A worked example with `art`: the same series, two objectives¶

Spanish national CPI (INE IPC290751), monthly, 2002-01…2019-12, 216 observations. Every number below is from the real guided analysis recorded in SF_MEG/empirical/cases/ES_CPI/RESULTS.md; nothing here is illustrative.

The point of reading both halves is the contrast. The nodes of the guided pipeline are the same in both; the criterion at some of them is not, because what the model is FOR changes what counts as a good decision. That is the part of Box-Jenkins-Treadway that is hardest to learn from a manual and easiest to show on one series.

Part I — the objective is FORECASTING¶

Node 1 · The transformation (λ)¶

Evidence: describe_boxcox gives the range-mean plot and the gap between the log and level fits.

Decision: λ = 0 (logs). And it is taken on a theoretical criterion, not on the test: this is a number index with an arbitrary base, so its level carries no information and only relative changes do. Had the test disagreed, the criterion would still have won.

Worth stating plainly for a newcomer: the first node is one where the evidence supports a decision the theory had already made. Not every node is a contest.

Node 2 · The order of integration (d)¶

Evidence: ADF and KPSS both lean toward d = 2. Shin-Fuller on the AR side gives Φ̂₁ᵤ = 37.5, far above its critical value, with ρ̂ = 0.40 — nowhere near a unit root.

Decision: d = 1, against two of the three tests.

The reasoning is the one an analyst has to internalise: ADF and KPSS are testing whether another difference is admissible; Shin-Fuller is testing whether the remaining root is one. A series that is I(1) with strong inertia looks "insufficiently stationary" to ADF while being plainly not I(2) to a test that looks at ρ. Over-differencing is not a conservative error — it manufactures a unit MA root and, as Part I's third model shows, it costs 43 % of forecast accuracy at two years.

Node 3 · The seasonality, frequency by frequency¶

Evidence: eleven deterministic components fit — five cos/sin pairs for f = 1…5, plus the (−1)ᵗ alternator for the Nyquist f = 6. f = 2 dominates (the Spanish January-February sales pattern). The MEG sweep then asks, per frequency, deterministic or stochastic.

Result: every frequency deterministic except f = 3, which comes out stochastic — but marginal: LR = 2.30 against a critical value of 2.07.

Decision: the frontier is acknowledged, not resolved. The guided pipeline does not pretend 2.30 > 2.07 settles it, and this is where an analyst benefits from an assistant that argues rather than reports: the honest reading is that the two representations are nearly indistinguishable by this test, so the decision has to be made on other grounds — and Part I ends by making it on parsimony.

Node 4 · The confirmatory pair, and a test that says nothing¶

The MEG works on the MA side. The confirmatory pair runs Shin-Fuller on the AR side, and it comes out inconclusive: the free AR_f collapses to ρ̂ ≈ 0.07 when even a weak deterministic sinusoid would give ρ̂ ≈ 0.43.

This is expected and it is not a failure. The AR side lives on the ρ axis; what separates deterministic from stochastic seasonality is σ_S², a different axis. A test that cannot see the axis the question lives on will be silent whatever the answer.

The teaching point: an inconclusive test is information about the test, not about the series. Reading it as evidence for the null is a mistake, and it is exactly the kind of thing a guided assistant should say out loud.

Node 5 · The canonical airline (D = 1), and a tie broken on economics¶

This node deserves saying plainly: ∇∇₁₂ with a regular and a seasonal MA is the canonical Box-Jenkins model for a seasonal series — the airline model of the 1970 book — and it is what most practitioners would build first. It is not a straw man set up to lose. It is the reference against which the frequency-selective alternative has to justify itself, and if it wins there is nothing to discuss.

Built through the guided B2 branch on ∇∇₁₂, it needs a regular term. The textbook answer is MA(1). The decision was AR(1), on the grounds that inflation has inertia and the deterministic model had already estimated φ ≈ 0.40 for exactly that.

(1 − 0.3631·B)(∇∇₁₂ ln y + 0.0145) = (1 − 0.8575·B¹²) aₜ
    (0.0658)          (0.00673)          (0.0586)
σ̂ₐ = 0.2633 %   ℓ = −25.17   AIC = 56.34

All three coefficients significant, residuals white, normality fine.

