The collection

Every essay — page 9

Essays 193 to 216 of 613, in the same order.

Splitting the units

The only design decision that costs nothing: the same units, the same measurements, the same analysis, and a different variance. Sample the noisier arm more, the expensive one less, and the shared control by the square root of the number of arms — and then notice that every one of those rules is a function of a quantity nobody has, and measure what happens when it is estimated instead.

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.

8 figures · Allocation, part 1
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.

7 figures · Allocation, part 2
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.

8 figures · Multiplicity, part 4
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.

8 figures · Allocation, part 3
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.

4 figures · Allocation, part 8
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.

4 figures · Allocation, part 9
How often reporting whichever contrast is most significant rejects when the arms do not differ, by the split. 20,000 trials a point. With both arms at 0.05, 0.1, 0.3: at shares 0.3, 0.35, 0.4, 0.45, 0.5, 0.55, 0.6, 0.65, 0.7 the chosen contrast rejects 6.5%, 5.6%, 5.0%, 5.0%, 5.1%, 5.1%, 5.2%, 5.2%, 6.3%; 6.0%, 5.4%, 5.3%, 4.9%, 4.8%, 4.9%, 5.2%, 5.4%, 5.8%; 5.8%, 5.6%, 5.2%, 5.1%, 5.2%, 5.1%, 5.3%, 5.4%, 5.7%.

A contrast chosen after the data

A two-arm trial with a binary outcome that reports whichever of the risk difference, the log risk ratio and the log odds ratio comes out most significant has tested three statistics that ask one question. At an equal split the choice is no choice at all: the difference has the largest z in every one of twenty thousand trials, so the selected test is the difference's, at its 5%. At a 30–70 split the choice is real and costs a point and a half of false positives, 6.5% at the rarest outcome. Pre-registering the difference and the odds ratio as a pair with Bonferroni's correction — the cautious alternative — costs eight to twelve points of power at every split.

4 figures · Allocation, part 10

Designs that change while they run

The stopping-rule field is a fixed design looked at more than once. Here the design itself is a function of the data — how many units, which arm the next one goes to, which arms survive the interim — and the question stops being what the rule spends and becomes what it leaves behind. The estimate from an arm chosen for being ahead is ahead by more than it should be, and the unbiased estimate is the one that throws away the data the choice was made on.

What the interim sees, at an effect of 1. The same 60 observations, estimated two ways. Keeping the arms separate gives 0.995, which is σ. Pooling them without separating the arms — the price of staying blind to the comparison — gives 1.114, against the identity √(1 + Δ²/4σ²) = 1.118. The sample size is proportional to the variance, so a blinded design at this effect asks for 25% more units than it needs, and it does so systematically rather than by chance.

Choosing n after looking

Re-estimating the sample size from an interim is the one adaptation with a defence, and the defence is exactly what it costs: an analyst kept blind to the arms measures a spread that contains the effect, so the design overshoots by 1 + Δ²/4σ². Re-estimating the effect instead breaks the error rate.

8 figures · Stopping, part 3
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.

8 figures · Randomisation, part 2
The best of 8 arms, tested as though it were the only one. 8,000 trials with no effect in any arm. Stage one runs 8 arms at 60 each, the best is carried forward, and stage two adds 30 more to it and to the control. The histogram is where the final statistic lands and the curve is the standard normal it is being read against — shifted right, because the arm was chosen for being ahead. 10.5% of these trials clear 1.96 against a claimed 5%, and the value that actually holds the rate for this design is 2.313.

Dropping the losers

Carrying the best of eight arms forward and testing it at 1.96 rejects a true null 10.3% of the time — the hypothesis was chosen by looking at the data, so the statistic is a maximum wearing a single comparison's clothes. The value that holds the rate is 2.313, and it has to be solved for.

8 figures · Stopping, part 4
Three estimates of the same effect, and the honest one is the worst. 8,000 trials in which every arm has an effect of exactly 0. The arm carried forward is reported by its first stage at +0.187 above the truth — it was chosen for being ahead — and by its second stage at -0.0032, which is unbiased by construction because the selection could not see it. The combination that gets published is at 0.123, exactly the share of the first stage's bias the first stage contributes. Root mean squared error: 0.240, 0.261, 0.182 — the unbiased estimate is the least accurate of the three.

The estimate after the choice

An arm chosen for being ahead is ahead by more than it should be, and the trial then publishes the average of the stage that chose it and the stage that did not. The unbiased estimate is the one built from a third of the data — and it is the least accurate of the three.

