Put $R=\lVert X_{(L)}-x\rVert_\infty$ and $\delta_0=(2L/(nM))^{1/d}$. If $\delta_0\geq1$, the claim follows from $R\leq1$. Otherwise, for $\delta\geq\delta_0$, part (c) gives
$$
\mathbb P(R>\delta)
\leq\frac4{nM\delta^d}leq\frac2L.
$$
The <tail-sum formula> then yields
$$
\mathbb ER
\leq\delta_0+\int_{\delta_0}^1\mathbb P(R>\delta)\,d\delta
\leq\left(\frac{2L}{nM}\right)^{1/d}+\frac2L.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 218 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-218/2/d/solution