[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*}[/step]

[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: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]