One-Sample Robust Tests of Multivariate Location

Testing a Specified Location Vector

A one-sample multivariate location test asks whether a population location equals a specified vector. In robust inference, the estimator is built to limit the effect of unusual observations, but the test still needs to account for the estimator’s variability and the dependence among its coordinates. The resulting Wald test uses a robust location estimate and its estimated sandwich covariance.

The target requires care. Under a specified model, it may be the model’s location parameter. Under contamination, it is generally the location functional defined by the robust estimating equation for the distribution actually observed. Robustness limits sensitivity to contamination; it does not, without further assumptions, make the target equal to the uncontaminated component’s location.

Definition: Let $X_1,\ldots,X_n$ be independent and identically distributed random vectors in $\mathbb{R}^p$. A robust location estimator $\hat{\theta}$ may be defined as a consistent solution of $$ \Psi_n(\theta)=\frac{1}{n}\sum_{i=1}^n \psi(X_i;\theta)=0, $$ where $\psi(X_i;\theta)$ is a $p$-dimensional estimating function. Its population target $\theta^*$ satisfies $E[\psi(X;\theta^*)]=0$. The one-sample null hypothesis is $H_0:\theta^*=\theta_0$, against $H_A:\theta^*\ne\theta_0$, for a specified vector $\theta_0\in\mathbb{R}^p$.

A common robust location equation uses a bounded radial score. With a fixed positive-definite scatter matrix $S$, one version is $\psi(x;\theta)=w(r)(x-\theta)$, where $r=\{(x-\theta)^\mathsf{T}S^{-1}(x-\theta)\}^{1/2}$ and $w(r)$ decreases for large $r$. This downweights observations far from the proposed location. The test construction below applies more generally, including when scatter or other nuisance quantities are estimated, provided their effect is included in the estimator’s asymptotic covariance.

Asymptotic Covariance and the Wald Statistic

Write $A=E[\partial\psi(X;\theta^*)/\partial\theta^\mathsf{T}]$ for the derivative matrix and $B=\operatorname{Var}(\psi(X;\theta^*))$. Under regularity conditions—among them a valid asymptotic linearization, finite second moments for the estimating function, and nonsingular $A$—the estimator has the expansion

\sqrt{n}(\hat{\theta}-\theta^*)= -A^{-1}\frac{1}{\sqrt{n}}\sum_{i=1}^n\psi(X_i;\theta^*)+o_p(1).

The multivariate central limit theorem then gives an asymptotic normal distribution with covariance $V=A^{-1}B A^{-\mathsf{T}}$. For an estimator that jointly estimates nuisance parameters, this covariance must reflect their estimation as well; using only the location block of an incomplete calculation can give the wrong uncertainty.

Theorem: Suppose $H_0:\theta^*=\theta_0$ holds, $\sqrt n(\hat{\theta}-\theta^*)\xrightarrow{d}N(0,V)$, and $V$ is positive definite. If $\hat V\xrightarrow{p}V$ and $\hat V$ is invertible with probability tending to one, then $$ W=n(\hat{\theta}-\theta_0)^\mathsf{T}\hat V^{-1}(\hat{\theta}-\theta_0) \xrightarrow{d}\chi^2_p. $$

Here $V$ is the covariance of the limiting distribution of $\sqrt n(\hat{\theta}-\theta^*)$; the approximate covariance of $\hat{\theta}$ itself is $V/n$. A consistent sandwich estimator is commonly formed from sample estimates of the derivative and score variability. For example, if $\hat A$ estimates $A$ and $\hat B$ estimates $B$, then $\hat V=\hat A^{-1}\hat B\hat A^{-\mathsf{T}}$, when these quantities are well-defined and the required nuisance adjustments are included.

Estimate the location.
Solve the robust estimating equation to obtain $\hat{\theta}$.

Estimate its asymptotic covariance.
Compute a consistent $\hat V$ for the limiting covariance of $\sqrt n(\hat{\theta}-\theta^*)$, including estimated nuisance quantities as needed.

