The Kleibergen-Paap rk LM Underidentification Test

What the Test Asks

Instrumental-variables identification is a rank question before it is a strength question. With multiple endogenous regressors, the excluded instruments must provide enough independent variation to identify each endogenous regressor after accounting for included exogenous covariates. The Kleibergen-Paap rk LM test evaluates the null that this rank condition fails, using a heteroskedasticity-robust test statistic.

Rejecting that null is useful evidence that the model is not underidentified. It does not establish that the instruments are strongly related to the endogenous regressors, that conventional 2SLS inference will be reliable, or that the instruments satisfy the exclusion restriction. Those are separate questions.

Definition: Let \(D\) be an \(n \times K\) matrix of endogenous regressors, \(Z\) an \(n \times L\) matrix of excluded instruments, and \(W\) the included exogenous covariates. In the reduced-form regression of \(D\) on \(W\) and \(Z\), let \(\Pi\) be the \(L \times K\) coefficient matrix on \(Z\). The order condition requires \(L \ge K\); the rank condition for identification is \(\operatorname{rank}(\Pi)=K\). Underidentification is the null hypothesis \(H_0:\operatorname{rank}(\Pi)<K\).

The coefficients in \(\Pi\) describe the instruments’ conditional relationships with the endogenous regressors, holding the included covariates fixed. The full-rank condition requires more than a nonzero first-stage relationship somewhere: the instrument-induced variation must span all \(K\) endogenous-regressor directions. A model may therefore have several relevant instruments overall and still fail the rank condition for a particular combination of endogenous regressors.

Null Hypothesis, Degrees of Freedom, and Construction

The Kleibergen-Paap rk LM test is a Lagrange-multiplier (LM), or score, test of underidentification. It assesses whether the data provide evidence against the rank-deficient null. Its heteroskedasticity-robust version uses a robust covariance estimate in constructing the score statistic; it is not simply an ordinary first-stage F statistic with a different label.

Under the null of rank deficiency, the relevant boundary is rank \(K-1\): ranks below that are also underidentified, while the largest deficient rank is the closest rank-deficient case to full identification. Under the usual regularity conditions, the test has an asymptotic chi-square reference distribution with \(L-K+1\) degrees of freedom. The degrees of freedom depend on the numbers of excluded instruments and endogenous regressors, not on the number of included covariates.

df=L-K+1,\qquad p=P\!\left(\chi^2_{L-K+1}\geq \text{rk LM statistic}\right)

A useful way to understand the degrees of freedom is to count the restrictions imposed by rank deficiency. An \(L \times K\) matrix of rank \(r\) has dimension \(r(L+K-r)\). At the largest deficient rank \(r=K-1\), that dimension is \((K-1)(L+1)\). The unrestricted \(L \times K\) matrix has \(LK\) entries, so the number of restrictions is

$$ LK-(K-1)(L+1)=L-K+1. $$

The robust score test evaluates whether the reduced-form coefficient matrix can be treated as rank deficient, accounting for the estimated covariance of the score. In general, the reported Kleibergen-Paap statistic is obtained from the test’s rank-restricted score construction; it should be taken from a procedure that implements the test rather than reconstructed as an arbitrary combination of separate first-stage statistics.

Interpretation: A small p-value is evidence against \(H_0:\operatorname{rank}(\Pi)<K\), under the test’s assumptions and asymptotic approximation. It supports the conclusion that the rank condition is satisfied; it does not measure how far the model is from underidentification or establish strong instruments.

Worked Interpretations

Worked Example: One Endogenous Regressor and Two Instruments

Consider a hypothetical cross-sectional model with \(n=600\) observations, one endogenous regressor (\(K=1\)), and two excluded instruments (\(L=2\)), along with included exogenous covariates. A heteroskedasticity-robust IV procedure reports a Kleibergen-Paap rk LM statistic of 8.40. Test underidentification at \(\alpha=0.05\).

