Control Variates from Known Diffusion Prices

Why a Known-Mean Control Helps

Monte Carlo estimates fluctuate because simulated payoffs vary from path to path. A control variate reduces that fluctuation by pairing the payoff with another random variable that is correlated with it but whose expectation is already known. When the payoff is high, a positively correlated control will usually be high as well; subtracting a suitably scaled deviation of the control from its known mean can remove some of the payoff's random variation without changing the target expectation.

For option pricing, expectations are taken under the risk-neutral measure $\mathbb{Q}$, not the physical measure $\mathbb{P}$. Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{Q})$ carrying a Brownian motion $B_t$. For a simulated diffusion, any Ito integral $\int_0^t H_s\,dB_s$ below is understood with a progressively measurable integrand $H$ satisfying $\mathbb{E}^{\mathbb{Q}}\int_0^t H_s^2\,ds<\infty$. The drift and diffusion coefficients determine the simulated paths; the control-variate construction itself is a statement about the joint distribution of the payoff and control.

Definition: Let $X$ be the target payoff and $Y$ a control random variable, both with finite second moments, and suppose $\mu_Y=\mathbb{E}^{\mathbb{Q}}[Y]$ is known. For a real coefficient $\beta$, the adjusted payoff is $X^{(\beta)}=X-\beta(Y-\mu_Y)$. Its expectation equals $\mathbb{E}^{\mathbb{Q}}[X]$; the corresponding estimate from $n$ paired independent paths is the sample average of the adjusted payoffs.

Pairing matters: $X$ and $Y$ in each adjustment must come from the same simulated path. The centering by $\mu_Y$ is also essential. It removes the control's known mean rather than its particular simulated value, preserving the target expectation.

Optimal Coefficient and Variance Reduction

Theorem: Suppose $X,Y$ have finite second moments under $\mathbb{Q}$. If $\operatorname{Var}^{\mathbb{Q}}(Y)>0$, the unique coefficient minimizing $\operatorname{Var}^{\mathbb{Q}}(X-\beta(Y-\mu_Y))$ is $$ \beta^*=\frac{\operatorname{Cov}^{\mathbb{Q}}(X,Y)}{\operatorname{Var}^{\mathbb{Q}}(Y)}. $$ The minimum variance is $$ \operatorname{Var}^{\mathbb{Q}}(X)-\frac{\operatorname{Cov}^{\mathbb{Q}}(X,Y)^2}{\operatorname{Var}^{\mathbb{Q}}(Y)}. $$ If $\operatorname{Var}^{\mathbb{Q}}(Y)=0$, then $Y=\mu_Y$ almost surely and every coefficient gives the unadjusted payoff.

Proof. Since $\mathbb{E}^{\mathbb{Q}}[Y-\mu_Y]=0$, linearity of expectation gives $\mathbb{E}^{\mathbb{Q}}[X-\beta(Y-\mu_Y)]=\mathbb{E}^{\mathbb{Q}}[X]$ for every real $\beta$. Expanding the variance gives $$ \operatorname{Var}^{\mathbb{Q}}(X-\beta(Y-\mu_Y)) =\operatorname{Var}^{\mathbb{Q}}(X)-2\beta\operatorname{Cov}^{\mathbb{Q}}(X,Y) +\beta^2\operatorname{Var}^{\mathbb{Q}}(Y). $$ When $\operatorname{Var}^{\mathbb{Q}}(Y)>0$, this is a strictly convex quadratic in $\beta$. Completing the square yields $$ \operatorname{Var}^{\mathbb{Q}}(X)-\frac{\operatorname{Cov}^{\mathbb{Q}}(X,Y)^2}{\operatorname{Var}^{\mathbb{Q}}(Y)} +\operatorname{Var}^{\mathbb{Q}}(Y)\left(\beta-\frac{\operatorname{Cov}^{\mathbb{Q}}(X,Y)}{\operatorname{Var}^{\mathbb{Q}}(Y)}\right)^2. $$ The squared term is minimized uniquely at the stated $\beta^*$, proving the minimum-variance formula. If $\operatorname{Var}^{\mathbb{Q}}(Y)=0$, then $\mathbb{E}^{\mathbb{Q}}[(Y-\mu_Y)^2]=0$, so $Y=\mu_Y$ almost surely; the adjustment is zero almost surely for every $\beta$. $\square$

