In many settings, the average treatment effect for the complier group, also known as the local average treatment effect (LATE), is estimated via two-stage least squares (TSLS).

Suppose $Y(d)$ is the potential outcome of interest, $D(z)$ the potential treatment status, and $Z$ a valid instrument. Both $D$ and $Z$ are binary. Hereafter, lowercase letters denote realized values of the corresponding random variables.

Furthermore, let $G = (D(0), D(1))$ be the tuple of potential treatment statuses, where CP denotes compliers, AT always-takers, and NT never-takers. The effect of $Z$ on $D$ is assumed to be monotonic, so there are no defiers.

Hence, $G = (0, 1) = \mathrm{CP}$, $G = (0, 0) = \mathrm{NT}$, and $G = (1, 1) = \mathrm{AT}$. The TSLS takes the following form:

\[\begin{align*} D(z) &= \eta_0 + \eta_1 Z + u \\ Y(d) &= \beta_0 + \beta_1 D + \varepsilon \end{align*}\]

where $u$ and $\varepsilon$ are, respectively, an unobserved resistance to treatment and an idiosyncratic shock.

If $Z$ is a valid instrument, then $\beta_1$ is the parametric estimate of the complier treatment effect:

\[\begin{align*} \beta_1 &= \frac{\mathbb{E}[Y|Z = 1] - \mathbb{E}[Y|Z = 0]}{\mathbb{E}[D|Z = 1] - \mathbb{E}[D|Z = 0]} \\ &= \frac{\mathrm{Cov}(Y, Z)}{\mathrm{Cov}(D, Z)} \\ &= \Delta^{\mathrm{CP}} \end{align*}\]

Yet, when the instrument is valid only under conditional ignorability given control variables $X$, TSLS does not recover a LATE-related parameter if the relationship between $Z$ and $X$ is not linear.

Blandhol, Bonney, Mogstad, and Torgovitsky (2026)—hereafter BBMT—show that a TSLS with controls produces a LATE-related quantity plus the treatment effects of other groups.

More precisely, let the TSLS with controls take the following form:

\[\begin{align*} D(z) &= \eta_0 + \eta_1 Z + \eta' X + u \\ Y(d) &= \beta_0 + \beta_1 D + \beta' X + \varepsilon \end{align*}\]

Instead of $\Delta^{\mathrm{CP}}$, BBMT demonstrate that $\beta_1$ gives the following quantity:

\[\begin{align*} \beta_1 = \bigg(\mathbb{E}[\Delta^{\mathrm{CP}}(X) \, \omega^{\mathrm{CP}}] + \mathbb{E}[\Delta^{\mathrm{AT}}(X) \, \omega^{\mathrm{AT}}]\bigg) \times \mathbb{E}[D\tilde{Z}]^{-1} \end{align*}\]

where $\Delta^{g}(X)$ is the conditional treatment effect of group $g \in {\mathrm{CP}, \mathrm{AT}, \mathrm{NT}}$ and $\omega^{g}$ is its weight, with the following form:

\[\begin{align*} \omega^{\mathrm{CP}} &= \pi_{\mathrm{CP}} \, \mathbb{E}[Z|X](1 - \mathbb{L}[Z|X]) \\ \omega^{\mathrm{AT}} &= \pi_{\mathrm{AT}} \, \mathbb{E}[\tilde{Z}|X] \\ \omega^{\mathrm{NT}} &= \pi_{\mathrm{NT}} \, \mathbb{E}[\tilde{Z}|X] \end{align*}\]

in which $\tilde{Z} = Z - \mathbb{L}[Z|X]$ is the residual from the linear projection of $Z$ on $X$:

\[\mathbb{L}[Z|X] = X'\mathbb{E}[XX']^{-1}\mathbb{E}[XZ]\]

and $\pi_{g} = \mathbb{P}(G = g|X)$ is the conditional probability of belonging to group $g$ given $X$. Obviously, when $\mathbb{E}[Z|X] = \mathbb{L}[Z|X]$, the second term in $\beta_1$ cancels out, leaving us with only the first term.

Furthermore, even when the linearity assumption holds, the equation implies that $\beta_1$ does not have the direct LATE interpretation it has in the just-identified case. Rather, it gives a convex combination of conditional LATEs, weighted by the variance of the instrument at each level of $X$.

BBMT’s findings demonstrate the importance of the Frisch–Waugh–Lovell (FWL) theorem for understanding the least-squares machinery. As Heckman and Vytlacil (2006, 671) ask, “[w]hat parameter is being identified by the instrument?”

It appears that the FWL theorem guides us by showing what parameter is being identified in the OLS. For example, in another setting, de Chaisemartin and D’Haultfoeuille (2020) show, via the FWL theorem, that the two-way fixed-effects estimator produces a biased result by comparing already-treated units with later-treated ones.