Form and calibrate the statistic.
Calculate $W$, then compare it with the $\chi^2_p$ distribution for a large-sample test.

The theorem follows by standardization and Slutsky’s theorem. Under the null, let $Z_n=\sqrt n(\hat{\theta}-\theta_0)$. By assumption $Z_n\xrightarrow{d}Z$, where $Z\sim N(0,V)$, and consistency gives $\hat V^{-1}\xrightarrow{p}V^{-1}$. Therefore $W=Z_n^\mathsf{T}\hat V^{-1}Z_n\xrightarrow{d}Z^\mathsf{T}V^{-1}Z$. Since $V$ is positive definite, $V^{-1/2}Z\sim N(0,I_p)$, so $Z^\mathsf{T}V^{-1}Z$ is the sum of squares of $p$ independent standard normal variables. That sum has a $\chi^2_p$ distribution. $\square$

At significance level $\alpha$, reject $H_0$ when $W$ exceeds the $1-\alpha$ quantile of $\chi^2_p$; equivalently, report the asymptotic p-value $P(\chi^2_p\ge W)$. This is a large-sample calibration, not an exact finite-sample guarantee.

Worked Tests

Worked Example: A Correlated Two-Dimensional Test

Suppose $n=80$, the robust estimate is $\hat{\theta}=(2.4,-1.1)^\mathsf{T}$, and the null vector is $\theta_0=(2,-1)^\mathsf{T}$. A consistent estimated asymptotic covariance is $$ \hat V=\begin{pmatrix}1.44&0.36\\0.36&0.81\end{pmatrix}. $$ Test the null at level $0.05$.

The determinant is $1.44(0.81)-(0.36)^2=1.0368>0$, so this matrix is invertible. Its inverse and the difference vector are $$ \hat V^{-1}=\frac{1}{1.0368}\begin{pmatrix}0.81&-0.36\\-0.36&1.44\end{pmatrix}, \qquad d=\hat{\theta}-\theta_0=\begin{pmatrix}0.4\\-0.1\end{pmatrix}. $$

The quadratic form before multiplying by $n$ is $$ d^\mathsf{T}\hat V^{-1}d =\frac{0.81(0.4)^2+2(-0.36)(0.4)(-0.1)+1.44(-0.1)^2}{1.0368} =\frac{0.1728}{1.0368}=\frac{1}{6}. $$ Thus $W=80/6\approx13.33$. The $0.95$ quantile of $\chi^2_2$ is about $5.99$, so the test rejects. The asymptotic p-value is $\exp(-13.33/2)\approx0.00127$. The off-diagonal covariance matters: the calculation uses the joint uncertainty, not separate coordinate tests.

Worked Example: A Robust Score Under Symmetric Contamination

Consider six two-dimensional observations $$ (1,0),\;(-1,0),\;(0,1),\;(0,-1),\;(10,10),\;(-10,-10). $$ Use the radial Huber score $\psi(u)=u\min(1,2/\|u\|)$, with scatter fixed at $I_2$. Test the robust location null $\theta_0=(0.5,0.5)^\mathsf{T}$. The data are symmetric about zero, and each opposite pair has cancelling scores at $\theta=0$; hence the estimating equation is solved at $\hat{\theta}=0$.

At the four unit observations, the score derivative with respect to the residual is $I_2$. At either outer observation, $r=\sqrt{200}=10\sqrt2$, and the derivative is $$ \frac{2}{r}\left(I_2-\frac{uu^\mathsf{T}}{r^2}\right) =\frac{1}{10\sqrt2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix}. $$

Using the empirical average derivative with respect to $\theta$ and score outer products gives $$ \hat A=-\begin{pmatrix}0.69024&-0.02357\\-0.02357&0.69024\end{pmatrix}, \qquad \hat B=\begin{pmatrix}1&2/3\\2/3&1\end{pmatrix}. $$ For example, the four inner observations contribute $-4I_2$ to the $\theta$-derivative sum and $\operatorname{diag}(2,2)$ to the score-product sum. Each outer observation contributes $-\frac{1}{10\sqrt2}\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$ to the $\theta$-derivative sum. Each outer score is $(\sqrt2,\sqrt2)^\mathsf{T}$ or its negative, contributing twice the matrix with every entry equal to $2$ to the score-product sum. Dividing these sums by six yields the displayed estimates.

