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  "body": "\n## Counterexample\n\nLet $X_1, \\dots, X_n$ 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]].\n*Hodges' estimator* is\n$$T_n = \\begin{cases} \\bar{X}_n \u0026 \\text{if } \\abs{\\bar{X}_n} \\ge n^{-1/4}, \\\\ 0 \u0026 \\text{if } \\abs{\\bar{X}_n} \u003c n^{-1/4}. \\end{cases}$$\nThe [[Fisher information|G1Z3WSTE8Q]] of one observation is $1$, and $\\sqrt{n}(\\bar{X}_n - \\theta) \\sim \\mathcal{N}(0, 1)$ exactly.\n\nCheck of the limit laws, with [[convergence in distribution|0JDAQHWA9N]]:\n\n1. If $\\theta \\ne 0$, then $P_\\theta(\\abs{\\bar{X}_n} \u003c n^{-1/4}) \\to 0$, because $\\bar{X}_n \\to \\theta$ while $n^{-1/4} \\to 0$. So $T_n = \\bar{X}_n$ with probability tending to one, and $\\sqrt{n}(T_n - \\theta) \\xrightarrow{d} \\mathcal{N}(0, 1)$.\n2. If $\\theta = 0$, then $P_0(\\abs{\\bar{X}_n} \u003c n^{-1/4}) = P(\\abs{Z} \u003c n^{1/4}) \\to 1$ for $Z$ standard normal. So $T_n = 0$ with probability tending to one, and $\\sqrt{n}\\, T_n \\xrightarrow{d} 0$.\n\nSo 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.\nThis is *superefficiency*: the usual bound of [[efficiency|8VDNRZG3R1]] is beaten at one point.\n\n## Remarks\n\nThe price appears near $0$.\nTake $\\theta_n = \\frac{1}{2} n^{-1/4}$.\nThen $\\bar{X}_n - \\theta_n$ 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 $1$.\n\n$T_n$ 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 $h$.\nThis is why the [[Hájek–Le Cam convolution theorem|E5H2GFPK42]] assumes regularity.\n\n## Explanations\n\n### L1 human\n\nHodges' estimator rounds the sample mean to $0$ whenever it is small.\nIf the true mean really is $0$, this makes the estimate much more accurate than the sample mean.\nBut if the true mean is small and not $0$, the rounding causes large errors.\nSo beating the usual efficiency bound at one point costs much more at nearby points.\nThis shows that efficiency must be judged on small neighbourhoods of parameter values, not at one value at a time.\n",
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