Reading Rank and Depth Configurations

Reading Ranks and Depth Together

A multivariate data set does not generally have a single natural ordering from smallest to largest. An observation can be farther from the center in one direction but closer in another, and two observations may be incomparable without choosing a direction or a particular summary. Spatial ranks and depth address different parts of this problem: spatial ranks retain directional information, while depth summarizes centrality. Reading them together is more informative than forcing either into a single total ranking.

A rank-and-depth display is descriptive, not by itself a test of a hypothesis. Its interpretation depends on the chosen depth function, the reference distribution used to calculate it, and the geometry and scale of the variables. The aim is to identify patterns worth investigating, while keeping those choices visible.

Definition: For a random vector $X$ in $\mathbb{R}^d$, the spatial-rank vector of a point $x$ is $$ R_F(x)=E\left[\frac{x-X}{\|x-X\|}\,1\{X\ne x\}\right]. $$ The indicator makes the contribution zero when $X=x$. Given observations $X_1,\ldots,X_n$, the empirical spatial rank at $x$ is $$ R_n(x)=\frac{1}{n}\sum_{j=1}^n\frac{x-X_j}{\|x-X_j\|}\,1\{X_j\ne x\}. $$ A depth function $D(x;F)$ assigns a centrality value to $x$, with larger values interpreted as more central under that function. Its upper contour at level $\alpha$ is $C_\alpha=\{x:D(x;F)\ge\alpha\}$. A depth rank orders observations by their depth values; equal values remain tied.

What the Spatial-Rank Vector Says

Each nonzero term in $R_n(x)$ is a unit vector from an observation toward the point $x$. Thus the vector combines directions, not distances: an observation one unit away and one a hundred units away contribute vectors of the same length. A small rank norm means that directional contributions largely balance. A large norm means they point more consistently in one direction. The vector’s direction indicates how $x$ is positioned relative to the data, but it is not a displacement measured in the original units.

At a center of symmetry, spatial rank is zero. More generally, the point where spatial rank vanishes is called a spatial median when it minimizes expected distance, subject to the usual existence conditions; in a finite sample, minimizing the sum of distances gives a sample spatial median. The rank vector at a candidate point and the distance-minimizing center are related, but a rank vector should not be read as an ordinary mean difference.

Result: If $X-\theta$ and $\theta-X$ have the same distribution, then $R_F(\theta)=0$. Indeed, writing $Z=X-\theta$, the rank integrand at $\theta$ is $-Z/\|Z\|$ when $Z\ne0$ and zero otherwise. Symmetry makes its expectation equal to the expectation of $Z/\|Z\|$, which is its own negative. The expectation exists because the integrand has norm at most one. Therefore $R_F(\theta)=-R_F(\theta)$, so $R_F(\theta)=0$. $\square$

There is also an exact finite-sample balance property. It is useful when checking a calculation and when interpreting a plot of rank vectors at the observed points.

Result: For any finite sample in $\mathbb{R}^d$, including samples with repeated observations, $$ \sum_{i=1}^n R_n(X_i)=0. $$ To prove this, expand the sum: $$ \sum_{i=1}^n R_n(X_i) =\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n \frac{X_i-X_j}{\|X_i-X_j\|}\,1\{X_i\ne X_j\}. $$ The term with $i=j$ is zero. For each pair $i\ne j$ with unequal values, the $(i,j)$ term is the negative of the $(j,i)$ term, because $X_j-X_i=-(X_i-X_j)$ and the denominators agree. Equal-value pairs contribute zero in both directions. All terms therefore cancel pairwise, proving the identity. $\square$

Worked Example: Balanced Directions and a Central Point

Consider the four observations $(-1,0),(1,0),(0,-1),(0,1)$ in the plane. Calculate the empirical spatial rank at the origin and at the observation $(1,0)$.

At the origin, the four unit-vector contributions, in the order listed, are $(1,0),(-1,0),(0,1),(0,-1)$. Their sum is $(0,0)$, so $$ R_4((0,0))=\frac{1}{4}(0,0)=(0,0). $$ The directions balance exactly.

At $(1,0)$, the observation at that same point contributes zero. The contributions from $(-1,0),(0,-1),(0,1)$ are respectively $(1,0)$, $(1,1)/\sqrt{2}$, and $(1,-1)/\sqrt{2}$. Hence $$ R_4((1,0)) =\frac{1}{4}\left((1,0)+\frac{(1,1)}{\sqrt{2}}+\frac{(1,-1)}{\sqrt{2}}\right) =\left(\frac{1+\sqrt{2}}{4},0\right). $$ The horizontal components reinforce, while the vertical components cancel. The positive horizontal rank indicates that this point lies on the right-hand side of the balanced configuration.

For comparison, the empirical halfspace depth at the origin is $2/4$: any closed halfspace containing the origin contains at least two of the four observations, and a halfspace with boundary through the origin can contain exactly two. At $(1,0)$, the halfspace $x_1\ge1$ contains just that observation, so its depth is $1/4$. Rank and depth agree that the origin is more central, but only the spatial rank conveys a direction.

Reading Depth Contours and Depth Ranks

A depth contour collects points with at least a specified centrality. Higher contour levels give more central regions, when the depth function has the usual nested-contour property. For example, halfspace depth contours are convex; other depth notions may have different geometry. The specific function matters: a contour is not an assumption-free boundary around the data, but a summary under a stated definition of depth.

