Testing a Proposed IV Coefficient
With an instrument that is weakly related to the endogenous regressor, the usual IV Wald test can give unreliable inference. The Anderson–Rubin test takes a different route: for a proposed value of the structural coefficient, it transforms the outcome and asks whether the instrument predicts that transformed outcome. Its key attraction is that, under the stated null and classical error conditions, its reference distribution does not require the instrument to be strong.
This tutorial treats one endogenous regressor, one scalar instrument, and no included covariates. Let \(y_i\) be the outcome, \(x_i\) the endogenous regressor, and \(z_i\) the instrument for observation \(i\), with structural equation \(y_i=\beta x_i+u_i\). The target is a test of \(H_0:\beta=\beta_0\), where \(\beta_0\) is a specified value. Instrument relevance and the conditions required for causal interpretation remain substantive requirements; the test does not verify them.
Under the structural null, \(r_i(\beta_0)=u_i\). If the instrument is exogenous in the relevant conditional-mean sense, the instrument should not systematically predict this transformed outcome. The auxiliary regression is a testing device: its slope is not a new estimate of the structural effect.
Statistic and Exact Reference Distribution
Fit two nested regressions to the transformed outcome. The unrestricted regression includes an intercept and \(z_i\); the restricted regression includes only an intercept. Let \(RSS_U\) and \(RSS_R\) denote their residual sums of squares. If \(z_i\) varies in the sample and \(n>2\), the Anderson–Rubin statistic is
The numerator has one degree of freedom because the unrestricted model adds one instrument coefficient. The denominator has \(n-2\) degrees of freedom because the unrestricted regression estimates an intercept and a slope. Equivalently, this statistic is the square of the usual \(t\) statistic for the slope in that auxiliary regression.
The exact finite-sample reference distribution requires more than an unconditional zero correlation between \(z_i\) and \(u_i\). A convenient set of assumptions is that, conditional on the observed instrument values, the structural null holds and the errors are independent and normally distributed with conditional mean zero and a common variance:
Here the instruments are treated as fixed after conditioning. With random instruments, the corresponding requirement is that the errors be conditionally independent and normal with \(E[u_i\mid z_1,\ldots,z_n]=0\) and a common conditional variance. A constant conditional mean can also be accommodated by the intercept, provided the regression errors around that mean satisfy the same normality and variance conditions. Merely having \(E[z_i u_i]=0\) does not ensure the required conditional mean structure.
Why the F Distribution Applies
Under the null, the transformed response is the structural error: \(r_i(\beta_0)=u_i\). Conditional on the instruments, the null model for this response is an intercept-only regression, and the alternative regression adds the instrument. Under the stated assumptions, the errors are normal with a common variance, so the usual nested-regression decomposition applies. The additional sum of squares and the unrestricted residual sum of squares, each divided by its degrees of freedom, are independent estimates of the same variance under the null. Their ratio therefore has the central \(F_{1,n-2}\) distribution. If the conditional mean varies with \(z_i\), that null regression is misspecified and this argument no longer establishes a central F reference distribution.
This derivation also clarifies an important distinction: the null hypothesis concerns \(\beta\), while the auxiliary-regression test is a way of assessing that hypothesis under the IV model. Rejection indicates that the transformed outcome is unusually related to the instrument under the null assumptions. It does not, by itself, show which assumption failed.
Worked Example: Calculating a Non-Rejection
Suppose six hypothetical observations have instrument values \(z=(-2,-1,0,0,1,2)\), endogenous-regressor values \(x=(1,2,1,3,2,4)\), and outcomes \(y=(1,1,3,7,3,11)\). Test \(H_0:\beta=2\). The transformed outcomes are \(r_i=y_i-2x_i\), giving \(r=(-1,-3,1,1,-1,3)\).
The instrument values sum to zero, and the transformed outcomes also sum to zero. Thus the fitted intercept is zero. The slope is \(\hat{\gamma}=\sum_i z_i r_i/\sum_i z_i^2=10/10=1\). The unrestricted fitted values are therefore \(z_i\), and the residuals are \(r_i-z_i=(1,-2,1,1,-2,1)\). Hence \(RSS_U=1+4+1+1+4+1=12\).
For the restricted intercept-only regression, the fitted value is the transformed-outcome mean, zero, so \(RSS_R=\sum_i r_i^2=1+9+1+1+1+9=22\). Therefore \(F_{AR}=[(22-12)/1]/(12/4)=10/3=3.333\), with \(1\) and \(4\) degrees of freedom. Its upper-tail p-value is approximately \(0.1419\). At \(\alpha=0.05\), do not reject \(H_0:\beta=2\). This result is not proof that \(\beta=2\); it means the data do not provide sufficient evidence against that value using this test.
For the exact p-value interpretation to apply, the hypothetical errors under the null must satisfy the conditional assumptions above. The arithmetic alone does not establish those assumptions.
Worked Example: A Rejection With the Same Design
Keep the same six \(z_i\) and \(x_i\), but suppose the hypothetical outcomes are \(y=(3,5,3,9,7,19)\), and test \(H_0:\beta=2\). Then \(r_i=y_i-2x_i=(1,1,1,3,3,11)\) would be an unnecessarily awkward set of values. To make the calculation transparent, instead consider hypothetical outcomes \(y=2x+r\) for \(r=2z+q\), where \(q=(1,-2,1,1,-2,1)\). This gives \(r=(-3,-4,1,1,0,5)\), and outcomes \(y=(-1,0,3,7,4,13)\).
