Spatial Rank Score Tests for Location

Why Score Directions to Test Location?

A location test asks whether a specified point is the center of a multivariate distribution. In one dimension, a signed-rank statistic combines the sign of each residual with a score determined by its rank in absolute value. A spatial rank score test uses the same broad ingredients in several dimensions: the direction of each residual and a score based on its radial rank. The resulting vector sum records whether the observations’ directions and radial ranks are collectively unbalanced around the proposed center.

The construction matters. Pairwise empirical spatial ranks, based only on comparisons such as \(U(X_i-X_j)\), are unchanged by translating every observation by the same vector. They therefore cannot by themselves test a location parameter. Here, ranks are instead formed from distances to the hypothesized location. We derive the null calibration under a spherical model, where the directions have a particularly simple geometry.

Radial Ranks and Scored Spatial Directions

The factor \(n+1\) keeps the scaled ranks strictly between zero and one. If radii tie, a tie-handling rule must be specified; the results below assume ties occur with probability zero. Two scores used below are the linear score \(J(u)=u\) and the bounded score \(J(u)=u/(u+c)\), where \(c>0\). For either choice, a large score magnitude is assigned to a large radial rank, while the direction retains the orientation of the residual from \(\theta_0\).

A quadratic test compares the observed score sum with its null covariance. Write \(I_p\) for the \(p\)-dimensional identity matrix and set

When at least one score is nonzero, \(A_n\) is positive definite. Equivalently, with \(\bar q_n=S_n/n\), \(Q_n=n\bar q_n^\top(A_n/n)^{-1}\bar q_n\). The scaling is important: \(A_n\) is the covariance of the sum, not the covariance of the average.

Null Distribution Under Spherical Symmetry

The assumptions for the calibration can be stated through a radial-direction representation. Under the null, suppose the observations are independent and identically distributed as \(X_i=\theta_0+R_iV_i\), where \(R_i>0\) have a continuous distribution and \(V_i\) are independent and uniform on the unit sphere in \(\mathbb R^p\), independently of the radii. This is the spherical model about \(\theta_0\); its radial distribution need not be specified. Suppose also that \(J\) is Riemann integrable on \((0,1)\) and \(c_J=\int_0^1J(u)^2\,du\) is finite and strictly positive. Bounded, piecewise continuous scores satisfy these score conditions.

The proof uses the exact covariance of the score sum and a multivariate central limit theorem. Conditional on the radii, the radial ranks—and hence all score magnitudes—are fixed. Ordering observations by radius merely permutes their directions. Since the directions are independent and uniform, the reordered directions remain independent and uniform. A uniform sphere direction \(V\) satisfies \(E[V]=0\) by symmetry and \(E[VV^\top]=I_p/p\): its covariance is invariant under rotations, hence a scalar multiple of \(I_p\), and its trace is \(E[\|V\|^2]=1\).

Here is the full asymptotic argument. Riemann-sum convergence gives \[ \frac1n\sum_{k=1}^n J\left(\frac{k}{n+1}\right)^2\longrightarrow c_J, \] so \(A_n/n\longrightarrow(c_J/p)I_p\). In particular, the eigenvalues of \(A_n\) grow proportionally to \(n\). The standardized sum \(A_n^{-1/2}S_n\) is a sum of independent, conditionally mean-zero vectors \(A_n^{-1/2}J(k/(n+1))V_k\), where \(V_k\) denotes the direction attached to the \(k\)th radial order. Their conditional covariance matrices sum to \(I_p\).

For bounded \(J\), the norm of every summand is at most \(\sup_u|J(u)|/\sqrt{\lambda_{\min}(A_n)}\), which tends to zero. Thus, for every \(\varepsilon>0\), eventually every summand has norm at most \(\varepsilon\); the conditional Lindeberg condition follows. The multivariate Lindeberg central limit theorem then gives \(A_n^{-1/2}S_n\xrightarrow{d}N_p(0,I_p)\). The same argument applies to unbounded scores if their weighted directions satisfy the corresponding Lindeberg condition. Squaring the Euclidean norm is a continuous operation, so \(Q_n=\|A_n^{-1/2}S_n\|^2\xrightarrow{d}\chi^2_p\), as asserted. \(\square\)

For a Riemann-integrable score with \(c_J>0\), the covariance has the simpler large-sample approximation \(A_n\approx n(c_J/p)I_p\). The finite-\(n\) definition of \(A_n\), however, uses the actual score values assigned to the ranks and avoids replacing their sum by an integral.

What the Result Does—and Does Not—Guarantee

Spherical symmetry is the key geometric assumption in the theorem. It makes the directions uniform and independent of the radial ordering, so the score sum has the stated isotropic covariance even though the radial distribution is unknown. The result does not assert the same calibration for every centrally symmetric or elliptically shaped distribution: in those settings, the directions need not be uniform or independent of the radii in the required way.

A frequent mistake is to use pairwise empirical spatial ranks and assume their sum tests location. For example, if each rank is built from \(n^{-1}\sum_j U(X_i-X_j)\), translating all \(X_i\) by \(-\theta_0\) leaves every difference unchanged. Such ranks contain no information about the proposed center. The correction is to rank distances from that center and attach each score to the residual direction, as done here.

A second pitfall is to use \(A_n\) as though it were the covariance of \(\bar q_n\). It is the covariance of \(S_n\); the average has covariance \(A_n/n^2\). Equivalently, the quadratic form for the average must include the factor \(n\): \(Q_n=n\bar q_n^\top(A_n/n)^{-1}\bar q_n\). Omitting that scaling changes the statistic and invalidates the chi-square calibration.