Antithetic Variates

Motivation: why this matters in quant finance

Plain Monte Carlo halves its standard error only by quadrupling the sample size. Antithetic variates is the cheapest of the variance-reduction techniques: a one-line change to the simulation code that, for monotone payoffs, can cut variance by a factor of 2 or more for free.

The idea: each draw $Z$ has an "antithetic twin" $-Z$ . If the payoff is a monotone function of $Z$ , then the values at $Z$ and $-Z$ are negatively correlated, so their average has lower variance than two independent samples.

The informal idea

Standard MC: take $N$ i.i.d. draws $Z_1, \dots, Z_N \sim N(0,1)$ , compute $f(Z_i)$ , average.

Antithetic MC: take $N/2$ draws, pair each $Z_i$ with $-Z_i$ , average over $N$ pairs:

\hat\theta_{AV} = \frac{1}{N}\sum_{i=1}^{N/2}\left(f(Z_i) + f(-Z_i)\right) = \frac{2}{N}\sum_{i=1}^{N/2}\frac{f(Z_i) + f(-Z_i)}{2}.

The key quantity is

\text{Var}\!\left(\frac{f(Z) + f(-Z)}{2}\right) = \frac{1}{4}\big(2\text{Var}(f(Z)) + 2\text{Cov}(f(Z), f(-Z))\big) = \frac{1}{2}\text{Var}(f(Z))(1 + \rho),

where

\rho = \text{Corr}(f(Z), f(-Z))

. If

\rho = 0

(independent), antithetic gives the same variance per pair as two independent samples — no benefit. If

\rho < 0

(negative correlation), antithetic reduces variance.

Formal statement

Let $Z \sim N(0, I_d)$ (or any distribution symmetric about 0), and consider the antithetic estimator

\hat\theta_{AV} = \frac{1}{N/2}\sum_{i=1}^{N/2}\frac{f(Z_i) + f(-Z_i)}{2}.

Variance.

\text{Var}(\hat\theta_{AV}) = \frac{2}{N}\cdot\frac{\text{Var}(f(Z))}{2}(1 + \rho) = \frac{\text{Var}(f(Z))}{N}(1 + \rho).

Compared to plain MC variance $\text{Var}(f(Z))/N$ , antithetic reduces variance by factor $(1 + \rho)$ . For maximum benefit, $\rho \to -1$ , giving 100% variance reduction (antithetic samples cancel exactly).

Cost.

N

payoff evaluations either way. Antithetic does

N/2

random draws and

N

payoff evaluations; plain MC does

N

random draws and

N

payoff evaluations. So antithetic is slightly cheaper in random-number generation but otherwise the same compute.

When $\rho < 0$ . A sufficient condition is

f

monotone in each component of

Z

Algorithm and example

import numpy as np

def mc_call_antithetic(S0, K, T, r, sigma, N, seed=42):
    rng = np.random.default_rng(seed)
    Z = rng.standard_normal(N // 2)
    drift = (r - 0.5*sigma**2)*T
    diff = sigma*np.sqrt(T)
    ST_plus = S0 * np.exp(drift + diff*Z)
    ST_minus = S0 * np.exp(drift - diff*Z)
    payoff = 0.5 * (np.maximum(ST_plus - K, 0) + np.maximum(ST_minus - K, 0))
    discounted = np.exp(-r*T) * payoff
    return discounted.mean(), discounted.std(ddof=1) / np.sqrt(N // 2)

S0, K, T, r, sigma = 100, 100, 1, 0.05, 0.2

p, se = mc_call_antithetic(S0, K, T, r, sigma, N=100_000)
print(f"Antithetic: {p:.4f} ± {1.96*se:.4f}")

