Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 221 / 1 / b / Solution

In the second causal directed acyclic graph, $Y(a_1,a_2)$ depends on $U_2$ but not on $U_1$, whereas $A_1$ depends on $U_1$ and the two latent roots are independent. HenceAlthough conditioning on $X$ conveys information about $U_2$, the assignment $A_2$ uses only $(A_1,X)$ and fresh randomization, soThese are the two sequential exchangeability conditions. Using them successively, together with consistency of potential outcomes, giveswhich is the formula from part a.

Adding $X\to Y$ invalidates the argument in general. The latent variable $U_1$ confounds $A_1$ and $X$, so $\mathbb{P}(X\mid A_1=a_1)$ need not equal the distribution of $X(a_1)$. When $X$ directly affects $Y$, that discrepancy no longer cancels after summing over $x$. The same observed distribution can then correspond to different intervention means, so the displayed formula need not identify the effect.