 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/29l/e/solution

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

The MLE error is rate-

n

exponential, hence

MSE (θ_{n}) = E (θ_{n} - θ)^{2} = \frac{2}{n ^{2}} .

Since

θ_{n} = \overset{ˉ}{X}_{n} - 1

is unbiased with variance

1/ n

\frac{MSE ( θ _{n} )}{MSE ( θ _{n} )} = \frac{2}{n} \to 0.

The boundary-based MLE is asymptotically much more efficient.

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