 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/28k/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Each leave-one-out statistic has the same distribution as

T_{n - 1}

, so

E B_{n} = (n - 1) (B_{n - 1} (θ) - B_{n} (θ)) .

Consequently the bias of the jackknife bias correction is

E T_{JACK} - θ = n B_{n} - (n - 1) B_{n - 1} .

The assumed expansion gives

n B_{n} = a + \frac{b}{n} + O (n^{- 2})

and

(n - 1) B_{n - 1} = a + \frac{b}{n - 1} + O (n^{- 2}) = a + \frac{b}{n} + O (n^{- 2}) .

Their difference is therefore

O (n^{- 2})

, as required.

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