The Weak-IV Limit of Two-Stage Least Squares

Why the 2SLS Limit Changes Under Weak Instruments

With a strong instrument, the first-stage relationship becomes increasingly well determined as the sample grows. Under a weak-instrument sequence, the first-stage coefficient shrinks as the sample grows, so its signal remains comparable to random variation in the first-stage sample cross-product. The denominator of just-identified two-stage least squares (2SLS) therefore does not settle to a nonzero constant. Instead, it has a normal limit shifted by the local first-stage signal.

This shift produces a ratio of correlated normal variables as the limiting distribution of 2SLS. The ratio form is the central feature: increasing local first-stage strength moves the denominator's mean, but it does not remove the denominator's random variation at this asymptotic scale.

Model and Assumptions

Definition: In the scalar just-identified IV model, the outcome $y_i$, endogenous regressor $x_i$, and instrument $z_i$ satisfy $$ y_i=\beta x_i+u_i,\qquad x_i=\pi_n z_i+v_i,\qquad i=1,\ldots,n. $$ The structural parameter $\beta$ is fixed. The first-stage coefficient follows a local-to-zero sequence $\pi_n=c/\sqrt n$, where $c$ is a fixed real number. The just-identified 2SLS estimator, when its denominator is nonzero, is $$ \hat{\beta}_{2SLS} =\frac{\sum_{i=1}^n z_i y_i}{\sum_{i=1}^n z_i x_i}. $$

For the result below, assume the triples $(z_i,v_i,u_i)$ are independent and identically distributed, with $E[z_i^2]=q_{zz}$ satisfying $0<q_{zz}<\infty$, $E[z_i v_i]=E[z_i u_i]=0$, and finite second moments for the vector $(z_i v_i,z_i u_i)$. Write its covariance matrix as $$ \Omega= \begin{pmatrix} \Omega_{vv}&\Omega_{vu}\\ \Omega_{vu}&\Omega_{uu} \end{pmatrix} = \operatorname{Var} \begin{pmatrix} z_i v_i\\ z_i u_i \end{pmatrix}, \qquad \Omega_{vv}>0. $$ These assumptions give the joint central limit theorem for the two cross-products and ensure that the limiting first-stage cross-product has a continuous normal distribution. They allow $u_i$ and $v_i$ to be correlated, as is usual when $x_i$ is endogenous.

Deriving the Joint Reduced-Form Limit

The sample cross-product between instrument and regressor decomposes into a local signal and a random disturbance: $$ \frac{1}{\sqrt n}\sum_{i=1}^n z_i x_i =c\left(\frac{1}{n}\sum_{i=1}^n z_i^2\right) +\frac{1}{\sqrt n}\sum_{i=1}^n z_i v_i. $$ By the law of large numbers, the term in parentheses converges in probability to $q_{zz}$. By the joint central limit theorem, the sums involving $z_i v_i$ and $z_i u_i$, divided by $\sqrt n$, converge jointly to a mean-zero normal vector. Thus the local first-stage signal is $c q_{zz}\sqrt n+o_p(\sqrt n)$ before scaling the cross-product by $1/\sqrt n$; it is zero when $c=0$. The disturbance term in the unscaled cross-product is of order $\sqrt n$. After scaling, the signal converges to $c q_{zz}$, while the disturbance has a nondegenerate normal limit.

The outcome cross-product follows from the structural equation: $$ \frac{1}{\sqrt n}\sum_{i=1}^n z_i y_i =\beta\frac{1}{\sqrt n}\sum_{i=1}^n z_i x_i +\frac{1}{\sqrt n}\sum_{i=1}^n z_i u_i. $$ The same first-stage disturbance appears in both cross-products. That shared term is why the two reduced-form limits, and hence the numerator and denominator of the estimator's limiting ratio, are generally correlated.