8 figures · Curse, part 4
How far three estimates of the chosen arm sit from its true effect, with all eight alike. Eight arms of sixty, the best carried forward with thirty more. Average error: both stages, as published +0.123; stage two alone −0.001; stage one shrunk, then both +0.020. The arms' spread estimates to zero on 57.1% of trials.

Shrinking the arm that was chosen

A trial that carries the best of eight arms forward reports it as better than it is — by +0.123 when the eight arms are alike — and its second stage alone is unbiased but the least accurate of the estimates. Pulling the chosen arm's first stage back towards the eight arms' mean before combining removes most of the bias when the arms are alike (+0.020) and gives the smallest error of any estimate, 0.133 against the published 0.182. When one arm really is ahead by half a standard deviation, the same shrinkage pulls the true winner back towards the losers and understates it by 0.103, with more error than the published estimate. The weight is estimated from eight numbers, and it is right when the selection was noise and wrong when the selection was right.

5 figures · Curse, part 5
How far four estimates of the chosen arm sit from its true effect, with one arm ahead by 0.5. Eight arms of sixty, the best carried forward with thirty more, 20,000 trials. Average error: both stages, as published +0.006; shrunk under a normal prior −0.103; shrunk under a t prior, 3 df −0.101; normal prior with a fifth vague −0.073. Root-mean-square error: 0.151, 0.189, 0.192, 0.176.

A prior that lets one arm stand out

Shrinking the chosen arm of eight under a normal prior understates a genuine winner by 0.103. A t prior on three degrees of freedom with the same spread understates it by 0.101 — the tail is too far out to reach a lead of three and a half standard errors. A prior with a fifth of its weight on a vague component understates it by 0.073, costs 0.010 of bias when the arms are alike, and still has more error than the unshrunk estimate. With sixteen arms the same mixture recovers three fifths of the loss, because a bell of fifteen shows its outlier; with eight, the arms put more than half their posterior weight on 'one stands out' in 0.4% of trials where one does.

5 figures · Curse, part 6
The error of the chosen arm's estimate against a fixed second stage's size, and a second stage sized by the interim, 8 arms, one ahead by half a standard deviation. Fixed second stages of 10, 25, 40, 55, 70 patients: both stages combined err by 0.172, 0.155, 0.143, 0.133, 0.126; the mixture-shrunk estimate by 0.208, 0.183, 0.164, 0.151, 0.140. The rule that sizes the second stage by the interim spends 41.4 on average and errs by 0.142 and 0.159; its second stage alone errs by 0.225.

A second stage sized by the first

A drop-the-losers trial can size its second stage by what the first stage says about the arms' shape — more patients when the chosen arm's lead looks like luck, fewer when it looks real. The rule does what it was built to do: at sixteen arms with a genuine winner it spends 51.0 patients a group when the interim chose the wrong arm and 24.2 when it chose the right one. And it buys nothing. Against a fixed second stage of the same average size, every estimate of the chosen arm errs by the same to within a few thousandths; the published combination's bias rises from 0.014 to 0.022, and the second stage alone, still unbiased, errs by 0.319 against 0.281.

4 figures · Curse, part 7
How far the published estimate of the chosen arm sits from the best arm's effect, by how six hundred patients are split between choosing and estimating. Eight arms and a control, six hundred patients: a stated number an arm in the first stage and the rest split equally between the chosen arm and the control in the second. Root mean squared error of the published estimate about the best arm's true effect, over 5,000 trials a split. All eight alike: best at 16 an arm (0.092), 0.183 at the familiar sixty; One modestly ahead: best at 56 an arm (0.151), 0.153 at the familiar sixty; One clearly ahead: best at 48 an arm (0.146), 0.154 at the familiar sixty; Two clearly ahead: best at 40 an arm (0.128), 0.151 at the familiar sixty; Spread evenly: best at 40 an arm (0.131), 0.146 at the familiar sixty. The split that guards every world puts 32 an arm before the choice and costs at most 21.3% over each world's best.

Patients spent before the choice

A trial that picks the best of eight arms and then confirms it splits its patients between choosing and estimating, and the familiar shape — sixty an arm before the choice, thirty after — spends ninety per cent of a six-hundred-patient budget on the choice. Measured against the effect a reader is told about, the best arm's, that split is nearly right when one arm is modestly ahead and twice as wrong as it needs to be when the arms are alike. A split of thirty-two an arm before the choice is never more than 21.3% worse than the best split for whichever world holds; sixty an arm can be 98.8% worse. The two targets an estimate could serve — the chosen arm's effect and the best arm's — pull the split in opposite directions, and only one of them is what gets published.

