Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 221 / 2

Conditionally on $X$, a valid instrumental variable $Z$ must satisfy three core conditions.

Together with consistency of potential outcomes and positivity in causal inference, these assumptions make variation in $A$ induced by $Z$ causally interpretable. In the displayed graph, relevance is the edge $Z\to A$, independence is the absence of a path from $Z$ to $U$ after conditioning on $X$, and exclusion is the absence of a direct $Z\to Y$ edge.

Retaining the estimator exactly as printed, define the probability limitThe empirical equations are linear in $(\beta ,\alpha )$. Their coefficient matrix converges toTherefore a sufficient condition, requiring neither parametric assumption, iswith finite moments sufficient for the weak law of large numbers. The empirical determinant then converges to a nonzero number, so the two linear equations have a unique solution with probability tending to one.

The homogeneous treatment effect and consistency implyInstrument validity gives $Z\perp R\mid X$.

If Assumption 1 holds, put $V=R-{\alpha }_{0}X$. Then $\mathbb{E}[V\mid X]=0$, so conditional instrument independence givesConsequently both population estimating equations vanish at $({\beta }_0,{\alpha }_0)$ for any probability limit of $\hat{\gamma }$.

If Assumption 2 holds, choose the linear-projection coefficientThen $\mathbb{E}[(R-{\alpha }_{\ast }X)X]=0$, whileThe first term is zero by conditional instrument independence and the second by the definition of ${\alpha }_{\ast }$. Thus $({\beta }_0,{\alpha }_{\ast })$ solves the population equations. Under the nonsingularity condition from part b, the root is unique, so standard estimating equation consistency proveswhenever either Assumption 1 or Assumption 2 holds. This is double robustness.

LetAssumption 2 implies ${\gamma }_0=q/d$. Although the printed $\hat{\gamma }$ need not converge to ${\gamma }_0$, its first-order effect vanishes because $\mathbb{E}[VX]=0$. Inverting the $2\times 2$ Jacobian of the remaining estimating equations gives the influence functionThe first-stage condition yieldsHence the asymptotic variance of $\sqrt{n}(\hat{\beta }-{\beta }_0)$ is the sandwich expression

There is a defect in the printed assumptions: $\mathrm{Var}(V\mid A,X)={\sigma }^2$ alone does not determine $\mathbb{E}[V^2{R}_0^2]$, because $R_0$ depends on $Z$ and $V$ can have a nonzero conditional mean given $(A,X)$ under unmeasured confounding. Under the standard intended strengthening $\mathbb{E}[V^2\mid X,Z]={\sigma }^2$, the formula simplifies toThe asymptotic variance of $\hat{\beta }$ itself is ${\mathcal{V}}_{\beta }/n$.