Weak Identification with Many Instruments

Why Instrument Count Changes the Weak-Instrument Problem

Adding valid instruments can seem like an uncomplicated way to improve a first stage: the fitted values can use more directions in the data. But each added direction also gives the first stage another opportunity to fit random variation. When the number of instruments is large relative to the sample size, this fitted noise can be substantial. If the structural error and first-stage error are correlated, the same noise can also enter the 2SLS numerator and produce a nonvanishing bias.

The key distinction is between a fixed number of weak instruments and a growing number of individually weak instruments. With a fixed count, local weakness can leave the 2SLS estimator with a nondegenerate weak-IV limit. With a growing count, the aggregate first-stage fit may be substantial even when no single instrument is strongly informative; at the same time, the growing projection of noise can shift the 2SLS probability limit.

A Model for Studying Instrument Count

Consider one endogenous regressor, no controls, and $K$ instruments. Write the structural and first-stage equations as

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

Here $Z$ is a nonrandom $n\times K$ instrument matrix with full column rank, $\pi$ is the first-stage coefficient vector, and $P_Z=Z(Z'Z)^{-1}Z'$ is the projection matrix onto the instrument space. Assume the rows of $(u,v)$ are independent and identically distributed bivariate normal with mean zero and covariance matrix

\begin{pmatrix} \sigma_u^2 & \sigma_{uv}\\ \sigma_{uv} & \sigma_v^2 \end{pmatrix}, \qquad \sigma_v^2>0.

The instruments are exogenous: they are fixed independently of the errors. Endogeneity of $x$ is represented by $\sigma_{uv}\ne 0$. The 2SLS estimator in this model is

\hat{\beta}_{2SLS} =\frac{x'P_Zy}{x'P_Zx} =\beta+\frac{x'P_Zu}{x'P_Zx},

provided the denominator is nonzero. The equality follows by substituting $y=\beta x+u$. It isolates the issue: instrument count affects both the projected first-stage variation in the denominator and the projected error in the numerator.

Definition: A many-instrument asymptotic sequence has $K=K_n$ increasing with $n$. The benchmark considered below assumes $K/n\to\alpha$, where $0<\alpha<1$, and the squared first-stage signal satisfies $(Z\pi)'(Z\pi)/n\to\mu^2$ for a finite $\mu^2\ge 0$. A fixed-count sequence keeps $K$ constant as $n$ grows. These are distinct asymptotic regimes, not two descriptions of the same limit.

First-Stage Fit Includes Projected Noise

Let $s=Z\pi$, the systematic part of the first stage. Because $s$ lies in the column space of $Z$, $P_Zs=s$. The uncentered first-stage $R^2$ in this no-intercept formulation is $x'P_Zx/(x'x)$. The numerator contains the signal $s's$, but it also contains $v'P_Zv$: the portion of the first-stage error that happens to lie in the instrument space.

Theorem: Suppose $K/n\to\alpha\in[0,1)$, $s's/n\to\mu^2<\infty$, and the error assumptions above hold. Then $$ \frac{x'P_Zx}{n}\xrightarrow{p}\mu^2+\alpha\sigma_v^2, \qquad \frac{x'x}{n}\xrightarrow{p}\mu^2+\sigma_v^2. $$ Consequently, the uncentered first-stage $R^2$ converges in probability to $$ \frac{\mu^2+\alpha\sigma_v^2}{\mu^2+\sigma_v^2}. $$

Proof. Expand the projected sum of squares:

$$ x'P_Zx=s's+2s'v+v'P_Zv. $$

Since $s'v$ has mean zero and variance $\sigma_v^2s's$, $(s'v)/n\xrightarrow{p}0$: its variance after dividing by $n$ is $\sigma_v^2s's/n^2\to0$. In an orthonormal basis for the $K$-dimensional instrument space, $v'P_Zv$ is a sum of $K$ independent squared $N(0,\sigma_v^2)$ variables. Its mean is $K\sigma_v^2$ and its variance is $2K\sigma_v^4$. Thus $v'P_Zv/n\xrightarrow{p}\alpha\sigma_v^2$, by the mean and variance limits and Chebyshev’s inequality. Combining these limits proves the first claim.

