Field

Counting what is independent

Akaike's penalty is 2q because the optimism of a fit is a trace and the trace collapses to the parameter count when the rows are independent. Repair the trace and the criterion gets worse, on either scale, because the row count entered twice and a penalty is the second place: whiten the fit and the ordinary penalty is correct again, which recovers 88% of what counting rows gives up. Estimating the dependence from the rows being selected on costs 8% of that. And a multiplier resampling cannot keep more dependence than the residuals have, which is a ceiling rather than a tuning problem.

The correction is not a property of the sample. tr(HΩ)/q for each of fifteen candidates, at ρ = 0.7. Two candidates that fit the same number of coefficients need corrections that differ by as much as 1.49, because one of them is fitting the persistent predictors and the other is not — so no single number can be right for both, and the scalar n/n_eff = 5.537 is above every one of them. The four predictors carry persistences 0.9, 0.6, 0.3, 0; at one persistence for every column the whole spread collapses and a scalar looks exactly as good as the trace.

A penalty is a trace

Akaike's 2q is not a count of coefficients. It is the answer a trace collapses to when the rows are independent — and once they are not, the trace is still the right object and is no longer the count.

The one candidate an effective sample size is right about. n/n_eff with the finite-sample inflation Σ(1 − |k|/n)ρ^|k| is not an approximation to tr(HΩ) for a fit with only an intercept — it is that trace, to machine precision, because the hat matrix of a constant column is 1/n everywhere and its trace against Ω is the mean of Ω. The quoted limit form n(1 − ρ)/(1 + ρ) is not even right about that one. And the average correction the table's fifteen candidates actually need is 3.318 per parameter, well below the scalar, so applying it to all of them over-charges every one.

One number for a table of candidates

An effective sample size is a real quantity, it is exactly right about one thing, and that thing is a mean. Substituted into Akaike's criterion it changes nothing at all, because the penalty it is meant to fix has no sample size in it.

The row count entered twice, and a penalty is one place. Regret against the best available model as the errors are made persistent. Counting rows more than quadruples; both penalty repairs — the trace, and the scalar effective sample size — are worse than it at every persistence measured; and whitening the sample and keeping the ordinary penalty falls, recovering 86.9% of what counting rows gives up at ρ = 0.85. Doing it at an estimated ρ recovers 80.6%, so having to estimate the dependence from the rows being selected on costs 7.3% of what knowing it is worth. Mallows' forms are drawn beside the logarithmic ones and behave the same, which is what rules the linearisation out.

The repair that was exact and made it worse

A penalty computed from the trace is exactly the optimism it estimates, and selecting with it gives up a fifth more than not correcting anything. The row count entered the criterion twice, and a penalty is the second place.

A fit takes the low frequencies out of what it leaves behind. The autocorrelation of the errors, of the residuals of a fitted benchmark, and of those residuals rescaled by their own leverage. (I − H) removes the component of the errors lying in a column space that is itself slow-moving, so the residuals are less persistent at every lag — by 5.9% at the first and 26.6% by the fourth. The leverage correction is the standard repair for what a fit does to a residual's size; drawn here against what it does to a residual's dependence, it does nothing.

The residuals are not the errors

A fit removes the part of the errors lying in its own column space, and a persistent design's column space is itself slow — so what is left behind is smoother than what went in, at every lag, by an amount that grows with the lag.

The ceiling a multiplier cannot reach past. A wild-type resampling forms e*_t = e_t·w_t with the multiplier independent of the residual, so what comes out has autocovariance γ_resid(k)·γ_w(k) — the residuals' own, multiplied by the multiplier's. Since |γ_w| ≤ 1 the reference distribution's dependence is bounded above by the residuals', and the residuals' is already below the errors'. The two shortfalls compose. For a block of ℓ the multiplier's autocorrelation is exactly the triangle (1 − k/ℓ)⁺, drawn here as the dashed prediction against the realised resamples at ℓ = 5; the bound is attained only at ℓ = n, where the reference distribution is built from one sign.

What a multiplier cannot keep

Two reasons were named for the quarter a blocked resampling falls short, and taking either away makes the gap larger. What is left is a bound — a multiplier can only take dependence out, and the residuals' own is already below the errors'.