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

Under the null, ${\epsilon }_1$ and ${\xi }_1$ have conditional independence given $z_1$. The law of total expectation gives

Conditional on the covariates and all responses used to construct $\hat{g}$, the residual ${\epsilon }_i$ has conditional second moment at most $C$; conditional independence under the null is what permits this conditioning. ConsequentlyThe Markov inequality proves

Next, the Cauchy-Schwarz inequality givesThe 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.

Two applications of the Cauchy-Schwarz inequality giveandHere the empirical residual second moment is again $O_p(1)$ by the weak law of large numbers, while the assumed product error and the conclusion of part b are $o_p(1)$.

WriteExpanding $R_i^2$ gives the leading term ${\epsilon }_i^2{\xi }_i^2$ and terms of the forms treated in parts b and c, together with their versions obtained by interchanging $(\epsilon ,F)$ and $(\xi ,G)$. For example, the pure error terms are ${\epsilon }_i^2{G}_i^2$, ${\xi }_i^2{F}_i^2$, and $F_i^2{G}_i^2$, and each cross term is controlled by Cauchy-Schwarz from these. HenceAssuming this limit is positive, the continuous mapping theorem yields ${\tau }_D\stackrel{p}{\to }\{\mathbb{E}({\epsilon }_1^2{\xi }_1^2){\}}^{1/2}$. Combining this with the assumed convergence in distribution of $\sqrt{n}{\tau }_N$ and applying the Slutsky theorem givesThis is the studentization of the generalized covariance measure statistic.