LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Expectation, variance, and MSE of a finite MLMC estimator

Statement

Use the finite hierarchy and estimator of Geometric MLMC hierarchy and coupled corrections. Put $V_{ℓ} = Var (Δ_{ℓ})$ . Then $E {\hat{μ}}_{L, N} = E P_{L}, Var ({\hat{μ}}_{L, N}) = \sum_{ℓ = 0}^{L} \frac{V_{ℓ}}{N_{ℓ}},$ and the mean square error for the target $E P$ is exactly $E [({\hat{μ}}_{L, N} - E P)^{2}] = \sum_{ℓ = 0}^{L} \frac{V_{ℓ}}{N_{ℓ}} + (E P_{L} - E P)^{2} .$

Facts & Assumptions

Given: A finite MLMC hierarchy with square-integrable level corrections, independent identically distributed samples within each level, and independent sample blocks at distinct levels.

[F1]

Geometric MLMC hierarchy and coupled corrections: each level uses independent identically distributed copies of its correction, and complete sample blocks from distinct levels are independent.

[F2]

Geometric MLMC hierarchy and coupled corrections: every correction has finite second moment because its two level approximations do.

[F3]

Unbiasedness and exact variance of the IID Monte Carlo mean: for an IID real sample with finite second moment, its mean is unbiased and has variance equal to the single-draw variance divided by the sample size.

[F4]

Linearity, monotonicity, and the modulus bound for expectation: expectation is linear for finite sums of integrable random variables.

[F5]

MSE of a sum of error components includes covariance terms: for square-integrable errors, the MSE of their sum is the squared total bias plus the sum of variances and pairwise covariance terms; pairwise-independent centered errors have zero covariance correction.

Proof

Given: The hypotheses above.

1.1F1F2F3

For each level $ℓ$ , the sample $Δ_{ℓ, 1}, \dots, Δ_{ℓ, N_{ℓ}}$ is IID by [F1] and square-integrable by [F2]. Applying [F3] to its sample mean gives $E {\overset{ˉ}{Δ}}_{ℓ} = E Δ_{ℓ}$ and $Var ({\overset{ˉ}{Δ}}_{ℓ}) = V_{ℓ} / N_{ℓ}$ .

2.1F1F4step 1.1algebra

The level-zero mean is $E Δ_{0} = E P_{0}$ . For $ℓ \geq 1$ , the coupling marginals in [F1] give $E Δ_{ℓ} = E P_{ℓ} - E P_{ℓ - 1}$ . Finite linearity [F4] and telescoping therefore yield $E {\hat{μ}}_{L, N} = \sum_{ℓ = 0}^{L} E {\overset{ˉ}{Δ}}_{ℓ} = E P_{L}$ .

2.2F1F5step 1.1

Set $ξ_{ℓ} = {\overset{ˉ}{Δ}}_{ℓ} - E {\overset{ˉ}{Δ}}_{ℓ}$ . Each $ξ_{ℓ}$ is square-integrable, centered, and measurable with respect to its own level sample block. The blocks are independent by [F1], so the $ξ_{ℓ}$ are pairwise independent. Applying [F5] to their finite sum gives $Var ({\hat{μ}}_{L, N}) = \sum_{ℓ = 0}^{L} Var ({\overset{ˉ}{Δ}}_{ℓ}) = \sum_{ℓ = 0}^{L} V_{ℓ} / N_{ℓ}$ , using step 1.1.

3.1F5step 2.1step 2.2∎

The single error $E = {\hat{μ}}_{L, N} - E P$ is square-integrable. By step 2.1 its mean is $E P_{L} - E P$ , and by step 2.2 its variance is $\sum_{ℓ = 0}^{L} V_{ℓ} / N_{ℓ}$ . The one-variable case of [F5] gives $E [E^{2}] = Var (E) + (E E)^{2}$ , which is the asserted MSE identity.

Depends on

Used by

MLMC work bound for bounded finite-state Poisson tau-leaping Corollary
Nondecaying correction variance can force cubic cost Counterexample
A three-level telescoping estimator and its remaining bias Example
MLMC cost complexity under bias, variance, and work rates Theorem

Dependency tree · two levels

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Sources

Michael B. Giles, Multilevel Monte Carlo Path Simulation, §2 (standard reference, not scraped)
Christoph Reisinger, Stochastic Simulation: Lecture 9a, Multilevel Monte Carlo Theory, slides 4–9 (standard reference, not scraped)