text:
\section{Introduction}

In this study, we examine the identification of a system of structural equations that takes the following form:
\begin{equation}\label{model}
\begin{aligned}
Y &=& g(X,\epsilon) \\
X &=& h(Z,\eta),
\end{aligned}
\end{equation}
where $Y\in\mathbb{R}$ is a scalar response variable, $X\in\mathbb{R}$ is a continuous endogenous variable, $Z \in \{0,1\}$ is a binary instrument, and $\epsilon$ and $\eta$ are unobservable scalar variables.
For simplicity, we suppose that $X$ is a scalar variable.
It is straightforward to extend the results to a case in which $X$ is a vector.
This specification is nonseparable in unobservable variable $\epsilon$ and captures unobserved heterogeneity in the effect of $X$ on $Y$.
Such models have also been considered by, for example, \cite{d2015identification} and \cite{torgovitsky2015identification}.

\cite{d2015identification} and \cite{torgovitsky2015identification} show that $g$ is point identified when $g(x,e)$ and $h(z,v)$ are strictly increasing in $e$ and $v$ and $Z$ is independent of $(\epsilon,\eta)$.
Their results are important for empirical analyses, in which many instruments are binary or discrete, such as the intent-to-treat in a randomized controlled experiment and quarter of birth used in \cite{angrist1991does}.
For nonseparable models, such as (\ref{model}) with continuously distributed $X$, other nonparametric point identification results require $Z$ to be continuously distributed.
See, for example, \cite{newey2003instrumental}, \cite{chernozhukov2007instrumental}, and \cite{imbens2009identification}.

\cite{d2015identification} and \cite{torgovitsky2015identification} use two key assumptions when establishing point identification for $g$.
First, $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ have intersections.
Second, $g(x,e)$ is strictly increasing in $e$.
There are, however, many empirically important models that do not satisfy these assumptions.
For example, $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ do not have an intersection when $Z$ has a strictly monotonic effect on $X$ such as linear models $X=\beta_0 + \beta_1 Z + \eta$.
Further, in many applications, instrumental variables have a strictly monotonic effect on endogenous variables such as LATE proposed by \cite{imbens1994identification}.
For example, as seen in \cite{baird2011cash}\if0 and \cite{haushofer2013household}\fi, cash transfer programs have been implemented in many countries.
Then, if we use treatment indicator $Z$ as the instrumental variable for income $X$, $Z$ has a strictly monotonic effect on $X$, which violates the intersection assumption.
Actually, in this example, we can write $X=\beta Z + \eta$, where $\beta$ is transfer amount and $\eta$ is initial household income.
Hence, $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ never have an intersection in this example.
When $Y$ is discrete or censored, $g(x,e)$ is not strictly increasing in $e$.
Many problems in economics involve dependent variables that are discrete or censored.
For example, development economists may want to analyze the effects of income changes on child education.
If school attendance is used as a dependent variable, then $Y$ is discrete.
As another example, suppose we want to analyze the effects of income changes on education expenditure.
Then, education expenditure is censored at zero when children do not attend school.

This study shows that when $g(x,e)$ is monotone or concave in $x$, we can partially identify $g$, even if $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ have no intersection and $g(x,e)$ is not strictly increasing in $e$.
$g(x,e)$ is monotone or concave in $x$ in many economic models.
For example, the Engel function of a normal good is monotonically increasing in income, the demand function is ordinarily decreasing in price, and economic analyses of production often suppose that the production function is monotone and concave in inputs.
\cite{manski1997monotone} assumes these assumptions and shows that the average treatment response is partially identified.
The partial identification approach using the concavity assumption introduced in this study is somewhat similar to that considered by \cite{d2013nonlinear}.

There is rich literature on the identification of nonseparable models with endogeneity.
For example, \cite{chernozhukov2005iv}, \cite{chesher2007instrumental}, \cite{chernozhukov2007instrumental}, and \cite{imbens2009identification} consider the identification of nonseparable models with endogeneity.
In particular, \cite{imbens2009identification} consider models similar to (\ref{model}).
Their study allows $\epsilon$ to be multivariate and shows that the quantile function of $g(x,\epsilon)$ is point identified, while in this analysis, $\epsilon$ is imposed to be scalar.
Their result needed continuous instruments, whereas \cite{d2015identification}, \cite{torgovitsky2015identification}, and the present study do not.

This analysis adopts the control function approach, which is also used by, for example, \cite{imbens2009identification}, \cite{d2015identification}, and \cite{torgovitsky2015identification}.
This approach requires the strict monotonicity of $h(z,v)$ in $v$ and strong exogeneity: $Z$ is independent of $(\epsilon,\eta)$.
The estimating equation approach followed by, for example, \cite{newey2003instrumental}, \cite{chernozhukov2005iv}, and \cite{chernozhukov2007instrumental}, does not require these assumptions.
\cite{chernozhukov2005iv} allow $\eta$ to be multivariate and correlated with $Z$.
However, they assume that independent and instrumental variables are discrete.

By contrast, we suppose that the instrumental variable $Z$ is binary.
\cite{d2015identification} consider the case in which the instrumental variable takes more than two values.
In this case, they show point identification can be achieved using group and dynamical systems theories, even when $F_{X|Z}(x|z)$ and $F_{X|Z}(x|z')$ have no intersection.

The remainder of this study is organized as follows. Section 2 introduces the assumptions employed in this study. Sections 3 and 4 demonstrate the partial identification of $g$ under the monotonicity and concavity assumption when conditional distributions have no intersections. Section 5 extends the result in Section 3 to a general case, where we allow $Y$ to be discrete or censored. Section 6 concludes.

\section{Model}

The following two assumptions are the same as those in \cite{d2015identification} and \cite{torgovitsky2015identification}.

\begin{Assumption}
The instrument is independent of the unobservable variables: $Z\mathop{\perp\!\!\!\perp}(\epsilon,\eta)$.
\end{Assumption}

\begin{Assumption}
(i) Function $g$ is continuous and $g(x,e)$ is strictly increasing in $e$ for all $x \in \mathcal{X}$. (ii) For all $z \in \{0,1\}$, $h(z,v)$ is continuous and strictly increasing in $v$.
\end{Assumption}

Assumption 1 is a strong exogeneity assumption and typically employed when using the control function approach.
See, for example, \cite{imbens2009identification}, \cite{d2015identification}, and \cite{torgovitsky2015identification}.
\cite{imbens2009identification} also need Assumption 2(ii) but allow $\epsilon$ to be multivariate.
\cite{chernozhukov2005iv} and \cite{chernozhukov2007instrumental} employ Assumption 2(i) but do not need Assumption 2(ii).
We relax a part of Assumption 2(i) in Section 5, where we assume $g(x,e)$ is nondecreasing in $e$.

The next assumption regarding the conditional distributions of $X$ conditional on $Z$ differs from that imposed by \cite{d2015identification} and \cite{torgovitsky2015identification}.
For any random variable $U$ and $W$, let $F_{U|W}$ denote the conditional distribution function of $U$ conditional on $W$.
Let $\mathcal{X}$, $\mathcal{X}_z$, and $\mathcal{Y}_{x,z}$ denote interiors of the support of $X$, $X|Z=z$, and $Y|X=x,Z=z$, respectively.

\begin{Assumption}
(i) $F_{X|Z}(x|z)$ is continuous in $x$ for all $z \in \{0,1\}$ and $F_{X|Z}(x|0) < F_{X|Z}(x|1)$ for all $x \in \mathcal{X}_0 \cap \mathcal{X}_1$. (ii) $\mathcal{X}_0 = (\underline{x}_0,\overline{x}_0)$, $\mathcal{X}_1 = (\underline{x}_1,\overline{x}_1)$, and $-\infty < \underline{x}_1<\underline{x}_0 < \overline{x}_1<\overline{x}_0 < \infty$.
\end{Assumption}

Conditions (i) and (ii) imply that $F_{X|Z}(x|z)$ is strictly increasing and continuous in $x$ conditional on $\mathcal{X}_{z}$.
This assumption implies that $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ do not have any intersection on the support of $X$ and $X|Z=0$ stochastically dominates $X|Z=1$.
Thus, $Z$ has a strictly monotonic effect on $X$.
\cite{d2015identification} and \cite{torgovitsky2015identification} rule out this case because they assume $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ have intersections.
\cite{torgovitsky2015identification} shows that the point identification of $g$ holds when $\underline{x}_1 = \underline{x}_0 = \xi$, that is, $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ intersect at $x=\xi$, and $g(\xi,e)$ is strictly increasing in $e$.
Even when $\mathcal{X}_0 = \mathcal{X}_1 = \mathcal{X} = (\underline{x},\overline{x})$, we can not identify $g$ unless $g(\underline{x},e)$ or $g(\overline{x},e)$ is strictly increasing in $e$.

Next, we impose restrictions on the conditional distributions of $Y$ conditional on $X$ and $Z$.

\begin{Assumption}
(i) For all $(z,x,y) \in \{0,1\} \times \mathcal{X}_z \times \mathcal{Y}_{x,z}$, $F_{Y|X,Z}(y|x,z)$ is continuous in $x$ and $y$. (ii) For all $(z,x) \in \{0,1\} \times \mathcal{X}_z$, we have $\mathcal{Y}_{x,z} = \mathcal{Y} = (\underline{y},\overline{y})$, where $- \infty \leq \underline{y}<\overline{y} \leq \infty$.
\end{Assumption}

\cite{d2015identification} and \cite{torgovitsky2015identification} also assume condition (i) but do not assume condition (ii).
Conditions (i) and (ii) imply that $F_{Y|X,Z}(y|x,z)$ is strictly increasing and continuous in $y$ conditional on $\mathcal{Y}$.
Hence, the conditional quantile function of $Y$ conditional on $X$ and $Z$ is the inverse of $F_{Y|X,Z}(y|x,z)$.
Condition (ii) is not necessary for this study's result but without it, deriving the results can be cumbersome.
We relax condition (i) in Section 5 and allow $Y$ to be discrete or censored.

Finally, we impose the normalization assumption on unobservable variables and support condition of $\epsilon|X=x,Z=z$.

\begin{Assumption}
(i) $\epsilon \sim U(0,1)$ and $\eta \sim U(0,1)$. (ii) For all $(z,x) \in \{0,1\} \times \mathcal{X}_z$, the interior of the support of $\epsilon|X=x,Z=z$ is $(0,1)$.
\end{Assumption}

Condition (i) is the usual normalization in a nonseparable model (see, \cite{matzkin2003nonparametric}).
\cite{torgovitsky2015identification} does not use this normalization but \cite{d2015identification} normalize $\epsilon$ to be uniformly distributed.
\cite{d2015identification} do not assume $\eta \sim U(0,1)$.
The theorems in this study hold without this assumption, but the proof becomes more complicated.
Condition (ii) implies that $g(x,e) \in (\underline{y},\overline{y}) = \mathcal{Y}$ for all $(x,e) \in \mathcal{X} \times (0,1)$.

\section{Partial Identification through Monotonicity}

Let $\overline{\mathcal{Y}}$ be the closure of $\mathcal{Y}$.
We establish the partial identification of $g$ by showing that we can identify functions $T^U_{x',x}(y):\overline{\mathcal{Y}}\rightarrow \overline{\mathcal{Y}}$ and $T^L_{x',x}(y):\overline{\mathcal{Y}}\rightarrow \overline{\mathcal{Y}}$ and that they are (i) strictly increasing in $y$ (ii) surjective, that is, $T^U_{x',x}\left( [\underline{y}, \overline{y}] \right)=T^L_{x',x}\left( [\underline{y}, \overline{y}] \right)=[\underline{y}, \overline{y}]$, and (iii) satisfy the following inequalities:
\begin{eqnarray}
g(x',e) &\leq & T_{x',x}^U \left( g(x,e) \right), \label{TU} \\
g(x',e) &\geq & T_{x',x}^L \left( g(x,e)\right). \label{TL}
\end{eqnarray}
$T^U_{x',x}(y)$ and $T^L_{x',x}(y)$ correspond to $Q_{x'x}$ in \cite{d2015identification}.
If $T_{x',x}^U(y)$ is identified for all $x,x' \in \mathcal{X} \equiv \mathcal{X}_0 \cup \mathcal{X}_1$, we can obtain the lower bound of the structural function $g(x,e)$ in the following manner.
Here, we define $G_x^L(u) \equiv \int F_{Y|X=x'} \left( T_{x',x}^U (u) \right) dF_{X}(x')$.
If $T_{x',x}^U(y)$ satisfying (\ref{TU}) is obtained for all $x,x' \in \mathcal{X}$, then we have
\begin{eqnarray}
 G_x^L\left( g(x,e) \right) &=&\int F_{Y|X=x'} \left( T_{x',x}^U \left(g(x,e)\right) \right) dF_{X}(x') \nonumber \\
 &\geq & \int F_{Y|X=x'}\left(g(x',e)\right) dF_{X}(x') \nonumber \\
 &=& \int P(g(x',\epsilon) \leq g(x',e)|X=x') dF_{X}(x') \nonumber \\
 &=& \int P(\epsilon \leq e |X=x') dF_{X}(x') = e, \label{approach_DF1}
\end{eqnarray}
where the first inequality follows from (\ref{TU}) and the monotonicity of $F_{Y|X}(y|x')$ in $y$ and the third equality follows from the strict monotonicity of $g(x,e)$ in $e$.
Furthermore, $G_x^L(u)$ is invertible because $T_{x',x}^U(y)$ is strictly increasing in $y$.
Because $T_{x',x}^U(y)$ is surjective, we have $G_x^L\left( [\underline{y},\overline{y}] \right) = [0,1]$.
Hence, for all $e \in (0,1)$, we have
\begin{equation}
g(x,e) \geq \left(G_x^L\right)^{-1}(e). \label{approach_DF2}
\end{equation}
Similarly, define $G_x^U(u) \equiv \int F_{Y|X=x'} \left( T_{x',x}^L (u) \right) dF_{X}(x')$, and thus, we have
\begin{equation}
g(x,e) \leq \left(G_x^U\right)^{-1}(e). \nonumber
\end{equation}

This idea is similar to that of \cite{d2015identification}.
They use function $T_{x',x}(y)$ that is strictly increasing in $y$ and satisfies $g(x',e) = T_{x',x} \left( g(x,e) \right)$.
Define $G_x(u) \equiv \int F_{Y|X=x'} \left( T_{x',x}(u) \right) dF_{X}(x')$.
Then, similar to the above inequality, we have $G_x \left(g(x,e)\right) = e$.
Hence, if $T_{x',x}(y)$ is strictly increasing in $y$, then $G_x(u)$ is invertible and we have $g(x,e) = \left( G_x \right)^{-1}(e)$.

