The covariate on its ranks

Everything a balancing rule is said to do to an interaction on a normal covariate rests on a symmetry. Balancing a skewed covariate found what happens without it: a rule holding the means of two covariates suddenly removes a large share of their product, since the zero depended on odd moments that a skewed marginal no longer has. A share at a hundred units then found that a trial mostly does not collect that share, because the share lives in a long tail that a hundred units barely samples.

Its last section proposed the next thing to try. A rule that balances a covariate’s ranks rather than its values balances a function whose law is uniform on any marginal, so every symmetry argument holds for it exactly. It gives up the tail, since a rank treats the largest value as one more unit. Whether that trade wins would say whether the skew is a property of the covariate or of the scale the rule was told to read it on.

The answer is that it is both, and that the two cases can be told apart by a closed form before any trial runs. It also turns out that the trade is not the best one on offer.

What a rank holds, and what it cannot #

Four marginals are run, each the latent normal read through a different transform: normal, a skew of 2.00 (the exponential), and two lognormal-shaped scales with skews of 2.26 and 4.75. Pairs of covariates correlate at 0.5 on the latent scale. The rule reads either the two values, or the two ranks among all the units, and takes the better arm with probability 0.8. What is scored is the variance of the imbalance left in each of six shapes, against a coin on the same units.

The figure at the head of this essay is two of those shapes at four hundred units. Read the lower pair first, a median split of one covariate.

On a rule reading values, the share of a coin’s imbalance left in the median split runs 37.4% on the normal, 58.7% on the exponential, 61.9% at a skew of 2.26 and 95.5% at a skew of 4.75. As the skew grows, the value of the covariate has less and less to do with which side of the median a unit falls on: most of a skewed value’s variance is in a few units in its tail, and balancing it says almost nothing about the middle. At the strongest skew, a rule balancing values has effectively stopped seeing the median split.

On a rule reading ranks, the same share is 25.4%, 25.0%, 25.0% and 25.1%. It does not move.

That is an exact consequence of what a rank is. A median split is a function of a rank alone, so its relationship to the rank is the same whatever the marginal. Correlating a uniform with its own median split gives a squared correlation of 3/4, and the rule therefore leaves a quarter. So for this shape, the skew was a property of the scale. On the ranks the question has one answer for all four covariates.

Now read the upper pair, a shape linear in the value. On a rule reading values it is 1.1%, 1.4%, 1.5% and 2.5% across the four marginals. The rule was told to balance this and it does, a little less well as the skew grows and the tail gets harder to hold. On a rule reading ranks it is 5.7%, 28.1%, 28.6% and 52.2%.

So on the strongest skew the rank has given up nearly everything on the most ordinary outcome a covariate could have, because a rank does not know how large a large value is. A patient whose biomarker is thirty times the median and a patient whose biomarker is twice the median sit one rank apart, and an outcome linear in that biomarker does not agree.

The cap, in closed form #

That number is not a sampling accident, and no number of units would bring it down. A rule reading a rank can balance only what is a function of the rank. So what it must leave of a shape linear in the value is one less the squared correlation between the value and its rank.

On the normal that correlation is well known. $1 - 3/\pi = 4.5\%$, and the rule at four hundred units leaves 5.7%. On the exponential the cap is 25.0% and it leaves 28.1%. On a skew of 2.26, the cap is 25.2% and it leaves 28.6%; and on 4.75, it is 45.6%, and it leaves 52.2%, on a trial’s noisiest ratio. From a hundred units to four hundred nothing changes materially, since the cap binds well before the rule’s own residual imbalance does: on the strongest skew, 49.5% at a hundred and 52.2% at four hundred, a difference well within sampling error.

So the essay on a hundred units now has a companion. It showed that values leave more than they should because the tail is hard to hold. Ranks leave more than they should because the tail is not there. These are not the same thing, and the second has a floor nothing can move, where the first keeps improving as the trial grows: 11.8% at a hundred units on the strongest skew falling to 2.5% at four hundred.