Proof

First step

The first step of the proof is to show that the coefficient of interest in the TSLS with controls can be represented by the following expression:

\[\begin{align*} \beta_1 &= \frac{\mathrm{Cov}(Y, \tilde{Z})}{\mathrm{Cov}(D, \tilde{Z})} \end{align*}\]

By the FWL theorem, $\eta_1$ and $\beta_1$ in the TSLS with controls can be obtained by regressing the residualized outcome on the residualized explanatory variable of interest with respect to $X$. Therefore, we can write

\[\begin{align*} \beta_1 &= \frac{\mathrm{Cov}(\tilde{Y}, \tilde{Z})}{\mathrm{Cov}(\tilde{D}, \tilde{Z})} \end{align*}\]

where

\[\begin{align*} \tilde{Y} &= Y - \mathbb{L}[Y|X] \\ \tilde{Z} &= Z - \mathbb{L}[Z|X] \\ \tilde{D} &= D - \mathbb{L}[D|X] \end{align*}\]

Let $\beta_V$ be the coefficient of the linear projection of $V$ on $X$, for $V \in {Y, D, Z}$. Then we have

\[\begin{align*} \beta_1 &= \frac{\mathrm{Cov}(Y - X'\beta_Y, Z - X'\beta_Z)}{\mathrm{Cov}(D - X'\beta_D, Z - X'\beta_Z)} \end{align*}\]

In the numerator, we have the following identity:

\[\begin{align*} \mathrm{Cov}(Y - X'\beta_Y, Z - X'\beta_Z) &= \mathrm{Cov}(Y, Z - X'\beta_Z) - \beta_Y' \, \mathrm{Cov}(X, Z - X'\beta_Z) \\ &= \mathrm{Cov}(Y, \tilde{Z}) \end{align*}\]

The last line follows from the fact that the covariance between $X$ and the projection residual $\varepsilon_Z = Z - X’\beta_Z$ is zero by construction.

Similarly, in the denominator we have:

\[\begin{align*} \mathrm{Cov}(D - X'\beta_D, Z - X'\beta_Z) &= \mathrm{Cov}(D, Z - X'\beta_Z) - \beta_D' \, \mathrm{Cov}(X, Z - X'\beta_Z) \\ &= \mathrm{Cov}(D, \tilde{Z}) \end{align*}\]

Hence,

\[\begin{align*} \beta_{1} &= \frac{\mathrm{Cov}(Y, \tilde{Z})}{\mathrm{Cov}(D, \tilde{Z})} \\ &= \frac{\mathbb{E}[Y\tilde{Z}]}{\mathbb{E}[D\tilde{Z}]} \end{align*}\]

since $\mathbb{E}[\tilde{Z}] = 0$ by construction.

Second step

The next step of the proof is to work on the numerator of $\beta_1$ in the previous equation. By the law of iterated expectations,

\[\begin{align*} \mathbb{E}[Y\tilde{Z}] &= \mathbb{E}[\mathbb{E}[Y\tilde{Z}|X]] \\ &= \mathbb{E}[\mathrm{Cov}(Y, \tilde{Z}|X)] + \mathbb{E}\bigg[\mathbb{E}[Y|X] \, \mathbb{E}[\tilde{Z}|X]\bigg] \end{align*}\]

The last line follows from the definition of covariance. Because $\mathbb{L}[Z|X]$ is a constant, $Y$ covaries only with $Z$ in the first term, giving us this identity:

\[\begin{align*} \mathbb{E}[\mathrm{Cov}(Y, \tilde{Z}|X)] + \mathbb{E}\bigg[\mathbb{E}[Y|X] \, \mathbb{E}[\tilde{Z}|X]\bigg] = \mathbb{E}[\mathrm{Cov}(Y, Z|X)] + \mathbb{E}\bigg[\mathbb{E}[Y|X] \, \mathbb{E}[\tilde{Z}|X]\bigg] \end{align*}\]

Third step

The third step is to work on the first term in the last equation. By the definition of covariance:

\[\begin{align*} \mathrm{Cov}(Y, Z|X) &= \mathbb{E}[YZ|X] - \mathbb{E}[Y|X] \, \mathbb{E}[Z|X] \end{align*}\]

Expanding the first term through iterated expectations:

\[\begin{align*} \mathbb{E}[YZ|X] &= \mathbb{E}[Y|X, Z = 1] \, \mathbb{P}(Z = 1|X) \\ &= \mathbb{E}[Y|X, Z = 1] \, \mathbb{E}[Z|X] \end{align*}\]

