Under $H_0$, <conditional independence> gives
$$
\mathbb E[(x_1-f(z_1))(y_1-g(z_1))\mid z_1]=0.
$$
Multiplication by the measurable sign $s(z_1)$ and the <law of total expectation> prove the first identity.

Write
$$
a_i=\widehat f(z_i)-f(z_i),
\qquad
b_i=\widehat g(z_i)-g(z_i).
$$
Then
$$
\tau_N
=\frac1n\sum_i\epsilon_i\xi_i s(z_i)
-\frac1n\sum_i\epsilon_i b_i s(z_i)
-\frac1n\sum_i\xi_i a_i s(z_i)
+\frac1n\sum_i a_ib_i s(z_i).
$$
Conditional orthogonality under $H_0$ and the variance bounds in (ii) give
$$
\sqrt n\,n^{-1}\sum_i\epsilon_i b_i s(z_i)
=O_p(\sqrt{\operatorname{MSPE}_g})=o_p(1),
$$
and similarly for the term containing $\xi_i a_i$. The <Cauchy-Schwarz inequality> bounds the last term by
$$
\left(n^{-1}\sum_i a_i^2\right)^{1/2}
\left(n^{-1}\sum_i b_i^2\right)^{1/2},
$$
which is $o_p(n^{-1/2})$ by $n\operatorname{MSPE}_f\operatorname{MSPE}_g\to0$.

The leading summands $\epsilon_i\xi_i s(z_i)$ are independent, centered, and have variance $\operatorname{Var}(\epsilon_1\xi_1)$ because $s^2=1$. The <central limit theorem> and consistency of $\tau_D$ therefore give, by the <Slutsky theorem>,
$$
T=\frac{\sqrt n\,\tau_N}{\tau_D}
\xrightarrow dN(0,1).
$$
This is the <generalized covariance measure statistic>.

Without the null, condition first on $(x_1,z_1)$. Since
$$
\mathbb E[\mathbb E(y_1\mid x_1,z_1)-\mathbb E(y_1\mid z_1)\mid z_1]=0,
$$
the term involving $\mathbb E(x_1\mid z_1)$ vanishes, giving
$$
\mathbb E[(x_1-\mathbb E(x_1\mid z_1))
(y_1-\mathbb E(y_1\mid z_1))s(z_1)]
=\mathbb E[x_1\{\mathbb E(y_1\mid x_1,z_1)
-\mathbb E(y_1\mid z_1)\}s(z_1)].
$$
For the specified alternative, independence and $\mathbb Ex_1=0$ make the right side
$$
\mathbb E(x_1^2)\mathbb E[h(z_1)s(z_1)].
$$
The constant choice $s=1$ gives zero because $\mathbb Eh(z_1)=0$, so no first-order power is expected. Taking
$$
s(z)=\operatorname{sgn}h(z)
$$
instead gives $\mathbb E(x_1^2)\mathbb E|h(z_1)|>0$, producing asymptotic power.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-205/3/solution