 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/28j/c/solution

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

S_{n} / n = \overline{X} \to 1/ θ

almost surely, and hence in probability. Therefore

θ_{π} = \frac{1 + α / n}{X + β / n} p θ

θ_{π} - θ_{MLE} = \frac{α S _{n} - n β}{S _{n} ( S _{n} + β )} .

Since

S_{n} / n \to 1/ θ

, the numerator is

O_{p} (n)

and the denominator is

O_{p} (n^{2})

, so this difference is

O_{p} (n^{- 1}) = o_{p} (n^{- 1/2})

. The limit in part a and Slutsky theorem now give

n (θ_{π} - θ) d N (0, θ^{2}) .

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