The cap also answers the question in a form that can be used. Before the trial, compute $1 - \operatorname{Corr}(\text{rank}, \text{value})^2$ for the covariate’s marginal. That is the fraction of any linear effect that a rank-based rule will leave unbalanced, whatever else it does. If the outcome is believed to be closer to linear in the covariate than to a step, that number is the case against ranks.

The other four shapes #

The hoped-for improvement does not materialise elsewhere. On the strongest skew at four hundred units, the square of one covariate is left at 72.0% by values and 101.3% by ranks. The product of the two covariates is 95.8% and 110.8%. The product of the two median splits, the shape the symmetry argument was about, is 94.8% and 105.9%; a coin is 100%, so neither rule touches it, which is the same exact zero the arithmetic has been promising. And one covariate times the other’s square is 102.1% and 106.0%.

So on the skewed scale the rule on values collects a modest share of the product and the square, by the skew’s own doing, and the rule on ranks gives that up too. The ranks restore the normal’s table on all of those shapes, zeros included. They do restore the symmetry. They just do it by removing what the symmetry’s absence was accidentally paying for.

A rule cannot have the ranks it wants #

A rule assigns as units arrive. A rank among the whole cohort is not known until the last unit, so a sequential rule has to read a running rank: each unit’s position among those already seen. On the first units that is nearly noise, and it gets better as the trial accumulates.

The same figure has that third bar, and at four hundred units the cost is small but not nothing. A running rank leaves 49.1% of the linear shape on the strongest skew, within noise of the cohort rank’s 52.2%, and 26.3% of the median split against 25.1%. On the normal, at a hundred units, it leaves 35.6% of the split against the cohort rank’s 30.3%, and 15.6% of the linear shape against 8.9%. So at a hundred units the running rank is noticeably worse and at four hundred mostly is not, which is the shape a learning curve should have. It is a cost that a trial with baseline measurements known before assignment does not pay, and any sequential enrolment does.

The rule that does not have to choose #

There is an alternative, and it is not subtle. The rank was being asked to carry two things: to reach the middle of the distribution, which the value does not, and to replace the value, which it cannot. A rule can simply hold both of them.

Add a median split of each covariate to a rule reading values, and to a rule reading ranks. On the strongest skew at four hundred units:

At a hundred units, it is 17.7% and 50.8% on the linear shape, 6.5% and 6.3% on the split, 79.4% and 102.4% on the product.

So a rule holding values and splits ties ranks on the shape ranks were brought in for, and it beats them by a factor of ten on the linear shape, and it collects a real share of the product besides. It holds four functions rather than two, and the essay on how many a trial can hold and the essay on what the list of functions decides price that: at four hundred units nothing, and at a hundred, 11.8% on the linear shape rising to 17.7%. Against a loss to 50.8%, that is cheap.

So the median split was what the rule needed, and the rank was an elaborate way of asking for it. A rank’s correlation with a median split on any marginal is $\sqrt{3}/2$. Most of what makes a rank attractive on a skewed covariate is that it is largely a median split, and the part that is not is only a coarse version of what the value does better. A split survives what a mean does not had already shown that the split’s parity is immune to the marginal. This essay finds that once a rule holds the split, no reason remains to change the scale.

A worked outcome makes the gap concrete. Suppose a biomarker with the strongest skew here enters the outcome half through its value and half through a threshold at its median, so a trial’s excess variance from imbalance is the average of the two shares. At four hundred units a rule on values leaves 49.0%, because it has lost the threshold. A rule on ranks leaves 38.7%, having partly lost the value. A rule on values and splits leaves 3.0%. That is a much larger difference than anything separating the first two, and it does not depend on knowing how the outcome divides between the two: at any mixture, the third beats both, or ties the one that happens to have held the only shape present.