Theorem: Under the assumptions above, let $(G_v,G_u)$ be a mean-zero bivariate normal vector with covariance matrix $\Omega$. Then $$ \left( \frac{1}{\sqrt n}\sum_{i=1}^n z_i x_i,\, \frac{1}{\sqrt n}\sum_{i=1}^n z_i y_i \right) \xrightarrow{d} \left( c q_{zz}+G_v,\, \beta(c q_{zz}+G_v)+G_u \right). $$ Consequently, $$ \hat{\beta}_{2SLS}\xrightarrow{d} \frac{\beta(c q_{zz}+G_v)+G_u}{c q_{zz}+G_v} =\beta+\frac{G_u}{c q_{zz}+G_v}. $$

Proof. The law of large numbers gives $n^{-1}\sum_i z_i^2\xrightarrow{p}q_{zz}$. The multivariate central limit theorem gives $$ \frac{1}{\sqrt n}\sum_{i=1}^n \begin{pmatrix} z_i v_i\\ z_i u_i \end{pmatrix} \xrightarrow{d} \begin{pmatrix} G_v\\ G_u \end{pmatrix}. $$ Since $\pi_n=c/\sqrt n$, substituting the first-stage equation yields $$ \frac{1}{\sqrt n}\sum_i z_i x_i =c\left(\frac{1}{n}\sum_i z_i^2\right) +\frac{1}{\sqrt n}\sum_i z_i v_i \xrightarrow{d}c q_{zz}+G_v, $$ by Slutsky's theorem. Substituting $y_i=\beta x_i+u_i$ then gives the stated joint limit for the two cross-products.

Because $\Omega_{vv}>0$, $c q_{zz}+G_v$ is a normal random variable with positive variance and hence has probability zero of equaling zero. The ratio function is continuous whenever its denominator is nonzero, so the continuous mapping theorem gives the ratio limit. In particular, the probability that the sample denominator is exactly zero tends to zero: for any $\epsilon>0$, that probability is bounded by the probability that its scaled version lies within $\epsilon$ of zero, whose limit can be made arbitrarily small as $\epsilon$ decreases to zero. The estimator can be assigned any value when its denominator is zero without changing its limiting distribution. $\square$

How First-Stage Strength Enters

The limiting first-stage cross-product is $c q_{zz}+G_v$. Its mean $c q_{zz}$ is the local signal; its standard deviation is $\sqrt{\Omega_{vv}}$. A useful standardized measure of the local shift is $$ \delta=\frac{c q_{zz}}{\sqrt{\Omega_{vv}}}, \qquad \lambda_{\mathrm{local}}=\delta^2 =\frac{c^2 q_{zz}^2}{\Omega_{vv}}. $$ The sign of $c$ determines the direction of the shift, while its magnitude relative to $\sqrt{\Omega_{vv}}$ determines how far the denominator's mean lies from zero in standard-deviation units.

In the common special case where $v_i$ is independent of $z_i$ and has variance $\sigma_v^2$, $\Omega_{vv}=E[z_i^2v_i^2]=q_{zz}\sigma_v^2$. Therefore $\lambda_{\mathrm{local}}=c^2q_{zz}/\sigma_v^2$. The general expression above is useful when the first-stage disturbance has a variance that depends on the instrument.

Worked Example: Zero Local First-Stage Signal

Suppose $z_i\sim N(0,1)$, independently of $(v_i,u_i)$, and let $(v_i,u_i)$ be bivariate normal with variances $1$ and $4$, and covariance $1$. Take $c=0$ and $\beta=2$. Independence and $E[z_i^2]=1$ give $$ q_{zz}=1,\qquad \Omega= \begin{pmatrix} E[z_i^2v_i^2]&E[z_i^2v_i u_i]\\ E[z_i^2v_i u_i]&E[z_i^2u_i^2] \end{pmatrix} = \begin{pmatrix} 1&1\\ 1&4 \end{pmatrix}. $$

The covariance matrix is valid: its determinant is $1(4)-1^2=3>0$. The theorem gives $$ \hat{\beta}_{2SLS}\xrightarrow{d}2+\frac{G_u}{G_v}, \qquad \begin{pmatrix}G_v\\G_u\end{pmatrix} \sim N\left( \begin{pmatrix}0\\0\end{pmatrix}, \begin{pmatrix}1&1\\1&4\end{pmatrix} \right). $$ The first-stage signal is identically zero, while the first-stage disturbance $G_v$ still has variance $1$. The limit is a ratio involving correlated normal variables, not a normal distribution centered at $2$.

