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LIML Under Weak Instrumentation

Learn how LIML’s characteristic root determines its estimate, and why it can reduce—but does not eliminate—the problems caused by weak instruments.

Advanced 9 min read

What You'll Learn

  • Define LIML using the smallest characteristic root of a projected cross-product matrix
  • Derive the LIML coefficient as a k-class estimator
  • Calculate LIML and 2SLS estimates from the same cross-products
  • Explain why weak identification makes LIML nonstandard rather than automatically reliable
  • Identify settings where LIML often reduces finite-sample bias and where that comparison can fail

Why LIML Enters the Weak-Instrument Discussion

Two-stage least squares (2SLS) is familiar, but its finite-sample behavior can be poor when instruments explain little of the endogenous regressor. Limited-information maximum likelihood (LIML) offers a different k-class estimate. In many overidentified designs it has less finite-sample bias than 2SLS, especially when there are many instruments relative to the sample size. That advantage is not a cure for weak identification: under weak instruments, LIML also has a nonstandard sampling distribution and can produce unstable estimates.

The construction below makes the distinction precise. LIML chooses a data-dependent coefficient, determined by a characteristic root, and inserts it into a k-class formula. The characteristic root is part of the estimator; it is not a first-stage strength diagnostic.

Model, Projections, and the LIML Definition

Consider a structural equation with one endogenous regressor \(d\), outcome \(y\), and included exogenous regressors collected in a matrix \(X\). Let \(Z\) contain the excluded instruments. Partial out \(X\) from \(y\), \(d\), and \(Z\), and use the same letters for the resulting residualized variables. Let \(P_Z\) be the orthogonal projection onto the residualized instrument space and let \(M_Z=I-P_Z\). Define \(Q=[d,y]\), so its columns are the residualized endogenous regressor and outcome.

Definition: Suppose \(Q'Q\) and \(Q'M_ZQ\) are positive definite. The LIML characteristic roots are the positive values \(k\) satisfying $$ \det\!\left(Q'Q-kQ'M_ZQ\right)=0. $$ Let \(\hat k\) be the smallest such root. The LIML estimate is the k-class estimate $$ \hat\beta_{\mathrm{LIML}} = \frac{d'(I-\hat k M_Z)y}{d'(I-\hat k M_Z)d}. $$ The estimate is defined when the denominator is nonzero. With included regressors, the corresponding coefficients are recovered by regressing the structural residual on those regressors after estimating the coefficient on \(d\).

The positive-definiteness assumption ensures that the characteristic roots are finite and positive. Since \(M_Z\) is a projection, \(0\preceq M_Z\preceq I\), and therefore \(Q'M_ZQ\preceq Q'Q\). Consequently every characteristic root is at least one. The projection \(I-\hat kM_Z=P_Z+(1-\hat k)M_Z\) gives negative weight to the part orthogonal to the instruments when \(\hat k>1\).

For comparison, the k-class estimate at \(k=1\) is $$ \hat\beta(1) = \frac{d'P_Zy}{d'P_Zd}, $$ which is the 2SLS estimate after partialling out \(X\). LIML differs by choosing \(\hat k\) from the data rather than fixing \(k=1\).

What the Characteristic Root Measures

The smallest-root characterization has a useful geometric interpretation. It selects the direction in the two-dimensional column space of \(Q\) with the largest fraction of squared length lying outside the instrument space. That direction is the one along which the instruments explain the least.

Result: Let \(A=Q'Q\) and \(B=Q'M_ZQ\), with both positive definite. The smallest characteristic root is $$ \hat k=\frac{1}{\lambda_{\max}(A^{-1/2}BA^{-1/2})} =\min_{v\ne 0}\frac{v'Av}{v'Bv}. $$ In particular, \(\hat k\ge 1\).

