Allocation — the series · Normal Distribution

Every split of 100 units, σ = 1 against 3. Each point is one integer split, with its variance computed exactly rather than simulated. The minimum is at 25:75, which is the ratio of the spreads 25:75, and equal allocation costs 25% more variance — the same as throwing away 20 of the 100 units. The shaded band is every split within 5% of the best, and it runs from 17% to 35%: sharp to state, flat to sit on.

Not half and half

The same units, the same measurements, the same analysis — and a different variance, decided before anything is measured. When the two arms have different spreads the best split is σ₁ : σ₂, equal allocation costs 2(σ₁²+σ₂²)/(σ₁+σ₂)², and at three to one that is a quarter of the experiment.

part 1 · allocation

A budget of 4,000, at 1 and 20 a unit. Every affordable pair, enumerated. The best is 280 cheap units and 186 expensive ones — a ratio of 1.51, against the σᵢ/√cᵢ rule's 1.49. The unit rule, which says buy in the ratio of the spreads, lands at 66:197 and costs 17% more variance for the same money. Both rules are right about their own constraint; only one of them was asked.

The cost of a unit

Change the constraint from units to money and the allocation rule changes with it — from σᵢ to σᵢ/√cᵢ, which can point the other way. An arm that is noisy and expensive gets fewer units than the same arm would if the money were not the thing running out.

part 2 · allocation

What a pilot buys, σ = 1 against 3. Each point is 6,000 two-stage experiments of 100 units: a pilot of m per arm, then the rest split by the pilot's own estimate of the two spreads. Above the line the pilot has made the experiment worse than not bothering. The best pilot here is 8 per arm at 0.809, against 0.800 for a designer who knew the spreads — so the rule recovers 96% of what knowing them is worth. A larger pilot estimates the ratio better and has less left to apply it to, which is why the curve turns.

Allocating on a guess

Every allocation rule in this field is a function of quantities the experiment is being run to find out. Fed a pilot's estimate of them, the rule that minimises the variance makes the experiment worse than not bothering — until the arms differ by about a factor of two, which is further than anyone would guess.

part 3 · allocation

Balancing towards unequal targets

A three-arm trial allocating two to one to one is the ordinary case, and a balancing rule built from raw counts does not know it. It balances the arms towards equality inside every factor level, delivers a third to each arm, and reports that it minimised imbalance.

part 4 · multiarm

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.

part 5 · basis

Free until the sums stop seeing what the differences see. Coverage with and without the block sums pooled into the interval's variance estimate. With one effect and one level they are free. With an effect that varies between blocks they are still free, because a block's sum picks that variation up exactly as its difference does. With a level that varies they make the interval 37% wider and conservative. And where the effect falls as the level rises — a ceiling, and not an exotic thing to suppose — the sums carry none of the between-block variation while the differences carry all of it, the pooled estimate is short, and the interval that uses it covers 88.75% on a width 20% narrower than the honest one.

What a two-arm rule may not pool

A spread computed "within the block" without the arm label carries a share of the effect, so the trial runs 173 observations at a null and 282 at an effect of 1.5. The stopping rule is reading the thing it exists to measure, and the phrase that produced it is one word long.

part 6 · contrast

The weights may not read the block they weight. A weighted least squares decomposition needs weights that are constants, or at least independent of the differences they multiply. One λ̂ pooled across the trial is estimated on hundreds of degrees of freedom and is effectively a constant; a λ̂ estimated inside each block is estimated on that block's own two or three, and is correlated with the difference it weights. Coverage falls from 94.68% to 82.76% — and the interval gets wider while doing it, 0.5163 against 0.3024, which is the signature of weights that are noise.

The condition that cannot be dropped

The weights may not read the block they weight. Estimate the variance ratio inside each block rather than across the trial and the coverage falls to 83% — on an interval that is at the same time seventy per cent wider.

part 7 · corner

What the guess is worth, when it is worth anything. The variance cost of an even split relative to the variance-minimising one for a risk difference, against the first arm's proportion, with the second at 0.3. The cost is a pure number: it does not depend on the trial's size. It is exactly zero at 0.3 and at 0.70, where the two arms have the same p(1 − p); it is 0.19% at a half and 4.36% at a tenth. Across the whole range from a tenth to nine tenths it never exceeds 4.36%, which is what the variance-minimising rule is worth here — and what it is worth is the reason it is safe to use with a guess.

The arm whose variance is its answer

With a binary outcome the allocation rule is a function of the proportions the trial exists to estimate. It costs at most 4.36% of variance to ignore it anywhere between a tenth and nine tenths, because √(p(1−p)) stays within a factor of two of its peak across 98% of the unit interval.

part 8 · allocation

Three contrasts on one dataset, three different splits. The variance-minimising allocation for each of three ways of reporting the same two-arm comparison, against the first arm's proportion, with the second at 0.1. A risk difference wants the arm with the larger p(1 − p) to get more units; a log odds ratio wants it to get fewer, and the two curves are exact reflections of each other in the half line. A log risk ratio wants something else again. At a first-arm proportion of 0.6 they ask for 62.0%, 21.4% and 38.0% of the units. A trial reporting more than one of them cannot be optimal for either.

Two contrasts, one split

A risk difference wants 62.0% of the units in the first arm, a log risk ratio wants 21.4% and a log odds ratio wants 38.0% — on one dataset, with one pair of proportions. The difference's rule and the odds ratio's are exact reflections of each other, so no split can be near-optimal for both.

part 9 · allocation