Why Multiple Instruments Change the Analysis
When a treatment has several proposed instruments, the central question is not whether each one looks persuasive in isolation. The relevant strength diagnostic asks whether the instruments jointly predict treatment after adjustment for included covariates. And if that joint prediction is weak, a conventional interval around the two-stage least squares estimate can be misleadingly reassuring. This worked analysis combines a joint first-stage diagnostic with Anderson–Rubin inference, which remains valid under weak identification when its model assumptions hold.
The example uses a continuous outcome and treatment. It is hypothetical, and its compact regression summaries are constructed to make the calculations transparent. The same logic applies when there are more than two instruments, although the joint tests then have more numerator degrees of freedom.
Models and Assumptions
The interpretation of \(\tau\) as a causal effect requires a correctly specified structural model and valid instruments. In particular, the instruments must be relevant conditional on \(X\), independent of unmeasured causes of the outcome under the design, and excluded from the outcome equation except through \(D\). In the linear model, the moment condition \(E[Z u\mid X]=0\) expresses the required instrument-outcome orthogonality. The relevant population rank condition is that \(\pi\) is not the zero vector. Weak instruments are instruments whose conditional predictive information about \(D\) is too limited for reliable conventional large-sample approximations; relevance in principle does not guarantee adequate strength in a particular sample.
For the exact finite-sample F calculations below, assume the linear models are correctly specified, errors are normally distributed and homoskedastic conditional on the regressors, and the regressors are fixed (or condition on their observed values). These assumptions are stronger than those needed for some asymptotic or robust versions of the procedures. If heteroskedasticity or clustering is present, use an appropriate robust test rather than treating the classical F reference distribution as exact.
With \(k\) columns in \(X\), including an intercept if one is fitted, the first-stage diagnostic tests \(H_0:\pi_1=\cdots=\pi_q=0\). It is the partial F test for adding all excluded instruments to a regression already containing \(X\). A large F is evidence against this joint null; it is not a test of instrument validity. There is no universal cutoff that makes instruments safe in every design.
Worked Example: A Joint First-Stage Diagnostic
Worked Example: A Joint First-Stage Diagnostic
Suppose a hypothetical sample has \(n=40\), an intercept and no other covariates, and two instruments. The instruments are centered and orthogonal, with \(\sum_i Z_{i1}^2=\sum_i Z_{i2}^2=100\). The first-stage coefficients are \(\hat\pi_1=0.12\) and \(\hat\pi_2=0.08\); the full first-stage residual mean square is 1.00. Thus the residual degrees of freedom are \(40-1-2=37\).
Because the instruments are orthogonal, the fitted sum of squares attributable to the two instruments is \(100(0.12)^2+100(0.08)^2=1.44+0.64=2.08\). The partial F statistic is the explained sum of squares per added instrument divided by the full-model residual mean square:
This is weak joint first-stage evidence in this hypothetical sample: the two instruments together explain little treatment variation relative to the residual variation. The calculation does not establish that either instrument is invalid, nor does it prove that the population first stage is exactly zero. It warns that conventional 2SLS standard errors and intervals may not have their usual accuracy.
Anderson–Rubin Tests With Several Instruments
For a candidate effect \(\tau_0\), rearrange the structural equation to obtain \(Y-\tau_0D=X'\beta+u\) under the null \(H_0:\tau=\tau_0\). Regress this transformed outcome on \(X\) and all \(q\) instruments, then test whether the coefficients on the instruments are jointly zero. The Anderson–Rubin test does not divide by an estimated first-stage coefficient. This is the key reason its null test remains usable when the first stage is weak.
The justification follows directly from the null model. If \(H_0\) is true, the transformed outcome is \(X'\beta+u\), and instrument validity gives no systematic \(Z\) contribution after conditioning on \(X\). Under the classical error assumptions, the partial sum of squares from adding \(q\) instruments, divided by \(q\), and the full-model residual mean square estimate the same error variance under the null. Their ratio therefore has the stated F reference distribution. Unlike a conventional t test for the 2SLS coefficient, this test does not require the first-stage estimate to be sufficiently far from zero.
The test’s robustness has a defined scope: it protects the test’s size against weak instrument strength, not against failure of exclusion, independence, or the structural model. A small p-value can reflect a false candidate effect, but it can also reflect invalid instruments or model misspecification. Instrument validity still requires substantive design arguments.
Worked Example: Testing a Candidate Effect
Worked Example: Testing a Candidate Effect
Continue with the same hypothetical instruments and sample. Suppose the reduced-form regression coefficients of \(Y\) on \(Z_1,Z_2\), conditional on the intercept, are \(\hat\rho_1=0.80\) and \(\hat\rho_2=0.40\). For this constructed example, the residual mean square in the regression of \(Y-\tau D\) on the intercept and instruments is \(1+\tau^2\). The instrument columns have sums of squares 100 and are orthogonal, so the partial sum of squares for adding the instruments is \(100\{(\hat\rho_1-\tau\hat\pi_1)^2+(\hat\rho_2-\tau\hat\pi_2)^2\}\).
Test \(H_0:\tau=0\). The two instrument coefficients in the transformed-outcome regression are \(0.80-0(0.12)=0.80\) and \(0.40-0(0.08)=0.40\). Their partial sum of squares is \(100(0.80^2+0.40^2)=100(0.64+0.16)=80\). The full-model residual mean square is \(1+0^2=1\), so
At significance level \(\alpha=0.05\), this statistic exceeds the \(F(2,37)\) critical value of approximately 3.25, so the zero-effect null is rejected. The result says that the observed transformed outcome is inconsistent with the joint instrument coefficient restriction at this level, under the model assumptions. It does not by itself establish instrument validity or identify the direction and magnitude of the effect.
