Superefficiency And Pointwise Variance Comparisons

Why Pointwise Variance Needs Care

An estimator can look exceptionally precise at a particular parameter value without being uniformly or locally precise. The Hodges estimator is the classical illustration: in a normal location model, it has asymptotic variance zero at zero, but has the usual variance at every fixed nonzero parameter. The apparent gain is possible because the estimator behaves discontinuously on the local scale relevant to regular asymptotic comparisons.

The distinction matters when interpreting an information bound. Such bounds concern estimators satisfying regularity conditions, not arbitrary pointwise comparisons. We will calculate the Hodges estimator's behavior directly, then examine what happens when the true parameter moves toward zero at the $n^{-1/2}$ scale.

The Model and the Hodges Estimator

Definition: Suppose $X_1,\ldots,X_n$ are independent observations from $N(\theta,\sigma^2)$, where $\sigma^2>0$ is known and $\theta\in\mathbb R$. Write $\bar X_n=n^{-1}\sum_{i=1}^n X_i$. The Hodges estimator with threshold $n^{-1/4}$ is $$ \hat\theta_{H,n}= \begin{cases} 0, & |\bar X_n|\le n^{-1/4},\\ \bar X_n, & |\bar X_n|>n^{-1/4}. \end{cases} $$

The threshold shrinks to zero, but more slowly than the standard error $n^{-1/2}$: their ratio is $n^{1/4}$. Thus, at $\theta=0$, the sample mean usually falls inside the threshold and the estimator is set to zero. At any fixed $\theta\ne0$, the sample mean eventually lies outside the threshold with probability approaching one.

For this model, the Fisher information per observation for $\theta$ is $1/\sigma^2$. The asymptotic information bound for a regular estimator of $\theta$ is therefore $\sigma^2$, in the usual variance scale for $\sqrt n(\hat\theta-\theta)$. The qualification “regular” is essential: the bound does not assert that every estimator has pointwise asymptotic variance at least $\sigma^2$.

Worked Example: The Exact Variance at Zero

Take $\theta=0$. Then $Z_n=\sqrt n\,\bar X_n/\sigma$ has the standard normal distribution. Set $a_n=n^{1/4}/\sigma$. The estimator is zero when $|Z_n|\le a_n$, and otherwise equals $\sigma Z_n/\sqrt n$. By symmetry its expectation is zero, so $$ n\operatorname{Var}_0(\hat\theta_{H,n}) =\sigma^2 E\!\left[Z_n^2\mathbf{1}\{|Z_n|>a_n\}\right]. $$

For a standard normal variable $Z$, integration by parts gives $$ E[Z^2\mathbf{1}\{|Z|>a\}] =2\int_a^\infty z^2\phi(z)\,dz =2\left[a\phi(a)+1-\Phi(a)\right], $$ where $\phi$ and $\Phi$ are the standard normal density and distribution function. Since $a_n\to\infty$, both terms in brackets tend to zero. Hence $n\operatorname{Var}_0(\hat\theta_{H,n})\to0$, rather than $\sigma^2$.

For a finite illustration, let $\sigma^2=1$ and $n=100$. Then $a_n=\sqrt{10}$, and $$ n\operatorname{Var}_0(\hat\theta_{H,n}) =2\left[\sqrt{10}\,\phi(\sqrt{10})+1-\Phi(\sqrt{10})\right] \approx 0.0186. $$ This is small but not zero at $n=100$; the limiting value is zero as $n$ increases.

Pointwise Asymptotic Behavior

Theorem: In the normal location model above, the Hodges estimator satisfies $$ \sqrt n(\hat\theta_{H,n}-\theta)\xrightarrow{p}0 \quad\text{when }\theta=0, $$ and, for each fixed $\theta\ne0$, $$ \sqrt n(\hat\theta_{H,n}-\theta)\xrightarrow{d}N(0,\sigma^2), \qquad n\operatorname{Var}_\theta(\hat\theta_{H,n})\longrightarrow\sigma^2. $$ Thus its pointwise asymptotic variance is zero at zero and $\sigma^2$ at every fixed nonzero $\theta$.

