<Instrumental-variable monotonicity> is
$$
A(1)\geq A(0)\quad\text{for every subject}.
$$
It excludes defiers who would quit without the incentive but continue smoking when offered it. The possible <principal strata> are then never-takers $(0,0)$, compliers $(0,1)$, and always-takers $(1,1)$.

By random assignment, consistency, and exclusion,
$$
\begin{aligned}
\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]
&=\mathbb E\{Y(A(1))-Y(A(0))\}\\
&=\mathbb E[(Y(1)-Y(0))(A(1)-A(0))].
\end{aligned}
$$
The second identity follows by checking the two possible binary exposure values. Under monotonicity, $A(1)-A(0)$ is the indicator of being a complier. Therefore the numerator is
$$
\mathbb P(\text{complier})
\mathbb E[Y(1)-Y(0)\mid\text{complier}],
$$
while
$$
\mathbb E[A\mid Z=1]-\mathbb E[A\mid Z=0]
=\mathbb E[A(1)-A(0)]
=\mathbb P(\text{complier}).
$$
Their ratio is the <complier average treatment effect>, also called the <local average treatment effect>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 221 / 4 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-221/4/ii/solution