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

Randomisation — where it appears

Assigning units to arms by a mechanism rather than by judgement, which is what makes a comparison of arms a comparison of treatments. It also supplies a reference distribution: the test can be built from the assignments the mechanism could have produced, with no distributional assumption at all.

Named by 36 essays across 14 fields — each of them below, with the objects they name alongside it.

What a median split can see. A standard normal covariate with its median marked, and the mean of each category as a vertical rule: -0.7979, 0.7979. A rule that balances the categories is balancing those numbers and nothing else, so the part of the covariate it can act on is the variance between them — 0.6366 of the total, which at two categories is exactly 2/π because the two half-normal means are ±√(2/π). The rest, 0.3634, is variation inside the categories that the rule cannot see and does not touch: the assignment within a category is still a coin. Everything the next figure measures is a consequence of this one, and it is available before any unit has arrived.

A covariate with no levels

Every balancing rule on this site reads a level. Age and blood pressure have none, so somebody cuts them into categories — and a median split can see exactly 2/π of a normal covariate, whatever the rule does with the halves.

continuous · Assignment The expansion that never terminates. The Hermite coefficients of a median split, in magnitude, against the reference j to the power −3/4, anchored at the first one. Every even order is exactly zero because sign is an odd function, and every odd order is not, so no truncation is exact — where a polynomial of degree d is exact at any order past d. Summed, the tail past J falls like 1/√J: sixty orders still leave 6.6% of the variance outside. That statement is what made a cut dictionary's geometry unavailable in closed form, and it is a statement about the function against itself. What it is not is the accuracy of an inner product between two correlated variables, where every term past J carries a factor of ρ^m as well.

A cut is not a polynomial, and it does not have to be

A threshold's expansion never terminates, which is why a balancing dictionary's geometry was closed for powers and taken to draws for cut points. Conditioning on the second variable closes it for both.

splits · Blocking 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 A proposal that moves more, refused more often. The two halves of the trade, both exact, on the 410 admissible assignments of twelve units. The integrated autocorrelation time of an imbalance the rule was never handed falls from 7.30 at one swap to 3.97 at three, and the acceptance rate falls with it, from 58.8% to 40.8%. A rejected proposal costs one evaluation and leaves the chain where it was, so acceptance is not the price of anything and the ranking by acceptance is the reverse of the ranking by cost. Past three the family folds: exchanging k of six from each arm is the complement of exchanging six − k, so k = 5 has the same 36 proposals as k = 1 and k = 6 has 1.

A proposal that moves more than two units

The walk's autocorrelation is a fact about its step size and not about its acceptance rate. Exchanging three units from each arm mixes nearly twice as fast as exchanging one, and is refused a third more often.

blocks · Randomisation Four rules, three shapes, and no ordering that survives. The variance of the unadjusted treatment estimate under each rule, as a fraction of the variance a coin gives, over 450 trials of 200 units each. Against a covariate that enters linearly the rule that reads the number nearly halves it. Against a threshold at 1 it removes about a fifth. Against a quadratic every rule here is at or worse than a coin — they are all optimising a criterion that is one over the variance of an estimate in a model this outcome does not obey, and a constraint that helps nothing still costs something. Nothing in a trial says which column it is in.

Balanced on the wrong function

A rule that reads a covariate's numbers halves the variance of the treatment estimate, if the covariate enters the outcome as a straight line. If it enters as a threshold the rule is worth a fifth of that, and if it enters as a curve every rule here is worse than a coin.

shape · Blocking What each rule leaves behind, at 120 patients. Four allocation rules over the same cohorts and the same seeds, each scored on three imbalances: the number of patients in each arm, the worst of the nine factor levels, and the worst of the 24 cells of the cross-classification. No rule holds all three. Permuted blocks hold the totals exactly and leave the margins near a coin's. Blocks inside every cell hold the cells and let the totals drift, because 24 part-filled blocks do not have to end level. Minimisation holds the margins and the totals and is at 83% of a coin's cell imbalance. Each of the three columns is somebody's definition of a balanced trial.

