Local-To-Zero Asymptotics for Instrumental Variables

Why Let Instrument Strength Change With Sample Size?

In a linear instrumental-variables model, an instrument can be valid yet only weakly related to the endogenous regressor. This creates a difficulty that does not disappear just by describing the sample as “large”: if the population first-stage relationship becomes weaker as the sample grows, the data may never accumulate a strong signal of relevance. Local-to-zero asymptotics formalize that situation by specifying how quickly the first-stage coefficient shrinks with sample size.

The key comparison is between the population first-stage signal and the sampling noise in its estimate. Under ordinary regularity conditions, the estimate’s noise is of order $1/\sqrt{n}$. A coefficient that also shrinks at that rate remains on the same scale as the noise. This tutorial defines that sequence and explains its meaning. It does not derive the resulting limiting distribution of 2SLS or develop weak-instrument tests.

Model and Local-to-Zero Definition

For clarity, consider one endogenous regressor $x_i$, one instrument $z_i$, and an outcome $y_i$, after any included exogenous controls have been partialled out. Write the structural and first-stage equations as

y_i=\beta x_i+u_i,\qquad x_i=\pi_n z_i+v_i,\qquad i=1,\ldots,n.

Here $\beta$ is the structural coefficient of interest, $u_i$ is the structural error, $v_i$ is the first-stage error, and $\pi_n$ is allowed to depend on sample size. Assume the observations are independent and identically distributed within each sample, with stable moments as $n$ changes. For this simple exposition, take $E[z_i]=0$, $E[z_i^2]=Q_{zz}>0$, $E[z_i v_i]=0$, and $E[z_i u_i]=0$. The last two conditions state the first-stage orthogonality and instrument orthogonality used here. Assume the relevant second moments are finite. These conditions are not, by themselves, enough to ensure nonzero sampling noise; positive-variance conditions will be stated whenever that noise is discussed.

Definition: A local-to-zero sequence for the first-stage coefficient sets $$ \pi_n=\frac{c}{\sqrt{n}}, $$ where $c$ is a fixed finite constant. If $c\ne 0$, the instrument is relevant at each finite sample size under $Q_{zz}>0$, since $\operatorname{Cov}(z_i,x_i)=\pi_n Q_{zz}\ne0$. But the population first-stage coefficient tends to zero as $n$ grows.

The phrase “local to zero” describes a sequence of models, not a single model with a fixed first-stage coefficient. The constant $c$ determines the signal’s scale in this sequence. A larger absolute value of $c$ means a stronger signal on the local-to-zero scale; $c=0$ gives no first-stage relationship at all in this specification.

Why the Inverse Square-Root Rate Matters

The sample first-stage slope through the origin is

\hat{\pi}_n=\frac{\sum_{i=1}^{n}z_i x_i}{\sum_{i=1}^{n}z_i^2} =\pi_n+\frac{\sum_{i=1}^{n}z_i v_i}{\sum_{i=1}^{n}z_i^2}.

The equality follows by substituting $x_i=\pi_n z_i+v_i$ into the numerator. The denominator divided by $n$ converges to $Q_{zz}>0$ under the stated sampling conditions. To characterize the random term in the numerator, add the nondegeneracy condition $\operatorname{Var}(z_i v_i)>0$. Since $E[z_i v_i]=0$, an iid central limit theorem gives a nondegenerate root-$n$ limit for $\sum z_i v_i$, provided the stated finite-variance condition holds. Thus the first-stage estimation noise is of order $1/\sqrt{n}$, not merely because of orthogonality, but because the variance is positive and the central limit theorem applies.

Under the local-to-zero sequence, multiplying the slope by $\sqrt{n}$ makes the comparison explicit:

\sqrt{n}\,\hat{\pi}_n =c+\frac{n^{-1/2}\sum_{i=1}^{n}z_i v_i}{n^{-1}\sum_{i=1}^{n}z_i^2}.

The population signal contributes the fixed quantity $c$, while the second term has a nondegenerate random scale under the conditions just stated. The estimate therefore does not become arbitrarily precise relative to the shrinking population coefficient. By contrast, if $\pi_n=\pi\ne0$ is fixed, its signal stays constant while the estimation noise shrinks toward zero. This distinction is the core of the local-to-zero representation.

