Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 205 / 3 / Solution

Under $H_0$, conditional independence givesMultiplication by the measurable sign $s(z_1)$ and the law of total expectation prove the first identity.

WriteThenConditional orthogonality under $H_0$ and the variance bounds in (ii) giveand similarly for the term containing ${\xi }_i{a}_i$. The Cauchy-Schwarz inequality bounds the last term bywhich is $o_p(n^{-1/2})$ by $n{\mathrm{MSPE}}_f{\mathrm{MSPE}}_g\to 0$.

The leading summands ${ϵ}_i{\xi }_{i}s(z_i)$ are independent, centered, and have variance $\mathrm{Var}({ϵ}_1{\xi }_1)$ because $s^2=1$. The central limit theorem and consistency of ${\tau }_D$ therefore give, by the Slutsky theorem,This is the generalized covariance measure statistic.

Without the null, condition first on $(x_1,z_1)$. Sincethe term involving $\mathbb{E}(x_1\mid z_1)$ vanishes, givingFor the specified alternative, independence and $\mathbb{E}{x}_1=0$ make the right sideThe constant choice $s=1$ gives zero because $\mathbb{E}h(z_1)=0$, so no first-order power is expected. Takinginstead gives $\mathbb{E}(x_1^2)\mathbb{E}|h(z_1)|>0$, producing asymptotic power.