[proofplan]
The proof rewrites both the treatment contrast and the outcome contrast in terms of potential outcomes. Treatment consistency identifies the denominator with $\mathbb E[A(1)-A(0)]$, and monotonicity turns this into the probability of the complier stratum $\{A(1)>A(0)\}$. Outcome consistency and exclusion identify the numerator with $\mathbb E[Y(A(1))-Y(A(0))]$. A principal-strata decomposition then shows that only compliers contribute to this difference, so dividing by the complier probability gives the local average treatment effect.
[/proofplan]
[step:Define the instrument-indexed treatment and outcome potential variables]
For $z\in\{0,1\}$, define the binary [random variable](/page/Random%20Variable)
\begin{align*}
A_z:\Omega\to\{0,1\}
\end{align*}
by
\begin{align*}
A_z=zA(1)+(1-z)A(0).
\end{align*}
Thus $A_1=A(1)$ and $A_0=A(0)$.
For $z\in\{0,1\}$, define the real-valued random variable
\begin{align*}
Y_z:\Omega\to\mathbb R
\end{align*}
by
\begin{align*}
Y_z=A_zY(1)+(1-A_z)Y(0).
\end{align*}
Treatment consistency gives $A=A_Z$ almost surely, and outcome consistency with exclusion gives $Y=Y_Z$ almost surely. In particular, on the event $\{Z=z\}$ we have
\begin{align*}
A=A_z
\end{align*}
almost surely on $\{Z=z\}$, and
\begin{align*}
Y=Y_z
\end{align*}
almost surely on $\{Z=z\}$.
[/step]
[step:Rewrite the treatment contrast as the complier probability]
Since $A_0$ and $A_1$ are bounded random variables and each is a [measurable function](/page/Measurable%20Function) of $(Y(1),Y(0),A(1),A(0))$, instrument independence implies $Z\perp A_0$ and $Z\perp A_1$. Therefore, for $z\in\{0,1\}$,
\begin{align*}
\mathbb E[A\mid Z=z]=\mathbb E[A_z\mid Z=z]=\mathbb E[A_z].
\end{align*}
Hence
\begin{align*}
\mathbb E[A\mid Z=1]-\mathbb E[A\mid Z=0]=\mathbb E[A_1]-\mathbb E[A_0]=\mathbb E[A(1)-A(0)].
\end{align*}
By monotonicity, $A(1)-A(0)$ is a $\{0,1\}$-valued random variable, and
\begin{align*}
A(1)-A(0)=\mathbb{1}_{\{A(1)>A(0)\}}
\end{align*}
almost surely. Therefore
\begin{align*}
\mathbb E[A\mid Z=1]-\mathbb E[A\mid Z=0]=\mathbb P(A(1)>A(0)).
\end{align*}
The relevance assumption says that the left-hand side is nonzero, while monotonicity makes it nonnegative. Thus
\begin{align*}
\mathbb P(A(1)>A(0))>0.
\end{align*}
[guided]
The denominator in the Wald ratio measures how much the instrument changes the observed probability of treatment. Treatment consistency lets us replace the observed treatment under $Z=z$ by the potential treatment $A_z$.
For $z\in\{0,1\}$, the equality $A=A_z$ holds almost surely on $\{Z=z\}$. Hence
\begin{align*}
\mathbb E[A\mid Z=z]=\mathbb E[A_z\mid Z=z].
\end{align*}
The random variable $A_z$ is bounded and is a measurable function of $(Y(1),Y(0),A(1),A(0))$. Since $Z$ is independent of that whole potential-outcome vector, $Z$ is independent of $A_z$. Therefore conditioning on $Z=z$ does not change its expectation:
\begin{align*}
\mathbb E[A_z\mid Z=z]=\mathbb E[A_z].
\end{align*}
Combining these equalities for $z=1$ and $z=0$ gives
\begin{align*}
\mathbb E[A\mid Z=1]-\mathbb E[A\mid Z=0]=\mathbb E[A_1]-\mathbb E[A_0].
\end{align*}
Because $A_1=A(1)$ and $A_0=A(0)$, this is
\begin{align*}
\mathbb E[A\mid Z=1]-\mathbb E[A\mid Z=0]=\mathbb E[A(1)-A(0)].
