Choosing Rank Scores for Tail Sensitivity

Why the Radial Score Matters

A multivariate rank test can use the same ordering of observations and still respond differently to a location shift, depending on how it weights each radial rank. A score that gives every observation the same weight treats an extreme point like a central one. A score that increases with radial rank gives more influence to observations farther from the center. This choice affects local sensitivity, especially when the distribution has substantial tail mass.

The comparison here concerns radial scores: functions of an observation’s percentile in distance from a specified center. It does not change the ordering rule itself. We examine how bounded, linear-percentile, and tail-emphasizing scores weight that ordering, then derive a criterion for comparing their sensitivity to small location shifts.

Radial Ranks and Score Choices

Let $X$ be a random vector in $\mathbb{R}^p$, with center at zero and a spherically symmetric density $q(x)=c\,g(\lVert x\rVert)$, where $c$ is a normalizing constant. Write $R=\lVert X\rVert$, $S=X/R$ when $R>0$, and let $F_R$ be the distribution function of $R$. Assume $F_R$ is continuous. Then $U=F_R(R)$ is uniform on $(0,1)$; $U$ is the radial percentile, and $S$ is the direction.

Definition: A radial score function is a measurable function $J:(0,1)\to\mathbb{R}$ used to weight the direction $S$ according to radial percentile $U$. The corresponding population score vector is $A_J=J(U)S$. In a rank procedure, the unknown percentile $U_i$ is replaced by a radial rank percentile, such as $R_i/(n+1)$, where $R_i$ is the rank of the $i$th radial distance.

Three representative choices clarify what score selection does. A bounded sign score $J_B(u)=1$ weights all directions equally, regardless of radial percentile. A linear-percentile score $J_L(u)=u$ increases proportionally with percentile, but remains bounded on $(0,1)$. A Gaussian radial score $J_G(u)=F_{\chi^2_p}^{-1}(u)^{1/2}$ assigns increasing, unbounded weights toward the outer ranks; $F_{\chi^2_p}$ is the distribution function of a chi-squared random variable with $p$ degrees of freedom. The Gaussian score is tail-emphasizing because its weight grows without bound as $u$ approaches one.

“Linear” here means linear in percentile, not linear in the original radius. Also, a score’s shape alone does not determine its sensitivity: the distribution of $R$ determines which radii correspond to each percentile and how strongly a location shift changes the score-vector mean.

Worked Example: Weights Assigned to Four Radial Percentiles

Compare the three scores at radial percentiles $u=0.25, 0.50, 0.90,$ and $0.99$ in dimension $p=2$. For $p=2$, the Gaussian radial score is $J_G(u)=\sqrt{-2\log(1-u)}$, because the $\chi^2_2$ distribution has quantile $-2\log(1-u)$.

At $u=0.25$, the bounded score is $1$, the linear score is $0.25$, and the Gaussian score is $\sqrt{-2\log(0.75)}\approx0.76$.

At $u=0.50$, the scores are $1$, $0.50$, and $\sqrt{-2\log(0.50)}\approx1.18$.

At $u=0.90$, they are $1$, $0.90$, and $\sqrt{-2\log(0.10)}\approx2.15$. At $u=0.99$, they are $1$, $0.99$, and $\sqrt{-2\log(0.01)}\approx3.03$.

The bounded score does not distinguish radial ranks. The linear score increases steadily but cannot assign a weight above one. The Gaussian score gives much more weight to the most extreme ranks. Its increasing weight does not by itself guarantee greater sensitivity; that depends on how its weights align with the model’s location score.

Local Sensitivity Under a Location Shift

Suppose the density under location parameter $\theta\in\mathbb{R}^p$ is $q_\theta(x)=q(x-\theta)$. Assume $g$ is positive and differentiable, and define the radial location score $\psi(r)=-g'(r)/g(r)$. The likelihood score for location at $\theta=0$ is $\psi(R)S$. For the derivation below, assume the model is differentiable under the integral sign for the score-vector expectation, $E[\psi(R)^2]<\infty$, and the score $J$ satisfies $0<E[J(U)^2]<\infty$ and $E[|J(U)\psi(R)|]<\infty$. Also assume the relevant expectations are differentiable at $\theta=0$. These conditions make the mean derivative and covariance used in the comparison well-defined.

