Conditional Likelihood-Ratio Tests for Weak Instruments

Why condition in a weak-instrument test?

When an instrument is weak, the usual large-sample approximation behind a Wald test can be unreliable. The Anderson–Rubin and Kleibergen LM tests provide ways to test structural coefficients without relying on strong identification. The conditional likelihood-ratio (CLR) test is another weak-instrument-robust procedure. Its construction uses reduced-form information about both the outcome and the endogenous regressor, while conditioning on a statistic that carries information about the unknown strength of the first stage.

The focus here is inference on one structural coefficient attached to one endogenous regressor, with one or more excluded instruments. The CLR procedure is not a first-stage diagnostic: it tests a specified structural coefficient. Its conditional reference distribution is what protects the test from depending on an unknown, possibly small, first-stage effect.

Model and assumptions: After removing any included exogenous controls, consider \(y=\beta x+u\), where \(x\) is a scalar endogenous regressor and the excluded instruments form an \(n\times q\) matrix \(Z\), with \(q\geq 1\). The reduced-form errors for \(y\) and \(x\) are assumed jointly Gaussian, independent across observations, and homoskedastic, with a nonsingular covariance matrix. The instruments are exogenous, and the instrument design has full column rank. The canonical CLR calculation uses the corresponding covariance-normalized reduced-form statistics. These assumptions specify the classical homoskedastic CLR setting; they do not justify using its critical values for heteroskedastic errors.

Reduced-form statistics and the CLR statistic

To test \(H_0:\beta=\beta_0\), the reduced form supplies information about the instrument-related variation in both \(y\) and \(x\). After covariance normalization and an orthogonal transformation appropriate to the null value, this information can be represented by two \(q\)-dimensional statistics, \(S\) and \(T\). The normalization is chosen so that, under the null, their canonical distribution has the form

S\sim N(0,I_q),\qquad T\sim N(\mu,I_q),\qquad S\ \text{independent of }T.

Here \(I_q\) is the \(q\times q\) identity matrix and \(\mu\) is an unknown nuisance parameter reflecting reduced-form strength. In words, under the null, \(S\) has a known standard normal distribution, whereas \(T\) contains the nuisance strength information. The precise covariance-normalizing transformation depends on the reduced-form covariance matrix and the null value; \(S\) and \(T\) refer to the canonical statistics after that transformation, not to unscaled sample correlations.

Write \(A=S'S\), \(B=T'T\), and \(C=S'T\). The CLR statistic is

\operatorname{CLR} = \frac{1}{2}\left[ A-B+\sqrt{(A+B)^2-4(AB-C^2)} \right].

The radicand is nonnegative: \(C^2\leq AB\) by the Cauchy–Schwarz inequality, so \((A+B)^2-4(AB-C^2)\geq(A-B)^2\geq0\). An equivalent expression is

\operatorname{CLR} = \frac{1}{2}\left[ A-B+\sqrt{(A-B)^2+4C^2} \right].

This form shows that the statistic depends on the lengths of \(S\) and \(T\), and on their alignment through \(C\). It is nonnegative. A large value is evidence against the tested coefficient. The inner product is signed, but the statistic uses \(C^2\), so changing the sign of either statistic does not change the result.

What the conditioning step does

The nuisance parameter \(\mu\) makes an unconditional reference distribution for a likelihood-ratio statistic depend on unknown instrument strength. The CLR test conditions on the observed \(T\). In the canonical null distribution, \(S\) is independent of \(T\) and has the same \(N(0,I_q)\) distribution for every \(\mu\). Thus, once \(T\) is held fixed, the null distribution of the CLR statistic no longer depends on the nuisance strength parameter.

For a chosen significance level \(\alpha\), let \(c_\alpha(T'T)\) be the upper-tail conditional critical value: under \(H_0\), it is the \(1-\alpha\) quantile of the CLR statistic calculated with \(T\) fixed at its observed value and \(S\sim N(0,I_q)\). Reject \(H_0\) if the observed CLR statistic exceeds this value. Equivalently, compute the conditional p-value

p_{\mathrm{CLR}} = P\!\left( \operatorname{CLR}(S^*,T_{\mathrm{obs}}) \geq \operatorname{CLR}_{\mathrm{obs}} \right), \qquad S^*\sim N(0,I_q),

where the probability is calculated over \(S^*\), holding \(T_{\mathrm{obs}}\) fixed. A numerical implementation can evaluate this conditional probability or its quantile by numerical integration or simulation from \(N(0,I_q)\). For multiple instruments, the cutoff generally depends on \(T_{\mathrm{obs}}'T_{\mathrm{obs}}\); it is not generally a single chi-square critical value.

Why this is a conditional likelihood-ratio test: The likelihood-ratio construction profiles out the nuisance reduced-form strength, while the conditional reference distribution holds fixed the statistic carrying information about that strength. The resulting statistic is compared with a distribution conditional on \(T\), rather than with an unconditional reference distribution that would require knowing \(\mu\). The conditioning is what removes dependence of the null calibration on the unknown first-stage strength in the canonical model.

Worked calculations and decisions

Worked Example: One Instrument and an Exact Conditional Cutoff

Suppose there is one excluded instrument, so \(q=1\), and the canonical statistics are \(S=1.90\) and \(T=0.40\). Test \(H_0:\beta=\beta_0\) at \(\alpha=0.05\).

