MWU is over 75 years old. It may be used when an approximately normal distribution of sample scores cannot be relied upon. A literature search for MWU for the present article gave over 50,000 articles on PubMed [2].

MWU was not designed for ties but for continuous data [3]. Continuous data can have any value and expects no ties [4]. Small or tied samples may occur while analyzing restraint or other clinical problems on wards, which may feature “small” population sizes less than twenty and/or rating scales smaller than that. Indeed, as proven below, ties must occur when the sample size exceeds the scoring options, see theorem at 1.3. It is problematic and motivating that a canonical test is provably inexact.

The literature discusses how to adapt MWU to handle ties. Options (ignore ties, jitter ties) were compared in mental health care by McGee, who concluded empirically that it is least bad to ignore ties [5]. In response, Neuhäuser and Ruxton, while also using empirical approaches, raised concerns that ignoring ties can violate test assumptions. They particularly recommend exact combinatorial methods [6].

The present manuscript aims to support such debate at least for small samples. This new test is called “Node-Crossing Vectors,” or NXV, due to its mechanism and visual appearance of lines crossing dots across a lattice. NXV is contrived to spell MWU shifted forward, one letter alphabetically: (M → N; W → X; U → V).

NXV defines a “tied element.” NXV creates a “local” distribution for all of the mutually exclusive predicates regarding how many tied elements there are per sample. The range of such predicates is constrained by sample size. Each mutually exclusive predicate is a statement, “the sample has τ tied elements.” NXV proves a distribution of each τ-predicate called E which turns out to be a truncated Poisson distribution, see proof at 1.4.5.

NXV then uses combinatorics derived from further proofs to “solve” each local distribution. There are three such novel combinatorial algorithms. They agree with each other and with the MWU statistical tables for small values, which were predicated on τ₀. They are Napkin-shifting (see below), bitstring partition, and constrained Stern-Brocot path generation (see Sections 2.4.1–2). They are open source. These make sense in a visually accessible lattice paradigm, and all paths with any τ cross a lattice of size s×t as will be seen. The local combinatorics are tightly constrained by τ, because τ affects path length, see discussion at Sections 1.4.1–1.4.4.

To bridge the longstanding gap, this manuscript describes MWU concepts in NXV terms. These include: U_N, the “order score” of a path N; and 𝕌_{s, t}, the “null distribution” of the order score for a given pair of sample sizes s and t. A real a priori problem is analyzed, and ties are shown to be decisive. This is placed in the context of simulation data.

Imagine truly continuous data, or pragmatically accept methods of jittering, or ignoring ties. Hence, confidently expect τ₀. Certainty in τ₀ implies 100% weight to τ₀, 0% to τ₁, ..., until the options for τ run out. Denote such expectation (with foreshadowing) as λ₀. This predicate λ₀ allows acceptable use of MWU if MWU does not handle ties.

There is a visually accessible way to consider orders of two samples given λ₀ see Figure 1. It has been discussed by authorities such as Bucchianico [12], Knuth [13], and Stanley [14]. NXV calls the smaller sample S with s elements and the other T with t elements. Call the total sample N with n = s+t elements. Any order has its order score. Denote the order score U of a total sample N using U_N. U_N tends toward s×t as S comes first in order stochastically.

On a lattice of width s and height t, paths encoding all distinct orders with regard to U may be comprised by vectors going right for s and up for t. The ordered set of all possible paths across this lattice encodes 𝕌_st. Paths must always go from the Cartesian origin (0, 0) to the terminus (s, t). For each path, the size of the area around (0, t), cut off by the path, is U_N. The number of possible untied paths is calculable using the binomial. That is simpler than generating paths, see Technical Appendix A. The binomial works for the same reason Pascal's triangle relates to the binomial: each node enjoys the sum of its parent nodes as paths to the terminus, see Technical Appendix B.

To illustrate, take a small non-trivial stochastic order of two samples over a 𝕌_s3t4λ0 lattice, N = [t, t, t, s, t, s, s] . As encoded in Figure 1, it has an order score U_N = 1. See that |𝕌 ≤ U_N| = 2 in the dotted box. See that |𝕌| = 35. The exact probability of this local s₃t₄λ₀U₁ problem is hence pλ0=235=0.0571(3sf).

Now, only the tail needs to be generated to give |𝕌 ≤ U_N|. The present author wrote a simple algorithm for any s, t, U, λ₀ situation, see Technical Appendix C. The algorithm is called Napkin-shifting. Iterating from U₀ to U_N gives the numerator of the probability equation. If the tail is small, i.e., near α, this is efficient. Napkin-shifts resemble napkins on a table being moved while remaining pushed into the corner of a table. The intent was to reduce the need for approximation, especially for small probabilities, but it is a lemma to the tied problem.

