Why Compare These Intervals?
Instrumental-variable estimates can look precise even when the instrument provides little information about treatment. Under strong identification, the conventional 2SLS Wald interval is often a useful summary. Under weak identification, its usual approximation can fail: the interval may be too narrow, poorly calibrated, or misleading about how much the data identify. Anderson–Rubin (AR) and conditional likelihood ratio (CLR) confidence sets are designed to maintain inference that is robust to weak first-stage strength under their stated model assumptions.
The purpose of comparison is not to choose whichever interval looks most attractive. Each set answers a different inferential question under different approximations. A conventional interval is typically easy to interpret and efficient when identification is strong. AR offers a transparent weak-IV-robust benchmark, sometimes at the cost of substantial width. CLR can improve power relative to AR while retaining weak-IV robustness under its model. A confidence set’s shape and width are therefore part of the result, not cosmetic details.
What the Three Procedures Report
A conventional 2SLS Wald interval has the form \(\hat{\beta}_{2SLS}\pm z_{1-\alpha/2}SE(\hat{\beta}_{2SLS})\), or uses an appropriate \(t\) critical value when that is the specified approximation. Its center is the 2SLS estimate and its standard error is computed from the IV model. Its familiar symmetric shape comes from approximating the estimator’s sampling distribution by a normal distribution centered at the true parameter. With a weak first stage, that approximation need not describe the estimator’s distribution well.
For AR inference, test each candidate value \(\beta_0\) by testing whether the instruments predict \(Y-\beta_0D\). The AR confidence set contains precisely those candidate values not rejected by the AR test at level \(\alpha\). Its validity does not require the first-stage coefficient to be large, but its power can be low. Depending on the data, inversion can produce a bounded interval, an unbounded set, disconnected pieces, or the entire real line.
CLR inference also tests candidate values and inverts the test, but uses the conditional likelihood ratio procedure to account for nuisance information about the first stage. Under its specified model, it is weak-IV-robust and can be more powerful than AR. The resulting confidence set need not be symmetric around 2SLS, nor is it guaranteed to be narrower in every sample. The introductory material on the conditional likelihood ratio test and its confidence-set construction gives the test and inversion framework; here the emphasis is on comparing the resulting inference.
Worked Comparisons as First-Stage Strength Changes
Worked Example: Strong First Stage and Similar Conclusions
Suppose hypothetical reduced-form estimates are \(\hat r=0.640\) for the effect of the instrument on the outcome and \(\hat g=0.800\) for its effect on treatment. Their standard errors are \(0.060\) and \(0.050\), and their estimated covariance is zero. The ratio estimate is \(0.640/0.800=0.800\) outcome units per treatment unit. For illustration, use the delta-method standard error for this ratio.
For a ratio \(\hat r/\hat g\), the first-order variance is \(\operatorname{Var}(\hat r)/\hat g^2+\hat r^2\operatorname{Var}(\hat g)/\hat g^4-2\hat r\operatorname{Cov}(\hat r,\hat g)/\hat g^3\). Here the covariance term is zero, so
Using \(1.96\) as the 95% normal critical value gives the Wald interval \(0.800\pm1.96(0.09014)=0.800\pm0.1767\), or approximately \([0.623,0.977]\) outcome units per treatment unit.
The AR set can be calculated in this one-instrument illustration by retaining \(\beta\) when \((\hat r-\beta\hat g)^2\leq1.96^2\{SE(\hat r)^2+\beta^2SE(\hat g)^2\}\). Substitution gives \[ (0.640-0.800\beta)^2\leq3.8416(0.0036+0.0025\beta^2), \] or \(0.630396\beta^2-1.024\beta+0.39577024\leq0\). The roots are approximately \(0.6338\) and \(0.9906\), so the AR set is \([0.634,0.991]\) in the same units. The sets are similar in this example, as expected when the first-stage estimate is large relative to its standard error (\(0.800/0.050=16\)). This agreement is reassuring, but it does not establish instrument validity.
Worked Example: Weak First Stage and an Unbounded AR Set
Now consider a different hypothetical sample with \(\hat r=0.040\), \(\hat g=0.050\), standard errors \(0.060\) and \(0.050\), and zero estimated covariance. The ratio remains \(0.800\), but the first-stage estimate is only one standard error from zero. Using the same delta-method calculation,
The nominal 95% Wald interval is \(0.800\pm1.96(1.442)\), or approximately \([-2.027,3.627]\) outcome units per treatment unit. The calculation is arithmetically correct, but weak identification makes the usual Wald calibration questionable.
For AR inversion, substitute the same values into the candidate-value inequality: \[ (0.040-0.050\beta)^2\leq3.8416(0.0036+0.0025\beta^2). \] After moving the right side to the left, this becomes \[ -0.007104\beta^2-0.004\beta-0.01222976\leq0. \] The discriminant is \[ (-0.004)^2-4(-0.007104)(-0.01222976) =0.000016-0.000347\text{ (approximately)}<0. \] The quadratic opens downward and has no real roots, so it is negative for every real \(\beta\). Thus no candidate is rejected by this AR test, and the 95% AR confidence set is the entire real line. This is not a computational failure: the data provide too little information for this test to exclude any treatment-effect value.
