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{\bf Proof of Theorem 3 from Section~\ref{sec:s4}. Identification of LATES.}
{\bf Part 1. Case }$\mathbf{m = 1}\text{ }${\bf .}
We follow very closely the structure of the proof of Theorem 1 from \citep{Frolich} and extend it to the case of MSSM.
Consider $(x, p, w, z) \in \mathrm{supp}(X_i, \bar{P}_i^{(z)}, W_i^{(D)}, Z_i)$.
For brevity, denote $\tilde{X}_i = (X_i, \bar{P}_i^{(z)}, Z_i)$ and $\tilde{x} = (x, p, z)$.
By using the law of total expectation, we obtain:
\begin{equation}
\begin{aligned}
\mathbb{E} &\left( Y_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = w, Z_i = z \right) = \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w \right) = \\
&= \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w, \mathrm{compiler}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w \right) +\\
&\quad + \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w, \mathrm{defier}_i = 1 \right) \times \\
&\quad \times \underbrace{\mathbb{P} \left( \mathrm{defier}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w \right)}_{\text{Equals 0 by Assumption 2F}} + \\
&\quad + \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w, \mathrm{always{\text -}taker}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w \right) + \\
&\quad + \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w, \mathrm{never{\text -}taker}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{never{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = w \right).
\end{aligned}
\end{equation}
Note that if $W_i^{(D)} = 1$, then we observe $Y_{1i}$ for compliers.
Similarly, if $W_i^{(D)} = 0$, then we observe $Y_{0i}$ for compliers.
For always-takers and never-takers we always observe $Y_{1i}$ and $Y_{0i}$, respectively. By using these facts, we obtain:
\begin{equation}
\begin{aligned}
\mathbb{E} &\left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) - \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) = \\
&= \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1, \mathrm{compiler}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) + \\
&\quad + \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1, \mathrm{always{\text -}taker}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) + \\
&\quad + \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1, \mathrm{never{\text -}taker}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{never{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) - \\
&\quad - \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0, \mathrm{compiler}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) - \\
&\quad - \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0, \mathrm{always{\text -}taker}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) - \\
&\quad - \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0, \mathrm{never{\text -}taker}_i = 1 \right) \times \\
&\quad \times \mathbb{P} \left( \mathrm{never{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right).
\end{aligned}
\end{equation}
Because of Assumptions 1F, 4F, and 5F, we may apply Lemma 2, which allows us to remove the conditioning on $W_i^{(D)}$ in the expectations and probabilities:
\begin{equation}
\begin{aligned}
\mathbb{E} &\left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) - \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) = \\
&= \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, \mathrm{compiler}_i = 1 \right) \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x} \right) + \\
&\quad + \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, \mathrm{always{\text -}taker}_i = 1 \right) \mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x} \right) + \\
&\quad + \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, \mathrm{never{\text -}taker}_i = 1 \right) \mathbb{P} \left( \mathrm{never{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x} \right) - \\
&\quad - \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, \mathrm{compiler}_i = 1 \right) \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x} \right) - \\
&\quad - \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, \mathrm{always{\text -}taker}_i = 1 \right) \mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x} \right) - \\
&\quad - \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, \mathrm{never{\text -}taker}_i = 1 \right) \mathbb{P} \left( \mathrm{never{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x} \right) = \\
&= \left[ \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, \mathrm{compiler}_i = 1 \right) - \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, \mathrm{compiler}_i = 1 \right) \right] \times \\
&\quad \times \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x} \right).
\end{aligned}
\end{equation}
Because of Assumption 3F, we may divide both sides of the last expression by the conditional probability of complying:
\begin{equation}
\label{eq:T3-1}
\begin{aligned}
& \text{E}\left( {{Y}_{1i}}|{{{\tilde{X}}}_{i}}=\tilde{x},\text{complie}{{\text{r}}_{i}}=1 \right)-\text{E}\left( {{Y}_{0i}}|{{{\tilde{X}}}_{i}}=\tilde{x},\text{complie}{{\text{r}}_{i}}=1 \right)= \\
& =\frac{\text{E}\left( {{Y}_{i}}|{{{\tilde{X}}}_{i}}=\tilde{x},W_{i}^{(D)}=1 \right)-\text{E}\left( {{Y}_{i}}|{{{\tilde{X}}}_{i}}=\tilde{x},W_{i}^{(D)}=0 \right)}{\text{P}\left( \text{complie}{{\text{r}}_{i}}=1|{{{\tilde{X}}}_{i}}=\tilde{x} \right)}.
