The <no unmeasured confounding assumption>, or <conditional exchangeability>, is
$$
(Y(0),Y(1))\mathrel\perp A\mid X.
$$
Also assume <consistency of potential outcomes>, no <interference in causal inference>, <positivity in causal inference>, and finite expectations. For $a\in\{0,1\}$, the <law of total expectation>, exchangeability, and consistency give
$$
\begin{aligned}
\mathbb E[Y(a)]
&=\mathbb E\{\mathbb E[Y(a)\mid X]\}\\
&=\mathbb E\{\mathbb E[Y(a)\mid A=a,X]\}\\
&=\mathbb E\{\mathbb E[Y\mid A=a,X]\}.
\end{aligned}
$$
Positivity ensures that the observed conditional means exist on the covariate support being averaged. Subtracting the two cases identifies the <average treatment effect> as
$$
\beta_1
=\mathbb E\{\mathbb E[Y\mid A=1,X]\}
-\mathbb E\{\mathbb E[Y\mid A=0,X]\}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 221 / 3 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-221/3/i/solution