 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/28l/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Put

d_{i} = \overset{ˉ}{X}_{n - 1, i} - \overset{ˉ}{X}_{n} = (\overset{ˉ}{X}_{n} - X_{i}) / (n - 1)

T_{n - 1, i} - T_{n} = g^{'} (\overset{ˉ}{X}_{n}) d_{i} + R_{n, i}, ∣ R_{n, i} ∣ \leq M d_{i}^{2} .

Since

\sum_{i} d_{i} = 0

, centering the leave-one-out values and expanding the square gives

v_{JACK} = \frac{n - 1}{n} (A_{n} + B_{n} + 2 C_{n}),

with the stated

A_{n}, B_{n}

and the cross term satisfying

∣ C_{n} ∣ \leq A_{n} B_{n}

by Cauchy--Schwarz.

Now

A_{n} = \frac{g ^{'} ( X ˉ _{n} ) ^{2}}{( n - 1 ) ^{2}} \sum_{i} (X_{i} - \overset{ˉ}{X}_{n})^{2} \sim \frac{g ^{'} ( μ ) ^{2} σ ^{2}}{n} .

The fourth-moment assumption gives

B_{n} \leq \frac{M ^{2}}{( n - 1 ) ^{4}} \sum_{i} (X_{i} - \overset{ˉ}{X}_{n})^{4} = O (n^{- 3}),

and hence

C_{n} = O (n^{- 2})

. Since

σ_{n}^{2} = g^{'} (μ)^{2} σ^{2} / n

\frac{v _{JACK}}{σ _{n}^{2}} \to 1.

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