Two functions a covariate

What the extra function buys read five balancing dictionaries by their worst case over seven outcome shapes, and found the answer set by parity: a mean and a median split are both odd, the product of two odd functions is even, and no number of odd functions touches an even shape. The fix was one square, which moved the floor from exactly zero to 6.8%.

Those five dictionaries were built by adding functions, so each one’s span held the last one’s and the widest won every comparison it entered. The essay noted that this is not the comparison a trial makes. Each constraint a rule holds thins the set of assignments it admits at the same rate whatever the function is — at two hundred units, two constraints admit 45.7% of assignments and four admit 19.8% — so the question is not whether another function helps but which functions to spend a fixed number of constraints on. And it ended on the caution that bites hardest in that setting: a worst case is a minimum over a list, and a floor quoted without its list is not a number.

This essay spends the budget. Two functions of each covariate, chosen from the eight that anybody would consider: the covariate itself, its square, cube and fourth power, and cuts at −1, 0, 1 and 2. That is twenty-eight pairs, every one of them four constraints, every one priced the same. The seven shapes and the closed-form shares are the earlier essay’s, unchanged.

The pair the worst case picks #

At a correlation of a half, the best floor among the twenty-eight belongs to the square and a cut at one, at 7.00%. The mirror pair, the square and a cut at −1, ties it exactly. Then come the mean and the square at 6.84%, and the two off-median cuts together at 6.78%.

So the worst case’s choice of two functions for each covariate leaves out the covariate. It is worth reading that slowly, because it is the opposite of what any protocol does. Every trial that balances anything balances the mean; the maximin answer, at a fixed budget, is to spend neither constraint on it.

Eight of the twenty-eight pairs have a floor of exactly zero, and each for the same kind of reason: the pair has no component at all along one of the shapes. Three hold two odd functions — the mean with the cube, the mean with the median split, the cube with the median split — and so remove nothing of the square. Two hold the fourth power beside an odd function, and the fourth power is even but orthogonal to the square, so the square is untouched again. And three hold nothing correlated with the covariate itself — the square with the cube, the square with the fourth power, the cube with the fourth power — and remove nothing of a linear outcome. Orthogonality decides which pairs are in the contest. It does not decide the contest.

A contest held on one shape #

The shape beside the bars is the same almost all the way down: the product of two median splits. It holds the floor for the winner, for the mean and the square, for the two cuts together and for every pair in the top six. And it is the one shape that no pair can do much about. The most any of the twenty-eight removes of it at a correlation of a half is 8.3%, and the pair that manages that — the square with the fourth power — removes nothing of a linear outcome and so has a floor of zero.

That is what makes the floor a strange instrument here. The pairs at the top are ranked by how much of an untouchable shape they happen to graze — 7.00% against 6.84% against 6.78% — and the differences are hundredths of the shape’s variance. The earlier essay found the same shape holding down every dictionary from four functions to eight; at a fixed budget it holds the decision as well.

Why the mean is the function that loses #

The mechanism is parity again, read on the one shape that decides. The product of two median splits is even under the joint sign flip, and an odd function is orthogonal to an even one at every correlation. So the mean removes exactly nothing of the deciding shape, alone or in any company: the square alone removes 6.84% of it, and the mean and the square together remove 6.84% too, to every digit. In a pair that holds the mean, the mean’s half of the budget is spent on shapes that never decide the floor.

A cut at one is not odd. A cut away from the median is a mixture of the two parities, and its even part can graze an even shape. Alone it removes 2.46% of the product of median splits; beside the square it lifts the square’s 6.84% to 7.00%. The sixteen hundredths that win the contest are the even part of a function placed to do something else.

So the worst case does not drop the mean because a cut at one is a better function. It drops the mean because, on the shape the worst case is reading, the mean is worth exactly zero and almost anything else is worth a little more. Any function with an even part beats it there, and the one with the most even part on offer wins. It is the same arithmetic that made the zero the earlier field found a statement about a premise rather than a weakness: a parity zero is exact, and an exact zero loses every comparison with a small positive number.

What the pick gives up #

A worst case reads one number off each row. The rest of the row is where the choice is paid for.

