Why a Multivariate Rank Needs a Rule
For a univariate sample, observations can be sorted from smallest to largest, with a convention for ties. In two or more dimensions, “smaller” is not automatic. One observation may be lower on one coordinate and higher on another, so the data themselves do not always determine which observation should come first. A multivariate rank therefore depends on an additional choice: what feature of the points defines their order?
That choice is geometric. One may compare coordinates, distance from a chosen center, or position along a chosen direction. Each rule can answer a useful question, but different rules can assign different ranks to the same observations. The point is not that one construction is universally correct; it is that a rank has to be interpreted in light of the geometry used to define it.
The distinction between \(\preceq_c\) and \(\prec_c\) matters. The first relation is reflexive: every point is coordinatewise at most itself. The second is irreflexive: a point does not strictly dominate itself. The non-strict relation is a partial order, while the strict relation is its associated strict order. Neither relation compares every pair of points when \(p\geq 2\).
Coordinatewise Order Does Not Give a Total Rank
A partial order permits some pairs to remain incomparable. For example, a patient with a lower biomarker value but a higher blood-pressure value cannot be declared smaller on both measurements. Sorting such observations into a single list requires a further convention; it cannot be obtained from coordinatewise dominance alone.
Proof. A partial order must be reflexive, antisymmetric, and transitive. Reflexivity holds because \(x_k\leq x_k\) for every coordinate. If \(x\preceq_c y\) and \(y\preceq_c x\), then \(x_k=y_k\) for every coordinate, so \(x=y\); this proves antisymmetry. If \(x\preceq_c y\) and \(y\preceq_c z\), then \(x_k\leq y_k\leq z_k\) for every coordinate, so \(x\preceq_c z\); this proves transitivity.
To see why the order is not total, consider \(a=(1,0,0,\ldots,0)\) and \(b=(0,1,0,\ldots,0)\). The first coordinate prevents \(a\preceq_c b\), while the second prevents \(b\preceq_c a\). Thus \(a\) and \(b\) are incomparable. A total order must compare every pair, so the coordinatewise partial order is not total. Moreover, either \(a\) before \(b\) or \(b\) before \(a\) can be chosen when extending the order to these two points; the partial order itself selects neither choice. \(\square\)
Worked Example: Identifying Incomparable Observations
Four observations have two measured outcomes:
For each successive point, the first coordinate increases and the second decreases. For example, \(A\preceq_c B\) fails because \(3\not\leq 2\), and \(B\preceq_c A\) fails because \(1\not\leq 0\). The same opposing-coordinate pattern shows that every pair among \(A,B,C,D\) is incomparable.
Consequently, coordinatewise order gives no basis for saying which of these four observations has the highest or lowest rank. A rule that ranks them by the first coordinate gives \(A,B,C,D\); a rule that ranks them by the second coordinate gives \(D,C,B,A\). Both are possible orderings, but neither follows from coordinatewise dominance.
Distance Creates Radial Ranks
One way to make a scalar ordering is to choose a center \(c\) and rank observations by their distance from it. Given scores \(s_i=d(X_i,c)\), a low radial rank means close to the selected center and a high radial rank means far from it. This turns multivariate points into scalar scores, but the result depends on the center and the distance function.
For a fixed set of scores, ties also need a convention. A common choice is the midrank: if a score has \(L\) observations strictly below it and \(t\) observations tied at that score, assign each tied observation rank \(L+(t+1)/2\). Any rank calculation should state how it handles ties, since different tie conventions can affect later statistics.
Worked Example: Rescaling Reverses a Distance Ranking
Take the center to be \(c=(0,0)\), and compare \(a=(2,0)\) with \(b=(0,1.5)\) using Euclidean distance, \(d_E(x,c)=\sqrt{(x_1-c_1)^2+(x_2-c_2)^2}\).
The distances are \(d_E(a,c)=\sqrt{2^2+0^2}=2\) and \(d_E(b,c)=\sqrt{0^2+1.5^2}=1.5\). Thus \(a\) is ranked farther from the center than \(b\).
Now measure the first coordinate in units that are half as large, using \(T=\operatorname{diag}(1/2,1)\). The transformed points are \(Ta=(1,0)\) and \(Tb=(0,1.5)\), so their Euclidean distances from the transformed center are \(1\) and \(1.5\), respectively. The ordering reverses: \(b\) is now farther out. The data points did not change in substance; the coordinate scaling, and therefore the geometry used to compare them, changed.
A related choice is to account for the shape of the data with a positive-definite matrix \(\Sigma\). The squared Mahalanobis distance from \(x\) to \(c\) is \((x-c)^\mathsf{T}\Sigma^{-1}(x-c)\). Directions with larger scale under \(\Sigma\) contribute less to this distance than equally sized deviations in directions with smaller scale. As a result, Euclidean and shape-adjusted radial ranks need not agree.
Worked Example: Euclidean and Shape-Adjusted Ranks Differ
Use the same center \(c=(0,0)\) and observations \(a=(2,0)\), \(b=(0,1.5)\), but now take the specified shape matrix \(\Sigma=\operatorname{diag}(4,1)\). Its inverse is \(\Sigma^{-1}=\operatorname{diag}(1/4,1)\).
The squared Mahalanobis distance for \(a\) is \((2,0)\Sigma^{-1}(2,0)^\mathsf{T}=4/4=1\), so its distance is \(1\). For \(b\), the squared distance is \((0,1.5)\Sigma^{-1}(0,1.5)^\mathsf{T}=1.5^2=2.25\), so its distance is \(1.5\). Under this shape adjustment, \(b\) is farther from the center than \(a\), even though its Euclidean distance is smaller.
The ranking reflects the selected shape: variation along the first coordinate is treated as four times as large in variance units, so a two-unit displacement there is only one unit of Mahalanobis distance. The matrix is part of the ranking rule, not a detail that can be omitted from its interpretation.
