Why condition for weak-instrument inference?
In instrumental-variables (IV) regression, a weak first stage makes a conventional Wald interval unreliable: the usual normal approximation to the ratio-based estimator can be poor, even in large samples. The earlier material on Fieller geometry shows why the resulting confidence sets need not behave like ordinary intervals. The conditional likelihood ratio (CLR) test offers another way to test a candidate structural coefficient while allowing the first-stage strength to be weak.
The key idea is to use two statistics with different roles. One captures the reduced-form discrepancy implied by a candidate coefficient; the other contains information about the unknown first-stage parameter. Under the null, conditioning on the latter removes that nuisance parameter from the reference distribution. The resulting test adapts its calibration to the observed first-stage statistic rather than relying on a strong-instrument normal approximation.
Canonical model and conditioning strategy
Consider one endogenous regressor, \(X\), and \(k\) excluded instruments, collected in \(Z\). Included exogenous covariates can first be partialled out of the outcome, endogenous regressor, and instruments. For the basic CLR result, assume a linear structural equation with coefficient \(\beta\), valid instruments (relevance, exclusion, and exogeneity), and a full-rank instrument design. Use the usual Gaussian, homoskedastic reduced-form model with a common covariance matrix for the outcome and endogenous-regressor disturbances. The instrument design is treated as fixed, and the disturbance covariance is positive definite. These conditions give a finite-sample canonical normal representation; they are stronger than conditions often used for asymptotic or robust inference.
For a candidate null value \(\beta_0\), the reduced-form statistics can be whitened and rotated to form \(k\)-vectors \(S\) and \(T\). Under \(H_0:\beta=\beta_0\), their canonical representation is
where \(I_k\) is the \(k\)-dimensional identity matrix and \(\mu\) is an unknown nuisance parameter determined by first-stage strength and direction. The transformation uses the reduced-form covariance to account for correlation between the outcome and endogenous-regressor disturbances. Thus \(S\) is not simply an unadjusted outcome residual, and \(T\) is not an ordinary first-stage F statistic. Their normalization is part of the canonical construction.
Conditioning is not the same as asserting that the first stage is known or strong. The observed \(T\) is held fixed when obtaining the conditional reference distribution, and its magnitude and direction influence the test statistic and its conditional calibration. The unknown mean of \(T\), which otherwise complicates null inference, does not enter the conditional law of \(S\).
The CLR statistic
For the observed \(S\) and \(T\), define the Anderson–Rubin component \(A\), the first-stage magnitude \(Q\), and the score component \(L\) as follows:
Here \(A\) measures the squared length of the null-implied reduced-form discrepancy, \(Q\) is the squared length of the first-stage statistic, and \(L\) measures the part of \(S\) aligned with \(T\). When \(Q=0\), define \(L=0\). The CLR statistic is
This expression can be computed stably through an equivalent nonnegative-root representation. Its radicand is nonnegative because
The conditional likelihood-ratio calculation profiles over the nuisance first-stage parameter and conditions on \(T\). In the canonical Gaussian model, the resulting statistic is the larger root of a quadratic equation, and this root is nonnegative. When \(QL>0\), it is the unique positive root; when \(QL=0\), there can be two nonnegative roots. Solving that equation explains the displayed form:
The larger root increases with the evidence in \(A\) and \(L\), while incorporating the observed first-stage information through \(Q\). For \(k=1\) and \(Q>0\), \(L=A\), so \(\operatorname{CLR}=A\): with one instrument, this statistic equals the Anderson–Rubin statistic. With multiple instruments, the alignment of \(S\) and \(T\) matters, and CLR is generally not just \(A\).
Worked Example: One Instrument
Suppose a hypothetical analysis has one instrument and, for a candidate coefficient \(\beta_0\), yields \(S=1.60\) and \(T=2.00\). The canonical model is assumed to hold, including the normal, homoskedastic reduced-form conditions. Here \(k=1\), \(A=S^2\), \(Q=T^2\), and \(L=A\).
Calculate the components. \(A=1.60^2=2.56\), \(Q=2.00^2=4.00\), and \(L=(1.60\times2.00)^2/4.00=10.24/4.00=2.56\).
Calculate CLR. \(\operatorname{CLR}=\tfrac12\{2.56-4.00+\sqrt{(2.56+4.00)^2-4(4.00)(2.56-2.56)}\}\). The radicand is \(6.56^2=43.0336\), so \(\operatorname{CLR}=\tfrac12(-1.44+6.56)=2.56\).
For one instrument, the conditional null distribution is that of \(S^2\), a chi-square distribution with one degree of freedom. The p-value is \(P(\chi^2_1\geq2.56)\approx0.1096\). At significance level \(\alpha=0.05\), this test does not reject \(H_0\). This conclusion concerns the candidate \(\beta_0\); it does not establish that the null is true.
Conditional calibration in practice
For multiple instruments, the conditional p-value is obtained by comparing the observed CLR statistic with its distribution under \(S^*\sim N_k(0,I_k)\), holding the observed \(T\) fixed. In other words, repeatedly draw \(S^*\), calculate the CLR statistic from \((S^*,T)\), and estimate the fraction at least as large as the observed value. Because the null law of \(S\mid T\) is known in the canonical model, numerical integration or conditional simulation can perform this calibration. A simulation-based p-value should report its simulation size and account for Monte Carlo error; the finite-simulation estimate is not itself an exact p-value.
The following example works through the statistic but does not assign a p-value without carrying out that conditional calibration.
Worked Example: Two Instruments and the Alignment Component
Suppose a hypothetical two-instrument analysis gives \(S=(1.20,0.80)'\) and \(T=(2.00,1.00)'\) for a proposed \(\beta_0\). Assume these are the correctly standardized canonical statistics under the stated Gaussian model.