Proof. At $\theta=0$, the scaled estimator equals $Z_n\sigma\mathbf{1}\{|Z_n|>a_n\}$. For any $\varepsilon>0$, $$ P_0\!\left(\left|\sqrt n\hat\theta_{H,n}\right|>\varepsilon\right) \le P(|Z_n|>a_n)\longrightarrow0. $$ The probability tends to zero because $a_n\to\infty$ and $Z_n$ has the standard normal distribution. The exact variance calculation in the worked example also shows $n\operatorname{Var}_0(\hat\theta_{H,n})\to0$.

Now fix $\theta\ne0$, and let $A_n=\{|\bar X_n|\le n^{-1/4}\}$, the event on which the estimator is set to zero. For all sufficiently large $n$, $n^{-1/4}<|\theta|/2$. On $A_n$, therefore, $|\bar X_n-\theta|\ge|\theta|/2$, so $$ P_\theta(A_n)\le P_\theta(|\bar X_n-\theta|\ge|\theta|/2)\longrightarrow0. $$ The estimator equals $\bar X_n$ off $A_n$. Consequently, for any $\varepsilon>0$, $$ P_\theta\!\left(\left|\sqrt n(\hat\theta_{H,n}-\bar X_n)\right|>\varepsilon\right) \le P_\theta(A_n)\longrightarrow0. $$ Since $\sqrt n(\bar X_n-\theta)\sim N(0,\sigma^2)$ for every $n$, this proves the asserted distributional limit.

To verify the variance limit rather than infer it from convergence in distribution, put $D_n=\sqrt n(\hat\theta_{H,n}-\bar X_n)$. On $A_n$, $D_n=-\sqrt n\bar X_n$, and off $A_n$, $D_n=0$. Hence $$ E_\theta[D_n^2]\le n^{1/2}P_\theta(A_n)\longrightarrow0. $$ Indeed, the normal tail bound gives $P_\theta(A_n)\le 2\exp\{-n\theta^2/(8\sigma^2)\}$ for all sufficiently large $n$: $A_n$ implies $|\bar X_n-\theta|\ge|\theta|/2$, and the two-sided normal tail is at most $2\exp\{-x^2/(2\sigma^2/n)\}$ at distance $x=|\theta|/2$. Thus $D_n\to0$ in mean square. The scaled sample mean has bounded second moment, so Cauchy–Schwarz implies that adding $D_n$ changes its variance by a quantity tending to zero. Therefore $n\operatorname{Var}_\theta(\hat\theta_{H,n})\to\sigma^2$. $\square$

Worked Example: A Fixed Nonzero Parameter

Let $\sigma^2=1$ and fix $\theta=0.5$. The information bound for a regular estimator is $1$. For $n$ large enough that $n^{-1/4}<0.25$, the event that the Hodges estimator is set to zero implies $|\bar X_n-0.5|\ge0.25$. Since $\bar X_n\sim N(0.5,1/n)$, $$ P_{0.5}(A_n)\le P_{0.5}(|\bar X_n-0.5|\ge0.25) \le 2\exp(-n/32)\longrightarrow0. $$

The theorem’s variance argument also applies: on $A_n$, $n(\hat\theta_{H,n}-\bar X_n)^2=n\bar X_n^2\le n^{1/2}$, while the probability of $A_n$ decreases exponentially. Thus the scaled difference from the sample mean tends to zero in mean square. It follows that $n\operatorname{Var}_{0.5}(\hat\theta_{H,n})\to1$, and the scaled estimation error converges to $N(0,1)$. The pointwise variance gain at zero does not occur at this fixed nonzero value.

