Considering the first term, it is straightforward to verify that
∬S×X∇t=0Et(Y∣E=1,M=m,X=x)fM∣E,X(m∣E=0,X=x)fX(x)dμ(m,x)=E[UfE∣X(E∣X)I(E=1){Y−E(Y∣E,M=m,X=x)fM∣E,X(M∣E=1,X)fM∣E,X(M∣E=0,X)}].
Similarly, one can easily verify that
∬S×XE(Y∣E=1,M=m,X=x)∇t=0fM∣E,X;t(m∣E=0,X=x)fX(x)dμ(m,x)=E[UfE∣X(E∣X)I(E=0){E(Y∣E=1,M=m,X=x)−η(1,0,X)}],
and finally, one can also verify that
∬S×XE(Y∣E=1,M=m,X=x)fM∣E,X(m∣E=0,X=x)∇t=0fX;t(x)dμ(m,x)=E[U{η(1,0,X)−θ0}].
Thus we obtain
∇t=0θt=E{Sθ0eff,nonpar(θ0)U}.
Given $S_{\delta e}^{\text{eff,nonpar}}(\delta_e)$, the results for the direct and indirect effect follow from the fact that the influence function of a difference of two functionals equals the difference of the respective influence functions. Because the model is nonparametric, there is a unique influence function for each functional, and it is efficient in the model, leading to the efficiency bound results. $\square$
PROOF OF THEOREM 2. We begin by showing that
E{Sθ0eff,nonpar(θ0;βm∗,βe∗,βy∗)}=0(7)
under model $\mathcal{M}_{\text{union}}$. First note that $(\beta_y^*, \beta_m^*) = (\beta_y, \beta_m)$ under model $\mathcal{M}_a$. Equality (7) now follows because $\mathbb{E}^{\text{par}}(Y|\tilde{X}, M, E=1; \beta_y) = \mathbb{E}(Y|\tilde{X}, M, E=1)$ and $\eta(1,0,X; \beta_y, \beta_m) = \mathbb{E}[{\mathbb{E}^{\text{par}}(Y|\tilde{X}, M, E=1; \beta_y)}|E=0, X] = \eta(1,0,X):$
\begin{align*}
& \mathbb{E}\{S_{\theta_0}^{\text{eff,nonpar}}(\theta_0; \beta_m, \beta_e^*, \beta_y)\} \\
& = \mathbb{E}\left[\frac{I\{E=1\} f_{M|E,X}^{\text{par}}(M|E=0, X; \beta_m)}{f_{E|X}^{\text{par}}(1|X; \beta_e^*) f_{M|E,X}^{\text{par}}(M|E=1, X; \beta_m)}
\end{align*}
