Why Pair Opposite Brownian Endpoints?
A Monte Carlo call-price estimate is an average of discounted payoffs evaluated at simulated stock prices. The simulation error comes from the randomness of those payoffs. Antithetic variates use that randomness in a structured way: for each standard normal draw \(Z\), also evaluate the payoff at \(-Z\). For a call, the payoff rises with the stock price, so the two payoffs tend to move in opposite directions. Averaging them can reduce variance without changing the expected price.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{Q})\) carrying a Brownian motion \(B_t\) under the risk-neutral measure \(\mathbb{Q}\). Consider a non-dividend-paying stock with initial price \(S_0>0\), constant interest rate \(r\), volatility \(\sigma>0\), and maturity \(T>0\). Its risk-neutral model is \(dS_t=rS_t\,dt+\sigma S_t\,dB_t\). The drift coefficient is \(rS_t\), and the diffusion coefficient is \(\sigma S_t\). The Ito integrand \(\sigma S_t\) is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite horizon, since \(\mathbb{E}^{\mathbb{Q}}[S_t^2]=S_0^2e^{(2r+\sigma^2)t}\) and its integral over \([0,T]\) is finite.
The exact terminal stock price is \(S_T=S_0\exp((r-\tfrac12\sigma^2)T+\sigma\sqrt{T}Z)\), where \(Z\) is standard normal under \(\mathbb{Q}\). For completeness, set \(f(t,b)=S_0\exp((r-\tfrac12\sigma^2)t+\sigma b)\). Its derivatives are \(f_t=(r-\tfrac12\sigma^2)f\), \(f_b=\sigma f\), and \(f_{bb}=\sigma^2f\). Ito’s formula gives the drift \((r-\tfrac12\sigma^2)f+\tfrac12\sigma^2f=rf\) and diffusion \(\sigma f\), verifying the stated model. This uses \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\); it is not the ordinary chain rule.
Each pair uses two payoff evaluations, so \(n\) pairs and \(2n\) ordinary draws have the same evaluation cost. The two payoffs within a pair are dependent. The pair averages \(A_1,\ldots,A_n\), however, are independent because they are functions of independent draws \(Z_1,\ldots,Z_n\).
Unbiasedness, Variance, and Standard Error
Proof. A standard normal variable has the same distribution as its negative. Therefore \(\mathbb{E}[g(-Z)]=\mathbb{E}[g(Z)]\), and linearity of expectation gives \(\mathbb{E}[A]=\mathbb{E}[g(Z)]\). Both terms have finite second moment, so their covariance is defined. The variance-of-a-sum identity gives \(\operatorname{Var}(A)=\tfrac14\big(\operatorname{Var}(g(Z))+\operatorname{Var}(g(-Z))+2\operatorname{Cov}(g(Z),g(-Z))\big)\). Since \(g(-Z)\) and \(g(Z)\) have the same distribution, their variances agree, yielding the displayed formula. Independence of the \(A_i\) then gives \(\mathbb{E}[\overline A_n]=\mathbb{E}[A]\) and \(\operatorname{Var}(\overline A_n)=\operatorname{Var}(A)/n\). \(\square\)
For a call payoff, \(g\) is nondecreasing: increasing \(z\) increases the terminal stock price and cannot decrease the call payoff. The opposite input \(-z\) therefore moves in the other direction. This gives a nonpositive covariance, as the next result establishes.
Proof. Let \(Z'\) be an independent copy of \(Z\). For square-integrable random variables \(U,V\), expansion and independence give \(\mathbb{E}[(U-U')(V-V')]=2\operatorname{Cov}(U,V)\), where \((U',V')\) is an independent copy of \((U,V)\). Apply this identity with \(U=g(Z)\), \(V=g(-Z)\), \(U'=g(Z')\), and \(V'=g(-Z')\). If \(Z\ge Z'\), then \(g(Z)-g(Z')\ge0\), while \(-Z\le-Z'\) gives \(g(-Z)-g(-Z')\le0\). If \(Z\le Z'\), the first difference is nonpositive and the second is nonnegative. Thus their product is nonpositive in either case, so its expectation and the covariance are nonpositive. The variance formula in the Theorem now implies \(\operatorname{Var}(A)\le\tfrac12\operatorname{Var}(g(Z))\). With \(2n\) independent ordinary draws, the ordinary estimator has variance \(\operatorname{Var}(g(Z))/(2n)\); the paired estimator has variance \(\operatorname{Var}(A)/n\), which is no larger. \(\square\)
The inequality compares theoretical variances at equal cost; it does not guarantee that every finite antithetic sample has a smaller observed standard error. With \(n\ge2\) independent pair averages, estimate the variance of \(A\) by \(s_A^2=\frac1{n-1}\sum_{i=1}^n(A_i-\overline A_n)^2\). The estimated standard error of the paired price is \(s_A/\sqrt n\). Use the pair averages in this formula, not the \(2n\) individual payoffs. With only one pair, a sample standard error cannot be calculated from the data.
Worked Calculations
Worked Example: One Pair and Its Discounted Payoff
Take \(S_0=K=100\), \(T=1\), \(\sigma=\ln 2\), and \(r=\tfrac12(\ln 2)^2\). For one draw \(Z=1\), evaluate the antithetic pair and its price estimate.
Here \((r-\tfrac12\sigma^2)T=0\), so the terminal prices at \(z=1\) and \(z=-1\) are \(100e^{\ln2}=200\) and \(100e^{-\ln2}=50\). The undiscounted call payoffs are \((200-100)^+=100\) and \((50-100)^+=0\). The discount factor is \(e^{-\frac12(\ln2)^2}\), which is approximately \(0.7864\). Thus the two discounted payoffs are approximately \(78.64\) and \(0\), and the pair average is approximately \(39.32\). The one-pair estimator is \(39.32\).