For the total sum of squares, $x'x=s's+2s'v+v'v$. The same variance argument gives $s'v/n\xrightarrow{p}0$, while $v'v/n\xrightarrow{p}\sigma_v^2$ by the law of large numbers. Therefore $x'x/n\xrightarrow{p}\mu^2+\sigma_v^2$. This limit is positive, so the continuous mapping theorem gives the stated ratio. $\square$

Worked Example: Fit From Instruments With No Signal

Suppose $K/n\to0.20$ and $\pi=0$, so $\mu^2=0$. Let $\sigma_v^2=1$. The limiting first-stage $R^2$ is

\frac{0+0.20(1)}{0+1}=0.20.

There is no population first-stage signal in this example. The limiting fit comes entirely from projecting the noise $v$ onto a space whose dimension is about one fifth of the sample size. A first-stage fit of this size therefore cannot, by itself, be read as evidence that the instruments carry corresponding population signal.

The Many-Instrument Limit of 2SLS

The same projected noise that raises first-stage fit can affect the 2SLS numerator. In particular, $v$ and $u$ have covariance $\sigma_{uv}$. Each instrument-space direction contributes, on average, $\sigma_{uv}$ to $v'P_Zu$. As the number of directions grows in proportion to $n$, this contribution need not vanish after division by $n$.

Theorem: Under the model assumptions above, if $K/n\to\alpha\in[0,1)$ and $s's/n\to\mu^2<\infty$, then $$ \hat{\beta}_{2SLS}\xrightarrow{p} \beta+\frac{\alpha\sigma_{uv}}{\mu^2+\alpha\sigma_v^2}, $$ provided the limiting denominator is positive.

Proof. Since $P_Zs=s$, expand the numerator of the estimation error as

$$ x'P_Zu=s'u+v'P_Zu. $$

The first term has mean zero and variance $\sigma_u^2s's$, so $s'u/n\xrightarrow{p}0$. To evaluate the second term, take an orthonormal basis for the instrument space. In that basis, $v'P_Zu$ is a sum of $K$ independent products $V_jU_j$, where each pair has the bivariate error distribution above. Each product has mean $\sigma_{uv}$ and finite variance $\sigma_v^2\sigma_u^2+\sigma_{uv}^2$. The variance of the sum is therefore $K(\sigma_v^2\sigma_u^2+\sigma_{uv}^2)$. Dividing by $n$, its mean tends to $\alpha\sigma_{uv}$ and its variance tends to zero because $K/n^2\to0$. Hence $(x'P_Zu)/n\xrightarrow{p}\alpha\sigma_{uv}$.

The first theorem gives $(x'P_Zx)/n\xrightarrow{p}\mu^2+\alpha\sigma_v^2$. If this limit is positive, taking the ratio yields the claimed probability limit. $\square$

The limit displays two ingredients: the aggregate first-stage signal $\mu^2$, and the instrument-count contribution $\alpha\sigma_v^2$. When $\sigma_{uv}\ne0$ and $\alpha>0$, the latter enters both numerator and denominator. Thus, even if the instruments are exogenous, the 2SLS estimator need not converge to $\beta$ in this many-instrument regime. If instead $K/n\to0$ while $\mu^2>0$, the displayed bias term vanishes. This does not guarantee good behavior under every weak-signal sequence: if the signal also shrinks, the relative sizes of signal and projected noise matter.

Worked Example: Calculating the Many-Instrument Limit

Suppose $K/n\to0.10$, $\mu^2=0.04$, $\sigma_v^2=1$, and $\sigma_{uv}=0.60$. The limiting 2SLS bias is

\frac{0.10(0.60)}{0.04+0.10(1)} =\frac{0.06}{0.14} =\frac{3}{7} \approx0.429.

The corresponding probability limit is $\beta+3/7$, not $\beta$. The assumptions are compatible with a valid covariance matrix: for example, $\sigma_u^2=1$ gives covariance matrix with diagonal entries 1 and off-diagonal entries 0.60, whose determinant is $1-0.60^2=0.64>0$. The positive bias here follows from the positive covariance between the structural and first-stage errors.

Fixed-Count Weakness Is a Different Limit

For a fixed number of instruments, a local-to-zero first stage can leave only a finite amount of signal in the instrument space as $n$ grows. In that case, projected signal and projected errors can remain on the same scale. The estimator’s limit is then generally a random ratio rather than the deterministic ratio in the many-instrument theorem.