Worked Example: A Nonzero Local Signal

Keep the same distributions and take $c=2$, still with $q_{zz}=1$, $\Omega_{vv}=1$, and $\beta=2$. The limiting denominator is $2+G_v$, with $G_v\sim N(0,1)$, and the estimator limit is $$ 2+\frac{G_u}{2+G_v}. $$ The standardized shift is $\delta=2/\sqrt{1}=2$, so $\lambda_{\mathrm{local}}=4$. The probability that the limiting denominator is positive is $$ P(2+G_v>0)=P(G_v>-2)=\Phi(2)\approx 0.9772, $$ where $\Phi$ is the standard normal distribution function. The denominator is more often positive than in the zero-signal case, but it remains random and can still be close to zero. The local shift changes the ratio's distribution; it does not turn that ratio into a normal variable.

Worked Example: Instrument Scale and Disturbance Covariance

Let $z_i$ equal $2$ or $-2$, each with probability $1/2$, and independently let $(v_i,u_i)$ have mean zero, variances $9$ and $16$, and covariance $6$. This error covariance matrix is positive definite because $9(16)-6^2=108>0$. Here $q_{zz}=4$ and $$ \Omega=E[z_i^2] \begin{pmatrix}9&6\\6&16\end{pmatrix} = \begin{pmatrix}36&24\\24&64\end{pmatrix}. $$ For $c=3$, the first-stage signal is $c q_{zz}=12$, and the first-stage disturbance standard deviation is $\sqrt{36}=6$. Thus $\delta=12/6=2$ and $\lambda_{\mathrm{local}}=4$. For $\beta=1$, the estimator limit is $$ 1+\frac{G_u}{12+G_v}, \qquad \operatorname{Corr}(G_v,G_u)=\frac{24}{\sqrt{36\cdot64}}=\frac{1}{2}. $$ Also, $P(12+G_v>0)=\Phi(12/6)=\Phi(2)\approx0.9772$. This example verifies that the instrument's second moment and the disturbance covariance both enter the limit: the signal is $12$, not merely $c=3$.

Why the Ratio Matters

For every fixed $c$, the limiting denominator retains nonzero variance. Even when the local signal is positive, realizations near zero can make the ratio highly variable. The same reduced-form noise $G_v$ appears in the numerator through $\beta(cq_{zz}+G_v)$ and in the denominator, while $G_u$ is correlated with $G_v$ whenever $\Omega_{vu}\ne0$. Thus neither the numerator nor denominator can generally be treated as fixed when describing the limit.

This limit also shows why a strong-instrument normal approximation cannot simply be carried over to the local-to-zero sequence. The ratio is the asymptotic distribution itself under these assumptions; the local strength parameter shifts its denominator but does not eliminate its randomness. This distributional fact is important for understanding why conventional 2SLS inference needs separate analysis under weak identification.

Common Mistake

Wrong reasoning: “Because $\pi_n=c/\sqrt n$, the first-stage signal is always of order $\sqrt n$ in the unscaled cross-product.” This treats $c$ as necessarily nonzero and confuses the signal with the random disturbance.
Correction: The unscaled signal is $(c/\sqrt n)\sum_i z_i^2=cq_{zz}\sqrt n+o_p(\sqrt n)$. It is identically zero when $c=0$. The disturbance $\sum_i z_i v_i$ is of order $\sqrt n$ under the stated central limit theorem, including when $c=0$. After dividing both cross-products by $\sqrt n$, the signal becomes $cq_{zz}$, and the disturbance remains random in the limit.
Another tempting shortcut: “The numerator and denominator have normal limits, so their ratio is normal.” A ratio of normal variables is not generally normal. The theorem gives the joint normal pair first, then applies the ratio map; it does not assert that the resulting ratio has a normal distribution.

Key takeaway: Under $\pi_n=c/\sqrt n$, just-identified 2SLS converges to $\beta+G_u/(cq_{zz}+G_v)$. The local first-stage signal shifts the denominator by $cq_{zz}$, while first-stage sampling noise remains in that denominator and is generally correlated with the reduced-form outcome noise.

Check Your Understanding

Use the model and assumptions in this tutorial to answer the following questions.

In the first worked example, why does the denominator retain a random component even though the local signal is zero?
For the second example, calculate the standardized local shift if $c=1$ instead of $c=2$.
Explain why $G_v$ appears in both the numerator and denominator of the limiting estimator.
In the third example, verify the value of $\operatorname{Corr}(G_v,G_u)$ from the covariance matrix.
What assumption ensures that the limiting denominator equals zero with probability zero?