Normal Distribution
nothing is called 95% until it has been counted
Field

Choosing what the rule reads

A balancing rule cannot tell one basis from another with the same span, so choosing what to hand it is choosing a subspace, and everything about it is exact geometry: what a rule removes of a shape is that shape's squared multiple correlation on the span. The maximin basis over a list of shapes is then a finite problem with an exact answer — and over a whole subspace the guarantee is exactly zero rather than small, which is why the list cannot be avoided. Randomising which functions the rule reads is worth twice the best fixed choice.
The six best bases of 2 functions, and what each protects. Every cell is R²(g | span B) — the share of the imbalance in that shape a rule balancing that basis removes — computed from exact inner products between Hermite functions and indicators, with nothing simulated. The rows are ordered by their worst cell, which is the number an experimenter who does not know the shape is exposed to. The best row here guarantees 26.8% against every shape in the list, and the worst of the six guarantees 15.1%: the difference between them is entirely which subspace was picked, at the same cost per arrival.

A basis is a subspace

A balancing rule cannot tell one basis from another with the same span, so choosing what to hand it is choosing a subspace — and then what it removes of any outcome shape is a projection, computable exactly, with no trial anywhere in it.

 The guarantee, as the basis is allowed more functions. The lower line is the best worst case over the six named shapes for a basis of each size, found by scoring every subset of the dictionary — an exact answer, since the problem is finite. One function guarantees 2.3%, which is nearly nothing; three guarantee 59.0% and the basis that does it is the covariate, its square and its cube, with no indicator in it. The upper line is the same problem with the basis drawn rather than fixed, which is worth 2.09 times as much at two functions and 1.32 at three. The two lines converge because a basis large enough to protect everything has nothing left to randomise over.

Which shapes are worth protecting

Choosing a basis by its worst case is a finite problem with an exact answer. The answer has no tie in it, which a maximin optimum is supposed to have — and the tie comes back, along with twice the guarantee, when the basis is drawn rather than chosen.

 A class that is a subspace has no guarantee below its own dimension. Each cell is the worst case over every unit-variance function in a class of dimension m, for a rule reading k functions: the smallest squared principal-angle cosine between the two subspaces. Wherever k is less than m the number is zero to machine precision, and that is not a weak guarantee but the absence of one — some direction of the class is orthogonal to the entire basis, and against an outcome in that direction the rule does exactly what a coin does. An experimenter who declines to name the shapes and asks instead to be protected against everything smooth is asking for the cells above the diagonal.

Where the guarantee is exactly zero

An experimenter who declines to name the shapes, and asks instead to be protected against anything in a class, is asking for a number that is not small but zero. Bounding the class is unavoidable, and the two ways of doing it choose different bases.

 Where the constraints exhaust the randomisation. At 16 units there are 12,870 equal splits, so the ones meeting a stated tolerance can be counted rather than estimated. With each of the first k standardised imbalances required to be within 0.4 of a coin's own spread, the admissible count runs 3874 → 1006 → 314 → 0 → 0 → 0 — and at 4 functions there is no admissible assignment at all. The count is the number of distinct answers a randomisation test can give: at 3 functions its finest attainable p-value is 1 in 314. Balance improves with every constraint and the reference distribution shrinks with it, and the two run out at different rates.

When the constraints run out

Every function added to a basis is a constraint the assignment has to satisfy with the same units. At sixteen units and a stated tolerance the admissible assignments run 3,874, then 1,006, then 314, then none — and the count is exact, because the assignment space is finite.

 The same number of assignments, held less tightly. For each basis size the box admits at a tolerance of 0.4, the Mahalanobis rule is given exactly the same number of admissible assignments — the ones with the smallest Mahalanobis distance — so the two differ only in the shape of the region. Each bar is the worst named function's remaining imbalance variance as a share of a coin's, with the mean over the named functions beside it. At one function the regions are the same interval: 0.0523 both ways. At 3 functions and 314 assignments, the box leaves 0.0771 in its worst function and the ellipse 0.1545; on average 0.0633 against 0.0927. The ellipse spends its assignments on a direction no function in the basis names.

What the ellipse spends its assignments on

Rerandomising on Mahalanobis distance never runs out of assignments, where a box of per-function tolerances does at four functions. But at the same number of admissible assignments it holds the worst named function half as tightly, because it spends its constraint on a combination of the basis nobody named.

 Three regions at the box's own count. Across 24 sets of 16 units, on the sets where a box at a tolerance of 0.4 has any admissible assignment, the share of its own variance left in the worst of the k named functions at the box's count. the ellipse: 0.070, 0.134, 0.196, 0.194, 0.182. a sphere on the standardised functions: 0.062, 0.076, 0.091, 0.086, 0.080. the box: 0.062, 0.072, 0.081, 0.073, 0.069, at 2, 3, 4, 5, 6 functions, on 24, 20, 9, 4, 4 of the sets. The box's mean admissible count is 1045, 354, 115, 32, 13.

A box with a count

The ellipse never runs out of assignments and holds the named functions loosely; the box holds them tightly and can come up empty. Dropping the ellipse's thin directions was meant to give both, and it does not: no cutoff comes within a third of the box. But the box's emptiness was never a property of the box. It belongs to a tolerance fixed in advance. Set the tolerance by a count instead, and at that count the region is the box, on every set of units, with nothing left to run out.

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