For a direct illustration, take one instrument, $\sigma_v^2=\sigma_u^2=1$, and $\sigma_{uv}=\rho$ with $|\rho|<1$. Choose a sequence of first stages whose projected signal has length $\sqrt{\lambda}$, where $\lambda\ge0$ is fixed. In the one-dimensional instrument space, write the projected first-stage error as $V\sim N(0,1)$, and the projected structural error as $U\sim N(0,1)$, with $\operatorname{Cov}(U,V)=\rho$. The weak-IV limit of the estimation error in this setup is

\hat{\beta}_{2SLS}-\beta \ \xrightarrow{d}\ \frac{U}{\sqrt{\lambda}+V}.

This limit does not collapse to zero for fixed $\lambda$. The denominator can also be close to zero, which explains why the ratio can be unstable. The fixed-count case is not obtained by simply substituting a small value for $\alpha$ in the many-instrument probability limit: its random denominator remains part of the limiting distribution.

Worked Example: A Fixed-Count Local Weak Instrument

Let there be one instrument and set $\lambda=1$, $\rho=0.50$. The limiting estimation error is

\frac{U}{1+V}, \qquad \begin{pmatrix}U\\V\end{pmatrix} \sim N\!\left( \begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}1&0.50\\0.50&1\end{pmatrix} \right).

The denominator is $1+V$, where $V\sim N(0,1)$, so it has positive probability of being arbitrarily close to zero. For example, $P(|1+V|<0.1)=P(-1.1<V<-0.9)>0$. The numerator is not identically zero, so the ratio does not converge to a constant. This calculation illustrates why a fixed-count weak-IV limit must retain the joint randomness of the reduced-form errors rather than replacing numerator and denominator by their expectations.

What “Many Individually Weak” Means

A growing instrument set can contain many instruments, none of which has a first-stage contribution that increases without bound on its own. For example, with columns normalized so that $z_j'z_j=n$, suppose $K$ is proportional to $n$, and the signal is spread roughly evenly across the columns. If $s's/n\to\mu^2$, then $\pi_j^2$ is on the order of $\mu^2/K$, so each instrument’s individual concentration contribution $n\pi_j^2$ is on the order of $n\mu^2/K$, a constant. The aggregate signal can still be of order $n$. “Individually weak” therefore does not imply that the collection has no aggregate first-stage signal.

At the same time, a large aggregate first-stage fit does not remove the many-instrument effect. The projected error contributes roughly $K\sigma_v^2$ to the first-stage sum of squares, while the correlated projected error contributes roughly $K\sigma_{uv}$ to the 2SLS numerator. These are different from the fixed-$K$ local-weak mechanism, where a finite-dimensional random ratio persists.

Common Mistake and Practical Interpretation

Wrong reasoning: “Adding valid instruments can only improve first-stage fit, so a larger fitted $R^2$ shows stronger identification.”
Correction: More instruments expand the projection space and can raise fitted $R^2$ by capturing noise. When $K/n$ is not negligible, the limiting first-stage fit includes the term $\alpha\sigma_v^2$, even with no first-stage signal.
Wrong reasoning: “Each instrument is weak, so their number does not matter.”
Correction: Many weak contributions can accumulate. Instrument count also changes the scale of projected noise, and under endogeneity this can create a nonvanishing 2SLS bias.
Wrong reasoning: “The fixed-number weak-IV limit applies unchanged when the instrument count grows.”
Correction: The asymptotic regimes differ. A fixed-count local-weak limit retains a random ratio; the many-instrument result above has a deterministic probability limit under its stated proportional-growth conditions.

The practical point is not that a growing instrument count must always produce severe bias. The size of the effect depends on the instrument-to-sample-size ratio, the aggregate signal, and the covariance between structural and first-stage errors. Rather, instrument count is part of the asymptotic specification. Interpreting first-stage fit or 2SLS behavior without considering that count can be misleading.

Check Your Understanding

Check that you can distinguish the source of fit from the source of signal.

If $K/n\to0.15$, $\mu^2=0$, and $\sigma_v^2=2$, what is the limiting uncentered first-stage $R^2$ in the stated model?
In the many-instrument probability limit, which covariance determines the sign of the asymptotic 2SLS bias when the denominator is positive?
Why can the projected first-stage error contribute order $n$ to $x'P_Zx$ when $K/n\to\alpha>0$?
What is the main asymptotic distinction between fixed-count local weakness and many instruments with $K/n\to\alpha>0$?
With $K/n\to0$ and a fixed positive signal limit $\mu^2$, what happens to the bias term in the many-instrument formula?