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

Allocation — where it appears

How many of an experiment's units go to each arm. It decides the variance of the treatment estimate through (m_A + m_B)(1/m_A + 1/m_B), which is exactly four at balance and grows without limit as one arm shrinks.

Named by 21 essays across 9 fields — each of them below, with the objects they name alongside it.

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.

product · Blocking The construction survives a difference of two weighted means. Coverage of δ̂ ± t√(S_D²/H) on b − 1 degrees of freedom, over 900 runs at a requirement of 0.3, where δ̂ is the block differences weighted by h_b = (1/m_A + 1/m_B)⁻¹ and H is their total. The theorem the one-mean field rests on goes through with h_b in place of the block size, and the reason is that the weights a weighted least squares decomposition needs are the inverse variances — which is exactly what h_b is. The stopping rule reads only within-arm within-block contrasts, so it is a function of nothing the interval reports, whatever it does with the block sizes. Each bar is within 2.9% of the level it claims.

A width promised for a difference

The exact fixed-width interval was built for one mean. Two arms make the target 42.7 units of effective size and each unit costs four observations, so the same promise about a difference costs 169.4 rather than 42.7 — and the theorem survives untouched with the harmonic size in place of the block size.

contrast · Width 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.

allocation · Allocation Which allocations reverse the overall comparison. The per-group success rates are held fixed; only the split of each group between treatment and control changes. 32% of the allocations reverse, and the worst reverses by 13.1 percentage points.

Simpson's reversal is a region, not a table

The treatment wins in both groups and loses overall. That is normally shown with one famous table, which cannot answer the two questions a reader has — how often, and how large. Swept, it turns out to occupy 31% of the allocation space.

paradox · Simpson The same 40 units, arranged two ways. Both designs estimate the same effect of 0.5 and both are unbiased — 0.488 and 0.497. The blocked design's estimate has standard deviation 0.318 against 0.692, a variance ratio of 0.21 where the model predicts 0.20.

The variance removed before the data

Arranging forty units in pairs rather than assigning them at random cuts the variance of the estimated effect to a fifth — and the fifth is knowable in advance, because it is exactly the share of the variance the pairs do not carry.

design · Blocking Where one rule becomes three. Every arrival in 200 simulated trials is put to all three scores, and the picture is how often they would send that patient to different arms. The range and the pairwise sum are the same rule at two arms and at three — for sorted counts the pairwise sum is twice the range, so the arm that minimises one minimises the other — and they part company at four, where the pairwise sum is 3(d − a) + (c − b) and the range still sees only d − a. The variance disagrees with both from two arms onwards, on 5.4% of arrivals at two and 27.0% at five, because the scores are summed over 3 factors and a sum of squares does not order the candidates the way a sum of absolute values does. All three are called minimisation.

Three arms and three scores

Minimisation balances a trial by keeping the arms' counts even inside every prognostic factor. With two arms there is one way to measure how uneven two counts are. With three there are several, they are all called minimisation, and they send different patients to different arms.

multiarm · Assignment How often a randomised trial reports the reversal, advantage 6 points. Simple randomisation against randomisation stratified by group, 4,000 trials at each size. The simple design reverses on 3.40% of trials at its worst size and 0.50% at 1280 units; the stratified design reverses on none of them, at any size.

The reversal a coin cannot prevent

Randomisation removes Simpson's reversal in expectation, which is not the same as removing it. A correctly randomised trial of eighty units, on a population where the treatment helps in both groups, reports it losing overall on 3.40% of trials — and stratifying the randomisation takes that to zero at every size.

reversal · Simpson A trial designed 2:1:1, and what two scores deliver. 500 trials of 180 patients, three arms, a target of 2:1:1. The shaded bars are a minimisation score that divides each arm's count by the share that arm is supposed to receive before measuring the spread; it delivers 49.9% : 25.1% : 25.1%. The others are the same rule with the counts left raw, which delivers 33.4% : 33.3% : 33.3% — the balance it enforces inside every factor level is equality, and equality is what it gets. The marks are the shares that were asked for.

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.

multiarm · Allocation Every way of splitting 16 units into two halves. All 12,870 assignments, enumerated. The spread of the standardised imbalance is exactly 2/√n = 0.500, whatever the covariate's own distribution, and 33.3% of assignments differ by more than 0.5 standard deviations. Randomisation does not deliver balance; it delivers a known distribution of imbalance.

