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

When the two are not independent

The dictionary that is an outer product needs the covariates independent, and the count that is a rate times a binomial coefficient needs the draws independent. Neither holds in a real trial. A linearisation turns the missing fourth-order expectation into Mehler's formula applied to products, which makes the whole geometry closed at any correlation — and the result it was built to test does not survive: what a rule removes of a pure interaction is exactly nothing at independence and 4ρ²/(1+ρ²)² everywhere else. A walk on the admissible set is exactly uniform, needs a burn-in, and is dearer than hunting until almost nothing is admissible.
The fourth-order expectation, by two routes. Every inner product in an eight-term slice of the dictionary at ρ = 0.5, computed from the linearisation and Mehler's formula and counted from two hundred thousand draws of a correlated pair. The entries that matter are the ones off the main effects: ⟨f(X)u(Y), g(X)v(Y)⟩ is a fourth-order expectation, which the independent-covariate field could not write down. The worst departure is 1.99 standard errors over 36 pairs, measured in each pair's own error because the entries differ in size by two orders of magnitude.

The fourth moment that was missing

Mehler's formula makes the main effects exact at any correlation and stops there, because the interactions need an expectation of four Hermite functions rather than two. A linearisation turns the four into two, and the whole geometry becomes closed again.

 The zero was a fact about independence. What a balancing rule handed every main effect of both covariates removes of a pure interaction, as the covariates are allowed to move together. At ρ = 0 it is exactly nothing — at machine precision, at any number of main effects — which is the independent-covariate result and is correct. It is not small anywhere else: the product of the two covariates loses 64.0% of itself by ρ = 0.5, because h₁h₁ = h₀ + √2·h₂ and Mehler pairs h₂ with h₂ at ρ². Four interactions are drawn and none of them keeps the zero.

A zero that was an assumption

A rule handed every main effect of both covariates removes exactly none of a pure interaction. That is true at machine precision, it is a fact about independence, and it dies as the square of the correlation.

 How long a walk has to be given. Every equal split of twelve units is enumerated, the 410 admissible ones are found, the transition matrix is built, and the distance from uniform is computed exactly at each step — no simulation anywhere. The walk is started at the least balanced admissible assignment, which is the state a rejection sampler is least likely to have handed it and the one a burn-in has to cover. It is 0.0849 away after twenty steps and 0.00008 after ninety. A real cost, and a small one, and naming it is what stops it being assumed to be zero.

Walking the admissible set

A rerandomisation test hunts for admissible assignments and throws away the rest. A walk visits them instead — and it is exactly uniform only because it stands still when a proposal fails, which is the step that looks like waste.

 Where a walk is cheaper than a hunt. Both costs in the same unit. A rejection sampler evaluates 1/p assignments per independent draw and does not care how large the trial is; a walk evaluates one per step and yields an effective draw every τ steps, and τ is a property of the constraint and the statistic together. They cross at a tolerance of 0.194 standard deviations, where about one assignment in 396 is admissible — far tighter than any trial is designed at. And the walk does not remove the acceptance cost; it pays it once, hunting for somewhere to start.

Draws that repeat each other

A hunt costs 1/p evaluations per independent draw. A walk costs one per step and yields an effective draw every τ steps. Both are counted in the same unit, and the walk is dearer at every tolerance a trial is designed at.

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