# Web Appendix for “Causal effects of intervening variables in settings with
unmeasured confounding”

##  Appendix A Example from Fulcher et al., (2020)

######  Example 1 (The safer deliveries program, (Fulcher et al.,, 2020)).

The ‘Safer deliveries’ program was designed to reduce the relatively high
rates of maternal and neonatal mortality in Zanzibar, Tanzania. The program
provided counselling to pregnant women preparing for delivery. Women deemed to
be in a high pregnancy “risk category”, based on a mobile device algorithm,
were instructed to deliver at a referral hospital, a specialty healthcare
resource that generally incurred higher expenses for the women’s family. Then,
given the recommended delivery location, the mobile device algorithm also
calculated an amount that they recommended the family should save in
anticipation of future obstetric care costs (i.e. a “tailored savings
recommendation”). At a later point in the study, the amount that the families
actually saved for this purpose was recorded in the data (i.e. the “actual
savings”).

Using data from the ‘Safer deliveries’ program, Fulcher et al., (2020) aimed
to evaluate “the effectiveness of this tailored savings recommendation by risk
category on actual savings”. They reported estimates of the PIIE of delivery
risk (high risk versus low/medium risk exposure) on actual savings at delivery
(outcome), mediated by a recommended savings amount calculated by the mobile
device algorithm. As noted in Fulcher et al., (2020), there may be unmeasured
confounding between a participant’s recommended risk category and actual
savings at delivery, for example by socioeconomic factors and individual’s
health-seeking behaviour.

Fulcher et al., (2020) argued that the PIIEs in Example A.1 was an appropriate
estimand to study “the effectiveness of this tailored savings recommendation”
for pregnant women. However, it is not clear that the plain English
justification translates to a PIIE defined by interventions on a woman’s
recommended risk category, an exposure that is non-interveneable or of limited
scientific interest. To interpret the results of their analysis we either have
to consider:

  1. (1)

obstetric risk category to be defined as a composite of various embodied
socio-demographic and clinical features, in which case intervention on
obstetric risk category can only be defined as interventions on the
constituent components used to characterize the exposure. Such an intervention
would be difficult to imagine as all of these embodied socio-demographic and
clinical features may be hard to identify; or

  2. (2)

obstetric risk category to be defined as a conceptually distinct feature from
the measured socio-demographic and clinical features used to compute it, and
thus possibly manipulable separately from these features. In this case the
“risk category” variable would simply be no more or less than the computed
“risk category” that appears on the screens of mobile devices, and these risk
categories could be manipulated simply by intervening on the software run on
these devices. However, such interventions would be of a substantively
different nature with profound differences in interpretation and will have
different implications for policy-makers. Moreover, the exposure “risk
category” would not be susceptible to unmeasured confounding.

Considering effects of intervening variables ameliorates this ambiguity and
also clarifies assumptions. Specifically, an intervention that avoids these
difficulties would be to fix the output value from the algorithm, so that it
recommends a delivery location as usual, but the patient’s recommended savings
amount is based on the delivery location the original algorithm would
recommend if that patient had been deemed to be at low risk for obstetric
complications. We can explicitly define the algorithm’s computed risk category
as $A_{M}$ (a modifiable intervening variable) that is distinct from the
patient’s non-modifiable embodied risk category ($A$). In the observed data,
$A=A_{M}$ with probability 1. However, we could conceive an intervention that
modifies this intervening variable $A_{M}$ without changing the exposure $A$.

##  Appendix B Proofs of identification the Frontdoor Formula

The proof for the frontdoor identifying formula
(LABEL:eq:_simple_frontdoor_formula) is given as follows:

$\displaystyle E(Y^{a^{\dagger}})$
$\displaystyle=\sum_{m}E(Y^{a^{\dagger}}\mid
M^{a^{\dagger}}=m)P(M^{a^{\dagger}}=m)$
$\displaystyle=\sum_{m}E(Y^{a^{\dagger}}\mid M^{a^{\dagger}}=m)P(M=m\mid
A=a^{\dagger})\leavevmode\nobreak\ \leavevmode\nobreak\ \text{(By A5, A6)}$
$\displaystyle=\sum_{m}E(Y^{a^{\dagger},m}\mid M^{a^{\dagger}}=m)p(m\mid
a^{\dagger})\leavevmode\nobreak\ \leavevmode\nobreak\ \text{(By A6)}$
$\displaystyle=\sum_{a,m}E(Y^{a^{\dagger},m}\mid
A=a,M^{a^{\dagger}}=m)p(a)p(m\mid a^{\dagger})\leavevmode\nobreak\
\leavevmode\nobreak\ \text{(By A5)}$ $\displaystyle=\sum_{m}p(m\mid
a^{\dagger})\sum_{a}E(Y\mid A=a,M=m)p(a)\leavevmode\nobreak\
\leavevmode\nobreak\ \text{(By A6, A7, A8)}$

Alternatively, we also provide a slightly different proof:

$\displaystyle E(Y^{a^{\dagger}})$
$\displaystyle=\sum_{u}E(Y^{a^{\dagger}}\mid U=u)f(u)$
$\displaystyle=\sum_{u}E(Y^{a^{\dagger}}\mid A=a^{\dagger},U=u)f(u)$
$\displaystyle=\sum_{u}E(Y\mid A=a^{\dagger},U=u)f(u)$
$\displaystyle=\sum_{u,m}E(Y\mid A=a^{\dagger},U=u,M=m)f(m\mid
a^{\dagger},u)f(u)$ $\displaystyle=\sum_{m}f(m\mid
a^{\dagger}){\sum_{u}E(Y\mid A=a^{\dagger},U=u,M=m)f(u)}$
$\displaystyle=\sum_{m}f(m\mid a^{\dagger})\underbrace{\sum_{u}E(Y\mid
U=u,M=m)f(u)}_{(\ast)}$ $\displaystyle=\sum_{m}f(m\mid
a^{\dagger})\sum_{a}E(Y\mid A=a,M=m)f(a).$

Aside from probability laws, we note the following conditions that are used in
the proof above: line 2 follows by conditional exchangeability of
$Y^{a^{\dagger}}$ and $A$ conditional on $U$ seen in the SWIG in Figure
LABEL:fig:front2a and follows from Assumption LABEL:ass:exchAM111Since there
is no unmeasured common causes of exposure-mediator by Assumption
LABEL:ass:exchAM, the only path from $A$ to $Y^{a^{\dagger}}$ is a backdoor
path via $U$.; line 3 follows by consistency that is implied from recursive
substitution of underlying NPSEMs; line 5 follows from Assumption
LABEL:ass:exchAM222If $U$ has a direct arrow to $M$ not through $A$, then this
will violate Assumption LABEL:ass:exchAM.; line 6 follows from Assumptions
LABEL:ass:noDirect and LABEL:ass:exchMY and can be seen from the conditional
independence of $Y$ and $A$ given $U$ and $M$ as seen in DAG
LABEL:fig:front1333Assumption LABEL:ass:noDirect ensures that there is no
direct path from $A$ to $Y$ not mediated by $M$. In addition, since there is
no unmeasured common causes of mediator-outcome by Assumption
LABEL:ass:exchMY, the only path from $A$ to $Y^{a^{\dagger}}$ is a backdoor
path via $U$ and a frontdoor path via $M$.; and line 7 follows from the SWIG
in Figure LABEL:fig:front2b where it can be seen that
$E(Y^{m})=\sum_{u}E(Y\mid U=u,M=m)f(u)=\sum_{a}E(Y\mid A=a,M=m)f(a)$ as $U$ or
$A$ blocks the backdoor back from $M$ to $Y^{m}$. Alternatively, line 7 holds
by algebraically by realizing the following:

$\displaystyle\sum_{a}E(Y\mid A=a,M=m)f(a)$ $\displaystyle=\sum_{a,u}E(Y\mid
A=a,M=m,U=u)f(u\mid a,m)f(a)$ $\displaystyle=\sum_{u}E(Y\mid
U=u,M=m)\sum_{a}f(u\mid a)f(a)$ $\displaystyle=\sum_{u}E(Y\mid U=u,M=m)f(u).$

######  Remark 1 (The front door formula is a weighted average).

Consider a binary treatment $A$ taking values $a^{\dagger}$ and $a^{\circ}$.
It can be trivially shown that $E(Y^{a^{\dagger}})$ is a weighted average of
$E(Y\mid A=a^{\dagger})$ and a separable estimand of treatment on $Y$ denoted
by $E(Y^{a_{M}=a^{\dagger},a_{Y}=a^{\circ}})$ that can be identified from the
observed data that e.g., follow an extended DAG seen in Figure D.2. This
extended DAG results from splitting the treatment node $A$ into two sub-
components, namely $A_{M}$ and $A_{Y}$. The bolded arrows from $A$ to $A_{M}$
and $A_{Y}$ indicate a deterministic relationship.444Specifically in the
observed data, $A=A_{M}=A_{Y}=1$ with probability 1, and $A=A_{M}=A_{Y}=0$
with probability 1. More specifically, the aforementioned two estimands are
weighted by the probability of receiving treatment $a^{\dagger}$ and
$a^{\circ}$ such that $E(Y^{a^{\dagger}})$ equals

$\displaystyle P(A=a^{\dagger})E(Y\mid
A=a^{\dagger})+P(A=a^{\circ})\underbrace{\sum_{m}E(Y\mid
A=a^{\circ},M=m)f(m\mid
a^{\dagger})}_{E(Y^{a_{M}=a^{\dagger},a_{Y}=a^{\circ}})},$ (B.1)

as stated in the main text (Equation (C.2)). We utilize this decomposition in
deriving efficient influence functions for the frontdoor formula. We note that
that both estimands $E(Y\mid A=a^{\dagger})$ and
$E(Y^{a_{M}=a^{\dagger},a_{Y}=a^{\circ}})$ allow for a direct path from $A$ to
$Y$ not mediated through $M$. Moreover, $E(Y\mid A=a^{\dagger})$ is identified
even if there are unmeasured common causes of $A$ and $Y$. This decomposition
implies that when all individuals in the observed data take treatment
$a^{\dagger}$, then $E(Y^{a^{\dagger}})=E(Y\mid A=a^{\dagger})$ and when all
individuals in the observed data take treatment $a^{\circ}$, then
$E(Y^{a^{\dagger}})=E(Y^{a_{M}=a^{\dagger},a_{Y}=a^{\circ}})$ (although this
would not be identified from observed data unless $f(m\mid a^{\dagger})$ is
known a priori for all $m$). Note that when
$E(Y^{a^{\dagger}})=E(Y^{a_{M}=a^{\dagger},a_{Y}=a^{\circ}})$,
$A=A_{Y}=a^{\circ}$ for all observations, and thus intervening on $A_{Y}$ (and
creating a mediated path from $A$black$\rightarrow$ $A_{Y}\rightarrow Y$) is
unnecessary.

##  Appendix C Identification and estimation of new causally manipulable
estimand in absence of $L$

In this section, we will assume that $L$ is the empty set.

######  Theorem C.1.

The average counterfactual outcome under an intervention on $A_{M}$ is
identified by the frontdoor formula LABEL:eq:_simple_frontdoor_formula under
Assumptions LABEL:ass:pos, LABEL:ass:_delta_1 and LABEL:ass:dismiss, that is,

$\displaystyle E$ $\displaystyle(Y^{a_{M}=a^{\dagger}})=\sum_{m}P(M=m\mid
A=a^{\dagger})\underbrace{\sum_{a}E(Y\mid A=a,M=m)P(A=a)}_{(\ast\ast)}.$

Suppose that the observed data $\mathcal{O}=(A,M,Y)$ follow a law $P$ which is
known to belong to $\mathcal{M}=\\{P_{\theta}:\theta\in\Theta\\}$, where
$\Theta$ is the parameter space. The efficient influence function
$\varphi^{\operatorname*{eff}}(\mathcal{O})$ for a causal parameter
$\Psi\equiv\Psi(\theta)$ in a non-parametric model $\mathcal{M}_{\text{np}}$
that imposes no restrictions on the law of $\mathcal{O}$ other than positivity
is given by
${d\Psi(\theta_{t})}/{dt}|_{t=0}=E\\{\varphi^{\operatorname*{eff}}(\mathcal{O})S(\mathcal{O})\\}$,
where ${d\Psi(\theta_{t})}/{dt}\mid_{t=0}$ is known as the pathwise derivative
of the parameter $\Psi$ along any parametric submodel of the observed data
distribution indexed by $t$, and $S(\mathcal{O})$ is the score function of the
parametric submodel evaluated at $t=0$ (Newey,, 1994; Van Der Vaart,, 2000).

