Source
Granero-Belinchón, Roux and Garnier (2016), EPL 115, 58003, LIT-1243, read in full in NOTE-970 from arXiv:1607.05511v2. The conditioning procedure and its symmetry argument are the paper’s (“Conditioned entropy rate”, Fig. 3). The invariance statement and the counterexample are the record’s.
What was actually shown
Setting. x(t) is a stationary signal with a joint density for every finite block. The paper’s entropy rate is h_τ^(m) = H(X^(m+1,τ)) − H(X^(m,τ)), where X^(n,τ) = (x(t), x(t − τ), …, x(t − (n − 1)τ)).
Step 1. Invariance of the entropy rate. Differential entropy is unchanged by any map of unit absolute Jacobian determinant.
- Reflection. x → −x maps each block by −I, with |det| = 1. So every block entropy, and so every h_τ^(m), is the same for x and −x.
- Time reversal. The reversed signal y(t) = x(−t) has the block (y(t), y(t − τ), …, y(t − mτ)) = (x(−t), x(−t + τ), …, x(−t + mτ)). That is the x-block ending at s = −t + mτ with its coordinates in reverse order. By stationarity it has the law of a permuted x-block, and a permutation has |det| = 1. So y and x have the same block entropies and the same entropy rates, for every m and τ.
The longitudinal increment skewness, and the third-order structure function S₃ that the 4/5 law equates to −(4/5)εr, change sign under both maps. Under Taylor’s hypothesis, temporal reversal of the probe signal is spatial reflection of the frozen field. So no entropy rate, at any embedding, can distinguish a signal from the one with the opposite sign of S₃. The paper’s “cannot probe the energy cascade because it is not sensitive enough to the asymmetry” understates this. The blindness is exact, and it comes from symmetry, not from sensitivity.
Step 2. What the conditioned rate tests. The paper splits the points x(t) by the sign of δ_τ(t) = x(t) − x(t − τ) and estimates h^(1) on each subset, giving h⁺ and h⁻. For m = 1 it says the test probes “the asymmetry of the joint PDF p(x(t), x(t − τ))”. Write a = x(t), b = x(t − τ). The subsets are the half-planes a > b and a < b, and the central reflection (a, b) → (−a, −b) maps one onto the other with |det| = 1. So if p(a, b) = p(−a, −b), then h⁺ = h⁻, which is the paper’s statement. The converse does not follow, and the paper says so: if the increments are skewed, the joint PDF is not centrally symmetric, but “the reciprocal may not hold”. So h⁺ ≠ h⁻ establishes that the joint law lacks central symmetry. That is necessary for skewed increments, and it is not sufficient.
Step 3. Counterexample. Let a and b be independent with the same skew-normal law, skewness 0.175. That is the one-point skewness the paper reports for its Modane signal, and it is what the joint law of any stationary signal tends to at lags beyond the correlation time. Since (a, b) is exchangeable, a − b has the law of b − a, so the increments are exactly symmetric and S₃ = 0. But p(a)p(b) ≠ p(−a)p(−b), and
- by quadrature (grids of 1401² and 2801² on [−8, 8]²), h⁺ − h⁻ = H(a | b, a > b) − H(a | b, a < b) = 0.0657 nats;
- by the paper’s estimator family (Kozachenko–Leonenko, k = 5, maximum norm) at its sample size N = 2¹⁷, on standardized samples, ten replicates: 0.0645 ± 0.0066 nats, with a measured increment skewness of 0.002.
With a Gaussian copula of correlation c and the same marginal, the difference falls with c: 0.046 nats at c = 0.5, 0.021 at c = 0.9 and 0.0065 at c = 0.99. A centrally symmetric (Gaussian) marginal gives 0 in every case, as Step 2 requires.
Bearing on the paper’s result. The paper reports h⁺ − h⁻ “almost constant around 0.1 ± 0.01” across all scales, with a standard deviation of h^(1)(X^±) near 0.03. Increment skewness in turbulence grows toward small scales and vanishes beyond the integral scale, where samples decorrelate. A difference that stays roughly constant into the integral range is what one-point skewness predicts, by Step 3. Nothing in the paper separates the two sources. Its two robustness checks, equalizing the subset sizes and comparing with a random half, address estimator bias from sample size. They do not address this. Fig. 3’s lower axis is labelled with a normalization, (h⁺ − h⁻)/σ_h in extraction, so the stated 0.1 may not be in nats. The argument does not depend on its exact value.
What this does not say
- That the Modane signal has no skewed increments. It surely has them; the 4/5 law is well tested. The claim is that this test does not show it, because a signal with symmetric increments and the same one-point skewness would also pass.
- That no conditioned information measure can detect S₃. Conditioning on the sign of the increment while removing one-point asymmetry (for example by testing on a rank-transformed or symmetrized marginal), or measuring an asymmetry functional of the increments directly, could isolate it. The paper does neither.
- Anything about information transfer across scales. The paper defines no information flux, no transfer entropy and no mutual information between scale bands. Its only link from an information quantity to the energy balance is the chain h⁺ ≠ h⁻ → skewed increments → 4/5 law, and the first link fails as shown. It computes no ε, no S₃, and no sign.
Connections
- QUESTION-219. It sharpens A3’s caution. Energy transfer has a direction across scales because S₃ is odd. The entropy rates are even under the maps that flip that sign, so they assign no direction at all. The one asymmetric quantity the paper builds detects a different asymmetry.
- THEORY-750. The companion result: the unconditioned rate in these data is S₂ in nats.
- THEORY-754. The energy flux there is a signed, Galilean-invariant triple product with a budget. No information quantity in this paper has either property.