Spatial Signs and Their Null Geometry

Direction, Not Distance

A multivariate observation has both a direction and a distance from any chosen center. A spatial sign keeps the direction and discards the distance. That distinction is useful in multivariate rank methods: observations on the same ray from the center receive the same sign, however far apart they are.

The geometry depends on the center being specified. It also depends on the distributional null: when observations are spherically distributed about that center, their nonzero directions have a particularly simple law. Observations exactly at the center require separate treatment, because their direction is undefined.

Definition: Let $x,c\in\mathbb{R}^d$, where $c$ is a specified center. The spatial sign of $x$ relative to $c$ is $$ S_c(x)= \begin{cases} \dfrac{x-c}{\|x-c\|}, & x\ne c,\\[6pt] 0, & x=c. \end{cases} $$ For a random vector $X$, write $S_c=S_c(X)$. The zero-vector convention at $X=c$ makes the sign defined everywhere; it does not assign a direction to a zero-length displacement.

For $X\ne c$, the sign has norm one and lies on the unit sphere $\mathbb{S}^{d-1}=\{u\in\mathbb{R}^d:\|u\|=1\}$. For $X=c$, its norm is zero. Thus the sign records whether the observation is at the center and, otherwise, its direction from the center. It records no radial distance.

Spherical Symmetry and the Null Geometry

A distribution is spherically symmetric about $c$ if rotating an observation around $c$ does not change its distribution. Formally, for every orthogonal $d\times d$ matrix $Q$, the random vectors $X-c$ and $Q(X-c)$ have the same distribution. Here orthogonal means $Q^\mathsf{T}Q=I_d$. If a density exists, a common sufficient form is $f_X(x)=g(\|x-c\|)$ for a nonnegative function $g$; the invariance definition also allows distributions without densities.

Theorem: Suppose $X\in\mathbb{R}^d$ has a distribution spherically symmetric about $c$, and let $p_0=P(X=c)$. Then $S_c=0$ with probability $p_0$. Conditional on $X\ne c$, $S_c$ is uniform on $\mathbb{S}^{d-1}$, meaning its law is the rotation-invariant probability measure on that sphere. In particular, $S_c$ and $-S_c$ have the same distribution and $E[S_c]=0$.

The word “uniform” refers to surface-area measure normalized to have total mass one; when $d=1$, the sphere consists of the two points $-1$ and $1$, each with probability one half. The theorem makes no assumption that the radius has a density or that it is independent of the direction.

Rotation commutes with taking a sign.
For any orthogonal $Q$, $S_c(c+Q(x-c))=Q S_c(x)$. This holds at $x=c$ under the zero convention as well.

The sign law is rotation-invariant.
Spherical symmetry and Step 1 imply that $Q S_c$ has the same distribution as $S_c$. Also, $P(S_c=0)=P(X=c)=p_0$, so conditioning on $S_c\ne0$ preserves this invariance.

Rotation-invariant directions are uniform.
The rotation-invariant probability measure on the sphere is unique. To see why an invariant direction $V$ must have this law, take an independent uniformly distributed random orthogonal matrix $Q$. Invariance gives $QV\stackrel{d}{=}V$; conditional on any fixed $V=v\in\mathbb{S}^{d-1}$, $Qv$ is uniform on the sphere. Thus $V$ is uniform.

Symmetry gives a zero mean.
Uniform directions are unchanged by the rotation $Q=-I_d$, so $S_c\stackrel{d}{=}-S_c$. Each coordinate is bounded in absolute value by one and therefore has a finite expectation. Its distribution is symmetric about zero, so each coordinate has expectation zero and $E[S_c]=0$.

This proves the theorem, including when there is positive probability at the center. In the case $p_0=1$, the conditional statement given $X\ne c$ has no conditioning event of positive probability and is understood to be vacuous. The unconditional sign then equals zero with probability one.

Worked Directional Probabilities

Worked Example: Signs from a Bivariate Normal Distribution

Let $X=(X_1,X_2)^\mathsf{T}$ have a bivariate normal distribution with mean $c=(0,0)^\mathsf{T}$ and covariance $I_2$. Find the probability that the sign lies in the first quadrant, and the probability that it lies within $30^\circ$ of the positive horizontal axis.

This distribution is spherically symmetric about $c$, and it has a density, so $P(X=c)=0$. The theorem says the sign is uniform on the unit circle. The first quadrant occupies one quarter of the circle, giving $$ P(S_{c,1}>0,\ S_{c,2}>0)=\frac{\pi/2}{2\pi}=\frac14. $$

The arc within $30^\circ$ of the positive horizontal axis has total angle $60^\circ=\pi/3$. Its probability is therefore $$ \frac{\pi/3}{2\pi}=\frac16. $$ These are probabilities of directional regions, not of distances: moving any observation farther from $c$ along the same ray leaves its sign unchanged.

Worked Example: A Spherical Distribution with Center Ties

Suppose $X=c$ with probability $0.2$. With probability $0.8$, let $X$ be uniformly distributed on the circle of radius $2$ centered at $c$. Determine the distributional components of $S_c$, its mean, and the probability it lies in a specified arc of length $\pi/3$.