The frontdoor formula can be re-expressed as a weighted average,

$\displaystyle\Psi=P(A=a^{\dagger})E(Y\mid
A=a^{\dagger})+P(A=a^{\circ}){\sum_{m}E(Y\mid A=a^{\circ},M=m)f(m\mid
a^{\dagger})},$ (C.2)

and thus the efficient influence function can be broken into two components.
Using the chain rule, the efficient influence function of
$\Psi=\sum_{m}f(m\mid a^{\dagger})\sum_{a}E(Y\mid A=a,M=m)f(a)$ can be derived
by finding the efficient influence function of (1) $\psi_{1}=P(A=a)$, (2)
$\psi_{2}=E(Y\mid A=a^{\dagger})$ and (3) $\psi_{3}={\sum_{m}E(Y\mid
A=a^{\circ},M=m)f(m\mid a^{\dagger})}$. We will use that $\psi_{3}$ is an
established identifying formula for $E(Y^{A_{Y}=a^{\dagger},A_{D}=a^{\circ}})$
a term in established identification formula for separable effect, which is
equal to the same functional of the observed data law $p(o)$ as
$E(Y^{a^{\circ}}\mid A=a^{\dagger})$ in the average treatment effect on the
treated (ATT) if $a^{\circ}=0$ and $a^{\dagger}=1$ (Tchetgen and Shpitser,,
2012).

######  Theorem C.2.

The efficient influence function $\varphi^{\operatorname*{eff}}(\mathcal{O})$
of the frontdoor formula in $\mathcal{M}_{\text{np}}$ is given by

$\displaystyle\varphi^{\operatorname*{eff}}(\mathcal{O})=$
$\displaystyle{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\left[I(A=a^{\dagger})-P(A=a^{\dagger})\right]\psi_{2}+P(A=a^{\dagger})\frac{I(A=a^{\dagger})}{P(A=a^{\dagger})}(Y-\psi_{2})}+$
$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[I(A=a^{\circ})\frac{P(A=a^{\dagger}\mid
M)}{P(A=a^{\circ}\mid
M)}\\{Y-b_{0}(M)\\}+I(A=a^{\dagger})\\{b_{0}(M)-\psi_{3}\\}\right]+}$
$\displaystyle{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}[I(A=a^{\circ})-P(A=a^{\circ})]\psi_{3}},$

where the terms in red are the efficient influence function for
$P(A=a^{\dagger})\psi_{2}$, the terms in blue is the efficient influence
function for $P(A=a^{\circ})\psi_{3}$, and $b_{0}(M)=E(Y\mid A=a^{\circ},M)$.
The efficient influence function can be reduced to the following,

$\displaystyle\varphi^{\operatorname*{eff}}(\mathcal{O})=$ $\displaystyle
I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}+$ (C.3)
$\displaystyle{\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[I(A=a^{\circ})\frac{P(A=a^{\dagger}\mid
M)}{P(A=a^{\circ}\mid
M)}\\{Y-b_{0}(M)\\}+I(A=a^{\dagger})\\{b_{0}(M)-\psi_{3}\\}\right]}-\Psi.$

which can be re-expressed as:

$\displaystyle\varphi^{\operatorname*{eff}}(\mathcal{O})=$ $\displaystyle
I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}+$ (C.4)
$\displaystyle{\left[I(A=a^{\circ})\frac{f(M\mid a^{\dagger})}{f(M\mid
a^{\circ})}\\{Y-b_{0}(M)\\}+\frac{I(A=a^{\dagger})P(A=a^{\circ})}{P(A=a^{\dagger})}\\{b_{0}(M)-\psi_{3}\\}\right]}-\Psi,$

by realizing that ${I(A=a^{\circ})P(A=a^{\circ})P(A=a^{\dagger}\mid
M)}\\{P(A=a^{\dagger})P(A=a^{\circ}\mid M)\\}^{-1}=I(A=a^{\circ}){f(M\mid
a^{\dagger})}\\{f(M\mid a^{\circ})\\}^{-1}$.

After some algebra, it can be shown that (C.4) can be written as the form of
the efficient influence function for $\Psi$ given by Equation (5) in Theorem 1
in Fulcher et al., (2020) with $C=\emptyset$. Following Theorem 1 in Fulcher
et al., (2020), the semiparametric efficiency bound for $\Psi$ in
$\mathcal{M}_{\text{np}}$ is given by
var$\left(\varphi^{\operatorname*{eff}}\right)$.

####  C.0.1. On semiparametric estimators for the frontdoor formula

Writing the efficient influence function for the frontdoor formula ($\Psi$)
given by in Expressions (C.3) or (C.4) allows us to construct estimators that
guarantee sample-boundedness. A weighted iterative conditional expectation
(Weighted ICE) estimator that guarantee sample-boundedness is given in the
following algorihtm. In what follows, we let
$\mathbb{P}_{n}(X)=n^{-1}\sum_{i=1}^{n}X_{i}$ and let $g^{-1}$ denote a known
inverse link function555For instance, if $Y$ is dichotomous, the $g$ is the
logit link function.

Algorithm 1 Algorithm for Weighted ICE (frontdoor formula)

1:Non-parametrically compute $P(A=a^{\circ})$ and $P(A=a^{\dagger})$.

2:Compute the MLEs $\hat{\alpha}$ of ${\alpha}$ from the observed data for the
treatment model $P(A=a\mid M;\alpha)$, or compute the MLEs $\hat{\gamma}$ of
$\gamma$ from the observed data for the mediator model $P(M=m\mid A;\gamma)$.

3:In the individuals whose $A=a^{\circ}$, fit a regression model
$Q({M};\theta)=g^{-1}\\{\theta^{T}\phi(M)\\}$ for $b_{0}(M)=E(Y\mid
M,A=a^{\circ})$ where the score function for each observation is weighted by
$\hat{W}_{1}$ where $\hat{W}_{1}$ equals
$\frac{P(A=a^{\circ})P(A=a^{\dagger}\mid
M;\hat{\alpha})}{P(A=a^{\dagger})P(A=a^{\circ}\mid M;\hat{\alpha})}$ if
$\hat{\alpha}$ was estimated in the previous step, or $\hat{W}$ equals
$\frac{f(M\mid A=a^{\dagger};\hat{\gamma})}{f(M\mid
A=a^{\circ};\hat{\gamma})}$ if $\hat{\gamma}$ was estimated in the previous
step. Moreover, $\phi(M)$ is a known function of $M$. More specifically, we
solve for $\theta$ in the following estimating equations:
$\mathbb{P}_{n}\left[I(A=a^{\circ})\phi(M)\hat{W}_{1}\left\\{Y-Q({M};\theta)\right\\}\right]=0$

4:In those whose $A=a^{\dagger}$, fit an intercept-only model
$T(\beta)=g^{-1}(\beta)$ for $\psi_{3}=E\\{b_{0}(M)\mid A=a^{\dagger}\\}$,
where the score function for each observation is weighted by
$\hat{W}_{2}=\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}.$ More specifically, we
solve for $\beta$ in the following estimating equations:
$\mathbb{P}_{n}\left[I(A=a^{\dagger})\hat{W}_{2}\left\\{Q({M};\hat{\theta})-T(\beta)\right\\}\right]=0$

5:Compute $\hat{T}\equiv T(\hat{\beta})$ for all observations.

6:Estimate
$\hat{\Psi}_{WICE}=\mathbb{P}_{n}\\{I(A=a^{\dagger})Y+I(A=a^{\circ})\hat{T}\\}$

Steps 3 and 4 ensure that the estimates for $\psi_{3}=E\\{b_{0}(M)\mid
A=a^{\dagger}\\}$ are sample bounded. Step 6 confirms that $\hat{\Psi}_{WICE}$
is a convex combination of $Y$ and estimates for $\psi_{3}$, both of which are
bounded by the range of the outcome $Y$. Thus, $\hat{\Psi}_{WICE}$ will also
be sample-bounded. For instance if the outcome is binary, then
$\hat{\Psi}_{WICE}$ will always be bounded between 0 and 1. Note that
estimators based on Expression (C.3) are more convenient to construct than
estimators based on Expression (C.4) when (1) $M$ is continuous, and/or (2)
there are multiple mediator variables ($M_{1}$, $M_{2}$, $M_{3}$…).

In Appendix F, we prove that an estimator based on the efficient influence
function given by (C.3) is doubly robust in the sense that it will be
consistent as long as the model for $P(A=a\mid M)$ or the model for
$b_{0}(M)=E(Y\mid A=a^{\circ},M)$ is correctly specified, and an estimator
based on the efficient influence function given by (C.4) is doubly robust in
the sense that it will be consistent as long as the model for $P(M=m\mid A)$
or the model for $b_{0}(M)$ is correctly specified.

##  Appendix D Interventionist identification with the frontdoor formula

Consider an extended causal DAG, which includes $A$ and also the variable
$A_{M}$, where the bold arrow from $A$ to $A_{M}$ indicates a deterministic
relationship. That is, Figure D.2 is the extended DAG of such a split with
$V=(U,A,A_{M},M,Y)$, and in the observed data, with probability one under
$f(v)$, either $A=A_{M}=1$ or $A=A_{M}=A_{Y}=0$.

Here and henceforth we use of “$(G)$” to indicate that the variables refer to
the hypothetical trial where $A_{M}$ is randomly assigned, as illustrated in
D.2. Consider an intervention that sets $A_{M}(G)$ to $a_{M}=a^{\dagger}$. The
average counterfactual outcome under such an intervention is indeed identified
as shown in the following:

$\displaystyle E$
$\displaystyle(Y^{a_{M}=a^{\dagger}})=E(Y(G)^{a_{M}=a^{\dagger}})$
$\displaystyle=\sum_{a}E(Y(G)^{a_{M}=a^{\dagger}}\mid
A_{Y}(G)=a)P(A_{Y}(G)=a)$
$\displaystyle=\sum_{a}E(Y(G)^{a_{M}=a^{\dagger}}\mid
A_{Y}(G)=a,A_{M}(G)=a^{\dagger})P(A_{Y}(G)=a)$
$\displaystyle=\sum_{a}E(Y(G)\mid
A_{Y}(G)=a,A_{M}(G)=a^{\dagger})P(A_{Y}(G)=a)$
$\displaystyle=\sum_{a,m}E(Y(G)\mid A_{Y}(G)=a,A_{M}(G)=a^{\dagger},M(G)=m)$
$\displaystyle\hskip 25.00003ptP(M(G)=m\mid
A_{Y}(G)=a,A_{M}(G)=a^{\dagger})P(A_{Y}(G)=a)$
$\displaystyle=\sum_{m}P(M(G)=m\mid
A_{Y}(G)=a^{\dagger},A_{M}(G)=a^{\dagger})$ $\displaystyle\hskip
25.00003pt{\sum_{a}E(Y(G)\mid A_{Y}(G)=a,A_{M}(G)=a,M(G)=m)P(A_{Y}(G)=a)}$
$\displaystyle=\sum_{m}P(M=m\mid A=a^{\dagger}){\sum_{a}E(Y\mid
A=a,M=m)P(A=a)}$

Aside from probability laws, we note the following conditions that are used in
the proof above: equality 1 follows from consistency that is implied from
recursive substitution; equality 3 follows by definition of $G$ and can be
seen via d-separation that follows from intervention on $A_{M}(G)$ in Figure
D.2; equality 4 holds by consistency; equality 6 holds by the conditional
independence $M(G)\mathrel{\perp\mspace{-10.0mu}\perp}A_{Y}(G)\mid A_{M}(G)$
and by the conditional independence
$Y(G)\mathrel{\perp\mspace{-10.0mu}\perp}A_{M}(G)\mid A_{Y}(G),M(G)$, both of
which follows from d-separation in Figure D.2; and equality 7 holds by
definition of $G$, consistency and determinism such that the event
$\\{A_{Y}(G)=a\\}$ is the same as the event $\\{A(G)=a\\}$ in $G$ and that
$\\{A_{Y}=a,A_{M}=a\\}$ is the same as the event $\\{A=a\\}$ in the observed
data. Of course, since we are not intervening on $A_{Y}(G)$, we can remove it
from the extended DAG as below. This will only require us to define one
particular variable $A_{M}$ that is deterministically equal $A$ in the
observed data.

$A$$M$$Y$$U$

Figure D.1. DAG of the observed random variables.

$A$$A_{M}$$A_{Y}$$M$$Y$$U$ (a) DAG of an expanded graph from Figure D.1. Bold
arrow denotes deterministic relationship.

$A(G)$$A_{M}(G)$$A_{Y}(G)$$M(G)$$Y(G)$$U(G)$ (b) DAG corresponding to
hypothetical trial based on the expanded graph D.2. Bold arrow denotes
deterministic relationship.

Figure D.2. DAG and SWIG of expanded graph.

