causal inference exercise - “covariate-specific effect”

This graph and questions come from: CAUSAL INFERENCE IN STATISTICS A Primer - Pearl Glymour and Jewell (2016).

We are interested in the effect of $$X$$ on $$Y$$. In order to identify it we looking for the sets compliant with the backdoor criterion, for example $$[A Z]$$ and $$[A Z C]$$ seems me among them. For simplicity we assume that the SCM is linear and the path coefficients are named: $$\beta_{y,w}$$, $$\beta_{y,z}$$, …, $$\beta_{z,c}$$, … and so on.

So the causal effect of $$X$$ on $$Y$$ is: $$\beta_{y,w} \beta_{w,x}$$

I ask you If my solutions for the study question above (a bit extended/modified by me) are correct.

point (a)

We looking for $$P(Y=y|do(X=x),C=c)$$

I read that, given $$(X,Y)$$, we need a set of control that include $$C$$ and deal with backdoor criterion. So $$[A Z C]$$ is a compliant set. The expanded expression is given in the book but I want to understand if, given the linear simplification, the regression like:

$$Y = \theta_1 X + \theta_2 C + \theta_3 A + \theta_4 Z + r$$

Is what we need, and in particular if $$\theta_1 + \theta_2$$ represent the c-specific effect. In term of path coefficients we have $$\beta_{y,w} \beta_{w,x} + \beta_{y,d} \beta_{d,c}$$. It is correct?

Point (b)

$$[Z A B C D]$$ seems me a correct set and a regression similar to the the previous is what we need and the z-specifc effect is: $$\beta_{y,w} \beta_{w,x} + \beta_{y,z}$$. It is correct?

Point (c)

Let me say that, for example, $$\beta_{y,w} \beta_{w,x} = 3,6$$ and $$\beta_{y,z}=0,9$$

so $$E[Y|do(x),z] = 3,6 x + 0,9 z$$

if $$z = (1,2)$$ we have $$x = 0$$ , se $$z = (3,4,5)$$ we have $$x = 1$$

if $$z = 1$$

$$E[Y|do(x),z] = 3,6*0 + 0,9*1 = 0,9$$

if $$z = 2$$

$$E[Y|do(x),z] = 3,6*0 + 0,9*2 = 1,8$$

if $$z = 3$$

$$E[Y|do(x),z] = 3,6*1 + 0,9*3 = 6,4$$

if $$z = 4$$

$$E[Y|do(x),z] = 3,6*1 + 0,9*4 = 7,2$$

if $$z = 5$$

$$E[Y|do(x),z] = 3,6*1 + 0,9*5 = 8,1$$

It is correct?

If I'm wrong what are the correct solutions ?

EDIT: In order to clarify my doubts I add something. I’m sure that in linear models the quantity involved in the equation below (letters have no relations with the example above):

can be translated in regression term as follow. I’have to perform the regression:

$$Y=\theta_1 X + \theta_2 W + r$$

and $$\theta_1$$ give us the total effect of $$X$$ on $$Y$$. Therefore what I looked for.

Now the z-specific effect of $$X$$ on $$Y$$ is representable as:

But in linear regression terms what regression I have to perform? What coefficient I'm interested in?

In a bit different case, the w-specific effect of $$X$$ on $$Y$$ is also representable as:

But in linear regression terms what regression I have to perform? What coefficient I'm interested in?

• What if linearity is not a good assumption, and you just need to work with probabilities? What's a more general answer? – Adrian Keister May 11 at 15:49
• General tools are more powerful, I know. However I'm primarily focused on linear model. From your comment I understand that my solutions are acceptable. It is so? – markowitz May 11 at 19:37
• You do not need to condition on $A.$ Otherwise, your solution for a is correct in the linear case. Your answer to b is incorrect: you are supposed to have only four variables, and $X$ and $Y$ absolutely must be two of them. You're gonna need $Z$ as well, because it's clearly a confounder. I have not checked your answer to c, because your b is incorrect. – Adrian Keister May 11 at 21:57
• About point (a) – I understand that $A$ is not necessary in the set but It is admissible, indeed the effect measured above in term of path coefficients remain the same. So, in this sense, I would say that my reply is correct. Or It is wrong? – markowitz May 12 at 11:01
• About point (b) – I thought I had to add four variable at $Y,X,Z$ but probably I misunderstood the point. The set that your suggest is $Y,X,Z,A$ while the mine is $Y,X,Z,A,B,C,D$. The problem here is that running the regressions the coefficients for $X$ remain the same but for $Z$ it not. So my set is wrong. Why? For the solution, what combination of paths coefficients is the right? Said differently my main question is not about the selection of the minimal set but about the proper use of the regression (coefficients) for identification of covariate-specific effect in linear case. – markowitz May 12 at 11:21

1 Answer


a. To get the $$c$$-specific effect of $$X$$ and $$Y,$$ which is $$P(Y=y|\doop(X=x),C=c),$$ we analyze as follows. The set $$S$$ in Rule 2 is actually the $$\{Z,C\}$$ nodes, minimally. Hence, we have that $$P(Y=y|\doop(X=x),C=c)=\sum_z P(Y=y|X=x,Z=z,C=c)\,P(Z=z|C=c).$$

b. We would have to measure $$X,Y,Z,$$ and one of $$A,B,C,$$ or $$D.$$ I'll pick $$A$$, so that the expression becomes $$P(Y=y|\doop(X=x), Z=z)=\sum_aP(Y=y|X=x,A=a,Z=z)\,P(A=a|Z=z).$$

c. We have that $$X= \begin{cases} 0,&Z=1,2\\ 1,&Z=3,4,5. \end{cases}$$ Here $$Z\in\{1,2,3,4,5\}.$$ Now the desired quantity is $$E(Y)$$ under the $$Z$$ strategy. That is, we want \begin{align*} E(Y) &=\sum_y\left[y P(Y=y|\doop(X=g(Z)))\right]\\ &=\sum_y\left[y \sum_zP(Y=y|\doop(X=x),Z=z)|_{x=g(z)}\,P(Z=z)\right]\\ &=\sum_y\left[y \sum_z\left[\sum_aP(Y=y|X=x,A=a,Z=z)\,P(A=a|Z=z)\right]_{x=g(z)}\,P(Z=z)\right]\\ &=\sum_{a,y}\sum_z\left[y\,P(Y=y|X=g(z),A=a,Z=z)\,P(A=a|Z=z)\,P(Z=z)\right]\\ &=\sum_{a,y}\Bigg\{ \sum_{z=1}^2\left[y\,P(Y=y|X=0,A=a,Z=z)\,P(A=a|Z=z)\,P(Z=z)\right]\\ &\qquad+\sum_{z=3}^5\left[y\,P(Y=y|X=1,A=a,Z=z)\,P(A=a|Z=z)\,P(Z=z)\right]\Bigg\}. \end{align*} That's about as far as we can get without knowing the probability distributions more exactly.

• Your reply is not precisely what I looked for, nevertheless it is useful (+1) – markowitz May 12 at 14:58