Introduction: A Diagnostic Is Not a Verdict
A linear instrumental-variables analysis can produce a precise-looking coefficient even when the instruments provide little information about the endogenous regressor. Adding many instruments can make this problem worse: they may fit the sample first stage well while also overfitting noise. A sound analysis therefore checks instrument strength and instrument count, then asks whether the inference remains informative when strength is uncertain.
This workshop uses one endogenous regressor and a set of excluded instruments. Its aim is not to turn a diagnostic statistic into a pass-or-fail rule. Instead, it shows how to organize evidence and how weak-instrument-robust confidence sets can qualify, or overturn, a conventional interval.
The first stage summarizes how much instrument-based variation in \(x_i\) is available. Weak-instrument inference asks whether conclusions about \(\beta\) remain reliable even if that variation is limited. Conventional 2SLS Wald inference can be misleading when instruments are weak; tests such as Anderson–Rubin provide inference that does not require strong first-stage identification, given valid instruments and the stated sampling conditions.
Step 1: Assess First-Stage Strength
In a homoskedastic first-stage regression with an intercept and \(q\) excluded instruments, the conventional joint F statistic tests \(H_0:\pi_1=\cdots=\pi_q=0\). If the unrestricted and restricted first-stage \(R^2\) values are \(R^2_U\) and \(R^2_R\), respectively, then
This statistic tests whether the instruments are jointly relevant in the sample under the homoskedastic linear-model calculation. Its magnitude is informative, but a rule such as “F must exceed 10” is not a universal guarantee. What counts as concerning depends on the design, the number of instruments, the desired control of bias or test size, and the error structure. Under heteroskedasticity, the conventional F calculation is not the appropriate robust diagnostic; use a heteroskedasticity-robust weak-instrument diagnostic, such as the effective F statistic, with its relevant calibration.
Worked Example: A Modest First-Stage F Statistic
Suppose a hypothetical sample has \(n=200\), one endogenous regressor, and \(q=3\) excluded instruments. The restricted first stage includes only an intercept, with \(R^2_R=0\); the unrestricted first stage has \(R^2_U=0.12\). Assume homoskedastic errors for this calculation.
The unrestricted regression estimates an intercept and three instrument coefficients, so the residual degrees of freedom are \(200-4=196\). The conventional joint statistic is
The instruments explain 12% of the sample variation in the first stage, and their joint F statistic is about 8.91. This is a reason to investigate weak-instrument concerns, not a proof that the instruments are unusable or invalid. If heteroskedasticity is plausible, the homoskedastic calculation does not settle the question: a robust diagnostic should also be reported. For example, a separately computed effective F statistic of 6.4 would signal that the robust strength assessment is also concerning; it should not be inferred from the \(R^2\) calculation above.
A first-stage \(R^2\) alone can also mislead. It measures total sample fit, while the F statistic accounts for how many instrument coefficients are being estimated. A high \(R^2\) generated by a large instrument set can coexist with weak average information per instrument.
Step 2: Check the Many-Instrument Risk
When the number of instruments is large relative to the sample size, the first stage can fit idiosyncratic variation in \(x_i\). The fitted values may then reflect noise as well as useful instrument-driven variation. In finite samples, this overfitting can pull 2SLS toward the ordinary least-squares estimate, which is problematic when the regressor is endogenous. The risk depends on the design and instrument quality; the ratio \(q/n\) is a warning indicator, not a complete diagnosis.
Worked Example: Many Instruments Despite a Large First-Stage R-squared
Consider a hypothetical sample with \(n=240\) observations and \(q=48\) excluded instruments. With an intercept, the unrestricted first stage has \(240-49=191\) residual degrees of freedom. Suppose the instruments have a partial first-stage \(R^2\) of 0.30, compared with a restricted \(R^2\) of zero.
The conventional homoskedastic F statistic is
The first stage explains 30% of the variation in the sample, but the joint F statistic is only about 1.71 because that fit is spread across 48 coefficients. Also, \(q/n=48/240=0.20\), so instruments make up a substantial share of the sample size. Both facts warrant scrutiny. Neither number alone proves that 2SLS is biased by a particular amount; the concern is that many estimated first-stage coefficients can overfit, making the usual finite-sample behavior less reassuring.
A useful report should state \(n\), \(q\), the first-stage measure of fit, and an appropriate strength diagnostic. It should also explain how the instruments were selected. Methods designed to reduce many-instrument bias, including jackknife IV, address a different part of the problem; they do not establish instrument validity or remove the need for weak-instrument-robust inference.
The strength statistic and the instrument count answer different questions. A strength diagnostic assesses how much predictive information is available under its calibration. The count and design prompt concern about overfitting and finite-sample behavior. A large first-stage fit does not erase the many-instrument concern, and a small instrument count does not guarantee strong identification.
Step 3: Compare Conventional and Robust Inference
A conventional Wald interval is usually built around the 2SLS estimate and its standard error. Its usual justification relies on sufficiently informative instruments and an approximation in which the estimator behaves approximately normally. Under weak identification, the estimate is a ratio with a potentially unstable denominator, and that approximation can be poor.
The Anderson–Rubin (AR) approach instead tests a candidate value \(b\) for \(\beta\) by asking whether the instruments predict \(y_i-bx_i\). Under \(H_0:\beta=b\), the structural equation implies \(y_i-bx_i=u_i\), which should have no relationship with valid instruments. Inverting the test—collecting every \(b\) not rejected—gives an AR confidence set. Its validity does not depend on the instruments being strongly predictive of \(x_i\), although it still depends on instrument validity and appropriate sampling and variance assumptions.