Result: For any depth function, its upper contours are nested: if $\alpha_2\ge\alpha_1$, then $C_{\alpha_2}\subseteq C_{\alpha_1}$. If $x\in C_{\alpha_2}$, then $D(x;F)\ge\alpha_2\ge\alpha_1$, which implies $x\in C_{\alpha_1}$. This proves the inclusion. $\square$

A depth rank turns depth values into a centrality ordering: observations with larger depth rank as more central, and those with smaller depth rank as more peripheral. But this is not a total ordering of multivariate locations in any geometric sense. Distinct observations can have equal depth, and two observations with unequal depth can lie in very different directions. Even a strict ordering by depth discards the direction that separates them. Ties are often substantive: symmetry, repeated values, or a finite sample can give several observations the same depth.

When plotting contours, state which distribution supplied the reference depth. In a group comparison, pooled depth describes how central each observation is relative to the pooled sample. Group-specific depth instead describes centrality relative to each group and does not put the groups on the same reference scale automatically. These are different questions.

Worked Example: Centrality Levels Do Not Give Direction

Use the four-point configuration from the preceding example and empirical halfspace depth. The origin has depth $2/4$, while each observed point has depth $1/4$. These values identify the origin as more central than any of the four observations. They do not say whether a particular point lies to the right, left, above, or below the origin.

The upper contour at level $1/4$ includes all four observed points, while the level $2/4$ identifies the more central region containing the origin. The spatial rank at $(1,0)$, calculated above, supplies the missing directional information: it points horizontally to the right. For an interpretation, the depth level answers “how central?” and the rank vector answers “in what direction is this location positioned?” Neither answer replaces the other.

This distinction also explains why depth-ranked observations with the same depth need not be close to one another. Equal centrality is not equality of location.

Using Configurations to Assess Asymmetry and Group Separation

Directional asymmetry can appear as rank vectors that do not balance around a proposed center, or as observations and groups concentrated in particular directions across depth contours. A rank vector is a local directional summary at its evaluation point; it is not a complete description of the shape of a distribution. A useful reading considers its direction and norm alongside the locations of points, the contour geometry, and the chosen center.

Worked Example: A Rank Vector Under Directional Imbalance

Take the four observations $(1,0),(0,1),(-1,0),(2,0)$, and evaluate their empirical spatial rank at the origin. The four contributions are $(-1,0),(0,-1),(1,0),(-1,0)$. Therefore $$ R_4((0,0))=\frac{1}{4}(-1,-1)=(-1/4,-1/4). $$

The rank has norm $\sqrt{2}/4$ and points down and left. This records the imbalance of unit directions from the observations toward the origin: the contributions do not cancel. It does not mean that the sample mean is $(-1/4,-1/4)$, nor does its magnitude estimate a distance to a center. In particular, moving the observation at $(2,0)$ farther to the right would not change its contribution at the origin, which remains $(-1,0)$. The display suggests directional imbalance, but further summaries or formal inference are needed to characterize its cause.

For group separation, compare configurations using a common reference, such as pooled depth contours or pooled spatial ranks. Ask whether the groups occupy different directions, whether one group tends to lie in deeper or shallower regions, and whether apparent differences persist across the configuration rather than being driven by a few points. A depth-only comparison can conceal separation: two groups can have the same distribution of depth values while occupying opposite sides of the center.

Worked Example: Equal Depth Distributions but Separated Groups

Place six observations on the horizontal axis, at first coordinates $-3,-2,-1,1,2,3$, and use their pooled empirical halfspace depth in the plane. Assign Group A the first three negative points and Group B the three positive points. For distinct collinear points ordered by first coordinate, the depth at position $i$ is $\min(i,n-i+1)/n$: a halfspace whose boundary is perpendicular to the line can contain either the $i$ points on or to one side or the $n-i+1$ points on or to the other side, and the smaller count is attainable. Here $n=6$.

The depths of Group A, in order from $-3$ to $-1$, are $1/6,2/6,3/6$. The depths of Group B, from $1$ to $3$, are $3/6,2/6,1/6$. Thus the two groups have identical depth distributions. A comparison using only depth ranks would not reveal that the groups occupy opposite sides.

Their spatial ranks do distinguish the directions. For these ordered points on a line, the rank at position $i$ is $(2i-n-1)/n$, since $i-1$ observations lie to its left and $n-i$ lie to its right. Group A therefore has ranks $-5/6,-3/6,-1/6$, with mean $-3/6=-1/2$. Group B has ranks $1/6,3/6,5/6$, with mean $3/6=1/2$. The opposite mean rank directions expose the separation that the depth distributions alone miss. This is a descriptive contrast, not a significance test.

Common Pitfalls

Treating depth rank as a complete multivariate order. Two points can tie in depth while lying in different directions. Report their locations or directional summaries as well.
Reading rank magnitude as distance. Spatial rank averages unit directions, so its norm is not a distance from a center and is insensitive to how far each observation lies along its direction.
Calling a low-depth observation anomalous without context. Depth depends on the chosen function, reference distribution, dimension, and sample. Low depth is a relative centrality statement, not by itself evidence of an error or an outlier-generating mechanism.
Comparing group-specific depth values as if they shared a scale. Depths calculated against different group distributions answer different questions. Use a common reference when the aim is a direct pooled comparison.
Inferring group separation from depth values alone. As the collinear example shows, identical depth distributions can accompany directional separation.

Key takeaway: Read depth as centrality and spatial rank as direction. Together they can reveal central, peripheral, asymmetric, or separated configurations, but ties and lost directional information mean neither summary creates a total ordering or substitutes for formal inference.

Check Your Understanding

For each question, explain both the interpretation and any limitation that applies.

What does a spatial-rank vector with small norm indicate, and why does it not necessarily indicate a nearby observation?
Why is spatial rank zero at a center of symmetry? State the role of symmetry in the argument.
If two observations have equal depth, what can and cannot be concluded about their locations?
Why might pooled depth contours be preferable to separate group-specific contours when comparing centrality?
How can two groups have identical pooled depth-rank distributions and still be directionally separated?