Balancing what is known in advance

Four allocation rules, three definitions of balance, and no rule that holds more than one of them. Minimisation keeps the worst factor margin near three patients whether the trial has forty or six hundred and forty — and lets the imbalance in the cross-classified cells climb to 86% of a coin's, because the cells are not what it is watching.

covadapt · Assignment 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 threshold in the tail is a threshold nothing balances. The share of a coin's imbalance in an indicator 1{x > c} that survives a rule which balances the covariate itself. The smooth curve is 1 − ρ² with ρ = φ(c)/√(p(1−p)), a closed form with no trial in it; the points are counted over 500 trials of 200 units at each threshold. At the median the two agree that about a third survives — the removed share is exactly 2/π — and by two standard deviations 86.9% survives. The closed form is exact in the limit and optimistic by a few points at this many units, because the rule balances the sample's mean rather than the population's.

A threshold in the tail

How much of a threshold's imbalance a balanced covariate removes is a correlation, and the correlation is a closed form. At the median it is exactly 2/π — the same 2/π a median split throws away — and two standard deviations out it is an eighth.

shape · Blocking 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 Stationary is not the same as convergent. How far each k-swap walk is from uniform after t steps, started at the least balanced admissible assignment of 410. Every one of these chains has a symmetric proposal and rejects by standing still, so every one of them is doubly stochastic and every one preserves the uniform distribution exactly. Only five of the six get there. Exchanging all six units of each arm is a single proposal — the complement — and the admissible set is closed under complement, so the walk takes it every time and oscillates between two assignments for ever: after 160 steps it has visited 1 state and sits 0.9976 from uniform. Its stationary distribution is a fact about the matrix; its limit does not exist.

Stationary is not convergent

A walk that exchanges every unit in each arm preserves the uniform distribution exactly and never gets near it. Every doubly stochastic matrix has the same stationary distribution; only some of them have a limit.

blocks · Randomisation What a rule gives away by being predictable. Minimisation run at every probability from a coin to fully deterministic, over 500 cohorts of 120 at each. The upper line is the share of assignments an investigator who knows the rule and the enrolled patients can name in advance: 49.7% at p = 0.5, which is a coin and cannot be beaten, and 87.6% at p = 1 — short of everything only where the two arms tie and the rule falls back on a coin. The lower line is what that is worth: an investigator who enrols a patient 0.5 of a standard deviation better than average whenever they predict their favoured arm produces a treatment effect of 0.75 where the truth is zero. Nothing about the randomisation was broken; the bias entered through who was enrolled, which is the one thing an allocation rule cannot control. The dashed line is the closed form 2δ(2g − 1).

The rule that can be guessed

A balancing rule improves as it becomes more deterministic, and a deterministic rule can be worked out in advance from information the person enrolling the patient already has. At full determinism 87.6% of assignments are guessable, and an investigator who acts on the guess produces a treatment effect of three quarters of a standard deviation where the truth is zero.

covadapt · Assignment Two rates, not a factor. The standard deviation of the covariate imbalance under three rules, at five trial sizes, 260 trials each, on log axes. The upper line is a coin: its slope is -0.489, against a closed form of exactly −½. The middle line is minimisation on a median split; its slope is -0.519 — the same rate — because inside a category the assignment is still a coin, and what it buys is the constant, 0.654 of a coin's at n = 200. The lower line is the rule that reads x and maximises the information about the treatment effect: slope -0.987, nearly twice as steep. Its advantage is therefore not a number that can be quoted — it is 0.258 of a coin's at n = 50 and 0.065 at n = 800, and it keeps going.

The rule that reads the number

Stop categorising and let the rule read the covariate itself. What it should minimise is not an invented distance but the variance of the effect being estimated — and what comes back is not a better constant but a different rate.

continuous · Assignment 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.

basis · Blocking One odds ratio in every stratum, and a different one marginally. Five strata with baseline risks from 5% to 85%, a treatment allocated by a coin in each, and a conditional odds ratio of exactly 2.5 throughout. The marginal odds ratio is 1.789. Nothing is confounded; the odds ratio is simply not a weighted average of odds ratios.

