Conditional on the covariates and all responses used to construct $\widehat g$, the residual $\varepsilon_i$ has conditional second moment at most $C$; <conditional independence> under the null is what permits this conditioning. Consequently
$$
\mathbb E\!\left[\frac1n\sum_{i=1}^n\varepsilon_i^2G_i^2\right]
\leq C\,\mathbb E\!\left[\frac1n\sum_{i=1}^nG_i^2\right]\longrightarrow0.
$$
The <Markov inequality> proves
$$
\frac1n\sum_i\varepsilon_i^2G_i^2\xrightarrow{p}0.
$$

Next, the <Cauchy-Schwarz inequality> gives
$$
\left|\frac1n\sum_i\xi_i\varepsilon_i^2G_i\right|
\leq
\left(\frac1n\sum_i\varepsilon_i^2G_i^2\right)^{1/2}
\left(\frac1n\sum_i\varepsilon_i^2\xi_i^2\right)^{1/2}.
$$
The first factor converges to zero in probability. The <weak law of large numbers> and part a make the second $O_p(1)$, so the product converges to zero in probability.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 205 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-205/3/b/solution