What a mean split leaves, with both halves varying
The share of a mean split's interaction that survives the rule balancing it, at every copula and every marginal, matched at a Spearman correlation of 0.40. The three radially symmetric copulas leave exactly nothing with a symmetric covariate and rise steeply with the skew. The two asymmetric ones start at 7.707% and go opposite ways: the lower-tail copula falls to 0.002% at a skewness of 0.95 — the two failures cancel almost exactly, and a guarantee both fields report as broken is restored — while the upper-tail one climbs to 40.288%. And the heavy-tailed symmetric covariate, which leaks exactly nothing on its own, doubles what the asymmetric copulas leak: 14.229% against 7.707%.
Both halves of the dependence at onceslider: which rule the trial balances, 2 positionswide4 views
What else it draws
The same object, drawn to answer the other questions the essays put to it.
A median split's interaction leak at all 30 combinations of copula and marginal, on a log scale. Every one is under 10⁻¹⁶ and the largest is 1.74e-20, which is the quadrature's own noise rather than a leak. The reason is arithmetic and it is short: a centred median split takes the values ±½, so its square is a quarter identically — for every unit, on every draw, whatever the covariate's scale is and whatever joint law the ranks have. The interaction is then orthogonal to both main effects by construction, and there is nothing for either half of the dependence to break. Both of the fields this one joins report this zero holding under their own variation; running both variations at once is what establishes that it is not two coincidences.
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%.
The share of the product of three median splits that a rule balancing the three splits removes, by Spearman rank correlation from 0.1 to 0.7. At 0.1: Gaussian 1.18%, t 1.26%, Frank 1.47%, Clayton either way up 1.11%; At 0.2: Gaussian 4.25%, t 4.51%, Frank 5.25%, Clayton either way up 4.08%; At 0.3: Gaussian 8.76%, t 9.26%, Frank 10.70%, Clayton either way up 8.58%; At 0.4: Gaussian 14.47%, t 15.23%, Frank 17.48%, Clayton either way up 14.45%; At 0.5: Gaussian 21.26%, t 22.31%, Frank 25.40%, Clayton either way up 21.68%; At 0.6: Gaussian 29.20%, t 30.50%, Frank 34.41%, Clayton either way up 30.39%; At 0.7: Gaussian 38.48%, t 40.00%, Frank 44.61%, Clayton either way up 40.85%. For two covariates the corresponding number is exactly zero at every correlation on every copula.
Where it is used
7 essays draw this figure, each at the numbers its own argument is about, so the same picture answers 7 different questions.
- Two failures that cancel Both halves of the dependence at once
- A symmetry that was not enough Both halves of the dependence at once
- The zero that was a crossing The same table at seven correlations
- A copula that halves a marginal Both halves of the dependence at once
- The zero that survives both Both halves of the dependence at once
- A median with ties Both halves of the dependence at once
- The zero that needed two Both halves of the dependence at once