For the second term of the covariance, we have:

\[\begin{align*} \mathbb{E}[Y|X] \, \mathbb{E}[Z|X] &= \mathbb{E}[Z|X] \bigg(\mathbb{E}[Y|X, Z = 1] \, \mathbb{P}(Z = 1|X) + \mathbb{E}[Y|X, Z = 0] \, \mathbb{P}(Z = 0|X) \bigg) \\ &= \mathbb{E}[Z|X] \bigg(\mathbb{E}[Y|X, Z = 1] \, \mathbb{E}[Z|X] + \mathbb{E}[Y|X, Z = 0] \, \mathbb{P}(Z = 0|X) \bigg) \\ &= \mathbb{E}[Z|X] \bigg(\mathbb{E}[Y|X, Z = 1] \, \mathbb{E}[Z|X] + \mathbb{E}[Y|X, Z = 0] \, (1 - \mathbb{E}[Z|X]) \bigg) \\ &= \mathbb{E}[Z|X] \bigg(\mathbb{E}[Y|X, Z = 0] + \mathbb{E}[Z|X] \times (\mathbb{E}[Y|X, Z = 1] - \mathbb{E}[Y|X, Z = 0]) \bigg) \end{align*}\]

Combining the two:

\[\begin{align*} \mathrm{Cov}(Y, Z|X) &= \mathbb{E}[Z|X]\bigg(\mathbb{E}[Y|X, Z = 1] - \mathbb{E}[Y|X, Z = 0] - \mathbb{E}[Z|X] \times (\mathbb{E}[Y|X, Z = 1] - \mathbb{E}[Y|X, Z = 0]) \bigg) \\ &= (\mathbb{E}[Y|X, Z = 1] - \mathbb{E}[Y|X, Z = 0]) \times \mathbb{E}[Z|X](1 - \mathbb{E}[Z|X]) \end{align*}\]

Now consider the definition of the conditional LATE:

\[\begin{align*} \Delta^{\mathrm{CP}}(X) &= \frac{\mathbb{E}[Y|Z = 1, X] - \mathbb{E}[Y|Z = 0, X]}{\pi_{\mathrm{CP}}} \end{align*}\]

This gives the following equation:

\[\begin{align*} \mathrm{Cov}(Y, Z|X) &= \Delta^{\mathrm{CP}}(X) \, \pi_{\mathrm{CP}} \times \mathbb{E}[Z|X](1 - \mathbb{E}[Z|X]) \end{align*}\]

Fourth step

We now have the numerator of $\beta_1$ represented by the following expression:

\[\begin{align*} \mathbb{E}[Y\tilde{Z}] &= \mathbb{E}\bigg[\Delta^{\mathrm{CP}}(X) \, \pi_{\mathrm{CP}} \times \mathbb{E}[Z|X](1 - \mathbb{E}[Z|X])\bigg] + \mathbb{E}\bigg[\mathbb{E}[Y|X] \, \mathbb{E}[\tilde{Z}|X]\bigg] \end{align*}\]

The fourth step of the proof is to work on the second term of the above equation. Expanding the conditional expectation by iterating over $G$, we have the following identity:

\[\begin{align*} \mathbb{E}[Y|X] &= \mathbb{E}[Y(1)|X, G = \mathrm{AT}] \, \pi_{\mathrm{AT}} + \mathbb{E}[Y(0)|X, G = \mathrm{NT}] \, \pi_{\mathrm{NT}} \\ &\quad + \mathbb{E}[Y(1)Z + Y(0)(1 - Z)|X, G = \mathrm{CP}] \, \pi_{\mathrm{CP}} \end{align*}\]

Simplify the last term above as follows:

\[\begin{align*} \pi_{\mathrm{CP}} \, \mathbb{E}[Y(1)Z + Y(0)(1 - Z)|X, G = \mathrm{CP}] &= \pi_{\mathrm{CP}} \, \mathbb{E}[Y(0)|X, G = \mathrm{CP}] + \pi_{\mathrm{CP}} \, \mathbb{E}[Z(Y(1) - Y(0))|X, G = \mathrm{CP}] \\ &= \pi_{\mathrm{CP}} \, \mathbb{E}[Y(0)|X, G = \mathrm{CP}] + \pi_{\mathrm{CP}} \, \mathbb{E}[Y(1) - Y(0)|X, G = \mathrm{CP}, Z = 1] \, \mathbb{E}[Z|X] \\ &= \pi_{\mathrm{CP}} \, \mathbb{E}[Y(0)|X, G = \mathrm{CP}] + \pi_{\mathrm{CP}} \, \mathbb{E}[Y(1) - Y(0)|X, G = \mathrm{CP}] \, \mathbb{E}[Z|X] \\ &= \pi_{\mathrm{CP}} \, \mathbb{E}[Y(0)|X, G = \mathrm{CP}] + \pi_{\mathrm{CP}} \, \Delta^{\mathrm{CP}}(X) \, \mathbb{E}[Z|X] \end{align*}\]

Where the second and third lines of equation above are obtained by the fact that $Z$ is conditionally independent on the potential outcome given $X$.

