 Solution

Write $p_x=\mathbb{P}(X=x)$, ${\pi }_x=\mathbb{E}(A\mid X=x)$, and ${\mu }_x=\mathbb{E}(Y\mid X=x)$. The functional isThe first term has influence function $AY-\mathbb{E}(AY)$. Applying the product rule to the second term, including perturbations of $p_x$, ${\pi }_x$, and ${\mu }_x$, givesSubtracting and using $\beta =\mathbb{E}(AY)-\mathbb{E}[\pi \mu ]$ yieldsIts expectation is zero because the first product has expectation $\mathbb{E}\{\mathrm{Cov}(A,Y\mid X)\}=\beta$.