Efficiency and Power of Robust Location Tests

Why Efficiency and Power Must Be Considered Together

A robust location test is not automatically preferable to a classical one. Under an exact Gaussian model, the sample mean uses the distributional information efficiently; a robust estimator may have greater sampling variability and lower power against small alternatives. Under contamination, however, a single distant observation can move the mean substantially, while a bounded-influence estimator limits that movement. Comparing tests therefore requires asking two questions: how much power is lost under the reference model, and how much sensitivity to contamination is gained?

We make this comparison through local alternatives, whose departures from the null shrink at the $n^{-1/2}$ rate, and through contamination that has the same asymptotic order. This gives a precise account of how both effects enter the limiting distribution of a robust Wald statistic. To keep the comparison focused on location, we first treat the scatter matrix as known.

Definition: A sequence of location alternatives is local if the true location is $\theta_n=\delta/\sqrt{n}$, where $\delta\in\mathbb{R}^p$ is fixed. A test has local asymptotic power equal to the limiting rejection probability along this sequence, when that limit exists.

Local Power of a Wald Test

Let $T(F)$ be a location functional, with $T(F)=0$ under the null distribution $F$. Suppose its estimator $\hat\theta$ is asymptotically linear at $F$: for independent observations from $F$,

\sqrt{n}\,\hat\theta = \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\operatorname{IF}(X_i;T,F)+o_p(1) \xrightarrow{d}N_p(0,V).

Here $\operatorname{IF}(x;T,F)$ is the influence function and $V=\operatorname{Var}_F\{\operatorname{IF}(X;T,F)\}$ is assumed positive definite and finite. A Wald statistic using a consistent estimator of $V$ is $W_n=n\hat\theta^{\mathsf T}\hat V^{-1}\hat\theta$. Under the null it converges in distribution to a central chi-square random variable with $p$ degrees of freedom.

Theorem: Suppose the asymptotic linear expansion above holds, and its central limit theorem also holds under the local distributions considered. If the true location is $\delta/\sqrt{n}$, then $$ W_n\xrightarrow{d}\chi^2_p(\lambda), \qquad \lambda=\delta^{\mathsf T}V^{-1}\delta. $$ At asymptotic significance level $\alpha$, the local power is $P\{\chi^2_p(\lambda)>c_{p,\alpha}\}$, where $c_{p,\alpha}$ is the upper $1-\alpha$ quantile of the central $\chi^2_p$ distribution.

To prove the result, translation equivariance gives $T(F_{\delta/\sqrt n})=\delta/\sqrt n$. The assumed local central limit theorem therefore yields $\sqrt n\hat\theta\xrightarrow{d}N_p(\delta,V)$. Consistency of $\hat V$ and Slutsky’s theorem imply that $W_n$ converges to the quadratic form $Y^{\mathsf T}V^{-1}Y$, where $Y\sim N_p(\delta,V)$. Writing $Y=V^{1/2}Z+\delta$, with $Z\sim N_p(0,I)$, identifies that quadratic form as noncentral chi-square with $p$ degrees of freedom and noncentrality $\delta^{\mathsf T}V^{-1}\delta$. This proves the claim. $\square$

The matrix $V$ determines local power: a smaller asymptotic covariance in a direction gives greater noncentrality for alternatives in that direction. A test’s finite-sample calibration is a separate issue; the comparison here concerns limiting power when the Wald statistic is calibrated using a consistent covariance estimate.

Gaussian Efficiency: Mean and Spatial Sign

For a concrete comparison, standardize the observations so the null model is $N_p(0,I)$, and take $p=2$. The spatial-sign location estimator solves $\sum_{i=1}^n\psi(X_i-\hat\theta)=0$, with $\psi(x)=x/\|x\|$ for $x\ne0$. It gives each observation a score of unit length, rather than a score whose size increases with its distance from the center.

For a differentiable estimating equation, its influence function is $A^{-1}\psi(x)$, where $A=E[D\psi(X)]$, and its asymptotic covariance is $A^{-1}BA^{-\mathsf T}$, where $B=E[\psi(X)\psi(X)^{\mathsf T}]$. These expressions apply here because the bivariate normal distribution assigns probability zero to $X=0$, and $E(1/\|X\|)$ is finite.