\end{aligned}
\end{equation}
To expand the denominator in the last expression, note that:
\begin{equation}
\begin{aligned}
\mathbb{E} &\left( D_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) = \mathbb{P} \left( D_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) = \\
&= \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) + \\
&\quad + \mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) = \\
&= \underbrace{\mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x} \right)}_{\text{Lemma 2}} + \underbrace{\mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x} \right)}_{\text{Lemma 2}}.
\end{aligned}
\end{equation}
and:
\begin{equation}
\begin{aligned}
\mathbb{E} &\left( D_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) = \mathbb{P} \left( D_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) = \\
&= \mathbb{P} \left( \mathrm{defier}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) + \\
&\quad + \mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) = \\
&= \underbrace{\mathbb{P} \left( \mathrm{defier}_i = 1 \mid \tilde{X}_i = \tilde{x} \right)}_{\text{Lemma 2}} + \underbrace{\mathbb{P} \left( \mathrm{always{\text -}taker}_i = 1 \mid \tilde{X}_i = \tilde{x} \right)}_{\text{Lemma 2}}.
\end{aligned}
\end{equation}
Hence, by taking the difference, we obtain:
\begin{equation}
\label{eq:T3-2}
\begin{aligned}
\mathbb{E} &\left( D_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) - \mathbb{E} \left( D_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right) = \\
&= \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x} \right) - \underbrace{\mathbb{P} \left( \mathrm{defier}_i = 1 \mid \tilde{X}_i = \tilde{x} \right)}_{\text{Equals 0 by Assumption 2F}} = \\
&= \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i = \tilde{x} \right).
\end{aligned}
\end{equation}
By plugging equation (\ref{eq:T3-2}) into the denominator of equation (\ref{eq:T3-1}), we get the expression for the conditional local average treatment effect:
\begin{equation}
\begin{aligned}
\mathrm{CLATES}(x,p &\mid Z_i = z) = \mathbb{E} \left( Y_{1i} \mid \tilde{X}_i = \tilde{x}, \mathrm{compiler}_i = 1 \right) - \\
&\quad - \mathbb{E} \left( Y_{0i} \mid \tilde{X}_i = \tilde{x}, \mathrm{compiler}_i = 1 \right) = \\
&= \frac{\mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) - \mathbb{E} \left( Y_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right)}{\mathbb{E} \left( D_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 1 \right) - \mathbb{E} \left( D_i \mid \tilde{X}_i = \tilde{x}, W_i^{(D)} = 0 \right)} = \\
&= \frac{\bar{\mu}_Y \left( 1, x, p, z \right) - \bar{\mu}_Y \left( 0, x, p, z \right)}{\bar{\mu}_D \left( 1, x, p, z \right) - \bar{\mu}_D \left( 0, x, p, z \right)},
\end{aligned}
\end{equation}
where the conditional expectations used in this expression exist by Assumption 6F.