Here \(A=S'S=1.90^2=3.61\), \(B=T'T=0.40^2=0.16\), and \(C=S'T=(1.90)(0.40)=0.76\). Substitution gives

\begin{aligned} \operatorname{CLR} &=\frac{1}{2}\left[3.61-0.16+ \sqrt{(3.61+0.16)^2-4\{(3.61)(0.16)-0.76^2\}}\right]\\ &=\frac{1}{2}\left[3.45+\sqrt{3.77^2-4(0.5776-0.5776)}\right]\\ &=\frac{1}{2}(3.45+3.77)=3.61. \end{aligned}

For \(q=1\), the formula simplifies to \(\operatorname{CLR}=S^2\), whatever the observed \(T\). Since \(S\sim N(0,1)\) under the null, the conditional statistic has a \(\chi^2_1\) distribution. Its 5% upper-tail cutoff is approximately 3.841, and the p-value is \(P(\chi^2_1\geq3.61)\approx0.0574\). Because \(3.61<3.841\), equivalently \(0.0574>0.05\), this test does not reject \(H_0\) at the 5% level. This is not evidence that \(H_0\) is true; it means the statistic does not cross the specified rejection threshold.

Worked Example: Two Instruments and the Conditional Cutoff

For two instruments, suppose the observed canonical statistics satisfy \(S'S=7.45\), \(T'T=0.40\), and \(S'T=1.18\). At the 5% level, suppose numerical conditional calibration at the observed \(T'T=0.40\) gives a critical value of 5.90. This cutoff is an input from the conditional calibration, not a universal CLR cutoff.

First calculate the determinant-like term in the statistic:

$$ AB-C^2=(7.45)(0.40)-(1.18)^2 =2.98-1.3924 =1.5876. $$

Then the CLR statistic is

\begin{aligned} \operatorname{CLR} &=\frac{1}{2}\left[ 7.45-0.40+ \sqrt{(7.45+0.40)^2-4(1.5876)} \right]\\ &=\frac{1}{2}\left[ 7.05+\sqrt{61.6225-6.3504} \right]\\ &=\frac{1}{2}\left[7.05+\sqrt{55.2721}\right]\\ &\approx\frac{1}{2}(7.05+7.4345) =7.2423. \end{aligned}

Rounded to three significant figures, the statistic is 7.24. Since \(7.2423>5.90\), reject \(H_0\) at the 5% level using the stipulated conditional cutoff. This decision is specific to the stated calibration at \(T'T=0.40\); comparing 7.2423 with a generic \(\chi^2_2\) cutoff would not implement the two-instrument CLR test.

Worked Example: A Smaller Statistic Does Not Reject

Again take \(q=2\). Suppose \(S'S=1.40\), \(T'T=2.00\), and \(S'T=0.50\). Suppose the 5% conditional calibration at \(T'T=2.00\) reports a critical value of 5.85.

The intermediate quantity and statistic are

\begin{aligned} AB-C^2&=(1.40)(2.00)-(0.50)^2=2.80-0.25=2.55,\\ \operatorname{CLR} &=\frac{1}{2}\left[ 1.40-2.00+\sqrt{(1.40+2.00)^2-4(2.55)} \right]\\ &=\frac{1}{2}\left[-0.60+\sqrt{11.56-10.20}\right]\\ &=\frac{1}{2}(-0.60+\sqrt{1.36}) \approx\frac{1}{2}(-0.60+1.1662) \approx0.2831. \end{aligned}

Because \(0.2831<5.85\), this test does not reject \(H_0\) at the 5% level. The two-instrument critical value is again conditional on the observed \(T\); the example uses the stated calibration value to make the decision explicit.

Why the CLR test matters—and a common pitfall

Weak-instrument-robust inference is useful because it can maintain its test size without treating the first-stage estimate as strong. The CLR test uses more of the reduced-form information than a test based only on the null-implied residual, while conditioning to eliminate the nuisance strength from the reference distribution. It therefore offers a distinct route to inference alongside Anderson–Rubin and Kleibergen LM testing. Its advantages do not remove the need to state the model, use the appropriate conditional calibration, or respect the assumptions behind that calibration.

Pitfall: using an unconditional cutoff. A tempting shortcut is to calculate the CLR statistic and compare it with a fixed chi-square cutoff for \(q>1\). That ignores the dependence of the conditional reference distribution on the observed \(T\). Use the conditional critical value or conditional p-value for the observed \(T\).
Pitfall: interpreting non-rejection as acceptance. Not rejecting \(H_0\) does not establish that the tested coefficient equals \(\beta_0\). It means the evidence was insufficient for rejection at the chosen significance level.
Pitfall: treating robustness as assumption-free. Classical CLR calibration relies on the specified reduced-form distribution. The homoskedastic Gaussian version should not automatically be used when errors are heteroskedastic; that setting requires an appropriate robust procedure.
Pitfall: confusing a coefficient test with a strength diagnostic. The CLR test evaluates \(H_0:\beta=\beta_0\). It is not a standalone test that the instruments are strong.

Key takeaway: In the scalar-endogenous-regressor setting, calculate the CLR statistic from \(S'S\), \(T'T\), and \(S'T\), then calibrate it conditional on the observed \(T\). Conditioning removes the unknown first-stage strength from the canonical null reference distribution; it does not make the test assumption-free or give it one universal cutoff for every number of instruments.

Check Your Understanding

Answer each question by referring to the model and conditional calibration described above.

Under the canonical null distribution, which statistic has a standard normal distribution independent of the nuisance strength parameter?
For \(q=1\), why does the CLR statistic simplify to \(S^2\), and what reference distribution follows under the null?
For \(q>1\), why is it generally incorrect to use one fixed chi-square cutoff without conditioning on the observed \(T\)?
If a calculated CLR statistic is below its conditional 5% critical value, what is the correct test decision, and what does that decision not establish?
Name one assumption of the classical homoskedastic CLR calibration that should not be silently carried over to heteroskedastic data.