Why Pair Simulations?
Monte Carlo estimates can be noisy even when the simulation model is correct. Antithetic variates are a way to reduce that noise by deliberately pairing inputs that tend to produce outputs on opposite sides of the target. The method does not change the quantity being estimated: it changes how the simulation draws are organized and averaged.
The key is covariance. If the two outputs in each pair have the same variance as an ordinary simulated output but negative covariance with each other, their average has lower variance than an average of two independent outputs. Negative covariance is not guaranteed, however. It must follow from the output function or be checked from the paired results.
The pair members are generally dependent: once \(U\) is known, \(1-U\) is determined. To use the usual independent-replication variance formula across pairs, generate \(U_1,\ldots,U_m\) independently and form \((U_j,1-U_j)\) for each \(j\). The \(m\) pair averages are then independent and identically distributed, even though the two outputs within a pair are not independent.
Variance of the Antithetic Estimator
Write \(\theta=E[Y]=E[Y']\), \(\sigma^2=\operatorname{Var}(Y)=\operatorname{Var}(Y')\), and \(C=\operatorname{Cov}(Y,Y')\). Since one pair average is \(Z=(Y+Y')/2\), its variance is
The estimator averages \(m\) independent pair averages. Its expectation and variance are therefore
This derivation shows why covariance controls precision. With correlation \(\rho=C/\sigma^2\), the variance is \(\sigma^2(1+\rho)/(2m)\). At the same simulation cost, \(2m\) independent outputs have estimator variance \(\sigma^2/(2m)\). Thus the ratio of antithetic variance to independent variance is \(1+\rho\). A negative \(\rho\) gives a reduction; a positive \(\rho\) makes the antithetic estimator less precise. Since \(-1\leq\rho\leq1\), this equal-cost ratio must lie between 0 and 2.
When Does Pairing Tend to Give Negative Covariance?
A useful sufficient condition is that \(g\) is monotone on \((0,1)\). If \(g\) is nondecreasing, a larger \(U\) tends to give a larger \(g(U)\), while the corresponding \(1-U\) is smaller and tends to give a smaller \(g(1-U)\). If \(g\) is nonincreasing, the directions reverse for both outputs, again tending to make them move oppositely.
For a direct justification, let \(U\) and \(V\) be independent uniform variables. The two pair members have equal marginal means, so
For monotone \(g\), the two differences in the product have opposite signs or one is zero: if \(U>V\), then \(1-U<1-V\), and monotonicity makes the corresponding changes in the two function values oppose one another. The product is therefore nonpositive for every \(U,V\), so its expectation and the covariance are nonpositive. The covariance can be zero for some functions, and a nonmonotone output can have positive covariance. Pairing is not a guarantee of variance reduction.
Worked Example: Antithetic Inputs for a Squared Uniform Output
A hypothetical simulation uses \(U\sim\operatorname{Uniform}(0,1)\) to estimate \(\theta=E[U^2]\). Compare 40 antithetic pairs with 80 independent simulations, so each design uses 80 output evaluations.
Method. Use \(g(u)=u^2\), pair \(U\) with \(1-U\), and calculate the variance of each design’s estimator. The integral \(E[U^2]=\int_0^1u^2\,du=1/3\), so both estimators target \(1/3\).
For one antithetic pair, the pair average is
Let \(X=U-1/2\), which is uniform on \((-1/2,1/2)\). Then \(E[X^2]=1/12\) and \(E[X^4]=1/80\). Consequently, \(E[Z]=1/12+1/4=1/3\), and
The average of 40 independent pair averages has variance \((1/180)/40=1/7200\), and standard error \(\sqrt{1/7200}=0.0118\), rounded to four decimals.
For the independent design, \(\operatorname{Var}(U^2)=E[U^4]-(E[U^2])^2=1/5-1/9=4/45\). The variance of the mean of 80 independent outputs is \((4/45)/80=1/900\), with standard error \(1/30=0.0333\), rounded to four decimals. The ratio of estimator variances is \((1/7200)/(1/900)=1/8\). Thus, in this example and at equal output-evaluation cost, the antithetic estimator has one-eighth the variance. The standard error is reduced by a factor of \(\sqrt{8}\), not by a factor of 8.
Check. Each estimator has expectation \(1/3\), and the antithetic variance is smaller than the independent variance. The comparison uses 80 output evaluations for each method. The common mistake this avoids is comparing the variance of one pair average with the variance of a mean of many independent outputs.
Worked Example: Antithetic Pairing for a Simulated Event Probability
A hypothetical model estimates \(P(U<0.20)\), where \(U\sim\operatorname{Uniform}(0,1)\). Use 100 pairs \((U_j,1-U_j)\) and compare the estimator with 200 independent event indicators.
Method. Set \(Y=I(U<0.20)\) and \(Y'=I(1-U<0.20)\). The second event is \(U>0.80\). These events cannot occur together, so \(E[YY']=0\). Each indicator has mean \(0.20\) and variance \(0.20(0.80)=0.16\); their covariance is \(0-(0.20)^2=-0.04\).