Invariance Is a Choice, Too
A geometry can be chosen partly for how it behaves when coordinates are changed. A distance is invariant under a transformation if applying the transformation to the data leaves the relevant distances unchanged. Euclidean distance is preserved by translations and orthogonal transformations, but generally not by unequal rescalings. Mahalanobis distance has a useful affine invariance when the shape matrix is transformed together with the observations.
Proof. Since \(B\) is nonsingular and \(\Sigma\) is positive definite, \(B\Sigma B^\mathsf{T}\) is positive definite. The inverse identity \[ (B\Sigma B^\mathsf{T})^{-1}=(B^\mathsf{T})^{-1}\Sigma^{-1}B^{-1} \] follows by multiplying the right-hand side by \(B\Sigma B^\mathsf{T}\) and obtaining the identity matrix. Set \(z=x-c\). The transformed squared distance is \[ (Bz)^\mathsf{T}(B\Sigma B^\mathsf{T})^{-1}(Bz) =z^\mathsf{T}B^\mathsf{T}(B^\mathsf{T})^{-1}\Sigma^{-1}B^{-1}Bz =z^\mathsf{T}\Sigma^{-1}z. \] This is exactly the original squared Mahalanobis distance. \(\square\)
This invariance statement is conditional: the shape matrix must be transformed along with the points. If coordinates are changed but the shape matrix is held fixed, the equality need not hold. For sample-based procedures, a corresponding invariance claim also requires the estimated shape matrix to transform appropriately. These qualifications matter because different invariance requirements can motivate different rank constructions.
Worked Example: Checking Affine Invariance Numerically
Continue with \(\Sigma=\operatorname{diag}(4,1)\), \(a=(2,0)\), and \(b=(0,1.5)\). Apply \(B=\operatorname{diag}(2,1)\). The transformed shape matrix is \(B\Sigma B^\mathsf{T}=\operatorname{diag}(16,1)\), whose inverse is \(\operatorname{diag}(1/16,1)\). The transformed points are \(Ba=(4,0)\) and \(Bb=(0,1.5)\).
For \(Ba\), the transformed squared distance is \(4^2/16=1\), matching the original squared distance for \(a\). For \(Bb\), it is \(1.5^2=2.25\), matching the original squared distance for \(b\). Their radial ordering remains the same when both data and shape are transformed. This calculation illustrates why the transformation rule for the shape matrix is essential.
Direction Creates Another Kind of Rank
Instead of asking how far a point is from a center, one can ask how far it lies along a chosen direction. For a unit vector \(u\) and center \(c\), the projection score is \(u^\mathsf{T}(x-c)\). Sorting these scores gives a univariate ordering along that direction; ties can again be handled by midranks or an explicitly specified tie-breaking rule. A different direction can give a different ordering, or turn a difference between two points into a tie.
Worked Example: Direction Changes Projection Ranks
Let \(a=(1,0)\), \(b=(0,1)\), and \(c=(0,0)\). Along \(u_1=(1,1)^\mathsf{T}/\sqrt{2}\), the scores are \(u_1^\mathsf{T}a=1/\sqrt{2}\) and \(u_1^\mathsf{T}b=1/\sqrt{2}\). The two observations tie in this direction.
Along \(u_2=(1,-1)^\mathsf{T}/\sqrt{2}\), the scores are \(u_2^\mathsf{T}a=1/\sqrt{2}\) and \(u_2^\mathsf{T}b=-1/\sqrt{2}\), so \(a\) ranks above \(b\). Reversing the direction to \(-u_2\) reverses that ordering. The rank is therefore not a property of the unordered pair of points alone; it also depends on the direction chosen for comparison.
What a Multivariate Rank Communicates
Coordinatewise, radial, and directional rankings encode different questions. Coordinatewise order records whether one point is no larger on every measured coordinate. Radial ranks summarize distance from a specified center under a specified metric. Directional ranks summarize position along a specified axis. None is a geometry-free substitute for the univariate ordering, because multivariate data do not have a single natural “smallest to largest” path through all points.
This does not make multivariate ranks arbitrary or unhelpful. It makes the construction part of the statistical method. When interpreting a rank-based procedure, identify the center, distance or direction, treatment of ties, and transformations under which the rule is intended to behave consistently. These choices determine what “more extreme” or “higher ranked” means. Different rank methods make different choices to serve different inferential aims.
Common Mistake
- Wrong reasoning: “Coordinatewise comparison is a partial order, so it assigns every pair of observations an order and therefore gives each point a unique rank.” This confuses a partial order with a total order. A partial order is reflexive, antisymmetric, and transitive, but it need not compare every pair.
- Correction: Coordinatewise order leaves mixed-coordinate pairs incomparable. A single ranking requires an added rule, such as distance from a chosen center or projection along a chosen direction. The resulting rank depends on that rule, and ties may require a further convention.
- Another pitfall: Calling strict coordinatewise dominance itself a non-strict partial order. Strict dominance requires at least one strict coordinate inequality, so no point strictly dominates itself. The reflexive partial order is \(\preceq_c\); its associated irreflexive relation is \(\prec_c\).
Check Your Understanding
Answer these questions by identifying the ordering rule and the geometry it uses.
- For \(x=(1,4)\) and \(y=(2,3)\), are the points comparable under the non-strict coordinatewise order? Explain.
- Why is strict coordinatewise dominance not a reflexive partial order?
- Two points have different Euclidean distances from a center. Describe one coordinate transformation that could change their distance ordering.
- What must be transformed along with the data for the stated affine invariance of Mahalanobis distance to apply?
- For a fixed center, what feature of the observations is summarized by projection ranks, and why can changing the direction alter their ordering?