Node 6 · The verdict, out of sample¶

Estimate once on 2002-2019, freeze the parameters, roll the origin one month at a time from 12/2019 and forecast 24 ahead — 54 balanced origins per horizon.

h	RMSE_D	HSM/D	SARIMA/D	de-biased
1	0.602	1.002	1.002	0.983
6	1.725	0.999	1.189	1.066
12	2.931	0.998	1.272	1.110
24	4.889	0.997	1.435	1.314

Two conclusions, and the second is the one that teaches:

D and the frequency-selective S forecast identically — the ratio is 1.00 at every horizon. The f = 3 witness sits at ≈ −1, so the stochastic seasonal almost cancels and the two share a forecast function. D is retained by parsimony, NOT because it forecasts better. Saying that correctly matters: an earlier draft of the note claimed D won, and that was an artifact of forecasting with the non-invertible representation.
The blunt ∇₁₂ degrades monotonically, to +43 % at two years — and the Diebold-Mariano test is not significant at any horizon (p 0.16–0.78). Fifty-four overlapping 24-step origins are not enough power. So the case against the airline rests on the structural argument, not on the number: ∇₁₂ over-differences frequencies that are deterministic, and the persistently significant mean it needs is a symptom of that, not a virtue.

This is the honest shape of an empirical conclusion, and it is worth an analyst seeing it once: a difference that is economically large and statistically undemonstrated, reported as exactly that.

Part II — the objective is a TRANSFER MODEL¶

Now the same series is not the end product. It is going to be the output of a transfer model whose input is the oil price (WTI), estimated in mtram. Same data, same evidence at every node — and two of the decisions change.

What does NOT change¶

Nodes 1 and 2. λ = 0 and d = 1 are properties of the series and of what it measures; no downstream use alters them.

Node 3 changes: prefer DETERMINISTIC seasonality, and know why¶

In Part I the frontier at f = 3 was resolved by parsimony and it could have gone either way. Here it cannot.

The identification of a transfer model prewhitens: it filters the output through the INPUT's ARMA and reads the cross-correlation. WTI carries nothing seasonal. So if the output keeps a stochastic seasonal component, the filter does not remove it, the filtered series is still non-stationary at that frequency, and the CCF does not come out empty — it comes out with structure everywhere, and the heuristic reads a delay and an order off it anyway.

That is a silent failure: a plausible-looking (b, r, s) obtained from a contaminated correlogram. Deterministic seasonality has no such effect, because subtracting fixed harmonics leaves nothing behind to survive the filter.

This is not a workaround. The Box-Jenkins-Treadway tradition specifies seasonality as provisionally deterministic anyway and only afterwards resolves it frequency by frequency. An analyst heading for mtram should choose it at the start: re-doing it later costs far more.

Node 3b, new: the operators must AGREE with the input's¶

This node does not exist in Part I, and it is the one that would have silently corrupted the answer until 2026-08.

The transfer relates the levels, with the differencing carried by the noise. So the input has to enter differenced by the OUTPUT's operator. When the two operators agree, that is the input's own column and nothing has to happen. When they differ, what gets fitted is not ν(B) but ν(B)·Δ(B), with Δ = op_output / op_input, and the reported gain is wrong by Δ(1).

For this pair:

model of ES_CPI	operator	vs WTI's `∇`	consequence
the Part I model (`d=1, D=0`, harmonics)	`∇`	Δ(1) = 1	nothing to do
the airline variant (`d=1, D=1`)	`∇∇₁₂`	Δ(1) = 0	gain annihilated

Measured, on this very series: eight variants of ES_CPI give a gain of 0.0264–0.0274 once the dispatch is in place. Before it, the five airline variants said the pass-through was ≈ 0.006 while the three harmonic ones said ≈ 0.027 — a factor of five that an analyst would have attributed to economics and that was arithmetic.

mtram's check_operators answers this before anything is estimated, and it costs one call.

Node 6 changes: the verdict is no longer about this model¶

In Part I the univariate model was judged on its own forecasts. Here it is judged on whether the transfer model built on top of it earns its place:

the diagonal gate — with a diagonal structure the exact likelihood factorises, so the joint fit must reproduce the sum of the univariate ones. It is a complete check that the crossing preserved everything, and it is cheap enough to run first;
the likelihood-ratio test of the transfer against that diagonal rung;
and the rule from FORECAST_DIAGNOSIS.md: a transfer model is kept over its univariate iff it forecasts better out of sample.

The univariate model is, in Treadway's phrase, the measuring stick. If the transfer model cannot beat it, the transfer model is what needs rethinking.

What the two halves are meant to leave behind¶

Same series, same evidence, same nodes — and at nodes 3 and 6 the right decision depends on what the model is for. A guided assistant is worth having precisely there: not to compute the tests, which any program does, but to say which question this node is answering right now and why this criterion and not the other one. Every decision above is defensible, and none of them is automatic.

A worked example with art: the same series, two objectives¶