Cut Monte Carlo variance with antithetic pairs

Question

You are estimating \int0^1 e^x\,dx = e - 1 by averaging e^{U} over uniform draws. A colleague suggests that for each draw U you also use its mirror 1 - U and average the pair. Write the antithetic estimator and quantify how much variance it saves versus plain Monte Carlo, for the same number of e^x evaluations.

Accepted Answer

For each draw form Y = 	frac{1}{2}\left(e^{U} + e^{1-U}ight). Because U and 1-U move oppositely, e^U and e^{1-U} are negatively correlated, and averaging cancels much of the noise while keeping the estimator unbiased (E[Y] = e - 1). The variance gain. With \operatorname{Var}(e^U) \approx 0.242 and \operatorname{Cov}(e^U, e^{1-U}) = E[e^{U}e^{1-U}] - (e-1)^2 = e - (e-1)^2 \approx -0.234, \operatorname{Var}(Y) = 	frac14\big(0.242 + 0.242 + 2(-0.234)\big) \approx 0.00391. Each pair spends 2 evaluations of e^x. For a budget of n evaluations you get n/2 pairs, so the estimator variance is 0.00…

Cut Monte Carlo variance with antithetic pairs

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