Thinking in Waves Episode 52 / Part XVIII — Putting Error Bars On It

How much of Episode 42's \(f(\alpha)\) curve can actually be measured?

Extracting the Order's Distribution from Finite Data —
Only the tails are out of reach Episode 42 said "the order is not one number but a distribution." ★ But can the shape of that distribution actually be measured?
★ Multiply the number of boxes by a thousand (1,024 → 1,048,576) and the reachable \(q\) grows only from 2 to 4.
★★ The peak of \(f(\alpha)\) is always measurable; only the two tails are out of reach in principle — and the tails are the interesting part.

Tools needed: Episodes 22, 42, 49, 51Computation: kenshou/multi.js

Episode 51 put error bars on the decade count.
★ This episode does the same in the \(q\) direction — "how far out in \(q\) may you believe the answer?"

01Why sweep \(q\)? — recalling Episode 42

\(q\) is the dial selecting which strength of region you look at $$Z(q,r)=\sum\mu^q\propto r^{\tau(q)},\qquad \alpha=\frac{d\tau}{dq},\qquad f(\alpha)=q\alpha-\tau(q)$$
\(q\)\(\alpha\) (exact)\(f(\alpha)\) (exact)What it picks up
−41.6970730.207439the weakest regions
01.1257691.000000★ the dimension of the support
10.8812910.881291the representative regions
80.5159630.012758the strongest regions

★ The exact range of \(\alpha\) is 0.514573 to 1.736966 (the p-model at \(p=0.7\)).

02From a single realization, large \(q\) goes wrong

\(q\)exact \(\tau\)\(n=12\) (4,096)\(n=16\) (65,536)\(n=20\) (1M)
−2−3.71720−3.66379−3.67331−3.68473
0−1.00000−1.00000−1.00000−1.00000
20.785880.751420.814980.78468
84.114943.851034.401793.96955
126.174825.835276.633946.00266

The point of §02

★ Small \(|q|\) agrees well, but larger \(q\) departs — and not even with a consistent sign (high at \(n=16\), low at \(n=20\)).
★★★ The cause: large \(q\) is decided by a single strongest box. Whether that one box happens to be in your sample decides the answer.
★ Only \(q=0\) is exactly \(-1\) — because it is just counting boxes.

03★ There is a ceiling on the reachable \(q\)

We measure the range of \(q\) that keeps \(|\tau_{\rm emp}-\tau_{\rm exact}|\le0.1\), over five seeds, taking the median.

levels \(n\)boxes★ \(q^*\) (+)\(q^*\) (−)reachable \(\alpha\) widthof the full range
101,0242.0−1.00.6660 / 1.222454.5 %
124,0963.0−2.00.9435 / 1.222477.2 %
1416,3843.0−3.01.0440 / 1.222485.4 %
1665,5364.0−2.00.9928 / 1.222481.2 %
18262,1444.0−3.01.0933 / 1.222489.4 %
201,048,5764.0−3.01.0933 / 1.222489.4 %

The point of §03 — ★★★ the core of this episode

A thousandfold increase in boxes (1,024 → 1M) moves \(q^*\) only from 2 to 4.
The \(\alpha\) width goes 54.5 % → 89.4 % and stops.

★ The remaining 10 % is the very edge, and it does not improve with more boxes.
★★ \(q^*\) grows logarithmically — reaching the tails requires exponentially more data.
★ So "collect a longer record and you will see the ends of \(f(\alpha)\)" does not hold.

04★ Which part of \(f(\alpha)\) is measurable

At \(n=18\) (262,144 boxes), the reachable range is \(q=-3.0\) to \(4.0\).

\(q\)\(\alpha\)\(f(\alpha)\)Measurable?Meaning
−81.7355760.012758× out of reachthe weakest regions
−41.6970730.207439× out of reachthe weakest regions
01.1257691.000000★ yes★ dimension of the support
20.7042550.622634★ yes
40.5544660.207439★ yesthe strongest regions
80.5159630.012758× out of reachthe strongest regions

The point of §04

★ The peak of \(f(\alpha)\) (\(q=0\), \(f=1\)) is always measurable — it is the dimension of the support.
★★ What is not measurable is both tails — the strongest and the weakest regions.
★★★ And those are the interesting part. Intermittency is about rare, violent events — and being rare is exactly why they are not in your sample.

