[smallnamespace]

Something to note about this formulation is the explicit assumption that in p(y|do(x)), the 'do' operation is supposed to be completely independent of prior observed variables, e.g. the doers are 'unmoved movers' [1].

That fits the model where you randomly 'do' one thing or another (e.g. blinded testing); however this is not the same thing as p(y|do'(x)), where do' is your empirical observation of when you yourself have set X=x in a more natural context.

E.g. let's say you will always turn on the heat when it's cold outside. P(cold outside | do(turn on heat)) = P(cold outside), because turning on the heat does not affect the temperature outdoors.

However, P(cold outside | do'(turned on heat)) > P(cold outside), because empirically, you actually only choose to turn on the heat when it's cold outdoors.

These two are also different from P(cold outside | heat was turned on) (since someone else might have access to the thermostat).

In reality our choices and actions are also products of the initial states (including our own beliefs, and our own knowledge of what would happen if we did x). Our actions both move the world, but we are also moved by the world.

Does do-calculus have a careful treatment of 'mixed' scenarios where actions are both causes and effects of other causes?

[1] https://en.wikipedia.org/wiki/Unmoved_mover

[Darmani]

You can write P(cold outside | do(turned on heat=X)), where X is another random variable. So P(cold outside | do(turned in heat=X=1)) will be equal to P(cold outside | do(turned on heat=1))P(X=1 | cold outside) = P(cold outside)P(X=1 | cold outside).

But, you might want to consider making "turned on heat" part of the system in this case, and go back to using the classic conditioning operator instead of the do operator.

This is covered in chapter 4 of Pearl's Causality.