A split that depends on the order

The decomposition this field is built on runs the first search, pins it at the answer it gives, and searches the second against it. Nothing in that says which search is first.

Run it the other way and the same draw gives a different overlap and a different interaction. The difference between them is identical, because the excess is order-free by construction and its two components are not.

The two orders #

Over three hundred draws under a first-order autoregression, on the four pairs where both orders exist:

Two of the four move by more than three paired standard errors, and one of them changes sign.

What stays out, and one thing that does not #

Two limits before the readings, because both change how the table above should be taken.

Three hundred draws. The paired comparisons are made on the same draws in both orders, so the draw-to-draw variation cancels and the standard errors are small for the count — which is why a difference of six thousandths in the break-and-column pair reads at 8.70 paired standard errors. An unpaired comparison of the same two numbers would say nothing at all.

One law. Everything in the table is a first-order autoregression. The split is re-run under all four laws the collection works in and the containment rung’s interaction is exactly zero under every one, which is the structural claim; whether the order-dependence of the break-and-column pair survives a change of law is measured only in the first order.

And one thing that is not a limit: the two orders are computed from the same six suprema per draw, so nothing about the comparison depends on two runs agreeing about anything. The base fit, the two solo searches and the joint search are shared between them, and the orders differ only in which of the two pinned runs is used.

What the order-dependence is #

It is not a defect in the construction, and it is worth being precise about what it is instead.

Both orders compute a legitimate quantity. How much does the second search lose by having the first already run? is a question with an answer, and so is the same question with the roles swapped. The two answers differ because the two searches do different things to the sample.

A break search that has run has fitted two segments and centred them separately, so the residual series the column search reads is not the one it would have read alone. A column search that has run has taken a direction out of the design, so the residual series the break search reads is not the one it would have read alone.

There is no order-free way to say which of two searches “found” ground both of them can reach, because the ground is not divisible. The two orders bracket it, and the bracket is a fact about the pair rather than a shortcoming of the measurement.

The pair that changes sign #

The break-and-column pair is the one worth reading, because its overlap is negative one way and positive the other.

Break first. The break search runs on the untouched series and picks its position. The column search then runs with the break pinned there — and finds more than it found alone, because a series split into two separately centred segments has different residuals for a column to explain. The overlap is −0.004395: a loss that is a gain.

Column first. The column search runs and picks its column. The break search then runs with the column pinned — and finds a little less than it found alone, because the column has taken some of the residual sum the break was going to take. The overlap is +0.002163: an ordinary loss.

Both readings are correct about their own order, and the excess is the same −0.006759 either way, so the pair is complementary in the net whichever way it is read. What differs is the attribution: with the break first, the complementarity is mostly in the overlap; with the column first, it is mostly in the interaction, which reads 0.008922 against 0.002364.

The joint search also moves more often in the second order — 54% of draws against 29% — which is the same fact seen from the other side, and is the pattern the field that built the ladder had no way to see. Pinning a column leaves the break free to be moved by the joint search; pinning the break leaves less for it to move.

The identity the two orders have to satisfy #

The claim that the excess is order-free while its components are not is checkable to every digit printed, and it checks.

The excess is the overlap less the interaction. Break first: −0.004395 − 0.002364 = −0.006759. Column first: 0.002163 − 0.008922 = −0.006759. The same six decimal places, from four numbers of which no two agree.

That is the strongest evidence available that the two orders are readings of one object rather than two experiments. Nothing about the arithmetic forces it if the two orders had been run as separate simulations — they would agree to the paired noise and no further — and it holds exactly here because both orders are built from the same six suprema on each draw. The base fit, the two solo searches and the joint search are shared; only the two pinned runs differ.

So the pair really is bracketed rather than disputed. Two attributions of one quantity, summing to the same total, in the way a variance decomposition’s components do.

Where the factor of four in the interaction comes from #

The break-and-column interaction reads 0.002364 in the first order and 0.008922 in the second, a factor of 3.77, and the field already reports the other half of it: the joint search moves on 54% of draws in the second order against 29% in the first.

Dividing one by the other separates the two effects. Per draw on which the joint search moves at all, the interaction is 0.002364 / 0.29 = 0.0082 in the first order and 0.008922 / 0.54 = 0.0165 in the second.

A factor of 1.9 in how often the joint search finds something, and a factor of 2.0 in how much it finds when it does. The two are nearly equal and they multiply to the 3.8 observed.