###  D.1. Removing $A_{Y}$ from the extended DAG

The above results still hold if we remove $A_{Y}$ from the extended DAG given
in Figure D.2. The modified extended DAG with $A_{Y}$ removed is presented in
Figure LABEL:fig:front6a, where again we use of “$(G)$” to indicate that the
variables refer to the hypothetical trial where $A_{M}$ is randomly assigned.
In particular, conditioning sets that include $A(G)=a,A_{M}(G)=a^{\dagger}$
refer to the hypothetical trial $G$ in which $A_{M}$ is randomly assigned666or
imagine a trial where the only arrow into $A_{M}$ is from $A$ (which may not
be deterministic).. Then, the proof proceeds as follows:

$\displaystyle E(Y^{a_{M}=a^{\dagger}})$
$\displaystyle=E(Y(G)^{a_{M}=a^{\dagger}})$
$\displaystyle=\sum_{a}E(Y(G)^{a_{M}=a^{\dagger}}\mid A(G)=a)P(A(G)=a)$
$\displaystyle=\sum_{a}E(Y(G)^{a_{M}=a^{\dagger}}\mid
A(G)=a,A_{M}(G)=a^{\dagger})P(A(G)=a)$ $\displaystyle=\sum_{a}E(Y(G)\mid
A(G)=a,A_{M}(G)=a^{\dagger})P(A(G)=a)$ $\displaystyle=\sum_{a,m}E(Y(G)\mid
A(G)=a,A_{M}(G)=a^{\dagger},M(G)=m)$ $\displaystyle\hskip
25.00003ptP(M(G)=m\mid A(G)=a,A_{M}(G)=a^{\dagger})P(A(G)=a)$
$\displaystyle=\sum_{a,m}P(M(G)=m\mid A(G)=a^{\dagger},A_{M}(G)=a^{\dagger})$
$\displaystyle\hskip 25.00003ptE(Y(G)\mid A(G)=a,A_{M}(G)=a,M(G)=m)P(A(G)=a)$
$\displaystyle=\sum_{m}P(M=m\mid A=a^{\dagger})\sum_{a}E(Y\mid A=a,M=m)P(A=a)$

Aside from probability laws, we note the following conditions that are used in
the proof above: equality 1 holds by consistency that follows by recursive
substitution; equality 3 holds by definition of $G$ and can be seen via
d-separation –
$Y(G)^{a_{M}=a^{\dagger}}\mathrel{\perp\mspace{-10.0mu}\perp}A_{M}(G)\mid
A(G)$ – that follows by the intervention on $A_{M}(G)$ in DAG for $G$;
equality 4 holds by consistency; equality 6 holds by d-separation following
Assumption (LABEL:ass:dismiss) where
$Y(G)\mathrel{\perp\mspace{-10.0mu}\perp}A_{M}(G)\mid A(G),M(G)$ and
$M(G)\mathrel{\perp\mspace{-10.0mu}\perp}A(G)\mid A_{M}(G)$ (since we are
assuming for now that $L(G)=\emptyset$); and equality 7 holds by definition of
$G$, consistency and determinism in the observed data such that the event
$\\{A=a,A_{M}=a\\}$ is the same as the event $\\{A=a\\}$.

###  D.2. A causally manipulable interpretation of the Population Intervention
Indirect Effect in the general scenario

As before, consider splitting the treatment node $A$ on a DAG given in Figure
LABEL:fig:front1 into two sub-components, namely $A_{M}$ and $A_{Y}$. Figure
D.3 is the extended DAG of such a split with $V=(L,U,A,A_{M},A_{Y},M,Y)$. This
extended DAG is analogous to the extended DAG described in Section
LABEL:sec:manip_simple but generalized to include measured common causes ($L$)
of $A$, $M$ and $Y$. Analogous to Section D.1, we can also remove $A_{Y}$ in
the extended DAG and the results above would still hold to identify the
average counterfactual outcome under an intervention on $A_{M}$ that sets it
equal to $a^{\dagger}$. Again, consider an intervention on $A_{M}(G)$ that
sets it equal to $a_{M}=a^{\dagger}$ as shown in Figure D.3777we can also
imagine a trial where there are non-deterministic arrows from $A$ and $L$ into
$A_{M}$.. The average counterfactual outcome under such an intervention is
indeed identified in the following:

$\displaystyle E(Y^{a_{M}=a^{\dagger}})$
$\displaystyle=E(Y(G)^{a_{M}=a^{\dagger}})$
$\displaystyle=\sum_{l,a}E(Y(G)^{a_{M}=a^{\dagger}}\mid
L(G)=l,A(G)=a)P(A(G)=a\mid L(G)=l)P(L(G)=l)$
$\displaystyle=\sum_{l,a}E(Y(G)^{a_{M}=a^{\dagger}}\mid
A_{M}(G)=a^{\dagger},L(G)=l,A(G)=a)P(A(G)=a\mid L(G)=l)P(L(G)=l)$
$\displaystyle=\sum_{l,a}E(Y(G)\mid
A_{M}(G)=a^{\dagger},L(G)=l,A(G)=a){P(A(G)=a\mid L(G)=l)P(L(G)=l)}$
$\displaystyle=\sum_{m,l,a}E(Y(G)\mid
M(G)=m,A_{M}(G)=a^{\dagger},L(G)=l,A(G)=a)$ $\displaystyle\hskip
25.00003ptP(M(G)=m\mid A_{M}(G)=a^{\dagger},A(G)=a,L(G)=l)P(A(G)=a\mid
L(G)=l)P(L(G)=l)$ $\displaystyle=\sum_{m,l}P(M(G)=m\mid
A_{M}(G)=a^{\dagger},A(G)=a^{\dagger},L(G)=l)P(L(G)=l)$ $\displaystyle\hskip
40.00006pt{\sum_{a}E(Y(G)\mid M(G)=m,L(G)=l,A(G)=a,A_{M}(G)=a)P(A(G)=a\mid
L(G)=l)}$ $\displaystyle=\sum_{m}P(M=m\mid
A=a^{\dagger},L=l)P(L=l){\sum_{a}E(Y\mid M=m,L=l,A=a)P(A=a\mid L=l)}$

Aside from probability laws, we note the following conditions that are used in
the proof above: equality 1 holds by consistency that follows by recursive
substitution; equality 3 follows by definition of $G$ and can be seen via
d-separation that follows from intervention on $A_{M}(G)$ in $G$ shown in
Figure D.3 (which also holds if there are arrows from $L(G)$ to $A_{M}(G)$ and
$A(G)$ to $A_{M}(G)$); equality 4 holds by consistency; equality 6 holds by
the the dismissible component conditions such that conditional independence
$M(G)\mathrel{\perp\mspace{-10.0mu}\perp}A(G)\mid A_{M}(G),L(G)$ and
$Y(G)\mathrel{\perp\mspace{-10.0mu}\perp}A_{M}(G)\mid A(G),M(G),L(G)$ hold;
and equality 7 holds by definition of $G$, consistency and determinism in the
observed data such that the event $\\{A=a,A_{M}=a\\}$ is the same as the event
$\\{A=a\\}$.

$A$$A_{M}$$A_{Y}$$M$$Y$$U$$L$ (a) DAG of an extended graph from Figure
LABEL:fig:front1.

$A(G)$$A_{M}(G)$$M(G)$$Y(G)$$U(G)$$L(G)$ (b) SWIG corresponding to an
intervention on the extended graph in D.3.

Figure D.3. extended DAG and SWIG.

##  Appendix E Derivation of efficient influence function in Section
LABEL:sec:estimators

###### Proof.

Note that for binary treatment variables, the generalized frontdoor formula
$\Psi=\sum_{m}f(m\mid a^{\dagger})\sum_{a}E(Y\mid A=a,M=m)f(a)$ is equivalent
to the following:

$\Psi=P(A=a^{\dagger})E(Y\mid A=a^{\dagger})+P(A=a^{\circ})\sum_{m,l}E(Y\mid
M=m,L=l,A=a^{\circ})f(m\mid a^{\dagger},l)f(l\mid a^{\circ}).$

The efficient influence function in the nonparametric model $\mathcal{M}_{NP}$
is defined as the unique mean zero, finite variance random variable
$\varphi^{\operatorname*{eff}}(\mathcal{O})$ such that

$\frac{d\Psi(\theta_{t})}{dt}\Big{|}_{t=0}=E\\{\varphi^{\operatorname*{eff}}(\mathcal{O})S(\mathcal{O})\\}$

where $\mathcal{O}=(L,A,M,Y)$, ${d\Psi(\theta_{t})}/{dt}\mid_{t=0}$ is known
as the pathwise derivative of parameter $\Psi$ along a parametric submodel
indexed by $t$, and $S(\mathcal{O})$ is the score function of the parametric
submodel evaluated at $t=0$.

The efficient influence function of $\Psi$ can be realized by finding the
efficient influence function of (1) $\psi_{1}=P(A=a)$, (2) $\psi_{2}=E(Y\mid
A=a^{\dagger})$ and (3) $\psi_{3}=\sum_{m,l}E(Y\mid
M=m,L=l,A=a^{\circ})f(m\mid a^{\dagger},l)f(l\mid a^{\circ})$. In particular,
using differentiation rules, the efficient influence function is given by:

$\displaystyle\varphi^{\operatorname*{eff}}(\mathcal{O})=\underbrace{[I(A=a^{\dagger})-P(A=a^{\dagger})]}_{(\ast)}\psi_{2}+\psi_{2}^{\text{eff}}P(A=a^{\dagger})+\psi_{3}^{\text{eff}}P(A=a^{\circ})+\underbrace{[I(A=a^{\circ})-P(A=a^{\circ})]}_{(\ast\ast)}\psi_{3}$

where expression $(\ast)$ is the efficient influence function of
$P(A=a^{\dagger})$ and expression $(\ast\ast)$ is the efficient influence
function of $P(A=a^{\circ})$. Moreover,

$\displaystyle\frac{d\psi_{2}(\theta_{t})}{dt}\Big{|}_{t=0}$
$\displaystyle=E\\{YS({Y\mid A=a^{\dagger}})\mid A=a^{\dagger}\\}$
$\displaystyle=E\left[\left\\{Y-E(Y\mid A=a^{\dagger})\right\\}S({Y\mid
A=a^{\dagger}})\mid A=a^{\dagger}\right]$
$\displaystyle=E\left[\frac{I(A=a^{\dagger})\left(Y-\psi_{2}\right)}{P(A=a^{\dagger})}S(\mathcal{O})\right]$

and

$\displaystyle\frac{d\psi_{3}(\theta_{t})}{dt}\Big{|}_{t=0}=$
$\displaystyle\underbrace{E\left(E\left[E\\{YS(Y\mid A=a^{\circ},M,L)\mid
A=a^{\circ},M,L\\}\mid A=a^{\dagger},L\right]\mid
A=a^{\circ}\right)}_{\raisebox{-0.05em}{\scalebox{0.85}{A}}⃝}+$
$\displaystyle\underbrace{E\left[E\left\\{E(Y\mid A=a^{\circ},M,L)S(M\mid
A=a^{\dagger},L)\mid A=a^{\dagger},L\right\\}\mid
A=a^{\circ}\right]}_{\raisebox{-0.05em}{\scalebox{0.85}{B}}⃝}+$
$\displaystyle\underbrace{E\left[E\left\\{E(Y\mid A=a^{\circ},M,L)\mid
A=a^{\dagger},L\right\\}S(L\mid A=a^{\circ})\mid
A=a^{\circ}\right]}_{\raisebox{-0.05em}{\scalebox{0.85}{C}}⃝}$

We look at each expression separately. First, we consider Expression
$\raisebox{-0.05em}{\scalebox{0.85}{A}}⃝$:

$\displaystyle\raisebox{-0.05em}{\scalebox{0.85}{A}}⃝$
$\displaystyle=E\left(E\left[E\\{YS(Y\mid A=a^{\circ},M,L)\mid
A=a^{\circ},M,L\\}\mid A=a^{\dagger},L\right]\mid A=a^{\circ}\right)$
$\displaystyle=E\left(E\left[E\left\\{Y\frac{I(A=a^{\circ})}{P(A=a^{\circ}\mid
L,M)}S(Y\mid A,M,L)\mid M,L\right\\}\mid A=a^{\dagger},L\right]\mid
A=a^{\circ}\right)$
$\displaystyle=E\left(E\left[E\left\\{Y\frac{I(A=a^{\circ})}{P(A=a^{\circ}\mid
L,M)}S(Y\mid A,M,L)\mid
M,L\right\\}\frac{I(A=a^{\dagger})}{P(A=a^{\dagger}\mid L)}\mid
L\right]\frac{I(A=a^{\circ})}{P(A=a^{\circ})}\right)$
$\displaystyle=E\left(E\left[E\left\\{Y\frac{I(A=a^{\circ})}{P(A=a^{\circ}\mid
L,M)}S(Y\mid A,M,L)\mid M,L\right\\}\frac{P(A=a^{\dagger}\mid
L,M)}{P(A=a^{\dagger}\mid L)}\mid L\right]\frac{P(A=a^{\circ}\mid
L)}{P(A=a^{\circ})}\right)$
$\displaystyle=E\left(E\left[E\left\\{\Big{(}Y-b_{0}(M,L)\Big{)}\frac{I(A=a^{\circ})}{P(A=a^{\circ}\mid
L,M)}S(Y\mid A,M,L)\mid M,L\right\\}\frac{P(A=a^{\dagger}\mid
L,M)}{P(A=a^{\dagger}\mid L)}\mid L\right]\frac{P(A=a^{\circ}\mid
L)}{P(A=a^{\circ})}\right)$
$\displaystyle=E\left[\Big{\\{}Y-b_{0}(M,L)\Big{\\}}\frac{I(A=a^{\circ})P(A=a^{\dagger}\mid
L,M)P(A=a^{\circ}\mid L)}{P(A=a^{\circ}\mid L,M)P(A=a^{\dagger}\mid
L)P(A=a^{\circ})}S(\mathcal{O})\right]$

which also equals

$E\left[\Big{\\{}Y-b_{0}(M,L)\Big{\\}}\frac{I(A=a^{\circ})f(M\mid
A=a^{\dagger},L)}{P(A=a^{\circ})f(M\mid A=a^{\circ},L)P(A=a^{\dagger}\mid
L)}S(\mathcal{O})\right].$

Next, we consider $\raisebox{-0.05em}{\scalebox{0.85}{B}}⃝$:

$\displaystyle\raisebox{-0.05em}{\scalebox{0.85}{B}}⃝$
$\displaystyle=E\left[E\left\\{\underbrace{E(Y\mid
A=a^{\circ},M,L)}_{b_{0}(M,L)}S(M\mid A=a^{\dagger},L)\mid
A=a^{\dagger},L\right\\}\mid A=a^{\circ}\right]$
$\displaystyle=E\left[E\left\\{b_{0}(M,L)\frac{I(A=a^{\dagger})}{P(A=a^{\dagger}\mid
L)}S(M\mid A,L)\mid L\right\\}\frac{P(A=a^{\circ}\mid
L)}{P(A=a^{\circ})}\right]$
$\displaystyle=E\left[\Big{\\{}b_{0}(M,L)-E\Big{(}b_{0}(M,L)\mid
L,A\Big{)}\Big{\\}}\frac{I(A=a^{\dagger})P(A=a^{\circ}\mid
L)}{P(A=a^{\dagger}\mid L)P(A=a^{\circ})}S(\mathcal{O})\right]$

Finally, we consider $\raisebox{-0.05em}{\scalebox{0.85}{C}}⃝$:

$\displaystyle\raisebox{-0.05em}{\scalebox{0.85}{C}}⃝$
$\displaystyle=E\left[\underbrace{E\left\\{E(Y\mid A=a^{\circ},M,L)\mid
A=a^{\dagger},L\right\\}}_{h_{\dagger}(L)}S(L\mid A=a^{\circ})\mid
A=a^{\circ}\right]$
$\displaystyle=E\left\\{h_{\dagger}(L)\frac{I(A=a^{\circ})}{P(A=a^{\circ})}S(L\mid
A)\right\\}$
$\displaystyle=E\left\\{\Big{(}h_{\dagger}(L)-\psi_{3}\Big{)}\frac{I(A=a^{\circ})}{P(A=a^{\circ})}S(\mathcal{O})\right\\}$

Thus, putting everything together and after some further algebraic
simplification, we can see that the efficient influence function is indeed
given by:

$\displaystyle\varphi^{\operatorname*{eff}}(\mathcal{O})=$ $\displaystyle
I(A=a^{\dagger})Y+I(A=a^{\dagger})\psi_{3}+$
$\displaystyle{\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[I(A=a^{\circ})\frac{P(A=a^{\dagger}\mid
M)}{P(A=a^{\circ}\mid
M)}\\{Y-b_{0}(M)\\}+I(A=a^{\dagger})\\{b_{0}(M)-\psi_{3}\\}\right]}-\Psi$

It is trivial to realize that this Expression of the efficient influence
function can also be re-expressed as the following:

$\displaystyle\varphi^{\operatorname*{eff}}(\mathcal{O})=$ $\displaystyle
I(A=a^{\dagger})Y+I(A=a^{\dagger})\psi_{3}+$
$\displaystyle{\left[I(A=a^{\circ})\frac{f(M\mid a^{\dagger})}{f(M\mid
a^{\circ})}\\{Y-b_{0}(M)\\}+\frac{I(A=a^{\dagger})P(A=a^{\circ})}{P(A=a^{\dagger})}\\{b_{0}(M)-\psi_{3}\\}\right]}-\Psi$

∎

##  Appendix F Robustness against model misspecification

###  F.1. Non-generalized front door formula

We show that an estimator based on Equation (C.3) is doubly robust in the
sense that it will be consistent as long as

  1. (1)

the model for $P(A=a\mid M)$ is correctly specified, or

  2. (2)

the model for $E(Y\mid A=a^{\circ},M)$ is correctly specified.

and that an estimator based on Equation (C.4) is doubly robust in that it will
be consistent as long as

  1. (1)

the model for $P(M=m\mid A)$ is correctly specified, or

  2. (2)

the model for $E(Y\mid A=a^{\circ},M)$ is correctly specified.

We consider an estimator based on Equation (C.3). Suppose that
$\alpha^{\ast}$, $\theta^{\ast}$ and $\beta^{\ast}$ are probability limits of
$\alpha$, $\theta$ and $\beta$, respectively. Furthermore, let
$b_{0}^{\ast}(M)=Q(M;\theta^{\ast})$ where as before $Q(M;\theta)$ is a model
for $b_{0}(M)=E(Y\mid M,A=a^{\circ})$, and let
$\psi_{3}^{\ast}=T(\beta^{\ast})$ where $T(\beta)$ is a non-parametric model
for $\psi_{3}=E\\{b_{0}(M)\mid A=a^{\dagger}\\}$. Under Equation (C.3)
suffices to show that

$\displaystyle E\Bigg{(}I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}^{\ast}+$
$\displaystyle\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[I(A=a^{\circ})\frac{P(A=a^{\dagger}\mid
M;\alpha^{\ast})}{P(A=a^{\circ}\mid
M;\alpha^{\ast})}\\{Y-b_{0}^{\ast}(M)\\}+I(A=a^{\dagger})\\{b_{0}^{\ast}(M)-\psi_{3}^{\ast}\\}\right]-\Psi\Bigg{)}=0$

under scenario (1) where $\alpha^{\ast}=\alpha$ and thus $P(A=a^{\dagger}\mid
M;\alpha^{\ast})=P(A=a^{\dagger}\mid M)$, or under scenario (2) where
$\theta^{\ast}=\theta$ and thus $b_{0}^{\ast}(M)=b_{0}(M)$ and
$\psi_{3}^{\ast}=\psi_{3}$.

###### Proof.

Suppose first that only the model for $P(A=a\mid M)$ is correctly specified.
Then,

$\displaystyle E\Bigg{(}I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}^{\ast}+$
$\displaystyle\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[I(A=a^{\circ})\frac{P(A=a^{\dagger}\mid
M;\alpha^{\ast})}{P(A=a^{\circ}\mid
M;\alpha^{\ast})}\\{Y-b_{0}^{\ast}(M)\\}+I(A=a^{\dagger})\\{b_{0}^{\ast}(M)-\psi_{3}^{\ast}\\}\right]-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}P(A=a^{\dagger}\mid M)E(Y\mid
A=a^{\dagger},M)+P(A=a^{\circ}\mid M)\psi_{3}^{\ast}+$
$\displaystyle\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[\cancel{P(A=a^{\circ}\mid
M)}\frac{P(A=a^{\dagger}\mid M;\alpha^{\ast})}{\cancel{P(A=a^{\circ}\mid
M;\alpha^{\ast})}}\\{b_{0}(M)-b_{0}^{\ast}(M)\\}+P(A=a^{\dagger}\mid
M)\\{b_{0}^{\ast}(M)-\psi_{3}^{\ast}\\}\right]-\Psi\Bigg{)}$ $\displaystyle=$
$\displaystyle\sum_{m}\Big{(}P(A=a^{\dagger}\mid M=m)E(Y\mid
A=a^{\dagger},M=m)+P(A=a^{\circ}\mid M=m)\psi_{3}^{\ast}+$
$\displaystyle\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\Big{[}P(A=a^{\dagger}\mid
M=m;\alpha^{\ast})b_{0}(m)-P(A=a^{\dagger}\mid
M=m)\psi_{3}^{\ast}\\}\Big{]}\Bigg{)}P(M=m)-\Psi$ $\displaystyle=$
$\displaystyle E(Y\mid
A=a^{\dagger})P(A=a^{\dagger})+\sum_{m}P(A=a^{\circ})P(M=m\mid
A=a^{\dagger})b_{0}(M)-\Psi$ $\displaystyle=$ $\displaystyle 0$

Next, suppose that only the model for $E(Y\mid A=a^{\circ},M)$ is correctly
specified. Then,

$\displaystyle E\Bigg{(}I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}^{\ast}+$
$\displaystyle\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[I(A=a^{\circ})\frac{P(A=a^{\dagger}\mid
M;\alpha^{\ast})}{P(A=a^{\circ}\mid
M;\alpha^{\ast})}\\{Y-b_{0}^{\ast}(M)\\}+I(A=a^{\dagger})\\{b_{0}^{\ast}(M)-\psi_{3}^{\ast}\\}\right]-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}P(A=a^{\dagger}\mid M)E(Y\mid
A=a^{\dagger},M)+P(A=a^{\circ}\mid M)\psi_{3}^{\ast}+$
$\displaystyle\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[P(A=a^{\circ}\mid
M)\frac{P(A=a^{\dagger}\mid M;\alpha^{\ast})}{P(A=a^{\circ}\mid
M;\alpha^{\ast})}\\{\underbrace{b_{0}(M)-b_{0}^{\ast}(M)}_{=0}\\}+P(A=a^{\dagger}\mid
M)\\{b_{0}^{\ast}(M)-\psi_{3}^{\ast}\\}\right]-\Psi\Bigg{)}$ $\displaystyle=$
$\displaystyle\sum_{m}\Bigg{(}P(A=a^{\dagger}\mid M=m)E(Y\mid
A=a^{\dagger},M=m)+P(A=a^{\circ}\mid M=m)\psi_{3}^{\ast}+$
$\displaystyle\frac{P(A=a^{\circ})}{P(A=a^{\dagger})}\left[P(A=a^{\dagger}\mid
M=m)\\{b_{0}^{\ast}(M)-\psi_{3}^{\ast}\\}\right]\Bigg{)}P(M=m)-\Psi$
$\displaystyle=$ $\displaystyle E(Y\mid
A=a^{\dagger})P(A=a^{\dagger})+\sum_{m}P(A=a^{\circ})P(M=m\mid
A=a^{\dagger})b_{0}(m)-\Psi$ $\displaystyle=$ $\displaystyle 0$

∎

The proof of double robustness for estimators based on Equation (C.4) follows
analogously as the proof shown above.

###  F.2. Generalized front door formula

Our proposed estimator based on the efficient influence function given by
(LABEL:eq:eif_gen1) and (LABEL:eq:eif_gen2) is robust against 3 classes of
model misspecification scenarios. Specifically, the weighted ICE estimator
where a model for $P(A=a\mid M,L)$ is specified will be consistent when at
least one of the following holds:

  1. (1)

the models for $P(A=a\mid M,L)$ and $P(A=a\mid L)$ are correctly specified, or

  2. (2)

the models for $b_{0}(M,L)$ and $h_{\dagger}(L)$ are correctly specified, or

  3. (3)

the models for $b_{0}(M,L)$ and $P(A=a\mid L)$ are correctly specified.

The weighted ICE estimator where a model for $P(M=m\mid A,L)$ is specified
will be consistent when at least one of the following holds

  1. (1)

the models for $P(M=m\mid A,L)$ and $P(A=a\mid L)$ are correctly specified, or

  2. (2)

the models for $b_{0}(M,L)$ and $h_{\dagger}(L)$ are correctly specified, or

  3. (3)

the models for $b_{0}(M,L)$ and $P(A=a\mid L)$ are correctly specified.

We will prove robustness for an estimator based on (LABEL:eq:eif_gen2) where a
model for $P(M=m\mid A,L)$ is specified. Proof of robustness for estimator
based on (LABEL:eq:eif_gen1) where a model for $P(A=a\mid M,L)$ is specified
follows analogously.