For comparison, test the candidate \(\tau_0=2\). The transformed-outcome instrument coefficients are \(0.80-2(0.12)=0.56\) and \(0.40-2(0.08)=0.24\). The partial sum of squares is \(100(0.56^2+0.24^2)=100(0.3136+0.0576)=37.12\). The residual mean square is \(1+2^2=5\), giving \(F_{\mathrm{AR}}(2)=(37.12/2)/5=3.712\). Since \(3.712>3.25\), this candidate is also rejected at the 5% level.
Inverting the Test: A Disconnected Confidence Set
A 95% Anderson–Rubin confidence set consists of every candidate \(\tau_0\) that is not rejected by the two-sided level-0.05 test. It is obtained by evaluating the test over candidate values and retaining those for which \(F_{\mathrm{AR}}(\tau_0)\leq F_{0.95}(q,n-k-q)\). It need not be a single bounded interval. The geometry of such sets, including disconnected and unbounded cases, is part of the earlier material on weak-IV confidence sets.
Worked Example: Inverting the Multiple-Instrument Test
For the same hypothetical summaries, the partial sum of squares at candidate \(\tau\) is \(100\{(0.80-0.12\tau)^2+(0.40-0.08\tau)^2\}\). Expanding the squares gives \(100(0.80-0.256\tau+0.0208\tau^2)=80-25.6\tau+2.08\tau^2\). Dividing by two and by the residual mean square \(1+\tau^2\) gives
Using the 5% critical value \(c\approx3.25\), retain \(\tau\) when \((40-12.8\tau+1.04\tau^2)/(1+\tau^2)\leq3.25\). Since \(1+\tau^2\) is positive for every real \(\tau\), multiplying both sides preserves the inequality:
The quadratic has roots \(\{ -12.8\pm\sqrt{12.8^2+4(2.21)(36.75)}\}/(2(2.21))\). The discriminant is \(163.84+324.87=488.71\), whose square root is approximately 22.11. The roots are therefore approximately \((-12.8-22.11)/4.42=-7.90\) and \((-12.8+22.11)/4.42=2.11\). Because the quadratic opens upward, it is nonnegative outside the roots. The approximate 95% confidence set is
This set is disconnected and unbounded: it excludes effects between the two endpoints but includes arbitrarily large positive and negative values in its two rays. It does not say that the true effect has a 95% probability of lying in this particular set. In repeated samples under the assumptions, the procedure has 95% coverage. Here the shape warns that the data do not pin down the effect in a conventional, finite interval. The reported \(F\) critical value is rounded, so the endpoints are approximate.
Diagnostics, Alternatives, and Limitations
A multiple-instrument analysis should make the instrument set and the diagnostic explicit. Report the number of instruments, the included covariates, the joint partial first-stage statistic and its degrees of freedom, and whether the reference calculation is classical or robust. If adding or removing instruments materially changes the first stage or the inferential result, investigate why rather than selecting the set that produces the preferred answer. A strong joint F does not establish that every instrument is valid; nor does it rule out near-collinearity or a poorly chosen instrument set.
Conventional 2SLS inference is often more concise when identification is strong and its approximation is credible. It typically produces an estimate with a standard error and a finite interval. But a reassuring conventional interval should not be treated as decisive merely because it is narrow, particularly after selecting an analysis based on a strength pretest. Anderson–Rubin inference directly tests candidate effects and is robust to weak strength under its assumptions, but it can have low power and yield a wide, unbounded, or disconnected set. Conditional likelihood ratio methods are another weak-IV-robust option; their construction and comparison with Anderson–Rubin inference are covered in the earlier material on weak-IV tests.
The main failure modes are substantive as well as numerical. A classical first-stage F test can be inappropriate under heteroskedasticity or clustering unless a suitable robust version is used. A large first-stage statistic does not repair a violation of exclusion or independence. And an awkward Anderson–Rubin confidence set is not a computational error to be silently converted to a convenient bounded interval: it is information about how weakly the data constrain the effect under the specified model. State the full set and explain its implications.
Common Mistake
- Wrong reasoning: “There are two instruments, so I will check each first-stage coefficient separately; if one is significant, the instrument set is strong.”
- Correction: Strength is assessed by the joint partial test of all excluded instruments conditional on the included covariates. Individual coefficient tests can miss joint predictive information or misrepresent how much the set contributes.
- Wrong reasoning: “An Anderson–Rubin confidence set with two rays is unusable, so I will report only the central gap as the interval.”
- Correction: The retained set is the union of the two rays, not the gap. Its unusual shape and unboundedness are substantive features of weakly informative data, not a reason to replace it with a conventional interval.
Check Your Understanding
Try these questions, then compare your reasoning with the brief solutions.
- With three excluded instruments and an intercept, what are the numerator and denominator degrees of freedom for the classical joint first-stage F statistic when \(n=60\)?
- Why does the Anderson–Rubin test remain useful when the first-stage coefficients are small?
- In the worked inversion, why is it valid to multiply the inequality by \(1+\tau^2\)?
- Does a large joint first-stage F establish that all instruments satisfy exclusion? Explain.
- What should be reported if inversion gives two unbounded rays rather than one finite interval?
Solutions. (1) There are three added instrument coefficients, so the numerator degrees of freedom are 3; with an intercept and three instruments, the denominator degrees of freedom are \(60-1-3=56\). (2) The test evaluates whether the instruments jointly predict \(Y-\tau_0D\) under each candidate; it does not divide by the first-stage estimate. (3) \(1+\tau^2\) is strictly positive for every real \(\tau\), so multiplication cannot reverse the inequality. (4) No. Strength concerns prediction of treatment; exclusion is a separate substantive validity condition. (5) Report the complete union of rays and explain that the data do not provide a bounded confidence region under this procedure.