Result: Under these assumptions, $E[A_J]=0$ and $\operatorname{Cov}(A_J)=E[J(U)^2]I_p/p$, where $I_p$ is the $p$-dimensional identity matrix. Under the local alternative $\theta=t h$, the derivative of the score-vector mean at $t=0$ is $E[J(U)\psi(R)]h/p$. Thus the squared mean shift measured relative to the null covariance, per unit squared shift $\lVert h\rVert^2$, is $$ \mathcal{S}(J)=\frac{\{E[J(U)\psi(R)]\}^2}{p\,E[J(U)^2]}. $$

The quantity $\mathcal{S}(J)$ is a local sensitivity index for the population score vector. It compares the first-order mean response to a small shift with the null variability. Rank tests use sample radial ranks rather than the population $U$; the formula explains the population weighting principle behind their score choices. Under standard conditions for a rank statistic’s local asymptotics, this is the corresponding first-order comparison.

Deriving the Sensitivity Index

Spherical symmetry gives two directional identities: $E[S]=0$ and $E[SS^\mathsf{T}]=I_p/p$. The second follows because the matrix is unchanged by every rotation, so it is a scalar multiple of $I_p$; its trace is $E[\lVert S\rVert^2]=1$, making that scalar $1/p$.

First, $J(U)$ depends only on $R$. Conditional on $R$, the direction is uniform on the unit sphere, so $E[A_J]=E[J(U)E(S\mid R)]=0$. The same conditioning yields the covariance:

\operatorname{Cov}(A_J) =E[J(U)^2SS^\mathsf{T}] =E[J(U)^2]\,\frac{I_p}{p}.

Next, differentiate $E_\theta[A_J]$ at $\theta=0$. By the assumed differentiability under the integral sign, its derivative is the expectation of $A_J$ multiplied by the likelihood score for location. That score is $\psi(R)S$. Therefore:

\left.\frac{d}{dt}E_{th}[A_J]\right|_{t=0} =E[J(U)\psi(R)SS^\mathsf{T}]h =\frac{E[J(U)\psi(R)]}{p}\,h.

The squared length of this derivative is $\lVert h\rVert^2\{E[J(U)\psi(R)]\}^2/p^2$. The null covariance has inverse $pI_p/E[J(U)^2]$. Multiplying the squared derivative by this inverse in the corresponding quadratic form gives $\lVert h\rVert^2\mathcal{S}(J)$. This proves the stated sensitivity formula. $\square$

Worked Example: Comparing Scores Under a Bivariate Gaussian Model

Let $X\sim N(0,I_2)$. The radial variable $R$ has a chi distribution with two degrees of freedom, so $U=1-\exp(-R^2/2)$, $\psi(R)=R$, and $J_G(U)=R$. We calculate the sensitivity index for the bounded sign score $J_B=1$, the linear score $J_L(u)=u$, and the Gaussian radial score $J_G$.

For $J_B$, $E[J_B^2]=1$, and $E[J_B\psi(R)]=E[R]=\sqrt{\pi/2}$. Hence $\mathcal{S}(J_B)=(\pi/2)/2=\pi/4\approx0.785$.

For $J_L$, $E[U^2]=1/3$. To obtain $E[UR]$, use $t=1-U$, so $R=\sqrt{-2\log t}$:

E[UR] =\int_0^1(1-t)\sqrt{-2\log t}\,dt =\sqrt{\frac{\pi}{2}}-\frac{\sqrt{\pi}}{4} \approx0.810.

Consequently, $\mathcal{S}(J_L)\approx0.810^2/\{2(1/3)\}\approx0.985$.

For $J_G(U)=R$, $E[J_G^2]=E[R^2]=2$ and $E[J_G\psi(R)]=E[R^2]=2$. Thus $\mathcal{S}(J_G)=2^2/(2\cdot2)=1$. In this Gaussian model, the Gaussian radial score has the largest of these three sensitivity indices. It is proportional to the radial likelihood score and is locally optimal among square-integrable scalar radial scores, as the next result shows.

Result: If $\psi(R)$ is square-integrable and the score may be any nonzero square-integrable function of $U$, then $\mathcal{S}(J)\leq E[\psi(R)^2]/p$. Equality holds when $J(U)$ is proportional to $\psi(F_R^{-1}(U))$, provided this function is an admissible score.