The vector \(q\) sums to zero and is orthogonal to \(z\): \(\sum_i q_i=0\) and \(\sum_i z_iq_i=(-2)(1)+(-1)(-2)+(1)(-2)+(2)(1)=0\). The transformed-outcome regression consequently has slope \(2\), with residuals \(q_i\). Thus \(RSS_U=\sum_i q_i^2=12\). Since \(r=2z+q\), the transformed-outcome mean is zero, \(RSS_R=\sum_i r_i^2=4\sum_i z_i^2+\sum_i q_i^2=4(10)+12=52\); the cross-product term is zero.
It follows that \(F_{AR}=[(52-12)/1]/(12/4)=40/3=13.33\). With \(1\) and \(4\) degrees of freedom, the upper-tail p-value is approximately \(0.022\). At the \(5\%\) level, reject \(H_0:\beta=2\), conditional on the test assumptions. In context, the transformed outcome is too strongly associated with the instrument to be consistent with this proposed coefficient under the null model.
Conditional Exogeneity Is Not Just Zero Correlation
A common mistake is to treat \(E[z_i u_i]=0\) as sufficient for the exact F test. That moment condition constrains an average relationship; it does not guarantee that the conditional mean \(E[u_i\mid z_i]\) is constant. The distinction matters because the auxiliary regression must have the specified conditional-mean structure for the central-F justification to hold.
Worked Example: Zero Covariance but a Varying Conditional Mean
Consider a hypothetical instrument \(z\) taking values \(0,1,2\) with equal probabilities, and let \(u=g(z)+\varepsilon\), where \(g(0)=1\), \(g(1)=-2\), \(g(2)=1\), and \(\varepsilon\mid z\sim N(0,1)\). The conditional means of \(u\) are \(1,-2,1\), which are not constant.
Nevertheless, \(E[z]=1\) and \(E[u]=(1-2+1)/3=0\). Also \(E[zu]=(0\cdot1+1\cdot(-2)+2\cdot1)/3=0\), so \(\operatorname{Cov}(z,u)=E[zu]-E[z]E[u]=0\). Thus zero covariance holds even though the conditional mean varies with \(z\). A linear regression with an intercept and \(z\) does not represent this conditional mean pattern. The classical central-F result cannot be justified from the covariance condition alone.
The correction is to state and assess the conditional-mean assumption needed for the classical result, rather than silently replacing it with a weaker unconditional moment. In applied work, that assumption is part of the IV model and is not proved merely by observing a small sample correlation.
Interpretation, Diagnostics, and Limitations
The Anderson–Rubin test is robust to weak instrument relevance in a particular sense: under its null and error assumptions, its size does not rely on a strong first stage. A weak instrument can nevertheless make the test have low power against false coefficient values. In practical terms, a large p-value may reflect little information about \(\beta\), not close agreement with \(\beta_0\). The first-stage diagnostics studied earlier address instrument strength; they do not replace an assessment of the assumptions needed for the Anderson–Rubin reference distribution.
Before relying on the classical statistic, check that the instrument varies, the sample size permits residual degrees of freedom, and the structural and conditional error assumptions are credible. Also inspect the auxiliary regression for unusual observations or patterns that might signal a poor error model. With one instrument and no covariates, the auxiliary regression has only an intercept and one slope; a linear slope test cannot diagnose every possible dependence of the transformed outcome on the instrument. In particular, a nonlinear conditional mean can violate the exact result even if a linear association is near zero.
The usual IV Wald test instead estimates the structural coefficient and tests it using an estimated standard error. Its conventional approximation can fail when the first stage is weak, as discussed in the earlier material on weak-instrument Wald inference. The Anderson–Rubin test avoids that particular weak-first-stage approximation but depends, in its classical form, on stronger distributional and conditional-mean assumptions. Neither procedure establishes instrument validity. Robust versions and extensions with covariates or multiple instruments require different formulations and are outside this single-instrument, no-covariate calculation.
Write the null \(H_0:\beta=\beta_0\) and calculate \(r_i=y_i-\beta_0x_i\) for every observation.
Regress \(r_i\) on an intercept alone and on an intercept plus \(z_i\); record \(RSS_R\) and \(RSS_U\).
Use \(F_{AR}=(RSS_R-RSS_U)/(RSS_U/(n-2))\), compare with \(F_{1,n-2}\), and state the p-value, significance level, and contextual conclusion.
The central pitfall is to report the F p-value as exact simply because the sample correlation between \(z\) and the structural residual appears small. The correction is to distinguish the test calculation from the assumptions that make its reference distribution valid: conditional mean structure, independent errors, normality, and a common variance are part of the classical finite-sample result. If those conditions are doubtful, do not present the classical p-value as guaranteed exact.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- For \(n=8\), one nonconstant instrument, and no included covariates, what are the degrees of freedom of the classical Anderson–Rubin F statistic?
- In a hypothetical test, \(RSS_R=30\), \(RSS_U=18\), and \(n=8\). Calculate \(F_{AR}\).
- Why is a non-rejection of \(H_0:\beta=\beta_0\) not evidence that the null is true?
- Does \(E[zu]=0\) alone establish the exact central-F reference distribution? Explain briefly.
Solutions. (1) The unrestricted regression estimates an intercept and one slope, so the degrees of freedom are \(1\) and \(8-2=6\). (2) \(F_{AR}=[(30-18)/1]/[18/(8-2)]=12/3=4.000\), with \(1\) and \(6\) degrees of freedom; the p-value must be calculated from the \(F_{1,6}\) distribution and interpreted using a stated \(\alpha\). (3) Non-rejection means the data do not provide sufficient evidence against that candidate under the test assumptions; weak instruments can leave the test with limited power. It does not assign a probability to the null being true. (4) No. Zero covariance is an unconditional moment and does not ensure a constant conditional mean given the instrument. The classical result requires the appropriate conditional-mean structure as well as its conditional error assumptions.