Why the Instrument’s Sequence Matters
Instrument strength is not determined by whether a first-stage coefficient is nonzero in one particular sample. In asymptotic analysis, the coefficient may itself depend on the sample size \(n\). The key question is how quickly the first-stage signal changes relative to sampling noise. A coefficient that stays fixed as \(n\) grows leads to a different inferential regime from one that shrinks at rate \(1/\sqrt{n}\).
We use a scalar instrumental-variable model, with controls understood to have been partialled out if present: \(y_i=\beta x_i+u_i\) and \(x_i=\pi_n z_i+v_i\). Here \(z_i\) is the instrument, \(x_i\) is the endogenous regressor, and \(\pi_n\) is its first-stage coefficient. The errors \(u_i\) and \(v_i\) may be correlated; validity requires \(E[z_i u_i]=0\), while the first-stage equation requires \(E[z_i v_i]=0\). For simplicity, suppose observations are iid within each sample, the distribution of \((z_i,v_i,u_i)\) does not change with \(n\), and the moments used below are finite. Assume \(E[z_i]=0\), \(q=E[z_i^2]>0\), and \(0<\omega=E[z_i^2v_i^2]<\infty\).
The first-stage least-squares estimator is \(\hat{\pi}_n=\sum_{i=1}^n z_i x_i/\sum_{i=1}^n z_i^2\). Its sampling error is governed by the same noise term under both asymptotic regimes. What differs is the size of the signal \(\pi_n\).
What Happens to the First-Stage Estimator?
Under the stated assumptions, the first-stage estimator can be written as \(\hat{\pi}_n=\pi_n+\{\sum_{i=1}^n z_i v_i\}/\{\sum_{i=1}^n z_i^2\}\). The law of large numbers gives \(n^{-1}\sum z_i^2\xrightarrow{p}q\). The central limit theorem gives \(n^{-1/2}\sum z_i v_i\xrightarrow{d}N(0,\omega)\), since \(E[z_i v_i]=0\) and \(\operatorname{Var}(z_i v_i)=\omega\).
Proof. From the estimator identity, \[ \sqrt{n}(\hat{\pi}_n-\pi_n) = \frac{n^{-1/2}\sum_{i=1}^n z_i v_i}{n^{-1}\sum_{i=1}^n z_i^2}. \] The numerator converges in distribution to \(N(0,\omega)\), and the denominator converges in probability to the positive constant \(q\). Slutsky’s theorem therefore gives \(\sqrt{n}(\hat{\pi}_n-\pi_n)\xrightarrow{d}N(0,\omega/q^2)\). For fixed \(\pi_n=\pi\), this also implies \(\hat{\pi}_n-\pi\xrightarrow{p}0\), so the estimator is consistent. For \(\pi_n=c/\sqrt{n}\), write \(\sqrt{n}\hat{\pi}_n=c+\sqrt{n}(\hat{\pi}_n-\pi_n)\). Adding the constant \(c\) gives the stated shifted normal limit. \(\square\)
The distinction is visible in the scale of \(\hat{\pi}_n\). With fixed strength, its standard error shrinks at rate \(1/\sqrt{n}\) while its target remains away from zero. With a local-to-zero coefficient, the target itself is of that same order. Thus, multiplying the estimate by \(\sqrt{n}\) leaves both signal and noise in the limit; the estimate does not become large relative to its sampling uncertainty.
Worked Example: A Fixed First-Stage Coefficient
Suppose \(q=2\), \(\omega=1.5\), and \(\pi_n=0.3\) for every \(n\). The theorem gives \(\hat{\pi}_n\xrightarrow{p}0.3\), with asymptotic variance for \(\sqrt{n}(\hat{\pi}_n-0.3)\) equal to \[ \frac{\omega}{q^2}=\frac{1.5}{2^2}=0.375. \] At \(n=500\), the approximate standard error of \(\hat{\pi}_n\) is \(\sqrt{0.375/500}=\sqrt{0.00075}\approx0.0274\). The first-stage coefficient is about \(0.3/0.0274\approx10.95\) standard errors from zero. Increasing \(n\) makes this ratio grow: the signal stays at \(0.3\), while the standard error decreases. This is the fixed-strength pattern.
Worked Example: A Local-to-Zero Coefficient
Suppose \(c=2\), \(q=1\), and \(\omega=4\), so \(\pi_n=2/\sqrt{n}\). At \(n=100\), \(\pi_n=2/10=0.2\); at \(n=400\), \(\pi_n=2/20=0.1\). The limiting distribution is the same at both sample sizes in the local asymptotic description: \[ \sqrt{n}\hat{\pi}_n\xrightarrow{d}N\left(2,\frac{4}{1^2}\right)=N(2,4). \] Equivalently, \(\sqrt{n}(\hat{\pi}_n-\pi_n)\) has limiting standard deviation \(2\), while the signal on the \(\sqrt{n}\) scale is \(2\). The signal-to-noise ratio therefore remains finite rather than growing with \(n\). The coefficient is nonzero at each of these sample sizes, but that fact alone does not make the sequence strongly identifying.
