set.seed(1848)
n <- 1000
z <- runif(n)
S <- matrix(c(1, 0.5,
0.5, 1), 2, 2, byrow = TRUE)
library(mvtnorm)
errors <- rmvnorm(n, mean = c(0, 0), sigma = S)
u <- errors[, 1]
v <- errors[, 2]
pi <- 0.7
beta <- 1.2
x <- pi * z + v
y <- beta * x + u
library(ivreg)
library(tidyverse)
library(broom)
get_coef_se <- function(results, varname) {
results |>
tidy() |>
filter(term == varname) |>
select(estimate, std.error)
}
iv <- ivreg(y ~ x | z) |>
get_coef_se('x')
first_stage <- lm(x ~ z) |>
get_coef_se('z')
reduced_form <- lm(y ~ z) |>
get_coef_se('z')Puzzler No. 4 14 August 2026 5 min
Econometrics Puzzler #4: Rescaling the Reduced Form
Consider the simplest textbook instrumental variables (IV) model with one endogenous regressor \(X\) and a single valid instrument \(Z\). You are given the estimated slope coefficient \(\hat{\gamma}\) from the reduced form regression of \(Y\) on \(Z\) along with its standard error \(\text{SE}(\hat{\gamma})\). You are also given the estimated slope coefficient \(\hat{\pi}\) from the first stage regression of \(X\) on \(Z\). The IV estimate equals \(\hat{\gamma}/ \hat{\pi}\). Does its standard error equal \(\text{SE}(\hat{\gamma})/\hat{\pi}\)?
1 Solution
NoteSolution
Here’s why it’s tempting to answer yes. If \(Y\) is measured in dollars and \(Z\) is measured in grams, then \(\hat{\gamma}\) and \(\text{SE}(\hat{\gamma})\) are both measured in dollars per gram. Now suppose that you wanted to express both quantities in dollars per kilogram rather than dollars per gram. No problem: simply multiply each by 1000. The IV logic from above looks analogous: the IV estimator is simply a rescaled version of the reduced form. So does the same relationship hold for the standard error? Well, there’s an equally tempting reason to say no. Since \(\hat{\pi}\) is estimated, rather than a known constant, dividing by it should add sampling variability of its own. This reasoning suggests that \(\text{SE}(\hat{\gamma})/\hat{\pi}\) should be too small. So which is it? When in doubt, simulate:
The IV slope estimate is exactly equal to the ratio of reduced form and first stage slopes, as it should be:
est_by_hand est_ivreg
1.160068 1.160068 but the IV standard error does not share the same proportionality relationship to the reduced form standard error:
se_by_hand se_ivreg
0.3407198 0.1861929 So the answer is no, but it seems that both of the tempting intuitions have failed us here: the naive standard error isn’t too small; it’s nearly twice as large as the correct one. What’s going on?
To answer this question, we’ll start by writing down the structural equation \(Y = \beta X + U\) and the first stage equation \(X = \pi Z + V\).1 Combining them gives the reduced form: \[ \begin{aligned} Y &= \beta X + U = \beta(\pi Z + V) + U\\ &= (\beta\pi) Z + (\beta V + U)\\ &= \gamma Z + \varepsilon \end{aligned} \] where \(\gamma \equiv \beta \pi\) and \(\varepsilon \equiv \beta V + U\). Now, the IV estimator can be expanded according to \[ \begin{aligned} \hat{\beta} &= (\underline{Z}'\underline{X})^{-1}(\underline{Z}'\underline{Y}) = (\underline{Z}'\underline{X})^{-1}\left[\underline{Z}'(\beta \underline{X} + \underline{U})\right]\\ &= \beta + (\underline{Z}'\underline{X})^{-1}(\underline{Z}'\underline{U}) \end{aligned} \] where \(\underline{Z}'\) denotes the row vector \((Z_1, \dots, Z_n)\) and analogously for \(\underline{Y}, \underline{X}\) and \(\underline{U}\). Thus, subtracting \(\beta\) from both sides and multiplying by \(\sqrt{n}\), \[ \sqrt{n}(\hat{\beta} - \beta) = \left(\frac{\underline{Z}'\underline{X}}{n}\right)^{-1}\left(\frac{\underline{Z}'\underline{U}}{\sqrt{n}}\right) \xrightarrow{d} \frac{1}{\mathbb{E}(ZX)} \times N\left(0, \mathbb{E}[Z^2U^2]\right) \] by the central limit theorem. Proceeding similarly for the reduced form estimator: \[ \begin{aligned} \hat{\gamma} &= (\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{Y}) = (\underline{Z}'\underline{Z})^{-1}\left[\underline{Z}'(\gamma \underline{Z} + \underline{\varepsilon})\right]\\ &= \gamma + (\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{\varepsilon}) \end{aligned} \] and therefore, \[ \begin{aligned} \sqrt{n}(\hat{\gamma} - \gamma) = \left( \frac{\underline{Z}'\underline{Z}}{n}\right)^{-1}\left( \frac{\underline{Z}'\underline{\varepsilon}}{\sqrt{n}}\right) \xrightarrow{d} \frac{1}{\mathbb{E}(Z^2)} \times N(0, \mathbb{E}[Z^2\varepsilon^2]) \end{aligned} \] To keep the math simple, I’ll assume that \(Z\) is independent of both \(U\) and \(V\), as in the simulation above.2 Under this assumption \(Z\) is also independent of \(\varepsilon\) and thus \[ \begin{aligned} \mathbb{E}(Z^2 \varepsilon^2) &= \mathbb{E}(Z^2)\mathbb{E}(\varepsilon^2) = \sigma_z^2 \sigma_\varepsilon^2\\ \mathbb{E}(Z^2 U^2) &= \mathbb{E}(Z^2)\mathbb{E}(U^2) = \sigma_z^2 \sigma_u^2 \end{aligned} \] since \(Z, \varepsilon\) and \(U\) are mean zero. Thus, the asymptotic variances of the IV and reduced form estimators are given by \[ \begin{aligned} \text{AVAR}(\hat{\beta}) &= \left(\frac{1}{\mathbb{E}[ZX]}\right)^2 \mathbb{E}(Z^2 U^2) = \left(\frac{1}{\pi \sigma_z^2}\right)^2 \sigma_z^2 \sigma_u^2 = \frac{\sigma_u^2}{\pi^2 \sigma_z^2}\\ \text{AVAR}(\hat{\gamma}) &= \left(\frac{1}{\mathbb{E}[Z^2]}\right)^2 \mathbb{E}(Z^2 \varepsilon^2) = \left( \frac{1}{\sigma_z^2}\right)^2 \sigma_z^2 \sigma_\varepsilon^2 = \frac{\sigma_\varepsilon^2}{\sigma_z^2} \end{aligned} \] since the first stage slope coefficient \(\pi\) equals \(\mathbb{E}[ZX]/ \text{Var}(Z)\). And now we have our answer! Taking square roots gives \[ \sqrt{\text{AVAR}(\hat{\beta})} = \frac{\sigma_u}{\pi \sigma_z}, \quad \sqrt{\text{AVAR}(\hat{\gamma})} = \frac{\sigma_\varepsilon}{\sigma_z} \] The respective standard errors are estimators of these quantities, divided by \(\sqrt{n}\). Fundamentally, the standard error of the IV estimator is governed by the variation in the structural error \(U\) while that of the reduced form estimator is governed by the variation in the reduced form error \(\varepsilon\). It follows that \[ \sqrt{\text{AVAR}(\hat{\beta})} = \left(\frac{\sigma_u}{\sigma_\varepsilon}\right) \cdot \frac{\sqrt{\text{AVAR}(\hat{\gamma})}}{\pi} \] so there really is a factor of \(\pi\) in the relationship between the IV and reduced form standard errors! The trouble is that there is a second factor, \(\sigma_u/\sigma_\varepsilon\), and this does not equal one. Instead, \[ \sigma_\varepsilon^2 = \text{Var}(\beta V + U) = \beta^2 \sigma_v^2 + \sigma_u^2 + 2\beta \,\text{Cov}(U,V) \] so that \(\sigma_\varepsilon^2 - \sigma_u^2 = \beta^2 \sigma_v^2 + 2\beta \text{Cov}(U,V)\). The first term in the difference, \(\beta^2 \sigma_v^2\), is always positive but the second is negative whenever the sign of \(\beta\) disagrees with that of \(\text{Cov}(U,V)\). This means that \(\sqrt{\text{AVAR}(\hat{\beta})}\) can be either larger or smaller than \(\sqrt{\text{AVAR}(\hat{\gamma})}/\pi\) depending on \(\beta\) and on the variances and covariance of \((U,V)\). In our simulation \(\beta\) and \(\text{Cov}(U,V)\) are both positive, so \(\sigma_\varepsilon > \sigma_u\). This is why the naive standard error was too large.
What about our second intuition? Expand \(\hat{\gamma}\) again, but this time substitute the structural equation \(Y = \beta X + U\) in place of the reduced form: \[ \begin{aligned} \hat{\gamma} &= (\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{Y}) = (\underline{Z}'\underline{Z})^{-1}\left[\underline{Z}'(\beta \underline{X} + \underline{U})\right]\\ &= \beta (\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{X}) + (\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{U})\\ &= \beta \hat{\pi} + (\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{U}) \end{aligned} \] where \(\hat{\pi} = (\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{X})\) is the first stage OLS slope. This identity holds in every sample. Whatever sampling error \(\hat{\pi}\) contains is carried along in the term \(\beta \hat{\pi}\), so dividing through by \(\hat{\pi}\) replaces it with the constant \(\beta\): \[ \hat{\beta} = \frac{\hat{\gamma}}{\hat{\pi}} = \beta + \frac{(\underline{Z}'\underline{Z})^{-1}(\underline{Z}'\underline{U})}{\hat{\pi}} = \beta + \frac{\underline{Z}'\underline{U}}{\underline{Z}'\underline{X}} \] The first stage error is now in the denominator, \(\underline{Z}'\underline{X} = \pi \underline{Z}'\underline{Z} + \underline{Z}'\underline{V}\), which is of course still random. But the rescaling that sets up the central limit theorem divides it by \(n\), and \(\underline{Z}'\underline{X}/n\) converges in probability to \(\mathbb{E}(ZX)\), a constant. Its reciprocal is precisely the \(1/\mathbb{E}(ZX)\) multiplying the normal limit in our expression for \(\sqrt{n}(\hat{\beta} - \beta)\) above.
Finally, we can check to see that our calculations are correct by comparing them to the simulation results from above. The standard errors reported by lm and ivreg are finite-sample counterparts of our asymptotic standard errors, so the same relationship should hold with \(\hat{\sigma}_u\) and \(\hat{\sigma}_\varepsilon\) in place of \(\sigma_u\) and \(\sigma_\varepsilon\). The reduced form residuals \(\hat{\varepsilon}_i\) and the first stage residuals \(\hat{V}_i\) give us the structural residuals, since \(\hat{\varepsilon}_i - \hat{\beta}\hat{V}_i = Y_i - \hat{\beta}X_i = \hat{U}_i\):
2 For more on the various notions of “independence” in econometrics, see this post.
1 To keep the math simple, I implicitly assume here that \((Y,X,Z)\) are mean zero so that there is no need to include an intercept in either equation. Equivalently, we could simply subtract the mean from each random variable before proceeding.