For $n$ independent paired paths, the variance of the sample-average estimator is the variance of one adjusted payoff divided by $n$. Thus the variance-reduction factor does not depend on $n$. When both variances are positive, the fraction of variance removed at the optimum is $\operatorname{Corr}^{\mathbb{Q}}(X,Y)^2$. Indeed, substituting $\operatorname{Cov}(X,Y)^2=\operatorname{Corr}(X,Y)^2\operatorname{Var}(X)\operatorname{Var}(Y)$ into the theorem gives this identity. A control uncorrelated with the payoff has no effect; a perfectly linearly correlated control can remove all variance.

Worked Example: A Simple Pair with a Known Control Mean

Let $U$ and $\varepsilon$ be independent, mean-zero random variables with variances $4$ and $1$, respectively. Set $X=2+U+\varepsilon$ and $Y=5+U$. Find the optimal coefficient and compare the variances.

The means are $\mathbb{E}[X]=2$ and $\mathbb{E}[Y]=5$. Independence and the zero means give $\operatorname{Var}(X)=4+1=5$, $\operatorname{Var}(Y)=4$, and $\operatorname{Cov}(X,Y)=\operatorname{Cov}(U+\varepsilon,U)=4+0=4$. Therefore $\beta^*=4/4=1$.

The adjusted payoff is $X-(Y-5)=2+U+\varepsilon-U=2+\varepsilon$, with variance $1$. The variance falls from $5$ to $1$, so the fraction removed is $(5-1)/5=4/5$. Check: the theorem gives $5-4^2/4=1$, matching the direct calculation.

Geometric Average as an Asian-Option Control

Consider a geometric Brownian stock under $\mathbb{Q}$, with constant interest rate $r$, volatility $\sigma>0$, and initial value $S_0>0$. Its dynamics are $dS_t=rS_t\,dt+\sigma S_t\,dB_t$, and its solution is $S_t=S_0\exp((r-\tfrac12\sigma^2)t+\sigma B_t)$. Here $rS_t$ is the drift coefficient and $\sigma S_t$ is the diffusion coefficient. For two monitoring times $t_1=1$ and $t_2=2$, define the geometric average $G=(S_1S_2)^{1/2}$. It is lognormal because its logarithm is a linear combination of jointly normal Brownian values.

Write $m=\mathbb{E}^{\mathbb{Q}}[\log G]$ and $v=\operatorname{Var}^{\mathbb{Q}}(\log G)$. From the stock solution and $\operatorname{Var}(B_1+B_2)=1+2+2(1)=5$, $$ m=\log S_0+\frac32\left(r-\frac12\sigma^2\right),\qquad v=\frac54\sigma^2. $$ In particular, $\mathbb{E}^{\mathbb{Q}}[G]=\exp(m+v/2)$, a known expectation. Use $Y=G$ as the control for the geometric Asian call payoff $X=(G-K)^+$, where $K>0$ and $x^+=\max(x,0)$. The option's discounted value at time zero is $e^{-2r}\mathbb{E}^{\mathbb{Q}}[X]$; multiplying the payoff and its adjusted version by this fixed discount factor scales their variances equally and does not change the percentage reduction.

For explicit calculations, let $\Phi$ denote the standard normal cumulative distribution function. If $\log G$ is normal with mean $m$ and variance $v>0$, then for $p=0,1,2$, $$ \mathbb{E}^{\mathbb{Q}}\!\left[G^p\mathbf{1}_{\{G>K\}}\right] =\exp\left(pm+\frac12p^2v\right) \Phi\left(\frac{m+pv-\log K}{\sqrt v}\right). $$ To obtain this formula, write $G^p=\exp(p\log G)$ in the normal-density integral over $\log G>\log K$. Completing the square combines the exponential factor with the normal density to give $\exp(pm+\tfrac12p^2v)$ times a normal density with mean $m+pv$ and variance $v$. Integrating that density above $\log K$ gives the displayed $\Phi$ term. The strict tail event is correct because the payoff is positive exactly when $G>K$; the equality event has probability zero for $v>0$.

