Let
$$
d=\mathbb E[X^2],\qquad q=\mathbb E[XZ],\qquad
R_0=Z-\gamma_0X.
$$
Assumption 2 implies $\gamma_0=q/d$. Although the printed $\widehat\gamma$ need not converge to $\gamma_0$, its first-order effect vanishes because $\mathbb E[VX]=0$. Inverting the $2\times2$ Jacobian of the remaining <estimating equations> gives the influence function
$$
\operatorname{IF}_\beta
=\frac{V\{dW-\mathbb E[XW]X\}}
{d\mathbb E[AW]-\mathbb E[XW]\mathbb E[AX]}
=\frac{VR_0}{\mathbb E[AR_0]}.
$$
The first-stage condition yields
$$
\mathbb E[AR_0]
=\mathbb E[\mathbb E[A\mid X,Z]R_0]
=\lambda\mathbb E[R_0^2].
$$
Hence the asymptotic variance of $\sqrt n(\widehat\beta-\beta_0)$ is the <sandwich covariance matrix>[sandwich] expression
$$
\boxed{
\mathcal V_\beta
=\frac{\mathbb E[V^2R_0^2]}
{\lambda^2\{\mathbb E[R_0^2]\}^2}}.
$$

There is a defect in the printed assumptions: $\operatorname{Var}(V\mid A,X)=\sigma^2$ alone does not determine $\mathbb E[V^2R_0^2]$, 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 to
$$
\boxed{
\mathcal V_\beta=\frac{\sigma^2}{\lambda^2\mathbb E[(Z-\gamma_0X)^2]}}.
$$
The asymptotic variance of $\widehat\beta$ itself is $\mathcal V_\beta/n$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-221/2/d/solution