05Adding noise — and a prediction that failed

Each box's measure is multiplied by relative noise (log-normal, sd \(=s\)) and re-measured.

noise \(s\)\(q=-2\)\(q=0\)\(q=2\)\(q=4\)\(q=6\)
(exact)−3.71720−1.000000.785882.010423.07853
0−3.65544−1.000000.800592.095303.21716
0.3−3.67155−1.000000.795652.112653.30442
1.0−3.89711−1.000000.71160★ 1.72184★ 2.59398

The point of §05 — ★ where our prediction failed

★ \(q=0\) is completely unaffected by noise (it only counts boxes). For small noise (\(s\le0.3\)) nothing moves much.
★★★ We initially predicted "negative \(q\) will be the most fragile" — and were wrong. At \(s=1.0\) the shifts are 0.24 (6 %) at \(q=-2\), 0.37 (18 %) at \(q=4\), 0.62 (19 %) at \(q=6\) — the positive side moves more.
★ Why: with log-normal noise the large values scatter more in absolute terms, so positive \(q\) breaks first.
★★ Either way it is large \(|q|\) that suffers — the same direction as §03's conclusion.

06What can and cannot be done

GoalVerdictWhy
Say "it is a distribution"◎ yes★ the bend in \(\tau(q)\) suffices
Decide the \(q\) range from the data◎ yesmeasure §03's \(q^*\) every time
Report only near \(q=0\)◎ yesthe peak and the curvature are measurable
Average many realizations△ limited★ the tails are set by one sample's extremes
Use large \(|q|\)× not advised★ fragile to both sampling and noise
Publish the whole \(f(\alpha)\)× no★ the ends are unreachable in principle

The point of §06

★★ Episode 42's claim, "the order is not a single number," holds using only a narrow range of \(q\). All you need is to see \(\tau(q)\) bend away from a straight line, and \(q=0\)–2 is enough.
★★ What cannot be said is what shape the distribution has — that is decided by the size of your sample.

07Amendments to Episodes 49 and 51

StepNote
★ ⑥-a measure \(q^*\) before sweeping \(q\)★ test on synthetic data of the same sample size
⑥-b report only inside \(q^*\)outside it you are watching your sample's maximum
⑥-c treat large \(|q|\) separately★ if you report it, report the sensitivity too
★ ⑥-d stop at "it bends"★ naming the shape usually needs more sample than you have
⑥-e treat it like the decade count (Ep. 51)the \(q\) range is also a "how far can you speak" indicator

08What wave language can and cannot say

ItemVerdictBasis
Small \(|q|\) agrees well◎ numerics\(q=0\)–2 nearly exact at 1M boxes
★ Large \(q\) departs◎ numerics★ decided by a single strongest box
★ A ceiling on reachable \(q\)◎ numerics★ 1000× boxes moves \(q^*\) only 2 → 4
★★ The ends of \(f(\alpha)\) are unreachable◎ numerics★ 89 % of the range even at 1M boxes
\(q=0\) is noise-proof◎ numericsit only counts boxes
★ The positive side was the fragile one◎ numerics★ our prediction failed (log-normal tails)
"It is a distribution" is sayable◎ logicthe bend in \(\tau(q)\) suffices
★ Naming the distribution's shape× no★ the sample size sets the ceiling
★ Validation on real data× not done★ synthetic data only, here too

