Antithetic Sampling for Diffusion Payoffs

Why Pair a Brownian Path with Its Reverse?

Monte Carlo estimates fluctuate because different simulated paths produce different payoffs. Antithetic sampling tries to make two payoffs fluctuate in opposite directions: whenever one path gives a relatively high value, its paired path should tend to give a relatively low one. The construction is simple for Brownian simulation—use the same increments with every sign reversed—but whether it reduces variance depends on the payoff's covariance across the pair.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. For pricing, the same construction is applied under the risk-neutral measure $\mathbb{Q}$, with $B_t$ Brownian under $\mathbb{Q}$; expectations for prices are then $\mathbb{E}^{\mathbb{Q}}$, not $\mathbb{E}$. Fix a time grid $0=t_0<t_1<\cdots<t_N=T$, and write $h_j=t_j-t_{j-1}$. The standardized increments $Z_j=(B_{t_j}-B_{t_{j-1}})/\sqrt{h_j}$ are independent standard normal random variables. Their sign-reversed vector $(-Z_1,\ldots,-Z_N)$ has the same distribution as $(Z_1,\ldots,Z_N)$.

Definition: Let $F:\mathbb{R}^N\to\mathbb{R}$ be the simulated payoff as a function of the standardized Brownian increments, and assume $\mathbb{E}[F(Z_1,\ldots,Z_N)^2]<\infty$. An antithetic pair is $H=F(Z_1,\ldots,Z_N)$ and $H^-=F(-Z_1,\ldots,-Z_N)$. Its paired payoff is $A=(H+H^-)/2$. If discounting is required, multiply the payoff and estimator by the deterministic discount factor.

For an Euler–Maruyama simulation, the first path is built with increments $\sqrt{h_j}Z_j$, and the second with $-\sqrt{h_j}Z_j$, using the same initial value, coefficients, and time grid. For a path-dependent payoff, the payoff is computed from each entire simulated path before averaging the two results. The pair is not two independent simulations: its members are deliberately dependent.

The Variance Formula and Its Consequence

Theorem: Suppose $H$ and $H^-$ are square-integrable and have the same distribution, as they do for a sign-reversed Gaussian increment vector. For $m$ independent antithetic pairs, the estimator $\widehat{V}_m=m^{-1}\sum_{i=1}^m A_i$ is unbiased for $\mathbb{E}[H]$, and $$ \operatorname{Var}(\widehat{V}_m)=\frac{1}{m}\operatorname{Var}(A) =\frac{1}{2m}\left(\operatorname{Var}(H)+\operatorname{Cov}(H,H^-)\right). $$ Against $2m$ independent payoff evaluations, antithetic sampling has lower variance exactly when $\operatorname{Cov}(H,H^-)<0$, equal variance when the covariance is zero, and higher variance when it is positive.

Proof. The equality in distribution gives $\mathbb{E}[H^-]=\mathbb{E}[H]$, so $\mathbb{E}[A]=\mathbb{E}[H]$. Thus $\widehat{V}_m$ is unbiased. The pairs are independent across $i$, and each $A_i$ has a finite second moment because $H$ and $H^-$ do. Therefore the variance of their average is $\operatorname{Var}(A)/m$. Since the two members have equal variance, expansion of the variance of a sum gives $\operatorname{Var}(A)=(\operatorname{Var}(H)+\operatorname{Cov}(H,H^-))/2$, proving the formula. The average of $2m$ independent evaluations has variance $\operatorname{Var}(H)/(2m)$. Subtracting this from the antithetic variance gives $\operatorname{Cov}(H,H^-)/(2m)$; its sign proves all three comparisons. $\square$

The theorem also explains why a pair's average is not automatically better. Antithetic sampling changes the dependence between payoff evaluations, not their marginal distributions. For a decreasing covariance, the estimator improves; for a payoff unchanged by sign reversal, it does not.

Worked Example: Exact Cancellation for a Linear Diffusion Payoff

Let $X=3+2B_1$, and estimate $\mathbb{E}[X]$ from Brownian paths at time $1$. Here $Z=B_1$ is standard normal, and the payoff function is $F(Z)=3+2Z$. Its antithetic value is $F(-Z)=3-2Z$.