By NXV's novel definition, any “tied element” is directionally tied to one, and only one, other measure. In graph-theory terms, NXV considers elements, the nodes of tying rather than “ties” which are the edges. A tied element is never tied to two or more. There are obvious base cases that neither an empty sample s₀t₀n₀, nor a single measure s₀t₁n₁, have tied elements. Iff a second element is tied, it is tied to the first, and so on by induction. The stochastic mapping of tied elements to the elements they are tied to, amongst a shared value, is injective; but the distribution does not “need to know” the specific mapping, which would not affect U or U. Each measure is either tied or not, so τ is a whole number. Because each tied measure exhausts one element to be “tied to,” by induction, there cannot be more than n−1 tied measures. Hence, in NXV, τ < n and 0 ≤ τ ≤ n−1 also denoted τ∈[0:n−1], i.e. tau is in the range zero to n−1.

In λ₀, each element encoded a step which, for s, is a vector going right in the U-lattices in Figure 1; and the t is a vector going up. Under non-λ₀, in NXV, the vector sum of st-tied steps gives an angled vector. Where there could be st-tied elements, but there happen not to be, in a given step, due to either no s or no t, the step is still a vector; but simply a vector wth a zero value for the relevant axis. This gives MWU-style vertical or horizontal steps as in Figure 1.

Sloping steps have a number of consequences. Any possible combination of steps will still cross the lattice, even a complete tie, which will be a long diagonal line, due to the associative nature of addition. These include null distributions with half-scores for U_N, not just whole numbers. The number of possible paths grows. For example, s₁ and t₁ can have three orders: [s, t], [t, s] and [ts], rather than two. This 3 is the same three near the top of Figure 2: it has 3 paths home to the top.

The present author contributed a peer-reviewed method to compute all paths across a lattice if tied elements and sloped steps are allowed [15]. This creates McMeekin Hill, an elaboration of Pascal's Triangle named for his doctoral supervisors, Figure 2 [16]. This gives a visual proof that the total number of paths across an s₃t₄ lattice is 266 if the lattice can allow slopes to model ties. Each number in that triangular structure is the sum of all visible nodes, not just the two parents. For example, see the path below in Figure 3. This use of slopes to manage the complexity of tied elements is the core innovation of NXV.

Any MWU or NXV path encoding an order of measures has steps. Each step encodes the presence of one or more measures from S or T. Denote the number of steps q. q has limits derivable from s and t as follows.

The lower limit of q is 1, because any path has steps/measures. One upper limit of q is n, because steps encode elements in n. When there are no ties, all paths can be made from n steps as in Figure 1. This q = n attribute is true of all paths with no tied elements and hence all orders handled by MWU.

Tied elements shorten paths, reducing the steps. The pigeonhole principle implies n = q+τ or, from q's point of view, q = n−τ. This is consistent with λ₀ having paths of length n.

Furthermore, if a rating scale has a countable number of possible scores r, as in CGI r₇, the path cannot have more steps than r. So there have to be at least as many tied elements as the difference between n and r. This limit is denoted τ≥r−n. NXV needs a new null U_st, or family of them, that handles λ_>0. It must correctly weigh the expectation of each mutually exclusive predicate regarding τ. These are [τ₀, τ₁...τ_n−1] if r≥n; or [τ_0+n−r, τ_1+n−r...τ_n−1] if r<n.

Two useful axioms seem fair in the context of constructing this null.

The null distribution of tied elements is hence modeled by a known distribution. That is the Poisson distribution, the assumptions for which are: constancy of λ; and independence of tied elements in the null. Poisson is used for independent events occurring over a defined space, like the number of calls a resident doctor gets per on-call. The distribution will be bounded by the limits on τ, which are 0 or n−r at the bottom and n−1 at the top. Bringing this together, the null distribution of tied elements is a “truncated Poisson distribution.” □

Sam's cohort had CGI scores of S = [4, 4, 5, 6]. Terri's had T = [1, 2, 3, 3, 3, 4, 4, 4, 5]. Terri's come first, but to keep U below s×t2, aiding tail calculation, paths can be reversed, to keep the U-area top left without affecting extremity, in a two-tailed test. If round brackets mean steps with tied measures and the superscripts are counts of tied elements in each step, the overall order may be written:

There are τ = 2+4+1 = τ₇ tied elements in total, so this is an s₄t₉n₁₃τ₇ situation.

See Figure 3 for the path encoding, which features slopes for steps with st-tied elements. The order score is U_5.5, picked out in gray dashes, analogous to the black parts of Figure 1. As with the more blocky MWU lattices, there is a decidable order of all possible lattices that comprise the null distribution. The problem's path length is thus n₁₃−τ₇ = q₆, though the first steps have the same angle. The appearance of four actual congruent lines may be called an appearance of four “limbs.”