Next, we propose how to construct functions $T^U_{x',x}(y)$ and $T^L_{x',x}(y)$ that satisfy (\ref{TU}) and (\ref{TL}).
As in \cite{torgovitsky2015identification}, define $\pi(x):\mathcal{X}_0\rightarrow\mathcal{X}_1$ and $\pi^{-1}(x):\mathcal{X}_1\rightarrow\mathcal{X}_0$\footnote{These functions correspond to $s_{ij}$ in \cite{d2015identification}.} as
\begin{equation}\label{pi}
\begin{aligned}
 \pi(x) &\equiv & Q_{X|Z}\left( F_{X|Z}(x|0) | 1 \right), \\
 \pi^{-1}(x) &\equiv & Q_{X|Z}\left( F_{X|Z}(x|1) | 0 \right).
\end{aligned}
\end{equation}
Here, for any random variables $U$ and $W$, let $Q_{U|W}(\tau|w)$ denote the conditional $\tau$-th quantile of $U$ conditional on $W=w$, that is, $Q_{U|W}(\tau|w) \equiv \inf\{u:F_{U|W}(u|w) \geq \tau \}$.
The following result is essentially proven by \cite{d2015identification} (Theorem 1).
We state this result as a proposition because it plays a central role in the following and our assumptions somewhat differ from those in \cite{d2015identification}.

\begin{Proposition}
\begin{eqnarray}
\tilde{T}_x^{(1)}(y) &\equiv & Q_{Y|X,Z} \left( F_{Y|X,Z} \left( y | x,0 \right)|\pi(x), 1 \right), \nonumber \\
\tilde{T}_{x}^{(-1)}(y) &\equiv & Q_{Y|X,Z} \left( F_{Y|X,Z} \left( y|x,1 \right)|\pi^{-1}(x),0 \right). \nonumber
\end{eqnarray}
Then, under Assumptions 1--5, we have
\begin{eqnarray}
 g\left( \pi(x),e \right) &=& \tilde{T}_{x}^{(1)} \left( g(x,e) \right), \nonumber \\
 g\left( \pi^{-1}(x),e \right) &=& \tilde{T}_{x}^{(-1)} \left( g(x,e) \right). \nonumber
\end{eqnarray}
\end{Proposition}

\begin{proof}
\underline{Step.1}\,
We show that, for all $e \in (0,1)$,
\begin{eqnarray}
P(\epsilon \leq e | X=x, Z=0) = P(\epsilon \leq e | X=\pi(x), Z=1). \label{dist_e}
\end{eqnarray}

First, we examine variable $V \equiv F_{X|Z}(X|Z)$.
This is called ``control variable" in \cite{imbens2009identification}.
Let $h^{-1}(z,x)$ be the inverse function of $h(z,v)$ with respect to $v$.
We have for all $(z,x) \in \{0,1\} \times \mathcal{X}_z$,
\begin{eqnarray}
 F_{X|Z}(x|z) &=& P\left( h(z,\eta) \leq x | Z=z\right) \nonumber \\
 &=& P\left( \eta \leq h^{-1}(z,x) | Z=z\right) \nonumber \\
 &=& P\left( \eta \leq h^{-1}(z,x) \right) = h^{-1}(z,x), \nonumber
\end{eqnarray}
where the second equality follows from the strict monotonicity of $h(x,v)$ in $v$ and the third equality follows from $Z \mathop{\perp\!\!\!\perp} (\epsilon,\eta)$.
Therefore, we obtain
$$
V = h^{-1}(Z,X)=\eta.
$$

Next, we show that the conditional distribution of $\epsilon$ conditional on $(X,Z)=(x,z)$ is the same as that of $\epsilon$ conditional on $V=F_{X|Z}(x|z)$.
Because $(x,z) \rightarrow (F_{X|Z}(x|z),z)$ is one-to-one and $F_{X|Z}(x|z)$ is continuous in $x$, $Z \mathop{\perp\!\!\!\perp} (\epsilon, \eta)$, and $V=\eta$, we have
\begin{eqnarray}
 P\left( \epsilon \leq e | X=x, Z=z \right) &=& P\left( \epsilon \leq e | V=F_{X|Z}(x|z), Z=z \right) \nonumber \\
 &=& P\left( \epsilon \leq e | V=F_{X|Z}(x|z) \right). \label{V_property}
\end{eqnarray}
Hence, the conditional distribution of $\epsilon$ conditional on $X$ and $Z$ solely depends on $V=F_{X|Z}(X|Z)$.

By definition, functions $\pi(x)$ and $\pi^{-1}(x)$ satisfy that
\begin{equation}\label{pi_property2}
\begin{aligned}
F_{X|Z}(\pi(x)|1) &=& F_{X|Z}(x|0), \\
F_{X|Z}(\pi^{-1}(x)|0) &=& F_{X|Z}(x|1).
\end{aligned}
\end{equation}
Hence, events $\{X=x,Z=0\}$ and $\{X=\pi(x),Z=1\}$ have the same $V=F_{X|Z}(X|Z)$ and (\ref{dist_e}) follows from (\ref{V_property}). \\
\underline{Step.2}\,
We show that (\ref{dist_e}) implies $g\left( \pi(x),e \right) = \tilde{T}_{x}^{(1)} \left( g(x,e) \right)$.
For all $(x,e) \in \mathcal{X}_0 \times (0,1)$, we have
\begin{eqnarray}
\tilde{T}_{x}^{(1)}\left( g(x,e) \right) &=& Q_{Y|X=\pi(x),Z=1} \left( F_{Y|X=x,Z=0} \left( g(x,e) \right) \right) \nonumber \\
 &=& Q_{Y|X=\pi(x),Z=1}\left( P(\epsilon \leq e | X=x, Z=0) \right) \nonumber \\
 &=& Q_{Y|X=\pi(x),Z=1}\left( P(\epsilon \leq e | X=\pi(x), Z=1) \right) \nonumber \\
 &=& Q_{Y|X=\pi(x),Z=1}\left( F_{Y|X=\pi(x),Z=1}\left( g(\pi(x),e) \right) \right) \nonumber \\
 &=& g(\pi(x),e), \nonumber
\end{eqnarray}
where the third equality follows from (\ref{dist_e}).

Similarly, we can prove that $g\left( \pi^{-1}(x),e \right) = \tilde{T}_{x}^{(-1)} \left( g(x,e) \right)$.
\end{proof}

By definition, $\tilde{T}_{x}^{(1)}(y)$ and $\tilde{T}_{x}^{(-1)}(y)$ are strictly increasing, $\tilde{T}_{x}^{(1)}([\underline{y},\overline{y}]) = [\underline{y},\overline{y}]$, and $\tilde{T}_{x}^{(-1)}([\underline{y},\overline{y}]) = [\underline{y},\overline{y}]$.
Define $\pi^n(x) \equiv \pi \circ \cdots \circ \pi(x)$ for $n \in \mathbb{N}$ if $\pi^{n}(x)$ exists.
Then, we obtain $g(\pi^n(x),e) = \tilde{T}_{\pi^{n-1}(x)}^{(1)} \circ \cdots \circ \tilde{T}_{x}^{(1)} \left( g(x,e) \right)$.
Define $\tilde{T}_{x}^{(n)} (y) \equiv \tilde{T}_{\pi^{n-1}(x)}^{(1)} \circ \cdots \circ \tilde{T}_{x}^{(1)} \left( y \right)$ for $n \in \mathbb{N}$, and thus, we have
$$
 g(\pi^n(x),e) = \tilde{T}_{x}^{(n)} \left( g(x,e) \right).
$$
In addition, $\tilde{T}_{x}^{(n)} (y)$ is strictly increasing in $y$ and $\tilde{T}_{x}^{(n)} ([\underline{y},\overline{y}]) = [\underline{y},\overline{y}]$.
Similarly, define $\pi^{-n}(x) \equiv \pi^{-1} \circ \cdots \circ \pi^{-1}(x)$ and $\tilde{T}_{x}^{(-n)} (y) \equiv \tilde{T}_{\pi^{-(n-1)}(x)}^{(-1)} \circ \cdots \circ \tilde{T}_{x}^{(-1)} \left( y \right)$ for $n \in \mathbb{N}$; then, we have
$$
 g(\pi^{-n}(x),e) = \tilde{T}_{x}^{(-n)} \left( g(x,e) \right).
$$
$\tilde{T}_{x}^{(-n)} (y)$ is strictly increasing in $y$ and $\tilde{T}_{x}^{(-n)} ([\underline{y},\overline{y}]) = [\underline{y},\overline{y}]$.
For all $x \in \mathcal{X}$, define $\tilde{T}_x^{(0)}(y) \equiv y$.

Here, we examine the properties of $\pi(x)$ and $\pi^{-1}(x)$.
Because $F_{X|Z}(x|0) < F_{X|Z}(x|1)$ conditional on $\mathcal{X}$, we have
\begin{equation}\label{pi_property1}
\begin{aligned}
\pi(x) &=& Q_{X|Z}\left( F_{X|Z}(x|0) | 1 \right)  \\
 &<& Q_{X|Z}\left( F_{X|Z}(x|1) | 1 \right) = x,  \\
 \pi^{-1}(x) &=& Q_{X|Z}\left( F_{X|Z}(x|1) | 0 \right) \\
 &>& Q_{X|Z}\left( F_{X|Z}(x|0) | 0 \right) = x.
\end{aligned}
\end{equation}
Figure 1 illustrates this intuitively.
Because $X|Z=0$ stochastically dominates $X|Z=1$ and functions $\pi(x)$ and $\pi^{-1}(x)$ satisfy (\ref{pi_property2}), these inequalities hold.

To facilitate the illustration of our identification result, we first review the identification approach employed by \cite{d2015identification} and \cite{torgovitsky2015identification} when $\underline{x}_0 = \underline{x}_1 = \xi$, even though Assumption 3 rules out the case of $\underline{x}_0 = \underline{x}_1 = \xi$.
Pick an initial point $x_0 \in \mathcal{X}$ (i.e., $x_0 > \xi$) and form a recursive sequence $x_{n+1} = \pi(x_n)$ for $n > 0$.
Because $\underline{x}_0 = \underline{x}_1 = \xi$ implies $\mathcal{X}_1 \subset \mathcal{X}_0$, we have $\pi(x) \in \mathcal{X}_0$ for all $x \in \mathcal{X}$ and there exists a sequence $\{\pi^n(x)\}_{n=1}^{\infty}$.
The sequence $\{x_n\}$ is decreasing by (\ref{pi_property1}) and $x_n > \xi$ for all $n \geq 0$ by the definition of $\pi(x)$.
Hence, sequence $\{x_n\}$ converges to a limiting point.
Because (\ref{pi_property2}) implies
$$
 F_{X|Z}(x_{n+1}|1) = F_{X|Z}(x_n|0)
$$
and $F_{X|Z}(x|z)$ is continuous in $x$, we have $F_{X|Z}(\lim_{n \rightarrow \infty}x_n|1) = F_{X|Z}(\lim_{n \rightarrow \infty}x_n|0)$.
Because $F_{X|Z}(x|0) < F_{X|Z}(x|1)$ is conditional on $(\xi, \overline{x}_0)$ and $F_{X|Z}(\xi|0) = F_{X|Z}(\xi|1) = 0$, the sequence $\{x_n\}$ converges to $\xi$ for any initial point $x_0 \in \mathcal{X}$.
Figure 2 illustrates this intuitively.
Define $\tilde T^{(\infty)}_x(y) \equiv \lim_{n \to \infty} \tilde T^{(n)}_x(y)$, which is strictly increasing and invertible in $y$.
From the continuity of $g$, we obtain, for all $x \in \mathcal{X}$,
$$
\tilde T^{(\infty)}_x(g(x,e)) = \lim_{n \to \infty} g(\pi^n(x),e) = g(\xi,e).
$$
A similar argument provides $\tilde T^{(\infty)}_{x'}(g(x',e)) =  g(\xi,e)$.
Therefore, $\tilde T^{(\infty)}_x(g(x,e)) = \tilde T^{(\infty)}_{x'}(g(x',e))$ holds for any $x,x'$ and
$$
 g(x',e) = \left( \tilde{T}_{x'}^{(\infty)} \right)^{-1} \left( \tilde{T}_{x}^{(\infty)} \left( g(x,e) \right) \right).
$$
Define $T_{x',x}(y) \equiv \left( \tilde{T}_{x'}^{(\infty)} \right)^{-1} \left( \tilde{T}_{x}^{(\infty)}(y) \right)$, and thus, $T_{x',x}(y)$ is strictly increasing and satisfies $g(x',e)=T_{x',x}\left( g(x,e) \right)$.
Hence, as discussed above, $g$ is point identified.

Under Assumption 3, this approach is not available because a convergent sequence $\{\pi^n(x)\}_{n=1}^{\infty}$ does not exist.
When $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ have no intersections, $\pi^n(x)$ lies in $\mathcal{X}_1 \cap \mathcal{X}_0^c = (\underline{x}_1,\underline{x}_0]$ when $n$ is sufficiently large.
If $\pi^n(x)$ is in $\mathcal{X}_1 \cap \mathcal{X}_0^c$, then $\pi^{n+1}(x)$ does not exist.
As seen in Lemma 1, for all $x \in \mathcal{X}$, $\{n: \pi^n(x)\ \text{exists.} \}$ is a finite set under Assumption 3.
For example, in Figure 1, $\pi(x)$, $\pi^{-1}(x)$, and $\pi^{-2}(x)$ exist but $\pi^2(x)$ and $\pi^{-3}(x)$ do not.

If we do not impose additional restrictions, then we cannot construct functions $T^U_{x',x}(y)$ and $T^L_{x',x}(y)$ that satisfy (\ref{TU}) and (\ref{TL}).
First, we show that a set $\Pi_{x',x}^M$ defined below is nonempty and finite, even when $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ have no intersections.
Next, we show that we can partially identify $g(x,e)$ using $\Pi_{x',x}^M$ when $g(x,e)$ is nondecreasing in $x$.

