Why Conventional IV Wald Inference Fails

Why a Conventional Wald Result Can Mislead

A conventional 2SLS output may show a coefficient, a standard error, and a familiar Wald confidence interval. Those calculations are useful when the instrument is sufficiently informative about the endogenous regressor. When that information is weak, however, the usual approximation behind the test and interval can fail: a nominal 5% test need not reject a true null only 5% of the time, and a nominal 95% interval need not cover the true effect 95% of the time.

This is not the same as saying that every weak-instrument estimate is imprecise, or that every conventional interval is too narrow. Weak instruments make the estimator's sampling distribution difficult for ordinary Wald calculations to approximate. In some samples, the reported standard error is large; in others, the estimate and interval can look deceptively decisive. The central issue is that the usual reference distribution may be wrong.

Setup: The IV Ratio and Its Usual Approximation

For clarity, consider one endogenous regressor \(D\), one instrument \(Z\), and no additional covariates. With covariates, the same argument applies after accounting for them in the first-stage and structural equations. Write the structural equation and first stage as

Y_i=\beta D_i+u_i,\qquad D_i=\pi Z_i+v_i.

Here \(\beta\) is the structural effect of interest, \(\pi\) describes how the instrument predicts \(D\), and \(u_i\) and \(v_i\) may be correlated. Instrument validity requires that \(Z_i\) be exogenous to the structural error and excluded from the outcome equation except through \(D_i\). Instrument relevance requires \(\pi\ne 0\) in the population. These are distinct requirements: a strong instrument can still be invalid, and a valid instrument can be weak.

Let \(\hat{\pi}\) be the estimated first-stage coefficient from regressing \(D\) on \(Z\), and let \(\hat{\gamma}\) be the coefficient from regressing \(Y\) on \(Z\), the reduced-form coefficient. In this just-identified model, 2SLS is the ratio of those estimates:

\hat{\beta}_{2SLS}=\frac{\hat{\gamma}}{\hat{\pi}}.

If the instrument is valid, the population reduced-form coefficient is \(\gamma=\pi\beta\). With a sufficiently strong first stage, standard large-sample theory treats \(\hat{\pi}\) as concentrated near a nonzero value. The ratio can then be approximated by a linear expansion around \((\gamma,\pi)\), yielding an approximately normal estimator and the familiar Wald statistic.

Key condition: Conventional 2SLS Wald inference relies on the first-stage signal not fading toward zero as the sample grows. Under that strong-identification condition, the estimator is approximately normal and its estimated standard error can support usual Wald tests and confidence intervals. Instrument validity alone does not establish this condition.

What Changes When the Instrument Is Weak

A useful way to describe weak identification is a sequence of samples in which the first-stage effect shrinks as the sample size increases. For the simple model, write \(\pi_n=c/\sqrt{n}\), where \(c\) is a fixed constant. This is called a local-to-zero or weak-instrument asymptotic sequence. The first-stage signal does not become large relative to its sampling noise.

Under standard regularity conditions, the estimates then have joint limiting behavior of the form

\sqrt{n}\hat{\pi}\ \Rightarrow\ c+V,\qquad \sqrt{n}\hat{\gamma}\ \Rightarrow\ \beta c+U,

where \(U\) and \(V\) are jointly distributed limiting random variables. They are generally correlated because the structural and first-stage errors may be correlated. Substituting these limits into the ratio gives

\hat{\beta}_{2SLS}\ \Rightarrow\ \frac{\beta c+U}{c+V}.

This is a ratio of random variables, not the usual normal approximation centered at \(\beta\). The denominator can land close to zero, making the ratio very large in either direction. In this weak-instrument sequence, the estimator need not converge to \(\beta\), even as \(n\) increases. A conventional Wald statistic that treats the estimate as approximately normal and uses a standard normal or \(t\) reference distribution therefore has no general reason to maintain its nominal rejection rate.

Worked Examples: Calculating and Interpreting the Failure

Worked Example: A Decisive Wald Test With a Weak First Stage

Suppose a hypothetical analysis reports a first-stage coefficient \(\hat{\pi}=0.10\) with standard error \(0.08\). Its first-stage \(t\)-statistic is

t_{\text{first stage}}=\frac{0.10}{0.08}=1.25.