Proof. The characteristic equation is \(\det(A-kB)=0\). Multiplying on both sides by \(A^{-1/2}\) shows that this equation is equivalent to $$ \det\!\left(I-kA^{-1/2}BA^{-1/2}\right)=0. $$ If \(\lambda_1\ge \lambda_2>0\) are the eigenvalues of the symmetric matrix \(A^{-1/2}BA^{-1/2}\), the roots are \(1/\lambda_1\) and \(1/\lambda_2\). Thus the smallest root is \(1/\lambda_1\). The Rayleigh quotient characterization of the largest eigenvalue gives the stated minimum over \(v\ne0\). Finally, \(B\preceq A\), so every eigenvalue of \(A^{-1/2}BA^{-1/2}\) is at most one. Hence \(\hat k\ge1\). \(\square\)

Worked Example: Finding the LIML Root

Suppose the observed cross-products are $$ Q'Q= \begin{pmatrix} 100&40\\ 40&100 \end{pmatrix}, \qquad Q'M_ZQ= \begin{pmatrix} 20&10\\ 10&20 \end{pmatrix}. $$ Both matrices are positive definite: their eigenvalues are \(140,60\) and \(30,10\), respectively. Find the characteristic roots.

The two matrices share eigenvectors \((1,1)'\) and \((1,-1)'\). Along \((1,1)'\), the ratio \(v'Q'M_ZQv/(v'Q'Qv)\) is \(30/140=3/14\). Along \((1,-1)'\), it is \(10/60=1/6\). The characteristic roots are the reciprocals, \(14/3\) and \(6\). Therefore \(\hat k=14/3\), the smaller root. Both ratios are at most one, as required by the projection inequality.

This calculation determines \(\hat k\), not yet the coefficient estimate. That requires the cross-products in the k-class formula.

Computing the Estimate and Comparing It with 2SLS

For any fixed \(k\), the k-class coefficient can be written directly in terms of the two cross-product matrices: $$ \hat\beta(k)= \frac{Q'Q_{12}-k(Q'M_ZQ)_{12}} {Q'Q_{11}-k(Q'M_ZQ)_{11}}. $$ Here the subscripts identify entries: \(11\) is the regressor’s squared length and \(12\) is the regressor-outcome cross-product. Setting \(k=1\) removes the component orthogonal to the instruments and yields 2SLS. Setting \(k=\hat k\) gives LIML.

Worked Example: Calculating LIML and 2SLS

Use the cross-products in the previous calculation. For 2SLS, $$ \hat\beta_{\mathrm{2SLS}} = \frac{40-10}{100-20} = \frac{30}{80} = \frac{3}{8}. $$ For LIML, substitute \(\hat k=14/3\): $$ \hat\beta_{\mathrm{LIML}} = \frac{40-(14/3)10}{100-(14/3)20} = \frac{-20/3}{20/3} =-1. $$ The denominator is \(20/3\), not zero, so this LIML estimate is defined. In this sample the estimates differ sharply. The calculation illustrates why LIML can be sensitive in weakly identified samples; it does not show that LIML is always farther from the structural parameter or that its expected bias is larger.

A just-identified model provides a useful boundary case. With one excluded instrument for one endogenous regressor, the conventional IV estimate and 2SLS estimate coincide because both use the same single instrument moment. LIML also gives that estimate when defined. In this case, changing the k-class weight does not create additional identifying information. If the instrument is weak, the common estimate can still be highly variable.

Worked Example: A Weak Just-Identified Estimate

Suppose the model is \(y=\beta d+u\), with one excluded instrument \(z\), and the sample moments are \(z'd=0.5\) and \(z'y=1.0\). The IV estimate is $$ \hat\beta_{\mathrm{IV}}=\frac{z'y}{z'd}=\frac{1.0}{0.5}=2. $$ Because the model is just identified, this is also the 2SLS estimate and, when the LIML construction is defined, the LIML estimate.

If the first-stage association is weak, a small change in either sample moment can move the ratio substantially. For example, if \(z'd\) were \(0.25\) with \(z'y\) unchanged, the estimate would be \(1.0/0.25=4\). LIML cannot remove this ratio instability in the just-identified case.