Worked Comparison: A Wald Interval and an Unbounded AR Set
Worked Example: Inverting a Robust AR Test
Suppose a hypothetical reduced-form analysis gives estimated instrument coefficients \(\hat{\delta}_y=0.40\) for \(y_i\) and \(\hat{\delta}_x=0.20\) for \(x_i\), for a single instrument. Suppose their estimated covariance matrix is
Assume this covariance estimate is heteroskedasticity-robust and that the large-sample normal approximation is appropriate. For a candidate \(b\), the estimated instrument association with \(y_i-bx_i\) is \(\hat{\delta}_y-b\hat{\delta}_x\). Its estimated variance is
With one instrument, the 5% two-sided AR test rejects when the squared standardized statistic exceeds \(3.8416\), the 95th percentile of a chi-squared distribution with one degree of freedom. The inverted 95% confidence set therefore consists of values satisfying
The denominator is a variance and is positive here. Multiplying through and collecting terms gives
The equality has roots approximately \(-0.2347\) and \(0.0883\). Because the coefficient on \(b^2\) is negative, the left-hand side is nonpositive outside the two roots. Thus the 95% AR confidence set is approximately
This unbounded, disconnected set is not a calculation error. It says that, under the stated assumptions, the data do not rule out very large positive or negative values of \(\beta\) at the 5% level. A precise finite interval would overstate what these weakly informative reduced forms establish.
Worked Example: Why the Wald Interval Looks More Decisive
Using the same hypothetical reduced-form estimates, the ratio estimate is \(\hat{\beta}=\hat{\delta}_y/\hat{\delta}_x=0.40/0.20=2.00\). A delta-method approximation uses the gradient \((1/\hat{\delta}_x,\ -\hat{\delta}_y/\hat{\delta}_x^2)=(5,-10)\). With the covariance matrix above, its estimated variance is
The approximate standard error is \(\sqrt{8.5}\approx2.915\), giving the conventional 95% Wald interval
This interval is finite and includes the point estimate, but it uses a local normal approximation to a ratio whose denominator estimate is only \(0.20\). The AR set is much wider in a qualitatively important way: it is unbounded. Under weak identification, the conventional interval should not be presented as the sole reliable summary. The robust set makes the lack of information visible.
Derivation: Why Weak Instruments Can Produce Unbounded Sets
The AR inequality also gives a direct explanation for the set’s shape. For a single instrument, write \(a=\hat{\delta}_y\), \(c=\hat{\delta}_x\), and let \(V_{yy}\), \(V_{xx}\), and \(V_{yx}\) be the estimated variances and covariance of the two reduced-form estimates. The confidence-set condition is
where \(k\) is the chosen chi-squared critical value. Subtracting the right-hand side gives a quadratic inequality in \(b\). Its coefficient on \(b^2\) is \(c^2-kV_{xx}\). If this coefficient is negative, the quadratic opens downward; when it has two real roots, the inequality holds outside the roots, producing two unbounded rays. In the worked comparison, \(c^2=0.04\), while \(kV_{xx}=3.8416(0.09)=0.345744\), so the coefficient is negative. The set’s shape follows from the relationship between the estimated first-stage signal and its uncertainty.
This derivation does not say that every weak-instrument problem yields an unbounded set. Depending on the reduced-form estimates and covariance, inversion can give a bounded interval or another shape. The essential point is that robust inference allows the data to display their limited identifying information rather than forcing a finite interval.
How to Report the Workshop
A defensible report separates diagnostics from inferential conclusions. State the number of instruments and observations, the first-stage statistic and its covariance assumptions, and any robust strength diagnostic used. For a many-instrument design, explain why the instrument count raises overfitting concerns and report any sensitivity analysis or estimator choice that addresses them.
Then report the conventional estimate and interval as such, rather than implying that they are robust to weak identification. Give the weak-instrument-robust test or confidence set, including its shape and units. If the set is unbounded or disconnected, say explicitly that the data do not exclude values in the unbounded regions at the stated confidence level. Neither a favorable first-stage statistic nor a robust confidence set verifies the exclusion restriction: validity remains a substantive assumption.
Common Mistakes
- Treating a cutoff as a proof. “The F statistic exceeds 10, so the instruments are strong” ignores the design, error structure, number of instruments, and inferential goal. Report the statistic and use a calibration appropriate to the setting.
- Equating first-stage fit with strength. A high \(R^2\) can result from fitting many coefficients. Consider the instrument count and a strength statistic, not fit alone.
- Calling strength a validity test. A first stage can be predictive even if an instrument is correlated with the structural error. Relevance and validity are distinct requirements.
- Reporting only a conventional Wald interval. A finite interval can look informative even when the first stage is weak. Compare it with weak-instrument-robust inference.
- Calling an unbounded set a failed calculation. An unbounded confidence set is a legitimate inferential outcome. It communicates that the data do not tightly locate the coefficient under the maintained assumptions.
Check Your Understanding
Answer each question using the definitions and calculations in this workshop.
- Why is a conventional first-stage F statistic of 8.91 a reason for scrutiny rather than a definitive pass-or-fail result?
- In the many-instrument example, how can a first-stage \(R^2\) of 0.30 coexist with an F statistic of about 1.71?
- What does a first-stage diagnostic assess, and what important instrument assumption does it not establish?
- In the reduced-form example, why can the AR confidence set be unbounded even though the ratio estimate is finite?
- How should a report describe an unbounded robust confidence set without claiming that the parameter is known to lie outside its bounded region?