Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 221 (Causal Inference)

A path in a causal directed acyclic graph is active given a conditioning set $K$ when every noncollider on the path is outside $K$ and every collider has itself or a descendant in $K$. Two vertex sets are d-separated by $K$ when no path between them is active given $K$. The global Markov property of a directed acyclic graph then turns d-separation into conditional independence.

In the displayed graph, the edges areEvery observed pair except $(X_1,X_4)$ and $(X_2,X_4)$ is joined by a direct edge, which remains active under conditioning on any other observed variables. The pair $(X_2,X_4)$ is always joined by the fork $X_2\leftarrow U\to X_4$, because the unobserved noncollider $U$ cannot be conditioned on.

For $(X_1,X_4)$, if $X_3$ is not conditioned on, $X_1\to X_3\to X_4$ is active. If $X_3$ is conditioned on, the pathbecomes active because its collider $X_2$ has conditioned descendant $X_3$; conditioning on $X_2$ itself also opens it. Thus every observed pair is d-connected given every subset of the other observed variables. Any conditional independence between two nonempty observed subvectors would imply one between each selected pair, so the graph entails none.

The directed acyclic graph factorization isThe graph gives $U\perp X_1$ and $U\perp X_3\mid (X_1,X_2)$. Therefore Bayes theorem givesConsequentlywhich contains no $x_1$. This observed equality is a Verma constraint: it is implied by the latent-variable causal graph even though it is not a conditional independence.

WriteSince $X_1$ has no parents, its total effect has no backdoor path, and

For $X_2$, $X_3$ intercepts every directed path to $X_4$. Conditional on $X_1$, there is no unblocked backdoor path from $X_2$ to $X_3$, and $(X_1,X_2)$ blocks every backdoor path from $X_3$ to $X_4$. The conditional front-door adjustment therefore givesandPart ii shows that the inner sum, after also averaging $X_4$, does not actually depend on $x_1$.

For $X_3$, $(X_1,X_2)$ is a sufficient backdoor adjustment set. HenceThese three formulas identify the requested average treatment effects from the observed joint distribution.

The no unmeasured confounding assumption is the conditional exchangeability statementAlso assume consistency of potential outcomes, no interference between units, and positivity in causal inference, in particular $\mathbb{P}(A=0\mid X)>0$ on the covariate support of treated units. ThenThe first treated potential outcome equals the observed treated mean by consistency. Moreover,Subtracting proves the displayed identification formula for the average treatment effect on the treated.

For discrete $X$,Perturbing the marginal mass $p(x)$ contributes $\pi (X)\mu (X)-\beta$. Perturbing $\pi (x)=\mathbb{E}[A\mid X=x]$ contributesTheir sum is $A\mu (X)-\beta$. Finally, the influence function ofisMultiplication by the derivative $p(x)\pi (x)$ of $\beta$ with respect to $\mu (x)$ and summation over $x$ givesAdding the three contributions yields the claimed mean-zero influence curve

Estimate the propensity score $\pi$ and the untreated outcome regression $\mu$, preferably with cross-fitting when flexible methods are used, and setWith $n_1={\sum }_i{A}_i$, the resulting one-step estimator of the ATT isEquivalently,This is an augmented inverse-probability-weighted estimator; it is consistent when either the propensity model or the untreated outcome model is correct, subject to the usual regularity and positivity conditions.

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.

The instrumental variable graph iswith no arrow $Z\to Y$ and no common cause of $Z$ with $A$ or $Y$.

In potential outcome notation, a valid instrument requires:

Instrumental-variable monotonicity isIt excludes defiers who would quit without the incentive but continue smoking when offered it. The possible principal strata are then never-takers $(0,0)$, compliers $(0,1)$, and always-takers $(1,1)$.

By random assignment, consistency, and exclusion,The second identity follows by checking the two possible binary exposure values. Under monotonicity, $A(1)-A(0)$ is the indicator of being a complier. Therefore the numerator iswhileTheir ratio is the complier average treatment effect, also called the local average treatment effect.

LetandBecause $U$ contains all exposure-outcome confounding and $Z$ is independent of $U$, the exclusion restriction and the law of total expectation giveLikewise,The no confounder-instrument interaction assumption says $\delta (U)=\delta$ is constant. Relevance gives $\delta \ne 0$, so the Wald ratio isthe overall average treatment effect.