# Compare with plain MC at the same total samples N=100_000:
def mc_call_plain(S0, K, T, r, sigma, N, seed=42):
    rng = np.random.default_rng(seed)
    Z = rng.standard_normal(N)
    ST = S0 * np.exp((r - 0.5*sigma**2)*T + sigma*np.sqrt(T)*Z)
    discounted = np.exp(-r*T) * np.maximum(ST - K, 0)
    return discounted.mean(), discounted.std(ddof=1) / np.sqrt(N)

p_plain, se_plain = mc_call_plain(S0, K, T, r, sigma, N=100_000)
print(f"Plain:      {p_plain:.4f} ± {1.96*se_plain:.4f}")
print(f"Variance ratio (plain/AV): {(se_plain*np.sqrt(100_000))**2 / (se*np.sqrt(50_000))**2:.3f}")
# Antithetic: 10.4521 ± 0.0529
# Plain:      10.4287 ± 0.0901
# Variance ratio (plain/AV): ~3.0 (varies)

For an ATM call: variance reduction by factor $\sim 3$ . Equivalent to running plain MC with 3x the sample size — a significant win for one extra line of code.

Key properties

Free for symmetric payoffs. If $f(Z) = f(-Z)$ (e.g., a straddle), antithetic gives zero variance: every pair averages to the same value. Of course in this case the answer is also trivially the constant $f(Z)$ , so we'd notice. More usefully: for a symmetric payoff like a butterfly, antithetic eliminates almost all the variance.
Effective for monotone payoffs. Calls (monotonically increasing in $Z$ ) and puts (monotonically decreasing in $Z$ ) get $\rho \approx -0.5$ to $-0.9$ depending on moneyness, giving 50-90% variance reduction.
Useless for non-monotone payoffs. Asian options on log-stock paths are not directly monotone in any single $Z_k$ — antithetic gives small benefit. Barrier options can have positive correlation between $f(Z)$ and $f(-Z)$ (both knock out for extreme paths), which actually increases variance.
Pairs only. Antithetic naturally generates pairs. Higher-order generalisations (using rotations of $Z$ in higher dimensions) exist but rarely outperform simpler control variates.
Compatible with other techniques. Antithetic stacks with control variates, importance sampling, and stratified sampling. Often used together for compound variance reduction.

Worked example: deep ITM vs OTM

ATM call $S_0 = K = 100$ : $\rho \approx -0.85$ , $\sim 6\times$ variance reduction.

Deep ITM $S_0 = 100, K = 50$ : payoff is essentially $S_T - K$ (linear), $\rho \approx -1$ , near-perfect cancellation. Variance reduction $> 100\times$ .

Deep OTM

S_0 = 100, K = 200

: most paths yield 0 from both

Z

and

-Z

— high correlation

\rho \approx +0.7

(both members of a pair are usually

0

). Antithetic increases variance. Use importance sampling instead.

Asian call: payoff depends on the average path. Antithetic helps but with $\rho \approx -0.4$ to $-0.5$ (less negative). $\sim 2\times$ variance reduction.

Barrier knock-out: complicated. Sometimes helps, sometimes doesn't. Empirical check is required before deploying.

Common confusions and pitfalls

Variance estimator must use pairs. When estimating SE from antithetic samples, treat the pair average $(f(Z_i) + f(-Z_i))/2$ as a single sample. Don't compute SE over $N$ individual values; use $N/2$ pair averages.
$Z$ must come from a symmetric distribution. The trick uses $-Z \overset{d}{=} Z$ for $Z \sim N(0, I)$ . For non-symmetric base distributions (e.g., simulating an exponential), the equivalent is $1 - U$ for uniforms, then transform.
Doesn't always help. Profile your specific payoff. If $\rho > 0$ , antithetic is worse than plain MC. The simplest check: compute sample correlation between $f(Z)$ and $f(-Z)$ on a small pilot run.
Not a substitute for high-quality RNG. Antithetic gives variance reduction; it does not fix bias, model error, or bad RNGs. Get the basics right first.
Multi-dimensional caution. For vector $Z \in \mathbb{R}^d$ , antithetic flips all components: $-Z = (-Z_1, \dots, -Z_d)$ . For some payoffs, partial flips can be more effective (rotated antithetic), though this is more intricate.

Where this goes next

Control variates — a more flexible variance-reduction tool.
Importance sampling — best for OTM options where antithetic fails.
Quasi-Monte Carlo — replaces randomness entirely.