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

When the set is too large to walk

Two constructions that were each measured by enumerating everything, at the size where enumerating everything stops being possible. A balancing dictionary over two covariates is an outer product, every inner product in it is still closed form, and a rule holding every main effect of both removes exactly nothing of any interaction. A trial of two hundred units has assignments that cannot be walked — and the acceptance rate turns out not to depend on the size of the trial at all, so the exhaustion a sixteen-unit count runs into is a fact about sixteen units.
Two covariates make the dictionary an outer product. Four functions of each covariate, and everything a balancing rule may be handed. The margins are the 8 main effects and the block between them is the 16 interactions, which are 66.7% of the dictionary. Every inner product in it is closed form — ⟨f₁g₁, f₂g₂⟩ = ⟨f₁,f₂⟩⟨g₁,g₂⟩ when the covariates are independent — so nothing about the geometry gets harder. What gets harder is the counting: choosing k of 24 is C(24, k), which is 10,626 at four and 735,471 at eight.

A dictionary that is a product

Two covariates make what a balancing rule may read an outer product — eight main effects and sixteen interactions — and every inner product in it is still closed form. What a rule holding all eight main effects removes of a pure interaction is not small. It is zero.

 One curve is a binomial coefficient and the other is a line. The number of subsets a maximin over this dictionary would have to score, against the number the exchange algorithm actually scores. At three functions the walk is 2,024 subsets and is the honest answer; at eight it is 735,471 and the exchange algorithm has looked at 421. The warrant for the second curve is the four sizes where both exist and agree, which is a weak warrant — it says the algorithm has not yet been wrong, not that it cannot be — and it is the only one available past the point the first curve leaves the page.

Where the enumeration stops

A maximin over an eight-function dictionary is a walk over seventy subsets. Over twenty-four it is 735,471 at eight functions, and the exchange algorithm that replaces the walk scores 421. What licenses the second curve is four sizes where both exist and agree, which is a weaker warrant than it looks.

 A rate that does not know how large the trial is. The share of equal splits admitted by a tolerance of 1 coin-spreads on 3 functions, at six trial sizes. The first two are exact — 12,870 and 184,756 splits, walked, averaged over eight draws of the units — and the rest are sampled. From a hundred units on, the rate sits on (2Φ(1) − 1)^3 = 0.3182, which contains no n at all. The two small trials are 29.2% and 27.7% short of it, so the sixteen-unit measurement understates the rate rather than bracketing it. Meanwhile the admissible count — the rate times C(n, n/2) — goes from 2^11.5 to 2^393.7: the exhaustion a small trial runs into is a fact about small trials.

A count that has to be estimated

At sixteen units the admissible assignments can be counted by walking all 12,870 of them. At four hundred there are about 2^393.70, and the share admitted is 0.31885 against a closed form of 0.31818 that has no trial size in it at all. The exhaustion a small trial runs into is a fact about small trials.

 The rate falls geometrically; the count does not fall at all. The share of equal splits of two hundred units that a tolerance of 1 coin-spreads admits, against the number of functions the tolerance is stated for, with the closed form (2Φ(1) − 1)^k drawn beside it. The rate falls by about two thirds with every constraint. The admissible count is that rate times C(200, 100), and it goes from 2^195 to 2^192 — it does not fall in any sense a trial cares about. What the falling rate costs is sampling: 9,878 draws to collect a thousand admissible ones at six constraints, against 1,465 at one.

What a reference distribution costs to sample

A randomisation test on a trial too large to enumerate has to sample its reference distribution, at 1/p attempts per draw and a p-value resolved to 1/(B + 1). Six constraints cost 9,878 attempts per thousand draws, and a thousand draws resolve p to 9.99·10⁻⁴ and not one digit finer.

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