Write $m_z(X)=\mu_z(X,\beta_z)$ and $\widetilde e(X)=e(X;\alpha)$. Conditional on $X$,
$$
\widetilde\mu_1^{\mathrm{dr}}(X)
=m_1(X)+\frac{e(X)}{\widetilde e(X)}\bigl(\mu_1(X)-m_1(X)\bigr),
$$
with the analogous control expression
$$
\widetilde\mu_0^{\mathrm{dr}}(X)
=m_0(X)+\frac{1-e(X)}{1-\widetilde e(X)}\bigl(\mu_0(X)-m_0(X)\bigr).
$$
If the propensity model is correct, both ratios are one and these equal $\mu_1(X)$ and $\mu_0(X)$, so their difference is the <conditional average treatment effect>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 221 / 4 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-221/4/b/i/solution