Worked Example: Comparing Fixed and Shrinking First Stages

Suppose the first-stage coefficient is either fixed at $0.06$, or follows a local-to-zero sequence with $c=1.5$. Compare the population coefficients at $n=400$ and $n=1{,}600$.

For the local-to-zero sequence,

\pi_{400}=\frac{1.5}{\sqrt{400}}=\frac{1.5}{20}=0.075,\qquad \pi_{1600}=\frac{1.5}{\sqrt{1600}}=\frac{1.5}{40}=0.0375.

When the sample size quadruples, the local-to-zero coefficient is halved. Its rescaled value stays constant: $\sqrt{400}(0.075)=1.5$ and $\sqrt{1600}(0.0375)=1.5$. The fixed coefficient, in contrast, remains $0.06$ at both sample sizes, so its signal becomes increasingly large relative to root-$n$ sampling noise. These are hypothetical population coefficients, not estimates from a real study.

What Happens to the Sample First-Stage Signal?

Another way to see the scaling is through the sample cross-product between the instrument and endogenous regressor:

\sum_{i=1}^{n}z_i x_i =\pi_n\sum_{i=1}^{n}z_i^2+\sum_{i=1}^{n}z_i v_i.

Under the local-to-zero sequence, the first term is approximately $(c/\sqrt{n})(nQ_{zz})=cQ_{zz}\sqrt{n}$. The second term is also of order $\sqrt{n}$ if $\operatorname{Var}(z_i v_i)>0$ and the relevant central limit theorem holds. Consequently, the first-stage cross-product’s signal and sampling variation are on comparable scales. With a fixed nonzero $\pi$, its population component instead grows at order $n$, while the centered noise remains on the root-$n$ scale under the same nondegeneracy condition.

This does not mean every random sum in an IV model must fluctuate at root-$n$ order. For example, $\sum z_i u_i$ has a nondegenerate root-$n$ limit only if $E[z_i u_i]=0$, $\operatorname{Var}(z_i u_i)>0$, and a suitable central limit theorem applies. If $u_i=0$ for every observation, that sum is identically zero. Similarly, the condition $E[z_i v_i]=0$ alone does not guarantee root-$n$ fluctuations of $\sum z_i v_i$; positive variance is needed. Keeping these conditions explicit prevents a zero-variance case from being mistakenly described as random noise of a particular order.

Worked Example: Signal and Noise in a First-Stage Slope

Suppose $n=900$, $Q_{zz}=1.5$, $c=1.2$, and $\operatorname{Var}(z_i v_i)=2.25$. Assume the iid and finite-moment conditions above, including the positive variance needed for a nondegenerate central limit theorem. What are the population slope and the approximate standard deviation of its first-stage sampling noise?

The local-to-zero coefficient is

\pi_{900}=\frac{1.2}{\sqrt{900}}=\frac{1.2}{30}=0.04.

The exact slope decomposition is $\hat{\pi}_n-\pi_n=(\sum z_i v_i)/(\sum z_i^2)$. Using $\sum z_i^2$ approximately $nQ_{zz}$, the noise variance is approximately

\operatorname{Var}(\hat{\pi}_n-\pi_n) \approx \frac{n\operatorname{Var}(z_i v_i)}{(nQ_{zz})^2} =\frac{\operatorname{Var}(z_i v_i)}{nQ_{zz}^2}.

Substituting the hypothetical values gives

SE(\hat{\pi}_{900})\approx \sqrt{\frac{2.25}{900(1.5)^2}} =\sqrt{\frac{2.25}{2025}} =\sqrt{0.001111\ldots} \approx 0.03333.

The population slope $0.04$ is on roughly the same scale as this approximate sampling standard deviation. This calculation illustrates the local-to-zero signal-noise comparison; it is not a formal first-stage test or a diagnostic decision rule.

