Worked Weak-Instrument Analysis of a Just-Identified Model

Why Work Through a Numerical Example?

In a just-identified instrumental-variables model, one instrument is used for one endogenous regressor. The estimator is easy to express as a ratio of two reduced-form estimates. That simplicity can be misleading: when the first stage is weak, the ratio can be highly variable, and conventional Wald inference can give a very different impression from inference based on the Anderson–Rubin test.

This worked analysis follows one numerical example from first-stage diagnosis to confidence-set interpretation. The calculations use a large-sample normal approximation for the reduced-form estimates. They illustrate the contrast between the methods; they do not establish that an instrument is valid. Instrument relevance, instrument exogeneity, and the exclusion restriction are distinct assumptions.

Model and Numerical Summary

Consider the structural equation \(y_i=\alpha+\beta x_i+u_i\), where \(x_i\) is endogenous, and use an instrument \(z_i\). The first-stage equation relates \(x_i\) to an intercept and \(z_i\). There is one excluded instrument and one endogenous regressor, so the model is just identified. For the calculations below, the intercept is included and all reported reduced-form estimates are understood to use the same sample.

Let \(\hat\pi\) be the estimated first-stage coefficient of \(z_i\) in the regression for \(x_i\). Let \(\hat\gamma_y\) be the reduced-form coefficient of \(z_i\) in the regression for \(y_i\). In a just-identified model, the IV estimate is the ratio \(\hat\gamma_y/\hat\pi\), provided \(\hat\pi\ne 0\).

Suppose \(n=100\), with no included regressors other than an intercept. The first-stage estimate is \(\hat\pi=0.20\), with standard error \(0.10\). The reduced-form estimates and their estimated covariance matrix are

Thus the reduced-form standard errors are \(0.12\) and \(0.10\), and the estimated correlation is \(0.006/(0.12\times0.10)=0.5\). The covariance matrix is positive definite: its determinant is \(0.0144(0.0100)-0.0060^2=0.000108>0\). This summary will be used consistently in both forms of inference.

First-Stage Strength

An F statistic of 4 is a warning that the first stage is not strong in this example. It is useful to report the number and the partial R-squared rather than converting either into a claim that identification has been established or disproved. A weak first stage matters especially because the IV estimate divides by \(\hat\pi\).

Conventional Wald Inference

The just-identified IV estimate is the ratio of the reduced-form coefficients:

A conventional Wald interval uses an approximately normal distribution for the estimator. Applying the delta method to the ratio \(g(a,b)=a/b\) gives gradient \((1/b,-a/b^2)\). Evaluated at the estimates, this is \((1/\hat\pi,-\hat\beta_{\mathrm{IV}}/\hat\pi)\). Therefore the estimated variance is

This expression uses both variances and their covariance; omitting the covariance would not be the delta-method variance for the ratio. In the example, the numerator is \(0.0144+(1.50)^2(0.0100)-2(1.50)(0.0060)=0.0189\). Hence

Anderson–Rubin Test and Inversion

For a proposed structural value \(\beta_0\), form the outcome \(y_i-\beta_0x_i\) and test whether the instrument's reduced-form coefficient for that outcome is zero. In terms of the estimates above, that coefficient is \(\hat\gamma_y-\beta_0\hat\pi\). Its estimated variance is obtained from the same covariance matrix:

The Anderson–Rubin statistic in this one-instrument example is the absolute value of that coefficient divided by its standard error. Under the null hypothesis that \(\beta=\beta_0\), and assuming the instrument is valid, the reduced-form coefficient being tested is zero. If the reduced-form estimates are approximately jointly normal with the stated covariance, this standardized linear combination is approximately standard normal under the null. Rejecting when its absolute value exceeds \(1.96\) gives an approximately 5% two-sided test.

This explains the weak-instrument robustness of the test: its null statistic tests a reduced-form restriction and does not divide by \(\hat\pi\). Under the stated approximation, its null calibration does not require a strong first stage. Inverting the test means retaining every value \(\beta_0\) that is not rejected.

What the Comparison Tells Us

The Anderson–Rubin set is much wider here. That width is not a computational defect: it reflects the limited information in a weak first stage about the structural coefficient. In other data, inversion can produce an unbounded or disconnected set; the shape follows from the test inequality and should be reported rather than forced into the form of a familiar symmetric interval.

Neither calculation establishes the causal interpretation of \(\hat\beta_{\mathrm{IV}}\). A weak-instrument-robust test addresses the reliability of inference about \(\beta\) under the model's assumptions; it does not repair an invalid instrument. Likewise, a first-stage diagnostic describes relevance, not exogeneity. The substantive justification for the instrument remains essential.

Common Mistake

Wrong reasoning: “The first-stage t statistic is 2, so the instrument is strong enough; the Wald interval is narrower, so it must be the more informative answer.” This treats one diagnostic as a pass/fail guarantee and mistakes narrowness for reliable precision.

Correction: The first-stage evidence here is \(F=4\) and partial \(R^2\approx0.0392\). These numbers indicate limited relevance in the sample, so the ratio-based Wald approximation deserves caution. The Anderson–Rubin set, calculated by testing each candidate value without dividing by the first-stage estimate, reveals that uncertainty is much larger. Report the method and assumptions, and interpret the contrast rather than selecting the narrower result by default.