That is a more specific statement than “pinning a column leaves the break free”. Pinning the column does not merely leave the break movable more often; it leaves it movable further, because the break positions the joint search can reach from a pinned column are a wider set than the column choices it can reach from a pinned break. The asymmetry is in the geometry of the two feasible sets, and it shows up equally in both factors.

The two that do not move #

Two of the four pairs give the same overlap either way, and it is worth saying why rather than treating agreement as the default.

Two independent dictionaries read 0.000514 and 0.000420, at 0.68 paired standard errors. Neither search changes what the other reads in any structured way: both are searches over columns drawn from their own streams, both leave the sample’s own structure alone, and what one takes out of the residual sum is a scalar rather than a shape. So the order in which they run is nearly irrelevant, and the small difference is noise.

Two step dictionaries read 0.044030 and 0.042184, at 0.52. These do share a great deal — their cut points interleave along the same range, so each one’s best column has a near-twin in the other — but they share it symmetrically. Neither dictionary is a special case of the other; they are two samples from the same family of features, cut a few rows apart, and the pair is invariant under swapping them by construction.

Symmetry of construction is what makes an order irrelevant, and neither of the two pairs that move has it. The control’s symmetry is the reason its zero survives every dictionary size as well: a pair invariant under swapping its two searches has an overlap that cannot depend on which one is pinned, whatever else is true of it. A break and a column are different kinds of object; a break and a window are different kinds of object; and in both cases one of them changes the series the other reads.

And one pair has no second order #

The fifth pair is missing from the table above, and the reason is not that it was left out.

A break paired with a dictionary of step columns cannot be run with the step column pinned. A centred step column is a constant inside whichever segment lies wholly on one side of its cut, so a break search with a step column always in the design finds one of its two segments rank-deficient at every break position. There is no admissible configuration.

The joint search escapes that by taking no column at all — which is a point in its feasible set and is the point it takes — and that is exactly why this pair’s interaction is zero on every draw. A run pinned to a specific column has had the take-nothing option removed and has nowhere to go.

So the containment rung is a rung with one order, and its number in the first order is the whole of its excess: 0.136647, all overlap, interaction exactly zero.

That is a more satisfying answer than a missing cell. The earlier field fixes this rung by construction — a break spans every direction a centred step column spans, so the step dictionary adds nothing — and here the same fact arrives twice over: once as an interaction of exactly zero, and once as a second order that does not exist. Two independent routes to one structural claim is the shape this collection prefers, and neither of them was arranged: the first is a measurement that came out at zero on every draw, and the second is a run that returned nothing.

What pinning removes #

The general form of that is worth stating because it applies to every dictionary search and to nothing else.

Pinning a break fixes a position but leaves every other choice available. Pinning a window fixes a width and leaves every other choice available. Pinning a dictionary column removes an option, because a dictionary search’s feasible set includes taking no column, and a pinned column is a commitment to take one.

Most of the time that is harmless: the pinned column is the one the solo search chose, so it is a column worth taking, and the joint search’s own optimum usually includes some column. It bites exactly when the joint search’s optimum includes no column, which is the containment case and only the containment case among these five pairs.

A reader building this decomposition on other searches should expect the same asymmetry. Any search whose feasible set contains a “do nothing” option will have an order in which it cannot be pinned without changing the problem.

The two quiet pairs are not equally quiet #

The pairs that agree are read off their paired t statistics, and those statistics are made of two things. Separating them says how much each non-result is worth.

Dividing each pair’s difference by its t gives the paired standard error of the comparison:

• two independent dictionaries: 0.000094 at t = 0.68, so 0.00014
• two step dictionaries: 0.001846 at t = 0.52, so 0.0036
• a break and a window: 0.002886 at t = 4.68, so 0.00062
• a break and an independent column: 0.006558 at t = 8.70, so 0.00075

The step-dictionary pair’s comparison is twenty-five times noisier than the independent dictionaries’ and six times noisier than the break-and-window pair’s, on the same three hundred draws.

That changes what its quiet reading means. Its two orders differ by 0.001846, which is nearly two thirds of the difference the break-and-window pair shows — and that pair reads at 4.68 standard errors. Measured with the break-and-window pair’s precision, the step dictionaries’ difference would have read at 3.0 and been reported as a real movement.