The change that is not confounding

Five strata, a treatment allocated by a coin in every one, and an odds ratio of exactly 2.5 in all five. The odds ratio computed on the pooled table is 1.789. Nothing is confounded — an odds ratio is not a weighted average of odds ratios, and the risk difference, on the same table, is exactly its own stratum value.

reversal · Simpson 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 The guarantee that survives a correlation, and the one that does not. What a balancing rule handed both main effects removes of the pure interaction between them, as the covariates become dependent. For median splits it is exactly zero at every correlation, because sign(x)² = 1: the interaction sign(X)sign(Y) is orthogonal to sign(X) and to sign(Y) whatever ρ is. For the product of the raw covariates it is 4ρ²/(1+ρ²)² — 64.00% by ρ = 0.5, rising to all of it at perfect correlation. A cut away from the median sits between them and is not small: 23.01% at a cut of one. The zero is not a fact about interactions. It is a fact about a dictionary whose functions square to a constant, which a polynomial one does not.

The zero that survives a cut

A rule holding both main effects removes half of a pure interaction between correlated powers and exactly none between correlated median splits. The guarantee that a correlation destroyed was never about interactions.

splits · Criterion Three analyses of the same trials, none of them wrong about the data. 320 trials at n = 60 with no treatment effect at all, so every rejection counted is a false one, and a covariate that drives the outcome with coefficient 1. The unadjusted comparison is at 5.94% after a coin — its level — and at 0.00% after the rule that reads the covariate: the design removed the imbalance and the analysis is still pricing it. Adjusting for the covariate gives 4.06%, and the rule's own reference distribution — hold the outcomes, re-run the rule 199 times, count — gives 3.13% against the 4.5% that 199 draws can deliver. The last of the three has to be told the assignment rule and nothing else, which is the one thing the experimenter certainly knows.

What the balanced trial is worth

A rule that reads the covariate removes three quarters of the imbalance. An analysis that does not know it happened prices the imbalance anyway, rejects one true null in two hundred instead of one in twenty, and finds a real effect less often than a coin-tossed trial does.

continuous · Randomisation The crossing barely moves. Both methods' costs in one unit — assignments evaluated per usable draw — as the tolerance tightens. A hunt costs 1/p and rises without limit: from 2.22 at a tolerance of 1.2 to 357.14 at 0.18. A walk costs its autocorrelation time and barely moves. The two cross at a tolerance of 0.190 at one swap and 0.195 at eight — the whole family of proposal sizes crosses inside a band of about two hundredths, because where the crossing is, the large proposal has already lost its advantage. A multi-swap proposal is worth a factor of 5.65 in the regime where the walk should not be used at all.

Where the gain is, and where the decision is

A bigger proposal is worth a factor of six at a loose tolerance and nothing at a tight one. The tolerances where it helps are the ones where a hunt costs two evaluations a draw, and the crossing barely moves.

blocks · Assignment What balancing several numbers at once costs each of them. The criterion generalises without a word changing — the covariate imbalance becomes a vector and the correction a quadratic form — so the question is what it is worth rather than whether it can be done. At n = 200 with 200 trials per point, a rule balancing one covariate leaves 12.7% of a coin's imbalance in it; balancing eight leaves 23.2% in each. The assignment has a fixed amount of freedom and every covariate added takes a share of it. The rule degrades rather than failing: at eight covariates it is still four times better balanced than a coin, and the eight are being held simultaneously rather than in turn.

