Bekker Asymptotics for Many Instruments

Why proportional instrument growth changes the limit

With a fixed number of instruments, the noise projected onto the instrument space becomes negligible per observation as the sample grows. That intuition changes when the instrument count itself grows with the sample. The projection can then capture a non-negligible amount of first-stage noise, and its contribution to the 2SLS denominator remains visible asymptotically. If the first-stage disturbance is correlated with the structural error, the same projection also carries a non-vanishing error covariance into the numerator.

Bekker asymptotics make this regime precise: the number of instruments grows in proportion to the sample size, while the aggregate first-stage signal remains of order the sample size. The result is generally a nonzero asymptotic bias for 2SLS when the endogenous-regressor disturbance and structural error are correlated. The estimate can have a well-defined limiting distribution, but that distribution is centered away from the structural coefficient.

Model and assumptions

Consider one endogenous regressor $x$, one outcome $y$, and a vector of $K$ excluded instruments collected in the $n\times K$ matrix $Z$. After partialling out any included exogenous regressors, write the structural and first-stage equations as

y = \beta x + u,\qquad x = Z\pi + v.

Here $\beta$ is the structural coefficient, $u$ is the structural error, $v$ is the first-stage disturbance, and $Z\pi$ is the systematic component of $x$. The instruments are exogenous in the sense that their columns are orthogonal to the structural error in the population; the asymptotic framework below conditions on the instrument design.

Definition: In the single-endogenous-regressor Bekker framework, $K=K_n$ increases with $n$ so that $K/n\to\alpha$, where $0<\alpha<1$. The instrument projection is $P_Z=Z(Z'Z)^{-1}Z'$, with rank $K$. The first-stage signal $s=Z\pi$ lies in the column space of $Z$, and its normalized strength satisfies $s's/n\to Q>0$. The row errors $(v_i,u_i)$ are independent and identically distributed, jointly normal, have mean zero, and have covariance matrix with entries $\operatorname{Var}(v_i)=\sigma_{vv}$, $\operatorname{Var}(u_i)=\sigma_{uu}$, and $\operatorname{Cov}(v_i,u_i)=\sigma_{uv}$.

The joint normality assumption gives a compact, explicit limiting variance below. The central probability-limit argument needs less than normality, but a distributional result requires suitable moment and regularity conditions. Relevance is represented by $Q>0$, while $\sigma_{uv}\ne 0$ represents endogeneity. The 2SLS estimator in this setup is

\hat\beta_{\mathrm{2SLS}} =\frac{x'P_Zy}{x'P_Zx} =\beta+\frac{x'P_Zu}{x'P_Zx}.

Probability limit and many-instrument bias

Because $s$ is in the instrument space, $P_Zs=s$. Substituting $x=s+v$ into the numerator and denominator of the estimation error gives

x'P_Zu=s'u+v'P_Zu,\qquad x'P_Zx=s's+2s'v+v'P_Zv.

The signal term $s's$ is of order $n$. Under the stated assumptions, $s'u/n$ and $s'v/n$ vanish, while $v'P_Zu/n$ and $v'P_Zv/n$ converge to $\alpha\sigma_{uv}$ and $\alpha\sigma_{vv}$, respectively. The latter limits arise because the projection has $K$ dimensions and $K/n\to\alpha$, rather than from the first-stage signal.

Theorem: Under the many-instrument assumptions, the 2SLS probability limit is $$ \operatorname{plim}\hat\beta_{\mathrm{2SLS}} =\beta+B,\qquad B=\frac{\alpha\sigma_{uv}}{Q+\alpha\sigma_{vv}}. $$ Thus 2SLS is generally inconsistent for $\beta$ when $\alpha>0$ and $\sigma_{uv}\ne0$. The bias disappears if the instrument count is asymptotically negligible relative to $n$, if $\sigma_{uv}=0$, or in the limiting case where the signal $Q$ dominates the projection-noise term.

The denominator $Q+\alpha\sigma_{vv}$ combines the normalized first-stage signal and the noise captured by the instrument projection. The numerator’s $\alpha\sigma_{uv}$ is the corresponding projected error covariance. Consequently, proportional instrument growth does not simply increase uncertainty around the usual 2SLS limit: it can change the limit itself.

