Wald Inference Under Weak Identification

Why Weak Identification Changes Wald Inference

A conventional 2SLS Wald test compares an estimate with a null value after dividing their difference by an estimated standard error. Under strong identification, this studentized statistic is often approximately standard normal, which supports familiar p-values and confidence intervals. When the first stage is weak, however, the estimate and its standard error can both have nonstandard sampling behavior. Their ratio need not be approximately standard normal, even in a large sample.

The key issue is not simply that a weak instrument makes the standard error larger. Under a local-to-zero first stage, the random first-stage score remains important relative to its population signal as the sample grows. It can make the 2SLS denominator unusually small, and the coefficient estimate need not concentrate around the structural parameter. Consequently, a conventional Wald interval need not achieve its nominal repeated-sampling coverage.

Definition: In this tutorial, weak identification refers to a setting in which the first-stage coefficient approaches zero as the sample size grows quickly enough that its population signal and sampling variation remain comparable. Conventional Wald inference uses a normal critical value, such as 1.96 for a two-sided 5% test or 95% interval, together with the usual 2SLS standard error.

Model and Local-to-Zero Sequence

Consider a scalar structural equation with one endogenous regressor $X_i$ and one excluded instrument $Z_i$. Assume variables have been centered, or partialled out with respect to included controls, so the following expressions omit an intercept and controls:

Y_i=\beta X_i+U_i,\qquad X_i=\pi_n Z_i+V_i.

The structural parameter is $\beta$. The instrument satisfies $E[Z_iU_i]=0$, and the usual first-stage specification has $E[Z_iV_i]=0$; thus $\pi_n$ is the population first-stage coefficient. Normalize $E[Z_i^2]=1$. Assume independent, identically distributed observations within each sample (or suitable triangular-array conditions), finite moments sufficient for the laws of large numbers and central limit theorem below, and a nonsingular covariance matrix for the limiting score pair. The covariance between $U_i$ and $V_i$ may be nonzero.

For a weak-instrument asymptotic sequence, let $\pi_n=c/\sqrt n$, where $c$ is fixed. Also suppose the joint score limits are

\left( \frac{1}{\sqrt n}\sum_{i=1}^n Z_iU_i,\, \frac{1}{\sqrt n}\sum_{i=1}^n Z_iV_i \right) \Rightarrow (A,B), \qquad \frac{1}{n}\sum_{i=1}^n Z_i^2\overset{p}{\longrightarrow}1.

Here $(A,B)$ is a mean-zero bivariate normal random vector under the stated central limit conditions. Its components need not be independent: the dependence is determined by the population moments of $Z_iU_i$ and $Z_iV_i$.

Key result: Under this local-to-zero sequence, the 2SLS coefficient deviation has the limit $$ \hat\beta_{2SLS}-\beta = \frac{n^{-1/2}\sum_{i=1}^n Z_iU_i} {n^{-1/2}\sum_{i=1}^n Z_iX_i} \Rightarrow \frac{A}{c+B}, $$ provided the limiting denominator is nonzero with probability one. The deviation is generally order one, not order $n^{-1/2}$.

Why the Coefficient Does Not Concentrate at the Usual Rate

With one instrument and one endogenous regressor, the just-identified 2SLS estimate is the IV ratio. Subtracting $\beta$ from that ratio gives the sample moment involving $Z_iU_i$ divided by the sample first-stage moment involving $Z_iX_i$. The denominator decomposes as

\frac{1}{\sqrt n}\sum_{i=1}^n Z_iX_i = \frac{1}{\sqrt n}\sum_{i=1}^n Z_i(\pi_n Z_i+V_i) = c\left(\frac{1}{n}\sum_{i=1}^n Z_i^2\right) + \frac{1}{\sqrt n}\sum_{i=1}^n Z_iV_i \Rightarrow c+B.

The numerator converges to $A$, so the continuous mapping theorem gives the ratio limit above. There is no extra factor of $\sqrt n$: multiplying both the numerator and denominator of the IV ratio by $n^{-1/2}$ leaves the ratio unchanged. Because both scaled sums have nondegenerate limits, their ratio does not generally shrink toward zero.

Worked Example: A Local First Stage and an Order-One Deviation

Consider a hypothetical sample with $n=100$, $c=0.4$, and illustrative scaled score values $n^{-1/2}\sum Z_iU_i=0.60$ and $n^{-1/2}\sum Z_iV_i=0.10$. Suppose $n^{-1}\sum Z_i^2=1$. These values are chosen to illustrate the local-limit calculation, not to represent an observed study. The first-stage coefficient is $\pi_n=0.4/\sqrt{100}=0.04$.