The degrees of freedom are

$$ df=L-K+1=2-1+1=2. $$

For a chi-square variable with 2 degrees of freedom, the upper-tail probability at 8.40 is \(e^{-8.40/2}\). Thus,

p=P(\chi^2_2\geq 8.40)=e^{-4.20}\approx 0.0150.

Because \(0.0150<0.05\), reject the underidentification null at the 5% significance level. In this hypothetical model, the result is evidence that the two instruments satisfy the rank condition for the single endogenous regressor. It does not show that their first-stage relationship is strong: even a nonzero relationship can be weak.

Worked Example: Two Endogenous Regressors

Suppose a hypothetical analysis has \(n=900\) observations, two endogenous regressors (\(K=2\)), and three excluded instruments (\(L=3\)). The robust rk LM statistic is 1.20. The model includes exogenous controls, and the stated statistic is calculated conditional on them.

The asymptotic degrees of freedom are

$$ df=3-2+1=2. $$

Therefore,

p=P(\chi^2_2\geq 1.20)=e^{-1.20/2}=e^{-0.60}\approx 0.5488.

Since \(0.5488>0.05\), do not reject the underidentification null at the 5% level. The data do not provide sufficient evidence against rank deficiency by this test. This is not proof that the model is underidentified: low power, limited sample information, or a weak relationship may contribute to a non-rejection. The finding calls for careful assessment rather than a claim that the rank is definitely deficient.

Worked Example: Robust and Nonrobust Conclusions Differ

In a hypothetical study with \(n=500\), two endogenous regressors and three excluded instruments, the degrees of freedom are \(3-2+1=2\). Suppose an analysis reports a conventional, nonrobust rank-test statistic of 8.90 and a heteroskedasticity-robust Kleibergen-Paap rk LM statistic of 5.10. For illustration, calculate the upper-tail probabilities using the two-degree-of-freedom chi-square reference distribution.

For the robust statistic,

p_{\text{robust}}=P(\chi^2_2\geq 5.10) =e^{-5.10/2}=e^{-2.55}\approx 0.0781.

For the nonrobust statistic,

p_{\text{nonrobust}}=P(\chi^2_2\geq 8.90) =e^{-8.90/2}=e^{-4.45}\approx 0.0117.

At \(\alpha=0.05\), the robust test does not reject underidentification, whereas the nonrobust test does. If heteroskedasticity is plausible, the robust result is the relevant one among these two calculations; the difference reflects how the estimated score covariance affects the test. It is not appropriate to choose whichever result is more favorable. Neither result directly answers whether the instruments are strong.

Assumptions and Diagnostics

The test’s interpretation depends on the model and the covariance estimator being appropriate. In particular, the included covariates and excluded instruments must be specified consistently in the reduced forms, and the statistic must use the intended set of instruments and endogenous regressors. The asymptotic chi-square reference also relies on regularity conditions, such as a sufficiently informative sample and a suitable covariance estimate.

The heteroskedasticity-robust version allows the conditional error variance to vary across observations, under the relevant large-sample conditions. That protection is not universal. It does not automatically account for dependence among observations, such as within-cluster correlation; a covariance estimator suited to the sampling dependence is needed. It also does not repair a misspecified model, invalid instruments, or an inappropriate sampling design.

Before interpreting a reported result, check the following:

Confirm \(L\) and \(K\) from the excluded instruments and endogenous regressors actually included in the model. Verify that \(L\geq K\); otherwise the order condition fails.
Confirm that the reported statistic is the Kleibergen-Paap rk LM underidentification statistic, not the Kleibergen-Paap rk Wald F statistic or a conventional first-stage test.
Check whether the covariance estimator matches the plausible error structure. A heteroskedasticity-robust covariance estimate alone is not a cluster-robust one.
Read the statistic together with its degrees of freedom and p-value. The numerical statistic by itself has no universal rejection threshold.
Assess strength separately, using diagnostics intended for instrument strength and for the model’s number of endogenous regressors. Do not treat rejection of underidentification as a strength certificate.