Balancing more than one number

The criterion generalises to several covariates without a word changing, which makes the question what it is worth rather than whether it can be done. Each one added takes a share of the assignment's freedom, and the imbalance left in every one of them rises.

continuous · Blocking 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 Two groups, a baseline and a follow-up, and nothing happening in between — baseline reliability 0.6. 600 units in two pre-existing groups whose true means are 1.00 apart, read once at baseline and once at follow-up, with no change for anybody. The two groups' mean changes are −0.075 and −0.032, so the change-score analysis reports a group difference of 0.043. The regression of follow-up on baseline and group reports 0.409, against a closed form of (1 − λ) × 1.00 = 0.400: at any one baseline reading the two groups' lines sit that far apart, because each group's units regress towards their own group's mean. The pooled slope in this sample is 0.614, the baseline's reliability.

Two analyses of one baseline

Two groups read at baseline and again at follow-up, with no change for anybody. Subtracting the baseline reports a group difference of −0.0014 and adjusting for it reports 0.4008 — and each analysis is exactly right about one reason the groups started apart and wrong by 0.40 about the other.

paradox · Rtm The ratio of the arms' hazards over time, when treatment halves every patient's hazard. Every patient's hazard ratio is 0.5. With two risk groups — half at 0.1 a year, half at 1 — the ratio of the arms' hazards rises from 0.5 to 0.929 at 3.5 years and falls back towards 0.5. With a gamma frailty of variance 1 it rises steadily, to 0.857 at ten years.

Every patient at one half

A randomised trial in which the treatment halves every patient's hazard, and patients differ only in their baseline risk. The unadjusted Cox hazard ratio is 0.524 after six months of follow-up, 0.635 after three years and 0.679 after seven, and the ratio of the arms' hazards at a single moment climbs from 0.50 to 0.929 and falls back. Nothing is confounded and nothing is wrong. Survival removes high-risk patients from the control arm faster, and a hazard ratio compares whoever is left.

reversal · Simpson The sample a trial needs for the same power, as a share of the unadjusted analysis's, by how strongly a baseline covariate predicts the outcome. Adjusting for the covariate as measured needs one minus the squared correlation of the unadjusted sample; cut at the median, one minus 2/π of the squared correlation. At a correlation of 0.7 those are 0.510 and 0.688; cut in three, 0.611; the change from baseline, 0.600. The change score needs fewer patients than the median-cut adjustment once the correlation passes 0.624.

A baseline cut in two

Adjusting a trial's result for a baseline measurement that predicts the outcome shrinks the sample it needs to 1 − ρ² of the unadjusted one. Adjusting for whether that measurement was above its median shrinks it only to 1 − (2/π)ρ² — the cut keeps 63.7% of what the covariate could remove, exactly the fraction a cut outcome keeps. In patients the loss grows with the covariate's strength: a cut costs 12% more patients at a correlation of 0.5, 35% at 0.7 and 2.55 times as many at 0.9, where a simple change from baseline would have done better than the cut adjustment.

planned · Power How often a randomised trial of eighty shows a Simpson reversal on some baseline variable, against how many were tabulated. The treatment helps every patient equally. On one nominated, strongly prognostic variable the reversal appears in 4.1% of trials. Tabulating 2, 5, 10, 20, 50, 100 variables of mixed prognostic value, some variable reverses in 4.1%, 6.5%, 10.8%, 16.1%, 26.1%, 34.1% of trials; stratifying the randomisation on the most prognostic one gives 0.2%, 2.9%, 7.7%, 12.2%, 23.4%, 31.9%.

Twenty variables nobody stratified on

A randomised trial of eighty patients in which the treatment helps every patient by the same six points shows Simpson's reversal on one nominated, strongly prognostic baseline variable in 4.07% of trials. Tabulate twenty baseline variables of mixed prognostic value and some variable reverses in 16.07% of trials; a hundred, in 34.13%. Stratifying the randomisation on the strongest variable removes its reversal and leaves 12.17% and 31.93%: the protection is exactly as wide as the list it was given.

