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

Every essay — page 5

Essays 97 to 120 of 436, in the same order.

Stopping rules

A p-value is defined relative to a sampling plan, so two experiments with identical data and different stopping rules have different p-values. That sounds philosophical and it is arithmetical: testing five times at the nominal level rejects a true null 14% of the time.

The chance of crossing later from each interim z, under six schedules with O'Brien–Fleming-type spending boundaries. Exact. At an interim |z| of 1.0: end only 3.64%, +0.6 3.41%, +0.75 3.20%, +0.9 3.41%, every 0.125 3.00%, every 0.05 2.88%. At 2.5: end only 38.30%, +0.6 45.80%, +0.75 45.36%, +0.9 41.70%, every 0.125 50.08%, every 0.05 53.24%. The heavy line is the largest of the six at each z.

A look the trend asked for

Under an O'Brien–Fleming-type spending function, every schedule of looks fixed in advance spends exactly 5.0000%. A committee that adds a look at three quarters of the trial whenever the interim z is 1.5 or more spends 5.2323% — 5.315% counted over a hundred thousand trials — and the most a committee choosing among six schedules could spend is 5.4390%.

6 figures · Stopping, part 14

Groups that borrow

Eight hospitals are neither one hospital nor eight unrelated problems, and the estimate between those two answers is not a compromise. It is a weighted average whose weight — se²/(se² + τ²) — is decided by how well each group is measured and by nothing else, and the population spread in it is estimated from the data rather than assumed.

Eight groups, τ = 1 against a within-group spread of 3. Each row is a group. The hollow circle is the group's own mean, the filled one is the estimate after pooling, and the small mark is the truth the data was generated from. The group of 3 moves 75% of the way to the population mean of 0.10; the group of 40 moves 18%.

Eight groups, one population

Eight hospitals are neither one hospital nor eight unrelated problems. The two obvious answers cost 2.23 and 1.15 in squared error; the estimate between them costs 0.88, and the weight it uses is not a matter of taste.

7 figures · Pooling, part 1 The weight on the population, σ = 3. Each curve is one population spread τ. A group's estimate moves B = se²/(se² + τ²) of the way to the population mean, where se = σ/√n is what the group's own mean does not know. At τ = 1 a group of 9 observations sits halfway.

The weight that decides

B = se²/(se² + τ²) is not a compromise between two answers. It is exactly the posterior mean's weight, it agrees with a numerical integration to ten digits, and an argument that mentions no population at all arrives at almost the same estimator.

7 figures · Shrinkage, part 2 τ̂ across 2,000 datasets of 12 groups, true τ = 1.5. The population spread is not supplied to a hierarchical model — it is estimated from how far apart the group means are, after subtracting the noise that would separate them anyway. It averages 1.41 here against a true 1.5, and comes out exactly zero on 5% of datasets.

The prior the data estimates

A hierarchical model needs a population spread, and it does not ask for one. It reads τ off the distance between the group means — biased six per cent low, exactly zero on 53% of datasets where the groups are identical — and the prior stops being a belief.

7 figures · Prior, part 2 One group 6 population widths from the rest. Squared error for each group under partial pooling, with each group's own mean beside it. Seven of the eight are estimated better by pooling. The eighth, which was never from the population, is estimated 6.1 times worse — 13.9 against 2.3.

When borrowing goes wrong

Partial pooling wins on the total and can lose badly on one group. Placed six population widths out, the group that was never from the population is estimated six times worse than by its own mean — and nothing in the output says so.

7 figures · Pooling, part 3 What each estimator costs, against how many groups there are. Each point is 4,000 datasets, with the population spread estimated from the data rather than supplied. Partial pooling first beats BOTH of the estimators it sits between at 5 groups; below that, complete pooling — which estimates nothing at all — is the better answer. Its own cost falls from 1.751 at 2 groups to 0.795 at 40.

The fewest groups that can borrow

At three groups the estimator that shrinks towards its own data's mean returns the group means untouched, on every dataset, because its constant is J − 3. At two it expands instead of shrinking. And the number of groups at which partial pooling starts to be worth doing is five, or two, or never — it depends on how far apart the groups are.