So the two independent dictionaries genuinely do not move: their difference is small in absolute terms and measured precisely. The two step dictionaries have not been shown to move, which is a weaker statement, and the symmetry argument is what carries the claim rather than the measurement. The argument is a good one — a pair invariant under swapping its two searches cannot depend on which is pinned — but it should be read as the reason to believe the zero, with the t statistic as a non-refutation rather than as evidence.

The reason the comparison is noisy is legible in the pair itself. Two step dictionaries cut a few rows apart have near-twin columns, so which of the twins each order’s pinned run happens to select varies from draw to draw, and that selection noise does not cancel in the pairing the way the sample’s own variation does.

What a bracket is worth #

The two orders give two numbers, and it is fair to ask what a reader does with a pair of them.

For three of the four pairs the bracket is narrow enough to ignore. Two independent dictionaries: 0.000514 to 0.000420. Two step dictionaries: 0.044030 to 0.042184. A break and a window: 0.197114 to 0.194228 — a spread of one and a half per cent on a quantity whose own standard error across the ladder is larger than that.

For the fourth it spans zero, and there the bracket is the answer. A pair whose overlap is −0.004395 under one attribution and +0.002163 under the other has no defensible single overlap, and reporting either alone would be reporting a sign that the other order reverses.

That is the practical rule this essay leaves: report both when they disagree, report either when they do not, and check which case a pair is in rather than assuming. The check is one extra supremum per draw on top of the one the split already costs, so it is a fifth run rather than a sixth — the two sequential runs are the fifth and sixth of the six this field computes.

The excess is order-free and is asserted to be #

The one thing that must not move between the orders is their difference, and it is checked rather than argued.

$$\text{overlap}(A\!\to\!B) - \text{interaction}(A\!\to\!B) = \delta_A + \delta_B - \delta_C = \text{overlap}(B\!\to\!A) - \text{interaction}(B\!\to\!A)$$

The pinned supremum cancels in each order, so both differences are the same three suprema arranged the same way. It is asserted at a part in 10⁹ for every pair that has two orders, and it holds — which is what says the two orders are two attributions of one quantity rather than two quantities.

A decomposition whose total moved with the order would be a decomposition of nothing. That check is cheap, it can only fail if the construction has changed underneath it, and it is the reason the order-dependence above can be read as a finding rather than as a bug.

What the interaction does between the orders #

The overlap is the column reported above because it is the one with a free sign, but the interaction moves too, and it moves by more.

On the break-and-column pair it is 0.002364 with the break pinned and 0.008922 with the column pinned — nearly four times larger. On the break-and-window pair it is 0.005991 and 0.003106, about twice.

The direction is the one the construction predicts. Pinning a search removes its freedom, so the joint search’s advantage over the sequential one is whatever the pinned search would have done differently. Pin the break — a sweep over a hundred and more admissible rows — and the sequential run has already searched a large space, so the joint search has little left to add. Pin the column — one choice from six — and the sequential run has searched a small space, so the joint search has more to add.

The interaction measures what the pinned search gave up, which is why it is larger when the pinned search is the cheaper one. Its non-negativity is guaranteed in both orders and its size is not comparable across them, which is one more reason the two orders are two attributions rather than two estimates of one thing.

The same fact is visible in how often the joint search moves at all: 29% of draws with the break pinned and 54% with the column pinned, against 78% and 36% on the break-and-window pair — where the ordering runs the other way because there the pinned break is what the joint search most wants to move.

Which order to report #

A reader wanting one number has to choose, and this field does not choose for them. What it does is say what each choice means.

The first order — pin the search that runs first in practice. For the break-and-window pair that is the break, and it is the order a practitioner would actually work in: fit the break, then decide how wide a whitening window the residuals want. The overlap it reports, 0.197114, is then the honest answer to “what does the window search lose by the break search having already run?”.

The second order — pin the search that is cheaper to run. For a pair where one search is a sweep over a grid and the other is a supremum over a dictionary, pinning the cheap one and searching the dear one is the cheaper decomposition, and it answers the other half of the question.

Or report both, which is what the figures here do, and treat the pair as bracketed. That is the reading the first essay of this field uses in its own table, where the first order is printed because it is the practitioner’s order and the second is a figure away. For three of the four pairs the bracket is narrow — 0.000514 against 0.000420, 0.044030 against 0.042184, 0.197114 against 0.194228 — and for one it spans zero.

The rule that follows is short: a split reported in one order is a split conditional on that order, and the pairs where it matters are the pairs whose two searches change what the other one is reading. Those are also the interesting pairs.