proofs
Provedattentioninnovationexternal-review-3

Exact deadline completion: variance recurrence solution

For all p,ν∈R\htmlData{mx=1}{p, \nu \in \htmlData{mx=0}{\mathbb{R}}} and V:N→R\htmlData{mx=3}{V : \htmlData{mx=0}{\mathbb{N}} \mathrel{\htmlData{mx=2}{\to}} \htmlData{mx=1}{\mathbb{R}}}, if p≠0\htmlData{mx=0}{p} \mathrel{\htmlData{mx=1}{\neq}} 0, V(0)=0\htmlData{mx=1}{\htmlData{mx=0}{V}(0)} \mathrel{\htmlData{mx=2}{=}} 0 and ∀L, V(L+1)=ν+(1−p)⋅V(L)\htmlData{mx=13}{\forall L,\ \htmlData{mx=3}{\htmlData{mx=0}{V}(\htmlData{mx=2}{\htmlData{mx=1}{L} + 1})} \mathrel{\htmlData{mx=12}{=}} \htmlData{mx=11}{\htmlData{mx=4}{\nu} + \htmlData{mx=10}{(\htmlData{mx=6}{1 - \htmlData{mx=5}{p}}) \cdot \htmlData{mx=9}{\htmlData{mx=7}{V}(\htmlData{mx=8}{L})}}}}, then

∀L, V(L)=(1−(1−p)L)⋅(νp)\htmlData{mx=13}{\forall L,\ \htmlData{mx=2}{\htmlData{mx=0}{V}(\htmlData{mx=1}{L})} \mathrel{\htmlData{mx=12}{=}} \htmlData{mx=11}{(\htmlData{mx=7}{1 - \htmlData{mx=6}{(\htmlData{mx=4}{1 - \htmlData{mx=3}{p}})^{\htmlData{mx=5}{L}}}}) \cdot \left(\htmlData{mx=10}{\frac{\htmlData{mx=8}{\nu}}{\htmlData{mx=9}{p}}}\right)}}
What the notation means · or point at any symbol
RReal numbers
The real numbers: every point on the number line. (ℝ)
x ∈ TVariables
p, ν range over real numbers: the statement must hold for every choice. ((p ν : ℝ))
NNatural numbers
The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
⟹Implies
If ℕ, then ℝ. (Between types, A → B is the type of functions from A to B.) (ℕ → ℝ)
≠Not equal
p and 0 are different. (p ≠ 0)
f(x)Function application
V applied to 0. Lean writes f x; papers write f(x). (V 0)
=Equality
V(0) and 0 are the same. (V 0 = 0)
∀For all
The statement after the comma holds for every L. (∀ L, V (L + 1) = ν + (1 - p) * V L)
+Addition
L plus 1. (L + 1)
aᵇPower
1 − p raised to the power L. ((1 - p) ^ L)
·Multiplication
1 − p times V(L). ((1 - p) * V L)
-Subtraction
1 minus p. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (1 - p)
a/bDivision
ν divided by p. In Lean, dividing by 0 gives 0. (ν / p)

Solves the finite-deadline variance recurrence when its base and step are assumed; the separate finite-history theorem derives those moments from actual probability weights.

How it was proved

Proved
6 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
15 lines of Lean
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