The scaled denominator is $c+B=0.40+0.10=0.50$. The coefficient deviation is therefore approximately

\hat\beta_{2SLS}-\beta \approx \frac{0.60}{0.50}=1.20.

Equivalently, the unscaled numerator score is 6, while the denominator score is $\sum Z_iX_i=4+1=5$, giving $6/5=1.20$. The same ratio results. It would be incorrect to divide 1.20 by $\sqrt{100}$: the local-to-zero calculation has an order-one coefficient deviation.

What Happens to the Conventional Standard Error?

For this just-identified model, the conventional homoskedastic 2SLS variance estimate has the form $\hat{s}^{\,2}(X'P_ZX)^{-1}$, where $P_Z$ projects onto the instrument and $\hat{s}^{\,2}$ is the residual variance estimate. In scalar notation,

X'P_ZX=\frac{\left(\sum_{i=1}^n Z_iX_i\right)^2}{\sum_{i=1}^n Z_i^2}.

Write $\Delta_n=\hat\beta_{2SLS}-\beta$. The fitted structural residual is $Y_i-\hat\beta_{2SLS}X_i=U_i-\Delta_n X_i$. Since $\pi_n$ tends to zero, the first-stage component $\pi_n Z_i$ vanishes in the average residual second moment under the stated moment conditions. For a fixed value $\Delta$, that limiting second moment is

q(\Delta)=E[(U_i-\Delta V_i)^2] =\sigma_u^2-2\Delta\sigma_{uv}+\Delta^2\sigma_v^2,

where $\sigma_u^2=E[U_i^2]$, $\sigma_v^2=E[V_i^2]$, and $\sigma_{uv}=E[U_iV_i]$. Since $\Delta_n$ itself has a random order-one limit, the residual variance estimate has the corresponding random limit $q(\Delta)$, with $\Delta=A/(c+B)$. Meanwhile, $X'P_ZX$ converges to $(c+B)^2$. Thus the conventional standard error is also generally order one:

SE(\hat\beta_{2SLS})\Rightarrow\frac{\sqrt{q(\Delta)}}{|c+B|}, \qquad T\Rightarrow \frac{A\,\operatorname{sign}(c+B)}{\sqrt{q(\Delta)}}, \qquad \Delta=\frac{A}{c+B}.

This limit is not generally standard normal. In particular, the numerator, denominator, and residual variance can all depend on the same random scores. Studentization does not remove the weak-identification problem.

Worked Example: A Conventional Wald Interval on the Correct Scale

Continue the hypothetical calculation above, and suppose the structural moments are $\sigma_u^2=1$, $\sigma_{uv}=0.20$, and $\sigma_v^2=1$. These moments are feasible: the covariance matrix has positive determinant $1-(0.20)^2=0.96$. With $\Delta=1.20$,

$$ q(1.20)=1-2(1.20)(0.20)+(1.20)^2 =1-0.48+1.44=1.96. $$

The limiting conventional standard error is $\sqrt{1.96}/0.50=1.40/0.50=2.80$. The Wald statistic for testing $H_0:\beta=2$, given $\hat\beta_{2SLS}\approx 2+1.20=3.20$, is

T=\frac{3.20-2}{2.80}\approx0.4286.

Its two-sided standard-normal p-value is approximately $2\Phi(-0.4286)=0.668$, so this conventional calculation does not reject at the 5% level. The corresponding nominal 95% Wald interval is

3.20\pm1.96(2.80) = 3.20\pm5.488 \approx[-2.288,\ 8.688].

The interval is calculated on the coefficient scale: neither the deviation nor its standard error is divided by $\sqrt n$. Its nominal 95% label does not establish that such intervals cover $\beta$ in 95% of repeated samples under weak identification.

A Concrete Size Distortion

The limiting Wald statistic can differ sharply from a standard normal statistic. An extreme but informative case sets $c=0$: the instrument has no first-stage signal even asymptotically. Suppose $A$ and $B$ are independent standard normal variables, and let $\sigma_u^2=\sigma_v^2=1$ and $\sigma_{uv}=0$. The limit formulas give $\Delta=A/B$ and

T= \frac{A\,\operatorname{sign}(B)} {\sqrt{1+(A/B)^2}} = \frac{AB}{\sqrt{A^2+B^2}}.