U_5.5 does not exist in the MWU distribution. MWU only makes whole numbers of U. An approximation would be to use the efficient methods above applied to U_5.5. By factorials, |𝕌_{3, 9}| = 715. By Napkin-shifting |𝕌_{4, 9} ≤ 5| = 18. The exact probability of this incorrect question is pλ0=18715=0.0252(3sf). In this two tailed test, α was set as 0.025. Hence, by a small margin, the null hypothesis remains that neither clinician has easier patients: the author reaffirms that the data were collected a priori, though the narrowness of the result becomes pedagogically useful. Figure 4 shows the null 𝕌_{4, 9}λ₀ distribution with the lower tail picked out in blue. Though U_5.5 would be a gray scatter point, it does not arise.

There are thirteen possible s₄t₉τ_i distributions, each predicated on a particular possible τ. See Technical Appendix D. Sam and Terri ground λ on τ₇ and fix λ₇. The Poisson distribution predicts a probability of various τ being found in null distributions of similar samples. The lowest possible τ is n₁₃−r₇ = τ₆ which could be used if the assumption of lowest possible tied elements were seen as parsimonious. The highest is n₁₃−1 = τ₁₂. So all possible counts of tied elements are in that range, called “tau in range 6 to 12,” denoted τ∈[6:12]. In similar terms, the MWU distribution expecting no ties is a truncated Poisson distribution 𝔼_λ0[0:n−1].

The only conditions are the constancy of λ, independence of tied elements, and the constraints of truncation. Sam and Terri denote their truncated Poisson distribution 𝔼_λ7[6:12]. 𝔼 is a bridge to predicting path lengths. The claim was made that 𝔼 is easy to calculate. Each probability for each τ should have the standard Poisson formula:

This can be simplified as follows. The values e and λ are constant. e is a constant and λ does not vary due to constancy of λ. So, taking out e^−λ gives each τ a simpler formula while keeping their proportional values intact. Using ∝ for “proportional to” instead of =:

This is just λ times itself τ times, divided by the factorial of τ, and can be done by hand for small values. The sum of probabilities for all possible τ is p = 1.0, so they can all be divided by their sum to get a series of probabilities which add up to p = 1.0. This gives a formula for a list of probabilities in the truncated Poisson distribution 𝔼λ_7[6:12].

The link between τ and q is crucial. Each τ-predicate, iff true, implies one and only one q because of the rule n = q+t. These probabilities are exact when written as fractions, but are rounded here for readability. So, rounded to 3 significant figures, the probabilities of the various numbers τ of tied elements are:

It is convenient to reverse these values and leave just the q label before building all paths. The reversed form of the distribution could be denoted ℚ. ℚ is 𝔼 in reverse order, so q7=16328972073714=0.222(3sf) has gone from the front to the back.

Every path has at least one s or t element in every step in its length q. This means each path can be represented by a figure like this T 111110S 000111, of length q, which is called a “presence bit-string.” Indeed, this figure T 111110S 000111 represents N, the order of Sam and Terri's path. The 1s represent the presence of some elements in T and S over the whole path, and the length is still q₆. The present author has written another algorithm to generate all such figures. Then s = 4 and t = 9 can be shared out among the ones. Each presence bit-string represents a family of orders that are different, sharing of s and t amongst its figure 1s on each line. This generates all paths that can arise for a given length. There is some redundancy. In s₄t₄q₂, for example, the bit-string T 11S 11 could encode both T 2,2S 2,2 and T 3,1S 3,1 which result in the same limbs.

This approach, based on Figure 2, uses the Stern-Brocot tree to generate all vectors that have co-prime s and t vectors, allowing for extension through limbs. All paths of length q have q such vectors and meet at the terminus. With inputs of s, t, q, it generates all paths across a lattice of length q. Constrained Stern-Brocot path generation is efficient enough to have solved and cached all s, t, q distributions up to a certain size in a “output.txt” file on a domestic computer. The file in the shared source code is a Julia program called “stqU_iterator.jl” which writes the output, including Unicode graphs of distributions per Appendix D. Once output is cached, the execution is trivial on a domestic computer as a separate program called “jupyter-st_base.ipynb” does the lambda weighting and presentation. This was used for the simulations.

Therefore, each mutually exclusive q-set of paths has two calculable attributes shown in Table 1. Each has its U_stq which gives its own probability, based on a fraction tailtotal. Each also has its own probability under ℚ. For clarity, the left-hand column is the probabilities of each predicate, the right-hand column is the local probability if the predicate is true. The left is built using Poisson assumptions about τ. The right is combinatorially built using constraints of q. The bridge between them is the pigeonhole principle n = q+t.

Each local p_{nxv q} can be multiplied by the probability of the q path length arising to give a weighted sum. That is the exact test score, p_NXV, written with capital subscripts above to reflect it is the weighted sum of all the circumstances arising from this U, s, t, λ situation. Applying this method with λ₀ gives p_MWU test results, which is p = 0.252, as seen at 2.2.3. That is the probability if ties are impossible, and if the p_nxvqn tailtotal=18252 is weighted at p = 1.00 of occurring. That is out of the question, because there is no likelihood of λ₀ in this problem given n and CGI's small r.