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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

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1 Answer 1

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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 $z$.

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  • $\begingroup$ 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? $\endgroup$
    – Astrid
    Commented May 7 at 0:00

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