How It Differs From Strength and Validity Checks

Underidentification and weak identification are related but distinct. Underidentification means the reduced-form coefficient matrix lacks full column rank. Weak identification means the relevant directions may technically have full rank but are estimated imprecisely or have very little identifying variation. The rk LM test targets the rank-deficient null; its rejection does not quantify the size of the smallest identifying signal.

The Kleibergen-Paap rk Wald F statistic, discussed in the earlier material on weak-instrument diagnostics, is a different statistic used in assessing strength. The two should not be substituted for one another: the rk LM test is an underidentification test with a chi-square p-value, while the rk Wald F statistic addresses first-stage strength. Critical values developed for a different statistic do not automatically apply to the rk LM test.

Nor does a rank test establish instrument validity. The test does not determine whether excluded instruments are independent of the structural error or whether they affect the outcome only through the endogenous regressors. A model can pass the rank test and still have invalid instruments. Conversely, weak or inconclusive rank-test evidence is not itself proof that the exclusion restriction fails.

Common pitfall: “The rk LM test rejects underidentification, so the instruments are strong and 2SLS is safe.” The conclusion goes too far. Rejection is evidence against deficient rank, not evidence that the instruments have enough strength for reliable conventional inference. Follow it with a suitable strength assessment and evaluate instrument validity using substantive design arguments and appropriate diagnostics.

Why the Distinction Matters

A rank test is important because estimation and inference require the instruments to identify the endogenous regressors. But a binary reject-or-not-reject decision compresses information: it does not report the magnitude of the weakest identifying direction, and its p-value depends on sample size as well as the underlying relationship. With a large sample, a small but nonzero association can be statistically detectable without being strong enough to make ordinary 2SLS inference dependable.

The reverse limitation matters too. Failure to reject does not establish that the model is underidentified. A test can fail to reject because the sample contains little information, including when the underlying rank is full but the evidence is imprecise. Report the statistic, degrees of freedom, p-value, covariance estimator, and instrument/endogenous-regressor counts, then describe the conclusion as evidence for or against underidentification rather than as an absolute declaration.

Specify the rank problem.
Identify the excluded instruments, endogenous regressors, and included exogenous covariates. The target is whether the \(L \times K\) coefficient matrix \(\Pi\) has rank below \(K\).

Check the test setup.
Verify \(L\geq K\), select a covariance estimator suited to the data dependence, and confirm that the procedure reports the robust rk LM statistic.

Interpret the reference test.
Use \(df=L-K+1\) and compare the reported p-value with the prespecified \(\alpha\). A rejection is evidence against underidentification; a non-rejection is inconclusive rather than proof of deficient rank.

Continue the diagnostic assessment.
Evaluate strength separately and consider instrument validity and the substantive design. The rank-test result alone does not settle these questions.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

A model has \(K=2\) endogenous regressors and \(L=4\) excluded instruments. What are the rk LM test’s asymptotic degrees of freedom?
A test reports \(df=2\) and an rk LM statistic of 6.00. Calculate the chi-square upper-tail p-value and state the decision at \(\alpha=0.05\).
What does a rejection of underidentification establish, and name one conclusion it does not establish.
Why can a heteroskedasticity-robust test differ from a conventional nonrobust test?
A model fails to reject underidentification. Is that proof that the coefficient matrix has rank less than \(K\)? Explain.

Solutions. (1) \(df=4-2+1=3\). (2) For 2 degrees of freedom, \(p=e^{-6.00/2}=e^{-3}\approx0.0498\), so reject at the 5% level, narrowly. (3) It is evidence against a rank-deficient coefficient matrix; it does not establish strong instruments or instrument validity. (4) The robust test uses a covariance estimate that allows heteroskedasticity, which can change the score statistic and its p-value. (5) No. Non-rejection means the test did not provide sufficient evidence against the null; limited power or imprecise evidence can produce the same result.