Provedproposalenvelopevariational-cost-lawreview-response

Variance–envelope trade-off: minimising Σ_j A_j/r_j subject to r_j ≥ u_j/K̄ is attained by r*_j = max(u_j/K̄, γ√A_j) with γ normalising

For all $\htmlData{mx=1}{J \in \htmlData{mx=0}{\operatorname{Type}}}$ , $\htmlData{mx=3}{A, u, r, \mathit{rs} : \htmlData{mx=0}{J} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathbb{R}}}$ and $\htmlData{mx=1}{\mathit{Kbar}, \gamma \in \htmlData{mx=0}{\mathbb{R}}}$ , if $\htmlData{mx=4}{\forall j,\ 0 \mathrel{\htmlData{mx=3}{\leq}} \htmlData{mx=2}{\htmlData{mx=0}{A}(\htmlData{mx=1}{j})}}$ , $0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{\gamma}$ , $\htmlData{mx=16}{\forall j,\ \htmlData{mx=2}{\htmlData{mx=0}{\mathit{rs}}(\htmlData{mx=1}{j})} \mathrel{\htmlData{mx=15}{=}} \htmlData{mx=14}{\max(\htmlData{mx=7}{\frac{\htmlData{mx=5}{\htmlData{mx=3}{u}(\htmlData{mx=4}{j})}}{\htmlData{mx=6}{\mathit{Kbar}}}}, \htmlData{mx=13}{\htmlData{mx=8}{\gamma} \cdot \htmlData{mx=12}{\sqrt{\htmlData{mx=11}{\htmlData{mx=9}{A}(\htmlData{mx=10}{j})}}}})}}$ , $\htmlData{mx=4}{\forall j,\ 0 \mathrel{\htmlData{mx=3}{<}} \htmlData{mx=2}{\htmlData{mx=0}{\mathit{rs}}(\htmlData{mx=1}{j})}}$ , $\htmlData{mx=3}{\sum_{j} \htmlData{mx=2}{\htmlData{mx=0}{\mathit{rs}}(\htmlData{mx=1}{j})}} \mathrel{\htmlData{mx=4}{=}} 1$ , $\htmlData{mx=4}{\forall j,\ 0 \mathrel{\htmlData{mx=3}{<}} \htmlData{mx=2}{\htmlData{mx=0}{r}(\htmlData{mx=1}{j})}}$ , $\htmlData{mx=3}{\sum_{j} \htmlData{mx=2}{\htmlData{mx=0}{r}(\htmlData{mx=1}{j})}} \mathrel{\htmlData{mx=4}{=}} 1$ and $\htmlData{mx=9}{\forall j,\ \htmlData{mx=4}{\frac{\htmlData{mx=2}{\htmlData{mx=0}{u}(\htmlData{mx=1}{j})}}{\htmlData{mx=3}{\mathit{Kbar}}}} \mathrel{\htmlData{mx=8}{\leq}} \htmlData{mx=7}{\htmlData{mx=5}{r}(\htmlData{mx=6}{j})}}$ (given $\htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fintype}}\,\htmlData{mx=1}{J}}$ ), then

\htmlData{mx=7}{\sum_{j} \htmlData{mx=6}{\frac{\htmlData{mx=2}{\htmlData{mx=0}{A}(\htmlData{mx=1}{j})}}{\htmlData{mx=5}{\htmlData{mx=3}{\mathit{rs}}(\htmlData{mx=4}{j})}}}} \mathrel{\htmlData{mx=16}{\leq}} \htmlData{mx=15}{\sum_{j} \htmlData{mx=14}{\frac{\htmlData{mx=10}{\htmlData{mx=8}{A}(\htmlData{mx=9}{j})}}{\htmlData{mx=13}{\htmlData{mx=11}{r}(\htmlData{mx=12}{j})}}}}

What the notation means · or point at any symbol

Type: Type: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Type)
Fintype: Fintype: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Fintype)
f(x)Function application: Fintype applied to J. Lean writes f x; papers write f(x). (Fintype J)
⟹Implies: If J, then ℝ. (Between types, A → B is the type of functions from A to B.) (J → ℝ)
x : TVariables: A, u, r, rs range over J → ℝ: the statement must hold for every choice. ((A u r rs : J → ℝ))
x ∈ TVariables: Kbar, γ range over real numbers: the statement must hold for every choice. ((Kbar γ : ℝ))
∀For all: The statement after the comma holds for every j. (∀ j, 0 ≤ A j)
<Less than: 0 is strictly smaller than γ. (0 < γ)
∑Summation: Add up rs(j) for every j. (∑ j, rs j)
RReal numbers: The real numbers: every point on the number line. (ℝ)
≤At most: 0 is less than or equal to A(j). (0 ≤ A j)
=Equality: rs(j) and max(u(j)/Kbar, γ · √(A(j))) are the same. (rs j = max (u j / Kbar) (γ * Real.sqrt …)
maxMaximum: The larger of u(j)/Kbar and γ · √(A(j)). (max (u j / Kbar) (γ * Real.sqrt (A j)))
a/bDivision: u(j) divided by Kbar. In Lean, dividing by 0 gives 0. (u j / Kbar)
·Multiplication: γ times √(A(j)). (γ * Real.sqrt (A j))
√Square root: The non-negative number whose square is A(j). (Real.sqrt (A j))

theorem vcl-variance-envelope-constrained-optimum {J : Type} [Fintype J] (A u r rs : J → ℝ) (Kbar γ : ℝ) (hA : ∀ j, 0 ≤ A j) (hγ : 0 < γ) (hrs_def : ∀ j, rs j = max (u j / Kbar) (γ * Real.sqrt (A j))) (hrs0 : ∀ j, 0 < rs j) (hrs_sum : ∑ j, rs j = 1) (hr : ∀ j, 0 < r j) (hr_sum : ∑ j, r j = 1) (hfeas : ∀ j, u j / Kbar ≤ r j) :
    ∑ j, A j / rs j ≤ ∑ j, A j / r j

The constrained convex problem behind a proposal that is both variance-optimal and safe for rejection: allocate first the mass each key needs for the envelope K̄, then distribute the remainder by the square-root rule. Proof: the tangent inequality of x ↦ A/x at r* plus a multiplier argument (free coordinates have A/r² = 1/γ², constrained ones have A/r² ≤ 1/γ² and r ≥ r*). Suggested by an external reviewer (2026-10-05). Feasibility needs Σ_j u_j ≤ K̄, stated through Σ r* = 1.

How it was proved

Proved: 5 October 2026
By: Ajax, with the agent “Baguette math & proofs (Claude Code, 2026-10-03)”
Checked by: Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms: Classical.choice, Quot.sound, propext
Proof: 52 lines of Lean

Variance–envelope trade-off: minimising Σ_j A_j/r_j subject to r_j ≥ u_j/K̄ is attained by r*_j = max(u_j/K̄, γ√A_j) with γ normalising

How it was proved

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