Suppose that $\gamma^{\ast}$, $\kappa^{\ast}$, $\theta^{\ast}$, $\eta^{\ast}$
and $\beta^{\ast}$ are probability limits of $\gamma$,$\kappa$, $\theta$
$\eta$ and $\beta$, respectively. Furthermore, let
$b_{0}^{\ast}(M,L)=Q(M,L;\theta^{\ast})$ where as before $Q(M,L;\theta)$ is a
model for $b_{0}(M)=E(Y\mid M,L,A=a^{\circ})$, let
$h_{\dagger}^{\ast}(L)=R(L;\eta^{\ast})$ where $R(L;\eta)$ is a model for
$h_{\dagger}(L)$, and let $\psi_{3}^{\ast}=T(\beta^{\ast})$ where $T(\beta)$
is a non-parametric model for $\psi_{3}=E\\{h_{\dagger}(L)\mid
A=a^{\circ}\\}$.

Under Equation (LABEL:eq:eif_gen1) suffices to show that

$\displaystyle
E\Bigg{(}I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}^{\ast}+\frac{I(A=a^{\circ})f(M\mid
A=a^{\dagger},L;\gamma^{\ast})}{f(M\mid
A=a^{\circ},L;\gamma^{\ast})}\\{Y-b_{0}^{\ast}(M,L)\\}+$
$\displaystyle\frac{I(A=a^{\dagger})P(A=a^{\circ}\mid
L;\kappa^{\ast})}{P(A=a^{\dagger}\mid
L;\kappa^{\ast})}\\{b_{0}^{\ast}(M,L)-h_{\dagger}^{\ast}(L)\\}+I(A=a^{\circ})\\{h_{\dagger}^{\ast}(L)-\psi_{3}^{\ast}\\}-\Psi\Bigg{)}=0$

under scenario (1) where $\gamma^{\ast}=\gamma$ and $\kappa^{\ast}=\kappa$ and
thus $P(M=m\mid A,L;\gamma^{\ast})=P(M=m\mid A,L)$ and $P(A=a^{\dagger}\mid
L;\kappa^{\ast})=P(A=a^{\dagger}\mid L)$, or under scenario (2) where
$\theta^{\ast}=\theta$ and $\eta^{\ast}=\eta$ and thus
$b_{0}^{\ast}(M,L)=b_{0}(M,L)$, $h_{\dagger}^{\ast}(L)=h_{\dagger}(L)$ and
$\psi_{3}^{\ast}=\psi_{3}$, or under scenario (3) where $\theta^{\ast}=\theta$
and $\kappa^{\ast}=\kappa$ and thus $b_{0}^{\ast}(M,L)=b_{0}(M,L)$ and
$P(A=a^{\dagger}\mid L;\kappa^{\ast})=P(A=a^{\dagger}\mid L)$.

###### Proof.

Suppose first that only the models for $P(M=m\mid A,L)$ and $P(A=a\mid L)$ are
correctly specified. Then,

$\displaystyle E$
$\displaystyle\Bigg{(}I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}^{\ast}+\frac{I(A=a^{\circ})f(M\mid
A=a^{\dagger},L;\gamma^{\ast})}{f(M\mid
A=a^{\circ},L;\gamma^{\ast})}\\{Y-b_{0}^{\ast}(M,L)\\}+$
$\displaystyle\frac{I(A=a^{\dagger})P(A=a^{\circ}\mid
L;\kappa^{\ast})}{P(A=a^{\dagger}\mid
L;\kappa^{\ast})}\\{b_{0}^{\ast}(M,L)-h_{\dagger}^{\ast}(L)\\}+I(A=a^{\circ})\\{h_{\dagger}^{\ast}(L)-\psi_{3}^{\ast}\\}-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}\sum_{m}P(A=a^{\dagger}\mid L)E(Y\mid
A=a^{\dagger},M=m,L)f(m\mid a^{\dagger},L)+P(A=a^{\circ}\mid
L)\psi_{3}^{\ast}+$ $\displaystyle\sum_{m}\frac{P(A=a^{\circ}\mid L)f(m\mid
A=a^{\dagger},L;\gamma^{\ast})}{\cancel{f(m\mid
A=a^{\circ},L;\gamma^{\ast})}}\\{b_{0}(m,L)-b_{0}^{\ast}(m,L)\\}\cancel{f(m\mid
a^{\circ},L)}+$ $\displaystyle\frac{\cancel{P(A=a^{\dagger}\mid
L)}P(A=a^{\circ}\mid L;\kappa^{\ast})}{\cancel{P(A=a^{\dagger}\mid
L;\kappa^{\ast})}}\left\\{\sum_{m}b_{0}^{\ast}(m,L)f(m\mid
a^{\dagger},L)-h_{\dagger}^{\ast}(L)\right\\}+$ $\displaystyle
P(A=a^{\circ}\mid L)\\{h_{\dagger}^{\ast}(L)-\psi_{3}^{\ast}\\}-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}\sum_{m}P(A=a^{\dagger}\mid L)E(Y\mid
A=a^{\dagger},M=m,L)f(m\mid a^{\dagger},L)+\sum_{m}{P(A=a^{\circ}\mid
L)f(m\mid A=a^{\dagger},L;\gamma^{\ast})}b_{0}(m,L)\Bigg{)}$
$\displaystyle=P(A=a^{\dagger})E(Y\mid
A=a^{\dagger})+P(A=a^{\circ})\sum_{m,l}b_{0}(m,l)f(m\mid a^{\dagger},l)f(l\mid
a^{\circ})$ $\displaystyle=$ $\displaystyle 0$

Next, suppose that only the models for $b_{0}(M,L)$ and $h_{\dagger}(L)$ are
correctly specified. Then,

$\displaystyle E$
$\displaystyle\Bigg{(}I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}^{\ast}+\frac{I(A=a^{\circ})f(M\mid
A=a^{\dagger},L;\gamma^{\ast})}{f(M\mid
A=a^{\circ},L;\gamma^{\ast})}\\{Y-b_{0}^{\ast}(M,L)\\}+$
$\displaystyle\frac{I(A=a^{\dagger})P(A=a^{\circ}\mid
L;\kappa^{\ast})}{P(A=a^{\dagger}\mid
L;\kappa^{\ast})}\\{b_{0}^{\ast}(M,L)-h_{\dagger}^{\ast}(L)\\}+I(A=a^{\circ})\\{h_{\dagger}^{\ast}(L)-\psi_{3}^{\ast}\\}-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}\sum_{m}P(A=a^{\dagger}\mid L)E(Y\mid
A=a^{\dagger},M=m,L)f(m\mid a^{\dagger},L)+\cancel{P(A=a^{\circ}\mid
L)\psi_{3}^{\ast}}+$ $\displaystyle\sum_{m}\frac{P(A=a^{\circ}\mid L)f(m\mid
A=a^{\dagger},L;\gamma^{\ast})}{f(m\mid
A=a^{\circ},L;\gamma^{\ast})}\\{\underbrace{b_{0}(m,L)-b_{0}^{\ast}(m,L)}_{=0}\\}f(m\mid
a^{\circ},L)+$ $\displaystyle\frac{P(A=a^{\dagger}\mid L)P(A=a^{\circ}\mid
L;\kappa^{\ast})}{P(A=a^{\dagger}\mid
L;\kappa^{\ast})}\left\\{\underbrace{\sum_{m}b_{0}^{\ast}(m,L)f(m\mid
a^{\dagger},L)-h_{\dagger}^{\ast}(L)}_{=0}\right\\}+$ $\displaystyle
P(A=a^{\circ}\mid
L)\\{h_{\dagger}^{\ast}(L)-\cancel{\psi_{3}^{\ast}}\\}-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}\sum_{m}P(A=a^{\dagger}\mid L)E(Y\mid
A=a^{\dagger},M=m,L)f(m\mid a^{\dagger},L)+\sum_{m}{P(A=a^{\circ}\mid
L)f(m\mid A=a^{\dagger},L;\gamma^{\ast})}b_{0}(m,L)\Bigg{)}$ $\displaystyle=$
$\displaystyle P(A=a^{\dagger})E(Y\mid
A=a^{\dagger})+P(A=a^{\circ})\sum_{m,l}b_{0}(m,l)f(m\mid a^{\dagger},l)f(l\mid
a^{\circ})$ $\displaystyle=$ $\displaystyle 0$

Finally, suppose that only the models for $b_{0}(M,L)$ and $P(A=a\mid L)$ are
correctly specified. Then,

$\displaystyle E$
$\displaystyle\Bigg{(}I(A=a^{\dagger})Y+I(A=a^{\circ})\psi_{3}^{\ast}+\frac{I(A=a^{\circ})f(M\mid
A=a^{\dagger},L;\gamma^{\ast})}{f(M\mid
A=a^{\circ},L;\gamma^{\ast})}\\{Y-b_{0}^{\ast}(M,L)\\}+$
$\displaystyle\frac{I(A=a^{\dagger})P(A=a^{\circ}\mid
L;\kappa^{\ast})}{P(A=a^{\dagger}\mid
L;\kappa^{\ast})}\\{b_{0}^{\ast}(M,L)-h_{\dagger}^{\ast}(L)\\}+I(A=a^{\circ})\\{h_{\dagger}^{\ast}(L)-\psi_{3}^{\ast}\\}-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}\sum_{m}P(A=a^{\dagger}\mid L)E(Y\mid
A=a^{\dagger},M=m,L)f(m\mid a^{\dagger},L)+\cancel{P(A=a^{\circ}\mid
L)\psi_{3}^{\ast}}+$ $\displaystyle\sum_{m}\frac{P(A=a^{\circ}\mid L)f(m\mid
A=a^{\dagger},L;\gamma^{\ast})}{f(m\mid
A=a^{\circ},L;\gamma^{\ast})}\\{\underbrace{b_{0}(m,L)-b_{0}^{\ast}(m,L)}_{=0}\\}f(m\mid
a^{\circ},L)+$ $\displaystyle\frac{\cancel{P(A=a^{\dagger}\mid
L)}P(A=a^{\circ}\mid L;\kappa^{\ast})}{\cancel{P(A=a^{\dagger}\mid
L;\kappa^{\ast})}}\left\\{\sum_{m}b_{0}^{\ast}(m,L)f(m\mid
a^{\dagger},L)-h_{\dagger}^{\ast}(L)\right\\}+$ $\displaystyle
P(A=a^{\circ}\mid
L)\\{h_{\dagger}^{\ast}(L)-\cancel{\psi_{3}^{\ast}}\\}-\Psi\Bigg{)}$
$\displaystyle=$ $\displaystyle E\Bigg{(}\sum_{m}P(A=a^{\dagger}\mid L)E(Y\mid
A=a^{\dagger},M=m,L)f(m\mid a^{\dagger},L)+$ $\displaystyle P(A=a^{\circ}\mid
L;\kappa^{\ast})\left\\{\sum_{m}b_{0}^{\ast}(m,L)f(m\mid
a^{\dagger},L)-\cancel{h_{\dagger}^{\ast}(L)}\right\\}+\cancel{P(A=a^{\circ}\mid
L)h_{\dagger}^{\ast}(L)}-\Psi\Bigg{)}$ $\displaystyle=$ $\displaystyle
P(A=a^{\dagger})E(Y\mid
A=a^{\dagger})+P(A=a^{\circ})\sum_{m,l}b_{0}(m,l)f(m\mid a^{\dagger},l)f(l\mid
a^{\circ})$ $\displaystyle=$ $\displaystyle 0$

∎

##  Appendix G Other relevant estimators

###  G.1. Inverse probability weighted estimator

We describe one class of inverse probability weighted estimator that was used
in the simulation and data analysis (see Fulcher et al.,, 2020 for other
inverse probability weighted estimators). Specifically, we can solve for
$\Psi_{IPW}$ in the following IPW estimator to estimate $\Psi$:

$\mathbb{P}_{n}\left[\frac{I(A=a^{\dagger})}{f(A\mid
L;\hat{\kappa})}\left\\{\sum_{a}{E}(Y\mid A=a,M,L;\hat{\theta})f(a\mid
L;\hat{\kappa})-\Psi_{IPW}\right\\}\right]=0$

where $\mathbb{P}_{n}(X)=n^{-1}\sum_{i=1}^{n}X_{i}$ and ${E}(Y\mid
A,M,L;\hat{\theta})$ is an estimate of ${E}(Y\mid A,M,L)$ such that ${E}(Y\mid
A=a^{\circ},M,L)=b_{0}(M,L)$.

###  G.2. Iterative conditional expectation estimator

The ICE estimator that was used in the simulation and data analysis follows
from the weighted ICE procedure, whereby we set $\hat{W}=1$ for all regression
steps.

###  G.3. Iterative TMLE

Herein, we describe an iterative TMLE algorithm that is doubly robust in the
sense of Fulcher et al., (2020) for binary mediators.