\end{align*}
Now monotonicity is the step that removes defiers. Since $A(1)$ and $A(0)$ are binary and $A(1)\ge A(0)$ almost surely, the difference $A(1)-A(0)$ can only be $0$ or $1$. It equals $1$ exactly on the complier event $\{A(1)>A(0)\}$ and equals $0$ otherwise. Thus
\begin{align*}
A(1)-A(0)=\mathbb{1}_{\{A(1)>A(0)\}}
\end{align*}
almost surely, so
\begin{align*}
\mathbb E[A(1)-A(0)]=\mathbb P(A(1)>A(0)).
\end{align*}
Relevance states that the original treatment contrast is not zero. Since the contrast has just been identified with a probability, that probability must be strictly positive.
[/guided]
[/step]
[step:Rewrite the outcome contrast using instrument independence]
For $z\in\{0,1\}$, the random variable $Y_z$ is a measurable function of $(Y(1),Y(0),A(1),A(0))$. Hence instrument independence gives $Z\perp Y_z$ whenever the relevant expectation is finite.
We first verify integrability. Since $Y=Y_z$ almost surely on $\{Z=z\}$ and $\mathbb P(Z=z)>0$,
\begin{align*}
\mathbb E[|Y_z|\mid Z=z]=\mathbb E[|Y|\mid Z=z]<\infty.
\end{align*}
Because $Z\perp Y_z$, this implies $\mathbb E[|Y_z|]<\infty$. Therefore
\begin{align*}
\mathbb E[Y\mid Z=z]=\mathbb E[Y_z\mid Z=z]=\mathbb E[Y_z].
\end{align*}
Taking the difference for $z=1$ and $z=0$ yields
\begin{align*}
\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]=\mathbb E[Y_1-Y_0].
\end{align*}
By the definitions of $Y_1$ and $Y_0$,
\begin{align*}
Y_1=A(1)Y(1)+(1-A(1))Y(0)
\end{align*}
and
\begin{align*}
Y_0=A(0)Y(1)+(1-A(0))Y(0).
\end{align*}
Subtracting gives
\begin{align*}
Y_1-Y_0=(A(1)-A(0))(Y(1)-Y(0)).
\end{align*}
[/step]
[step:Reduce the outcome contrast to the complier stratum]
By monotonicity,
\begin{align*}
A(1)-A(0)=\mathbb{1}_{\{A(1)>A(0)\}}
\end{align*}
almost surely. Combining this with the preceding step gives
\begin{align*}
Y_1-Y_0=(Y(1)-Y(0))\mathbb{1}_{\{A(1)>A(0)\}}
\end{align*}
almost surely. Since $Y_1$ and $Y_0$ are integrable, this also proves that $(Y(1)-Y(0))\mathbb{1}_{\{A(1)>A(0)\}}$ is integrable. Hence
\begin{align*}
\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]
=\mathbb E[(Y(1)-Y(0))\mathbb{1}_{\{A(1)>A(0)\}}].
\end{align*}
[/step]
[step:Divide by the complier probability to obtain the local average treatment effect]
Let
\begin{align*}
C=\{A(1)>A(0)\}
\end{align*}
be the complier event. The previous steps show
\begin{align*}
\mathbb E[A\mid Z=1]-\mathbb E[A\mid Z=0]=\mathbb P(C)
\end{align*}
with $\mathbb P(C)>0$, and
\begin{align*}
\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]=\mathbb E[(Y(1)-Y(0))\mathbb{1}_C].
\end{align*}
By the definition of [conditional expectation](/page/Conditional%20Expectation) given an event of positive probability,
\begin{align*}
\mathbb E[Y(1)-Y(0)\mid C]=\frac{\mathbb E[(Y(1)-Y(0))\mathbb{1}_C]}{\mathbb P(C)}.
\end{align*}
Substituting the two identified numerator and denominator expressions gives
\begin{align*}
\frac{\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]}{\mathbb E[A\mid Z=1]-\mathbb E[A\mid Z=0]}
=\mathbb E[Y(1)-Y(0)\mid A(1)>A(0)].
\end{align*}
This is the claimed local average treatment effect identity.
[/step]
