proofs
Provedmlmcsnisreview-response

Antithetic SNIS identity: Y_{2m} − (Y₁+Y₂)/2 = (B₁−B₂)(Y₁−Y₂)/(2(B₁+B₂)), where B_h are the half-sample mean weights and Y_h the half-sample SNIS outputs

For all B1,B2,Y1,Y2∈R\htmlData{mx=1}{B_{1}, B_{2}, Y_{1}, Y_{2} \in \htmlData{mx=0}{\mathbb{R}}}, if 0<B10 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{B_{1}} and 0<B20 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{B_{2}}, then

B1⋅Y1+B2⋅Y2B1+B2−Y1+Y22=(B1−B2)⋅(Y1−Y2)2(B1+B2)\begin{aligned} & \htmlData{mx=16}{\htmlData{mx=11}{\frac{\htmlData{mx=7}{\htmlData{mx=3}{\htmlData{mx=1}{B_{1}} \cdot \htmlData{mx=2}{Y_{1}}} + \htmlData{mx=6}{\htmlData{mx=4}{B_{2}} \cdot \htmlData{mx=5}{Y_{2}}}}}{\htmlData{mx=10}{\htmlData{mx=8}{B_{1}} + \htmlData{mx=9}{B_{2}}}}} - \htmlData{mx=15}{\frac{\htmlData{mx=14}{\htmlData{mx=12}{Y_{1}} + \htmlData{mx=13}{Y_{2}}}}{2}}} \\ & \quad \htmlData{mx=0}{=} \htmlData{mx=28}{\frac{\htmlData{mx=23}{(\htmlData{mx=19}{\htmlData{mx=17}{B_{1}} - \htmlData{mx=18}{B_{2}}}) \cdot (\htmlData{mx=22}{\htmlData{mx=20}{Y_{1}} - \htmlData{mx=21}{Y_{2}}})}}{\htmlData{mx=27}{2(\htmlData{mx=26}{\htmlData{mx=24}{B_{1}} + \htmlData{mx=25}{B_{2}}})}}} \end{aligned}
What the notation means · or point at any symbol
RReal numbers
The real numbers: every point on the number line. (ℝ)
x ∈ TVariables
B1, B2, Y1, Y2 range over real numbers: the statement must hold for every choice. ((B1 B2 Y1 Y2 : ℝ))
<Less than
0 is strictly smaller than B1. (0 < B1)
a/bDivision
B1 · Y1 + B2 · Y2 divided by B1 + B2. In Lean, dividing by 0 gives 0. ((B1 * Y1 + B2 * Y2) / (B1 + B2))
-Subtraction
(B1 · Y1 + B2 · Y2)/(B1 + B2) minus (Y1 + Y2)/2. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. ((B1 * Y1 + B2 * Y2) / (B1 + B2) - (Y1 +…)
=Equality
(B1 · Y1 + B2 · Y2)/(B1 + B2) − (Y1 + Y2)/2 and (B1 − B2) · (Y1 − Y2)/(2 · (B1 + B2)) are the same. ((B1 * Y1 + B2 * Y2) / (B1 + B2) - (Y1 +…)
·Multiplication
B1 times Y1. (B1 * Y1)
+Addition
B1 · Y1 plus B2 · Y2. (B1 * Y1 + B2 * Y2)

The exact identity behind the external reviewer's antithetic multilevel route (2026-10-05): the level difference is a product of two first-order fluctuations, so under a bounded weight envelope its second moment is O(m⁻²) and randomised dyadic levels with probabilities ∝ 2^{−γℓ}, 1 < γ < 2, give exactly unbiased SNIS without a dense terminal level at O_{K,F}(ε⁻²) expected draws. The identity is proved here; the moment bounds and the level randomisation are the reviewer's paper argument, with constants depending on the envelope K rather than on the cost factor alone.

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
13 lines of Lean
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