For $(x,x') \in \mathcal{X} \times \mathcal{X}$, define $\Pi_{x',x}^M$ as
\begin{eqnarray}
 \Pi_{x',x}^M &\equiv & \left\{ (n,m) : \pi^n(x') \leq \pi^m(x),\right. \nonumber \\
 & & \left. n,m \in \mathbb{Z}, \ \pi^n(x')\ \text{and} \ \pi^m(x) \ \text{exist.} \right\}, \label{Pi}
\end{eqnarray}
where $\pi^{0}(x) \equiv x$.
In Figure 1, $\Pi^{M}_{x',x}=\{(-1,-2),(0,-2), (0,-1), (1,-2), (1,-1), (1,0) \}$.
The following lemma shows that $\Pi_{x',x}^M$ is nonempty and finite even when $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ have no intersection.

\begin{Lemma}
Under Assumptions 1--5, $\Pi_{x',x}^M$ defined by (\ref{Pi}) is nonempty and finite for all $(x,x') \in \mathcal{X} \times \mathcal{X}$.
\end{Lemma}

\begin{proof}
Observe that if $\pi^n(x)$ exists and $\pi^n(x) \in \mathcal{X}_0$, then $\pi^{n+1}(x)$ also exists from (\ref{pi}).
Suppose that $n \in \mathbb{N}\cup \{0\}$ does not exist such that $\pi^n(x)\in \mathcal{X}_1 \cap \mathcal{X}_0^c = (\underline{x}_1,\underline{x}_0]$.
Then, there exists sequence $\{ x_n \}_{n=0}^{\infty}$ such that $x_n = \pi^n(x)$.
By (\ref{pi_property1}), $\{ x_n \}_{n=0}^{\infty}$ is a decreasing sequence.
Because $x_n > \underline{x}_0$, $\{x_n\}_{n=0}^{\infty}$ converges to $x_{\infty} \in [\underline{x}_0, \overline{x}^0)$.
Because
$$
 F_{X|Z}(x_{n+1}|1) = F_{X|Z}(x_n|0)
$$
holds from (\ref{pi_property2}) and $F_{X|Z}(x_{\infty}|1) = F_{X|Z}(x_{\infty}|0)$ from the continuity of $F_{X|Z}$.
However, this equation violates Assumption 3.
Hence, for all $x \in \mathcal{X}$, $n \in \mathbb{N}\cup\{0\}$ exists such that $\pi^n(x) \in \mathcal{X}_1 \cap \mathcal{X}_0^c$.
Consequently, $\pi^{n'}(x)$ does not exist for $n'>n$.
Similarly, for all $x \in \mathcal{X}$, $n \in \mathbb{N}\cup\{0\}$ exists such that $\pi^{-n}(x) \in \mathcal{X}_0 \cap \mathcal{X}_1^c$.
Then, $\pi^{-n'}(x)$ does not exist for $n'>n$.
Therefore, $\Pi_{x',x}^M$ is finite for all $(x,x') \in \mathcal{X} \times \mathcal{X}$ because a set $\{(n,m)\in \mathbb{Z} \times \mathbb{Z}: \pi^n(x')\ \text{and} \ \pi^m(x) \ \text{exist.} \}$ is finite.

We proceed to show the nonemptyness of $\Pi_{x',x}^M$.
For all $x,x' \in \mathcal{X}$, $(n,m)\in \mathbb{Z}\times\mathbb{Z}$ exists such that $\pi^{n}(x') \in \mathcal{X}_1 \cap \mathcal{X}_0^c = (\underline{x}_1,\underline{x}_0]$ and $\pi^{m}(x) \in \mathcal{X}_0 \cap \mathcal{X}_1^c = [\overline{x}_1,\overline{x}_0)$.
It follows from Assumption 3(ii) that $\pi^{n}(x') < \pi^{m}(x)$.
\end{proof}

Under Assumptions 1--5, a set $\{n \in \mathbb{Z}: \pi^n(x)\ \text{exists.} \}$ is finite from Lemma 1.
Hence, $g$ cannot be point identified using the method proposed by \cite{d2015identification} and \cite{torgovitsky2015identification}).

We impose the following assumption:

\begin{AssumptionM}[Monotonicity]
For all $e \in (0,1)$, $g(x,e)$ is nondecreasing in $x$.
\end{AssumptionM}
The monotonicity assumption holds in many economic models.
For example, the Engel function for a normal good is monotonically increasing in income, the demand function is ordinarily monotonically decreasing in price, and economic analyses of production often suppose that the production function is monotonically increasing in input.
Monotonicity assumptions of this type have been employed in many studies.
\cite{manski1997monotone} imposes a monotonicity assumption on a response function and shows that the average treatment response is partially identified.

If $(n,m) \in \Pi_{x',x}^M$, then Assumption M implies that
$$
 \tilde{T}_{x'}^{(n)} \left( g(x',e) \right) = g(\pi^n(x'),e) \leq g(\pi^m(x),e) = \tilde{T}_{x}^{(m)} \left( g(x,e) \right).
$$
Because $\tilde{T}_{x'}^{(n)}(y)$ is strictly increasing in $y$ and $\tilde{T}_{x'}^{(n)}\left( [\underline{y},\overline{y}] \right) = [\underline{y},\overline{y}]$, we have $g(x',e) \leq \left( \tilde{T}_{x'}^{(n)} \right)^{-1} \left( \tilde{T}_{x}^{(m)} \left( g(x,e) \right) \right)$.
Hence, we have
\begin{equation}
 g(x',e) \leq \min_{(n,m)\in \Pi_{x',x}^M} \left( \tilde{T}_{x'}^{(n)} \right)^{-1} \left( \tilde{T}_{x}^{(m)} \left( g(x,e) \right) \right). \nonumber
\end{equation}
Define
\begin{equation}
\label{def_TMU_TML}
\begin{aligned}
T_{x',x}^{MU}(y) & \equiv \min_{(n,m)\in \Pi_{x',x}^M} \left( \tilde{T}_{x'}^{(n)} \right)^{-1} \left( \tilde{T}_{x}^{(m)} \left( y \right) \right), \\
T_{x',x}^{ML}(y) & \equiv \max_{(n,m)\in \Pi_{x,x'}^M} \left( \tilde{T}_{x'}^{(m)} \right)^{-1} \left( \tilde{T}_{x}^{(n)} \left( y \right) \right).
\end{aligned}
\end{equation}
Then, $T_{x',x}^{MU}(y)$ is strictly increasing and satisfies
\begin{equation}
g(x',e) \leq T_{x',x}^{MU}\left( g(x,e) \right). \label{TU_M}
\end{equation}
$T_{x',x}^{MU}(y)$ achieves the tightest lower bound of $g$ because function $\left( \tilde{T}_{x'}^{(n)} \right)^{-1} \left( \tilde{T}_{x}^{(m)} \left( y \right) \right)$ satisfies (\ref{TU}) for any $(n,m) \in \Pi_{x',x}^M$.
Similarly, $T_{x',x}^{ML}(y)$ is strictly increasing and satisfies
\begin{equation}
 g(x',e) \geq T_{x',x}^{MU}\left( g(x,e) \right). \label{TL_M}
\end{equation}

Define
\begin{eqnarray}
G_x^{ML}(u) &\equiv & \int F_{Y|X=x'} \left( T_{x',x}^{MU} (u) \right) dF_{X}(x'), \nonumber \\
G_x^{MU}(u) &\equiv & \int F_{Y|X=x'} \left( T_{x',x}^{ML} (u) \right) dF_{X}(x') , \nonumber \\
B^{ML}(x,e) &\equiv & \sup_{y:y \leq x} \left\{ \left(G_y^{ML}\right)^{-1}(e) \right\}, \nonumber \\
B^{MU}(x,e) &\equiv & \inf_{y:y \geq x} \left\{ \left(G_y^{MU}\right)^{-1}(e) \right\}. \nonumber
\end{eqnarray}
$G_x^{ML}(u)$ and $G_x^{MU}(u)$ provide lower and upper bounds on $g(x,e)$ on the basis of arguments (\ref{approach_DF1}) and (\ref{approach_DF2}).
$B^{ML}(x,e)$ and $B^{MU}(x,e)$ strengthen these bounds.
\begin{Theorem}
Under Assumptions 1--5 and M, for all $(x,e) \in \mathcal{X} \times (0,1)$, we have
$$
 B^{ML}(x,e) \leq g(x,e) \leq  B^{MU}(x,e).
$$
\end{Theorem}

\begin{proof}
\underline{Step.1}\,
First, we show that
\begin{equation}
\left(G_x^{ML}\right)^{-1}(e) \leq g(x,e) \leq \left(G_x^{MU}\right)^{-1}(e). \label{g_bound}
\end{equation}
Function $T_{x',x}^{ML}(y)$ defined in (\ref{def_TMU_TML}) is strictly increasing in $y$ because $\tilde{T}_{x}^{(n)}(y)$ is strictly increasing in $y$ for all $n \in \mathbb{Z}$ and $x \in \mathcal{X}$ and $\Pi_{x',x}^M$ is finite as per Lemma 1.
Similarly, $T_{x',x}^{MU}(y)$ is strictly increasing in $y$.
Because $\tilde{T}_{x}^{(n)}([\underline{y},\overline{y}])=[\underline{y},\overline{y}]$ for all $n \in \mathbb{Z}$ and $x \in \mathcal{X}$, we have $\left( \tilde{T}_{x'}^{(m)} \right)^{-1} \left( \tilde{T}_{x}^{(n)}([\underline{y},\overline{y}]) \right) = [\underline{y},\overline{y}]$.
Because $\left( \tilde{T}_{x'}^{(m)} \right)^{-1} \left( \tilde{T}_{x}^{(n)}([\underline{y},\overline{y}]) \right) = [\underline{y},\overline{y}]$ and $\Pi_{x',x}^M$ is finite, it follows that $G_x^{ML}\left( [\underline{y},\overline{y}] \right) = [0,1]$.
Similarly, $G_x^{MU}\left( [\underline{y},\overline{y}] \right) = [0,1]$.
Therefore, inequalities (\ref{g_bound}) hold from (\ref{approach_DF1}) and (\ref{approach_DF2}).\\
\underline{Step.2}\,
Next, we show that
\begin{equation}
B^{ML}(x,e) \leq g(x,e) \leq  B^{MU}(x,e). \label{b_bound}
\end{equation}
Because $g(x,e)$ is nondecreasing in $x$ and $g(x,e) \geq \left(G_x^{ML}\right)^{-1}(e)$, we have $g(x,e) \geq g(y,e) \geq \left(G_y^{ML}\right)^{-1}(e)$ for all $y \leq x$.
Hence, it follows that
$$
g(x,e) \geq \sup_{y:y \leq x} \left\{ \left(G_y^{ML}\right)^{-1}(e) \right\}.
$$
Similarly, $g(x,e) \leq g(y,e) \leq \left(G_y^{MU}\right)^{-1}(e)$ for all $y \geq x$ and it follows that
$$
g(x,e) \leq \inf_{y:y \geq x} \left\{ \left(G_y^{MU}\right)^{-1}(e) \right\}.
$$
Therefore, inequalities (\ref{b_bound}) hold.
\end{proof}

In the first step, we show $\left(G_x^{ML}\right)^{-1}(e) \leq g(x,e) \leq \left(G_x^{MU}\right)^{-1}(e)$.
In the second step, we strengthen these bounds to $B^{ML}(x,e) \leq g(x,e) \leq  B^{MU}(x,e)$.
Figure 3 illustrates this proof intuitively.
This idea is similar to that in \cite{manski1997monotone}, who considers a case in which response function $y(t)$ is increasing, where $y(t)$ is a latent outcome with treatment $t$.
He then uses the monotonicity of $y(t)$ to partially identify the average response function $E[y(t)]$ when outcome support is bounded.
Our approach does not require information on the infimum and supremum of the support of outcomes.

Theorem 1 shows that we can obtain nontrivial bounds of $g$ by using the monotonicity assumption.
However, without this assumption, we show that the sharp bounds of $g$ become trivial bounds, that is $\underline{y} < g(x,e) < \overline{y}$.
To see this, suppose that $g$ is the true structural function.
Let $\mathcal{G}$ be a set of functions that satisfy Assumption 2 (i).
Similarly to \cite{torgovitsky2015identification}, we show that the identified set is
$$
\mathcal{G}^* \equiv \{\tilde{g}\in \mathcal{G}:(\tilde{g}^{-1}(X,Y),V) \mathop{\perp\!\!\!\perp} Z, \, P(Y \leq \tilde{g}(X,e))=e \  \text{for all $e\in (0,1)$}. \}.
$$
First, we show that if $\tilde{T}^{(n)}_x(\tilde{g}(x,e)) = \tilde{g}(\pi^n(x),e)$ holds, then $\tilde{g}(x,e)$ satisfies $(\tilde{g}^{-1}(X,Y),V) \mathop{\perp\!\!\!\perp} Z$.
By the definition, $\tilde{T}^{(1)}_x(\tilde{g}(x,e)) = \tilde{g}(\pi(x),e)$ implies that
\begin{eqnarray}
& & F_{Y|X,Z}(\tilde{g}(x,e)|x,0) = F_{Y|X,Z}(\tilde{g}(\pi(x),e)|\pi(x),1) \nonumber \\
& \Leftrightarrow & P(\tilde{g}^{-1}(X,Y)\leq e | V = v, Z=1) = P(\tilde{g}^{-1}(X,Y)\leq e | V = v, Z=0), \nonumber
\end{eqnarray}
where $v = F_{X|Z}(x|1)$.
Therefore, if $\tilde{T}^{(n)}_x(\tilde{g}(x,e)) = \tilde{g}(\pi^n(x),e)$ for all $x$ and $e$, then $\tilde{g}(x,e)$ satisfies $(\tilde{g}^{-1}(X,Y),V) \mathop{\perp\!\!\!\perp} Z$.
Define $g_{\phi}(x,e) \equiv g(x,\phi(x,e))$, where $\phi:\mathcal{X}\times (0,1) \mapsto (0,1)$ is continuous and strictly increasing $e$.
Because we have $\tilde{T}^{(n)}_x(g(x,\phi(x,e))) = g(\pi^n(x),\phi(x,e))$ by Proposition 1, if $\phi$ satisfies
\begin{equation}
\phi(x,e) = \phi(\pi^n(x),e) \ \ \text{for all $x\in\mathcal{X}$ and $e\in (0,1)$}, \label{phi identified set 1}
\end{equation}
then $(g_{\phi}^{-1}(X,Y),V) \mathop{\perp\!\!\!\perp} Z$ holds.
Hence, if $\phi$ satisfies (\ref{phi identified set 1}) and
\begin{equation}
P(Y \leq g_{\phi}(X,e)) = \int P(\epsilon \leq \phi(x,e)|X=x) dF_X(x) \ \ \text{for all $e \in (0,1)$}, \label{phi identified set 2}
\end{equation}
then $g_{\phi} \in \mathcal{G}^*$.
Fix $x' \in \mathcal{X}$ and $e' \in (0,1)$.
Then, we can find the function $\phi$ such that $\phi(x',e')$ is close to 0 or 1 and satisfies (\ref{phi identified set 1}) and (\ref{phi identified set 2}).
Therefore, without the monotonicity assumption, the sharp bounds of $g(x',e)$ become $\underline{y}$ and $\overline{y}$.

To illustrate Theorem 1, we consider the following example:
\begin{eqnarray}
Y &=& X^{1/3} exp \left(\alpha + \beta \Phi^{-1}(\epsilon) \right) \nonumber \\
X &=& d + c (1-Z) + \eta , \nonumber
\end{eqnarray}
where $\Phi(\cdot)$ is the standard normal distribution function and $Z$ is a random Bernoulli variable with $p=0.5$.
Suppose that
\begin{eqnarray}
\epsilon &=& \Phi(U) \nonumber \\
\eta &=& \Phi(V) \nonumber \\
(U,V) &\sim & N\left( \left( \begin{array}{c}
0 \\ 0
\end{array} \right) ,\left( \begin{array}{cc}
1 & \rho \\ \rho & 1
\end{array} \right) \right). \nonumber
\end{eqnarray}
Thus, $\epsilon \sim U(0,1)$ and $\eta \sim U(0,1)$.
In this example, $F_{X|Z}(x|1) = x-d$ for $x \in \mathcal{X}_1 = (d,1+d)$ and $F_{X|Z}(x|0) = x-c-d$ for $x \in \mathcal{X}_0= (c+d,1+c+d)$.
Hence, the conditional distribution functions $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ do not have any intersection.
This example satisfies the assumptions in Theorem 1.
We calculate the bounds of $g(x,0.5)$ using Theorem 1 when $\alpha = 0.5, \beta = 0.5, c = 0.5, d= 0.05$, and $\rho = 0.3$.
These bounds are depicted in Figure 4.

