Hodges' superefficient estimator

Counterexample

Let $X_1, \dots, X_n$ be i.i.d. $\mathcal{N}(\theta, 1)$, the normal distribution with unknown mean $\theta \in \R$, and let $\bar{X}_n$ be the sample mean. Hodges' estimator is $$T_n = \begin{cases} \bar{X}_n & \text{if } \abs{\bar{X}_n} \ge n^{-1/4}, \\ 0 & \text{if } \abs{\bar{X}_n} < n^{-1/4}. \end{cases}$$ The Fisher information of one observation is $1$, and $\sqrt{n}(\bar{X}_n - \theta) \sim \mathcal{N}(0, 1)$ exactly.

Check of the limit laws, with convergence in distribution:

1. If $\theta \ne 0$, then $P_\theta(\abs{\bar{X}_n} < n^{-1/4}) \to 0$, because $\bar{X}_n \to \theta$ while $n^{-1/4} \to 0$. So $T_n = \bar{X}_n$ with probability tending to one, and $\sqrt{n}(T_n - \theta) \xrightarrow{d} \mathcal{N}(0, 1)$.
2. If $\theta = 0$, then $P_0(\abs{\bar{X}_n} < n^{-1/4}) = P(\abs{Z} < n^{1/4}) \to 1$ for $Z$ standard normal. So $T_n = 0$ with probability tending to one, and $\sqrt{n}\, T_n \xrightarrow{d} 0$.

So the asymptotic variance of $T_n$ is $1$ for $\theta \ne 0$ and $0$ for $\theta = 0$: at $\theta = 0$ it beats the bound $1/I(\theta) = 1$ that maximum likelihood attains. This is superefficiency: the usual bound of efficiency is beaten at one point.