8 figures · Pooling, part 5 What each group gains from being pooled, τ = 1. Eight groups whose sizes span a factor of 13.3. The smallest gains 2.076 of squared error, which is 36.8% of the total reduction; the largest gains 0.030, which is 0.5%. 2 of the eight account for half of everything pooling buys.

Where the borrowing goes

Pooling cuts the total squared error across eight groups by 56%. Two of the eight take 61% of that reduction, the four best-measured groups share 11% between them, and the largest group gets 1.5% of what the smallest does. The headline is a fact about the groups nobody was asking about.

7 figures · Pooling, part 6 Twelve groups from two clusters, τ = 1. Every group's truth is in one of two clusters, and the population's total spread is exactly τ = 1, so the analysis recovers τ̂ = 1.515 and every shrinkage weight is what it would be for a single normal population. The estimates are pulled towards the grand mean, which is the middle of the gap — a place 0 of the 12 truths are and 6 of the estimates end up.

One population, or two

Group effects from two clusters rather than one bell, with the same total spread. The analysis recovers the same population spread, uses the same weight for every group, and reports nothing unusual — while 46% of its estimates land in a region holding 6.6% of the truths.

7 figures · Pooling, part 7

Decided before the data

Every other field here repairs an analysis after the fact. These figures are about the arrangement that makes the repair unnecessary: blocking removes exactly the variance the blocks carry, randomisation buys a known reference distribution rather than balance, and varying every factor at once estimates each of them from every run.

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.

7 figures · Blocking, part 1 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.

7 figures · Randomisation, part 1 Two factors that interact — where each design looks. The true response at the four corners is -10, 6, 4, 0. One factor at a time visits three of them, sees that raising either factor alone helps, and recommends raising both — a corner it never ran, and one that is worse than either single change. It picks the best corner 0.0% of the time against the factorial design's 99.4%, on the same number of runs.

One factor at a time

Changing one thing per experiment estimates each effect from two conditions; changing everything at once estimates each from every run. The ratio is (k+1)/2 and it is exact — and when two factors interact, the one-at-a-time design recommends a setting it never tried.

7 figures · Factorial, part 1 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.

7 figures · Power, part 3 What a 8-run fraction of 4 factors confounds. The defining relation is I = ABCD, so the resolution is 4. A is estimated as A + BCD; B is estimated as B + ACD; C is estimated as C + ABD; D is estimated as D + ABC. Each of those is an identity about the design rather than an approximation about the data.

The word a fraction costs

A half fraction estimates each main effect as an exact sum of that effect and everything it is confounded with — no error term, no sample-size argument. With every interaction at 0.8 the design reports a true effect of −1 as −0.20, and the design cannot test the assumption that makes the number mean anything.

7 figures · Factorial, part 4 The Box–Behnken design in three factors: 15 runs, none at a corner. Twelve runs at the midpoints of the cube's edges and 3 at its centre. Every run holds one factor at zero, so no run puts all three factors at an extreme — which is what makes it runnable where a corner is not. The three panels are the design's coordinate projections, with repeated positions marked.

The design that refuses the corners

Box–Behnken runs three factors in fifteen runs and puts none of them at a corner, which is what makes it usable where a corner cannot be run. It predicts the corner 1.84 times worse than the seventeen-run design that goes there, and 1.31 times worse at the middle of a face, and all three numbers are matrix computations with no simulation in them.

7 figures · Factorial, part 5 What one lost run costs a 16-run factorial fitting 11 coefficients. Every run is worth the same: dropping any one multiplies every coefficient's variance by 1.2000, which is 1 + 1/(N − p) with N = 16 and p = 11, and gives every pair of coefficients a correlation of 0.1667 where the complete design had exactly zero.

The run that did not happen

Lose one run from any orthogonal design and every coefficient's variance is multiplied by exactly 1 + 1/(N − p), and every pair of coefficients acquires a correlation of exactly 1/(N − p + 1) where there was none. The price is set by the design's spare capacity and by nothing else, and a saturated design cannot survive it at all.

