A split survives what a mean does not

The parity argument gives a mean and a median split the same status. Both are odd, so a rule holding both has a worst case over seven outcome shapes of exactly zero, and adding the second to the first leaves it there.

On a covariate that is not symmetric the two come apart completely. One of the zeros is the same number under every marginal there is, to the last digit stored, and the other is gone at a skewness of one.

Two guarantees under one argument #

The parity result was one sentence covering two rules, and nothing in it distinguished them. That is the usual way a result comes apart: not because it was wrong, but because it was stated at a level of generality that hid a difference the conditions were doing.

The reason, in one line #

A monotone map preserves order. So the median of X = T(Z) is the image of the median of Z, and

$$\mathbf{1}\{X > \operatorname{median}(X)\} = \mathbf{1}\{Z > 0\}$$

whatever T is. Centred, that indicator is sign(Z)/2 — odd on the latent scale for every marginal there is.

A mean is not. The centred covariate is T(Z) − E[T(Z)], which is odd only when T is odd, and T is odd only when the marginal is symmetric.

That is the whole of the essay. The rest is what it is worth.

Two zeros, measured #

The invariance is not “small under every marginal”. It is the same number.

A dictionary of median splits of each covariate, against an outcome depending on the product of those splits, removes 1.08 × 10⁻³⁰ under the normal, under the hyperbolic sine, under all three Tukey transformations and under the exponential. Six marginals with skewnesses of 0, 0, 0.95, 2.26, 4.75 and 2.00, and one number.

The reason it is one number rather than six small ones is that both the dictionary and the target are functions of the signs of the latent normals. The joint law of (sign Z₁, sign Z₂) is a two-by-two table that depends on ρ and on nothing else, so every inner product in the calculation is literally unchanged by T. It is not that the transformation’s effect is small; it is that the transformation is not in the calculation.

Against that, the means: 6.5 × 10⁻³³, 7.9 × 10⁻³³, 11.63%, 26.30%, 29.73% and 28.51%.

Where the invariance comes from, exactly #

The claim that “the transformation is not in the calculation” deserves the arithmetic, because it is what separates an invariance from a coincidence.

Every quantity in the removed share is an expectation of a product of dictionary entries and target components. When both are median splits, every one of those is an expectation of a product of indicators 1{Z₁ > 0} and 1{Z₂ > 0} — so every entry of the Gram matrix, every covariance with the target, and the target’s own variance is a function of one number:

$$\Pr(Z_1 > 0,\ Z_2 > 0) = \tfrac14 + \tfrac{1}{2\pi}\arcsin\rho,$$

which the cut-point field derives and which depends on ρ alone. The covariate never appears. So the removed share is a fixed function of ρ, and 1.08 × 10⁻³⁰ is the numerical zero that function returns at ρ = 0.5 through the quadrature’s own rounding.

The same argument covers any dictionary and any target built entirely from median splits. It does not cover a dictionary of median splits against a target that is a product of the covariates themselves — that target changes with T, and its removed share does too.

Both rules are the same formula at two points #

The two guarantees can be written as one expression, and doing so says exactly what separates them.

A cut at the latent point c has an odd share of $1/(2\Phi(c))$ — the closed form the threshold field derives — and both rules here are cuts at a latent point.

A median split cuts at c = 0, for every marginal, because a monotone map takes the median to the median. $\Phi(0) = \tfrac12$, so the odd share is exactly 1, and it is 1 whatever T is.

A mean split cuts at $c_T = T^{-1}(\operatorname{E}[T(Z)])$, which is zero only when the marginal’s mean coincides with its median. For an exponential covariate the mean is 1 and $P(X \le 1) = 1 - e^{-1} = 0.632$, so $c_T = 0.338$ and the odd share is $1/(2 \times 0.632)$, which is 0.791.

So the two rules are one formula read at two places. The median split sits at the single point where that formula equals one, and it sits there for every marginal there is; the mean split sits wherever the gap between a distribution’s mean and its median puts it.

