Write
$$
m(x_1,x_2,x_3)=\mathbb E[X_4\mid x_1,x_2,x_3].
$$
Since $X_1$ has no parents, its total effect has no <backdoor path>, and
$$
\tau_1
=\mathbb E[X_4\mid X_1=1]
-\mathbb E[X_4\mid X_1=0].
$$

For $X_2$, $X_3$ intercepts every directed path to $X_4$. Conditional on $X_1$, there is no unblocked backdoor path from $X_2$ to $X_3$, and $(X_1,X_2)$ blocks every backdoor path from $X_3$ to $X_4$. The <conditional front-door adjustment> therefore gives
$$
\mu_2(x)=
\sum_{x_1,x_3}p(x_1)p(x_3\mid x_1,x)
\sum_{x_2'}m(x_1,x_2',x_3)p(x_2'\mid x_1),
$$
and
$$
\tau_2=\mu_2(1)-\mu_2(0).
$$
Part ii shows that the inner sum, after also averaging $X_4$, does not actually depend on $x_1$.

For $X_3$, $(X_1,X_2)$ is a sufficient <backdoor adjustment set>. Hence
$$
\mu_3(x)=
\sum_{x_1,x_2}m(x_1,x_2,x)
p(x_2\mid x_1)p(x_1),
\qquad
\tau_3=\mu_3(1)-\mu_3(0).
$$
These three formulas identify the requested <average treatment effects> from the observed joint distribution.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 221 / 1 / iii / Solution

Conditional front-door adjustment applies the front-door argument within strata of observed covariates and then averages over those covariates. The covariates must remove exposure-mediator confounding, while the exposure together with the covariates must remove mediator-outcome confounding.


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 Incoming links: Conditional front-door adjustment