Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 221 (Causal Inference)

In a standardized linear structural equation model, attach to every directed edge its path coefficient. Wright path tracing rule says that a covariance is the sum, over admissible unblocked paths between the two variables, of the product of the coefficients along each path; an admissible path does not pass through a collider and does not enter and later leave the same variable in a way that reverses direction twice. The total causal effect of $X$ on $Y$ is the sum of the products of edge coefficients over all directed paths from $X$ to $Y$.

The $X_1$ coefficient is the total causal effect of $X_1$ on $Y$. The node $X_1$ has no parents, so there is no backdoor path into it, and the unadjusted regression sums all directed paths from $X_1$ to $Y$.

The $X_2$ coefficient has no causal interpretation. The open backdoor path $X_2\leftarrow X_1\to Y$, together with paths through $X_3$, confounds the unadjusted association.

The $X_3$ coefficient has no causal interpretation. In particular, $X_3\leftarrow U\to Y$ is an open backdoor path, and $X_1$ and $X_2$ supply further confounding paths.

After adjusting for $X_1$, the $X_2$ coefficient is the total causal effect of $X_2$ on $Y$, consisting of $X_2\to Y$ and $X_2\to X_3\to Y$. The $X_1$ coefficient is the causal effect of $X_1$ with $X_2$ held fixed: it retains paths not passing through $X_2$, namely the direct path and $X_1\to X_3\to Y$.

Neither coefficient has a causal interpretation. Conditioning on $X_3$ opens the collider bias path $X_1\to X_3\leftarrow U\to Y$ for the $X_1$ coefficient, while the $X_3$ coefficient remains confounded by $U$ and by omitted $X_2$.

Neither coefficient has a causal interpretation. The $X_2$ coefficient is confounded by omitted $X_1$, and conditioning on the collider $X_3$ also opens $X_2\to X_3\leftarrow U\to Y$. The $X_3$ coefficient retains the open path through $U$.

None of the three coefficients has a causal interpretation. The $X_3$ coefficient is unavoidably confounded by unmeasured $U$. Conditioning on the collider $X_3$ opens paths through $U$ for both $X_1$ and $X_2$, so adding all measured regressors does not repair the bias.

Write $p=p^{\ast }$, ${\mu }_z=\mathbb{E}(Y\mid Z=z)$, and ${\tau }^{\ast }={\mu }_1-{\mu }_0$. The summand has finite variance, so the central limit theorem giveswhere

For $\vartheta =(q,t)$ use the estimating equationIts empirical mean vanishes exactly at $(\hat{p},\hat{\tau })$. The population equation has the unique root $(p,{\tau }^{\ast })$, so the Z-estimator is consistent. Linearizing the equation, or simplifying the corresponding sandwich covariance, gives the influence functionHence

Subtracting the two asymptotic variances and expanding the conditional second moments givesEstimating the randomized treatment probability therefore projects out the component of the known-probability influence function proportional to $Z-p$, weakly improving asymptotic efficiency.

A minimal causal directed acyclic graph has edges

Yes. Every path from $D$ to $G$ is blocked by a collider, such as $D\to A\leftarrow G$ or $D\to C\to M\leftarrow A\leftarrow G$. Thus D-separation gives $D\perp G$.

No. Conditioning on the collider $A$ opens $D\to A\leftarrow Z$, so $D$ and $Z$ are not conditionally independent given $A$.

Yes. Conditioning on $A$ blocks the directed paths from $Z$ to $Y$. Although conditioning on $A$ opens $Z\to A\leftarrow G\to Y$ and $Z\to A\leftarrow D\to C\to M\to Y$, conditioning additionally on $G$ and $D$ blocks those paths. Hence $Z\perp Y\mid (A,G,D)$.

Yes. Conditioning on $D$ blocks $C\leftarrow D\to A$, while $C\to M\leftarrow A$ is blocked at the collider $M$. Therefore $C\perp A\mid D$.

The independence fails after conditioning on $S=1$. Since $S$ is a descendant of collider $M$, selection opens $D\to C\to M\leftarrow A\leftarrow G$.

The independence already failed, and it still fails because conditioning on $A$ opens $D\to A\leftarrow Z$.

The independence still holds. Every path from $Z$ starts through $A$; paths on which $A$ is a noncollider are blocked by conditioning on $A$, while paths opened at collider $A$ are blocked by the conditioned variables $G$ or $D$. Conditioning on $S$ creates no route avoiding these blocks.

The independence fails. Conditioning on $S$, a descendant of $M$, opens the collider path $C\to M\leftarrow A$ even after conditioning on $D$.

Each factor $f_{j\mid \mathrm{pa}(j)}$ depends only on node $j$ and its parents. Those vertices form a clique in the moral graph, because moralization joins every pair of parents and removes arrow directions. The positive joint density therefore factors into clique potentials of the moral graph. The Hammersley-Clifford theorem then implies the global Markov property for that undirected graph.

By consistency of potential outcomes and conditional exchangeability,Taking conditional expectations given $X_1=x_1$ and subtracting the cases $z=1$ and $z=0$ proves the outcome-regression identification formula.

Conditional on $X$,Multiplying by $1\{X_1=x_1\}$, taking expectations, dividing by $\mathbb{P}(X_1=x_1)$, and using part (i) proves the inverse probability weighting formula.

Write $m_z(X)={\mu }_z(X,{\beta }_z)$ and $\tilde{e}(X)=e(X;\alpha )$. Conditional on $X$,with the analogous control expressionIf the propensity model is correct, both ratios are one and these equal ${\mu }_1(X)$ and ${\mu }_0(X)$, so their difference is the conditional average treatment effect.

If both outcome models are correct, each residual term in the preceding conditional expectations has mean zero regardless of the working propensity model. Thus ${\tilde{\mu }}_z^{\mathrm{dr}}(X)={\mu }_z(X)$ for $z=0,1$, and their difference is again $\tau (X)$. This proves double robustness.