By taking an expectation and applying the Bayes' theorem, we obtain:
\begin{equation}
\begin{aligned}
&\mathrm{LATES} = \mathbb{E} \left( \mathbb{E} \left( Y_{1i} - Y_{0i} \mid X_i, \bar{P}_i^{(z)}, Z_i = z, \mathrm{compiler}_i = 1 \right) \right) = \\
&= \mathbb{E} \left( \mathrm{CLATES}(X_i, \bar{P}_i^{(z)} \mid Z_i = z) \mid Z_i = z, \mathrm{compiler}_i = 1 \right) = \\
&= \int \int \mathrm{CLATES}(x, p \mid Z_i = z) \, f_{(X_i, \bar{P}_i^{(z)}) \mid \mathrm{compiler}_i, Z_i}(x, p \mid 1, z) \, dx \, dp = \\
&= \int \int \mathrm{CLATES}(x, p \mid Z_i = z) \frac{\mathbb{P}(\mathrm{compiler}_i = 1, X_i = x, \bar{P}_i^{(z)} = p, Z_i = z)}{\mathbb{P}(\mathrm{compiler}_i = 1, Z_i = z)} \, dx \, dp \\
&= \int \int \mathrm{CLATES}(x, p \mid Z_i = z) \times \\
&\quad \times \frac{\mathbb{P}(\mathrm{compiler}_i = 1 \mid X_i = x, \bar{P}_i^{(z)} = p, Z_i = z) f_{(X_i, \bar{P}_i^{(z)}) \mid Z_i}(x, p \mid z)}{\mathbb{P}(\mathrm{compiler}_i = 1 \mid Z_i = z) \mathbb{P}(Z_i = z)} \times \\
&\quad \times \mathbb{P}(Z_i = z)\, dx \, dp = \\
&= \int \int \mathrm{CLATES}(x, p \mid Z_i = z) \frac{\mathbb{P}(\mathrm{compiler}_i = 1 \mid X_i = x, \bar{P}_i^{(z)} = p, Z_i = z)}{\mathbb{P}(\mathrm{compiler}_i = 1 \mid Z_i = z)} \\
&\quad \times f_{(X_i, \bar{P}_i^{(z)}) \mid Z_i}(x, p \mid z) \, dx \, dp = \\
&= \mathbb{E} \left( \mathrm{CLATES}(X_i, \bar{P}_i^{(z)} \mid Z_i = z) \frac{\mathbb{P}(\mathrm{compiler}_i = 1 \mid X_i, \bar{P}_i^{(z)}, Z_i = z)}{\mathbb{P}(\mathrm{compiler}_i = 1 \mid Z_i = z)} \mid Z_i = z \right).
\end{aligned}
\end{equation}
By expanding $\mathrm{CLATES}(X_i, \bar{P}_i^{(z)} \mid Z_i = z)$ via equation (\ref{eq:T3-1}), we finally establish the result for $m = 1$:
\begin{equation}
\begin{aligned}
\mathrm{LATES} &= \mathbb{E} \left( \frac{\bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z)}, z \right) - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z)}, z \right)}{\mathbb{P} \left( \mathrm{compiler}_i = 1 \mid X_i, \bar{P}_i^{(z)}, Z_i = z \right)} \right. \times \\
&\quad \left. \times \frac{\mathbb{P}(\mathrm{compiler}_i = 1 \mid X_i, \bar{P}_i^{(z)}, Z_i = z)}{\mathbb{P}(\mathrm{compiler}_i = 1 \mid Z_i = z)} \mid Z_i = z \right) = \\
&= \mathbb{E} \left( \frac{\bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z)}, z \right) - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z)}, z \right)}{\mathbb{P}(\mathrm{compiler}_i = 1 \mid Z_i = z)} \mid Z_i = z \right) = \\
&= \frac{\mathbb{E} \left[ \bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z)}, z \right) - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z)}, z \right) \mid Z_i = z \right]}{\mathbb{P}(\mathrm{compiler}_i = 1 \mid Z_i = z)} = \\
&= \frac{\mathbb{E} \left[ \bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z)}, z \right) - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z)}, z \right) \mid Z_i = z \right]}{\mathbb{E} \left[ \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i, Z_i = z \right) \mid Z_i = z \right]} = \\
&= \frac{\mathbb{E} \left[ \bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z)}, z \right) - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z)}, z \right) \mid Z_i = z \right]}{\mathbb{E} \left[ \bar{\mu}_D \left( 1, X_i, \bar{P}_i^{(z)}, z \right) - \bar{\mu}_D \left( 0, X_i, \bar{P}_i^{(z)}, z \right) \mid Z_i = z \right]}.
\end{aligned}
\end{equation}
{\bf Part 2. Case }$\mathbf{m > 1}${\bf .}
We apply the law of total expectation:
\begin{equation}
\label{eq:T3-3}
\begin{aligned}
\mathrm{LATES} &= \mathbb{E} \left( Y_{1i} \mid \tilde{Z}_i = 1, \mathrm{compiler}_i = 1 \right) - \mathbb{E} \left( Y_{0i} \mid \tilde{Z}_i = 1, \mathrm{compiler}_i = 1 \right) = \\
&= \sum\limits_{t=1}^{m} \mathbb{P} \left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1, \mathrm{compiler}_i = 1 \right) \times\\
&\quad \times \left( \mathbb{E} \left( Y_{1i} \mid Z_i = z^{(t)}, \mathrm{compiler}_i = 1 \right) \right. - \\
&- \left. \mathbb{E} \left( Y_{0i} \mid Z_i = z^{(t)}, \mathrm{compiler}_i = 1 \right) \right).