Why the Estimator Is Not Regular at Zero

Pointwise behavior holds the parameter fixed as $n$ increases. Regularity asks for a stronger comparison: it examines parameter values within order $n^{-1/2}$ of the value under study. In a one-dimensional model, a standard local regularity requirement at $\theta_0$ is that, for every fixed $h$, the limiting distribution of $\sqrt n(\hat\theta_n-(\theta_0+h/\sqrt n))$ under $\theta=\theta_0+h/\sqrt n$ is the same as the limiting distribution at the center (and does not depend on $h$). Regular asymptotic efficiency results apply under such local stability conditions, together with their other model assumptions.

Worked Example: A Local Alternative to Zero

Take $\sigma^2=1$, $\theta_0=0$, and consider the sequence of true parameters $\theta_n=h/\sqrt n$, where $h$ is fixed. Write $Y_n=\sqrt n\bar X_n$. Under this sequence, $Y_n\sim N(h,1)$, and the threshold condition becomes $|Y_n|\le n^{1/4}$. The scaled estimation error is $$ \sqrt n(\hat\theta_{H,n}-\theta_n) =Y_n\mathbf{1}\{|Y_n|>n^{1/4}\}-h. $$

Because $Y_n$ has the same $N(h,1)$ distribution for every $n$, $P(|Y_n|>n^{1/4})\to0$. The first term on the right is therefore zero with probability tending to one, and the scaled error converges in probability to $-h$. At $h=0$, the limit is zero; at $h=1$, it is $-1$. The limiting distribution depends on the local displacement $h$, so the estimator fails the stated regularity requirement at zero.

The result is more than a change in variance: along local alternatives, the scaled error is not centered around a stable normal limit. The estimator sets nearby nonzero parameters to zero with high probability, producing a local bias that does not disappear on the $n^{-1/2}$ scale.

What Superefficiency Does—and Does Not—Mean

An estimator is called superefficient at a parameter value, in this classical example, when its pointwise asymptotic variance there is below the regular-estimator information bound. At zero, the Hodges estimator has pointwise asymptotic variance zero, strictly below $\sigma^2$. This does not contradict the bound: the estimator is not regular at zero. Nor does the pointwise calculation say that the estimator is uniformly better. At every fixed nonzero value, it has the same first-order variance as the sample mean, and near zero its behavior can be poor on the local scale.

A useful comparison is between the order of the threshold and the standard error. The threshold $n^{-1/4}$ is large relative to $n^{-1/2}$, so ordinary sampling fluctuations at zero are usually suppressed. But local parameter changes of size $n^{-1/2}$ are also small relative to this threshold, so they are suppressed too. This is exactly why the estimator can have an exceptionally small variance at the center while failing local regularity there.

Pointwise variance evaluates the estimator with one fixed true parameter.
Regular asymptotic comparisons also examine nearby parameter sequences and require stable local behavior.
The information bound is not a universal prohibition on every estimator having a low pointwise variance at an isolated value.

The main lesson is to report the scope of an efficiency comparison. A pointwise variance calculation alone does not establish regular asymptotic efficiency or uniform performance; local behavior and the conditions behind the information bound matter as well.

Common pitfall: “The Hodges estimator has asymptotic variance zero at $\theta=0$, so it violates the information bound.” The error is applying a bound for regular estimators to an estimator that is not regular at that point. The correction is to distinguish the fixed-parameter variance calculation from the local comparison: under $\theta_n=h/\sqrt n$, the scaled error converges to $-h$, which depends on $h$.

Check Your Understanding

Use the normal location model and threshold defined above.

Why does $n^{-1/4}$ eventually exceed the typical size of $\bar X_n$ when $\theta=0$?
What is the pointwise asymptotic variance of the Hodges estimator at a fixed $\theta\ne0$?
Why is convergence in distribution alone insufficient to establish convergence of variances?
Under $\theta_n=h/\sqrt n$, what is the limiting scaled estimation error, and how does it depend on $h$?
Why does the zero pointwise asymptotic variance at $\theta=0$ not contradict the regular-estimator information bound?