The estimated reduced-form coefficient is \(\hat{\gamma}=0.20\), so the just-identified 2SLS estimate is

\hat{\beta}_{2SLS}=\frac{0.20}{0.10}=2.00.

Assume, for illustration, that the conventional 2SLS output gives \(SE=0.60\). Testing \(H_0:\beta=0\) with a large-sample Wald statistic gives

z=\frac{2.00-0}{0.60}=3.33.

The two-sided standard-normal p-value is about \(0.0009\), and the nominal 95% Wald interval is \(2.00\pm1.96(0.60)\), or \([0.824,3.176]\). Those calculations would ordinarily be read as strong evidence against zero and an interval excluding zero. But the first-stage \(t\)-statistic of 1.25 signals little estimated instrument-to-treatment association. The arithmetic is correct; the weak point is the usual Wald reference approximation. This output alone does not justify the claimed rejection rate or coverage.

Worked Example: A Noisy Denominator Can Amplify the IV Ratio

Consider hypothetical first-stage estimates with approximate sampling distribution \(\hat{\pi}\sim N(0.05,0.05^2)\). The signal-to-noise ratio is \(0.05/0.05=1\). Under this approximation,

P(\hat{\pi}<0)=P\left(\frac{\hat{\pi}-0.05}{0.05}<-1\right)\approx0.1587.

Thus, the estimated first-stage coefficient is negative in about 15.9% of repeated samples under this approximation, despite a positive population coefficient. More importantly for the IV ratio, estimates near zero are quite plausible. Suppose one sample has \(\hat{\pi}=0.01\) and \(\hat{\gamma}=0.06\). Then

\hat{\beta}_{2SLS}=\frac{0.06}{0.01}=6.

A small change in the denominator can alter the ratio sharply: holding \(\hat{\gamma}=0.06\) fixed, a first-stage estimate of \(0.02\) gives \(0.06/0.02=3\), while \(0.005\) gives \(0.06/0.005=12\). In actual sampling, numerator and denominator are generally dependent, so these calculations are not a complete sampling distribution. They show why a weak first stage makes ratio estimates unstable and ordinary normal approximations unreliable.

Worked Example: A Small Reported Standard Error Is Not a Guarantee

Suppose a hypothetical sample has \(n=400\), an estimated first-stage coefficient \(\hat{\pi}=0.10\), and a conventional residual standard deviation estimate of \(1.20\). In the simple one-instrument model with an intercept and a standardized instrument, the fitted treatment variation is approximately \(\hat{\pi}^2\sum_i(Z_i-\bar Z)^2\). If \(\sum_i(Z_i-\bar Z)^2\approx400\), the conventional homoskedastic standard-error calculation is approximately

SE(\hat{\beta}_{2SLS})\approx \frac{1.20}{\sqrt{400(0.10)^2}} =\frac{1.20}{2} =0.60.

If the point estimate is \(2.00\), the conventional interval is again \(2.00\pm1.96(0.60)=[0.824,3.176]\). This interval looks fairly narrow relative to the point estimate. Yet the first-stage estimate is only \(0.10\); if its standard error were \(0.08\), the first-stage \(t\)-statistic would be \(1.25\). A standard error computed from the realized fitted variation does not, by itself, capture the unusual ratio behavior under weak identification. Nor does this example imply that weak instruments always produce small standard errors: a weak first stage often inflates them. The point is that neither a small nor a large conventional standard error alone certifies Wald validity.

Derivation: Why the Usual Wald Reference Distribution Breaks

The failure follows directly from the ratio representation. Under strong identification, \(\hat{\pi}\) converges to a fixed nonzero \(\pi\). A first-order expansion of \(g(\gamma,\pi)=\gamma/\pi\) around \((\gamma,\pi)\) is reasonable because the denominator remains away from zero. Its derivatives are \(1/\pi\) with respect to \(\gamma\) and \(-\gamma/\pi^2\) with respect to \(\pi\). This produces an approximately linear combination of the reduced-form and first-stage estimation errors, which under standard conditions is approximately normal.

