---
tag: 2E46CY6S3Z
kind: counterexample
name: Hodges' superefficient estimator
msc: [62F12]
uses: [E5H2GFPK42, 8VDNRZG3R1, 0QNRBHWV35, 07RXBVTVZB, 0AJN7WHBJZ, 0JDAQHWA9N, G1Z3WSTE8Q, 0V4VSCBXP3]
sources:
  - url: https://en.wikipedia.org/wiki/Hodges%27_estimator
same_as:
  wikidata: Q5876082
  wikipedia: Hodges' estimator
---

## Counterexample

Let $X_1, \dots, X_n$ be [[i.i.d.|0AJN7WHBJZ]] $\mathcal{N}(\theta, 1)$, the [[normal distribution|07RXBVTVZB]] with unknown mean $\theta \in \R$, and let $\bar{X}_n$ be the [[sample mean|0QNRBHWV35]].
*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|G1Z3WSTE8Q]] 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|0JDAQHWA9N]]:

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|8VDNRZG3R1]] is beaten at one point.

## Remarks

The price appears near $0$.
Take $\theta_n = \frac{1}{2} n^{-1/4}$.
Then $\bar{X}_n - \theta_n$ is of order $n^{-1/2}$, so $P_{\theta_n}(T_n = 0) \to 1$, and the scaled [[mean squared error|0V4VSCBXP3]] satisfies $n\, E_{\theta_n}[(T_n - \theta_n)^2] \ge n \theta_n^2 P_{\theta_n}(T_n = 0) \approx \frac{1}{4}\sqrt{n} \to \infty$, while for $\bar{X}_n$ it equals $1$.

$T_n$ is not a regular estimator at $\theta = 0$: under $\theta_n = h/\sqrt{n}$, the law of $\sqrt{n}(T_n - \theta_n)$ tends to the point mass at $-h$, which depends on $h$.
This is why the [[Hájek–Le Cam convolution theorem|E5H2GFPK42]] assumes regularity.

## Explanations

### L1 human

Hodges' estimator rounds the sample mean to $0$ whenever it is small.
If the true mean really is $0$, this makes the estimate much more accurate than the sample mean.
But if the true mean is small and not $0$, the rounding causes large errors.
So beating the usual efficiency bound at one point costs much more at nearby points.
This shows that efficiency must be judged on small neighbourhoods of parameter values, not at one value at a time.