\end{aligned}
\end{equation}
Note that for any $z^* \in \{z^{(1)}, \dots, z^{(m)}\}$, we have:
\begin{equation}
\label{eq:T3-4}
\begin{aligned}
\mathbb{P} &\left( Z_i = z^* \mid \tilde{Z}_i = 1, \mathrm{compiler}_i = 1 \right) = \\
&= \frac{\mathbb{P} \left( \tilde{Z}_i = 1, \mathrm{compiler}_i = 1 \mid Z_i = z^* \right) \mathbb{P}(Z_i = z^*)}{\mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{Z}_i = 1 \right) \mathbb{P}(\tilde{Z}_i = 1)} = \\
&= \frac{\mathbb{P} \left( \mathrm{compiler}_i = 1 \mid Z_i = z^* \right) \mathbb{P}(Z_i = z^*)}{\mathbb{P} \left( \mathrm{compiler}_i = 1 \mid \tilde{Z}_i = 1 \right) \mathbb{P}(\tilde{Z}_i = 1)} = \\
&= \frac{\mathbb{P} \left( \mathrm{compiler}_i = 1 \mid Z_i = z^* \right)}{\sum\limits_{t=1}^{m} \mathbb{P} \left( \mathrm{compiler}_i = 1 \mid Z_i = z^{(t)} \right) \mathbb{P} \left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1 \right)} \times \frac{\mathbb{P}(Z_i = z^*)}{\mathbb{P}(\tilde{Z}_i = 1)}.
\end{aligned}
\end{equation}
By applying equation (\ref{eq:T3-2}) to the denominator of equation (\ref{eq:T3-4}), we obtain:
\begin{equation}
\label{eq:T3-5}
\begin{aligned}
\mathbb{P} &\left( \mathrm{compiler}_i = 1 \mid Z_i = z^{(t)} \right) = \\
&= \mathbb{E} \left( \mathbb{E} \left( \mathrm{compiler}_i = 1 \mid \tilde{X}_i^{(t)}, Z_i = z^{(t)} \right) \mid Z_i = z^{(t)} \right) = \\
&= \mathbb{E} \left[ \bar{\mu}_D \left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_D \left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \mid Z_i = z^{(t)} \right].
\end{aligned}
\end{equation}
By plugging equation (\ref{eq:T3-5}) into equation (\ref{eq:T3-4}) and then this modified equation (\ref{eq:T3-4}) into equation (\ref{eq:T3-3}), we obtain:
\begin{equation}
\begin{aligned}
&\mathrm{LATES} = \sum\limits_{t=1}^{m} \mathbb{P}(Z_i = z^{(t)} \mid \tilde{Z}_i = 1) \left[ \mathbb{E} \left( Y_{1i} \mid Z_i = z^{(t)}, \mathrm{compiler}_i = 1 \right) \right. - \\
&\quad \left. - \mathbb{E} \left( Y_{0i} \mid Z_i = z^{(t)}, \mathrm{compiler}_i = 1 \right) \right] = \\
&= \sum\limits_{t=1}^{m} \frac{\mathbb{E} \left[ \bar{\mu}_D \left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_D \left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \mid Z_i = z^{(t)} \right]}{\sum\limits_{k=1}^{m} \mathbb{E} \left[\begin{aligned} \bar{\mu}_D \left( 1, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right) - \\ - \bar{\mu}_D \left( 0, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right) \end{aligned} \mid Z_i = z^{(k)}\right] \mathbb{P} \left( Z_i = z^{(k)} \mid \tilde{Z}_i = 1 \right)} \times \\
&\quad \times \frac{\mathbb{P}(Z_i = z^{(t)})}{\mathbb{P}(\tilde{Z}_i = 1)} \times \\
& \quad \times \frac{\mathbb{E} \left[ \bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \mid Z_i = z^{(t)} \right]}{\mathbb{E} \left[ \bar{\mu}_D \left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_D \left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \mid Z_i = z^{(t)} \right]}.