Under the local-to-zero sequence, however, \(\pi_n=c/\sqrt n\) and the estimated denominator has the same order of magnitude as its sampling error. The ratio's derivative with respect to the denominator, \(-\gamma/\pi^2\), is not controlled as \(\pi\) approaches zero. Applying the limiting expressions above makes the problem explicit:

\hat{\beta}_{2SLS}-\beta \ \Rightarrow\ \frac{U-\beta V}{c+V}.

The right-hand side is generally nonnormal and does not collapse to zero. A standard Wald test divides the estimate's distance from the null by a conventional estimated standard error and compares the result with a familiar normal or \(t\) cutoff. That calibration is derived for a different, strong-identification approximation. Under weak identification, the resulting test can reject a true null too often or too rarely; the direction and size of the error depend on the design and assumptions. Its nominal confidence interval can consequently have coverage far from its stated level.

Diagnostics, Alternatives, and Limits

First-stage strength diagnostics are important warning signals. The earlier material on partial \(R\)-squared, concentration, first-stage statistics, robust strength tests, and first-stage plots helps assess whether the instrument predicts the endogenous regressor and whether a conventional strength rule is informative in the design. Use the diagnostic appropriate to the model, number of instruments, error structure, and clustering. A strength diagnostic is not a test of instrument validity, and passing a rule of thumb does not prove that a Wald interval has its advertised coverage.

Several tempting responses do not solve the problem. Heteroskedasticity-robust or cluster-robust standard errors address particular variance misspecifications; they do not, by themselves, restore the strong-identification approximation. Overidentification checks concern restrictions involving instrument validity, not whether the instruments provide enough information for conventional Wald inference. Replacing IV with OLS is not a general remedy either: when \(D\) is endogenous, OLS may be biased for \(\beta\).

Weak-instrument-robust methods, including the Anderson–Rubin test and the conditional likelihood ratio approach, are designed to avoid relying on the same conventional Wald approximation. They have their own assumptions, power properties, and implementation choices. They do not repair an invalid instrument or automatically guarantee a narrow, informative interval. Their detailed construction is a separate subject.

A practical assessment therefore separates three questions: Is the instrument valid on substantive and design grounds? Is its first-stage information adequate for the intended analysis? And does the chosen inferential procedure remain valid at the strength actually supported by the data? Weak-IV problems matter because a conventional coefficient table can answer the last question incorrectly while appearing entirely routine.

Common Pitfall

Wrong reasoning: “The 2SLS coefficient is many standard errors from zero, so the result is decisive, even though the first stage is weak.”
Correction: The Wald statistic is decisive only if its reference distribution is appropriate. With a weak first stage, the ratio estimator can be nonnormal and unstable, so the nominal p-value and confidence interval may not have their advertised error rates.
Wrong reasoning: “A robust standard error, a strong validity argument, or a favorable overidentification result fixes weak identification.”
Correction: These address different issues. Robust variance estimation concerns the error structure; validity concerns whether the instrument is exogenous and excluded; and overidentification tests do not establish sufficient first-stage strength.

Key takeaway: Conventional 2SLS Wald inference is justified by strong identification, not merely by instrument validity or the fact that software reports a standard error. When the first stage is weak, evaluate the strength evidence and use inference designed to remain reliable under weak identification.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

In the just-identified model, what two estimated coefficients form the 2SLS ratio?
Why can a weak first stage make the 2SLS estimator unstable even when the instrument is valid?
A reported 2SLS estimate is 1.5 with a conventional standard error of 0.4. Why is the resulting Wald statistic not automatically trustworthy if the first stage is weak?
Does using a heteroskedasticity-robust standard error alone resolve weak identification? Explain.

Solutions. (1) The estimator is \(\hat{\beta}_{2SLS}=\hat{\gamma}/\hat{\pi}\), the reduced-form coefficient divided by the first-stage coefficient. (2) A weak first-stage estimate can be close to zero relative to its sampling noise; division by a noisy, small denominator can produce a highly variable, nonnormal ratio. (3) The conventional Wald calculation is \(1.5/0.4=3.75\), but its usual reference distribution relies on a strong-identification approximation. If that approximation fails, the p-value and interval may have incorrect error rates despite correct arithmetic. (4) No. A heteroskedasticity-robust standard error addresses variance estimation under heteroskedasticity, but does not by itself make conventional Wald inference valid with a weak instrument.