Dual certificate for the recursive maximum-entropy penalty
Provedmme_recursive_thin_split_entropy_penalty_dual_boundLet be the set of admissible left-half splits of a recursive parent of total grade : each is a triple of coordinates in with and . Let be a probability distribution on . The recursive maximum-entropy penalty of the More Asymmetry analysis is
where denotes Shannon entropy in bits. For every choice of dual potentials , , write . Then
The right-hand side equals for the Gibbs distribution , so the inequality is the Lagrange-dual form of the maximum-entropy problem and is tight when has the marginals of . It turns each recursive penalty term of the regional entropy rate into an explicit inequality certified by one dual vector : to evaluate regionalRate in a finite regional certificate, no optimisation over same-marginal distributions has to be formalised, only the arithmetic of a single exponential sum at a candidate .
Formalization Note entropyPenalty is the platform supremum of mme_modern_entropyBits over SameMarginalDistributions alpha from Def_mme_recursive_thin_split_data. The hypotheses only say that is a probability vector, which makes the supremum range over a nonempty set; the marginals are the platform's mme_modern_marginal along each coordinate .
import Definitions.Def_mme_recursive_thin_split_data open BigOperators MME.RecursiveThinSplit open scoped Classical set_option autoImplicit false
theorem mme_recursive_thin_split_entropy_penalty_dual_bound {half : ℕ} {parent : Fin 3 → ℕ}
(alpha : Split half parent → ℝ) (hpos : ∀ c, 0 ≤ alpha c) (hmass : ∑ c, alpha c = 1)
(lam : Fin 3 → Fin (half + 1) → ℝ) :
entropyPenalty alpha ≤
(Real.log (∑ a : Split half parent, Real.exp (∑ i, lam i (a.val i))) -
∑ a, alpha a * ∑ i, lam i (a.val i)) / Real.log 2 -
mme_modern_entropyBits alpha := by sorry