The pair average is $\big((3+2Z)+(3-2Z)\big)/2=3$, so its variance is $0$. Individually, $\operatorname{Var}(F(Z))=4\operatorname{Var}(Z)=4$, and the covariance is $\operatorname{Cov}(3+2Z,3-2Z)=-4\operatorname{Var}(Z)=-4$. The theorem gives pair variance $(4-4)/2=0$. By contrast, the average of two independent evaluations has variance $4/2=2$.

Check: the expected payoff is $\mathbb{E}[3+2Z]=3$, which agrees with the constant paired payoff. The covariance calculation also gives $\operatorname{Var}(A)=(4-4)/2=0$, as direct cancellation requires.

Worked Example: A Payoff for Which Pairing Does Not Help

Take $F(Z)=Z^2$, the squared terminal value of a Brownian motion at time $1$. Sign reversal leaves this payoff unchanged: $F(-Z)=Z^2$. Since a standard normal variable has second moment $1$ and fourth moment $3$, $\mathbb{E}[F(Z)]=1$ and $\operatorname{Var}(F(Z))=3-1^2=2$.

The covariance of the two paired payoffs is $\operatorname{Cov}(Z^2,Z^2)=2$, and their average is still $Z^2$, with variance $2$. With two independent evaluations the average would instead have variance $2/2=1$. Thus pairing is worse at equal evaluation cost in this example.

Check: the variance formula gives $(2+2)/2=2$ for one pair. The same result follows because the paired average equals $Z^2$ exactly; no fluctuation has been cancelled.

European Call Payoffs

Consider a stock under the risk-neutral measure $\mathbb{Q}$ satisfying the geometric Brownian motion equation $dS_t=rS_t\,dt+\sigma S_t\,dB_t$, with initial value $S_0>0$, constant interest rate $r$, and constant volatility $\sigma>0$. For maturity $T>0$, its terminal value is $S_T=S_0\exp((r-\sigma^2/2)T+\sigma\sqrt{T}Z)$, where $Z$ is standard normal under $\mathbb{Q}$. A European call with strike $K>0$ has payoff $G=(S_T-K)^+$, where $x^+=\max(x,0)$. Its price is the expectation $\mathbb{E}^{\mathbb{Q}}[e^{-rT}G]$.

Proposition: Under the assumptions $S_0>0$, $K>0$, $\sigma>0$, and $T>0$, the call payoffs on the paths driven by $Z$ and $-Z$ have strictly negative covariance. Hence antithetic pairing strictly reduces the variance of the discounted call-payoff estimator compared with the same number of independent payoff evaluations.

Proof. Set $f(z)=\big(S_0\exp((r-\sigma^2/2)T+\sigma\sqrt{T}z)-K\big)^+$. This function is nondecreasing, and is zero below the finite threshold $z_*=(\log(K/S_0)-(r-\sigma^2/2)T)/(\sigma\sqrt{T})$, then strictly increasing above it. The paired payoffs before discounting are $f(Z)$ and $f(-Z)$; the first is nondecreasing in $Z$, while the second is nonincreasing.

Let $Z'$ be an independent standard normal variable. For square-integrable random variables $U,V$, with independent copies $U',V'$ formed from the same independent copy of their underlying input, the covariance identity is $\operatorname{Cov}(U,V)=\tfrac12\mathbb{E}[(U-U')(V-V')]$. Here this gives $$ \operatorname{Cov}(f(Z),f(-Z)) =\tfrac12\mathbb{E}\left[(f(Z)-f(Z'))(f(-Z)-f(-Z'))\right]. $$ The integrand is nonpositive: if $Z>Z'$, the first difference is nonnegative and the second nonpositive, and the signs reverse together if $Z<Z'$. It is strictly negative with positive probability. Indeed, choose $Z$ in an open interval strictly below both $z_*$ and $-z_*$, and $Z'$ in an open interval strictly above both. Then $f(Z)=0<f(Z')$, while $f(-Z)>0=f(-Z')$. Both intervals have positive normal probability. Consequently the expectation is strictly negative. The discounted covariance is multiplied by $e^{-2rT}>0$, so remains strictly negative. The variance comparison follows from the theorem. $\square$

