Differentiating the residualized objective gives
$$
\beta_4
=\frac{\mathbb E[\widetilde A\widetilde Y]}
{\mathbb E[\widetilde A^2]}
=\frac{\mathbb E[(A-e(X))(Y-\mu(X))]}
{\mathbb E[(A-e(X))^2]}
=\beta_3.
$$
This is the population <Frisch–Waugh–Lovell theorem>.

For a <semiparametric estimator>, estimate $e(x)=\mathbb E[A\mid X=x]$ and $\mu(x)=\mathbb E[Y\mid X=x]$ flexibly. With <cross-fitting>, obtain held-out predictions $\widehat e_i,\widehat\mu_i$ and regress the residualized outcome on the residualized treatment through the origin:
$$
\widehat\beta_3
=\frac{\sum_{i=1}^n(A_i-\widehat e_i)(Y_i-\widehat\mu_i)}
{\sum_{i=1}^n(A_i-\widehat e_i)^2}.
$$
Cross-fitting limits overfitting bias and permits flexible nuisance estimators under the usual convergence and overlap conditions.

Solution

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

 Solution

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