Algorithm 2 Algorithm for iterative Targeted maximum likelihood (generalized
frontdoor formula)

1:Non-parametrically compute $P(A=a^{\circ})$ and $P(A=a^{\dagger})$.

2:Obtain estimates $\hat{P}(A=a\mid L)$, $\hat{P}(A=a\mid M,L)$ and
$\hat{R}^{(0)}(A,L)\coloneqq\hat{P}(M=m\mid A,L)$ of $P(A=a\mid L)$,
$P(A=a\mid M,L)$ and $P(M=m\mid A,L)$, respectively, possibly using machine
learning methods.

3:In the individuals whose $A=a^{\circ}$, compute $\hat{Q}^{(0)}({M,L})$ by
regressing $Y$ on $(M,L)$. Here, $\hat{Q}^{(0)}({M,L})$ is possibly estimated
using machine learning methods, and it denotes an initial estimate for
$b_{0}(M,L)=E(Y\mid M,L,A=a^{\circ})$.

4:Define $t=0$. Iteratively update the following until convergence (i.e.,
until parameters $\delta^{(K)}\approx 0$ and $\nu^{(K)}\approx 0$):

  1. (A)

Update the previous regression. Specifically, in the individuals whose
$A=a^{\circ}$, fit a intercept-only regression model
$Q^{(t+1)}({M,L};\delta^{(t+1)})=g^{-1}[g\\{\hat{Q}^{(t)}({M,L})\\}+\delta^{(t+1)}]$
where the score function for each observation is weighted by
$\hat{W}_{1}^{(t)}=\frac{\hat{f}^{(t)}(M\mid A=a^{\dagger},L)}{\hat{f}(M\mid
A=a^{\circ},L)}$. More specifically, we solve for $\delta$ in the following
estimating equations:

$\mathbb{P}_{n}\left[I(A=a^{\circ})\hat{W}_{1}^{(t)}\left\\{Y-Q^{(t+1)}({M,L};\delta^{(t+1)})\right\\}\right]=0$

  2. (B)

Define
$\hat{Q}^{(t+1)}_{\text{diff}}=Q^{(t+1)}({m=1,L};\hat{\delta}^{(t+1)})-Q^{(t+1)}({m=0,L};\hat{\delta}^{(t+1)})$.
In those whose $A=a^{\dagger}$, fit a single covariate regression model (with
no intercept) for the conditional distribution of $M$ given by

$R^{(t+1)}(a^{\dagger},L;\nu^{(t+1)})=g^{-1[}\\{g\\{\hat{R}^{(t)}(a^{\dagger},L)\\}+\nu^{(t+1)}\hat{Q}^{(t+1)}_{\text{diff}}]$

with observational weights given by $\hat{W}_{2}=\frac{\hat{P}(A=a^{\circ}\mid
L)}{\hat{P}(A=a^{\dagger}\mid L)}$. More specifically, we solve for
$\nu^{(t+1)}$ in the following estimating equations:

$\mathbb{P}_{n}\left[I(A=a^{\dagger})\hat{W}_{2}\hat{Q}^{(t+1)}_{\text{diff}}\left\\{M-R^{(t+1)}(a^{\dagger},L;\nu^{(t+1)})\right\\}\right]=0$

  3. (C)

Update $\hat{Q}^{(t+1)}({M,L})\coloneqq Q^{(t+1)}({M,L};\hat{\delta}^{(t+1)})$
and $\hat{R}^{(t+1)}(a^{\dagger},L)\coloneqq
R^{(t+1)}(a^{\dagger},L;\hat{\nu}^{(t+1)})$.

  4. (D)

t+=1

5:Upon convergence at iteration $K$, define
$\hat{b}_{0}^{(K)}(m,L)=\hat{Q}^{(K)}({m,L})$ for $m=0,1$ and
$\hat{f}^{(K)}(M\mid A=a^{\dagger},L)=\hat{R}^{(K)}(a^{\dagger},L)$. In those
whose $A=a^{\circ}$, fit another regression model $T(\beta)=g^{-1}(\beta)$ for
$\psi_{3}=E\\{h_{\dagger}(L)\mid A=a^{\circ}\\}$ with just an intercept. More
specifically, we solve for $\beta$ in the following estimating equations:
$\mathbb{P}_{n}\left[I(A=a^{\circ})\left\\{\sum_{m=0}^{1}\hat{b}_{0}^{(K)}(m,L)\hat{f}^{(K)}(m\mid
a^{\dagger},L)-T(\beta)\right\\}\right]=0$

6:Compute $\hat{T}\coloneqq T(\hat{\beta})$ for all observations.

7:Estimate
$\hat{\Psi}_{iTMLE}=\mathbb{P}_{n}\\{I(A=a^{\dagger})Y+I(A=a^{\circ})\hat{T}\\}$

##  Appendix H Extensions to discrete exposure variables with more than two
levels

Extensions to discrete exposure variables with more than two levels is
straightforward. To see this, we can show that our estimand can be written as
follows:

$\Psi=P(A=a^{\dagger})E(Y\mid A=a^{\dagger})+\sum_{\forall a^{\circ}\neq
a^{\dagger}}P(A=a^{\circ})\sum_{m,l}E(Y\mid M=m,L=l,A=a^{\circ})f(m\mid
a^{\dagger},l)f(l\mid a^{\circ}).$

It then follows that in this extension, the efficient influence function for
$\Psi$ is given by: The efficient influence function
$\varphi^{\text{eff}}(\mathcal{O})$ for $A_{Y}=(L,A,M,Y)$ is given by

$\displaystyle\varphi^{\text{eff}}(\mathcal{O})$
$\displaystyle=I(A=a^{\dagger})Y+\sum_{\forall a^{\circ}\neq
a^{\dagger}}I(A=a^{\circ})\psi_{3}+$ $\displaystyle\sum_{\forall a^{\circ}\neq
a^{\dagger}}\Bigg{[}\frac{I(A=a^{\circ})P(A=a^{\dagger}\mid
M,L)P(A=a^{\circ}\mid L)}{P(A=a^{\circ}\mid M,L)P(A=a^{\dagger}\mid
L)}\\{Y-b_{0}(M,L)\\}+$ $\displaystyle\hskip
30.00005pt\frac{I(A=a^{\dagger})P(A=a^{\circ}\mid L)}{P(A=a^{\dagger}\mid
L)}\\{b_{0}(M,L)-h_{\dagger}(L)\\}+$ $\displaystyle\hskip
30.00005ptI(A=a^{\circ})\\{h_{\dagger}(L)-\psi_{3}\\}\Bigg{]}-\Psi,$

which can also be re-expressed as

$\displaystyle\varphi^{\text{eff}}(\mathcal{O})$
$\displaystyle=I(A=a^{\dagger})Y+\sum_{\forall a^{\circ}\neq
a^{\dagger}}I(A=a^{\circ})\psi_{3}+$ $\displaystyle\sum_{\forall a^{\circ}\neq
a^{\dagger}}\Bigg{[}\frac{I(A=a^{\circ})f(M\mid A=a^{\dagger},L)}{f(M\mid
A=a^{\circ},L)}\\{Y-b_{0}(M,L)\\}+$ $\displaystyle\hskip
30.00005pt\frac{I(A=a^{\dagger})P(A=a^{\circ}\mid L)}{P(A=a^{\dagger}\mid
L)}\\{b_{0}(M,L)-h_{\dagger}(L)\\}+$ $\displaystyle\hskip
30.00005ptI(A=a^{\circ})\\{h_{\dagger}(L)-\psi_{3}\\}\Bigg{]}-\Psi.$

The weighted estimator will still be sample-bounded, but will need to be
slightly modified in the following way:

Algorithm 3 Algorithm for Weighted ICE (generalized frontdoor formula for
discrete exposure with more than two levels)

1:Non-parametrically compute $P(A=a)$ for all values of $a\in\mathcal{A}$.

2:Compute the MLEs $\hat{\kappa}$ of ${\kappa}$ from the observed data for the
treatment model $P(A=a\mid L;\kappa)$. In addition, compute the MLEs
$\hat{\alpha}$ of ${\alpha}$ from the observed data for the treatment model
$P(A=a\mid M,L;\alpha)$, or compute the MLEs $\hat{\gamma}$ of $\gamma$ from
the observed data for the mediator model $P(M=m\mid A,L;\gamma)$

3:For all levels of $a^{\circ}\in\mathcal{A}$ that is not equal to
$a^{\dagger}$, do the following:

  1. (A)

In the individuals whose $A=a^{\circ}$, fit a regression model
$Q_{a^{\circ}}({M,L};\theta_{a^{\circ}})=g^{-1}\\{\theta_{a^{\circ}}^{T}\phi_{a^{\circ}}(M,L)\\}$
for $b_{0,a^{\circ}}(M,L)=E(Y\mid M,L,A=a^{\circ})$ where the score function
for each observation is weighted by $\hat{W}_{1,a^{\circ}}$ where
$\hat{W}_{1,a^{\circ}}1$ equals

$\frac{P(A=a^{\circ}\mid L;\hat{\kappa})P(A=a^{\dagger}\mid
M,L;\hat{\alpha})}{P(A=a^{\dagger}\mid L;\hat{\kappa})P(A=a^{\circ}\mid
M,L;\hat{\alpha})}$

if $\hat{\alpha}$ was estimated in the previous step, or
$\hat{W}_{1,a^{\circ}}$ equals

$\frac{f(M\mid A=a^{\dagger},L;\hat{\gamma})}{f(M\mid
A=a^{\circ},L;\hat{\gamma})}$

if $\hat{\gamma}$ was estimated in the previous step. Moreover,
$\phi_{a^{\circ}}(M,L)$ is a known function of $M$ and $L$. More specifically,
we solve for $\theta$ in the following estimating equations:

$\mathbb{P}_{n}\left[I(A=a^{\circ})\phi_{a^{\circ}}(M,L)\hat{W}_{1,a^{\circ}}\left\\{Y-Q_{a^{\circ}}({M,L};\theta_{a^{\circ}})\right\\}\right]=0$

  2. (B)

In those whose $A=a^{\dagger}$, fit a regression model
$R_{a^{\circ}}(L;\eta_{a^{\circ}})=g^{-1}\\{\eta_{a^{\circ}}^{T}\Gamma_{a^{\circ}}(L)\\}$
for $h_{\dagger}(L)=E(b_{a^{\circ}}(M,L)\mid L,A=a^{\dagger})$ where the score
function for each observation is weighted by

$\hat{W}_{2,a^{\circ}}=\frac{P(A=a^{\circ}\mid
L;\hat{\kappa})}{P(A=a^{\dagger}\mid L;\hat{\kappa})}.$

Here, $\gamma_{a^{\circ}}(L)$ is a known function of $L$. More specifically,
we solve for $\eta$ in the following estimating equations:

$\mathbb{P}_{n}\left[I(A=a^{\dagger})\Gamma_{a^{\circ}}(L)\hat{W}_{2,a^{\circ}}\left\\{Q_{a^{\circ}}({M,L};\hat{\theta}_{a^{\circ}})-R_{a^{\circ}}(L;\eta_{a^{\circ}})\right\\}\right]=0$

  3. (C)

In those whose $A=a^{\circ}$, fit another regression model
$T(\beta_{a^{\circ}})=g^{-1}(\beta_{a^{\circ}})$ for
$\psi_{3}=E\\{h_{\dagger,{a^{\circ}}}(L)\mid A=a^{\circ}\\}$ with just an
intercept. More specifically, we solve for $\beta_{a^{\circ}}$ in the
following estimating equations:

$\mathbb{P}_{n}\left[I(A=a^{\circ})\left\\{R_{a^{\circ}}({L};\hat{\eta}_{a^{\circ}})-T(\beta_{a^{\circ}})\right\\}\right]=0$

  4. (D)

Compute $\hat{T}_{a^{\circ}}\equiv T(\hat{\beta}_{a^{\circ}})$ for all
observations.

4:Estimate
$\hat{\Psi}_{WICE}=\mathbb{P}_{n}\left\\{I(A=a^{\dagger})Y+\sum_{\forall
a^{\circ}\neq a^{\dagger}}I(A=a^{\circ})\hat{T}_{a^{\circ}}\right\\}$

##  Appendix I Additional simulation study

The data-generating mechanism for our second simulation study and model
specifications are provided in Table 1. We consider four scenarios to
illustrate the robustness of our proposed estimator to model misspecification.
We consider four model specification scenarios: (1) all models are correctly
specified, (2) only the models for $b_{0}(M,L)$ and $h_{\dagger}(L)$ are
correctly specified, (3) only the models for $b_{0}(M,L)$ and $P(A=a\mid L)$
are correctly specified, and (4) only the models for $P(M=m\mid A,L)$ and
$P(A=a\mid L)$ are correctly specified. The correct mediator model in the
specification scenarios is the one used in the data generation process, and
the exposure and outcome models are approximately correctly specified by
including pairwise interactions between all the variables to ensure
flexibility.