Write $R=\|X\|$ and $U=X/R$. Under the bivariate standard normal model, the direction $U$ is uniform on the unit circle, independent of $R$, so $E[UU^{\mathsf T}]=I/2$. Also $D\psi(x)=(I-uu^{\mathsf T})/\|x\|$, where $u=x/\|x\|$. The Rayleigh density of $R$ is $r\exp(-r^2/2)$ for $r>0$, giving

E(1/R)=\int_0^\infty \exp(-r^2/2)\,dr=\sqrt{\pi/2}, \qquad A=\frac{\sqrt{\pi/2}}{2}I, \qquad B=\frac12 I.

Consequently, the spatial-sign estimator has asymptotic covariance $4I/\pi$. In contrast, the sample mean has asymptotic covariance $I$ under this Gaussian model.

Worked Example: Gaussian Efficiency in Two Dimensions

Compare the asymptotic variance per coordinate of the sample mean and the spatial-sign estimator under $N_2(0,I)$.

For the sample mean, each coordinate has asymptotic variance $1$. For the spatial-sign estimator, the calculation above gives variance $4/\pi$ per coordinate, and

\frac{4}{\pi}\approx1.27324.

Thus the spatial-sign estimator has about $27.3\%$ more asymptotic variance per coordinate than the mean. Its Gaussian asymptotic relative efficiency, defined here as the mean’s variance divided by the spatial-sign estimator’s variance, is $\pi/4\approx0.7854$, or about $21.5\%$ lower efficiency. The percentage increase in variance and the percentage reduction in efficiency are different quantities.

The efficiency comparison also gives a local-power comparison. For the same alternative $\theta_n=\delta/\sqrt n$, the mean-based Wald statistic has noncentrality $\delta^{\mathsf T}\delta$. The spatial-sign Wald statistic, using its own covariance $4I/\pi$, has noncentrality $(\pi/4)\delta^{\mathsf T}\delta$. Because noncentral chi-square power increases with its noncentrality parameter at a fixed critical value, the spatial-sign test has lower Gaussian local power in this setting.

Worked Example: Comparing Local Power

Take $p=2$, significance level $\alpha=0.05$, and local alternative $\delta=(2,0)^{\mathsf T}$. The mean-based test has $\lambda=2^2=4$; the spatial-sign test has $\lambda=(\pi/4)4=\pi$. The central $\chi^2_2$ critical value is approximately $5.991$.

For $p=2$, the noncentral chi-square distribution has the Poisson-mixture representation

P\{\chi^2_2(\lambda)>c\} = \sum_{k=0}^{\infty} e^{-\lambda/2}\frac{(\lambda/2)^k}{k!} \left[ e^{-c/2}\sum_{j=0}^{k}\frac{(c/2)^j}{j!} \right].

With $c=5.991$, substitution of $\lambda=4$ gives power approximately $0.415$; substitution of $\lambda=\pi$ gives approximately $0.335$. For example, the $k=0$ term in either calculation uses the central tail $e^{-5.991/2}$, which is approximately $0.05$; the remaining terms are weighted central tails with $4+2k$ or $2+2k$ degrees of freedom, respectively. The robust test loses some local power in this Gaussian example, as its lower noncentrality predicts.

Contamination Changes the Comparison

The influence function describes the first-order effect of a small perturbation to the distribution. Suppose observations instead come from a sequence of distributions that, to first order, combines the local alternative with a fraction $\gamma/\sqrt n$ of contamination at a fixed point $z$, where $\gamma\ge0$. Assume the asymptotic linear expansion and central limit theorem remain valid for this triangular sequence, and that the covariance estimate still converges to $V$. The estimator’s limiting mean is then shifted by both the genuine location change and the contamination effect.