The center atom produces $S_c=0$ with probability $0.2$. On the circle, every observation is nonzero relative to $c$, and the direction is uniform. Thus the sign distribution is a point mass of $0.2$ at the zero vector plus a uniform-circle component of total probability $0.8$. In particular, the nonzero signs have norm one.

The uniform-circle component has mean zero by symmetry, and the zero component also contributes zero. Hence $$ E[S_c]=0.2(0)+0.8(0)=0. $$ An arc of length $\pi/3$ occupies fraction $(\pi/3)/(2\pi)=1/6$ of the circle. The probability that the sign is nonzero and lies in that arc is therefore $0.8(1/6)=2/15$. If the event is instead conditioned on a nonzero sign, its probability is $1/6$. The distinction matters: center ties are zero signs, not directions assigned to the arc.

Worked Example: An Elliptical Normal Is Not Spherically Symmetric

Let $X_1\sim N(0,4)$ and $X_2\sim N(0,1)$ be independent, and take $c=0$. Show that the sign is not uniform on the circle by calculating the probability that it is within $45^\circ$ of the positive horizontal axis.

The event is $X_1>0$ and $|X_2|<X_1$. Since $X_1$ has a continuous distribution, $P(X_1=0)=0$. On $X_1>0$, the inequality is equivalent to $|X_2/X_1|<1$. Write $X_1=2Z_1$ and $X_2=Z_2$, where $Z_1,Z_2$ are independent standard normal variables. The ratio $Z_2/Z_1$ has the standard Cauchy distribution, so $$ P\left(\left|\frac{X_2}{X_1}\right|<1\right) =P\left(\left|\frac{Z_2}{Z_1}\right|<2\right) =\frac{2}{\pi}\arctan(2). $$

The event also requires $X_1>0$. The ratio $Z_2/Z_1$ is independent of the sign of $Z_1$: in polar coordinates for $(Z_1,Z_2)$, the ratio depends on angle modulo $\pi$, while the sign of $Z_1$ distinguishes the two opposite half-planes. Therefore $P(X_1>0\mid |X_2/X_1|<1)=1/2$, and $$ P(X_1>0,\ |X_2/X_1|<1)=\frac{1}{\pi}\arctan(2)\approx 0.352. $$ Uniform directions would give an arc probability of $90^\circ/360^\circ=0.25$. The different result confirms that the direction is not uniform. Unequal coordinate variances give an elliptical, rather than spherical, distribution; central symmetry alone does not imply the spherical null geometry.

Consequences and Common Pitfalls

Under the spherical null, every rotation of a directional region has the same probability, and regions with the same spherical surface area have the same probability. For $d\ge2$, any specified great subsphere has probability zero for a nonzero sign under the uniform law. In particular, a nonzero sign falls in either open half-space bounded by a hyperplane through the center with probability one half. In one dimension, the corresponding statement is that the two possible nonzero signs have equal probability.

The sign also has a simple symmetry consequence in a sample: if observations occur in pairs $x$ and $2c-x$, their two nonzero signs cancel exactly. But a zero sample sum is not guaranteed merely by spherical symmetry; the theorem establishes a zero population mean, not exact cancellation in every finite sample.

Giving a center observation an arbitrary direction. If $X=c$, the quotient defining the sign has zero denominator. Use the zero-vector convention, or explicitly condition on $X\ne c$ when discussing directions.
Confusing central symmetry with spherical symmetry. The condition $X-c\stackrel{d}{=}-(X-c)$ ensures opposite directions are balanced, but does not make all directions equally likely. The elliptical normal example is centrally symmetric but has nonuniform signs.
Assuming distance is retained. Every nonzero observation on a ray from $c$ has the same sign. Spatial signs cannot distinguish observations that differ only in their distance from the center.
Ignoring the tie probability. If $P(X=c)>0$, the unconditional sign distribution has an atom at zero. The uniform-sphere conclusion applies to the signs conditional on being nonzero.

These facts give the geometric null benchmark for methods based on spatial signs: under spherical symmetry, nonzero directions are uniform, while center ties remain visible as zero vectors. The benchmark is specific to the stated center and spherical symmetry; changing the center or replacing spherical symmetry by a weaker assumption can change the directional law.

Check Your Understanding

Use the definitions and null geometry to answer the following.

What is $S_c(x)$ when $x=c$, and why is that convention needed?
If $X$ is spherically symmetric about $c$ and $P(X=c)=0$, what is the conditional distribution of $S_c$ given that it lies in a specified arc or spherical cap?
Why does spherical symmetry imply $E[S_c]=0$, even if the distribution has an atom at $c$?
Does central symmetry by itself imply that nonzero spatial signs are uniform? Explain using the elliptical normal example.
In the center-tie example, what is the probability of a specified arc conditional on a nonzero sign, and what is the unconditional probability of a nonzero sign in that arc?

Spatial Signs and Their Null Geometry

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