The bounds become tighter as the difference between $g(x',e)$ and $T_{x',x}^U \left( g(x,e) \right)$ (or $T_{x',x}^L \left( g(x,e) \right)$) reduces.
The following theorem shows that, if $g(x,e)$ is flat in $x$ in a given interval, inequalities (\ref{TU}) and (\ref{TL}) become equalities and structural function $g$ is point identified.

\begin{Theorem}
Under Assumptions 1--5 and M, if there exists $\tilde{x} \in \mathcal{X}$ such that $x \mapsto g(x,e)$ is constant on $[\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$ for each $e \in (0,1)$, then $g(x,e)$ is point identified for all $(x,e) \in \mathcal{X} \times (0,1)$.
This result holds even when interval $[\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$ is unknown.
\end{Theorem}

\begin{proof}
\underline{Step.1}\,
First, we show that for all $x \in \mathcal{X}$, $n \in \mathbb{Z}$ exists such that $\pi^n(x), \pi^{n+1}(x) \in [\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$.
Because $\pi^n(x)$ is strictly increasing in $x$, for all $n \in \mathbb{Z}$, we have
\begin{equation}
x \leq y \ \ \Rightarrow \ \ \pi^n(x) \leq \pi^n(y). \label{pi_property3}
\end{equation}
We consider the following four cases: (i) $\pi(\tilde{x})\leq x \leq \tilde{x}$ (ii) $\tilde{x} \leq x \leq \pi^{-1}(\tilde{x})$ (iii) $x < \pi(\tilde{x})$, and (iv) $x > \pi^{-1}(\tilde{x})$.
In case (i), it follows from (\ref{pi_property3}) that $\pi(\tilde{x})\leq x \leq \tilde{x} \leq \pi^{-1}(x) \leq \pi^{-1}(\tilde{x})$.
In case (ii), it follows from (\ref{pi_property3}) that $\pi(\tilde{x})\leq \pi(x) \leq \tilde{x} \leq x \leq \pi^{-1}(\tilde{x})$.
In case (iii), by the definition of $\pi$, we have $\pi(\tilde{x}) \in \mathcal{X}_1$.
It follows from the proof of Lemma 1 that $n' \in \mathbb{Z}$ exists such that $\pi(\tilde{x}) \leq \pi^{-n'}(x) \in \mathcal{X}_0 \cap \mathcal{X}_1^c$.
Hence, $n \in \mathbb{Z}$ exists such that $\pi^{n+2}(x) \leq \pi(\tilde{x}) \leq \pi^{n+1}(x)$ and we can obtain $\pi(\tilde{x}) \leq \pi^{n+1}(x) \leq \tilde{x} \leq \pi^{n}(x) \leq \pi^{-1}(\tilde{x})$ from (\ref{pi_property3}).
In case (iv), similar to case (iii), there exists $n \in \mathbb{Z}$ such that $\pi^n(x) , \pi^{n+1}(x) \in [\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$.\\
\underline{Step.2}\,
Next, we show that $g$ is point identified.
From step 1, for all $x, x' \in \mathcal{X}$, there exists $n,m \in \mathbb{Z}$ such that $\pi^n(x'),\pi^{n+1}(x'),\pi^m(x),\pi^{m+1}(x) \in [\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$.
Then, from (\ref{pi_property3}), we have either $\pi^{n+1}(x') \leq \pi^{m+1}(x) \leq \pi^{n}(x') \leq \pi^{m}(x)$ or $\pi^{m+1}(x) \leq \pi^{n+1}(x') \leq \pi^{m}(x) \leq \pi^{n}(x')$.
If $\pi^{n+1}(x') \leq \pi^{m+1}(x) \leq \pi^{n}(x') \leq \pi^{m}(x)$, then we have $(n+1,m+1),(n,m) \in \Pi_{x',x}^M$ and $(m+1,n) \in \Pi_{x,x'}^M$.
If $\pi^{m+1}(x) \leq \pi^{n+1}(x') \leq \pi^{m}(x) \leq \pi^{n}(x')$, then we have $(n+1,m) \in \Pi_{x',x}^M$ and $(m+1,n+1), (m,n) \in \Pi_{x,x'}^M$.
Because $g(x,e)$ is constant in $x$ conditional on $[\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$, $\pi^n(x'), \pi^m(x) \in [\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$ implies that
$$
\tilde{T}_{x'}^{(n)} \left( g(x',e) \right) = \tilde{T}_{x}^{(m)} \left( g(x,e) \right).
$$
Therefore, $g(x',e)= T_{x',x}^{MU} \left( g(x,e) \right)$ and $g(x',e)= T_{x',x}^{ML} \left( g(x,e) \right)$.
Hence, $g(x,e)$ is identified because inequalities (\ref{TU_M}) and (\ref{TL_M}) become equalities.
\end{proof}

In the first step, we show that for all $x \in \mathcal{X}$, $n \in \mathbb{Z}$ exists such that $\pi^n(x), \pi^{n+1}(x) \in [\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$.
In the second step, we show that $g$ is point identified.
Because $g(x,e)$ is constant in $x$ conditional on $[\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$, we have $g(x',e)= T_{x',x}^{MU} \left( g(x,e) \right)$, and $g(x',e)= T_{x',x}^{ML} \left( g(x,e) \right)$ for all $x, x' \in \mathcal{X}$ and $e \in (0,1)$.
Hence, $g(x,e)$ is point identified because inequalities (\ref{TU_M}) and (\ref{TL_M}) become equalities.

\begin{Example}[Engel curve]
We consider the case where we want to estimate the Engel function by using a cash transfer experiment.
Let $Y$ be household expenditure on a normal good, $X$ be household income, and $Z$ be a treatment indicator of the cash transfer experiment.
Let $\beta$ denote transfer amount.
Then, we write the model (\ref{model}) as the following form:
\begin{eqnarray}
Y &=& g(X,\epsilon) \nonumber \\
X &=& \beta Z + \tilde{\eta}, \nonumber
\end{eqnarray}
where $\epsilon$ represents household preference and $\tilde{\eta}$ is initial household income.
If participants are randomly assigned to either the treatment or the control group, $Z$ is independent of $(\epsilon,\tilde{\eta})$.

In the proof of Proposition 1, we show that $P(\epsilon \leq e|X=x,Z=0) = P(\epsilon \leq e|X=\pi(x),Z=1)$.
In this case, we can understand it intuitively.
Observe that $\pi(x) = x + \beta$.
Then, households with $X=x$ and $Z=0$ have the same initial household income $\tilde{\eta}$ as households with $X=\pi(x)=x+\beta$ and $Z=1$.
Because $Z$ is independent of household preference and initial household income, the household with $X=x$ and $Z=0$ have the same conditional distribution of $\epsilon$ as the household with $X=\pi(x)=x+\beta$ and $Z=1$.
Hence, we obtain $P(\epsilon \leq e|X=x,Z=0) = P(\epsilon \leq e|X=\pi(x),Z=1)$.

In this case, we cannot use the identification results of \cite{d2015identification} and \cite{torgovitsky2015identification} because $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ do not have an intersection.
However, because the Engel function $g(X,\epsilon)$ is generally nondecreasing in household income $X$, we can partially identify $g$ by using our identification result.
\end{Example}

\section{Partial Identification through Concavity}

In this section, we propose how to construct the lower and upper bounds of $g(x,e)$ when $g(x,e)$ is concave in $x$.

First, we show that a set $\Pi_{x',x}^C$ defined below is nonempty and finite.
Next, we show that we can partially identify $g$ using $\Pi_{x',x}^C$ when $g(x,e)$ is concave in $x$.

For $(x,x') \in \mathcal{X} \times \mathcal{X}$, define $\Pi^C_{x',x}$ as
\begin{eqnarray}
\Pi^C_{x',x} &\equiv & \left\{ (n,m) : \pi^n(x') \leq \pi^m(x) \leq \pi^{n-1}(x') ,\right. \nonumber \\
 & & \left. n,m \in \mathbb{Z}, \ \pi^n(x'),\ \pi^{n-1}(x) \ \text{and} \ \pi^m(x) \ \text{exist.} \right\}. \label{Pi^C}
\end{eqnarray}
In Figure 1, $\Pi^C_{x',x} = \{ (0,-1), (1,0) \}$.
The following lemma shows that $\Pi^C_{x',x}$ is nonempty and finite, similar to Lemma 1.

\begin{Lemma}
Under Assumptions 1--5, $\Pi^C_{x',x}$ defined by (\ref{Pi^C}) is nonempty and finite for all $(x,x') \in \mathcal{X} \times \mathcal{X}$.
\end{Lemma}

\begin{proof}
From the proof of Lemma 1, $\Pi^C_{x',x}$ is finite.
Hence, we show the nonemptyness of $\Pi^C_{x',x}$.
From the proof of Lemma 1, for all $x,x' \in \mathcal{X}$, there exist $n, m \in \mathbb{Z}$ such that $\pi^{m}(x), \pi^{n}(x') \in \mathcal{X}_1 \cap \mathcal{X}_0^c$.
Without loss of generality, we assume $\pi^{n}(x') \leq \pi^{m}(x)$.
Then, $\pi^{m-1}(x)$ and $\pi^{n-1}(x')$ exist because $\pi^{m}(x) \in \mathcal{X}_1$ and $\pi^{n}(x') \in \mathcal{X}_1$.
Because $\pi^m(x) \in \mathcal{X}_0^c$ and $pi^{n-1}(x') \in \mathcal{X}_1$, we have $\pi^{n}(x') \leq\pi^{m}(x) \leq \pi^{n-1}(x') \leq \pi^{m-1}(x)$ from (\ref{pi_property3}), and hence $(n,m) \in \Pi_{x',x}^C$.
Therefore, $\Pi_{x',x}^C$ is nonempty.
\end{proof}

Similar to Section 3, we impose the following assumption:

\begin{AssumptionC}[Concavity]
For all $e \in (0,1)$, $g(x,e)$ is concave in $x$.
\end{AssumptionC}

The concavity assumption holds in many economic models.
For example, economic analyses of production often suppose that the production function is concave in inputs.
\cite{manski1997monotone} assumes a concavity assumption and shows that the average treatment response is partially identified.
\cite{d2013nonlinear} achieves the partial identification of the average treatment on treated effect using a locally concavity assumption.

As seen in Section 3, if we identify functions $T_{x',x}^U(y)$ and $T_{x',x}^L(y)$ that are strictly increasing in $y$, surjective, and satisfy (\ref{TU}) and (\ref{TL}), we can obtain the lower and upper bounds of $g(x,e)$.
Hence, we consider how to construct functions $T_{x',x}^U(y)$ and $T_{x',x}^L(y)$ that are strictly increasing in $y$, surjective, and satisfy (\ref{TU}) and (\ref{TL}).

If $(n,m) \in \Pi^C_{x',x}$, from Assumption C, we have
$$
\left[ t_{x',x}(n,m) \tilde{T}_{x'}^{(n)} + (1-t_{x',x}(n,m)) \tilde{T}_{x'}^{(n-1)} \right] \left( g(x',e) \right) \leq \tilde{T}_{x}^{(m)} \left( g(x,e) \right),
$$
where $\left[ t_{x',x}(n,m) \tilde{T}_{x'}^{(n)} + (1-t_{x',x}(n,m)) \tilde{T}_{x'}^{(n-1)} \right](y) =  t_{x',x}(n,m) \tilde{T}_{x'}^{(n)}(y) + (1-t_{x',x}(n,m))$ $\tilde{T}_{x'}^{(n-1)}(y)$ and $t_{x',x}(n,m)= \left( \pi^{n-1}(x') - \pi^m(x) \right) / \left( \pi^{n-1}(x') - \pi^n(x') \right)$.
Because $\tilde{T}_{x'}^{(n)}(y)$ and $\tilde{T}_{x'}^{(n-1)}(y)$ are strictly increasing in $y$, $\tilde{T}_{x'}^{(n)}\left( [\underline{y},\overline{y}] \right)=[\underline{y},\overline{y}]$, and $\tilde{T}_{x'}^{(n-1)}\left( [\underline{y},\overline{y}] \right)=[\underline{y},\overline{y}]$, we have
$$
g(x',e) \leq \min_{(n,m)\in\Pi^C_{x',x}} \left[ t_{x',x}(n,m) \tilde{T}_{x'}^{(n)} + (1-t_{x',x}(n,m)) \tilde{T}_{x'}^{(n-1)} \right]^{-1} \left( \tilde{T}_{x}^{(m)} \left( g(x,e) \right) \right).
$$
Define
\begin{equation}\label{def_TCU_TCL}
\begin{aligned}
T_{x',x}^{CU}(y) & \equiv & \min_{(n,m)\in\Pi^C_{x',x}} \left[ t_{x',x}(n,m) \tilde{T}_{x'}^{(n)} + (1-t_{x',x}(n,m)) \tilde{T}_{x'}^{(n-1)} \right]^{-1} \left( \tilde{T}_{x}^{(m)} \left( y \right) \right), \\
T_{x',x}^{CL}(y) & \equiv & \max_{(n,m)\in\Pi^C_{x,x'}} \left( \tilde{T}_{x'}^{(m)} \right)^{-1} \left( \left[ t_{x,x'}(n,m) \tilde{T}_{x}^{(n)} + (1-t_{x,x'}(n,m))  \tilde{T}_{x}^{(n-1)} \right] \left( y \right) \right).
\end{aligned}
\end{equation}
Then, $T_{x',x}^{CU}(y)$ defined in (\ref{def_TCU_TCL}) is strictly increasing and satisfies
\begin{eqnarray}
 g(x',e) \leq T_{x',x}^{CU} \left( g(x,e) \right). \label{TU_C}
\end{eqnarray}
Similarly, $T_{x',x}^{CL}(y)$ defined in (\ref{def_TCU_TCL}) is strictly increasing and satisfies
\begin{eqnarray}
 g(x',e) \geq  T_{x',x}^{CL} \left( g(x,e) \right). \label{TL_C}
\end{eqnarray}

Define
\begin{eqnarray}
G_x^{CL}(u) &\equiv & \int F_{Y|X=x'} \left( T_{x',x}^{CU} (u) \right) dF_{X}(x'), \nonumber \\
G_x^{CU}(u) &\equiv & \int F_{Y|X=x'} \left( T_{x',x}^{CL} (u) \right) dF_{X}(x'), \nonumber
\end{eqnarray}
$$
 B^{CL}(x,e) \equiv \sup_{y,y':y < x < y'} \left\{ \left( \frac{x-y}{y'-y} \right) \left( G_{y'}^{CL} \right)^{-1}(e) + \left( \frac{y'-x}{y'-y} \right) \left( G_y^{CL} \right)^{-1}(e) \right\},
$$
\begin{eqnarray}
 B^{CU}(x,e) \equiv & \min \left[  \inf_{y,y':x < y < y'} \left\{ \left( \frac{x-y}{y'-y} \right)B^{CL}(y',e) + \left( \frac{y'-x}{y'-y} \right) \left( G_y^{CU} \right)^{-1}(e) \right\}, \right. \nonumber \\
 & \left. \inf_{y,y':y' < y < x} \left\{ \left( \frac{y-x}{y-y'} \right)B^{CL}(y',e) + \left( \frac{x-y'}{y-y'} \right) \left( G_y^{CU} \right)^{-1}(e) \right\} \right]. \nonumber
\end{eqnarray}
$G_x^{CL}(u)$ and $G_x^{CU}(u)$ provide the lower and upper bounds of $g(x,e)$ as per arguments (\ref{approach_DF1}) and (\ref{approach_DF2}).
$B^{CL}(x,e)$ and $B^{CU}(x,e)$ strengthen these bounds.