\end{aligned}
\end{equation}
Note that in the last expression, the numerator of the first factor cancels with the denominator of the second factor. Therefore, we finally obtain:
\begin{equation}
\begin{aligned}
&\mathrm{LATES} = \sum\limits_{t=1}^{m} \frac{\mathbb{P}(Z_i = z^{(t)})}{\mathbb{P}(\tilde{Z}_i = 1)} \times \\
&\quad \times \frac{\mathbb{E} \left[ \bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \mid Z_i = z^{(t)} \right]}{\sum\limits_{k=1}^{m} \mathbb{E} \left[\begin{aligned} \bar{\mu}_D \left( 1, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right) -\\ - \bar{\mu}_D \left( 0, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right)\end{aligned} \mid Z_i = z^{(k)} \right] \mathbb{P} \left( Z_i = z^{(k)} \mid \tilde{Z}_i = 1 \right)} = \\
&= \frac{\sum\limits_{t=1}^{m} \mathbb{P} \left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1 \right) \mathbb{E} \left[\begin{aligned} \bar{\mu}_Y \left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \\ - \bar{\mu}_Y \left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right)\end{aligned} \mid Z_i = z^{(t)} \right]}{\sum\limits_{k=1}^{m} \mathbb{P} \left( Z_i = z^{(k)} \mid \tilde{Z}_i = 1 \right) \mathbb{E} \left[\begin{aligned} \bar{\mu}_D \left( 1, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right) - \\ - \bar{\mu}_D \left( 0, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right)\end{aligned} \mid Z_i = z^{(k)} \right]}.
\end{aligned}
\end{equation}
$\hfill\blacksquare$
{\bf Proof of Theorem 4 from Section~\ref{sec:s4}. Identification of LATE.}
{\bf Part 1. Case }$\mathbf{m = 1}${\bf .}
Consider $\left( x, p, w, z \right) \in \text{supp}\left( X_i, \bar{P}_i^{(z)}, W_i^{(D)}, Z_i \right)$. By using Lemma 3 and Assumption 8F, we obtain:
\begin{equation}
\label{eq:T4-1}
\begin{aligned}
& \mathbb{P}(\text{complier}_i = 1 \mid X_i = x, \bar{P}_i^{(z)} = p, Z_i = z) = \\
& = \left(\underbrace{\begin{aligned}\mathbb{E}\left( D_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 1, Z_i = z \right) - \\ - \mathbb{E}\left( D_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 0, Z_i = z \right)\end{aligned}}_{\text{By equation (\ref{eq:T3-2})}}\right) = \\
& = \mathbb{E}\left( D_{1i} \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 1, Z_i = z \right) - \\
&\quad - \mathbb{E}\left( D_{0i} \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 0, Z_i = z \right) = \\
& = \underbrace{\mathbb{E}\left( D_{1i} \mid X_i = x, \bar{P}_i^{(z)} = p, Z_i = z \right)}_{\text{Lemma 3: } D_{1i} \bot W_i^{(D)} \mid (X_i = x, \bar{P}_i^{(z)} = p, Z_i = z)} - \underbrace{\mathbb{E}\left( D_{0i} \mid X_i = x, \bar{P}_i^{(z)} = p, Z_i = z \right)}_{\text{Lemma 3: } D_{0i} \bot W_i^{(D)} \mid (X_i = x, \bar{P}_i^{(z)} = p, Z_i = z)} = \\
& = \underbrace{\mathbb{E}\left( D_{1i} \mid X_i = x, \bar{P}_i^{(z)} = p \right) - \mathbb{E}\left( D_{0i} \mid X_i = x, \bar{P}_i^{(z)} = p \right)}_{\text{Assumption 8F}} \\
& = \underbrace{\mathbb{E}\left( D_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 1 \right)}_{\text{Lemma 3: } D_{1i} \bot W_i^{(D)} \mid (X_i = x, \bar{P}_i^{(z)} = p)} - \underbrace{\mathbb{E}\left( D_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 0 \right)}_{\text{Lemma 3: } D_{0i} \bot W_i^{(D)} \mid (X_i = x, \bar{P}_i^{(z)} = p)} = \\
& = \mathbb{P}(\text{complier}_i = 1 \mid X_i = x, \bar{P}_i^{(z)} = p),
\end{aligned}
\end{equation}
where the last equality is easy to derive by using the same steps as in the proof of Theorem 3.