The pair average \(Z=(Y+Y')/2\) has expectation \(0.20\) and variance
Averaging 100 independent pair averages gives estimator variance \(0.06/100=0.0006\), with standard error \(\sqrt{0.0006}=0.0245\), rounded to four decimals. The independent design has variance \(0.16/200=0.0008\) and standard error \(\sqrt{0.0008}=0.0283\). The variance ratio is \(0.0006/0.0008=0.75\): pairing reduces variance by 25% at the stated equal cost.
Check. The expected pair average remains \(0.20\), so pairing has not changed the target probability. The reduction follows from the disjoint events and negative covariance, not merely from using complementary uniforms.
Worked Example: Using a Reported Pair Correlation
In a hypothetical severity simulation, each output has variance \(0.04\). A pilot run estimates the correlation between the two outputs in a pair as \(-0.60\). A production run will use 250 independent pairs. Calculate its standard error and compare it with 500 independent outputs.
Method. Convert correlation to covariance: \(C=\rho\sigma^2=(-0.60)(0.04)=-0.024\). The variance of one pair average is
The estimator based on 250 pairs has variance \(0.008/250=0.000032\), so its standard error is \(\sqrt{0.000032}=0.0057\), rounded to four decimals. For 500 independent outputs, the estimator variance is \(0.04/500=0.00008\), and its standard error is \(\sqrt{0.00008}=0.0089\). The antithetic-to-independent variance ratio is \(0.000032/0.00008=0.40\), which also equals \(1+\rho\).
Check. The negative correlation produces a variance ratio below 1, as the covariance formula predicts. The common mistake this avoids is using the correlation itself, \(-0.60\), as a variance ratio; the correct ratio is \(1+\rho=0.40\).
Implementation and Interpretation
For inverse-transform simulation, a common procedure is to generate one uniform input \(U_j\), use \(U_j\) for one simulated output, and use \(1-U_j\) for its antithetic partner. For example, if a simulated severity is \(X=F_X^{-1}(U)\), the partner is \(F_X^{-1}(1-U)\), where \(F_X^{-1}\) is the model’s quantile function. Both severities have the intended marginal distribution. The downstream output function must still be assessed: a monotone output of the severity is favorable for negative covariance, but thresholds, nonlinear transformations, or multiple random inputs can change the relationship.
With observed pair averages \(Z_1,\ldots,Z_m\), estimate the variance of the Monte Carlo estimator using the sample variance of the pair averages divided by \(m\). Do not treat all \(2m\) outputs as independent observations: that ignores the within-pair dependence and gives the wrong standard error. For sufficiently many independent pairs and finite variance, a normal approximation for the estimator can be used in the usual way, subject to the limitations of that approximation.
The practical decision should be based on precision per computational cost, not on the label “antithetic.” A pilot simulation can estimate the covariance or the variance of the pair averages. If pair members are negatively correlated, the paired estimator can be more precise than independent simulation at equal output-evaluation cost. If covariance is positive, pairing may increase error; consider the covariance and model structure before adopting the design.
Common Mistake: Reversing a Variance Ratio
Suppose a comparison gives an antithetic estimator variance of \(1/180\) and an independent estimator variance of \(1/1125\). The ratio, with antithetic variance in the numerator, is
The wrong calculation reports \(0.125\), which is the reciprocal of \(8\), not the ratio of the two stated variances. The correction is \(6.25\): on these numbers, the antithetic variance is 6.25 times the independent variance, not one-eighth of it. In an asserted equal-cost comparison of the standard antithetic design with equal marginal output variances, this result should also trigger an audit: the ratio must equal \(1+\rho\) and cannot exceed 2. The stated figures therefore cannot both describe a correctly calculated, standard equal-cost comparison under those assumptions. Check the denominators, cost basis, and whether the same estimator variance is being compared before drawing a conclusion.
Check Your Understanding
For each question, compare estimators on the stated cost basis and keep pair-level variance distinct from estimator variance.
- Let \(Y=I(U<0.30)\) and \(Y'=I(1-U<0.30)\), with \(U\sim\operatorname{Uniform}(0,1)\). For 100 independent pairs, find the covariance within a pair and the variance of the estimator. (Answer: covariance \(-0.09\); estimator variance \(0.0006\).)
- For \(g(U)=U^2\), use 50 antithetic pairs to estimate \(E[U^2]\). Find the estimator variance and compare it with 100 independent outputs. (Answer: antithetic variance \(1/9000\); independent variance \(1/1125\); ratio \(1/8\).)
- Each output in a pair has variance 9, and the within-pair covariance is \(-3\). With 400 independent pairs, find the estimator variance and standard error. (Answer: variance \(0.0075\); standard error approximately \(0.0866\).)
- A paired design has output variance 16 and within-pair correlation \(0.25\). At equal cost, find the antithetic-to-independent variance ratio and state whether pairing helps. (Answer: ratio \(1.25\); it increases variance by 25% and does not help.)
- A report lists an antithetic estimator variance of \(1/180\) and an independent estimator variance of \(1/1125\). Calculate their ratio in that order and interpret it. (Answer: ratio \(6.25\); the antithetic variance is 6.25 times the independent variance, so it is not a variance reduction.)