reversal · Simpson A median split of a measured scale. The share of the interaction a rule balancing a split above the median of each covariate removes, when the covariate is measured on k levels of a symmetric binomial, on five copulas at a Spearman rank correlation of 0.4. With an even number of levels nothing sits on the median and the split is exactly balanced: the rule removes nothing, as for a continuous covariate. With an odd number, 3 levels (50.00% on the median): Gaussian 10.44%, Clayton 4.91%, its reflection 18.44%; 5 levels (37.50% on the median): Gaussian 6.39%, Clayton 3.69%, its reflection 10.06%; 7 levels (31.25% on the median): Gaussian 4.58%, Clayton 2.91%, its reflection 6.77%; 9 levels (27.34% on the median): Gaussian 3.57%, Clayton 2.41%, its reflection 5.05%; 11 levels (24.61% on the median): Gaussian 2.92%, Clayton 2.05%, its reflection 4.01%; 15 levels (20.95% on the median): Gaussian 2.15%, Clayton 1.59%, its reflection 2.82%; 21 levels (17.62% on the median): Gaussian 1.53%, Clayton 1.19%, its reflection 1.93%; 31 levels (14.45% on the median): Gaussian 1.04%, Clayton 0.85%, its reflection 1.26%.

A median with ties

The median split's interaction zero survives every marginal and every copula, because a split that is on for exactly half the units squares to a constant. A covariate measured on a few levels cannot be split exactly in half: on a five-point scale three in eight units sit on the median, and a rule balancing 'above the median' removes 6.39% of the interaction under a Gaussian copula and 10.06% under one of the two Claytons. The size is set by the share of units on the median, which is the formula's own factor. And a fair coin for the tied units restores the zero exactly, on every copula.

compound · Adjustment What balancing a recorded number does to the true one. The imbalance three rules leave in the true covariate X when what they read is a recording W of it, at six reliabilities and n = 200, 600 trials each, as a share of a coin's exact 2/√n. The lower curve is √(1 − λ), the closed-form floor for any rule that reads W: X = λW plus a residual of variance 1 − λ that W says nothing about, and a coin is all any rule can be on that residual. The rule that reads the number sits on it — 0.125 at 1, 0.345 at 0.9, 0.452 at 0.8, 0.612 at 0.64, 0.718 at 0.5, 0.834 at 0.3 — while balancing W itself to 0.125–0.133 of a coin's throughout. The upper curve is √(1 − 2λ/π), the same floor for minimisation on a median split of W, and minimisation sits on that one. At λ = 2/π the rule that reads the number is exactly as good as a median split of an error-free covariate: a cut is a measurement with reliability 0.637.

A number read with error

A rule that balances the recorded covariate balances the true one only as far as the recording reaches. At a reliability of 0.8 it leaves 45% of a coin's imbalance in the covariate that drives the outcome, back on the coin's 1/√n rate — and a median split turns out to be a recording with reliability 2/π.

continuous · Assignment What a guesser gets from less than everything. 1,000 three-arm trials of 150 patients under minimisation on three factors with the preferred arm taken with probability 1, and five investigators guessing each next assignment from different information. Knowing the rule, the factors and every assignment so far: 86.3%. Knowing everything about the patients at their own centre and nothing about the others, and applying the rule's own score to that: 52.8%. Knowing only how many of their own centre's patients went to each arm: 55.5%. Knowing only the trial's arm totals, which a drug-supply system shows anyone who reorders: 54.5%. A guess at random gets 33.3%. The cruder the information, the less the guesser gets — except that the rule's own score, applied to a partial record, does worse than simply counting.

A guesser who sees the totals

The bound on how guessable a three-arm minimisation is assumes an investigator who sees everything. One who sees only the arm totals — what a drug-supply screen shows — names the next arm 54.5% of the time against a chance 33.3%, and manufactures 45% of the full guesser's bias.

multiarm · Assignment A basis drawn rather than chosen. The share of the worst of seven shapes a balancing rule removes, over two covariates and a dictionary of twenty-four functions, for the best fixed basis of each size (lower line) and for a basis drawn at random from the least favourable mixture (upper line), at one to four functions. The fixed basis guarantees 0.0000, 0.0000, 0.4244, 0.6366: nothing at one function or two, because two of the shapes are pure interactions and two of them are main effects of different covariates, and no one or two functions touch all of them. The drawn basis guarantees 0.2042, 0.4084, 0.6133, 0.7920 in expectation over the draw — about a fifth of a shape for every function up to three. The marks above the upper line are the best reply to nature's mixture, 0.2051, 0.4098, 0.6142, 0.7931: the game's value lies between the two.