That is a more useful statement of the difference than “one is invariant and the other is not”. It says what the mean split’s guarantee is worth when it is not zero — 79.1% odd on an exponential, so 20.9% even and unprotected — and it says the single quantity a designer would have to look at to know: how far the covariate’s mean is from its median, measured as a quantile.

What a trial is actually balancing #

A rule holding both — a mean and a median split of each covariate, which is what a covariate-adaptive design usually holds — has a worst case that is the worse of what its parts can do.

On the normal covariate that worst case is exactly zero, and it is attained on the square of one covariate, which is even and which an odd dictionary cannot touch. On the skewed marginals it is 0.74%, 1.83% and 2.24%, and on the exponential 2.49% — and the shape it is attained on changes: the square is now partly reachable, and the binding shape becomes the product of the two median splits, which is the zero that survives.

So the guarantee’s binding case moves onto the invariant zero. Whatever else a skewed covariate does, the one thing a rule of odd functions definitely cannot protect against is an interaction between two median splits, on every marginal, exactly.

That is worth reading twice, because it is a rare shape: an exact result about a real trial’s blind spot that does not need the covariate’s distribution. It is a negative result, and negatives that survive transformation are the ones worth writing into a protocol.

Which functions survive, and which do not #

The rule generalises past means and splits and it is worth stating in its general form.

A dictionary entry is protected — keeps its parity, and with it every exact zero it participates in — exactly when it is a function of the latent normal’s sign, or more generally an odd function of the latent normal. That is a statement about the entry’s expression in copula coordinates, not in the covariate’s own.

Only one quantile split is protected, and it is the median. A split at the median is 1{Z > 0}; a split at the 80th percentile is 1{Z > z₀.₈}, which is not odd on any marginal. So the invariance is about the median split rather than about splits in general, and a rule that stratifies at a tertile or a quartile has no more claim on it than one that stratifies at a value.

No power of a skewed covariate is protected. X, X², X³ are all mixtures of parities once T is not odd.

A rank-based covariate is protected entirely. Replace the covariate by its own normal scores — the inverse normal transform of its empirical rank — and the marginal is normal by construction, so every zero the parity argument gives is back exactly. That is not a small remark: it is a rule a trial can adopt, and it costs nothing but a monotone relabelling of the covariate before the balancing rule sees it — the same move the continuous-covariate field makes when it has to give a continuous covariate levels, with the same caveat: the relabelling is free to the rule and is not free to the interpretation.

What the relabelling costs #

Nothing, in the geometry, and something in the interpretation.

The balancing rule operates on the design and never on the outcome, so transforming a covariate before balancing changes which assignments are admissible and changes nothing about what the trial estimates. And the outcome model is not assumed to be linear in anything — the shapes worth protecting is a list, and the list is stated on the covariate rather than on its transformation.

That last clause is where the cost sits. A rule balancing the normal scores of a covariate protects against shapes that are functions of the normal scores, and the outcome depends on the covariate. A linear outcome in a lognormal covariate is a wildly non-linear outcome in its normal scores, and a rule balancing the scores is holding the wrong function.

So the choice is between a guarantee stated in coordinates the outcome does not live in, and no exact guarantee at all. Neither is obviously right, and this collection cannot decide it: the answer depends on whether an outcome is more nearly a function of a covariate or of its rank, which is a subject-matter question.

What the measurement contributes is that the choice exists and that it is a real one, which nothing before it could say.

The two rules, priced against each other #

Read the invariance the other way and it says something about which rule to prefer, and the answer is not the obvious one.

A dictionary of median splits removes 65.65% of a linear shape at a correlation of a half, and 100% of a median-split shape. A dictionary of means removes 100% of a linear shape and 63.66% of a median-split shape. So on a normal covariate the two are near-mirror images and the choice between them is a statement about which shape the outcome is more likely to take.

Off the normal that symmetry goes. On a strongly skewed covariate the median-split dictionary removes only 33.68% of a linear shape, because the covariate’s linear part is now mostly in its long tail and a split at the median cannot see where in the tail a unit is. The mean dictionary still removes 100% of it — a linear shape is exactly in its span by construction, whatever the marginal.

