CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Finite correction variance cannot replace bias control

Statement refuted

Finite correction variances and per-sample work bounds $O (h_{ℓ}^{- 1})$ alone guarantee that increasing the number of levels and samples can make the MLMC MSE arbitrarily small for any integrable target.

Facts & Assumptions

Given: The finite MLMC hierarchy, and the universal guarantee stated above, which is to be refuted.

[F1]

Geometric MLMC hierarchy and coupled corrections: the estimator is the sum of level sample means, with $Δ_{0} = P_{0}$ and coupled fine-minus-coarse corrections at later levels.

Counterexample

Take $T = 1$ , $M = 2$ , target $P = 0$ , and $P_{ℓ} = 1$ deterministically for every $ℓ \geq 0$ . Use the deterministic coupling $(P_{ℓ}^{f}, P_{ℓ - 1}^{c}) = (1, 1)$ at every positive level. Charge one level- $ℓ$ sample at cost $h_{ℓ}^{- 1}$ .

Given: Any finite $L \geq 0$ and any positive integer sample counts $N_{0}, \dots, N_{L}$ .

1.1F1givenalgebra

By [F1], $Δ_{0} = 1$ and $Δ_{ℓ} = 0$ for every $ℓ \geq 1$ . All correction variances are zero and all sample costs satisfy the stated work bound.

2.1F1givenstep 1.1algebra∎

By [F1], every sample mean equals its constant correction, so ${\hat{μ}}_{L, N} = 1$ for every $L$ and every sample allocation. The target mean is zero, hence the MSE is exactly $1$ and cannot be made arbitrarily small. The missing hypothesis is convergence of $E P_{L}$ to $E P$ ; here the bias remains $1$ .

Depends on

Geometric MLMC hierarchy and coupled corrections

Used by

Nothing in the library uses this result yet.

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Sources

Michael B. Giles, Multilevel Monte Carlo Path Simulation, §3 (standard reference, not scraped)