The mean and the square remove 100% of a linear outcome, 100% of the square, 63.7% of a median split, 65.8% of a threshold at one, 64.0% of the product, 6.84% of the product of median splits and 25.0% of the mixed interaction.

The square and a cut at one remove 58.9% of a linear outcome, 100% of the square, 26.8% of a median split, 100% of the threshold at one, 64.0% of the product, 7.00% of the product of median splits and 14.7% of the mixed interaction.

So the pick wins its floor by 0.16 of a point and loses 41.1 points on the linear shape, 36.8 on the median split and 10.3 on the mixed interaction. It gains 34.2 on the threshold at one, which is the shape its cut was placed to hold. On the average over the seven shapes the mean and the square remove 60.8% and the pick 53.1%.

A linear outcome is not an exotic risk. It is the shape every analysis model assumes first, and the shape a balance table reports on. A rule chosen to protect the worst case gives up most of a protection a trial would be embarrassed to lack, and buys a sixth of a point on a shape it was never going to protect.

The correlation moves the answer #

The slider on the first figure carries the same table to a correlation of 0.2 and 0.8, and the answer moves each time.

At 0.2 the products are nearly invisible to everything — the most any pair removes of the product of median splits is 1.67% — and three pairs tie for the floor at 1.54%: the square with a cut at one, at −1 or at two. The mean and the square sit at 1.52%, two hundredths of a point behind. The contest is decided in the third significant figure of a shape almost nobody removes anything of.

At 0.8 the winner changes. The best floor is 11.44%, and it belongs to a cut at −1 and a cut at one — a pair with neither the mean nor the square in it. The mean and the square fall to sixth, at 8.74%. The two cuts remove 77.9% of a linear outcome, 56.8% of the square and 36.8% of a median split; the mean and the square remove 100%, 100% and 63.7%. Between them the two cuts give up two fifths of a square and a fifth of a linear outcome to raise the floor by 2.7 points.

So the maximin pair is the square and a cut at one when the covariates are nearly independent or moderately correlated, and two off-median cuts when they are strongly correlated, and it is never the mean and anything. The correlation is a fact about the covariates that a trial can measure before it starts, so this is not an argument that the answer is unknowable. It is an argument that the answer is not a property of the functions: the same budget, the same seven shapes and the same closed forms give two different recommendations at two correlations a trial might plausibly have.

The list moves it further #

The correlation was one thing the floor depended on. The list is the other, and it is the one the earlier essay warned about. Seven shapes give 127 non-empty lists — every subset a designer might write down as the shapes worth protecting. For each one the best pair can be read off the same table.

At a correlation of a half, twenty-three of the twenty-eight pairs are best on at least part of one list. The most frequent winner is the square and a cut at two, with 26.4 of the 127 — 26 of them outright — and it is not the pair that wins the full list; the full list’s winner, the square and a cut at one, collects 18.6 and wins only one list on its own. The mean and the square collect 15.9, ten of them outright. 52 of the 127 lists are tied between two or more pairs.

No pair is best on as many as a quarter of the lists. At 0.2 the picture is much the same — twenty-four pairs win somewhere, 80 lists are tied, and the most frequent winner, the square with a cut at one, just passes a quarter at 32.7. At 0.8 one pair finally stands out, the two off-median cuts, which win 56 of the lists outright; even there, twenty-three pairs win somewhere.

So “the best two functions” is a phrase with two missing arguments. Given the correlation and the list it has an answer, and the answer is often a pair nobody would have proposed; given neither, it has twenty-odd answers.

Two readings of the same table #

The worst case is one way to read a row. The average share over the shapes is another, and it is the reading a designer who expects each shape to be about equally likely is implicitly using.

The average chooses the mean and the square at every correlation, at 50.4%, 60.8% and 67.0%. The worst case chooses the square with a cut at one, the square with a cut at one, and the two off-median cuts.

The two readings disagree at every correlation, and the reason they disagree is not subtle. The worst case is decided by the product of two median splits, and every pair removes almost nothing of it; the average is decided by the shapes a pair can actually remove, where the mean and the square remove all or most of four of the seven. A criterion’s answer is driven by the place where its numbers differ most from pair to pair, and for the worst case that place is a shape where they barely differ at all.