A median split loses power as the covariate skews and keeps its guarantees; a mean keeps its power and loses its guarantees. That is a genuine trade rather than a dominance, and it is the shape the field’s last essay measures across the whole table.

A third way to see it #

There is a formulation of the invariance that needs no algebra at all, and it is the one to carry.

A balancing rule that reads only ranks cannot be affected by a monotone relabelling of what it reads. A median split is a rank statement — is this unit in the top half — and so is the target it fails to protect against. Everything in that comparison is invariant under relabelling because relabelling does not change ranks.

A mean is not a rank statement. It reads the values, and a monotone map changes the values.

That reading also says exactly where the invariance stops. The moment either side of the comparison stops being a rank statement — a linear outcome, a squared covariate, a threshold at a value — the marginal is back in the calculation. Which is why the invariance covers one cell of the table and not the table.

The one that never had it #

The third rule in the picture is a threshold at a value on the covariate’s own scale — a dose, a temperature, a clinical cut-off — and it removes between 4.88% and 22.49% across the six marginals, with the largest figure on the normal covariate.

It never had a zero to lose. A cut at a value sits at some point of the latent scale that is not zero, and a threshold anywhere but the median is neither odd nor even. That is its own essay, and its practical importance is that it and the median split read the same in a protocol: split the covariate at a threshold.

What the numbers say about size #

Everything above is about whether a guarantee holds. The last question is whether it matters, and the honest answer is that the quantities involved are small.

The worst case of the rule a trial actually runs moves from exactly zero to 2.49% at its worst across these six marginals. Two and a half per cent of a variance is not a large fraction of anything, and a trial designer told that their blind spot has gone from nothing to two and a half per cent would reasonably shrug.

The reason to have measured it is that the shrug was not available before. A guarantee that has become an unknown is not the same object as one that has become a small number, and the only way to know which it is is to compute it. The computation is exact, it takes seconds, and it needed a route the field it came from does not have — the Hermite series that is exact for polynomials converges slowly for a lognormal tail, and the arithmetic had to be rebuilt to answer the question at all.

What the smallness does license is a statement of the form the guarantees degrade gracefully, which is worth having and which is not what a reader would assume: a result that holds exactly under one condition and fails under another usually fails badly.

What is claimed here, and what is not #

This essay takes which of the two functions every trial balances survives a change of marginal. The claims are that a median split of a covariate is the median split of the latent normal under every monotone transformation, so its exact zero is the same number — 1.08 × 10⁻³⁰ — under all six marginals rather than six small numbers; that a mean is odd only when the marginal is symmetric, so its zero is 11.63% at a skewness of 0.95 and 29.73% at 4.75; that a rule holding both has a worst case of exactly zero on a symmetric covariate and 0.74% to 2.49% otherwise, with the binding shape moving onto the invariant zero; and that replacing a covariate by its normal scores restores every exact zero at the cost of stating the protected shapes in coordinates the outcome does not live in.

What stays out, and is named as a decision: whether to use normal scores. The essay establishes that it restores the geometry and says what it costs, and does not recommend it. The cost is a statement about the outcome model and this collection’s whole discipline is that an outcome model is the thing nobody knows.

Also out: splits at quantiles other than the median. They are named as unprotected and not measured, and the reason is that a split at the 80th percentile is the same object as a cut at a value once the marginal is fixed — the two differ in how the cut point is specified rather than in where it lands.

The boundary against the essay that broke the zero is that it asks what the guarantee needs and this one asks which parts of it have it. The answer is that the collection had two guarantees under one argument, and only one of them was ever about the normal law.

The checks, and the refusals that make them mean something #

Two claims are gated. The median split’s zero is required to hold under every marginal and to be the same number — the spread across the six is required to be below a millionth, because six small numbers and one number are different findings and only the second is an invariance. And every transformation is required to be increasing, checked on the values the quadrature evaluates it at, because a transformation that were not monotone would break the identity the whole essay rests on without breaking anything visible.

The refusal is the protocol sentence. A threshold at a value read as though it were a median split is refused, with the median split’s exact zero printed beside the threshold’s 22.49% on the same normal covariate: the exact zero is a fact about the cut being at the median, not about it being a cut.