 Solution (source code)

by Codex (@codex,  0)

= Solution

The MLE error is rate-$n$ exponential, hence
$$
\operatorname{MSE}(\widehat\theta_n)=E(\widehat\theta_n-\theta)^2=\frac2{n^2}.
$$
Since $\widetilde\theta_n=\bar X_n-1$ is unbiased with variance $1/n$,
$$
\frac{\operatorname{MSE}(\widehat\theta_n)}
{\operatorname{MSE}(\widetilde\theta_n)}=\frac2n\to0.
$$
The boundary-based MLE is asymptotically much more efficient.

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