By defining $\mathrm{CLATE}(x,p)$ and applying equations (\ref{eq:T3-1}), (\ref{eq:T4-1}), and Assumption 8F, we obtain:
\begin{equation}
\begin{aligned}
&\mathrm{CLATE}(x,p) = \mathbb{E}\left( Y_{1i} \mid X_i = x, \bar{P}_i^{(z)} = p, \mathrm{complier}_i = 1 \right) - \\
&\quad - \mathbb{E}\left( Y_{0i} \mid X_i = x, \bar{P}_i^{(z)} = p, \mathrm{complier}_i = 1 \right) = \\
&= \left(\underbrace{\begin{aligned}\mathbb{E}\left( Y_{1i} \mid X_i = x, \bar{P}_i^{(z)} = p, \mathrm{complier}_i = 1, Z_i = z \right) - \\ - \mathbb{E}\left( Y_{0i} \mid X_i = x, \bar{P}_i^{(z)} = p, \mathrm{complier}_i = 1, Z_i = z \right)\end{aligned}}_{\text{Assumption 8F}}\right) = \\
&= \mathrm{CLATES}(x,p \mid Z_i = z) = \\
&= \underbrace{\frac{\left[\begin{aligned}\mathbb{E}\left( Y_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 1, Z_i = z \right) - \\ - \mathbb{E}\left( Y_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 0, Z_i = z \right)\end{aligned}\right]}{\mathbb{P}\left( \mathrm{complier}_i = 1 \mid X_i = x, \bar{P}_i^{(z)} = p, Z_i = z \right)}}_{\text{By equation (\ref{eq:T3-1})}} = \\
&= \underbrace{\frac{\left[\begin{aligned}\mathbb{E}\left( Y_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 1, Z_i = z \right) - \\ -\mathbb{E}\left( Y_i \mid X_i = x, \bar{P}_i^{(z)} = p, W_i^{(D)} = 0, Z_i = z \right)\end{aligned}\right]}{\mathbb{P}\left( \mathrm{complier}_i = 1 \mid X_i = x, \bar{P}_i^{(z)} = p \right)}}_{\text{By equation (\ref{eq:T4-1})}}.
\end{aligned}
\end{equation}
Therefore, we have established that $\mathrm{CLATE}(x,p)$ equals $\mathrm{CLATES}(x,p \mid Z_i=z)$. In addition, we have simplified its expression. By using these results and the same approach as in the proof of Theorem 3, we obtain:
\begin{equation}
\begin{aligned}
\mathrm{LATE} &= \mathbb{E}\left( \mathrm{CLATE}(X_i,\bar{P}_i^{(z)}) \mid \mathrm{complier}_i=1 \right) = \\
&= \mathbb{E}\left( \mathrm{CLATES}(X_i,\bar{P}_i^{(z)} \mid Z_i=z) \mid \mathrm{complier}_i=1 \right) \\
&= \mathbb{E}\left( \mathrm{CLATES}(X_i,\bar{P}_i^{(z)} \mid Z_i=z) \frac{\mathbb{P}(\mathrm{complier}_i=1 \mid X_i,\bar{P}_i^{(z)})}{\mathbb{P}(\mathrm{complier}_i=1)} \right) \\
&= \mathbb{E}\left( \frac{\bar{\mu}_Y\left( 1,X_i,\bar{P}_i^{(z)},z \right)-\bar{\mu}_Y\left( 0,X_i,\bar{P}_i^{(z)},z \right)}{\mathbb{P}\left( \mathrm{complier}_i=1 \mid X_i,\bar{P}_i^{(z)} \right)} \right. \times \\
&\quad \times \left. \frac{\mathbb{P}(\mathrm{complier}_i=1 \mid X_i,\bar{P}_i^{(z)})}{\mathbb{P}(\mathrm{complier}_i=1)} \right) \\
&= \frac{\mathbb{E}\left[ \bar{\mu}_Y\left( 1,X_i,\bar{P}_i^{(z)},z \right)-\bar{\mu}_Y\left( 0,X_i,\bar{P}_i^{(z)},z \right) \right]}{\mathbb{E}\left[ \mathbb{P}(\mathrm{complier}_i=1 \mid X_i=x,\bar{P}_i^{(z)}=p) \right]} \\
&= \underbrace{\frac{\mathbb{E}\left[ \bar{\mu}_Y\left( 1,X_i,\bar{P}_i^{(z)},z \right)-\bar{\mu}_Y\left( 0,X_i,\bar{P}_i^{(z)},z \right) \right]}{\mathbb{E}\left[ \bar{\mu}_D\left( 1,X_i,\bar{P}_i^{(z)} \right)-\bar{\mu}_D\left( 0,X_i,\bar{P}_i^{(z)} \right) \right]}}_{\text{Useful if }D_i\text{ is not subject to non-random selection}} = \\
&= \underbrace{\frac{\mathbb{E}\left[ \bar{\mu}_Y\left( 1,X_i,\bar{P}_i^{(z)},z \right)-\bar{\mu}_Y\left( 0,X_i,\bar{P}_i^{(z)},z \right) \right]}{\mathbb{E}\left[ \bar{\mu}_D\left( 1,X_i,\bar{P}_i^{(z)},z \right)-\bar{\mu}_D\left( 0,X_i,\bar{P}_i^{(z)},z \right) \right]}}_{\text{Assumption 8F}}.
