We thank the reviewer for the detailed questions regarding this study.
We apologize for the delay in responding to your comment.
> The Equation (3.14) in Section 3.2.3 is very different from the definition in Eq. (1) of the paper. Could you write down both equations in the same notation and explain the correspondence precisely?
Herein, to avoid confusion, we use only the notations used in [15] for the following discussions, except for using the bold font type to write sets of variables and their values. In particular, $D\_{A}$ denotes the domain of $A$, which is the same as $\mathcal{X}_A$ in us.
First, please note that (3.14) in [15] defines the causal effects determined by the following $do$ operation:
$$
\begin{aligned}
&P(x_1,\ldots,x_n|\hat{\mathbf{s}}) \\\\
&= P(X_1=x_1,\ldots,X_n=x_n|do(\mathbf{S}=\mathbf{s})).
\end{aligned}
$$
Let $\mathbf{PA}\_{\mathbf{S}}= \bigcup\_{X_i \in \mathbf{S}} PA_i$. From Therom 3.2.2 in [15], we have
$$
\begin{aligned}
& P(x_1,\ldots,x_n|\hat{\mathbf{s}})\\\\
&= \sum\_{\mathbf{pa}\_{\mathbf{s}} \in D\_{\mathbf{PA}\_{\mathbf{S}}}} P(x_1,\ldots,x_n|\mathbf{s}, \mathbf{pa}\_{\mathbf{s}}) \cdot P(\mathbf{pa}\_{\mathbf{s}})\\\\
&= \sum\_{\mathbf{pa}\_{\mathbf{s}} \in D\_{\mathbf{PA}\_{\mathbf{S}}}}
\frac{P(x_1,\ldots,x_n, \mathbf{s}, \mathbf{pa}\_{\mathbf{s}})}{P(\mathbf{s} |\mathbf{pa}\_{\mathbf{s}})}.
\end{aligned}
$$
Here, the last equality is obtained using the following equation:
$$
\begin{aligned}
&P(x_1,\ldots,x_n|\mathbf{s}, \mathbf{pa}\_{\mathbf{s}}) \cdot P(\mathbf{pa}\_{\mathbf{s}})
=\frac{P(x_1,\ldots,x_n,\mathbf{s}, \mathbf{pa}\_{\mathbf{s}})}{P(\mathbf{s} |\mathbf{pa}\_{\mathbf{s}})}.
\end{aligned}
$$
Now, we note that our definition considers the general case in which other variables from $\mathbf{X}={X_1, X_2, \ldots, X_n}$ exist.
Then, we use the notation $\mathbf{V}$ for all the observed random variables: $\mathbf{X} \subset \mathbf{V}$, and write $\mathbf{V}'$ for $\mathbf{V} \setminus (\mathbf{X} \cup \mathbf{PA}\_{\mathbf{S}} \cup \mathbf{S})$.
Subsequently,
$$
\begin{aligned}
&P(X_1=x_1,\ldots,X_n=x_n,
\mathbf{V'}=\mathbf{v}'|do(\mathbf{S}=\mathbf{s}))\\\\
&=P(x_1,\ldots,x_n, \mathbf{V'}=\mathbf{v}'|\hat{\mathbf{s}}) \\\\
&=\sum\_{\mathbf{pa}\_{\mathbf{s}} \in D\_{\mathbf{PA}\_{\mathbf{S}}}}
\frac{P(x_1,\ldots,x_n, \mathbf{V'}=\mathbf{v}'|\mathbf{s}, \mathbf{pa}\_{\mathbf{s}})}{P(\mathbf{s} |\mathbf{pa}\_{\mathbf{s}})}.
\end{aligned}
$$
By marginalizing $\mathbf{V}'$, we obtain
$$
\begin{aligned}
&P(X_1=x_1,\ldots,X_n=x_n|do(\mathbf{S}=\mathbf{s}))\\\\
&=\sum\_{\mathbf{v'} \in D\_{\mathbf{V}}}
P(X_1=x_1,\ldots,X_n=x_n,
\mathbf{V'}=\mathbf{v}'|do(\mathbf{S}=\mathbf{s}))\\\\
&=\sum\_{\mathbf{v'} \in D\_{\mathbf{V}}} \sum\_{\mathbf{pa}\_{\mathbf{s}} \in D\_{\mathbf{PA}\_{\mathbf{S}}}} \frac{P(x_1,\ldots,x_n, \mathbf{V'}=\mathbf{v}'|\mathbf{s}, \mathbf{pa}\_{\mathbf{s}})}{P(\mathbf{s} |\mathbf{pa}\_{\mathbf{s}})}.
\end{aligned}
$$
From this, we see that (1) in the present study coincides with (3.14) in [15].
($\{X_1, X_2, \ldots, X_n\}$ in the above equality corresponds to $\mathbf{Y}$ in (1), and $\mathbf{S}$ to $\mathbf{X}$.)