Write
$$
R_i=(x_i-\widehat f(z_i))(y_i-\widehat g(z_i))
=(\varepsilon_i+F_i)(\xi_i+G_i).
$$
Expanding $R_i^2$ gives the leading term $\varepsilon_i^2\xi_i^2$ and terms of the forms treated in parts b and c, together with their versions obtained by interchanging $(\varepsilon,F)$ and $(\xi,G)$. For example, the pure error terms are $\varepsilon_i^2G_i^2$, $\xi_i^2F_i^2$, and $F_i^2G_i^2$, and each cross term is controlled by <Cauchy-Schwarz inequality>[Cauchy-Schwarz] from these. Hence
$$
\tau_D^2=\frac1n\sum_iR_i^2
=\frac1n\sum_i\varepsilon_i^2\xi_i^2+o_p(1)
\xrightarrow{p}\mathbb E(\varepsilon_1^2\xi_1^2).
$$
Assuming this limit is positive, the <continuous mapping theorem> yields $\tau_D\xrightarrow{p}\{\mathbb E(\varepsilon_1^2\xi_1^2)\}^{1/2}$. Combining this with the assumed <convergence in distribution> of $\sqrt n\tau_N$ and applying the <Slutsky theorem> gives
$$
T=\frac{\sqrt n\tau_N}{\tau_D}\xrightarrow{d}N(0,1).
$$
This is the <studentization of the generalized covariance measure statistic>.

Solution

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

If the numerator of the <generalized covariance measure statistic> obeys a <central limit theorem> with variance $\mathbb E(\varepsilon^2\xi^2)$ and its empirical residual-product second moment converges in probability to the same positive quantity, the <Slutsky theorem> makes the studentized statistic converge in distribution to $N(0,1)$.

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 Incoming links: Studentization of the generalized covariance measure statistic