A numerical inversion of the CLR test should be reported separately rather than inferred from the Wald interval or assumed to equal the AR set. In this weak-first-stage example, a CLR inversion may also yield a very wide or unbounded set; its actual shape depends on the specified model and the conditional test statistic. The essential conclusion is that the ratio estimate \(0.800\) alone does not establish precise identification.
Worked Example: Reading a CLR Output Alongside AR
Suppose a hypothetical analysis, using the same instruments and a stated homoskedastic normal reduced-form model, numerically inverts the CLR test and reports a 95% confidence set \([0.650,0.970]\) outcome units per treatment unit. Suppose the corresponding AR set is \([0.634,0.991]\), as in the strong-first-stage example above, and the Wald interval is \([0.623,0.977]\).
The CLR set excludes treatment effects below \(0.650\) and above \(0.970\) under that inversion and model; it is somewhat narrower than the AR set in this illustrative output. This is consistent with CLR often having greater power, but one output does not prove that CLR must always be narrower. These assumed CLR endpoints are an example of how to interpret a numerical inversion, not endpoints that can be recovered from the two ratio estimates alone: CLR calculations use the specified reduced-form information and conditional test.
All three sets include effects near \(0.800\), so they agree on a range of plausible values in this example. Their widths and constructions differ. The Wald interval’s apparent precision should be justified by adequate identification; the AR and CLR sets obtain weak-IV robustness under their respective assumptions, not protection from violations of exclusion, exogeneity, or the chosen variance model.
Deriving the AR Set and Comparing Its Geometry
The one-instrument AR inequality also shows why weak first stages can produce unusual sets. Write \(r=\hat r\), \(g=\hat g\), \(s_r=SE(\hat r)\), and \(s_g=SE(\hat g)\), assuming zero covariance for this illustration. Retaining \(\beta\) when the absolute standardized reduced-form residual is no larger than \(c=1.96\) is equivalent to
The coefficient of \(\beta^2\) is \(g^2-c^2s_g^2\). When the first stage is strong relative to its uncertainty, this coefficient is positive and the inequality may have a bounded solution between two real roots. When it is negative, or when the roots do not exist, the accepted set can instead be unbounded or all of the real line. The geometry is therefore informative about identification; it should not be forced into a conventional symmetric interval.
In an analysis with covariates, multiple instruments, or heteroskedastic or clustered errors, use the corresponding AR statistic and its appropriate variance estimator rather than this simplified one-instrument inequality. The comparison must be made under a common instrument set, outcome scale, confidence level, and variance specification. Otherwise differences between reported sets may reflect different analyses, not different inference procedures.
Diagnostics, Failure Modes, and Practical Choice
A useful diagnostic is to compare how conclusions change across procedures as well as to inspect first-stage information. A large difference between a narrow Wald interval and a broad or unbounded weak-IV-robust set is a warning that conventional normal approximation may be doing substantial work. It is not a reason to discard the robust set. Conversely, agreement between procedures does not verify the exclusion restriction or other identifying assumptions.
| Approach | Useful feature | Important limitation |
|---|---|---|
| 2SLS Wald interval | Simple, familiar, and often informative with strong identification. | Usual calibration can fail under weak identification; symmetric finite endpoints can overstate information. |
| AR confidence set | Weak-IV-robust under its assumptions and transparent to invert. | Can have low power and be extremely wide, unbounded, disconnected, or all real numbers. |
| CLR confidence set | Weak-IV-robust under its model and often more powerful than AR. | Requires the specified conditional calculation; shape and width depend on the data and model. |
Common mistakes include reporting only the ratio estimate, treating the first-stage diagnostic as a guarantee that Wald inference is valid, or calling the entire-real-line AR result “no effect.” The last interpretation is wrong: it means the test did not exclude any candidate value at the chosen level, not that all effects are zero. Another mistake is to treat a confidence set’s inclusion of zero as proof that the null is true. A 95% confidence procedure describes long-run coverage under its assumptions; it does not assign a 95% probability to a fixed parameter being inside the particular observed set.
A practical report should give the 2SLS estimate and conventional interval when relevant, state the instrument-strength context, and report the weak-IV-robust confidence set with its shape. Explain the variance and sampling assumptions used, and retain unbounded or disconnected endpoints rather than replacing them with a convenient finite interval. The pretesting consequences of using a first-stage threshold to choose an interval procedure require separate care.
Check Your Understanding
Answer each question, then compare with the solutions.
- In the strong-first-stage example, what is the ratio estimate and the approximate 95% Wald interval?
- Why does the weak-first-stage example’s AR set include every real treatment effect?
- If a CLR set is narrower than an AR set in one sample, does that establish CLR is always narrower? Explain.
- What does an AR confidence set equal to the real line mean, and what does it not mean?
- Name one assumption that weak-IV-robust inference does not fix.
Solutions. (1) The ratio is \(0.640/0.800=0.800\); the Wald interval is approximately \([0.623,0.977]\) outcome units per treatment unit. (2) The AR inequality reduces to a downward-opening quadratic with negative discriminant, so its left side is negative for all real \(\beta\); no candidate is rejected. (3) No. Relative width varies with the data and model; one numerical comparison does not establish a universal ordering. (4) At the chosen test level, the procedure excludes no candidate value. It does not mean the effect is zero or that every value is equally plausible. (5) For example, it does not fix a violation of instrument exclusion or exogeneity.