Table 2 shows the results from the simulation study. As expected by our
theoretical derivations, when all of the working models are correctly
specified, all of the estimators are nearly unbiased. The AIPW estimator and
our proposed weighted ICE estimator are also nearly unbiased in the three
model misspecification settings whereas the IPW and ICE estimators are not all
unbiased.

Data generating mechanism
---
$U\sim$ |  $\text{Ber}\\{0.3\\}$
$L_{1}\sim$ |  $\text{Ber}\\{0.6\\}$
$L_{2}\sim$ |  $\text{Ber}\\{\operatorname*{expit}(1+4L_{1})\\}$
$A\sim$ |  $\text{Ber}\\{\operatorname*{expit}(-1+L_{1}+2L_{2}+5L_{1}L_{2}+U)\\}$
$M\sim$ |  $\text{Ber}\\{\operatorname*{expit}(1+A-3L_{1}+2L_{2}-5L_{1}L_{2})\\}$
$Y\sim$ |  $\text{Ber}\\{\operatorname*{expit}(-2.25+2A-5M-2AM+2L_{1}-2L_{2}-5L_{1}L_{2}+U)\\}$
Model misspecification
Scenario 2 |  $P(M=m\mid L,A;\gamma)=\operatorname*{expit}(\gamma_{0}+\gamma_{1}A+\gamma_{2}L_{2})$
|  $P(A=a\mid L;\kappa)=\operatorname*{expit}(\kappa_{0}+\kappa_{1}L_{2})$
Scenario 3 |  $P(M=m\mid L,A;\gamma)=\operatorname*{expit}(\gamma_{0}+\gamma_{1}A+\gamma_{2}L_{2})$
|  $R(L;\theta)=\operatorname*{expit}(\eta_{0}+\eta_{1}L_{2})$
Scenario 4 |  $Q(M,L;\theta)=\operatorname*{expit}(\theta_{0}+\theta_{1}M+\theta_{2}L_{1}+\theta_{3}L_{2})$
|  $R(L;\theta)=\operatorname*{expit}(\eta_{0}+\eta_{1}L_{2}(1-L_{1}))$
Table 1. Data generating mechanism and model misspecifications for scenarios in simulation study. | Scenario 1 | Scenario 2 | Scenario 3 | Scenario 4
---|---|---|---|---
| Bias | SE |  Biass | Bias | SE |  Biass | Bias | SE |  Biass | Bias | SE |  Biass
$n=100$
IPW | 0.05 | 1.16 | 4.24 | 0.80 | 2.13 | 37.41 | 0.05 | 1.16 | 4.24 | 3.40 | 1.64 | 208.00
ICE | 0.05 | 1.10 | 4.70 | 0.05 | 1.10 | 4.70 | 0.64 | 2.02 | 31.51 | 0.82 | 2.45 | 33.33
AIPW | 0.11 | 1.20 | 9.56 | - | - | - | 0.12 | 1.23 | 9.64 | 0.11 | 1.22 | 9.16
WICE | 0.06 | 1.11 | 5.06 | 0.05 | 1.10 | 4.76 | 0.00 | 1.08 | 0.00 | 0.01 | 1.13 | 1.18
$n=250$
IPW | -0.03 | 0.77 | -4.24 | 0.92 | 1.98 | 46.55 | -0.03 | 0.77 | -4.24 | 3.43 | 1.01 | 338.39
ICE | -0.01 | 0.70 | -1.28 | -0.01 | 0.70 | -1.28 | 0.81 | 1.93 | 42.10 | 1.09 | 2.61 | 41.87
AIPW | 0.00 | 0.70 | 0.31 | - | - | - | 0.00 | 0.73 | 0.64 | 0.00 | 0.70 | -0.11
WICE | -0.01 | 0.70 | -1.32 | -0.01 | 0.70 | -1.07 | -0.04 | 0.70 | -6.09 | -0.04 | 0.69 | -6.40
$n=500$
IPW | 0.01 | 0.93 | 1.17 | 1.16 | 2.19 | 52.69 | 0.01 | 0.93 | 1.17 | 3.45 | 0.73 | 471.59
ICE | -0.01 | 0.48 | -1.46 | -0.01 | 0.48 | -1.46 | 0.95 | 1.91 | 49.67 | 1.21 | 2.65 | 45.61
AIPW | 0.00 | 0.49 | 0.77 | - | - | - | 0.02 | 0.57 | 4.17 | 0.00 | 0.48 | -0.66
WICE | -0.01 | 0.48 | -1.21 | -0.01 | 0.48 | -1.20 | -0.02 | 0.49 | -4.29 | -0.02 | 0.49 | -4.80
Table 2. Simulation results: Bias, standard error (SE), and standardized bias
(Biass) are multiplied by 100.

##  Appendix J Asymptotic properties

In observational studies, model misspecification in the estimation of nuisance
functions can induce biased estimates of the ACE. In recent years, there has
been an explosion in developing flexible data-adaptive methods (e.g. kernel
smoothing, generalized additive models, ensemble learners, random forest)
combined with doubly robust estimators that can reduce the risk of model
misspecification and provide valid causal inference. These machine learning
techniques offer more protection against model misspecficiation than the
parametric models.

From first order expansion of a singly-robust plug-in estimator (IPW and ICE
estimators), it can be shown that we require the nuisance parameter estimators
to converge to the truth at rate $n^{-1/2}$. However, this is not possible for
non-parametric conditional mean functions as this rate is not attainable for
these types of functions. However when doubly robust estimators are used with
data-adaptive methods this issue largely disappears are doubly robust
estimators enjoy the small bias property (Newey et al.,, 2004).

In this section we will examine the Remainder or Bias term from the following
decomposition. For notational brevity, we suppress $\mathcal{O}$ in the
equations below. For generality, suppose that $\Psi(\hat{P})$ is an estimator
that solves the estimating equations based on the efficient influence
function. We have that

$\displaystyle\sqrt{n}(\Psi(\hat{P})-\Psi(P))$
$\displaystyle=\sqrt{n}\left[\mathbb{P}_{n}(\varphi^{eff}(\hat{P}))-P(\varphi^{eff}(\hat{P}))\right]+\sqrt{n}\left[\Psi(\hat{P})+P(\varphi^{eff}(\hat{P}))-\Psi(P)\right]$
$\displaystyle=\mathbb{G}_{n}({\varphi}^{eff}({P}))+\mathbb{G}_{n}[{\varphi}^{eff}(\hat{P})-{\varphi}^{eff}({P})]+$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\
\sqrt{n}\left[\Psi(\hat{P})+P({\varphi}^{eff}(\hat{P}))-\Psi(P)\right]$
$\displaystyle=\underbrace{\mathbb{G}_{n}({\varphi}({P}))}_{T_{1}}+\underbrace{\mathbb{G}_{n}[{\varphi}(\hat{P})-{\varphi}({P})]}_{T_{2}}+$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\
\sqrt{n}\left[\underbrace{\Psi(\hat{P})+P({\varphi}^{eff}(\hat{P}))-\Psi(P)}_{R}\right]$

where $\mathbb{G}_{n}[X]=\sqrt{n}(\mathbb{P}_{n}-P)(X)$ for any $X$ and we
define
${\varphi}(\mathcal{O};\tilde{P})={\varphi}^{eff}(\mathcal{O};\tilde{P})+\Psi(O;\tilde{P})$
for any $\tilde{P}$. The first term given by $T_{1}$ is a centered sample
average which converges to a mean zero Normal distribution by the central
limit theorem. The second term is known as an empirical process term, which
can be shown to be $o_{p}(1)$ if we assume that nuisance functions and their
corresponding estimators are not too complex and belong to Donsker class.
Alternatively, one can use sample splitting and cross fitting to overcome
issues with overfitting (Chernozhukov et al.,, 2018).

Formally, we assume the following conditions for the first two terms:

  1. C1.

$E\left[\varphi^{\text{eff}}(\mathcal{O})^{2}\right]<\infty$

  2. C2.

$\varphi^{\text{eff}}(\mathcal{O})$ and
$\hat{\varphi}^{\text{eff}}(\mathcal{O})$ belong to a Donsker family.

  3. C3.

$\lVert\hat{\varphi}^{\text{eff}}(\mathcal{O})-\varphi^{\text{eff}}(\mathcal{O})\rVert_{2}^{2}\overset{p}{\longrightarrow}0$

where $\hat{\varphi}^{\text{eff}}(\mathcal{O})$ denotes an estimator of
$\varphi^{\text{eff}}(\mathcal{O})$ where all nuisance functions estimators
are exactly the same as those in the TMLE estimator. It is not hard to show
that $\mathbb{P}_{n}\\{\hat{\varphi}^{\text{eff}}(\mathcal{O})\\}=0$ by
construction.

The last term is known as the remainder or bias term. We will need to show
that $R=o_{p}(1)$ under some conditions about the convergence rates of the
nuisance functions.

$\displaystyle\Psi(\hat{P})+P({\varphi}^{eff}(\hat{P}))-\Psi(P)=$
$\displaystyle
E_{P}\left[\underbrace{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\frac{I(A=a^{\circ})\hat{f}(M\mid
a^{\dagger},L)}{\hat{f}(M\mid
a^{\circ},L)}(Y-\hat{b}_{0}(M,L))}}_{(A)}+\underbrace{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\frac{I(A=a^{\dagger})\hat{f}(a^{\circ}\mid
L)}{\hat{f}(a^{\dagger}\mid
L)}(\hat{b}_{0}(M,L)-\hat{h}_{\dagger}(L))+I(A=a^{\circ})\hat{h}_{\dagger}(L)-P(A=a^{\circ})\psi_{3}}}_{(B)}\right]$

We examine the terms in blue (henceforth denoted as (A)) and term in red
(henceforth denoted as (B)) in detail. Starting with the term in (B):

$\displaystyle(B)$
$\displaystyle=E_{P}\left[I(A=a^{\circ})(\hat{h}_{\dagger}(L)-h_{\dagger}(L))+\frac{I(A=a^{\dagger})\hat{f}(a^{\circ}\mid
L)}{\hat{f}(a^{\dagger}\mid L)}(\hat{b}_{0}(M,L)-\hat{h}_{\dagger}(L))\right]$
$\displaystyle=E_{P}\left[I(A=a^{\circ})(\hat{h}_{\dagger}(L)-h_{\dagger}(L))+\frac{I(A=a^{\dagger})\hat{f}(a^{\circ}\mid
L)}{\hat{f}(a^{\dagger}\mid L)}\left\\{E_{P}(\hat{b}_{0}(M,L)\mid
A=a^{\dagger},L)-\hat{h}_{\dagger}(L)\right\\}\right]$
$\displaystyle=E_{P}\Bigg{[}f(a^{\circ}\mid
L))(\hat{h}_{\dagger}(L)-h_{\dagger}(L))+\frac{f(a^{\dagger}\mid
L)\hat{f}(a^{\circ}\mid L)}{\hat{f}(a^{\dagger}\mid
L)}\left\\{E_{P}(\hat{b}_{0}(M,L)\mid
A=a^{\dagger},L)-\hat{h}_{\dagger}(L)\right\\}+$ $\displaystyle\hskip
27.00005ptf(a^{\circ}\mid L)\left\\{E_{P}(\hat{b}_{0}(M,L)\mid
A=a^{\dagger},L)-E_{P}(\hat{b}_{0}(M,L)\mid A=a^{\dagger},L)\right\\}\Bigg{]}$
$\displaystyle=E_{P}\left[\left\\{E_{P}(\hat{b}_{0}(M,L)\mid
A=a^{\dagger},L)-\hat{h}_{\dagger}(L)\right\\}\left\\{\frac{f(a^{\dagger}\mid
L)\hat{f}(a^{\circ}\mid L)}{\hat{f}(a^{\dagger}\mid L)}-f(a^{\circ}\mid
L)\right\\}\right]+$ $\displaystyle\hskip
27.00005ptE_{P}\left\\{E_{P}\Big{(}\hat{b}_{0}(M,L)-b_{0}(M,L)\mid
A=a^{\dagger},L\Big{)}I(A=a^{\circ})\right\\}$
$\displaystyle=E_{p}\left[\left\\{E_{P}(\hat{b}_{0}(M,L)\mid
A=a^{\dagger},L)-\hat{h}_{\dagger}(L)+h_{\dagger}(L)-h_{\dagger}(L)\right\\}\left\\{\hat{f}(a^{\circ}\mid
L)-f(a^{\circ}\mid L)\right\\}\frac{1}{\hat{f}(a^{\dagger}\mid L)}\right]$
$\displaystyle\hskip
27.00005ptE_{P}\left\\{E_{P}\Big{(}\hat{b}_{0}(M,L)-b_{0}(M,L)\mid
A=a^{\dagger},L\Big{)}I(A=a^{\circ})\right\\}$
$\displaystyle=E_{P}\left[\\{h_{\dagger}(L)-\hat{h}_{\dagger}(L)\\}\left\\{\hat{f}(a^{\circ}\mid
L)-f(a^{\circ}\mid L)\right\\}\frac{1}{\hat{f}(a^{\dagger}\mid L)}\right]+$
$\displaystyle\hskip
27.00005ptE_{P}\left[E_{P}(\hat{b}_{0}(M,L)-b_{0}(M,L)\mid
A=a^{\dagger},L)\left\\{\hat{f}(a^{\circ}\mid L)-f(a^{\circ}\mid
L)\right\\}\frac{1}{\hat{f}(a^{\dagger}\mid L)}\right]+$ $\displaystyle\hskip
27.00005pt\underbrace{{\color[rgb]{.75,0,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,0,.25}E_{P}\left\\{E_{P}\Big{(}\hat{b}_{0}(M,L)-b_{0}(M,L)\mid
A=a^{\dagger},L\Big{)}I(A=a^{\circ})\right\\}}}_{(B.2)}$