Theorem: Under these conditions, $$ \sqrt n\,\hat\theta\xrightarrow{d} N_p\!\left(\delta+\gamma\operatorname{IF}(z;T,F),V\right), \qquad W_n\xrightarrow{d}\chi^2_p(\lambda_c), $$ where $$ \lambda_c= \left(\delta+\gamma\operatorname{IF}(z;T,F)\right)^{\mathsf T} V^{-1} \left(\delta+\gamma\operatorname{IF}(z;T,F)\right). $$

Indeed, asymptotic linearity makes the estimator’s first-order mean the expectation of its influence function under the perturbed distribution. The clean local shift contributes $\delta$; contamination contributes $\gamma\operatorname{IF}(z;T,F)$. The stated triangular-array central limit theorem supplies covariance $V$. The same quadratic-form argument used in the previous proof then gives the noncentral chi-square limit. $\square$

This result describes contamination at a shrinking fraction of the sample. A fixed positive contamination fraction is a different asymptotic regime: it can shift the limiting target away from the null, so nominal null calibration need not be maintained as $n$ grows. The local result is useful for comparing first-order sensitivity, not for claiming immunity to arbitrary contamination.

Worked Example: A Distant Contaminating Point

Continue with the bivariate standard normal reference model, test the null location $\delta=0$, and place local contamination in direction $u=(1,0)^{\mathsf T}$ at $z=10u$, with $\gamma=1$. For the sample mean, $\operatorname{IF}(z)=z$ and $V=I$. Its contamination noncentrality is therefore

\lambda_{\text{mean}}=z^{\mathsf T}z=10^2=100.

For the spatial-sign estimator, $A=(\sqrt{\pi/2}/2)I$, so its influence function at any nonzero $z$ is $A^{-1}z/\|z\|=\sqrt{8/\pi}\,u$. Its covariance is $V=(4/\pi)I$. Hence

\lambda_{\text{sign}} = \left(\sqrt{8/\pi}\,u\right)^{\mathsf T} \left((4/\pi)I\right)^{-1} \left(\sqrt{8/\pi}\,u\right) =2.

The point at distance 10 produces a much smaller first-order shift for the sign-based test. This is the practical trade-off: under the clean Gaussian model it has lower local power, but its bounded influence limits the effect of a distant observation. The calculation compares contamination-induced noncentralities; it does not imply that every robust test has greater power under every contaminated distribution.

Choosing a Test and Avoiding Common Errors

Under a credible Gaussian model with little concern about outliers, the mean-based test’s higher efficiency can be valuable. When distant contamination is plausible, a robust test with bounded influence can better control how individual observations affect the result. The choice should also account for the alternative of interest: local power depends on the direction $\delta$ through $V^{-1}$, not just on one overall efficiency number.

The known-scatter standardization used here isolates the location comparison. In an application where scatter is estimated, the covariance of the location estimator and the calibration of its Wald statistic must account for that estimation. The efficiency and power values derived here should not be transferred unchanged to a procedure with estimated scatter.

Common pitfall: A variance increase of $27.3\%$ is not the same as an efficiency decrease of $27.3\%$. In this example the variance multiplier is $4/\pi\approx1.27324$, while relative efficiency is its reciprocal, $\pi/4\approx0.7854$. Thus variance is about $27.3\%$ higher, but efficiency is about $21.5\%$ lower. Also, contamination-induced rejection under a shifted limiting distribution is not evidence that a test retains its nominal size under contamination.

Key takeaway: Gaussian efficiency describes local power under the reference model; influence describes first-order sensitivity to perturbation. Robust test choice balances these properties rather than maximizing either one in isolation.

Check Your Understanding

Use the limiting distributions and calculations in this tutorial to answer the following.

For a Wald test based on an asymptotically linear location estimator, how does the covariance matrix $V$ enter the local noncentrality parameter?
In the bivariate standard normal example, what is the spatial-sign estimator’s variance multiplier relative to the mean, and what is its relative efficiency?
Why does the spatial-sign test have lower local power than the mean-based test for the same Gaussian local alternative in the worked example?
How does a bounded influence function change the effect of a distant point under contamination of order $n^{-1/2}$?
Why should the known-scatter efficiency calculations not automatically be applied when scatter is estimated?