Exercises

  1. §02 notes that the cascade was changed to a non-conservative one. Why?
    Show answer
    Because in the conservative version the sampling fluctuation vanishes entirely. If each level always hands out \(p\) and \(1-p\) as a pair, the final multiset of box values is identical every time (\(C(n,k)\) boxes at \(p^k(1-p)^{n-k}\)).
    ★ Then \(Z(q)=(p^q+(1-p)^q)^n\) holds exactly and \(\tau_{\rm emp}\) always equals the exact answer — indeed the first version did exactly that and showed zero departure.
    ★★ The departure exists only when the multiset of values fluctuates, not merely their arrangement. Hence the switch to drawing each child independently.
    ★ A practical lesson about choosing models — check that the effect you are studying has not been erased by how you built the model.
  2. §03 found \(q^*\) growing only logarithmically. What is the physical reason?
    Show answer
    Because the probability of the box that dominates the \(q\)-th moment falls exponentially. The larger \(q\), the more extreme the dominating box, and its chance of appearing rises in proportion to the number of boxes.
    ★ So reaching "one level deeper into the extremes" requires a fixed multiple more boxes — the logarithm of the box count sets \(q^*\).
    ★★ Indeed \(n\) (\(=\log_2\) of the count) going 10 → 20 takes \(q^*\) from 2 to 4, roughly proportional to \(n\).
    ★ This is exactly Episode 51's "decades" again — both have the logarithm of the available data as their ceiling.
  3. §04 says the interesting ends are unreachable. What does one do in practice?
    Show answer
    Three things.
    ★ ① Fit a model and extrapolate: fit \(\tau(q)\) over the reachable range and get the ends from the model. But that is "assumed," not "measured" — per Episode 43, say which.
    ★ ② Measure something else that is sensitive to the tails: extreme-value statistics (the distribution of maxima), for instance, reaches the tail with fewer samples than moments do.
    ★ ③ Do not claim the ends: stop at "\(\tau(q)\) bends" — as §06 says, that suffices for Episode 42's claim.
    ★★ This series' default is ③, and if ① is used it must be labelled "a model assumption."

Summary — "it is a distribution" is sayable; "which distribution" is not

We asked whether Episode 42's \(f(\alpha)\) curve can actually be extracted from finite data.

Measuring \(\tau(q)\) from a single realization, small \(|q|\) agrees and large \(q\) departs — with an inconsistent sign, because large \(q\) is decided by a single strongest box. Only \(q=0\) is exactly \(-1\) (it just counts boxes).

★★★ The core —
a thousandfold increase in boxes (1,024 → 1M) moves \(q^*\) only from 2 to 4.
The \(\alpha\) width goes 54.5 % → 89.4 % and stops.

★ \(q^*\) grows logarithmically; reaching the tails needs exponentially more data.
★★ The peak of \(f(\alpha)\) (the support dimension) is always measurable; only the tails are not — intermittency is about rare, violent events, and being rare is exactly why they are missing from the sample.

★ One prediction failed: we expected negative \(q\) to be the most noise-fragile, but the positive side moved more (at \(s=1.0\): 6 % at \(q=-2\), 19 % at \(q=6\)) — with log-normal noise the large values scatter more.

★ Conclusion — Episode 42's "the order was a distribution" survives finite data.
But only as far as "it is a distribution."
Which distribution is decided by the size of your sample.

★ As with Episode 51's decades, the ceiling here is set not by the measurement but by the logarithm of the data you have. ★ Honestly: synthetic data only here too; nothing was applied to a real measurement.

This document is Episode 52 of the "Thinking in Waves" series. Computations are in kenshou/multi.js. On the numbers here: the cascade generation, the estimation of \(\tau(q)\) from a single realization, the measurement of \(q^*\) (median over five seeds), the reachable portion of \(f(\alpha)\), and the effect of added noise were all computed here. The multifractal formalism (\(\tau(q)\), the Legendre transform, \(f(\alpha)\)), multiplicative cascades, and the fact that in a finite sample the high-order moments are dominated by extremes so that \(\tau(q)\) linearizes are all well-known results and none are discoveries of this article — what this article did was connect them to this series' handbook (Episodes 49 and 51) and write "the \(q\) range" as an indicator of the same kind as the decade count. As noted in §02, a conservative cascade (handing out \(p\) and \(1-p\) in pairs) produces an identical multiset of box values every time, erasing the sampling fluctuation; this article therefore draws each child's multiplier independently and normalizes. That choice affects the results. The criterion \(|\Delta\tau|\le0.1\) defining \(q^*\) is a convention adopted here, and a different criterion gives different values. The \(q^*\) figures in §03 are medians over five seeds; the scatter from one realization to the next is larger. This article too used only synthetic data and was not applied to any real measurement. — To print, use your browser's Print and "Save as PDF."

Printing / PDF: ⌘+P (Ctrl+P on Windows). "Show answer" opens each solution.