\begin{Theorem}
Under Assumptions 1--5 and C, for all $(x,e) \in \mathcal{X} \times (0,1)$, we have
$$
 B^{CL}(x,e) \leq g(x,e) \leq B^{CU}(x,e).
$$
\end{Theorem}
\begin{proof}
Similar to the proof of Theorem 1, we can obtain
$$
\left( G_x^{CL} \right)^{-1}(e) \leq g(x,e) \leq \left( G_x^{CU} \right)^{-1}(e).
$$
Because $g(x,e)$ is concave in $x$, if $x = ty' + (1-t)y$ and $t \in (0,1)$, then we have $g(x,e) \geq t g(y',e) + (1-t) g(y,e) \geq t \left( G_{y'}^{CL} \right)^{-1}(e) + (1-t) \left( G_{y}^{CL} \right)^{-1}(e)$.
Hence, we have
$$
g(x,e) \geq \sup_{y,y':y < x < y'} \left\{ \left( \frac{x-y}{y'-y} \right) \left( G_{y'}^{CL} \right)^{-1}(e) + \left( \frac{y'-x}{y'-y} \right) \left( G_y^{CL} \right)^{-1}(e) \right\}.
$$
Because $g(x,e)$ is concave in $x$, if $x = ty' + (1-t)y$ and $t < 0$, then we have $g(x,e) \leq t g(y',e) + (1-t) g(y,e)$.
Because $B^{CL}(x,e) \leq g(x,e) \leq \left( G_x^{CU} \right)^{-1}(e)$, $t<0$, and $1-t>0$, we have $g(x,e) \leq t  B^{CL}(y',e) + (1-t) \left( G_{y}^{CU} \right)^{-1}(e)$.
Similarly, if $x = ty + (1-t)y'$ and $t > 1$, then we have $g(x,e) \leq t g(y,e) + (1-t) g(y',e) \leq t\left( G_{y}^{CU} \right)^{-1}(e) + (1-t) B^{CL}(y',e)$.
Hence, we have
\begin{eqnarray}
g(x,e) \leq & \min \left[  \inf_{y,y':x < y < y'} \left\{ \left( \frac{x-y}{y'-y} \right)B^{CL}(y',e) + \left( \frac{y'-x}{y'-y} \right) \left( G_y^{CU} \right)^{-1}(e) \right\}, \right. \nonumber \\
 & \left. \inf_{y,y':y' < y < x} \left\{ \left( \frac{y-x}{y-y'} \right)B^{CL}(y',e) + \left( \frac{x-y'}{y-y'} \right) \left( G_y^{CU} \right)^{-1}(e) \right\} \right]. \nonumber
\end{eqnarray}
\end{proof}

Similar to Theorem 1, we can show that $\left( G_x^{CL} \right)^{-1}(e) \leq g(x,e) \leq \left( G_x^{CU} \right)^{-1}(e)$.
We strengthen the bounds to $B^{CL}(x,e) \leq g(x,e) \leq B^{CU}(x,e)$ using the concavity of $g(x,e)$ in $x$.
Figure 5 illustrates this proof intuitively.
A similar approach is used by \cite{manski1997monotone}.
\cite{manski1997monotone} uses the concavity of the response function to partially identify the average response function when the support of outcome is bounded.
Our approach does not require information on the infimum and supremum of the support of outcomes.

This identification approach is somewhat similar to that considered by \cite{d2013nonlinear}.
They study the identification of nonseparable models with continuous, endogenous regressors, using repeated cross sections.
They consider the following model:
$$
Y_t = g_t(X_t,A_t), \ \ t = 1, \cdots, T,
$$
where $A_t$ is an unobserved heterogeneous factor.
They show that under the assumptions that $A_t|V_t \equiv F_{X_t}(X_t)=v \sim A_s|V_s \equiv F_{X_s}(X_s)=v$ and $g_t(x,a) = m_t(g(x,a))$, the average treatment on treated effect $\Delta^{ATT}(x,x') \equiv E[g_T(x,A_T)-g_T(x',A_T)|X_T=x]$ is identified when $F_{X_T}(x) = F_{X_t}(x')$.
Under their assumption, $\Delta^{ATT}(x,x')$ is not identified if $F_{X_T}(x) \neq F_{X_t}(x')$ for all $t \in \{1, \cdots , T-1\}$.
However, they show that $\Delta^{ATT}(x,x')$ is partially identified if $x \mapsto g(x,a)$ is locally concave.

In several cases, such as production function, we may assume that both of Assumptions M and C hold.
Then, it follows from Theorem 1 and 3 that
\begin{equation}
\max \{B^{ML}(x,e),B^{CL}(x,e)\} \leq g(x,e) \leq \min \{B^{MU}(x,e),B^{CU}(x,e)\}. \label{bounds_M_C_1}
\end{equation}
However, in this case, we can obtain the tighter bounds in the following manner.
Define
\begin{eqnarray}
T_{x',x}^{MCU}(y) & \equiv & \min\{T^{MU}_{x',x}(y),T^{CU}_{x',x}(y)\}, \nonumber \\
T_{x',x}^{MCL}(y) & \equiv & \max\{T^{ML}_{x',x}(y),T^{CL}_{x',x}(y)\}, \nonumber \\
G_x^{MCL}(u) &\equiv & \int F_{Y|X=x'} \left( T_{x',x}^{MCU} (u) \right) dF_{X}(x'), \nonumber \\
G_x^{MCU}(u) &\equiv & \int F_{Y|X=x'} \left( T_{x',x}^{MCL} (u) \right) dF_{X}(x'). \nonumber
\end{eqnarray}
Similarly to the above arguments, we have $g(x',e) \leq T_{x',x}^{MCU}(g(x,e))$ and $g(x',e) \geq T_{x',x}^{MCL}(g(x,e))$, and hence we can obtain
\begin{eqnarray}
\left( G_x^{MCL} \right)^{-1}(e) \leq g(x,e) \leq \left( G_x^{MCU} \right)^{-1}(e). \nonumber
\end{eqnarray}
Define
\begin{eqnarray}
\tilde{B}^{MCL}(x,e) &\equiv & \sup_{y:y \leq x} \left\{ \left(G_y^{MCL}\right)^{-1}(e) \right\}, \nonumber \\
\tilde{B}^{MCU}(x,e) &\equiv & \inf_{y:y \geq x} \left\{ \left(G_y^{MCU}\right)^{-1}(e) \right\}, \nonumber
\end{eqnarray}
$$
 \hat{B}^{MCL}(x,e) \equiv \sup_{y,y':y < x < y'} \left\{ \left( \frac{x-y}{y'-y} \right) \left( G_{y'}^{MCL} \right)^{-1}(e) + \left( \frac{y'-x}{y'-y} \right) \left( G_y^{MCL} \right)^{-1}(e) \right\},
$$
\begin{eqnarray}
 \hat{B}^{MCU}(x,e) \equiv & \min \left[  \inf_{y,y':x < y < y'} \left\{ \left( \frac{x-y}{y'-y} \right)\hat{B}^{MCL}(y',e) + \left( \frac{y'-x}{y'-y} \right) \left( G_y^{MCU} \right)^{-1}(e) \right\}, \right. \nonumber \\
 & \left. \inf_{y,y':y' < y < x} \left\{ \left( \frac{y-x}{y-y'} \right)\hat{B}^{MCL}(y',e) + \left( \frac{x-y'}{y-y'} \right) \left( G_y^{MCU} \right)^{-1}(e) \right\} \right]. \nonumber
\end{eqnarray}
Then, from the above results, both of $\tilde{B}^{MCU}(x,e)$ and $\hat{B}^{MCU}(x,e)$ are upper bounds of $g(x,e)$.
Similarly, both of $\tilde{B}^{MCL}(x,e)$ and $\hat{B}^{MCL}(x,e)$ are also lower bounds of $g(x,e)$.
Therefore, we can obtain
\begin{equation}
\max\{\tilde{B}^{MCL}(x,e), \hat{B}^{MCL}(x,e)\} \leq g(x,e) \leq \min\{\tilde{B}^{MCL}(x,e),\hat{B}^{MCL}(x,e)\}. \label{bounds_M_C_2}
\end{equation}
Clearly, these bounds are tighter than (\ref{bounds_M_C_1}).

Similar to Theorem 2 in Section 3, the following theorem shows that if $g(x,e)$ is linear in $x$ in a particular interval, then the inequalities (\ref{TU_C}) and (\ref{TL_C}) become equalities and structural function $g$ is point identified.

\begin{Theorem}
Under Assumptions 1--5 and C, if $\tilde{x} \in \mathcal{X}$ exists such that $g(x,e)$ is linear in $x$ on $[\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$, then $g(x,e)$ is point identified.
This result holds even if interval $[\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$ is unknown.
\end{Theorem}
\begin{proof}
Similar to Theorem 2, for all $x,x' \in \mathcal{X}$, there exist $n,m \in \mathbb{Z}$ such that $\pi^n(x')$, $\pi^{n-1}(x')$, and $\pi^m(x)$ are in $[\pi(\tilde{x}),\pi^{-1}(\tilde{x})]$, and $\pi^n(x') \leq \pi^m(x) \leq \pi^{n-1}(x')$.
Because $g(x,e)$ is linear in $x$, we have
$$
\left[ t_{x',x}(n,m) \tilde{T}_{x'}^{(n)} + (1-t_{x',x}(n,m)) \tilde{T}_{x'}^{(n-1)} \right] \left( g(x',e) \right)= \tilde{T}_{x}^{(m)} \left( g(x,e) \right).
$$
Similarly, for all $x,x' \in \mathcal{X}$, there exist $n,m \in \mathbb{Z}$ such that
$$
\tilde{T}_{x'}^{(m)} \left( g(x',e) \right) = \left[ t_{x,x'}(n,m) \tilde{T}_{x}^{(n)} + (1-t_{x,x'}(n,m)) \tilde{T}_{x}^{(n-1)} \right] \left( g(x,e) \right).
$$
Hence, as described above, $g(x,e)$ is point identified because inequalities (\ref{TU_C}) and (\ref{TL_C}) become equalities.
\end{proof}

\begin{Example}[quantile regression]
Suppose $g(X,\epsilon)=\theta_0(\epsilon)+\theta_1(\epsilon)X$, where $X$ is positive and $\theta_0(e)$ and $\theta_1(e)$ are strictly increasing in $e$.
This model is a quantile regression model with endogeneity.
The $\tau$-th quantile function of $g(x,\epsilon)$ is $\theta_0(\tau)+\theta_1(\tau)x$.
In this case, structural function $g(x,e) = \theta_0(e) + \theta_1(e)x$ is linear in $x$.
Hence, Theorem 4 shows that $\theta_0(e)$ and $\theta_1(e)$ are identified if binary instruments are available.
\end{Example}

\section{Extension: General Models}

In this section, we extend the results in Section 3 to more general models and allow $Y$ to be discrete or censored.
If outcomes are discrete or censored, then assumptions 2 and 4 are not satisfied.
Hence, we replace these assumptions with the following ones:

\begin{Assumption2}
(i) Function $g(x,e)$ is nondecreasing in $e$ for all $x \in \mathcal{X}$. (ii) For all $z \in \{0,1\}$, $h(z,v)$ is continuous and strictly increasing in $v$.
\end{Assumption2}

\begin{Assumption4}
For all $(z,x) \in \{0,1\} \times \mathcal{X}_z$, we have $\overline{\mathcal{Y}}_{x,z} = \overline{\mathcal{Y}}$.
Here, define $\overline{y} \equiv \sup \{y:y \in \overline{\mathcal{Y}} \}$ and $\underline{y} \equiv \inf \{y:y \in \overline{\mathcal{Y}} \}$.
\end{Assumption4}

Assumption 2'(i) differs from Assumption 2(i).
Assumption 2(i) imposes the strict monotonicity of $g(x,e)$ in $e$, while Assumption 2'(i) requires only the weak monotonicity of $g(x,e)$ in $e$.
For example, if we consider
$$
g(x,e)=\textbf{1}\{e > (1+\exp(\beta_0+\beta_1 x))^{-1}\},
$$
then $g(x,e)$ is not strictly increasing in $e$.
\cite{chesher2010instrumental} also employs a weak monotonicity condition.
Assumption 4 implies that $Y$ is continuously distributed, but assumptions 2' and 4' allow outcomes that are discrete or censored.

\cite{d2015identification} and \cite{torgovitsky2015identification} do not consider the case in which outcomes are discrete or censored because they assume that $g(x,e)$ is strictly increasing in $e$.
\cite{chesher2010instrumental} considers instrumental variable models for discrete outcome.
He shows that the structural function is partially identified using instruments.

In this section, we show that $g(x,e)$ is partially identified under assumptions 1, 2', 3, 4', 5, and M.