6 figures · Factorial, part 6

Weighting one sample into another

A weight turns the sample that was assigned into the sample a coin would have assigned, and the exchange is exact: integrated over the population, the standardised difference on every covariate goes to machine zero whatever the assignment rule was. What it charges is observations — a treated arm worth 94% of itself where the assignment is nearly a toss-up and 12% of itself where it is nearly decidable — and past that point trimming does not repair the estimate, it replaces the question. Weighting by an estimate of the probability turns out twice as precise as weighting by the probability itself, and weights fitted to balance the covariates directly reach the precision no estimator can beat — while staying exactly balanced, and silent, on every moment they were not told about.

A weight that balances, and one that unbalances. The standardised difference between the arms on each covariate, integrated over the population rather than counted in a sample. Unweighted, the arms differ by 0.8310 on the first covariate and 0.6015 on the second, which is what makes the raw difference of arm means 2.7102 against a true average effect of 1.0000. Weighting each unit by one over its own assignment probability removes both differences exactly — -2.78e-17 and -5.69e-19, which is machine precision and not a small number — because the weighted density of the treated arm is the population's own whatever the propensity is. Weighting by a score fitted without the second covariate balances the first to 0.0035 and pushes the second out to 0.7057, further apart than doing nothing.

A score that balances

Weighting each unit by one over its own assignment probability drives the standardised difference between the arms from 0.8310 to 2.8×10⁻¹⁷ — exactly, not nearly. A score fitted without the second covariate leaves that covariate at 0.7057, further apart than doing nothing at all.

6 figures · Weighting, part 1 Three answers to how much sample is left. What a set of inverse-probability weights leaves of the treated arm, by three routes, at six settings of the assignment rule. The integral 1/(π∫φ/e) reads the whole covariate space and falls from 0.9392 to 2.655e-3. Kish's effective size counted in samples of 600 falls only to 0.2861, because almost all of the integral's fall is in a region a sample of six hundred never draws from. And the fraction the variance of the weighted mean actually delivers is lower again — 0.1155 — because the variance is the average of one over the effective size and the effective size averaged is not the same number. At the widest overlap all three agree to 0.05%.

How many observations a weight leaves

Kish's effective sample size is exact — for an outcome whose mean does not move with the covariates the weights are built from, the studentised variance reads 1.0680 where the formula says one. For the population's own outcome the same reading is 6.769, rising to 52.497.

7 figures · Weighting, part 2 The estimator has no upper bound on what it costs. What a thinning overlap does to a stabilised inverse-probability estimate of an average effect of 1.0000, over 600 samples of 600 at each of six settings. The spread rises from 0.1965 to 0.8122 and the root mean square error from 0.1964 to 0.9219, so at the thin end the error is very nearly the whole of the quantity being estimated. The lower line is the share of the arm's weighted total the single largest observation owns, averaged over the same draws: 0.69% to 9.78%, and in the worst single draw of the sweep 82.75%. Coverage of the 95% interval goes from 93.7% to 55.5%.

The region with no comparison

A trimmed interval covers the average effect over everybody 90.8% of the time at six hundred rows and 41.0% at nine thousand six hundred, while covering the average effect over the units it kept 94.3% and 96.0% throughout. An interval that gets worse as the sample grows is an interval about something else.

7 figures · Weighting, part 3 Either model is enough; neither is not. The bias of three estimators of an average effect of 1.0000, over 600 samples of 600 units with the assignment rule at strength 1, in each of the four cells made by getting each nuisance model right or wrong. The wrong model in both cases is one that omits the second covariate, which the outcome and the assignment both depend on. The outcome model alone is off by 0.8064 whenever it is the wrong one; weighting alone is off by 0.8190 whenever the propensity model is. The augmented estimator built from both is off by -0.0085, -0.0083 and -0.0016 in the three cells where at least one of them is right, and by 0.8118 in the fourth — which is between its two components rather than better than either.

Either model, but not neither

The augmented estimator's bias is −0.0085, −0.0083 and −0.0016 wherever one nuisance model is right, against components off by 0.8064 and 0.8190. One step past the overlap sweep it is the least biased estimator on the table at 0.0857 and the worst on it at 1.9265.