The equality follows because $\sqrt{1+(A/B)^2}=\sqrt{A^2+B^2}/|B|$ and $\operatorname{sign}(B)|B|=B$. Also, $|T|<|A|$ and $|T|<|B|$ whenever $A$ and $B$ are nonzero. Therefore, a two-sided 5% Wald rejection, which requires $|T|>1.96$, can occur only if both $|A|>1.96$ and $|B|>1.96$. Independence implies

P(|T|>1.96) \leq P(|A|>1.96,\ |B|>1.96) =[2\Phi(-1.96)]^2\approx0.00249958\approx0.0025.

Thus, in this constructed limiting case, the rejection probability is at most 0.25%, far below the nominal 5%. The calculation demonstrates severe size distortion; it does not imply that weak instruments always make Wald tests underreject. Distortion can depend on the first-stage sequence and the joint distribution of the scores and errors. At $c=0$, the instrument supplies no identifying information, so this example also marks an especially stark boundary of the problem.

Worked Example: A Near-Canceling First-Stage Score

Consider another hypothetical local-limit realization with $c=0.40$, $A=0.60$, and $B=-0.35$. The limiting first-stage score is $c+B=0.05$, so the coefficient deviation is

\Delta=\frac{0.60}{0.05}=12.

Using the same structural moments as above gives

$$ q(12)=1-2(12)(0.20)+12^2 =1-4.8+144=140.2. $$

The conventional standard error is approximately $\sqrt{140.2}/0.05=11.84/0.05=236.8$. If $\beta=2$, the corresponding estimate is 14, and the nominal 95% Wald interval is $14\pm1.96(236.8)$, or approximately $[-450.1,\ 478.1]$. A nearly canceled first-stage score can produce an extreme estimate together with an enormous standard error. Such a realization is not evidence that the conventional interval has correct coverage; coverage is a repeated-sampling property, and its validity requires the relevant approximation to hold across samples.

Interpretation, Use, and a Common Pitfall

Under weak identification, the conventional Wald p-value is not generally calibrated to the stated significance level, and a nominal 95% interval is not guaranteed to cover the parameter in 95% of repeated samples. A p-value remains a tail probability computed under a specified reference distribution; it is not the probability that the null hypothesis is true. Likewise, a confidence interval's nominal level describes a repeated-sampling procedure under its assumptions, not a 95% probability that the fixed parameter lies in the particular interval observed.

The practical importance is substantial. A weak first stage can make conclusions sensitive to small changes in the sample, and a reported standard error alone does not reveal whether the normal approximation for the Wald statistic is reliable. The earlier material on first-stage F statistics and Stock-Yogo critical values concerns instrument strength; diagnosing weakness and obtaining inference with reliable weak-identification properties are related but distinct tasks.

Common mistake: Treating a weak-first-stage estimate as if $\hat\beta_{2SLS}-\beta$ were order $n^{-1/2}$, then dividing both the estimate deviation and conventional standard error by $\sqrt n$.
Correction: Under $\pi_n=c/\sqrt n$, the scaled numerator and denominator of the IV ratio both have order-one limits. Their ratio, the coefficient deviation, is generally order one; the conventional standard error is generally order one as well.
Common mistake: Assuming that using the usual 2SLS standard error automatically makes the Wald statistic approximately standard normal.
Correction: The residual variance and first-stage denominator can share random weak-identification limits. Their ratio need not have a standard normal distribution, so conventional p-values and intervals require caution.

Key takeaway: With a local-to-zero first stage, the 2SLS coefficient deviation and conventional standard error need not shrink at the usual rate, and the studentized Wald statistic need not be standard normal. Nominal Wald significance levels and interval coverage therefore cannot be assumed under weak identification.

Check Your Understanding

Use the model and local-to-zero calculations above to answer these questions.

Why does $E[Z_iV_i]=0$ make $\pi_n$ the population first-stage coefficient in $X_i=\pi_n Z_i+V_i$?
Under $\pi_n=c/\sqrt n$, what is the probability limit or distributional limit of $n^{-1/2}\sum Z_iX_i$, and which two components determine it?
Explain why multiplying both sums in the just-identified IV ratio by $n^{-1/2}$ does not imply that the coefficient deviation is order $n^{-1/2}$.
In the near-canceling example, what makes the 2SLS coefficient deviation large, and why does that alone not establish a precise estimate?
What does a nominal 95% confidence level describe, and why is that interpretation not guaranteed for conventional Wald intervals under weak identification?