So a comparison between values and ranks, which is how this choice is usually framed in a statistical analysis plan, is between two rules of about the same quality, and the rule that dominates both is not on the menu.

So whose is the skew? #

The question was whether skew is a property of the covariate or of the scale the rule reads it on. There are two honest answers, and they are about different shapes.

For the shapes that are functions of rank, skew is a property of the scale. A median split, or any other threshold at a quantile, has the same relationship to its rank on every marginal, and a rule on ranks gets exactly the same answer across all of them. That is the content of the flat row in the first figure, 25.4% to 25.1%, and it holds to within the sampling error of the trials.

For the shapes that care how large a value is, skew is a property of the covariate. There is real information in the tail, and it is real to the outcome if the outcome is linear in the covariate. A rank has thrown it away, by construction, and the closed form says exactly how much.

A trial reading a biomarker with a long right tail can decide which case it is in, with no simulation, from a clinical judgement about how the biomarker enters the outcome. If the protocol has no view, hold the values and a median split together, because that covers both.

Nothing about this has touched an analysis. A trial that assigned on a rule holding values and splits can still analyse on whatever scale it chooses, and the analysis and the shape has shown how much that choice recovers. What this essay fixes is only what the assignment leaves behind for the analysis to deal with.

How the numbers were made #

Fifteen hundred trials at each of four marginals and two sizes, with all six designs and a coin assigned on the same units in each trial. The latent normal pair has a correlation of 0.5, and each marginal is that pair read through its own transform. Cohort ranks are centred and scaled to unit variance by the uniform’s own $\sqrt{12}$, running ranks likewise, and values are standardised by the sample. At this number of trials a variance ratio has a relative error of about 3.7%, so differences like 49.1% against 52.2% are not meaningful, and neither is anything about a coin’s imbalance reading 104%.

The closed-form cap comes from a midpoint rule on the latent scale. Two routes are required to meet on every marginal at four hundred units: the rule on ranks must leave between a point below its cap and eight points above it on the linear shape, and a quarter to within four points on the median split. The reading the essay set out to test, that a rule on ranks and splits holds the linear shape as well as a rule on values and splits, is refused on the strongest skew.

Still open: a rank of a sum, and a tail that matters less #

A combined score. A trial balancing several skewed covariates might rank their sum, or the sum of their ranks, and those are different. The first reads the tail of whichever covariate is largest in a given unit, and the second does not. Whether a single score does better than holding each covariate separately is the multi-covariate version of this comparison, and on skewed margins, the field’s own threshold tables suggest it will not be simple.

A partial transform. Between a value and a rank there are transforms that pull the tail in without flattening it: a logarithm on a positive scale, a square root. Each has its own cap on the linear shape and its own distance from the split’s hold. A protocol is far more likely to log-transform a biomarker than to rank it, so the question of what a logged rule leaves on each shape, on the original scale, is one a real trial is actually asking. It is a short sweep, and it is not here.

Which scale the rule should read #

This essay takes a balancing rule reading ranks of a skewed covariate, on four marginals at a hundred and four hundred units. Its claims are that a rank holds a median split at a quarter of a coin’s imbalance on every marginal, where a value’s hold falls to 95.5% on the strongest skew; that a rank is capped on a shape linear in the value at one less the squared correlation between the two, 4.5% on the normal and 45.6% on the strongest skew, and the count sits just above that cap at both sizes; that on the other four shapes ranks restore the symmetric table and give up the skew’s accidental gains; and that a rule holding values and median splits holds the linear shape at 4.6%, the split at 1.4% and the product at 72.8%, which ranks and splits reach only on the split.

What stays out, as a decision: partial transforms, named above; more than two covariates; and analysis, since every number here is imbalance, and a rank-based rule does not require an analysis on ranks. The boundary against a share at a hundred units is that it asked what a rule reading values collects on a skewed covariate, and this one changes what the rule reads, and finds the closed form that limits the change.