Which shapes are worth protecting made the general case for the worst case: an outcome model is not known in advance, and a dictionary that leaves some shape untouched is betting that nature will not choose it. That argument is right about the zeros. Eight of the twenty-eight pairs leave a shape entirely untouched, and the worst case correctly puts them last. Past the zeros it stops being an argument about exposure and becomes an argument about hundredths of an untouchable shape.

What a designer can take from the table #

Three readings survive the measurement, and none of them is a recommendation of a pair.

Parity is the screen. A pair with both functions of one parity has a floor of exactly zero against a shape of the other, whatever the correlation, and the eight such pairs here should not be held. That is the earlier essay’s result and it carries over to a fixed budget without change.

Past the screen, the worst case is the wrong instrument when one shape is out of reach. When a shape exists that no affordable dictionary can protect, it becomes the worst case for every candidate, and the ranking is decided by how much of it each candidate grazes. A floor that varies by a sixth of a point should not override a column that varies by forty. The repair is to take the worst case over the shapes some candidate can protect — over the six shapes without the product of median splits, the mean and the square tie for the top at 25.0% with the mean and a cut at one, both held down by the mixed interaction — or to read the average beside it and decide what the difference is worth.

And the list is part of the design. Balancing a skewed covariate found the floors moving with the marginal as well, so the list, the correlation and the covariate’s shape are three arguments a quoted floor leaves out. Twenty-three pairs are best on some list, and a protocol that says its balancing rule was chosen to protect the worst case has stated half of its criterion; the rule is part of the data, and so is the list the rule was chosen against.

Every share a projection, and nothing simulated #

Every share is the closed form the dictionary field built: a projection of each shape onto the span of the dictionary under a bivariate normal at the stated correlation, with the cross terms computed by conditioning one covariate on the other through a Hermite expansion. Nothing is simulated, so nothing here has a standard error, and the ties are ties to twelve decimal places — the square with a cut at one and the square with a cut at −1 remove identical shares of every shape but the threshold, because the two cuts are mirror images under the joint sign flip and every other shape is odd or even under it.

Exact ties on the floor are broken by the next-worst shape and so on, which never prefers a pair that is worse somewhere and better nowhere; that is what separates the square with a cut at one from its mirror. On the lists, a tie on the floor is shared equally among the tied pairs, because a designer handed a tie has no reason in the list to prefer either. The twenty-eight pairs are every pair from the eight functions, held identically on both covariates, which is the arrangement every rule in this field uses.

The worst case drops the mean, and the average keeps it #

The claims are that at a fixed budget of two functions per covariate the worst case picks a pair without the mean at correlations of 0.2, 0.5 and 0.8; that the floor it maximises is held by the product of two median splits for nearly every pair, and the winner beats the mean and the square by 0.02, 0.16 and 2.7 points; that at a half the pick gives up 41.1 points of a linear outcome; that across the 127 lists of the seven shapes between twenty-three and twenty-four pairs are best somewhere at every correlation; and that the average share picks the mean and the square at all three.

Not claimed: that the mean should be dropped, which is what the worst case recommends and what this essay argues against; that the average is the right criterion, since it treats seven shapes as equally likely and nothing measured here says they are; and anything about budgets other than two functions a covariate, or about pairs that differ between the two covariates.

Still open: the budget that changes the answer #

Two functions a covariate is one budget. At one function parity says the mean and the square each have a worst case of exactly zero, and only an off-median cut, which mixes the two parities, can have a floor above it — so the worst case’s pick at one function is a cut by default, whatever the cut is worth on the shapes a trial expects. At three, a dictionary can hold the mean, the square and one cut, and the product of median splits stops being the only shape that matters, because every three-function dictionary that holds both parities grazes it about equally. The measurement that would say whether the worst case becomes a usable criterion as the budget grows — and at what budget, given the thinning of the admissible set that each extra function costs at a hundred and at two hundred units — is the one that turns this table into a design rule rather than a caution.