Worked Example: One Antithetic Pair for a European Call

Take $S_0=100$, $K=100$, $T=1$, $r=\tfrac12$, and $\sigma=1$. The log-drift term is $(r-\sigma^2/2)T=(\tfrac12-\tfrac12)\cdot1=0$. For a simulated standardized increment $Z=1$, the terminal stock value is $100e$ and the call payoff is $100(e-1)$. For the antithetic increment $-1$, the terminal value is $100e^{-1}$, which is less than $100$, so the call payoff is $0$.

The paired payoff before discounting is therefore $\big(100(e-1)+0\big)/2=50(e-1)$. Discounting by $e^{-rT}=e^{-1/2}$ gives the paired price contribution $50e^{-1/2}(e-1)$. Reversing the two increments gives the same pair average.

Check: the two terminal values multiply to $(100e)(100e^{-1})=10000=100^2$, as the opposite log-increments require. The discounted pair contribution is exactly the average of the two discounted payoffs, $ \big(100e^{-1/2}(e-1)+0\big)/2$.

The proposition is strict only under its stated assumptions. In particular, if $S_0=0$, the stock remains zero in the displayed terminal-value formula, and both call payoffs are constantly zero; their covariance is zero, not negative. A zero volatility or zero maturity likewise removes the random variation needed for strict reduction. Such degenerate cases should not be hidden behind the general call argument.

Practical Use and Common Pitfalls

For a simulation with $m$ pairs, generate $m$ independent Gaussian increment vectors. For each vector, run the diffusion scheme once with its increments and once with their negatives, compute the two payoffs, and average within the pair. The $m$ pair averages—not the $2m$ individual payoffs—are the independent observations used to estimate the Monte Carlo standard error. If the pair averages have sample variance $s_A^2$, the estimated standard error of their mean is $s_A/\sqrt{m}$.

For a general diffusion payoff there is no guarantee that the covariance is negative. A payoff may be nonmonotone in the increments, or may respond in the same direction to both a path and its sign reversal. The useful diagnostic is the covariance in the theorem, not the fact that the paths look different. For a scalar payoff $F(Z)$ that is nondecreasing in $Z$, $F(-Z)$ is nonincreasing, so the covariance is nonpositive by the same independent-copy argument used above; strictness additionally requires that the opposing changes occur on a set of positive probability.

A common mistake is to treat the two payoffs in a pair as independent and write $\operatorname{Var}((H+H^-)/2)=\operatorname{Var}(H)/2$. That calculation omits their dependence. The correct expansion is $\operatorname{Var}((H+H^-)/2)=(\operatorname{Var}(H)+\operatorname{Cov}(H,H^-))/2$. In the squared-normal example the omitted covariance is $2$, so the mistaken answer $1$ is wrong; the actual pair variance is $2$.

Finally, antithetic sampling addresses sampling variance, not time-discretization bias. If the diffusion is approximated on a finite grid, both members of each pair may share the same discretization error. Increasing the number of pairs reduces Monte Carlo sampling error but does not by itself remove that bias. The paired construction can be applied to a finer simulation scheme, but the two error sources must still be assessed separately.

Check Your Understanding

Use the variance identity and the assumptions stated above when answering.

For $F(Z)=4-Z$, compute the antithetic pair average and the covariance of the two payoffs.
For $F(Z)=Z^2$, compare the variance of one antithetic pair average with that of two independent evaluations averaged together.
Explain why equal marginal distributions of $H$ and $H^-$ are enough for unbiasedness, but not enough to ensure variance reduction.
In the European call proposition, which assumptions ensure that the threshold $z_*$ is finite and that both sides of it have positive normal probability?
For $m$ independent pairs, state the variance of the antithetic estimator and identify the covariance condition for improvement over $2m$ independent evaluations.

Antithetic Sampling for Diffusion Payoffs

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