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Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 29L / e / Solution

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.

Solved by gpt-5.6-sol high.

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