\end{aligned}
\end{equation}
Note that if $D_i$ is not subject to non-random selection, then pre-last representation of LATE is preferable. However, for greater generality, we consider the last representation in which $D_i$ is conditioned on $Z_i=z$.
{\bf Part 2. Case }$\mathbf{m > 1}${\bf .}
Since $\sum\limits_{t=1}^{m} \mathbb{P}\left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1, \mathrm{complier}_i = 1 \right) = 1$, the expression for $m>1$ is as follows:
\begin{equation}
\label{eq:T4-2}
\begin{aligned}
\mathrm{LATE} &= \sum\limits_{t=1}^{m} \mathbb{P}\left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1, \mathrm{complier}_i = 1 \right) \times \\
&\quad \times \frac{\mathbb{E}\left[ \bar{\mu}_Y\left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_Y\left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \right]}{\mathbb{E}\left[ \bar{\mu}_D\left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_D\left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \right]}.
\end{aligned}
\end{equation}
By inserting equation (\ref{eq:T4-1}) into equation (\ref{eq:T3-4}), we obtain:
\begin{equation}
\label{eq:T4-3}
\begin{aligned}
&\mathbb{P}\left( Z_i = z^* \mid \tilde{Z}_i = 1, \mathrm{complier}_i = 1 \right) = \\
&= \frac{\mathbb{P}\left( \mathrm{complier}_i = 1 \mid Z_i = z^* \right)}{\sum\limits_{t=1}^{m} \mathbb{P}\left( \mathrm{complier}_i = 1 \mid Z_i = z^{(t)} \right) \mathbb{P}\left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1 \right)} \frac{\mathbb{P}\left( Z_i = z^* \right)}{\mathbb{P}\left( \tilde{Z}_i = 1 \right)} \times \\
&= \mathbb{P}\left( Z_i = z^* \mid \tilde{Z}_i = 1 \right) \\
&\quad \times \frac{\mathbb{E}\left[ \bar{\mu}_Y\left( 1, X_i, \bar{P}_i^{(z^*)}, z^* \right) - \bar{\mu}_Y\left( 0, X_i, \bar{P}_i^{(z^*)}, z^* \right) \right]}{\sum\limits_{t=1}^{m} \mathbb{E}\left[ \bar{\mu}_D\left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_D\left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \right] \mathbb{P}\left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1 \right)}.
\end{aligned}
\end{equation}
By plugging equation (\ref{eq:T4-3}) into equation (\ref{eq:T4-2}), we finally obtain:
\begin{equation}
\begin{aligned}
&\mathrm{LATE} = \\
&= \frac{\sum\limits_{t=1}^{m} \mathbb{P}\left( Z_i = z^{(t)} \mid \tilde{Z}_i = 1 \right) \mathbb{E}\left[ \bar{\mu}_Y\left( 1, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) - \bar{\mu}_Y\left( 0, X_i, \bar{P}_i^{(z^{(t)})}, z^{(t)} \right) \right]}{\sum\limits_{k=1}^{m} \mathbb{P}\left( Z_i = z^{(k)} \mid \tilde{Z}_i = 1 \right) \mathbb{E}\left[ \bar{\mu}_D\left( 1, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right) - \bar{\mu}_D\left( 0, X_i, \bar{P}_i^{(z^{(k)})}, z^{(k)} \right) \right]}.
\end{aligned}
\end{equation}
$\hfill\blacksquare$
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