For discrete $X$,
$$
\beta=\sum_xp(x)\pi(x)\mu(x).
$$
Perturbing the marginal mass $p(x)$ contributes $\pi(X)\mu(X)-\beta$. Perturbing $\pi(x)=\mathbb E[A\mid X=x]$ contributes
$$
\mu(X)\{A-\pi(X)\}.
$$
Their sum is $A\mu(X)-\beta$. Finally, the <influence function> of
$$
\mu(x)=\mathbb E[Y\mid A=0,X=x]
$$
is
$$
\frac{\mathbf1_{\{A=0,X=x\}}}{p(x)\{1-\pi(x)\}}
\{Y-\mu(x)\}.
$$
Multiplication by the derivative $p(x)\pi(x)$ of $\beta$ with respect to $\mu(x)$ and summation over $x$ gives
$$
(1-A)\frac{\pi(X)}{1-\pi(X)}\{Y-\mu(X)\}.
$$
Adding the three contributions yields the claimed mean-zero <influence curve>
$$
(1-A)\frac{\pi(X)}{1-\pi(X)}\{Y-\mu(X)\}
+A\mu(X)-\beta.
$$

Solution

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

 Solution

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