Calculate \(A\) and \(Q\). \(A=1.20^2+0.80^2=1.44+0.64=2.08\), and \(Q=2.00^2+1.00^2=5.00\).
Calculate \(L\). \(S'T=(1.20)(2.00)+(0.80)(1.00)=3.20\), so \(L=3.20^2/5.00=10.24/5.00=2.048\).
Calculate CLR. The expression under the square root is \((2.08+5.00)^2-4(5.00)(2.08-2.048)=7.08^2-20(0.032)=50.1264-0.6400=49.4864\). Its square root is approximately \(7.034657\). Therefore \(\operatorname{CLR}=\tfrac12(2.08-5.00+7.034657)\approx2.057\).
The statistic alone does not determine a decision: its p-value must be calculated from the conditional distribution with this observed \(T\) held fixed. Comparing \(2.057\) with an unconditional chi-square cutoff would not be the CLR procedure.
Why orientation matters, and how CLR compares
For multiple instruments, \(A\) alone does not tell the whole story. If \(S\) points in much the same direction as \(T\), then \(L\) is large; if they are nearly orthogonal, \(L\) is small. The CLR statistic incorporates that relationship. The next example isolates this point while keeping the lengths of both vectors fixed.
Worked Example: Same Lengths, Different Alignment
Consider two hypothetical canonical outcomes with \(k=2\), both having \(S'S=4\) and \(T'T=9\). In the first, \(S=(2,0)'\) and \(T=(3,0)'\), so \(S'T=6\) and \(L=36/9=4\). Then
In the second, \(S=(2,0)'\) and \(T=(0,3)'\). The lengths remain \(A=4\) and \(Q=9\), but \(S'T=0\), so \(L=0\). Now
The CLR statistic therefore distinguishes the two alignments, even though the AR component and first-stage magnitude are identical. To make a test decision in either case, use the corresponding conditional reference distribution with that observed \(T\); a statistic value is not itself a p-value.
| Approach | How it handles the null | Key strength and limitation |
|---|---|---|
| Conventional IV Wald | Uses an estimated coefficient and its standard error with an approximate normal reference distribution. | Convenient when instruments are sufficiently strong and regularity conditions support the approximation; can be unreliable under weak identification. |
| Anderson–Rubin | Tests whether the reduced-form outcome is consistent with a proposed coefficient, using the instruments. | Provides weak-IV-robust testing under its model assumptions, but can have limited power; with one instrument it coincides with CLR here. |
| Conditional likelihood ratio | Uses the CLR statistic and conditions on the observed first-stage statistic for null calibration. | Retains weak-IV-robust calibration in the canonical model and often uses the data more effectively than AR, but requires the model conditions and conditional calibration. |
CLR is not a guarantee of greater power in every sample, nor is it automatically valid under every error structure. Its advantage over conventional Wald inference is that it does not depend on the same strong-first-stage normal approximation. Its advantage over AR is that the conditional likelihood-ratio construction uses first-stage information in a way that can improve power while controlling size under the canonical assumptions. The AR test remains an important alternative, especially when a simpler weak-IV-robust test is preferred.
Diagnostics, limitations, and common mistakes
Before interpreting a CLR result, verify that the statistics were constructed using the appropriate included covariates and reduced-form covariance, and that the chosen conditional reference distribution matches the model. Check the instrument set, sample size, and reduced-form residual behavior for signs of model misspecification. A small \(Q\) is not by itself a reason to discard CLR; the method is designed to accommodate weak first stages. But misspecified exclusion restrictions, invalid instruments, or inappropriate covariance modeling are not repaired by conditioning.
- Gaussian and covariance assumptions: The canonical finite-sample calibration relies on the stated normal, homoskedastic model. Heteroskedasticity or clustering calls for methods with a justification for those settings; applying the basic CLR calibration unchanged does not make it robust.
- Multiple endogenous regressors: The formulas here are for one endogenous regressor. They cannot be transferred unchanged to models with several endogenous regressors.
- Conditional computation: For multiple instruments, a chi-square cutoff applied directly to the CLR statistic is generally not its conditional reference distribution. Use conditional integration, valid critical values, or simulation that holds \(T\) fixed.
- Interpretation: Failure to reject a candidate coefficient is not proof that it is correct. A test addresses compatibility with the null under the assumed model and instrument validity.
Common mistake: “The first-stage statistic is large enough, so the Wald interval must be valid; otherwise CLR cannot help.” This reasoning confuses a diagnostic threshold with a guarantee about the Wald approximation and treats CLR as if it required strong instruments. CLR is designed to allow weak first stages, but only within its assumptions and with its conditional calibration. Instrument validity and the error model still matter.
Check Your Understanding
Answer each question, then compare with the solutions below.
- In the canonical null model, which statistic contains the nuisance first-stage mean, and what distribution does \(S\) have conditional on \(T\)?
- For \(S=(1,0)'\) and \(T=(0,2)'\), calculate \(A,Q,L\), and CLR.
- Why is an unconditional chi-square cutoff generally not the correct calibration for CLR with multiple instruments?
- Name one reason the basic finite-sample CLR result may not apply unchanged to an analysis.
Solutions. (1) \(T\) has mean \(\mu\); under the canonical null, \(S\mid T\sim N_k(0,I_k)\). (2) \(A=1\), \(Q=4\), and \(S'T=0\), so \(L=0\). Hence \(\operatorname{CLR}=\tfrac12\{1-4+\sqrt{(1+4)^2-4(4)(1)}\}=\tfrac12(-3+3)=0\). (3) The null reference law is conditional on the observed \(T\), and its calibration is not generally the same as an unconditional chi-square law. (4) Examples include heteroskedastic errors, clustered dependence, invalid instruments, or multiple endogenous regressors with the one-regressor formula.)