To prove this, apply the Cauchy–Schwarz inequality to $J(U)$ and $\psi(R)$:

\{E[J(U)\psi(R)]\}^2 \leq E[J(U)^2]E[\psi(R)^2].

Dividing by $pE[J(U)^2]$ gives $\mathcal{S}(J)\leq E[\psi(R)^2]/p$. Equality in Cauchy–Schwarz holds precisely when the two random variables are proportional almost surely. Since $R=F_R^{-1}(U)$ when $F_R$ is continuous, the equality score is proportional to $\psi(F_R^{-1}(U))$. This proves the result. $\square$

Worked Example: Why Heavy Tails Change the Choice

Consider a bivariate Student $t$ distribution with four degrees of freedom and identity scale. Its density is proportional to $(1+\lVert x\rVert^2/4)^{-3}$. Differentiating the log density gives the radial location score $\psi(r)=6r/(4+r^2)$.

Near the center, $\psi(r)$ increases with $r$; its derivative is $6(4-r^2)/(4+r^2)^2$, so it peaks at $r=2$. Beyond that point it decreases, with $\psi(r)$ behaving like $6/r$ in the far tail. Thus an extremely large radius is not automatically the most informative observation about a small location shift in this model.

The radial survival function is $P(R>r)=(1+r^2/4)^{-2}$. Solving $1-u=(1+r^2/4)^{-2}$ gives the radial quantile $F_R^{-1}(u)=2\sqrt{(1-u)^{-1/2}-1}$, which grows like $2(1-u)^{-1/4}$ near one. A score equal to this radial quantile would heavily weight the most extreme ranks. Its second moment is finite here: the $t_4$ distribution has $E[R^2]=4$, so this score satisfies the required square-integrability condition. Finite variance, however, does not make the extreme-rank weight efficient for location; the model’s $\psi(r)$ decreases in the far tail.

The Gaussian radial score grows more slowly, at order $\sqrt{\log(1/(1-u))}$, but it is still unbounded. A bounded score limits the contribution of any one observation, while a model-matched score aligns its weights with $\psi(F_R^{-1}(u))$. Their relative sensitivity should be judged using $\mathcal{S}(J)$, not from the score’s growth alone.

Choosing and Interpreting a Score

A bounded score is useful when limiting the influence of extreme radial ranks is a priority. A linear-percentile score provides a simple, moderate increase in weight toward the outer ranks while remaining bounded. A tail-emphasizing score is appropriate when the assumed model says that larger radii carry more information about the shift and the resulting variability is acceptable. In particular, the local-sensitivity result identifies the model’s radial likelihood score—not “the most increasing score”—as the benchmark for local efficiency.

There is a trade-off between local sensitivity under a specified model and stability when that model is wrong. A score that gives very large weights to a few outer ranks can respond strongly to a shift under one distribution, yet react sharply to contamination or to a different tail shape. Bounded scores trade some sensitivity in favorable models for protection against such extreme contributions. The appropriate choice depends on the scientific model and on how much robustness to tail misspecification matters.

Common Pitfall

Common mistake: “The score that grows fastest toward the largest ranks must be the most powerful for location.” This is not generally true. Local sensitivity depends on both $E[J(U)\psi(R)]$, which measures alignment with the model’s location score, and $E[J(U)^2]$, which measures score variability. The Cauchy–Schwarz bound shows that the best alignment is achieved by a score proportional to $\psi(F_R^{-1}(U))$, not by arbitrary rapid growth.

A related check is whether $E[J(U)^2]$ is finite. For example, $J(u)=(1-u)^{-1/2}$ has infinite second moment under uniform $U$, because $\int_0^1(1-u)^{-1}\,du=\infty$. The null covariance formula then fails, so the usual sensitivity comparison cannot be used without a different limiting theory.

Check Your Understanding

Use the definitions and sensitivity formula to answer these questions.

What feature of a bounded sign score makes its treatment of radial ranks different from that of a linear-percentile score?
In the sensitivity formula, what roles do $E[J(U)\psi(R)]$ and $E[J(U)^2]$ play?
Why does the Gaussian radial score attain the upper bound in the bivariate Gaussian example?
For a heavy-tailed model whose radial location score decreases in the far tail, why might an unbounded radial score be poorly aligned with local location information?
Why is finite $E[J(U)^2]$ necessary for the stated covariance and sensitivity comparison?