How do antithetic variates reduce Monte Carlo variance?

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AcadiFi · Accepted Answer

Antithetic variates pair each random draw Z with its negative -Z, simulating two paths from each random number stream and averaging them. If the payoff is roughly monotonic in the driver, one path will tend to overshoot the true expectation and the other undershoot, reducing variance in the pair mean. For a European call on Ferndale Biotech spot $118, strike $120, vol 38%, you'd simulate N/2 primary paths with draws Z_i, and N/2 antithetic paths with -Z_i. The pair average Y_i = 0.5*(payoff(Z_i) + payoff(-Z_i)) has variance Var(Y) = 0.5*(Var(X) + Cov(X, X_anti)). If the payoffs are negatively correlated (as they typically are for monotonic payoffs), Cov is negative and variance drops. Typical reduction: 30-60% for vanilla options. Antithetic fails for non-monotonic payoffs like straddles or barrier options near the barrier, where the correlation isn't systematically negative. For those, use control variates or importance sampling. Also note the effective sample size is still N (pairs), not 2N, so don't double-count paths. ```mermaid graph TD A[Draw Z] --> B[Path 1] A --> C[Path 2
uses -Z] B --> D[Average
pair] C --> D ```

How do antithetic variates reduce Monte Carlo variance?

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