Worked Example: Calculating the Asymptotic Bias

Suppose a hypothetical design has $n=1{,}000$, $K=200$, first-stage signal strength $Q=1.00$, and error moments $\sigma_{vv}=1.00$ and $\sigma_{uv}=0.80$. Let the structural coefficient be $\beta=2.50$, measured in outcome units per unit of $x$. Calculate the 2SLS probability limit.

The instrument fraction is $\alpha=K/n=200/1{,}000=0.200$. Substituting into the bias formula gives

B=\frac{(0.200)(0.80)}{1.00+(0.200)(1.00)} =\frac{0.160}{1.200} =0.1333.

Therefore, the probability limit is $2.50+0.1333=2.633$ outcome units per unit of $x$. This is an asymptotic calculation under the hypothetical model, not a claim about a real study. The positive sign follows from the positive covariance $\sigma_{uv}$.

Limiting distribution around the shifted limit

A limiting distribution describes sampling variation around the many-instrument probability limit, not automatically around $\beta$. Define $B$ as above and let $T=Q+\alpha\sigma_{vv}$. Under the joint normality and regularity assumptions, with $K/n$ and $s's/n$ approaching their limits sufficiently quickly for the displayed centering, the limiting distribution is

\sqrt{n}\bigl(\hat\beta_{\mathrm{2SLS}}-\beta-B\bigr) \ \xrightarrow{d}\ N(0,V),

where

V= \frac{ Q\sigma_{uu} +\alpha(\sigma_{vv}\sigma_{uu}+\sigma_{uv}^{2}) +B^{2}(4Q\sigma_{vv}+2\alpha\sigma_{vv}^{2}) -2B(2Q\sigma_{uv}+2\alpha\sigma_{vv}\sigma_{uv}) }{(Q+\alpha\sigma_{vv})^{2}}.

The centering condition matters. More generally, for a sequence with $K/n$ not approaching $\alpha$ at the required rate, the finite-sample centering is $b_n=\beta+B_n$, where

B_n=\frac{(K/n)\sigma_{uv}}{(s's/n)+(K/n)\sigma_{vv}}.

A distributional approximation centered at $\beta+B$ is appropriate when the convergence of the design quantities is fast enough that their deviations do not contribute at the $1/\sqrt n$ scale. Otherwise, the finite-sequence centering must be retained.

Why the variance has these components

The key justification is a first-order expansion of a ratio. Write the numerator of the estimation error as $N=x'P_Zu$ and its denominator as $D=x'P_Zx$. Their normalized limits are $N/n\to\alpha\sigma_{uv}$ and $D/n\to T$, so the ratio’s limit is $B$. Expanding around these limits yields

\sqrt n\left(\frac{N}{D}-B\right) = \frac{\sqrt n\{(N/n)-\alpha\sigma_{uv}\} -B\sqrt n\{(D/n)-T\}}{T} +o_p(1).

The numerator fluctuation has limiting variance $Q\sigma_{uu}+\alpha(\sigma_{vv}\sigma_{uu}+\sigma_{uv}^{2})$. Its first term comes from the signal-weighted structural error; its second comes from products of projected disturbances across the $K$ instrument directions. The denominator fluctuation has limiting variance $4Q\sigma_{vv}+2\alpha\sigma_{vv}^{2}$, combining signal–disturbance variation and projected squared-noise variation. The covariance of these two fluctuations is $2Q\sigma_{uv}+2\alpha\sigma_{vv}\sigma_{uv}$. Applying the variance formula to their difference in the expansion gives $V$.

For completeness, the projected-noise terms follow by expressing the projection in an orthonormal basis of its $K$-dimensional column space. Under joint normality, the transformed disturbance pairs remain independent across orthogonal directions. Each direction contributes expected product $E(v_i u_i)=\sigma_{uv}$, expected square $E(v_i^2)=\sigma_{vv}$, and, by the Gaussian fourth-moment identities, $\operatorname{Var}(v_i u_i)=\sigma_{vv}\sigma_{uu}+\sigma_{uv}^2$, $\operatorname{Var}(v_i^2)=2\sigma_{vv}^2$, and $\operatorname{Cov}(v_i u_i,v_i^2)=2\sigma_{vv}\sigma_{uv}$. Summing over $K$ directions and dividing by $n$ produces the terms multiplied by $\alpha$. The signal direction adds the terms multiplied by $Q$.

Worked Example: Computing the Limiting Variance

Continue with a hypothetical design having $Q=1.00$, $\alpha=0.200$, $\sigma_{vv}=1.00$, and $\sigma_{uv}=0.80$. Suppose $\sigma_{uu}=4.00$. The bias is $B=0.1333$, and $T=1.200$. Calculate the asymptotic variance coefficient $V$.