Define
\begin{eqnarray}
F_{Y|X,Z}^{+}(y|x,z) &\equiv & P(Y \leq y | X=x, Z=z), \nonumber \\
F_{Y|X,Z}^{-}(y|x,z) &\equiv & P(Y < y | X=x, Z=z), \nonumber \\
Q_{Y|X,Z}^{+}(\tau |x,z) &\equiv & \sup \{y: F_{Y|X,Z}^{-}(y|x,z) \leq \tau \} \wedge \overline{y}, \nonumber \\
Q_{Y|X,Z}^{-}(\tau |x,z) &\equiv & \inf \{y: F_{Y|X,Z}^{+}(y|x,z) \geq \tau \} \vee \underline{y}. \nonumber
\end{eqnarray}
$F_{Y|X,Z}^{+}(y|x,z)$ and $Q_{Y|X,Z}^{+}(\tau |x,z)$ are right continuous in $y$ and $\tau$.
$F_{Y|X,Z}^{-}(y|x,z)$ and $Q_{Y|X,Z}^{-}(\tau |x,z)$ are left continuous in $y$ and $\tau$.
Under assumptions 2' and 4', Proposition 1 does not hold.
Instead, we show the following proposition.

\begin{Proposition}
Define
\begin{eqnarray}
\hat{T}_x^{(1)}(y) &\equiv & Q_{Y|X,Z}^{-}\left(F_{Y|X,Z}^{-}(y|x,0) \middle| \pi(x),1 \right), \nonumber \\
\check{T}_x^{(1)}(y) &\equiv & Q_{Y|X,Z}^{+}\left(F_{Y|X,Z}^{+}(y|x,0) \middle| \pi(x),1 \right), \nonumber \\
\hat{T}_x^{(-1)}(y) &\equiv & Q_{Y|X,Z}^{-}\left(F_{Y|X,Z}^{-}(y|x,1) \middle| \pi^{-1}(x),0 \right), \nonumber \\
\check{T}_x^{(-1)}(y) &\equiv & Q_{Y|X,Z}^{+}\left(F_{Y|X,Z}^{+}(y|x,1) \middle| \pi^{-1}(x),0 \right). \nonumber
\end{eqnarray}
Then, under Assumptions 1, 2', 3, 4', and 5, we have
\begin{eqnarray}
g\left(\pi(x),e\right) & \geq & \hat{T}_x^{(1)}\left( g(x,e) \right), \nonumber \\
g\left(\pi(x),e\right) & \leq & \check{T}_x^{(1)}\left( g(x,e) \right), \nonumber \\
g\left(\pi^{-1}(x),e\right) & \geq & \hat{T}_x^{(-1)}\left( g(x,e) \right), \nonumber \\
g\left(\pi^{-1}(x),e\right) & \leq & \check{T}_x^{(-1)}\left( g(x,e) \right). \nonumber
\end{eqnarray}
\end{Proposition}

\begin{proof}
Because $g(x,e)$ is nondecreasing in $e$, we have
\begin{eqnarray}
F_{Y|X,Z}^{-}\left(g(x,e)|x,0\right) &=& P\left( g(x,\epsilon) < g(x,e) | X=x, Z=0 \right) \nonumber \\
& \leq & P\left( \epsilon < e | X=x, Z=0 \right) \nonumber \\
& = & P\left( \epsilon \leq e | X=\pi(x), Z=1 \right) \nonumber \\
& \leq & P\left( g\left( \pi(x),\epsilon \right) \leq g\left( \pi(x),e \right) | X=\pi(x), Z=1 \right) \nonumber \\
& = & F_{Y|X,Z}^{+}\left(g(\pi(x),e)|\pi(x),1\right), \label{hat_T_1}
\end{eqnarray}
where the first inequality follows from $\{\epsilon: g(x,\epsilon) < g(x,e)\} \subset \{\epsilon: \epsilon < e\}$ and the second inequality follows from $\{\epsilon: \epsilon \leq e\} \subset \{\epsilon: g(x,\epsilon) \leq g(x,e)\}$.
From the definition of $Q_{Y|X,Z}^{-}(\tau|x,z)$, it follows that $Q_{Y|X,Z}^{-}\left(F_{Y|X,Z}^{+}(y|x,z) \middle| x,z \right) = \inf \{y': F_{Y|X,Z}^{+}(y'|x,z) \geq F_{Y|X,Z}^{+}(y|x,z) \} \vee \underline{y} \leq y$ for all $y \in \overline{\mathcal{Y}}$.
Hence, inequality (\ref{hat_T_1}) implies that
\begin{eqnarray}
\hat{T}_x^{(1)}\left( g(x,e) \right) &=& Q_{Y|X,Z}^{-}\left(F_{Y|X,Z}^{-}(g(x,e)|x,0) \middle| \pi(x),1 \right) \nonumber \\
& \leq & Q_{Y|X,Z}^{-}\left(F_{Y|X,Z}^{+}\left(g(\pi(x),e)|\pi(x),1\right) \middle| \pi(x),1 \right) \nonumber \\
& \leq & g(\pi(x),e). \nonumber
\end{eqnarray}

Similarly, because $g(x,e)$ is nondecreasing in $e$, we have
\begin{eqnarray}
F_{Y|X,Z}^{+}\left(g(x,e)|x,0\right) \geq F_{Y|X,Z}^{-}\left(g(\pi(x),e)|\pi(x),1\right). \nonumber
\end{eqnarray}
Because $Q_{Y|X,Z}^{+}\left(F_{Y|X,Z}^{-}(y|x,z) \middle| x,z \right) = \sup \{y': F_{Y|X,Z}^{-}(y'|x,z) \leq F_{Y|X,Z}^{-}(y|x,z) \} \wedge \overline{y} \geq y$ for all $y \in \overline{\mathcal{Y}}$, we have
$$
g(\pi(x),e) \leq \check{T}_x^{(1)}\left( g(x,e) \right).
$$

Similarly, we have two inequalities, $g(\pi^{-1}(x),e) \geq \hat{T}_x^{(-1)}\left( g(x,e) \right)$ and $g(\pi^{-1}(x),e) \leq \check{T}_x^{(-1)}\left( g(x,e) \right)$.
\end{proof}

This approach is similar to the identification approaches in \cite{athey2006identification} and \cite{chesher2010instrumental}.
\cite{athey2006identification} show that the counterfactual distribution is partially identified using right and left continuous quantile functions when outcomes are discrete.
\cite{chesher2010instrumental} uses the result in which the weak monotonicity of $h(x,u)$ in $u$ implies $\{u:h(x,u) \leq h(x,\tau)\}\supset \{u:u \leq \tau\}$ and $\{u:h(x,u) < h(x,\tau)\}\subset \{u:u < \tau\}$ and shows that structural function $h$ is partially identified.

Define $\hat{T}_x^{(n)}(y) \equiv \hat{T}_{\pi^{n-1}(x)}^{(1)} \circ \cdots \circ \hat{T}_x^{(1)}(y)$ and $\check{T}_x^{(n)}(y) \equiv \check{T}_{\pi^{n-1}(x)}^{(1)} \circ \cdots \circ \check{T}_x^{(1)}(y)$ for $n \in \mathbb{N}$, and then, we have
\begin{eqnarray}
g(\pi^n(x),e) & \geq & \hat{T}_x^{(n)}\left( g(x,e) \right), \nonumber \\
g(\pi^n(x),e) & \leq & \check{T}_x^{(n)}\left( g(x,e) \right). \nonumber
\end{eqnarray}
Similarly, define $\hat{T}_x^{(-n)}(y) \equiv \hat{T}_{\pi^{-(n-1)}(x)}^{(-1)} \circ \cdots \circ \hat{T}_x^{(-1)}(y)$ and $\check{T}_x^{(-n)}(y) \equiv \check{T}_{\pi^{-(n-1)}(x)}^{(-1)} \circ \cdots \circ \check{T}_x^{(-1)}(y)$ for $n \in \mathbb{N}$, and thus, we have
\begin{eqnarray}
g(\pi^{-n}(x),e) & \geq & \hat{T}_x^{(-n)}\left( g(x,e) \right), \nonumber \\
g(\pi^{-n}(x),e) & \leq & \check{T}_x^{(-n)}\left( g(x,e) \right). \nonumber
\end{eqnarray}
Define $\hat{T}_x^{(0)}(y)=y$ and $\check{T}_x^{(0)}(y)=y$.

If $(n,m) \in \Pi_{x',x}^M$, then Assumption M implies that
$$
\hat{T}_{x'}^{(n)} \left( g(x',e) \right) \leq  g(\pi^n(x'),e)
\leq  g(\pi^{m}(x),e) \leq \check{T}_{x}^{(m)} \left( g(x,e) \right).
$$
Define
$$
\left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}(u) \equiv \sup \{y: \hat{T}_{x'}^{(n)}(y) \leq u \} \wedge \overline{y},
$$
then we have $\left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}\left( \hat{T}_{x'}^{(n)}(y) \right) = \sup \{y': \hat{T}_{x'}^{(n)}(y') \leq \hat{T}_{x'}^{(n)}(y) \} \wedge \overline{y} \geq y$ for all $y \in \overline{\mathcal{Y}}$.
Hence, we obtain
\begin{eqnarray}
g(x',e) \leq \min_{(n,m)\in \Pi_{x',x}^M} \left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}\left( \check{T}_{x}^{(m)}(g(x,e)) \right). \nonumber
\end{eqnarray}
Define $T_{x',x}^{GU}(y) \equiv \min_{(n,m)\in \Pi_{x',x}^M} \left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}\left( \check{T}_{x}^{(m)}(y) \right)$, then $T_{x',x}^{GU}(y)$ satisfies
\begin{equation}
g(x',e) \leq T_{x',x}^{GU}\left( g(x,e) \right). \label{TU_G}
\end{equation}
Similarly, define $T_{x',x}^{GL}(y) \equiv \max_{(n,m)\in \Pi_{x,x'}^M} \left( \check{T}_{x'}^{(m)} \right)^{\leftarrow}\left( \hat{T}_{x}^{(n)}(y) \right)$ and $\left( \check{T}_{x'}^{(m)} \right)^{\leftarrow} (u) \equiv \inf\{y:\check{T}_{x'}^{(m)}(y) \geq u\} \vee \underline{y}$, then $T_{x',x}^{GL}(y)$ satisfies that
\begin{equation}
g(x',e) \geq T_{x',x}^{GL}\left( g(x,e) \right). \label{TL_G}
\end{equation}

Define
\begin{eqnarray}
G_x^{GL}(u) & \equiv & \int F_{Y|X}^{+}\left( T_{x',x}^{GU}(u)|x' \right)dF(x'), \nonumber \\
G_x^{GU}(u) & \equiv & \int F_{Y|X}^{-}\left( T_{x',x}^{GL}(u)|x' \right)dF(x'), \nonumber \\
B^{GL}(x,e) & \equiv & \sup_{y:y\leq x} \left\{ \inf\{u:G_y^{GL}(u) \geq e\} \right\} \vee \underline{y}, \nonumber \\
B^{GU}(x,e) & \equiv & \inf_{y:y\geq x} \left\{ \sup\{u:G_y^{GU}(u) \leq e\} \right\} \wedge \overline{y}, \nonumber
\end{eqnarray}
where $F_{Y|X}^{+}(y|x) \equiv P(Y\leq y| X=x)$ and $F_{Y|X}^{-}(y|x) \equiv P(Y< y| X=x)$.
$G_x^{GL}(u)$ and $G_x^{GU}(u)$ provide the lower and upper bounds on $g(x,e)$ as per the argument similar to (\ref{approach_DF1}) and (\ref{approach_DF2}).
$B^{GL}(x,e)$ and $B^{GU}(x,e)$ strengthen these bounds.

\begin{Theorem}
Under Assumptions 1, 2', 3, 4', 5, and M, for all $(x,e) \in \mathcal{X} \times (0,1)$, we have
$$
B^{GL}(x,e) \leq g(x,e) \leq B^{GU}(x,e).
$$
\end{Theorem}

\begin{proof}
First, we show that
\begin{equation}
\inf\{u:G_x^{GL}(u) \geq e\} \vee \underline{y} \leq g(x,e) \leq \sup\{u:G_x^{GU}(u) \leq e\} \wedge \overline{y}. \label{g_bound_G}
\end{equation}
Because $T_{x',x}^{GU}(y)$ satisfies (\ref{TU_G}), we have
\begin{eqnarray}
G_x^{GL}\left( g(x,e) \right) &=& \int F_{Y|X=x'}^{+}\left( T_{x',x}^{GU}(g(x,e)) \right)dF(x') \nonumber \\
&\geq & \int F_{Y|X=x'}^{+}\left( g(x',e) \right)dF(x') \nonumber \\
&=& \int P\left( g(x',\epsilon) \leq g(x',e)|X=x' \right)dF(x') \nonumber \\
&\geq & \int P\left( \epsilon \leq e|X=x' \right)dF(x') = e, \nonumber
\end{eqnarray}
where the second inequality follows from $\{\epsilon:\epsilon \leq e\} \subset \{\epsilon: g(x',\epsilon) \leq g(x',e)\}$.
Because $g(x,e) \geq \underline{y}$, we can obtain $g(x,e) \geq \inf\{u:G_x^{GL}(u) \geq e\} \vee \underline{y}$.
Similarly, because $T_{x',x}^{GL}(y)$ satisfies (\ref{TL_G}), we have
\begin{eqnarray}
G_x^{GU}\left( g(x,e) \right) &\leq & \int F_{Y|X=x'}^{-}\left( g(x',e) \right)dF(x') \nonumber \\
&=& \int P\left( g(x',\epsilon) < g(x',e)|X=x' \right)dF(x') \nonumber \\
&\leq & \int P\left( \epsilon < e|X=x' \right)dF(x') = e, \nonumber
\end{eqnarray}
where the second inequality follows from $\{\epsilon: g(x',\epsilon) < g(x',e)\} \subset \{\epsilon:\epsilon < e\}$.
Hence, we can obtain $g(x,e) \leq \sup\{u:G_x^{GU}(u) \leq e\} \wedge \overline{y}$.

Because $g(x,e)$ is nondecreasing in $x$ and (\ref{g_bound_G}) holds, similar to Theorem 1, we have $B^{GL}(x,e) \leq g(x,e) \leq B^{GU}(x,e)$.
\end{proof}

\if0
For $(x,x') \in \mathcal{X} \times \mathcal{X}$, define set $\Pi_{x',x}^A$ and $\Pi_{x',x}^B$ as
\begin{eqnarray}
\Pi_{x',x}^A & \equiv & \left\{ (n,m) : \pi^n(x') \leq \pi^{-m}(x), \right. \nonumber \\
& & \left. n,m \in \mathbb{N} \cap \{0\}, \ \pi^n(x') \ \text{and} \ \pi^{-m}(x) \ \text{exist}. \right\}, \label{Pi_A} \\
\Pi_{x',x}^B & \equiv & \left\{ (n,m) : \pi^{-n}(x') \leq \pi^{m}(x), \right. \nonumber \\
& & \left. n,m \in \mathbb{N} \cap \{0\}, \ \pi^{-n}(x') \ \text{and} \ \pi^{m}(x) \ \text{exist}. \right\}. \nonumber
\end{eqnarray}
In Figure 1, $\Pi_{x',x}^A = \{ (0,1),(1,0),(1,1) \}$.
Similar to Lemma 1, it is shown that set $\Pi_{x',x}^A$ is nonempty and finite.
On the other hand, set $\Pi_{x',x}^B$ can be empty.
In Figure 1, $\Pi_{x',x}^B$ is empty.

\begin{Lemma}
Under assumptions 1, 2', 3, 4', and 5, set $\Pi_{x',x}^A$ defined by (\ref{Pi_A}) is nonempty and finite for all $(x,x') \in \mathcal{X} \times \mathcal{X}$.
\end{Lemma}

\begin{proof}
This proof is straightforward from the proof of Lemma 1.
\end{proof}

