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I'm looking for a hint in understanding semantics of $\text{do}$ operator. Starting from the original distribution $P$, an intervention $\text{do}(X=x)$ takes us to another distribution $P_x$ - in general different than $P$. When defining semantics of $\text{do}$ operator Pearl uses truncated product formula: $$ P(X_1, \dots , X_n|\text{do}(X_j=x))=\prod_{i \neq j}P(X_i | \text{pa}(X_i)) \text{ if } x_j=x \text{ and otherwise } 0 $$ or equivalently: $$ P(X_1, \dots , X_n|\text{do}(X_j=x))= \frac{P(X_1, \dots , X_n)}{P(X_j=x|\text{pa}(X_j))} \text{ if } x_j=x\text{ and otherwise } 0 .$$

How does this definition cover extreme interventions like $x$ outside of the range of $X$ in the original distribution, since both expressions seem to use the original distribution $P$?
How is the post-intervention distribution $P_x$ formally derived?

EDIT: The assumption that for $\text{do}(X=x)$ we only consider $x \in \text{support}(X)$ is not made explicit in any sources I checked (Pearl, Janzig) and imposes an unexpected limitation on the framework. It is easy to show case where intervention would be outside of $\text{support}(X)$ but intuitively one can find the probabilities. Consider system defined by structural equations: $$Z=\text{Bernoulli}\left(\frac{1}{2}\right) \\ X=Z \\ Y=X+Z$$ Now let us make an intervention $\text{do}(X=10)$. $P(X=10|\text{pa}(X))=0$ and the formula in the definition would be undefined but on the other hand by looking at the structural equations and the distribution of $Z$ one would expect that $$P(Y=10|\text{do}(X=10))= P_{x}(Y=10)=\frac{1}{2}$$

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  • $\begingroup$ Given your tag selection, is your question specific to bayesian networks or graphical models? $\endgroup$ Mar 23 at 9:10
  • $\begingroup$ @RichardHardy I selected broad set of tags for visiblity - this question originated from study of Pearls SCM. $\endgroup$
    – borg
    Mar 23 at 9:15
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    $\begingroup$ What does $pa(\cdot)$ stand for? $\endgroup$ Mar 23 at 10:48
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    $\begingroup$ @RichardHardy this is commonly used notational convention to denote parents of $X$ in the graph associated with the model. $\endgroup$
    – borg
    Mar 23 at 10:50
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    $\begingroup$ Thank you. I did not recognize that even though I studied Pearl et al. "Causal Inference in Statistics: A Primer" quite thoroughly some 4 years ago. They used $\text{PA}$ (and I used it e.g. here), and so $pa$ was clearly not in my graphical memory... $\endgroup$ Mar 23 at 16:12

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$x$ cannot be outside the support of $X$

Both equations deal with the theoretical law of the random variables, so any concrete values play no role here. That said, perhaps the question concerns the range of possible interventions. An intervention cannot set a variable to a value outside of its support (although it can set it to a value not previously observed empirically).

Derivation of post-intervention distribution

It seems to me like you have basically already given the derivation. In an SCM, the joint distribution of $X$ is given as $$ P(X_1, \dots, X_d) = \prod_{i=1}^dP(X_i|\text{Pa}(X_i)). $$ Given $\text{do}(X_j=x)$, the key insight is to note that value of $X_j$ is no longer random and $P(X_j=x)=1$, which we can simply leave out of the product $$ P(X_1, \dots, X_d|\text{do}(X_j=x)) = \prod_{i\neq j}P(X_i|\text{Pa}(X_i)), $$ which is exactly your first equation. The second equation you have given can be obtained by $$ P(X_1, \dots, X_d|\text{do}(X_j=x)) = \frac{\prod_{i=1}^dP(X_i|\text{Pa}(X_i)}{P(X_j=x|\text{Pa}(X_j))} = \frac{P(X_1,\dots,X_d)}{P(X_j=x|\text{Pa}(X_j))}. $$


Response to Edit

A random variable can only take values within its support by definition. For example, a Bernoulli random variable is defined on the sample space ${0, 1}$, and thus cannot take any other values. If that seems surprising, it may help to recall that causal models extend statistical models and inherit their properties. If you feel that the support of a random variable does not adequately reflect the range of possible interventions, you can simply define the variables differently so they better reflect the system you have in mind.

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  • $\begingroup$ Thanks for the answer! I included my reply in the question since the derivation part of the question actually requires some context. $\endgroup$
    – borg
    Mar 23 at 12:55
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    $\begingroup$ I guess the sums here should be prods $\endgroup$
    – Firebug
    Mar 23 at 17:43
  • $\begingroup$ I have added a response to your edit @borg. $\endgroup$
    – Scriddie
    Mar 24 at 17:40
  • $\begingroup$ Thanks! But I don't think this is entirely relevant since application of $do(X=x)$ operator de facto sets $P(X=x)=1$ in new model and still I don't see why you would restrict yourself to the support of $X$ (or the distribution of $X$ from the original setting for that matter). In real world practice such an intervention would be mean testing a change of some part of the original mechanism and it seems reasonable to expect that $do$ calculus can address that somehow in a consistent way. Maybe there is some extra formalism to cover that... $\endgroup$
    – borg
    Mar 24 at 18:35
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    $\begingroup$ There is a simple solution - instead of going into uncharted waters by moving "out-of-support", why not simply define variables such that their support covers the interventions you have in mind? After all, these variables are models. It is in your power to define them as you please. $\endgroup$
    – Scriddie
    Mar 24 at 20:11

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