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Coupled finite differences have a covariance-controlled variance

Statement

Let Y+ and Y− be square-integrable real simulation outputs with any specified joint law, and let h>0. The central-difference estimator D=(Y+−Y−)/(2h) satisfies Var⁡(D)=Var⁡(Y+)+Var⁡(Y−)−2Cov⁡(Y+,Y−)4h2. The formula applies in particular when the two outputs use common random numbers as in Coupled finite-difference simulation-gradient estimator. Positive covariance reduces this variance relative to independent outputs with the same marginals; coupling alone does not guarantee positive covariance.

Facts & Assumptions

Given: Square-integrable real outputs Y+,Y− on one probability space and a scalar h>0.

[F1]

Moments, variance, and covariance on a probability space: variance and covariance are the expectations of centered squares and products.

[F2]

Variance and covariance identities for random variables: Var⁡(X)=E[X2]−E[X]2 and Cov⁡(X,Y)=E[XY]−E[X]E[Y].

[F3]

Coupled finite-difference simulation-gradient estimator: the central difference is (Y+−Y−)/(2h), with the joint law fixed by the simulation coupling.

[F4]

Expectations factor over finite products of independent random variables: independent integrable factors have product expectation equal to the product of their expectations.

Proof

1.1givenF1F3algebra

By [F3], D=(Y+−Y−)/(2h). Centering gives D−ED=(Y+−EY+)−(Y−−EY−)2h. All terms are square-integrable by the hypotheses.

2.1step 1.1F1F2algebra

Squaring step 1.1, taking expectations, and expanding the cross term by [F1]–[F2] yields Var⁡(D)=Var⁡(Y+)+Var⁡(Y−)−2Cov⁡(Y+,Y−)4h2.

3.1F2F4step 2.1algebra∎

For independent outputs, [F4] gives E[Y+Y−]=E[Y+]E[Y−], so [F2] makes their covariance zero. Thus the right side in step 2.1 is no larger than the independent-output variance exactly when the chosen coupling has nonnegative covariance; no such sign is forced by sharing a random input.

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