We keep in mind the term in purple (term (B.2)) as we expand upon term (A):

$\displaystyle(A)$ $\displaystyle=E_{P}\left[\frac{I(A=a^{\circ})\hat{f}(M\mid
a^{\dagger},L)}{\hat{f}(M\mid a^{\circ},L)}(Y-\hat{b}_{0}(M,L))\right]$
$\displaystyle=E_{P}\left[\frac{I(A=a^{\circ})\hat{f}(M\mid
a^{\dagger},L)}{\hat{f}(M\mid
a^{\circ},L)}\Big{(}b_{0}(M,L)-\hat{b}_{0}(M,L)\Big{)}\right]$
$\displaystyle=E_{P}\left(I(A=a^{\circ})E_{P}\left[\frac{\hat{f}(M\mid
a^{\dagger},L)f(M\mid a^{\circ},L)}{\hat{f}(M\mid a^{\circ},L)f(M\mid
a^{\dagger},L)}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right)$

Now, adding (A) and (B.2) (term in purple) we get the following:

$\displaystyle E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{\hat{f}(M\mid
a^{\dagger},L)f(M\mid a^{\circ},L)}{\hat{f}(M\mid a^{\circ},L)f(M\mid
a^{\dagger},L)}-1\right\\}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right)$
$\displaystyle=E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{f(M\mid
a^{\circ},L)}{f(M\mid a^{\dagger},L)}\frac{1}{\hat{f}(M\mid
a^{\circ},L)}\right\\}\left\\{\hat{f}(M\mid a^{\dagger},L)-f(M\mid
a^{\dagger},L)\right\\}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right)+$ $\displaystyle\hskip
9.00002ptE_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{1}{\hat{f}(M\mid
a^{\circ},L)}\right\\}\left\\{f(M\mid a^{\circ},L)-\hat{f}(M\mid
a^{\circ},L)\right\\}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right)$

Thus, together we have (A)+(B) equals:

$\displaystyle
E_{P}\left[\\{h_{\dagger}(L)-\hat{h}_{\dagger}(L)\\}\left\\{\hat{f}(a^{\circ}\mid
L)-f(a^{\circ}\mid L)\right\\}\frac{1}{\hat{f}(a^{\dagger}\mid L)}\right]+$
$\displaystyle E_{P}\left[E_{P}(\hat{b}_{0}(M,L)-b_{0}(M,L)\mid
A=a^{\dagger},L)\left\\{\hat{f}(a^{\circ}\mid L)-f(a^{\circ}\mid
L)\right\\}\frac{1}{\hat{f}(a^{\dagger}\mid L)}\right]+$ $\displaystyle
E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{f(M\mid
a^{\circ},L)}{f(M\mid a^{\dagger},L)}\frac{1}{\hat{f}(M\mid
a^{\circ},L)}\right\\}\left\\{\hat{f}(M\mid a^{\dagger},L)-f(M\mid
a^{\dagger},L)\right\\}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right)+$ $\displaystyle
E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{1}{\hat{f}(M\mid
a^{\circ},L)}\right\\}\left\\{f(M\mid a^{\circ},L)-\hat{f}(M\mid
a^{\circ},L)\right\\}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right)$

By an application of Cauchy-Schwartz, we can show that as long as:

  1. (1)

$\lVert\hat{h}_{\dagger}(L)-h_{\dagger}(L)\rVert\leavevmode\nobreak\
\lVert\hat{f}(a^{\circ}\mid L)-f(a^{\circ}\mid L)\rVert=O_{p}(n^{-\nu})$, and

  2. (2)

$\lVert\hat{b}_{0}(M,L)-b_{0}(M,L)\rVert\leavevmode\nobreak\
\lVert\hat{f}(a^{\circ}\mid L)-f(a^{\circ}\mid L)\rVert=O_{p}(n^{-\nu})$, and

  3. (3)

$\lVert\hat{b}_{0}(M,L)-b_{0}(M,L)\rVert\leavevmode\nobreak\
\lVert\hat{f}(M\mid a,L)-f(M\mid
a,L)\rVert=O_{p}(n^{-\nu}),\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
a$

for $\nu>1/2$ and where $\lVert
f(x)\rVert=\left\\{\int|f(x)|^{2}dP(x)\right\\}^{1/2}$, i.e. the $L_{2}(P)$
norm. Then,
$\sqrt{n}\Big{\\{}\Psi(\hat{P})+P({\varphi}^{eff}(\hat{P}))-\Psi(P)\Big{\\}}=o_{p}(1)$.
This can be accomplished, for example, if the nuisance functions are each
consistently estimated at a rate of $n^{-1/4}$ or faster.

Note that $h_{\dagger}(L)=\sum_{m}b_{0}(m,L)f(m\mid a^{\dagger},L)$. In our
estimators, we propose estimating $h_{\dagger}(L)$ by regressing $b_{0}(M,L)$
on $L$ in those whose $A=a^{\dagger}$ to ensure sample-boundedness. However,
if we estimate is $h_{\dagger}(L)$ by calculating
$\sum_{m}\hat{b}_{0}(m,L)\hat{f}(m\mid a^{\dagger},L)$ explicitly (as in
Fulcher et al.,, 2020). Under the AIPW estimator of Fulcher et al., (2020)
(and the iterative TMLE), the remainder term reduces to the following:

$\displaystyle E_{P}\left[\sum_{m}\hat{b}_{0}(m,L)\left\\{f(m\mid
A=a^{\dagger},L)-\hat{f}(m\mid
A=a^{\dagger},L)\right\\}\left\\{\hat{f}(a^{\circ}\mid L)-f(a^{\circ}\mid
L)\right\\}\frac{1}{\hat{f}(a^{\dagger}\mid L)}\right]+$ $\displaystyle
E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{f(M\mid
a^{\circ},L)}{f(M\mid a^{\dagger},L)}\frac{1}{\hat{f}(M\mid
a^{\circ},L)}\right\\}\left\\{\hat{f}(M\mid a^{\dagger},L)-f(M\mid
a^{\dagger},L)\right\\}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right)+$ $\displaystyle
E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{1}{\hat{f}(M\mid
a^{\circ},L)}\right\\}\left\\{f(M\mid a^{\circ},L)-\hat{f}(M\mid
a^{\circ},L)\right\\}\left\\{b_{0}(M,L)-\hat{b}_{0}(M,L)\right\\}\mid
A=a^{\dagger},L\right]\right).$

Then we can show that if the model for $f(M\mid A,L)$ is correctly specified,
then the remainder term reduces to the following asymptotically888Note that
this is really only possible if $M$ was discrete, otherwise we need to resort
to using numerical integration.:

$\displaystyle E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{f(M\mid
a^{\circ},L)}{f(M\mid a^{\dagger},L)}\frac{1}{f^{\ast}(M\mid
a^{\circ},L)}\right\\}\left\\{f^{\ast}(M\mid a^{\dagger},L)-f(M\mid
a^{\dagger},L)\right\\}\left\\{b_{0}(M,L)-{b}^{\ast}_{0}(M,L)\mid
A=a^{\dagger},L\right\\}\right]\right)+$ $\displaystyle
E_{P}\left(I(A=a^{\circ})E_{P}\left[\left\\{\frac{1}{f^{\ast}(M\mid
a^{\circ},L)}\right\\}\left\\{f(M\mid a^{\circ},L)-f^{\ast}(M\mid
a^{\circ},L)\right\\}\left\\{b_{0}(M,L)-{b}^{\ast}_{0}(M,L)\mid
A=a^{\dagger},L\right\\}\right]\right)+o_{p}(1)$

where $f^{\ast}(M\mid A,L)$ and ${b}^{\ast}_{0}(M,L)$ denote the limiting
values of $\hat{f}(M\mid A,L)$ and $\hat{b}_{0}(M,L)$. This gives intuition to
why the augmented inverse probability weighted estimator proposed in Fulcher
et al., (2020) is consistent when models for $b_{0}(M,L)$ and $P(A=a\mid L)$
are correctly specified, or when the model for $P(M=m\mid A,L)$ is correctly
specified.

Similarly, it can also be shown that as long as:

  1. (1)

$\lVert\hat{h}_{\dagger}(L)-h_{\dagger}(L)\rVert\leavevmode\nobreak\
\lVert\hat{f}(a^{\circ}\mid L)-f(a^{\circ}\mid L)\rVert=O_{p}(n^{-\nu})$, and

  2. (2)

$\lVert\hat{b}_{0}(M,L)-b_{0}(M,L)\rVert\leavevmode\nobreak\
\lVert\hat{f}(a\mid L)-f(a\mid L)\rVert=O_{p}(n^{-\nu}),\leavevmode\nobreak\
\leavevmode\nobreak\ \forall a$, and

  3. (3)

$\lVert\hat{b}_{0}(M,L)-b_{0}(M,L)\rVert\leavevmode\nobreak\
\lVert\hat{f}(a\mid M,L)-f(a\mid
M,L)\rVert=O_{p}(n^{-\nu}),\leavevmode\nobreak\ \leavevmode\nobreak\ \forall
a$

for $\nu>1/2$ and where $\lVert
f(x)\rVert=\left\\{\int|f(x)|^{2}dP(x)\right\\}^{1/2}$, i.e. the $L_{2}(P)$
norm. Then,
$\sqrt{n}\Big{\\{}\Psi(\hat{P})+P({\varphi}^{eff}(\hat{P}))-\Psi(P)\Big{\\}}=o_{p}(1)$.
This can be accomplished, for example, if the nuisance functions are each
consistently estimated at a rate of $n^{-1/4}$ or faster.

## References

  * Chernozhukov et al., (2018) Chernozhukov, V., Chetverikov, D., Demirer, M., Duflo, E., Hansen, C., Newey, W., and Robins, J. (2018).  Double/debiased machine learning for treatment and structural parameters.  The Econometrics Journal, 21:C1–C68.
  * Fulcher et al., (2020) Fulcher, I. R., Shpitser, I., Marealle, S., and Tchetgen Tchetgen, E. J. (2020).  Robust inference on population indirect causal effects: the generalized front door criterion.  Journal of the Royal Statistical Society: Series B (Statistical Methodology), 82:199–214.
  * Newey, (1994) Newey, W. K. (1994).  The asymptotic variance of semiparametric estimators.  Econometrica: Journal of the Econometric Society, pages 1349–1382.
  * Newey et al., (2004) Newey, W. K., Hsieh, F., and Robins, J. M. (2004).  Twicing kernels and a small bias property of semiparametric estimators.  Econometrica, 72:947–962.
  * Robins and Richardson, (2010) Robins, J. M. and Richardson, T. S. (2010).  Alternative graphical causal models and the identification of direct effects.  Causality and psychopathology: Finding the determinants of disorders and their cures, 84:103–158.
  * Stensrud et al., (2021) Stensrud, M. J., Hernán, M. A., Tchetgen Tchetgen, E. J., Robins, J. M., Didelez, V., and Young, J. G. (2021).  A generalized theory of separable effects in competing event settings.  Lifetime data analysis, 27:588–631.
  * Tchetgen and Shpitser, (2012) Tchetgen, E. J. T. and Shpitser, I. (2012).  Semiparametric theory for causal mediation analysis: efficiency bounds, multiple robustness, and sensitivity analysis.  Annals of statistics, 40(3):1816.
  * Van Der Vaart, (2000) Van Der Vaart, A. W. (2000).  Asymptotic statistics, volume 3.  Cambridge university press.