The <no unmeasured confounding assumption> is the conditional exchangeability statement
$$
(Y(0),Y(1))\perp A\mid X.
$$
Also assume <consistency of potential outcomes>, no interference between units, and <positivity in causal inference>, in particular $\mathbb P(A=0\mid X)>0$ on the covariate support of treated units. Then
$$
\begin{aligned}
\mathbb E[Y(0)\mid A=1]
&=\mathbb E\{\mathbb E[Y(0)\mid X,A=1]\mid A=1\}\\
&=\mathbb E\{\mathbb E[Y(0)\mid X,A=0]\mid A=1\}\\
&=\mathbb E\{\mu(X)\mid A=1\}.
\end{aligned}
$$
The first treated potential outcome equals the observed treated mean by consistency. Moreover,
$$
\mathbb E\{\mu(X)\mid A=1\}
=\frac{\mathbb E[A\mu(X)]}{\mathbb P(A=1)}
=\frac{\mathbb E[\pi(X)\mu(X)]}{\mathbb E[\pi(X)]}.
$$
Subtracting proves the displayed identification formula for the <average treatment effect on the treated>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 221 / 2 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-221/2/i/solution