Normal Distribution
nothing is called 95% until it has been counted
Field

Searching among fitted models

The set field compares forecasts nobody estimated. A specification search compares a benchmark with a table of variants that all contain it, and two things change at once: every variant is behind before the search begins, by an amount with a closed form in the shape of the table, and the reference distribution for the winner can no longer be resampled from the data — it has to be generated from a model. The repair the nested case asks for, applied row by row, takes a table from its nominal level to a quarter.
What each variant loses before anything has been searched for. The mean loss differential of each of the eight variants against the benchmark, over 600 tables of 60 origins, with every fit given 71 rows. The series is an AR(1) and every variant adds a lag whose coefficient is zero, so in population the two forecasts are the same forecast and the difference drawn here is estimation noise and nothing else. The marked line is σ²(q₁ − q₀)/n = -0.01408, which is an expression in how many coefficients each model has and how many rows it was fitted on — it knows nothing about the series, the persistence or which lag the variant added, and every bar is within a fifth of it. This is the amount a reference distribution recentred at each column's own sample mean believes the candidates are already behind by.

A table of nested models

A benchmark and eight variants of it, each adding one thing. Every variant is behind before the search begins, by an amount that can be written down before the data exists — and the two most natural ways of reading the table are wrong in opposite directions.

 The distribution of the largest statistic in the table. Fit the benchmark to the whole series, resample its residuals, simulate 199 series in which the null is true by construction, re-run the entire eight-variant search on each, and keep the largest statistic. That is the distribution drawn here, and it is the distribution of the thing a specification search actually reports. It is centred at 1.045 — the maximum of eight statistics is not centred at zero however well each of them behaves — and its 5% point is 2.536. A table read against 1.671 is reading the distribution of one statistic; a Bonferroni correction reads it against 2.577 and is nearly right here, because eight variants that each add a different lag are nearly eight separate chances.

A null with a model in it

The distribution to read the winner of a table against cannot be resampled from the data, because the data does not contain the null. It has to be generated from a model — which is the assumption the resampling was chosen to avoid.

 The ranking on the left, the weights on the right. Eight moving-average forecasts of an AR(1) at φ = 0.4895, the persistence at which the best of them exactly ties the 60-observation benchmark. On the left, each candidate's expected squared error in units of the series' own variance: the smallest belongs to L = 2, at 1.0156. On the right, the weight each carries in the variance-minimising combination of all eight — and the best of them carries 0.00000. The two ends of the family carry 1.0172 of the weight between them, and the combination they make is worth 0.7817, which is 23.0% below the best single forecast. Both columns are closed forms in φ. Which forecast to keep and which forecasts to use are different questions, and this is a set where the answers share nothing.

The weight that is a vector

Two forecasts have a best combination and one number describes it. Eight have a best combination too, and the vector describing it puts nothing at all on the forecast with the smallest mean squared error.

 Eight encompassing nulls, all of them true at once. The forecast under test is the variance-minimising combination of the eight candidates, and the first-order condition that defines its weights is cov(e_c, e_j) = var(e_c) for every j — so every one of the eight nulls is exactly true simultaneously and every rejection counted here is false. 800 draws of 60 origins. The largest of the eight statistics rejects 25.0% of the time at a nominal 5% and one comparison stated in advance rejects 3.1%, which makes the set worth about 9.1 independent comparisons — nearly the eight it has. Bonferroni, which is far inside its level on a search over accuracy comparisons of the same eight forecasters, is at 4.8% here.

When every null is true

A reality check assumes that every candidate in the set is exactly as good as the benchmark, which is a configuration nobody's data is ever in. Test a combination against its own parts and that configuration is not assumed — it is what the arithmetic makes true.

 How often each reading rejects when every null is true. At the optimal combination of eight forecasts, where all eight encompassing nulls are exactly true, the share of 1,200 draws on which each reading rejects at 5%, at 60, 120, 240, 480 origins. The joint Wald test against a chi-square on seven degrees of freedom: 26.6%, 16.4%, 9.8%, 8.6%; with Hotelling's finite-sample F: 15.1%, 12.4%, 8.3%, 8.0%; the largest of the eight read as a single comparison: 25.0%, 25.9%, 24.7%, 23.4%; the largest with Bonferroni over eight: 5.2%, 4.9%, 4.7%, 4.4%.

One test for eight nulls

A combination of eight forecasts can be tested against its parts one comparison at a time or all at once. All at once is a Wald test on the vector of eight encompassing means — which has seven degrees of freedom, not eight, because the means sum to zero on every origin when weighted by the combination. On sixty origins it rejects a true null 26.6% of the time, Hotelling's correction brings that to 15.1%, and handed its own true critical value it is less powerful than Bonferroni over the largest of the eight against both wrong combinations measured: 12% against 18%, and 47% against 96%.

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