Why Conventional 2SLS Wald Inference Fails

Why Strong-Instrument Wald Inference Can Fail

Conventional 2SLS Wald inference relies on the first stage being strong enough that the estimator is approximately normal after scaling by its estimated standard error. Under weak identification, the first-stage signal can be the same order as its sampling noise. The 2SLS estimator then behaves like a ratio of random quantities, rather than concentrating around the structural coefficient. Its usual residual-based standard error can also remain random in the limit. As a result, the familiar normal and chi-square reference distributions need not apply.

The issue is not simply that a weak first stage makes the standard error large. The estimator and its estimated standard error are both affected by the same random first-stage denominator. Their ratio need not have the distribution assumed by the conventional t or Wald test.

Scalar IV Model and Local Weakness

Consider a scalar structural equation and first stage, with no intercept:

The instrument \(z_i\) is valid: it is independent of \((u_i,v_i)\), and the errors have mean zero. They may be correlated with each other, so \(x_i\) may be endogenous. For clarity, take \(z_i\) to be independent Rademacher variables, with \(P(z_i=1)=P(z_i=-1)=1/2\). Thus \(n^{-1}\sum_i z_i^2=1\). Assume the observations are independent and identically distributed, the error covariance matrix is positive definite, and the moments needed for the central limit theorem and laws of large numbers below are finite.

The local-to-zero sequence \(\pi_n=c/\sqrt n\), for a fixed constant \(c\), describes a first stage whose signal does not dominate its sampling noise as \(n\) grows. Define

The joint central limit theorem gives \((U_n,V_n)\xrightarrow{d}(U,V)\), where \((U,V)\) is mean-zero bivariate normal with covariance matrix

What the Weak-IV Limit Implies for the Estimator

Under the null \(\beta_0=\beta\), the estimator’s error and its denominator have a direct expression in the reduced-form sums. Since \(z_i^2=1\), \(\sum_i z_i x_i=c\sqrt n+\sum_i z_i v_i=\sqrt n(c+V_n)\). The corresponding numerator for \(\hat\beta-\beta\) is \(\sum_i z_i u_i=\sqrt n U_n\). Therefore

The denominator of the limit is nonzero with probability one because \(V\) has a continuous normal distribution. But it can be arbitrarily close to zero with positive probability. Consequently, this limit is not generally normal and does not collapse to zero: increasing \(n\) does not make the 2SLS estimator concentrate around \(\beta\) along this sequence.

Deriving the Conventional t Limit

Under the null, the residual from the fitted structural equation is \(y_i-\hat\beta x_i=u_i-(\hat\beta-\beta)x_i\). Write \(\delta_n=\hat\beta-\beta\) and \(S_{ab}=n^{-1}\sum_i a_i b_i\). Because \(\delta_n\) is bounded in probability and \(\pi_n\to0\), the first-stage term \(\pi_n z_i\) vanishes in the residual sum of squares per observation. The law of large numbers for \(u_i^2,u_iv_i,v_i^2\), together with the continuous mapping theorem for the random coefficient \(\delta_n\), gives the joint limit

The covariance matrix of \((u_i,v_i)\) is positive definite, so \(Q(\delta)>0\) for every real \(\delta\). The conventional standard error has the exact form \(\sqrt{\hat\sigma^2 n}/|\sum_i z_i x_i|\). Substituting the expressions for the numerator and denominator of the estimator error yields the t statistic’s limit.

Proof. By the estimator calculation, \(\delta_n=U_n/(c+V_n)\xrightarrow{d}U/(c+V)=\delta\). Also, \(n^{-1/2}\sum_i z_i x_i=c+V_n\). Using \(z'z=n\), the standard error is \(\sqrt{\hat\sigma^2 n}/|\sum_i z_i x_i|\). Hence

The joint convergence of \((U_n,V_n,\hat\sigma^2)\) to \((U,V,Q(\delta))\), and the fact that \(c+V\ne0\) and \(Q(\delta)>0\) almost surely, allow the continuous mapping theorem to be applied. This gives the stated limit. Since the t limit need not be normal, its square need not have the conventional chi-square reference distribution. \(\square\)

Confidence-Interval Coverage and a Useful Caution

A conventional two-sided interval based on a critical value \(q\) is \(\hat\beta\pm q\,\widehat{\operatorname{se}}(\hat\beta)\). Under the true coefficient, it covers precisely when \(|t|\le q\). If the t statistic has a nonstandard limit, the limiting coverage is \(P(|T|\le q)\), not automatically the nominal probability associated with a standard normal critical value.

A useful caution is that not every simplified Wald calculation fails in every special case. For example, if one substitutes the true structural error variance rather than estimating it from the 2SLS residuals, the exact-identified statistic can simplify through cancellation of the first-stage denominator. That special calculation is not the residual-based statistic derived here, and it does not justify treating conventional 2SLS inference as generally valid under weak identification. The relevant limit depends on the statistic, the variance estimator, and the weak-instrument sequence.

The practical point is that a conventional standard error and a familiar critical value do not by themselves ensure reliable inference. Under weak identification, the estimator can have a random-ratio limit, its residual-based standard error can also have a nondegenerate limit, and the resulting confidence interval can have coverage different from its advertised level. Inference methods designed to remain valid under weak instruments address this problem without relying on the strong-first-stage normal approximation.