7 figures · Weighting, part 4 Estimating a weight you already know is worth doing. The variance of an inverse-probability estimate weighted by a propensity fitted from the sample, over the variance of the same estimate weighted by the true propensity, paired on the same 500 samples of 600 units at each of five settings. Every reading is below one: the stabilised estimator keeps 27.8% of its true-weight variance where the assignment is nearly a coin toss and 72.0% where it is nearly decidable, and the unstabilised one 30.0% and 49.3%. Neither estimator is materially biased, so this is a variance rather than a trade. The true weights are right about the population and know nothing about the draw; the fitted weights are the value that sets this draw's own imbalance to zero, and that imbalance was what the variance was made of.

The estimated weight is the better one

The propensity is known exactly here, so it can be weighted by — and estimating it from the same data and weighting by that gives a variance ratio of 0.4769 on paired draws. The reason is a projection: the draw's own imbalance explains 56.33% of the true-weight variance and 0.05% of the estimated-weight one.

7 figures · Weighting, part 5 Weights that balance a sample by construction. What three sets of weights leave of the standardised difference between the arms on each covariate, as a root mean square over 1200 samples of 600 units. The true propensity leaves 0.1317 and 0.1186 — a sampling error, since it is right about the population and knows nothing of the draw. A likelihood fit leaves 0.0770 and 0.0657, having absorbed part of the draw's imbalance as a side effect of fitting the treatment. Weights fitted so that each arm's weighted means are the sample's leave 1.4e-14 and 1.2e-14, which is the arithmetic's floor rather than a small number: the largest gap between a weighted arm mean and the sample mean in any draw is 9.8e-14.

A weight fitted to balance

Weights fitted so that each arm's weighted covariate means equal the sample's leave a difference of 1.4×10⁻¹⁴ between the arms and give the estimate a third of the variance of weights fitted by likelihood — 0.011883, within a relative 5.8% of the bound no estimator can beat. In the world where the assignment carries a square nobody named, the same exact balance leaves the square further apart than no weighting at all, and where the outcome carries it too the estimate is wrong by 0.6973 with an interval that covers 1.5%.

6 figures · Weighting, part 6 One weighting told the means and one told the second moments, in five worlds. The bias of the fit to balance over 600 samples of 600 units in each world, fitted to the covariates' means and fitted to their means, squares and product. Told the means it is off by -0.0004, -0.0020, 0.0103, 0.6973, 0.2698 in the worlds with no square, a square in the assignment, a square in the outcome, a square in both and a cube in both; told the second moments, by -0.0009, -0.0000, 0.0009, -0.0045, 0.3099. Its interval covers 94.0%, 94.7%, 95.0%, 1.5%, 51.0% and 93.7%, 89.8%, 94.0%, 91.0%, 48.3%.

The moments a balance is told

Weights fitted to balance the covariates' means were wrong by 0.6973 in the world where both the assignment and the outcome carry a square. Told the squares and the product as well, the same construction is off by −0.0045 there and its interval covers 91.0%. The failure moves up a moment rather than away: with a cube in both, the second-moment balance is off by 0.3099 and leaves the cube twice as far apart as no weighting. And where overlap is thin, 37.0% of samples have no such weights at all.

6 figures · Weighting, part 7

When the observations repeat each other

Every standard error on this site divides by √n, which claims the observations carry independent information. In time order they usually do not: at a lag-one correlation of 0.8 a fifty-point series is worth about six independent observations, its 95% interval covers 47%, and two series that wander are called related three times out of four.

Twenty series with a lag-one correlation of 0.8. Every series has a true mean of zero and 60 observations. The marks on the right are the twenty sample means. The variance of that mean is 8.3 times what 60 independent observations would give, so the series is worth about 7 of them.

The observations that repeat each other

Almost every standard error divides by √n, which claims the observations carry independent information. At a lag-one correlation of 0.8 a fifty-point series is worth about six independent observations, and its 95% interval covers 47%.

7 figures · Dependence, part 1 Two independent random walks, 100 steps. Nothing connects these two series: each is generated from its own independent draws. Regressing one on the other gives a slope with t = -10.9, R² = 0.55 and p = 0.0e+0 — a result that would be reported as a finding by any standard output.

Two walks and a finding

Regress one random walk on another, independently generated, and the slope is significant 76.7% of the time with a median R² of 0.17. Nothing connects the two series, nothing in the output says so, and more data makes it worse.

7 figures · Spurious, part 1

FieldsThreadsSeriesConceptsFigure librarySearch