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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2022

Paper 3

The given identities imply
$$
\frac1n\sum_{i=1}^n
\left(\overline X_{n-1,i}^2-\overline X_n^2\right)
=\frac1{n(n-1)^2}
\sum_{i=1}^n(X_i-\overline X_n)^2.
$$
Thus, writing
$$
s_n^2=\frac1{n-1}\sum_{i=1}^n(X_i-\overline X_n)^2,
$$
we have the exact identity
$$
\widetilde T_{\rm JACK}
=T_n-\frac{s_n^2}{n}.
$$
The <sample variance> satisfies $s_n^2\xrightarrow{P}1$, so
$$
\sqrt n(\widetilde T_{\rm JACK}-T_n)
=-\frac{s_n^2}{\sqrt n}\xrightarrow{P}0.
$$
The <Slutsky theorem> and part (c) now give
$$
\boxed{
\sqrt n(\widetilde T_{\rm JACK}-\mu^2)
\xrightarrow{d}N(0,4\mu^2)
}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 28K / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 28K / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 28K / d / Solution

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