Going back to $\mathbb{E}[Y|X]$, subtracting and adding $\mathbb{E}[Y(0)|X, G = \mathrm{AT}] \, \pi_{\mathrm{AT}}$, we obtain this simplified form:

\[\begin{align*} \mathbb{E}[Y|X] &= \Delta^{\mathrm{AT}}(X) \, \pi_{\mathrm{AT}} + \Delta^{\mathrm{CP}}(X) \, \mathbb{E}[Z|X] \, \pi_{\mathrm{CP}} + \mathbb{E}[Y(0)|X] \end{align*}\]

The second term in the numerator of $\beta_1$ then becomes:

\[\begin{align*} \mathbb{E}\bigg[\mathbb{E}[Y|X] \, \mathbb{E}[\tilde{Z}|X]\bigg] &= \mathbb{E}\bigg[\bigg(\Delta^{\mathrm{AT}}(X) \, \pi_{\mathrm{AT}} + \Delta^{\mathrm{CP}}(X) \, \mathbb{E}[Z|X] \, \pi_{\mathrm{CP}} + \mathbb{E}[Y(0)|X]\bigg) \times \mathbb{E}[\tilde{Z}|X]\bigg] \end{align*}\]

Distributing $\mathbb{E}[\tilde{Z}|X]$ gives the weight of the always-taker component:

\[\begin{align*} \mathbb{E}\bigg[\mathbb{E}[Y|X] \, \mathbb{E}[\tilde{Z}|X]\bigg] &= \mathbb{E}\bigg[\Delta^{\mathrm{AT}}(X) \, \omega^{\mathrm{AT}} + \bigg(\Delta^{\mathrm{CP}}(X) \, \mathbb{E}[Z|X] \, \pi_{\mathrm{CP}} + \mathbb{E}[Y(0)|X]\bigg) \times \mathbb{E}[\tilde{Z}|X]\bigg] \end{align*}\]

Fifth step

The last part of the proof is to return to the main equation and express the numerator of $\beta_1$ in a new form:

\[\begin{align*} \mathrm{Cov}(Y, \tilde{Z}) &= \mathbb{E}\bigg[\Delta^{\mathrm{CP}}(X) \, \pi_{\mathrm{CP}} \times \mathbb{E}[Z|X](1 - \mathbb{E}[Z|X])\bigg] \\ &\quad + \mathbb{E}\bigg[\Delta^{\mathrm{AT}}(X) \, \omega^{\mathrm{AT}} + \bigg(\Delta^{\mathrm{CP}}(X) \, \mathbb{E}[Z|X] \, \pi_{\mathrm{CP}} + \mathbb{E}[Y(0)|X]\bigg) \times \mathbb{E}[\tilde{Z}|X]\bigg] \end{align*}\]

Gathering the $\Delta^{\mathrm{CP}}(X) \, \pi_{\mathrm{CP}}$ terms, and using $\mathbb{E}[\tilde{Z}|X] = \mathbb{E}[Z|X] - \mathbb{L}[Z|X]$, gives the following:

\[\begin{align*} \mathrm{Cov}(Y, \tilde{Z}) = \mathbb{E}[\Delta^{\mathrm{AT}}(X) \, \omega^{\mathrm{AT}}] + \alpha \end{align*}\]

where

\[\begin{align*} \alpha = \mathbb{E}\bigg[\Delta^{\mathrm{CP}}(X) \, \pi_{\mathrm{CP}} \times \bigg(\mathbb{E}[Z|X](1 - \mathbb{E}[Z|X]) + \mathbb{E}[Z|X](\mathbb{E}[Z|X] - \mathbb{L}[Z|X])\bigg)\bigg] \end{align*}\]

Working on the inner term of the equation above, we have this identity:

\[\begin{align*} \mathbb{E}[Z|X](1 - \mathbb{E}[Z|X]) + \mathbb{E}[Z|X](\mathbb{E}[Z|X] - \mathbb{L}[Z|X]) &= \mathbb{E}[Z|X](1 - \mathbb{E}[Z|X] + \mathbb{E}[Z|X] - \mathbb{L}[Z|X]) \\ &= \mathbb{E}[Z|X](1 - \mathbb{L}[Z|X]) \end{align*}\]

Therefore,

\[\begin{align*} \mathrm{Cov}(Y, \tilde{Z}) = \mathbb{E}[\Delta^{\mathrm{CP}}(X) \, \omega^{\mathrm{CP}}] + \mathbb{E}[\Delta^{\mathrm{AT}}(X) \, \omega^{\mathrm{AT}}] \end{align*}\]

Multiplying by $\mathbb{E}[D\tilde{Z}]^{-1}$ gives the result. $QED$