A basis drawn rather than chosen

On two covariates no fixed basis of one or two functions protects every shape an outcome might have — its worst case is exactly zero. Draw the basis at random from the right mixture and the guarantee is a fifth of a shape for every function. The price is per trial — at three functions, two draws in five leave one shape wholly unprotected.

product · Optimum How often a trial of 80 shows a Simpson reversal on its strongest baseline variable, against how many cut points are tried. The treatment helps every patient equally. Cut at its median, the variable's table reverses in 4.4% of trials; trying 2, 3, 5, 9 cuts, in 6.4%, 8.7%, 10.8%, 13.2%. Independent chances would give 33.3% at nine.

Five places to cut one variable

A continuous baseline variable has to be cut before it makes a subgroup table, and the cut is chosen. In a randomised trial of eighty where the treatment helps every patient equally, the strongest baseline variable cut at its median shows a Simpson reversal in 4.4% of trials; tried at five candidate cuts, some cut reverses in 10.8%, and at nine, 13.2%. Nine correlated cuts are worth about five independent chances, not nine. Across twenty continuous variables cut at five points each, some table reverses in 32.2% of trials — twice the rate with each variable cut once, and about thirty-one independent chances where the table holds a hundred.

reversal · Simpson 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.

basis · Allocation How often a Simpson reversal appears on the strongest baseline variable when its cut is slid along it, by the trial's size. The treatment helps every patient by the same six points. At 40, 80, 160, 320, 640 patients: some cut in the middle 80% reverses in 18.8%, 23.4%, 24.3%, 19.2%, 10.6% of trials; some decile in 12.2%, 13.2%, 14.1%, 11.0%, 6.2%; the median in 3.9%, 4.4%, 4.2%, 3.1%, 1.7%.

A cut that slides

Slide the cut on a trial's strongest baseline variable along every gap between observed values in its middle eighty per cent, and a trial of eighty whose treatment helps everyone equally shows a Simpson reversal somewhere in 23.4% of trials — against 13.2% for the nine deciles and 4.4% for the median. Sixty-five cut positions are worth about 9.5 independent cuts, and at 640 patients, with 513 positions, about 7.2: a table that moves one patient at a time is a continuous scan, and its count is set by the range scanned rather than by the number of cuts. The reversing cuts spread along the whole range, in 2.62 separate runs at eighty patients and 8.36 at 640.

reversal · Simpson How often a trial with nothing to find shows a Simpson reversal somewhere, by how many variables are searched and how. Trials of eighty whose treatment helps every patient by six points, twenty baseline variables tracking the risk with correlations from 0.95 to 0.10; 4,000 trials. Searching the strongest 1, 2, 5, 10, 20: every cut, 23.4%, 30.7%, 41.0%, 46.9%, 49.0%; the deciles, 13.2%, 19.4%, 27.9%, 33.3%, 36.4%; the median, 4.4%, 6.9%, 11.3%, 14.4%, 16.2%.

Every variable, every cut

Slide the cut along every one of twenty baseline variables in a randomised trial of eighty whose treatment helps every patient equally, and some variable shows a Simpson reversal at some cut in 49.0% of trials — a coin toss with nothing to find. The twenty scans do not add like independent chances: if they did they would reverse 92.4% of the time. They read one patient risk between them, and the weak variables seldom reverse at all, so the strongest five supply 41.0 of the 49.0 points. A trial that shows one reversal usually shows several: 4.8 variables on average.

reversal · Simpson

Named alongside it

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

Covariate balanceMonte CarloExperimental designMinimisationAllocationCovariate-adaptive randomisationBlockingClosed formContinuous covariateInteractionThresholdConfounding

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