The matrices have eigenvectors in the $(1,1)^\mathsf{T}$ direction. In that direction $\hat A$ has eigenvalue $-0.69024+0.02357\approx-0.66667$, while $\hat B$ has eigenvalue $1+2/3=5/3$. Therefore $\hat V=\hat A^{-1}\hat B\hat A^{-\mathsf{T}}$ has eigenvalue $(5/3)/(-2/3)^2=3.75$ in that direction. The difference $d=(-0.5,-0.5)^\mathsf{T}$ has squared length $0.5$ and lies in that direction, so $d^\mathsf{T}\hat V^{-1}d=0.5/3.75=0.13333$. Thus $W=6(0.13333)=0.8$. This illustrates the statistic, but with only six observations the large-sample chi-square approximation may be unreliable.

Worked Example: A Three-Dimensional Non-Rejection

Suppose $n=50$, $\theta_0=(0,0,0)^\mathsf{T}$, $\hat{\theta}=(0.3,-0.2,0.1)^\mathsf{T}$, and $$ \hat V=\begin{pmatrix}0.25&0&0\\0&0.36&0\\0&0&0.16\end{pmatrix}. $$ The covariance is positive definite. Since it is diagonal, $$ d^\mathsf{T}\hat V^{-1}d =\frac{0.3^2}{0.25}+\frac{(-0.2)^2}{0.36}+\frac{0.1^2}{0.16} =0.36+0.11111+0.0625=0.53361. $$ Consequently $W=50(0.53361)\approx26.68$, which exceeds the $0.95$ quantile of $\chi^2_3$, about $7.815$. The null is rejected. The result is a joint test: it assesses the full vector rather than requiring each coordinate to pass a separate test.

Conditions, Interpretation, and a Common Pitfall

The nonsingularity conditions are essential. In particular, finite score variability and a nonsingular derivative matrix do not by themselves guarantee a full-rank covariance $V$. If $B$ is singular, then $V=A^{-1}BA^{-\mathsf{T}}$ may be singular; the inverse in the Wald statistic may fail to exist, and the limiting quadratic form need not be $\chi^2_p$. The theorem requires positive-definite $V$ and a consistent covariance estimate that is invertible with probability tending to one.

Mistake: Assuming that a nonsingular $A$ and finite $B$ are enough for a $\chi^2_p$ test. Correction: Check that the resulting $V$ is positive definite and that $\hat V$ is invertible with probability tending to one.
Mistake: Using $V$, the covariance for $\sqrt n(\hat{\theta}-\theta^*)$, without the factor $n$ in the Wald statistic. Correction: The statistic is $n(\hat{\theta}-\theta_0)^\mathsf{T}\hat V^{-1}(\hat{\theta}-\theta_0)$; alternatively use an estimated covariance $ \hat V/n$ for $\hat{\theta}$ and omit the outside factor $n$.
Mistake: Treating a robust test as automatically testing the clean population center under contamination. Correction: State the target. Without assumptions that identify the clean component, the estimating equation targets the location functional of the observed distribution.

These tests are useful when a joint location question matters and outlying observations could unduly influence classical procedures. Robust estimation changes how the location and its uncertainty are computed; it does not remove the need to justify the target, check the covariance rank, or assess whether the asymptotic approximation is adequate for the sample size.

Check Your Understanding

Use the definitions and results above to answer the following.

What does $V$ represent, and how does it differ from the approximate covariance of $\hat{\theta}$?
Why does the Wald statistic have a $\chi^2_p$ limit when $V$ is positive definite?
Which condition fails if a score has finite variability but its covariance matrix is singular?
In the symmetric-contamination example, why does the estimating equation have a root at zero?
What location is being tested when the observations come from a contaminated distribution?

One-Sample Robust Tests of Multivariate Location

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