Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 28L / c / Solution

Put $d_i={\bar{X}}_{n-1,i}-{\bar{X}}_n=({\bar{X}}_n-X_i)/(n-1)$. Taylor's theorem givesSince ${\sum }_i{d}_i=0$, centering the leave-one-out values and expanding the square giveswith the stated $A_n,B_n$ and the cross term satisfying $|C_n|\le \sqrt{A_n{B}_n}$ by Cauchy--Schwarz.

NowThe fourth-moment assumption givesand hence $C_n=O(n^{-2})$. Since ${\sigma }_n^2=g^{\prime }(\mu {)}^2{\sigma }^2/n$, it follows that