If $(n,m) \in \Pi_{x',x}^A$, then it follows that
$$
\hat{T}_{x'}^{(n)} \left( g(x',e) \right) \leq  g(\pi^n(x'),e)
\leq  g(\pi^{-m}(x),e) \leq \check{T}_{x}^{(-m)} \left( g(x,e) \right).
$$
Similarly, if $(n,m) \in \Pi_{x',x}^B$, then it follows that
$$
\check{T}_{x'}^{(-n)} \left( g(x',e) \right) \leq  g(\pi^{-n}(x'),e)
\leq  g(\pi^{m}(x),e) \leq \hat{T}_{x}^{(m)} \left( g(x,e) \right).
$$
Define
$$
\left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}(u) \equiv \sup \{y: \hat{T}_{x'}^{(n)}(y) \leq u \} \wedge \overline{y},
$$
and thus, we have $\left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}\left( \hat{T}_{x'}^{(n)}(y) \right) = \sup \{y': \hat{T}_{x'}^{(n)}(y') \leq \hat{T}_{x'}^{(n)}(y) \} \wedge \overline{y} \geq y$ for all $y \in \overline{\mathcal{Y}}$.
Similarly, define
$$
\left( \check{T}_{x'}^{(-n)} \right)^{\rightarrow}(u) \equiv \sup \{y: \check{T}_{x'}^{(-n)}(y) \leq u \} \wedge \overline{y},
$$
then we have $\left( \check{T}_{x'}^{(-n)} \right)^{\rightarrow}\left( \check{T}_{x'}^{(-n)}(y) \right) \geq y$ for all $y \in \overline{\mathcal{Y}}$.
Hence, we obtain
\begin{eqnarray}
g(x',e) & \leq & \left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}\left( \check{T}_{x}^{(-m)}(g(x,e)) \right) \ \ \ \text{if } (n,m) \in \Pi_{x',x}^A, \nonumber \\
g(x',e) & \leq & \left( \check{T}_{x'}^{(-n)} \right)^{\rightarrow}\left( \hat{T}_{x}^{(m)}(g(x,e)) \right) \ \ \ \text{if } (n,m) \in \Pi_{x',x}^B. \nonumber
\end{eqnarray}
Define
\begin{eqnarray}
T_{x',x}^{AU}(y) & \equiv & \min_{(n,m) \in \Pi_{x',x}^A} \left( \hat{T}_{x'}^{(n)} \right)^{\rightarrow}\left( \check{T}_{x}^{(-m)}(y) \right) \ \ \text{and} \nonumber \\
T_{x',x}^{BU}(y) & \equiv & \min_{(n,m) \in \Pi_{x',x}^B} \left( \check{T}_{x'}^{(-n)} \right)^{\rightarrow}\left( \hat{T}_{x}^{(m)}(y) \right), \nonumber
\end{eqnarray}
then $T_{x',x}^{AU}(y)$ and $T_{x',x}^{BU}(y)$ satisfy $g(x',e) \leq T_{x',x}^{AU}\left( g(x,e) \right)$ and $g(x',e) \leq T_{x',x}^{BU}\left( g(x,e) \right)$.
Here, if $\Pi^B_{x',x}$ is empty, define $T_{x',x}^{BU}(y) \equiv \overline{y}$.
Similarly, define
\begin{eqnarray}
\left( \check{T}_{x'}^{(-m)} \right)^{\leftarrow}(u) & \equiv & \inf \{y: \check{T}_{x'}^{(-m)}(y) \geq u \} \vee \underline{y}) \ \ \text{and} \nonumber \\
\left( \hat{T}_{x'}^{(m)} \right)^{\leftarrow}(u) & \equiv & \inf \{y: \hat{T}_{x'}^{(m)}(y) \geq u \} \vee \underline{y}, \nonumber
\end{eqnarray}
then we have $\left( \check{T}_{x'}^{(-m)} \right)^{\leftarrow}\left( \check{T}_{x'}^{(-m)}(y) \right) \leq y$ and $\left( \hat{T}_{x'}^{(m)} \right)^{\leftarrow}\left( \hat{T}_{x'}^{(m)}(y) \right) \leq y$.
Define
\begin{eqnarray}
T_{x',x}^{AL}(y) & \equiv & \min_{(n,m) \in \Pi_{x,x'}^A} \left( \check{T}_{x'}^{(-m)} \right)^{\leftarrow}\left( \hat{T}_{x}^{(n)}(y) \right) \ \ \text{and} \nonumber \\
T_{x',x}^{BL}(y) & \equiv & \min_{(n,m) \in \Pi_{x,x'}^B} \left( \hat{T}_{x'}^{(m)} \right)^{\rightarrow}\left( \check{T}_{x}^{(-n)}(y) \right), \nonumber
\end{eqnarray}
then we have $g(x',e) \geq T_{x',x}^{AL}\left( g(x,e) \right)$ and $g(x',e) \geq T_{x',x}^{BL}\left( g(x,e) \right)$.
Here, if $\Pi^B_{x',x}$ is empty, define $T_{x',x}^{BL}(y) \equiv \underline{y}$.
Therefore, define $T_{x',x}^{GU}(y) \equiv T_{x',x}^{AU}(y)\wedge T_{x',x}^{BU}(y)$ and $T_{x',x}^{GL}(y) \equiv T_{x',x}^{AL}(y)\vee T_{x',x}^{BL}(y)$; then, we can obtain
\begin{eqnarray}
g(x',e) & \leq & T_{x',x}^{GU} \left( g(x,e) \right), \nonumber \\
g(x',e) & \geq & T_{x',x}^{GL} \left( g(x,e) \right). \label{TGU_and_TGL}
\end{eqnarray}

Define
\begin{eqnarray}
G_x^{GL}(u) & \equiv & \int F_{Y|X}^{+}\left( T_{x',x}^{GU}(u)|x' \right)dF(x'), \nonumber \\
G_x^{GU}(u) & \equiv & \int F_{Y|X}^{-}\left( T_{x',x}^{GL}(u)|x' \right)dF(x'), \nonumber \\
B^{GL}(x,e) & \equiv & \sup_{y:y\leq x} \left\{ \inf\{u:G_y^{GL}(u) \geq e\} \right\} \vee \underline{y}, \nonumber \\
B^{GU}(x,e) & \equiv & \inf_{y:y\geq x} \left\{ \sup\{u:G_y^{GU}(u) \leq e\} \right\} \wedge \overline{y}, \nonumber
\end{eqnarray}
where $F_{Y|X}^{+}(y|x) \equiv P(Y\leq y| X=x)$ and $F_{Y|X}^{-}(y|x) \equiv P(Y< y| X=x)$.
$G_x^{GL}(u)$ and $G_x^{GU}(u)$ provide the lower and upper bounds of $g(x,e)$ on the basis of an argument similar to (\ref{approach_DF1}) and (\ref{approach_DF2}).
$B^{GL}(x,e)$ and $B^{GU}(x,e)$ strengthen these bounds.

\begin{Theorem}
Under assumptions 1, 2', 3, 4', 5, and M, for all $(x,e) \in \mathcal{X} \times (0,1)$, we have
$$
B^{GL}(x,e) \leq g(x,e) \leq B^{GU}(x,e).
$$
\end{Theorem}

\begin{proof}
First, we show that
\begin{equation}
\inf\{u:G_x^{GL}(u) \geq e\} \vee \underline{y} \leq g(x,e) \leq \sup\{u:G_x^{GU}(u) \leq e\} \wedge \overline{y}. \label{g_bound_G}
\end{equation}
Because $T_{x',x}^{GU}(y)$ satisfies (\ref{TGU_and_TGL}), we have
\begin{eqnarray}
G_x^{GL}\left( g(x,e) \right) &=& \int F_{Y|X=x'}^{+}\left( T_{x',x}^{GU}(g(x,e)) \right)dF(x') \nonumber \\
&\geq & \int F_{Y|X=x'}^{+}\left( g(x',e) \right)dF(x') \nonumber \\
&=& \int P\left( g(x',\epsilon) \leq g(x',e)|X=x' \right)dF(x') \nonumber \\
&\geq & \int P\left( \epsilon \leq e|X=x' \right)dF(x') = e, \nonumber
\end{eqnarray}
where the second inequality follows from $\{\epsilon:\epsilon \leq e\} \subset \{\epsilon: g(x',\epsilon) \leq g(x',e)\}$.
Because $g(x,e) \geq \underline{y}$, we can obtain $g(x,e) \geq \inf\{u:G_x^{GL}(u) \geq e\} \vee \underline{y}$.
Similarly, we have
\begin{eqnarray}
G_x^{GU}\left( g(x,e) \right) &\leq & \int F_{Y|X=x'}^{-}\left( T_{x',x}^{GL}(g(x,e)) \right)dF(x') \nonumber \\
&=& \int P\left( g(x',\epsilon) < g(x',e)|X=x' \right)dF(x') \nonumber \\
&\leq & \int P\left( \epsilon < e|X=x' \right)dF(x') = e, \nonumber
\end{eqnarray}
where the second inequality follows from $\{\epsilon: g(x',\epsilon) < g(x',e)\} \subset \{\epsilon:\epsilon < e\}$.
Hence, we can obtain $g(x,e) \geq \sup\{u:G_x^{GU}(u) \leq e\} \wedge \overline{y}$.

Because $g(x,e)$ is nondecreasing in $x$ and (\ref{g_bound_G}) holds, similar to Theorem 1, we can obtain
$$
B^{GL}(x,e) \leq g(x,e) \leq B^{GU}(x,e).
$$
\end{proof}
\fi

\section{Conclusions}

In this study, we consider the partial identification of nonseparable models using binary instruments.
We show that partial identification can be achieved when $g(x,e)$ is monotone or concave in $x$, even if $X$ is continuous and $Z$ is binary.
\cite{d2015identification} and \cite{torgovitsky2015identification} show that $g$ is point identified without monotonicity and concavity.
They use two key assumptions to establish the point identification of $g$.
First, $F_{X|Z}(x|0)$ and $F_{X|Z}(x|1)$ have intersections and second, $g(x,e)$ is strictly increasing in a scalar unobservable.
There are, however, many empirically important models that do not satisfy these assumptions. Thus, we provide bounds for structural functions without the use of these assumptions.

\clearpage

\section*{Appendix: Figures}

\begin{figure}
\centering
\includegraphics[width=15cm]{figure1.pdf}
\caption{The left function is $F_{X|Z}(x|1)$ and the right function is $F_{X|Z}(x|0)$.}
\end{figure}

\begin{figure}
  \centering
  \includegraphics[width=15cm]{figure2.pdf}
  \caption{The left function is $F_{X|Z}(x|1)$ and the right function is $F_{X|Z}(x|0)$.}
\end{figure}

\begin{figure}
  \centering
  \includegraphics[width=16cm]{figure3.pdf}
  \caption{The dashed lines denote $\left( G_{x}^{ML} \right)^{-1}(e)$ and $\left( G_{x}^{MU} \right)^{-1}(e)$. The solid lines represent $B^{ML}(x,e)$ and $B^{MU}(x,e)$.}
\end{figure}

\begin{figure}
  \centering
  \includegraphics[width=10cm]{figure4.pdf}
  \caption{The dashed line denotes $g(x,0.5)$. The solid lines are the bounds of $g(x,0.5)$, $B^{ML}(x,0.5)$, and $B^{MU}(x,0.5)$. $\alpha = 0.5, \beta = 0.5, c = 0.5, d= 0.05, and \rho = 0.3$.}
\end{figure}

\begin{figure}
  \centering
  \includegraphics[width=16cm]{figure5.pdf}
  \caption{The dashed lines denote $\left( G_{x}^{CL} \right)^{-1}(e)$ and $\left( G_{x}^{CU} \right)^{-1}(e)$. The solid lines denote $B^{CL}(x,e)$ and $B^{CU}(x,e)$.}
\end{figure}

\clearpage

\bibliographystyle{ecta}

redpajama_set_name: RedPajamaArXiv


Monday was a tense day at one of our Central Texas High Schools, as Groesbeck police stayed on campus through a lockdown of the campus.
KXXV.com reports that a threat came into 9-1-1 around 2 p.m. made by an unknown individual who said they intended to 'shoot up' the school. The caller then hung up the phone.
Police have looked into the number the call came from and found what appears to be a deactivated cell phone that was still able to access its 911 services. While the caller didn't specifically name Groesbeck' High School, the call was received on a Groesbeck cell tower. This was enough to put police into action.
As a precaution, the police notified GISD and locked down the school. After some investigating, the lockout was lifted after police felt the threat was no longer credible.


The Forefront Server Security Administrator user interface contains the Shuttle Navigator on the left and the work panes on the right.
There are four shuttles in the Shuttle Navigator: SETTINGS, FILTERING, OPERATE, and REPORT. The shuttles are discussed in the "Microsoft Forefront Security for SharePoint User Guide."


text:
Share this Story: Dellacamera, Wagner Fox's lead team for Women's World Cup
Dellacamera, Wagner Fox's lead team for Women's World Cup
FILE - In this May 30, 2018, file photo, Aly Wagner talks during an interview in New York. JP Dellacamera and former U.S. midfielder Aly Wagner will be Fox's lead broadcast team for the Women's World Cup in France. The announcement comes 100 days before the tournament opens June 7 with a matchup of France and South Korea. Fox said Wednesday, Feb. 27, 2019, that 22 matches will be broadcast on the main Fox network, 27 on FS1 and three on FS2. Photo by Mark Lennihan /AP
LOS ANGELES — JP Dellacamera and former U.S. midfielder Aly Wagner will be Fox's lead broadcast team for the Women's World Cup.
Making its announcement 100 days before the tournament opens June 7 with a matchup of France and South Korea, Fox said Wednesday that 22 matches will be broadcast on the main Fox network, 27 on FS1 and three on FS2.
Dellacamera, Wagner Fox's lead team for Women's World Cup Back to video
Fox's Paris studio will be in the Palais Chaillot at the Cafe de l'Homme at the Trocadero, with the Eiffel Tower as a backdrop.
The U.S. is defending champion at the Women's World Cup and will play Thailand on June 11 at 3 p.m. EDT, Chile five days later at noon and Sweden on June 20 at 3 p.m.
This will be the 15th World Cup on television and radio for the 67-year-old Dellacamera, including five women's tournaments. It's the third for the 38-year Wagner, including her second women's tournament.
A two-time Olympic gold medallist , Wagner became the first woman to broadcast an English-language men's World Cup last year, when she was paired with Derek Rae. They called 10 group stage matches from Fox's Los Angeles studio before travelling to Russia for a pair of round of 16 games and the England-Sweden quarterfinal.
Dellacamera was paired with former U.S. goalkeeper Tony Meola in Russia, broadcasting nine games during the group stage, two in the round of 16, the Uruguay-France quarterfinal and the Belgium-England third-place match.
Fox is planning more than 800 hours of broadcast and digital coverage, including two daily studio shows in Paris and a late-night show from Los Angeles that will cover both the women's tournament and the CONCACAF Gold Cup, the regional men's championship of North and Central America and the Caribbean that opens June 15.
Fox also is televising the Gold Cup and will have a day-night doubleheader on July 7, with the women's final from Lyon at 11 a.m. EDT and the CONCACAF final from Chicago later in the day. John Strong and Stu Holden, Fox's lead broadcast team at last year's World Cup, is likely to be the network's primary team for the Gold Cup.
The final of the Copa America also is July 7, in Rio de Janeiro at 4 p.m. EDT. U.S. English-language TV rights for that tournament have not been announced. NBC sister network Telemundo has U.S. Spanish-language TV rights to the Women's World Cup and the Copa America, and Univision has U.S. Spanish-language TV rights to the Gold Cup.
More AP soccer: https://apnews.com/apf-Soccer and https://twitter.com/AP–Sports

