Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 221 / 3

The graph factorizes asConditioning on all variables except $X_1$, terms not involving $x_1$ cancel, leavingThis depends only on $(x_2,x_3,x_4)$, soThus $(X_2,X_3,X_4)$ is a Markov blanket of $X_1$.

For a distribution faithful to a Directed acyclic graph, the smallest Markov blanket of a vertex consists of

Conditioning on this set blocks every path from the vertex to all remaining vertices. Each listed neighbor is necessary under faithfulness of a directed acyclic graph: omitting a parent or child leaves its direct edge active, while omitting a child's other parent leaves the collider path through that conditioned child active. Faithfulness rules out accidental cancellations that could otherwise make a smaller blanket sufficient.

Conditional independence $A\perp B\mid D$ means that, for almost every $d$,equivalently $p(a\mid b,d)=p(a\mid d)$ wherever the conditional probabilities are defined.

Now use the chain rule and both assumed independences:This is exactly $A\perp (B,C)\mid D$, proving the contraction axiom for conditional independence.

Let $M$ be a Markov blanket of the treatment $A$ inside $(A,X)$, and write the remaining adjustment variables as $N=X\setminus M$. By definition,The sufficiency of $X=(M,N)$ givesApply the contraction axiom for conditional independence with first variable $A$, second variable $N$, third variable $Y(a)$, and conditioning variable $M$. It givesThe decomposition axiom for conditional independence then yields $A\perp Y(a)\mid M$. Hence every Markov blanket of $A$ in $(A,X)$ is itself a sufficient adjustment set.