4 figures · Curse, part 8

The reference distribution the design supplies

Hold the outcomes fixed, re-run the rule that assigned them, and count. A p-value built that way needs no assumption about the outcomes, no large-sample argument and no nuisance parameter — it needs to be told the rule, which is the one thing the experimenter already knows. It repairs the adaptive design the previous field leaves broken, and it is not free.

The allocations the rule could have made, from these exact patients. One 200-patient trial allocated by response-adaptive randomisation, re-randomised 999 times. No outcome is redrawn anywhere in this figure: each re-randomisation runs the same rule over the same patients in the same order, so what is drawn is the set of experiments that could have happened rather than a sampling distribution. The observed |z| is 1.417, 258 of the 999 re-randomisations reach it, and the p-value is (1 + 258)/(1 + 999) = 0.2590. The curve is the half-normal the ordinary analysis reads the same statistic against; its 5% point is 1.96 and this distribution's is 2.101.

The experiments that could have happened

An adaptive trial's allocation is a function of the outcomes it will later be compared against, so the ordinary analysis rejects a true null 9.2% of the time. Hold the outcomes fixed, re-run the rule that assigned them, and count — the same statistic against a reference distribution the trial could actually have drawn from is back at 4.0%.

8 figures · Reference, part 1
The allocations the rule could have made, from these exact patients. One 200-patient trial allocated by response-adaptive randomisation, re-randomised 999 times. No outcome is redrawn anywhere in this figure: each re-randomisation runs a fair coin rule over the same patients in the same order, so what is drawn is the set of experiments that could have happened rather than a sampling distribution. The observed |z| is 1.417, 155 of the 999 re-randomisations reach it, and the p-value is (1 + 155)/(1 + 999) = 0.1560. The curve is the half-normal the ordinary analysis reads the same statistic against; its 5% point is 1.96 and this distribution's is 1.946.

The test that needs the rule

A randomisation test assumes almost nothing about the data and one thing about the experiment. Tell it a fair coin produced an allocation that an adaptive rule produced — which is what every off-the-shelf permutation routine does — and it rejects 8.0% of true nulls where knowing the rule gives 4.0%.

9 figures · Reference, part 2
Where the +1 matters, and why nobody has noticed that it does. The true size of the two rules at every B, computed rather than simulated: under the null the count of re-randomisations reaching the observed statistic is uniform over {0 … B}, so both sizes are integer arithmetic. With the +1 the size is (⌊α(B+1)⌋)/(B+1), which never exceeds 5%. Without it the size is (⌊αB⌋+1)/(B+1), which is larger except at B = 19, 39, 59 — the values with B + 1 a multiple of 1/α, and the values everybody uses. At B = 19 the two rules are the same rule; at B = 20 the uncorrected one is a 9.5% test. The marks are simulated on 500 trials of 120 patients, as the second route to the same numbers.

The plus one and the round number

A sampled randomisation test counts the observed allocation as one of its own reference draws, and the correction is invisible at B = 19, 39, 59 and 999 — every value anybody uses. At B = 20 the version without it is an 8.00% test where the corrected one is 3.80%, and the convention protecting everybody is a preference for round numbers minus one.

7 figures · Reference, part 3
The cheap repair needs a number nobody has. The obvious alternative to re-randomising is to simulate the design under its null once and use the critical value that comes out — which is what the arm-dropping design does, where the critical value has to be solved for and is 2.313. It does not transfer here. The rule chases outcomes, so how imbalanced the allocation gets depends on how often anything succeeds, and the critical value moves from 1.668 at a success rate of 0.05 to 2.718 at 0.8. Calibrated at 0.3 and used at 0.8 the test's real size is 12.4%; used at 0.05 it is 0.12%. The randomisation test needs none of this, because it conditions on the outcomes that happened rather than on a rate they were supposed to come from.

What the exactness buys

Against a z test calibrated to reject exactly 5% of true nulls on this design, the randomisation test loses nineteen points of power. What it buys is that the calibration needs the success rate — which moves the critical value from 1.668 to 2.718 and is the quantity the trial was run to find out.