redpajama_set_name: RedPajamaCommonCrawl


99 Ways to Get Chopped from "Chopped": A Handy Tips-'n'-Tricks Checklist « Midlife Crisis Crossover!
Who among us has never looked at a bag of Cheetos and thought, "I bet I could turn this into haute cuisine"?
For years my wife Anne and I have been addicted to the Food Network's cooking-competition series Chopped, in which four chefs must outcook each other under strictly timed conditions using four specific ingredients. Inside every Chopped basket of goods lurks a surprising combination of the rare, the delicate, the expensive, the complicated, the whimsical, the outlandish, and/or the thoroughly disgusting. Every substance can be used, though not every substance is very good.
Food Network continues gifting us with new episodes every week hosted by the amazing colossal Ted Allen, who presides over this fast-paced showcase for chefs of every conceivable demographic from various American restaurants, caterers, bakeries, or other private businesses, each of whom keeps their eyes on the $10,000 prize to be had if they're the last entrant remaining after three grueling courses of speed-heating, kitchen-racing, and power-serving.
After watching several dozen episodes, Anne and I began to notice recurring patterns and tried to capture those observations and our fandom back in 2014 with a previous MCC entry called "How Not to Get Chopped from 'Chopped': A Starter Guide". I've been meaning to overhaul that entry for a while now that we're four years and literally 200+ episodes later, which includes every episode of the kids-only spinoff Chopped Junior and a handful of episodes of Chopped Canada, which was an interesting effort with its own angle and demeanor but wasn't quite the same thing. I'm ashamed to confess it was tough to watch for more than a few minutes before I started poking fun in a goofy faux-Fargo accent.
The following compilation is our revised armchair-chef advice for future would-be competitors on how not to do Chopped from where we sit. This list is doubtlessly far from complete, and we welcome any additions in the comments below, especially from those among you who can truly cook. Though neither of us is a trained gourmet by any stretch, we hope this helps someone out there anyway. If you raise a skeptical eyebrow at any of these, well…it's positively flabbergasting how many of these downfalls we've seen happen in actual episodes at the hands of trained professionals who run fantastic eateries back home but who lose their poise in front of the cameras. Even the best can make mistakes or watch their plans spin out of control.
Unnecessary disclaimer: exceptions can happen, but are extremely rare. We've seen at least two chefs with lousy attitudes win. We've seen rare instances in which hacking up an ingredient without any further treatment was secretly the best possible option. We once — exactly once — saw a single chef use white truffle oil without being chopped. Once.
I love this. We're big chopped fans. I have to show this to my husband. Thanks.


•The Twin Valley Community Local School Board of Education Regular Board meetings will be held in the Twin Valley South Media Center at 6 p.m. November/December meeting will be held on Monday, Dec. 3, at 6 p.m.
•The Preble County Chamber of Commerce Board of Directors meets at 7:30 a.m. the fourth Tuesday of each month at 122 W. Decatur St., Eaton.
•The Blue Star Mothers of Ohio, Preble County Chapter meetings will be the last Wednesday of each month at 6:30 p.m. at the L&M Products, next to McDonalds on Barron Street in Eaton.
•The Preble Soil and Water Conservation District Board of Supervisors conducts monthly board meetings at 7 a.m. the third Wednesday of each month (except as noted with an asterisk) at the Preble SWCD Office, 1651 N. Barron St., Eaton. Remaining meetings for 2018 will be: Nov. 21, Dec. 12*. For additional information contact the district office at 937-456-5159.
Commodore Preble Chapter of the DAR meets monthly. Call Debra Crumbaker at 937-452-5531 for more information.


Some history to AD's and how they got that way.
SiteProNews :: Webmaster News and Advice :: The Clash Of Titans: Article Directory Owners Vs. Automated Article Submission Services: "The Clash Of Titans Things really began to change in the AD directory ownership game a few months into the project. Directory owners began to realize that in order for them to profit from their article directories, they must do something that brings readers to their websites and encourages loyalty from their site's visitors. Since most article directories rely on advertising to drive their revenue stream, the directory owners had to do something that the other directory owners were not doing. They had to distinguish themselves from the masses. The AD owners who have survived the early explosion of AD sites have generally taken the attitude that they should focus on "quality over quantity". This one step has set the goals of the software developers' article distribution systems and their customers (quantity over quality) in conflict with the best interests of the article directory owners (quality over quantity)."


23 Fabulous Vintage Teen Girls Bedroom Ideas With Vintage Room Ideas Renovation | Jenniferumphress.com vintage hipster room ideas. vintage bedroom ideas. vintage dining room lighting ideas.
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1. Gently fry the onion in olive oil until translucent. Add the red pepper and soften for a couple of minutes.
2. Add the chopped tomatoes, place the stock cube in the tomato tin and fill to the top with boiling water, add the tomato puree, chilli flakes and chilli powder to the tin and stir well.
3. Pour the stock onto the onions, add the chickpeas and kidney beans.
4. Using one of the empty tins, half fill with broad beans and add to the pan, repeat with the sweetcorn.
5. Give it a good stir and simmer for 30 mins.
Serve with rice, homemade tortilla chips, jacket potato, potato wedges etc. Oh yes and plenty of cheese and sour cream!
If you don't overload with cheese and sour cream and rice (like I do!) it's pretty healthy and 'clean' too!


Use Fiddes Teak Oil on Teak, Cedar and other bare woods to nourish.
Also available: Fiddes Teak Oil Dark.
Based on special vegetable oils and solvents to give a quick drying, clear penetrating seal. Fiddes Teak Oil gives an attractive sheen on interior or exterior surfaces.
Dark Teak Oil has a light brown tint and is suitable for exterior use.
Clear Teak Oil is suitable for interior or exterior use.
Easily applied with a brush or cloth.
Please see our Oils Spirits and Solvents section for alternative products and our Fiddes section for the full product range.
Ive got one very happy customer so enough said.


text:
Kildalton Discussion Group
The Kildalton Discussion Group, in operation for over 20 years has 18 dairy farmer members. They're mostly from Kilkenny, Tipperary, Waterford and Wexford. Philip Donohoe, Chairman, discusses the group performance and areas for future improvement as part of the Virtual Dairy Conference today
Group members have almost 5,000 dairy cows and run spring-calving, relatively low input systems of grass based milk production at a farm stocking rate of 2.5 livestock units per hectare.
Genetic trends
The current EBI of the dairy stock on the 'average' member's farm is presented in Figure 1
Figure 1. Average EBI of the dairy cows, first calved heifers, breeding heifers and heifer calves on Kildalton discussion group members' farms
Our herd EBI includes a milk sub index of €51 and a fertility sub index of €86, and is €47 higher than the national average. With higher EBI young stock we're going to keep increasing the genetic merit of our herds in the years ahead. To get an idea of the genetic progress being made in Figure 2, we compare the EBI of heifers entering the herd over the past decade.
Figure 2. Average EBI of the first calved heifers entering the Kildalton Discussion Group herds since 2009
The graph in Figure 2 shows that the EBI of every generation of heifers entering the herd since 2009 is on average €10 greater than the previous one and indicates that group members have been using high EBI bulls for more than a decade. Such improvement in EBI over the years is great, but the trends in milk production and fertility are the real payback for the years of genetic improvement.
Milk production and fertility trends
The data in Table 1 shows the change we have seen over the past decade in terms of milk production, milk composition and herd fertility performance.
Table 1. Change in average milk solids production, milk composition and fertility performance of the Kildalton group members' farms between 2010 and 2020 and its economic value
[1] Fertility metrics based on the 2012-2021 period.
The data in Table 1 shows that using a value of €4 per kg milk solids (the equivalent of approximately 30 c/litre), the increase in milk solids yield per cow of 115 kg is valued at €460 per cow in 2020. Certainly part of the increase in milk solids yield per cow can be attributed to milk quota removal, but some of it is due to improvements in genetics for milk production, milk composition and fertility. Milk fat and protein content has increased by a total of 0.75%
over the decade which represents an increase in the value of every litre of milk sold of at least 3.6 c/litre at a base price of 30 c/litre. We have also seen substantial improvements in herd fertility. The 16% improvement in 6-week calving rate and the 11-day reduction in calving interval are valued at €132 and €42 per cow, respectively. This fertility improvement is an important contributor to the increase in milk yield. When we reviewed the trends in meal feeding by group members (see Table 2), there has not been an increase in the quantity of meal fed per cow. Instead, we have seen a substantial increase in grass utilisation as we have focused a lot on increasing soil fertility, grazing management and reseeding over the past decade. Grassland management improvements and genetics are responsible for the increase in milk solids yield per cow.
Table 2. Trends in milk yield, purchased feeds and forage, farm stocking rate (LU/ha) and grass used (t DM/ha) for various years between 2010 and 2020 for the Kildalton group members' farms
The group identifies three areas for improvement in the years ahead. Firstly with the environment issue facing us all, breeding for greater milk production and fertility have taken on a new relevance. We need to breed cows that are both productive and retain the ability to quickly and easily go back in calf to minimise replacement rate and the number of replacement heifers we have to rear.
Secondly, with greater awareness of and a need to reduce antibiotic use, genetic improvement of cow health has become increasingly important. So we all need to step up and record incidences of lameness and mastitis as they occur within our own herds.
Thirdly, a big challenge for specialised spring calving herds is the ability to sell calves quickly and easily after they are born. Reducing the use of dairy AI by using more sexed semen is an approach that an increasing number of the group members will adopt next breeding season. We particularly welcome the opening of the lab in Moorepark for 2022. Finally, we all need to adopt the DBI to breed quality beef breed calves from the dairy herd.
Pictured in Moorepark, Fermoy, Co Cork at the Teagasc Virtual Dairy Conference 2021 - 'Meeting Challenges with Technology' are speakers Dr Donagh Berry, Director VistaMilk, Dr Nicky Byrne, Teagasc Research Officer, Dr Emma-Louise Coffey, Teagasc Co-ordinator - Professional Diploma in Dairy Farm Management, dairy farmer Jim White, Mullinahone, Co Tipperary, Dr Stephen Butler, Teagasc Reproductive Physiologist & Liam Herlihy, Teagasc Chairman. Photo O'Gorman Photography.
Due to updated Government guidelines the Teagasc National Dairy Conference has changed from an 'in person' event to being an online 'virtual event' only.
The Dairy Conference 2021 is on Zoom over two days. Registration link below
The complete Dairy Conference Proceedings will be published on www.teagasc.ie And the individual articles will be published on Teagasc Daily throughput the week. For more information see Dairy Conference


text:
Q: Android - Play a sound on achievement unlocked I have an Android game and I'd like to play a sound whenever a players unlocks an achievement,
but I can't seem to find a way to know when the user unlocks the achievement, since most of my achievements are incremental I have no way to code it on the client side, the only way would be some callback called when the Google API shows the achievement box, but I can't find any of those on the documentation... Only thing i found was this: Integrating Google Play services achievements and android notifications
@Override
public void onAchievementUnlocked(final String id)
{
    //
}

but I have no idea how to implement that or how to call that method and I had no luck on Google

A: Use the following code:
Games.Achievements.incrementImmediate(GoogleApiClient apiClient, String id, int numSteps).setResultCallback(new  ResultCallback<Achievements.UpdateAchievementResult>() {

            @Override
            public void onResult(UpdateAchievementResult result) {
                if (result.getStatus().getStatusCode() == GamesStatusCodes.STATUS_ACHIEVEMENT_UNLOCKED) {
                    // play your sound. achievement was unlocked
                }
            }

        });

This will increment your achievement by the amount of steps you want, and when it is done, it will return some of these codes. If the code is GamesStatusCodes.STATUS_ACHIEVEMENT_UNLOCKED, then it means that increment unlocked the achievement and you can play your sound.

redpajama_set_name: RedPajamaStackExchange


text:
Dovregrynsnäcka (Vertigo parcedentata) är en snäckart som först beskrevs av Alexander Karl Heinrich Braun 1847.  Dovregrynsnäcka ingår i släktet Vertigo, och familjen grynsnäckor. IUCN kategoriserar arten globalt som sårbar. Arten har ej påträffats i Sverige.

Bildgalleri

Källor

Externa länkar

Grynsnäckor

redpajama_set_name: RedPajamaWikipedia


text:
//
//  YLStockNoticesModel.h
//  YLSimulatedStockOperation
//
//  Created by Waterstrong on 2/11/16.
//  Copyright © 2016 hoggenWang. All rights reserved.
//

#import <Foundation/Foundation.h>

@interface YLStockNoticesModel : NSObject

/**
 *  新闻公告模型
 */

@property (nonatomic, copy) NSString *title;

@property (nonatomic, copy) NSString *url;

@property (nonatomic, copy) NSString *date;

@end

redpajama_set_name: RedPajamaGithub


Is Software Aktiengesellschaft (ETR:SOW) A Financially Strong Company?
Small and large cap stocks are widely popular for a variety of reasons, however, mid-cap companies such as Software Aktiengesellschaft (ETR:SOW), with a market cap of €2.4b, often get neglected by retail investors. Despite this, the two other categories have lagged behind the risk-adjusted returns of commonly ignored mid-cap stocks. Let's take a look at SOW's debt concentration and assess their financial liquidity to get an idea of their ability to fund strategic acquisitions and grow through cyclical pressures. Note that this commentary is very high-level and solely focused on financial health, so I suggest you dig deeper yourself into SOW here.
How much cash does SOW generate through its operations?
SOW has sustained its debt level by about €315m over the last 12 months including long-term debt. At this current level of debt, SOW's cash and short-term investments stands at €413m for investing into the business. On top of this, SOW has produced €177m in operating cash flow over the same time period, resulting in an operating cash to total debt ratio of 56%, meaning that SOW's current level of operating cash is high enough to cover debt. This ratio can also be a sign of operational efficiency as an alternative to return on assets. In SOW's case, it is able to generate 0.56x cash from its debt capital.
Does SOW's liquid assets cover its short-term commitments?
Looking at SOW's €457m in current liabilities, the company has been able to meet these obligations given the level of current assets of €661m, with a current ratio of 1.45x. For Software companies, this ratio is within a sensible range as there's enough of a cash buffer without holding too much capital in low return investments.
Does SOW face the risk of succumbing to its debt-load?
SOW's level of debt is appropriate relative to its total equity, at 27%. SOW is not taking on too much debt commitment, which can be restrictive and risky for equity-holders.
Future Outlook: What are well-informed industry analysts predicting for SOW's future growth? Take a look at our free research report of analyst consensus for SOW's outlook.
How Does Vedanta Limited (NSE:VEDL) Fare As A Dividend Stock?
Is Vikas EcoTech Limited's (NSE:VIKASECO) ROE Of 15% Impressive?


text:
Q: Type hint for two classes that have themselves as element from two diferent files I have two classes from two different files.
I want to use type hints to make it easier to use.
from b import ClassB

# a.py
class ClassA():
    b: ClassB
    def __init__(self):
        self.b = None

from a import ClassA

# b.py
class ClassB():
    a: ClassA
    def __init__(self):
        self.a = None

But I run into a circular dependency problem
this doesn't work either because ClassA is not defined in this file
# b.py
class ClassB():
    a: "ClassA"
    def __init__(self):
        self.a = None