causal effect in linear causal model I have the following linear causal model:
$B = \epsilon_B$
$C = \epsilon_C$
$A = \beta_1 B + \epsilon_A$
$Z = \beta_2 B + \beta_3 C + \epsilon_Z$
$D = \beta_4 C + \epsilon_D$
$X = \beta_5 A + \beta_6 Z + \epsilon_X$
$W = \beta_7 X + \epsilon_W$
$Y = \beta_8 W + \beta_9 Z + \beta_{10} D + e_Y$
all structural errors are independent each others. In terms of DAG the causal model is:

My questions are, how amount the following quantities?
$E[Y|do(X),B]$ (b-specific effect of $X$ on $Y$
$E[Y|do(X),C]$ (c-specific effect of $X$ on $Y$
$E[Y|do(X),Z]$ (z-specific effect of $X$ on $Y$
$E[Y|do(X),do(B)]$ (combined effect of $X$ and $B$ on $Y$)
$E[Y|do(X),do(C)]$ (combined effect of $X$ and $C$ on $Y$)
$E[Y|do(X),do(Z)]$ (combined effect of $X$ and $Z$ on $Y$)
Them should be expressed with structural coefficients, if any. Moreover should be explicitated the linear regressions needed for identified those effects/parameters, then the relations among structural and regression parameters involved.
 A: $\newcommand{\eps}{\varepsilon}\newcommand{\doop}{\operatorname{do}}$My other answer has some value, I think, but rather than edit it, I think it might be better to simply add another answer. Let's take the Structural Equation Model (SEM):
\begin{align*}
B&=\eps_B\\
C&=\eps_C\\
A&=\beta_1B+\eps_A\\
Z&=\beta_2B+\beta_3C+\eps_Z\\
D&=\beta_4C+\eps_D\\
X&=\beta_5A+\beta_6Z+\eps_Z\\
W&=\beta_7X+\eps_W\\
Y&=\beta_8W+\beta_9Z+\beta_{10}D+\eps_Y.
\end{align*}
We take the first computation: $E[Y|\doop(X=x),B=b].$ As mentioned in my other answer, a sufficient set that satisfies the backdoor criterion in this case is $\{Z\},$ so that we must compute
$E[Y|\doop(X=x),B=b,Z=z].$ The $\doop$ operator means that we replace the structural equation for $X$ with $X=x$ and substitute in everywhere:
\begin{align*}
B&=\eps_B\\
C&=\eps_C\\
A&=\beta_1B+\eps_A\\
Z&=\beta_2B+\beta_3C+\eps_Z\\
D&=\beta_4C+\eps_D\\
X&=x\\
W&=\beta_7x+\eps_W\\
Y&=\beta_8W+\beta_9Z+\beta_{10}D+\eps_Y.
\end{align*}
With a conditional like $B=b$ or $Z=z,$ while we replace the variable, we don't delete the equation. So, to compute:
\begin{align*}
E[Y|\doop(X=x),B=b,Z=z]
&=\beta_8E[W|\doop(X=x),B=b,Z=z]\\
&\quad+\beta_9E[Z|\doop(X=x),B=b,Z=z]\\
&\quad+\beta_{10}E[D|\doop(X=x),B=b,Z=z]+0\\
&=\beta_8\beta_7x+\beta_9z+\beta_{10}E[D|\doop(X=x),B=b,Z=z].
\end{align*}
Now this last expression is what we must work on. For $Z$ to equal $z,$ (and since this is not the $\doop,$ we do not delete the equation for $Z$) we must have
\begin{align*}
z&=\beta_2b+\beta_3C\\
C&=\frac{z-\beta_2b}{\beta_3},
\end{align*}
which forces
$$E[D]=\beta_4\cdot \frac{z-\beta_2b}{\beta_3}.$$
Hence
$$E[Y|\doop(X=x),B=b,Z=z]=\beta_8\beta_7x+\beta_9z+\beta_{10}\beta_4\, \frac{z-\beta_2b}{\beta_3}.$$
For the $c$-specific effect, the computations will be similar. For the $z$-specific effect, one "gotcha" is that you will need to condition on either $A,B,C,$ or $D$ to block the backdoor path $X\leftarrow A\leftarrow B\to Z\leftarrow C\to D\to Y,$ which is opened up because of conditioning on $Z.$
For a combined effect computation, the work is actually easier in some ways. Let's take $E[Y|\doop(X=x),\doop(B=b)].$ The difference between this and the conditioned version $E[Y|\doop(X=x),B=b]$ is that in the $\doop$ version, we replace $B=\eps_B$ with $B=b$ and substitute $b$ for $B$ everywhere in the SEM, thus:
\begin{align*}
B&=b\\
C&=\eps_C\\
A&=\beta_1b+\eps_A\\
Z&=\beta_2b+\beta_3C+\eps_Z\\
D&=\beta_4C+\eps_D\\
X&=x\\
W&=\beta_7x+\eps_W\\
Y&=\beta_8W+\beta_9Z+\beta_{10}D+\eps_Y.
\end{align*}
The result should be
$$E[Y|\doop(X=x),\doop(B=b)]=\beta_8\beta_7x+\beta_9\beta_2b.$$
A: Partial Solution:
According to Rule 2 on page 70 of Causal Inference in Statistics: A Primer, by Pearl, Glymour, and Jewell:

The $z$-specific effect $P(Y=y|\operatorname{do}(X=x), Z=z)$ is identified whenever we can measure a set $S$ of variables such that $S\cup Z$ satisfies the backdoor criterion. Moreover, the $z$-specific effect is given by the following adjustment formula
\begin{align*}
&\quad P(Y=y|\operatorname{do}(X=x),Z=z)\\
&=\sum_s P(Y=y|X=x,S=s,Z=z)\,P(S=s|Z=z).
\end{align*}
This modified adjustment formula is similar to Eq. (3.5) with two exceptions. First, the adjustment set is $S\cup Z,$ not just $S,$ and, second, the summation goes only over $S,$ not including $Z.$

Once you have the probability distribution computed, you can calculate the expected value with the usual formula:
$$E[Y=y|\operatorname{do}(X=x),Z=z]=\sum_y y\cdot P(Y=y|\operatorname{do}(X=x),Z=z).$$
So here is a worked-out example: your first quantity. We must first find
$$P(Y=y|\operatorname{do}(X=x),B=b).$$
The adjustment set $S$ must be one of the following:
\begin{align*}
S&=\{Z\}\\
S&=\{Z,C\}\\
S&=\{Z,D\}\\
S&=\{Z,C,D\}.
\end{align*}
We do not need to worry about the backdoor path starting with $A,$ because conditioning on $B$ already blocks it. That leaves the backdoor path beginning with $Z.$ You must condition on $Z,$ else the backdoor path $X\leftarrow Z\to Y$ is open. The reason just $\{Z\}$ can work is that $B$ is already conditioned on, blocking the collider that opens up at $Z$ when conditioning on it. Hence the probability distribution you can write as
$$P(Y=y|\operatorname{do}(X=x),B=b)=
\sum_z P(Y=y|X=x,B=b,Z=z)\,P(Z=z|B=b).$$
The expectation you would write in a $\operatorname{do}$-free fashion as
$$E[Y=y|\operatorname{do}(X=x),B=b]=
\sum_y y\cdot \sum_z P(Y=y|X=x,B=b,Z=z)\,P(Z=z|B=b).$$
The other cases can be worked out similarly. The reason I am calling this a partial solution is that I do not know how to express these probabilities in terms of the edge weights $\beta_i.$ But perhaps this partial solution will help. I will say this: if $Y\in\{0,1\},$ then you can dispense with the $y$-summation in the final expression.