Identification and the Concentration Parameter
The population IV moment at a candidate coefficient \(b\) is \(E[z_i(y_i-bx_i)]=(\beta-b)E[z_i x_i]\), because \(E[z_i u_i]=0\). From the first-stage equation, \(E[z_i x_i]=\pi_n q\). Thus, when \(\pi_n\ne0\), the population moment has a unique zero at \(b=\beta\). In that finite-sample-size population sense, the parameter is identified. But the sensitivity of the moment to \(b\) is \(-\pi_n q\), which approaches zero under a local-to-zero sequence. Identification becomes weak in the asymptotic sense: small changes in \(b\) produce increasingly small changes in the population moment.
In the homoskedastic first-stage setting, with \(\operatorname{Var}(v_i)=\sigma_v^2\), the concentration parameter has the familiar population-scale expression \(\lambda_n=n\pi_n^2q/\sigma_v^2\). This expression makes the rate distinction explicit: it measures the squared first-stage signal accumulated across \(n\) observations relative to first-stage noise.
Derivation. Substituting \(\pi_n=c n^{-\alpha}\) into the expression for \(\lambda_n\) gives \[ \lambda_n=\frac{n(c n^{-\alpha})^2q}{\sigma_v^2} =\frac{c^2q}{\sigma_v^2}n^{1-2\alpha}. \] The exponent \(1-2\alpha\) is positive, zero, or negative according as \(\alpha\) is below, equal to, or above \(1/2\). Since the leading constant is positive, these three cases imply divergence, a positive constant, or convergence to zero, respectively. \(\square\)
Worked Example: An Intermediate Shrinking Rate
Let \(\pi_n=0.5n^{-1/3}\), with \(q=1\) and \(\sigma_v^2=1\). Here \(\alpha=1/3\), so \[ \lambda_n=n(0.5n^{-1/3})^2=0.25n^{1/3}. \] At \(n=64\), \(n^{1/3}=4\), so \(\lambda_{64}=0.25(4)=1\). At \(n=512\), \(n^{1/3}=8\), so \(\lambda_{512}=0.25(8)=2\). The first-stage coefficient shrinks from \(0.5/4=0.125\) to \(0.5/8=0.0625\), yet the concentration parameter grows. Its growth is slower than under a fixed coefficient, but it still diverges because \(1/3<1/2\). This illustrates why “the coefficient shrinks” and “the instrument is weak asymptotically” are not equivalent statements: the rate of shrinking matters.
Why the Local-to-Zero Case Changes Inference
A useful way to see the boundary rate is to compare the first-stage signal with its sampling error. In the model above, \(\hat{\pi}_n-\pi_n\) is of order \(1/\sqrt{n}\). If \(\pi_n\) is fixed and nonzero, the signal is larger than this error as \(n\) increases. If \(\pi_n=c/\sqrt{n}\), the two have the same order. If \(\pi_n\) shrinks faster than \(1/\sqrt{n}\), the signal is smaller still. The local-to-zero case is therefore not merely a small-coefficient version of the fixed-strength case; it changes which terms remain relevant asymptotically.
The same issue appears in the sample IV denominator, \(n^{-1}\sum z_i x_i\). Its population counterpart is \(\pi_n q\). Under fixed strength, the sample denominator converges to the nonzero value \(\pi q\). Under local-to-zero strength, its population value is of order \(1/\sqrt{n}\), while sampling variation in \(n^{-1}\sum z_i v_i\) is also of order \(1/\sqrt{n}\). The denominator is no longer approximated by a stable nonzero constant. The resulting weak-instrument behavior is why conventional strong-instrument approximations cannot simply be carried over. The joint reduced-form limit and the resulting IV estimator limit require separate treatment.
Common Mistake
Incorrect reasoning: “For every \(n\), \(\pi_n\ne0\), so the instrument is strong and the usual large-sample IV approximation applies.” This confuses finite-\(n\) population identification with the strength of identification along a sequence.
Correction: A nonzero \(\pi_n\) gives a unique population moment zero at that \(n\), but it says nothing by itself about how sharply that moment distinguishes nearby values of \(\beta\). For \(\pi_n=c/\sqrt{n}\), the moment’s slope approaches zero and the signal remains comparable to sampling noise. Check the sequence’s rate, or equivalently the behavior of the concentration parameter, rather than checking only whether each coefficient is nonzero.
Check Your Understanding
Use the rate of the first-stage coefficient, not just whether it is nonzero, to answer each question.
- If \(\pi_n=0.4\) for every \(n\), what happens to the concentration parameter as \(n\) grows?
- For \(\pi_n=3/\sqrt{n}\), what is the limiting distribution of \(\sqrt{n}\hat{\pi}_n\) in terms of \(q\) and \(\omega\)?
- Suppose \(\pi_n=cn^{-2/3}\), with \(c\ne0\). Does the concentration parameter diverge, stay finite, or approach zero? Explain using its rate.
- Why does a nonzero first-stage coefficient at each sample size not by itself establish strong asymptotic identification?
- At what shrinking rate is the first-stage coefficient on the same order as its sampling error?