7 figures · Nuisance, part 1
An exact test rejecting a true hypothesis a fifth of the time. How often each analysis reports an effect when the average treatment effect is exactly zero and the effect varies between units, at 150 units with 25% treated. The permutation test on the difference in means reads 4.20% where the effect is constant — where the two nulls coincide and its exactness applies — and 22.93% where the effect varies with a standard deviation of 3. The same test on the studentised difference reads 6.27% there, and the ordinary large-sample t, which makes no exactness claim at all, reads 6.60%.

The null the exactness is for

A permutation test is exact under the hypothesis that the treatment changed nothing for anybody. Under the hypothesis it changed nothing on average, with a quarter of the units treated and the effect varying between them, it rejects a true null 22.93% of the time.

5 figures · Nuisance, part 7
One statistic that is right under both hypotheses. Rejection rates for both statistics under both nulls, at 25% of 150 units treated, with the weak-null readings taken at an effect spread of 3. The difference in means is exact under the sharp null and rejects 22.93% of true weak nulls. The studentised difference is exact under the sharp null — 4.07% — and reads 6.27% under the weak one. The repair is a change of statistic inside the same construction: the same re-randomisations, the same fixed outcomes, a different number compared across them.

A statistic that is exact twice

Dividing the difference in means by its own separate-variance standard error before permuting takes the rejection rate under a true weak null from 20.47% to 6.07%, keeps the exactness under the sharp null at 4.07%, and costs 0.8 points of power against a real effect. At an even split it changes nothing at all, in every draw.

6 figures · Nuisance, part 8
How often four permutation tests of a covariate-adjusted effect reject a true weak null, effect varying with the outcome, 25% of 150 treated. At an effect spread of 0, 0.5, 1, 1.5, 2: the adjusted coefficient's permutation test rejects 5.9%, 10.8%, 13.2%, 16.6%, 14.8%; over its sandwich standard error, 5.7%, 5.0%, 3.4%, 3.4%, 3.0%; the coefficient with a treatment-by-covariate interaction, 5.1%, 10.2%, 12.8%, 14.7%, 14.0%; the same over its sandwich standard error, 5.7%, 5.2%, 5.2%, 5.1%, 3.9%.

The repair, with a covariate in it

Studentising a permutation test's statistic made the difference in means exact under the sharp null and nearly right under the weak one. Carried to the covariate-adjusted estimate a trial actually reports, the repair works only if the adjustment includes the treatment-by-covariate interaction. With a quarter of 150 units treated and effects that vary with the outcome, the adjusted estimate's permutation test rejects 14.8% of true weak nulls; over its sandwich standard error, 3.0%; the interacted estimate over its sandwich, 3.9%. When the effect varies with the covariate itself, even an even split does not protect: the adjusted test rejects none of a thousand true weak nulls, and at an average effect of 0.2 its power is 5.2% against the interacted, studentised test's 40.7%.

4 figures · Nuisance, part 9
How often an adjusted permutation test rejects a true sharp null when the covariate is chosen after randomisation, held fixed or re-chosen inside every shuffle. 150 units, half treated, six baseline covariates of which three are prognostic; 2,000 experiments and 99 re-randomisations each, read on the adjusted coefficient over its sandwich standard error. Choosing the covariate most correlated with the outcome: 5.0% held fixed, 5.0% re-chosen. Choosing the covariate with the smallest residual variance: 5.5% held fixed, 5.3% re-chosen. Choosing the covariate most imbalanced between the arms: 3.0% held fixed, 4.8% re-chosen. Choosing the covariate with the smallest adjusted p-value: 10.4% held fixed, 5.1% re-chosen. Naming the most prognostic covariate in advance: 5.5%.

The covariate chosen afterwards

A trial that chooses which baseline covariate to adjust for after randomisation has two ways to run its permutation test: hold the chosen covariate fixed across the shuffles, which is what an analysis that reports adjusting for it actually does, or re-make the choice inside every shuffle, which is exact by construction. Holding it fixed, a choice made on significance rejects 10.4% of true sharp nulls, a choice made on imbalance only 3.0% — and pays seven and a half points of power for it — while a choice that never reads the arm labels is the same test either way. The interacted, studentised statistic keeps its weak-null behaviour only when the covariate chosen is the one the effect varies with.

4 figures · Nuisance, part 10

Coverage without a distribution

Every interval counted here so far has been found short of what it promised. This one is not, and the reason is that it is not a statement about the data at all: the rank of a future observation among a set of exchangeable calibration scores is uniform, so an interval built at the right order statistic covers a stated share of the time at any sample size, under any distribution, around any model however wrong. What it does not say is where that coverage sits — and everything the procedure does not know turns up in the answer to that rather than in the total.

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