First calculate the numerator-fluctuation variance:

Q\sigma_{uu}+\alpha(\sigma_{vv}\sigma_{uu}+\sigma_{uv}^2) =4.00+0.200(4.00+0.6400)=4.928.

The denominator-fluctuation variance and the covariance are

4Q\sigma_{vv}+2\alpha\sigma_{vv}^2=4.00+0.400=4.400,

2Q\sigma_{uv}+2\alpha\sigma_{vv}\sigma_{uv} =1.600+0.3200=1.920.

Hence

V=\frac{4.928+(0.1333)^2(4.400)-2(0.1333)(1.920)}{(1.200)^2} \approx 3.121.

For $n=1{,}000$, the limiting standard deviation around the shifted center is approximately $\sqrt{3.121/1{,}000}=0.05586$. The approximation is centered at $\beta+0.1333$, not at $\beta$. Its standard deviation quantifies sampling variation; it does not remove the asymptotic bias.

Instrument growth rate and interpretation

The proportional-growth assumption is consequential. If the number of instruments stays fixed while $n$ grows, then $K/n\to0$, so the many-instrument bias term vanishes under this framework. If instead $K/n$ approaches a positive constant, the noise contribution remains in the probability limit. This explains why instrument count should be considered relative to sample size, not just as an absolute number.

Worked Example: Comparing Two Instrument-Growth Regimes

Consider a hypothetical sequence with $\beta=1.80$, $Q=0.800$, $\sigma_{vv}=0.640$, and $\sigma_{uv}=0.300$. Compare $n=500,\ K=25$ with $n=5{,}000,\ K=25$, treating the other quantities as unchanged.

For $n=500$, $\alpha=25/500=0.0500$, so

B=\frac{(0.0500)(0.300)}{0.800+(0.0500)(0.640)} =\frac{0.0150}{0.832} =0.01803.

The probability limit is $1.818$ outcome units per unit of $x$. For $n=5{,}000$, the same fixed number of instruments gives $\alpha=25/5{,}000=0.00500$, and

B=\frac{(0.00500)(0.300)}{0.800+(0.00500)(0.640)} =\frac{0.00150}{0.8032} =0.001868.

The bias term is much smaller as the instrument fraction falls. In contrast, if $K$ rose in proportion to $n$ so that $\alpha$ stayed at $0.0500$, this formula would retain the same bias term, given unchanged $Q$ and error moments.

Common mistake and practical significance

A common mistake is to reason that 2SLS must converge to $\beta$ because each instrument is valid and the first-stage signal is relevant. Validity and relevance do not eliminate the projected-noise terms when the instrument dimension grows proportionally with the sample. In particular, instrument validity does not imply $\sigma_{uv}=0$: the structural error may be correlated with the endogenous-regressor disturbance even when it is uncorrelated with the instruments.

The correction is to check the asymptotic regime and distinguish the structural coefficient from the 2SLS probability limit. When $K/n\to\alpha>0$, the ordinary centered-at-$\beta$ large-sample interpretation is generally not justified by this result. Also, a normal approximation around $\beta+B$ is not, by itself, an inference procedure for $\beta$. The shifted center and the variance formula answer different questions: the former describes the estimator’s limiting location, and the latter describes its local sampling variation around that location.

Key takeaway: Under Bekker asymptotics, $K/n\to\alpha>0$ leaves projected first-stage noise in both the numerator and denominator of 2SLS. With endogenous errors, this produces the limiting shift $\alpha\sigma_{uv}/(Q+\alpha\sigma_{vv})$; the limiting distribution is centered at the shifted value unless that bias vanishes.

Check Your Understanding

Use the model and assumptions above to answer the following questions.

In the Bekker regime, what does $K/n\to\alpha>0$ imply about the number of instruments as the sample grows?
For $Q=2$, $\alpha=0.10$, $\sigma_{vv}=1$, and $\sigma_{uv}=-0.50$, calculate the asymptotic bias term and state its sign.
Why does $v'P_Zv/n$ have a nonzero limit when $K/n\to\alpha>0$?
What is the center of the limiting distribution of 2SLS, and how does it differ from $\beta$ when $B\ne0$?
If $K$ stays fixed while $n$ increases, what happens to the bias term under the stated framework?