A Strength Index in a Simple Special Case

A familiar way to summarize the scaling in a simple, homoskedastic one-instrument model is the concentration parameter

\lambda_n=\frac{n\pi_n^2 Q_{zz}}{\sigma_v^2},

where $\sigma_v^2=\operatorname{Var}(v_i)$. This expression is useful here as a measure of how the population first-stage signal scales with sample size; it is not a universal formula for diagnostics under arbitrary error structures. Under a local-to-zero sequence,

\lambda_n=\frac{n(c^2/n)Q_{zz}}{\sigma_v^2} =\frac{c^2 Q_{zz}}{\sigma_v^2}.

It stays constant as $n$ increases. Under a fixed nonzero $\pi$, the same expression grows proportionally to $n$. The index makes the essential distinction plain: local-to-zero asymptotics let the sample size grow without allowing this signal measure to grow without bound.

Worked Example: A Constant Concentration Parameter

Suppose, hypothetically, that $c=2$, $Q_{zz}=4$, and $\sigma_v^2=9$. Compute the concentration parameter for $n=500$ and $n=2{,}000$ under the local-to-zero sequence.

At $n=500$, $\pi_{500}=2/\sqrt{500}$, so

\lambda_{500} =\frac{500(2/\sqrt{500})^2(4)}{9} =\frac{500(4/500)(4)}{9} =\frac{16}{9} \approx 1.778.

At $n=2{,}000$, $\pi_{2000}=2/\sqrt{2{,}000}$, and

\lambda_{2000} =\frac{2{,}000(2/\sqrt{2{,}000})^2(4)}{9} =\frac{2{,}000(4/2{,}000)(4)}{9} =\frac{16}{9} \approx 1.778.

The sample size increases by a factor of four, but the index remains unchanged because the squared first-stage coefficient decreases by the reciprocal factor. This example illustrates scaling only; no threshold or classification is being applied.

Interpretation, Scope, and Common Pitfall

Local-to-zero asymptotics are a deliberate way to study IV behavior in sequences of increasingly difficult identification problems. At each finite $n$, a nonzero $c$ gives a nonzero population first-stage coefficient. Yet the coefficient approaches zero, and the information about it does not accumulate as it would under a fixed nonzero first stage. The setup is therefore useful for examining whether usual large-sample approximations remain reliable when instrument relevance is weak.

The sequence is a modeling device, not a claim that an actual instrument’s strength must literally change when researchers collect more observations. It asks what happens along a sequence of populations whose first-stage relationships get weaker at a specified rate. It also does not establish instrument validity: the moment conditions involving $u_i$ and the substantive exclusion argument are separate from the strength of the first stage.

Common mistake: “The first-stage coefficient is nonzero for every finite $n$, so the instrument becomes strongly identifying as the sample grows.”
Correction: In the local-to-zero sequence, nonzero finite-sample relevance and strong identification are not equivalent. Since $\pi_n=c/\sqrt{n}$, the population signal shrinks at the same rate that the first-stage estimation noise shrinks; their relative scale does not diverge.
Another mistake: Claiming a sum fluctuates at order $\sqrt{n}$ just because its summands have mean zero.
Correction: A root-$n$ fluctuation claim also needs appropriate moment and limit conditions, including positive variance for the summand when a nondegenerate limit is claimed.

Key takeaway: The local-to-zero specification $\pi_n=c/\sqrt{n}$ keeps the first-stage signal on the same scale as its sampling noise. It formalizes weak identification without implying that the instrument is invalid or that every finite-sample first stage is exactly zero.

Check Your Understanding

Use the model and conditions in this tutorial to answer the following.

If $c=3$, what is $\pi_n$ when $n=900$, and what is $\sqrt{n}\pi_n$?
Why does a fixed nonzero first-stage coefficient behave differently from $c/\sqrt{n}$ as the sample size grows?
What additional condition is needed, beyond $E[z_i v_i]=0$, to claim a nondegenerate root-$n$ limit for $\sum z_i v_i$?
Under the stated simple homoskedastic formula, how does $\lambda_n$ change with $n$ under a local-to-zero sequence?
Does nonzero relevance at every finite sample size imply that the first stage becomes strongly identifying asymptotically? Explain.