Randomisation is not balance

A third of all ways to split sixteen units leave the two halves more than half a standard deviation apart on a covariate. What randomisation delivers is not balance but a known reference distribution — and it makes a test exact with no assumption about the data's shape at all.

design · Randomisation 20 adaptive trials, 45% against 25%. Each line is one trial allocating patients one at a time by the arm's own posterior. The average final share on the better arm is 84.7%, with a standard deviation of 10.3 points across these 20 trials. The rule does not deliver a fixed advantage: it delivers one that depends on how the first few patients came out.

Randomising towards the winner

Allocating more patients to the arm that is doing better is the humane thing to want and it buys nothing statistically: at a fixed total it costs thirty points of power. And because the allocation is a function of the outcomes, the ordinary test on it rejects a true null 7.8% of the time before any time trend is applied — and 58% after one.

adaptive · Randomisation 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.

allocation · Allocation 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.

product · Optimum Three sets of weights, five designs, and no estimator that is exact everywhere. Coverage of the same interval under three weightings. h_b is the inverse variance when the arms share a variance or the allocation is constant; equal weights are right when every block has the same two counts; the estimated precision weights are right in the limit and exact nowhere, because the decomposition needs the weights to be the constants they are only estimating. In the corner — two variances, changing sizes, changing allocation — the two exact estimators are the ones that miss, at 98.45% and 95.65%, and the one with no theorem behind it is at 95.05%. That is the whole statement: there is an exact estimator under either condition, and none under both.

Which weights are the inverse variances

There is an exact estimator when the two arms share a variance and another when every block has the same two counts, and between them they cover every trial anybody designs on purpose. In the corner where neither holds, both cover 98.45% instead of 95%, and the only estimator at its level is the one with no theorem behind it.

contrast · Nuisance 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.

product · Randomisation 3 arms against one control, 360 units in all. Every control size, enumerated. The best is 132 on the control and 76 on each arm — a ratio of 1.74, against √3 = 1.73. Splitting the units evenly over all 4 groups costs 7.2%, which is small; what the larger control also does is lower the correlation between the comparisons, from 0.50 to 0.37, and that changes which multiplicity correction is right.

One control, many arms

The control appears in every comparison, so it is worth √k treatment arms — and the same sharing makes the k tests correlated at n/(n+n₀), which is the quantity Bonferroni ignores. Both facts come out of one design decision, and it is the size of the control.

allocation · Multiplicity Welch's test on skewed groups: the low and high rejection rates in every cell, with equal means throughout. Each cell should read 2.5 / 2.5. Two identical exponentials at 20 and 20 read 2.04 / 2.22; the worst cell, a wide exponential against a normal at 8 and 32, reads 9.79 / 0.47.

The skewness of a difference

Welch's test holds its size to within half a point when both groups are normal. Give both groups the same skewed population and it still balances at twenty and twenty — and at eight and thirty-two it rejects low on 7.16% of samples and high on 0.66%. One number decides which: the skewness of the difference of the two means, which ranks twenty-five cells by their imbalance with a correlation of 0.997.

tails · Student 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.

allocation · Allocation What a guesser gets, and what a guesser gets for nothing. 600 trials of 150 patients under a fully deterministic rule, with an investigator who knows the rule, the factors and every assignment so far. The guess rate falls with the number of arms — 87.8% at 2, 86.2% at 3, 81.0% at 4 — which reads like a trial getting safer and is not: what a guesser can trade on is the excess over the 50%, 33%, 25% they would get by naming an arm at random, and that goes the other way, from 1.76× chance at two arms to 3.24× at 4. The gap a guesser manufactures between the best and worst arm under a true null is 0.759, 0.733, 0.711 standard deviations — nearly unchanged.

Guessing one arm in three

A balancing rule is guessable because it is balancing. With three arms the next assignment is worked out less often than with two — and by more, relative to what a guesser gets for nothing, and the damage they can do is almost unchanged.

multiarm · Assignment Power at an effect of 0.5 standard deviations. The curve is the non-central t on 2n − 2 degrees of freedom with δ = d√(n/2); the dots are 4,000 experiments run at each size. Reaching 80% power needs 64 per arm.

How many subjects

Sixty-four per arm for 80% power at half a standard deviation — a power figure that could only be simulated, with nothing to disagree with, until the non-central t was written. Two routes now, agreeing to within the simulation's own error.

design · Power 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.

product · Reference 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.

contrast · Allocation

Named alongside it

The objects these essays reach for when they reach for this one.

RandomisationBlockingCovariate balanceExperimental designSample sizeAllocation ratioError rateMonte CarloNeyman allocationBlindingConfidence intervalStandard deviation

All concepts