Concept

Covariate imbalance — where it appears

How far apart the arms are on a baseline variable, usually reported as the difference in means divided by the variable's own standard deviation. Under a random split of n units into halves that standardised difference has a spread of exactly 2/√n, which is the anchor every balancing rule is reported against.

Named by 5 essays across 2 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
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.

weights · Weighting
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
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
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.

weights · Weighting

Named alongside it

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

Continuous covariateCovariate-adaptive randomisationExperimental designMonte CarloRandomisationStandardised differenceBlockingDₐ-optimalityInformation matrixInverse-probability weightingLogistic regressionMaximum likelihood

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