# Understanding how to evaluate the integral causal-effect expression

I have this expression

$$p( Y \mid \text{do}(Z=z)) = \int_{B, S, W, X} dBdSdWdX \ \ P(B | S) P(W | B, S) P(X | B, S, Z=z) \left[ \int_{Z'} dZ' P(Z'| B,S,W) P(Y | B, S, W, X, Z') P(S) \right]$$

which represents the causal effect on $$Y$$ from making an intervention on $$Z$$ in the causal graph $$\mathcal{G}$$ which is part of a structural causal model.

This was found through the ID algorithm for those who are familiar with the literature.

Problem is now, I am not quite sure what this result means, in particular the inner integral. What does $$Z'$$ represent here? How do I evaluate this? Say that the intervention domain for my $$Z$$ variable is $$[-2,2]$$ and I intervene on $$Z=1$$, does this expression mean that I integrate the inner integral over the whole domain $$[-2,2]$$ except $$Z=1$$?

Thanks

I believe the result is a sequential application of the conditional front-door criterion and the backdoor criterion, from the do-calculus, to achieve the identification of $$p(Y\mid\text{do}(Z=z))$$ (not the expected value), given a latent confounder between $$Z$$ and $$Y$$.

It seems like $$X$$ is a mediator of the effect of $$Z$$ on $$Y$$ satisfying some graphical conditions to be leveraged for the identication of $$p(Y\mid\text{do}(Z=z), W, B, S)$$ via the front-door criterion (conditional on $$W,B,S$$) as:

\begin{aligned} & p(Y\mid\text{do}(Z=z), W, B, S)\\ &\, = \int \text{d}X\, p(X\mid Z=z,B,S) \int \text{d}Z'\, p(Z'\mid W,B,S)\, p(Y\mid X, Z',W,B,S) \end{aligned}

The final result is obtained via marginalization of non-$$Z$$-descendants (confounders) $$W,B,S$$, as:

\begin{aligned} & p(Y\mid\text{do}(Z=z))\\ &\, = \iiint \text{d}W\,\text{d}B\,\text{d}S\, p(W\mid B,S)\, p(B\mid S)\,p(S)\, p(Y\mid\text{do}(Z=z), W, B, S) \end{aligned}

Thus, the inner-most integral $$\int \text{d}Z'p(Z'\mid W,B,S)\cdots$$ should be done in the whole support of $$Z$$: all possible values for the intervention, including the value to evaluate $$z$$.

• Thanks for this excellent answer, very helpful (I have also updated the question to remove the expectation). Perhaps if you will allow me probe you a bit further then; should I want to approximate this with Monte-carlo integration (essentially just sampling it a lot of times to get an estimate of aforementioned expectation) then I take lots of samples from the joint, my intervention, and plus it all in, eventually taking the average over all samples? Commented May 7 at 0:00