcome, so the restriction does not identify a treatment effect.

In applied work, check that the transformed regression uses the same sample and instrument set across the unrestricted and restricted specifications, that the instruments are not collinear, and that the test is joint rather than a collection of separate t tests. A design matrix without full column rank does not support the stated \(q\)-restriction calculation. Also assess whether the classical error assumptions are plausible: heteroskedasticity or clustered observations call for inference that accounts for those features, rather than mechanically using the homoskedastic F test.

The AR test does not establish instrument validity. A statistically insignificant joint slope test for one candidate is not proof of exclusion or exogeneity; these are design assumptions, not conclusions guaranteed by this regression. The test also does not make weak instruments informative: low power may leave a broad range of candidates unrejected.

How the AR Test Differs From Conventional Wald Inference

Conventional IV Wald inference typically uses an estimated IV coefficient and its estimated standard error to test \(H_0:\beta=\beta_0\). When the first stage is weak, the coefficient behaves roughly like a ratio with a noisy denominator, and its usual normal or t-based approximation can be poor. The AR test instead examines whether the instrument slopes in \(Y-\beta_0D\) are jointly zero. It avoids dividing by the estimated first-stage coefficient, which is why its null test can retain its stated size under weak relevance when its validity and error-distribution assumptions hold.

The procedures answer related but distinct inferential questions and can give different results in weak-instrument settings. A conventional Wald interval may look narrow even when its approximation is unreliable; an AR test may fail to reject many candidates when information about the effect is limited. Neither a failure to reject nor a small p-value assesses instrument validity by itself. Robust variants of the AR procedure require adjusted variance estimation and are treated separately.

Common Mistakes

Calling the reduced-form restriction equivalent to the structural null in every case. It is implied by the null, but equivalence requires \(\pi\ne0\). If \(\pi=0\), every candidate satisfies the restriction.
Interpreting a non-significant test as proof of the candidate effect. Failure to reject means the data do not provide sufficient evidence against the restriction at the chosen level; weak instruments may also mean low power.
Using separate tests for each instrument slope. The AR test is a joint test of all \(q\) slopes. The joint statistic accounts for their covariance.
Applying the ordinary F reference distribution automatically. Its finite-sample justification depends on the stated homoskedastic, independent, normal-error setup. Dependence or heteroskedasticity can require a different variance estimator and test.

Key takeaway: For each candidate \(\beta_0\), regress \(Y-\beta_0D\) on an intercept and the instruments, then jointly test that all instrument slopes are zero. This tests a restriction implied by the structural null, without dividing by the estimated first stage; it identifies the candidate only when the instruments are relevant.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

For \(q=3\) instruments and \(n=100\), what are the numerator and denominator degrees of freedom for the classical AR F test with an intercept?
If \(\gamma=(0.10,0.04)'\) and \(\pi=(0.05,0.02)'\), calculate \(\gamma-\beta_0\pi\) for \(\beta_0=2\).
Why does \(\gamma-\beta_0\pi=0\) fail to distinguish candidate effects when \(\pi=0\)?
A test gives \(p=0.18\) at \(\alpha=0.05\). What is the correct conclusion?
Name one condition under which the ordinary finite-sample F reference distribution may not be justified.

Solutions. (1) The numerator degrees of freedom are \(q=3\); the denominator degrees of freedom are \(n-q-1=100-3-1=96\), so the reference distribution is \(F_{3,96}\). (2) \(\gamma-2\pi=(0.10,0.04)'-(0.10,0.04)'=(0,0)'\). (3) If \(\pi=0\), then \(\gamma-\beta_0\pi=\gamma\) for every \(\beta_0\); when \(\gamma=0\), every candidate satisfies the restriction. (4) Since \(0.18>0.05\), do not reject the tested restriction at the 5% level; this is not proof that the candidate is true. (5) For example, heteroskedastic errors or clustered observations violate the classical variance assumptions, so the ordinary F reference distribution may not apply.