Check: averaging the two discounted payoffs gives \(\tfrac12(78.64+0)=39.32\). This is an unbiased estimator across repeated draws, not an assertion that one simulated pair equals the true option price.
Worked Example: Paired and Ordinary Standard Errors
Use the same parameters and discount factor, and suppose two independent pair draws are \(Z_1=1\) and \(Z_2=\tfrac12\). Compute both estimators and their sample standard errors using the same four payoff evaluations.
At \(z=\tfrac12\), the terminal stock price is \(100e^{(\ln2)/2}=100\sqrt2\), and at \(z=-\tfrac12\) it is \(100/\sqrt2<100\). Thus the positive-input discounted payoff is \(x=100e^{-r}(\sqrt2-1)\), approximately \(32.58\), and the negative-input payoff is zero. From the first example, the positive-input payoff is \(y=100e^{-r}\), approximately \(78.64\), and its negative-input payoff is zero. The two pair averages are therefore \(y/2\approx39.32\) and \(x/2\approx16.29\). Their mean is \((39.32+16.29)/2\approx27.80\).
For two observations \(a\) and \(b\), the sample variance is \((a-b)^2/2\), so the standard error of their mean is \(|a-b|/2\). Here it is \(|39.32-16.29|/2\approx11.52\). The ordinary outcomes from the same evaluations are \(y,0,x,0\), with mean \((78.64+32.58)/4\approx27.80\). Their sample standard error is \(\sqrt{(3y^2-2xy+3x^2)/48}\), approximately \(18.61\), using \(x=32.58\) and \(y=78.64\) in the displayed expression.
Check: the paired and ordinary sample means agree because they use the same four payoffs. The paired sample standard error is smaller for these particular draws. The proposition compares population variances; this sample result illustrates, but does not guarantee, the finite-sample outcome.
Worked Example: Standard Error from Three Pairs
Let \(S_0=K=1\), \(T=1\), \(\sigma=1\), and \(r=\tfrac12\). Suppose the three independent normal draws are \(1,\tfrac12,0\). Calculate the paired standard error and compare it with the ordinary standard error from the same six payoff evaluations.
The terminal price is \(e^z\), and the discount factor is \(e^{-1/2}\). At \(z=1\), the discounted payoff is \(e^{-1/2}(e-1)\approx1.042\), so the pair average with \(z=-1\), whose call payoff is zero, is \(A_1\approx0.5211\). At \(z=\tfrac12\), the discounted payoff is \(e^{-1/2}(e^{1/2}-1)=1-e^{-1/2}\approx0.3935\); the opposite input has zero payoff, giving \(A_2\approx0.1967\). At \(z=0\), both payoffs are zero, so \(A_3=0\). The paired estimate is \((0.5211+0.1967+0)/3\approx0.2393\).
Using the unrounded expressions \(A_1=\tfrac12e^{-1/2}(e-1)\), \(A_2=\tfrac12(1-e^{-1/2})\), and \(A_3=0\), the sample variance formula gives \(s_A^2=\frac12\sum_{i=1}^3(A_i-\overline A_3)^2\approx0.06924\). The estimated standard error is \(\sqrt{s_A^2/3}\approx0.1519\). For ordinary simulation the six outcomes are approximately \(1.042,0,0.3935,0,0,0\). Their mean is also approximately \(0.2393\). Applying the sample variance formula with denominator \(6-1=5\), then dividing by six for the variance of the mean, gives an ordinary standard error of approximately \(0.1730\).
Check: the paired and ordinary calculations use the same six payoffs and return the same estimate. The standard errors differ because the paired calculation first groups opposite draws and estimates variability across three independent pair averages.
Implementation and Common Pitfalls
Draw \(Z_1,\ldots,Z_n\) independently from the standard normal distribution under the risk-neutral pricing model.
For each \(i\), calculate both \(g(Z_i)\) and \(g(-Z_i)\), then form \(A_i=(g(Z_i)+g(-Z_i))/2\).
Report \(\overline A_n\); when \(n\ge2\), calculate \(s_A/\sqrt n\) from the sample of pair averages.
Do not treat the two outcomes in a pair as independent. They share the same underlying draw, and their dependence is the point of the method. Also distinguish the pricing measure from the physical measure: this estimator targets the risk-neutral expectation \(\mathbb{E}^{\mathbb{Q}}[e^{-rT}(S_T-K)^+]\), not an expectation under \(\mathbb{P}\) unless the two measures happen to agree for the relevant calculation.
Antithetic variates are especially natural when the payoff is monotone in the normal input, as for a vanilla call. The covariance proof supplies the variance guarantee under that condition. It does not establish variance reduction for every payoff: nonmonotone payoffs may not produce a nonpositive covariance, and the result should not be assumed without checking the payoff’s dependence on the simulated input.
Check Your Understanding
Use the definitions and variance comparison above to answer these questions.
- Why is the estimator based on \(n\) pair averages an average of \(n\) independent observations rather than \(2n\)?
- Write the sample standard error formula for \(n\ge2\) antithetic pairs. Which observations enter its sample variance?
- At equal payoff-evaluation cost, compare the variance of \(2n\) ordinary draws with the variance of \(n\) pair averages.
- Which monotonicity property of a call payoff makes the covariance between \(g(Z)\) and \(g(-Z)\) nonpositive?
- Does the variance inequality guarantee a smaller observed standard error for every finite set of simulated draws? Explain.