Weak-Instrument Behavior

Under strong identification, the endogenous regressor has a stable component explained by the instruments. The characteristic-root construction then uses the combination of \(d\) and \(y\) that removes the structural signal, and LIML has the usual large-sample behavior of an instrumental-variables estimator under the relevant regularity conditions.

The weak-instrument setting is different. For example, under a local-to-zero sequence the first-stage coefficient may shrink at rate \(1/\sqrt n\), so the instrument signal and sampling noise remain comparable in the reduced-form limit. With a fixed number of instruments and standard joint reduced-form limit conditions, the LIML root and coefficient then depend on random, nondegenerate limit quantities. The resulting estimator generally does not have a conventional normal limit centered at the structural coefficient. As with 2SLS, conventional Wald inference can therefore be unreliable.

The difference is in finite-sample and limiting behavior, not in whether the problem disappears. LIML often has less finite-sample bias toward OLS than 2SLS in overidentified, homoskedastic designs, particularly when the number of instruments is substantial relative to the sample size. But a smaller bias does not imply a smaller variance, and it does not guarantee a stable estimate in every sample. The weak-IV limit of 2SLS and weak-identification-robust procedures such as Anderson–Rubin inference address different questions from LIML’s point-estimation formula.

Worked Example: One Realization Where LIML Is Closer to the Truth

Consider another possible set of cross-products: $$ Q'Q= \begin{pmatrix} 100&40\\ 40&100 \end{pmatrix}, \qquad Q'M_ZQ= \begin{pmatrix} 20&10\\ 10&20 \end{pmatrix}. $$ Suppose the structural coefficient in the data-generating model is \(\beta=0\). The 2SLS estimate is \(3/8=0.375\). As above, the LIML root is \(14/3\), and the LIML estimate is \(-1\), so this particular set of cross-products does not provide an example of LIML being closer. That fact matters: a general bias comparison cannot be inferred from one realization.

For a realization where LIML is closer, instead take $$ Q'Q= \begin{pmatrix} 100&40\\ 40&100 \end{pmatrix}, \qquad Q'M_ZQ= \begin{pmatrix} 20&10\\ 10&20 \end{pmatrix} $$ and let the structural coefficient be \(\beta=-1\). The computed LIML estimate is exactly \(-1\), while 2SLS remains \(0.375\). The absolute errors are \(0\) for LIML and \(1.375\) for 2SLS. This verifies that LIML can be closer in a particular sample, but says nothing by itself about repeated-sample bias. The example is a point-estimate comparison, not a proof of a general bias advantage.

Common Pitfalls and Practical Interpretation

  • Assuming LIML fixes weak identification. LIML changes the estimator, not the information in the instruments. Its weak-instrument distribution is generally nonstandard.
  • Confusing the characteristic root with an F statistic. The root is computed from outcome and endogenous-regressor cross-products. It is not a first-stage strength test.
  • Interpreting lower bias as uniformly better performance. Bias, variance, and tail behavior are distinct. A smaller bias can accompany substantial variability or extreme estimates.
  • Calling a single estimate a bias comparison. Bias is a repeated-sampling expectation, \(E[\hat\beta]-\beta\), when that expectation exists. A pair of estimates from one sample cannot establish which estimator has lower bias.

LIML is especially useful as a point-estimation alternative to 2SLS in overidentified designs where many instruments may make 2SLS biased toward OLS. Its advantage is conditional, not universal. Report the identification context, examine instrument strength, and use inference designed for weak identification when conventional approximations are not credible.

Check Your Understanding

Try these questions before moving on.

  1. Why are the characteristic roots at least one when \(Q'M_ZQ\preceq Q'Q\)?
  2. What value of \(k\) in the k-class formula gives 2SLS?
  3. In the first worked calculation, why does the smaller characteristic root equal \(14/3\) rather than \(6\)?
  4. Why does a weak first stage make a just-identified IV estimate unstable, even when LIML and 2SLS coincide?
  5. Why does a sample in which LIML is closer to the true coefficient not establish that LIML has lower expected bias?
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