Worked Example: Variance Reduction for a Geometric Asian Call

Take $S_0=100$, $r=0.08$, $\sigma=0.4$, $K=100$, and monitoring times $1$ and $2$. Use $X=(G-100)^+$ and the control $Y=G$. Calculate the optimal coefficient, the payoff variance before and after adjustment, and the fraction of variance removed.

First, $r-\tfrac12\sigma^2=0.08-\tfrac12(0.16)=0$, so $m=\log(100)$. Also $v=\tfrac54(0.16)=0.2$. Define $a=\sqrt{0.2}$. The truncated-moment formula gives $$ \mathbb{E}^{\mathbb{Q}}[X] =100\left(e^{0.1}\Phi(a)-\frac12\right),\quad \mathbb{E}^{\mathbb{Q}}[X^2] =10000\left(e^{0.4}\Phi(2a)-2e^{0.1}\Phi(a)+\frac12\right), $$ and $$ \mathbb{E}^{\mathbb{Q}}[XG] =10000\left(e^{0.4}\Phi(2a)-e^{0.1}\Phi(a)\right). $$ For the control, $\mathbb{E}^{\mathbb{Q}}[G]=100e^{0.1}$ and $\operatorname{Var}^{\mathbb{Q}}(G)=10000(e^{0.4}-e^{0.2})$. These expressions use $XG=G^2\mathbf{1}_{\{G>100\}}-100G\mathbf{1}_{\{G>100\}}$, and $X^2=G^2\mathbf{1}_{\{G>100\}}-200G\mathbf{1}_{\{G>100\}}+10000\mathbf{1}_{\{G>100\}}$.

Substitution gives $e^{0.1}\Phi(\sqrt{0.2})\approx0.743381$ and $e^{0.4}\Phi(2\sqrt{0.2})\approx1.215022$. Hence $\mathbb{E}^{\mathbb{Q}}[X]\approx24.3381$, $\mathbb{E}^{\mathbb{Q}}[X^2]\approx2282.592$, and $\operatorname{Var}^{\mathbb{Q}}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X])^2\approx1690.248$. The control variance is $10000(e^{0.4}-e^{0.2})\approx2704.219$. Using $\operatorname{Cov}(X,G)=\mathbb{E}[XG]-\mathbb{E}[X]\mathbb{E}[G]$ gives $\operatorname{Cov}^{\mathbb{Q}}(X,G)\approx2026.628$.

Therefore the optimal coefficient for the undiscounted payoff is $\beta^*\approx2026.628/2704.219\approx0.7494$. The optimized variance is $$ 1690.248-\frac{(2026.628)^2}{2704.219}\approx171.43. $$ The variance removed is approximately $1690.248-171.43=1518.82$, or $1518.82/1690.248\approx0.8986$, about $89.86\%$. Check: the correlation implied by these moments has squared value approximately $0.8986$, agreeing with the theorem's variance-reduction identity. The discounted option estimator uses $e^{-0.16}X$; its optimal coefficient is $e^{-0.16}\beta^*$, and its variance is multiplied by $e^{-0.32}$, so the reduction percentage remains $89.86\%$.

Using the Method in a Simulation

For each simulated path $i$, compute both the target payoff $X_i$ and the control $Y_i$. With a specified coefficient $\beta$, estimate the desired expectation by $$ \widehat{\mu}_X=\frac1n\sum_{i=1}^n\left(X_i-\beta(Y_i-\mu_Y)\right). $$ If the pairs are independent copies of $(X,Y)$, the estimator is unbiased and its variance is $\operatorname{Var}(X-\beta(Y-\mu_Y))/n$. In practice, the optimal coefficient is not usually known in advance because the covariance and variance are unknown. One can estimate it from an independent pilot sample and then use it on the main sample; using an independent pilot avoids treating the fitted coefficient as fixed when analysing the main estimator. The known value $\mu_Y$, however, must come from the model or an analytic calculation, not from the same Monte Carlo average.

Worked Example: A Small Paired Sample and the Sign of the Adjustment

Consider a distribution with four equally likely paired outcomes $(X,Y)=(0,1),(2,2),(4,3),(6,4)$. Its control mean is $\mu_Y=2.5$. Compute the optimal coefficient and the adjusted outcomes.

The means are $\mathbb{E}[X]=(0+2+4+6)/4=3$ and $\mathbb{E}[Y]=(1+2+3+4)/4=2.5$. The centered values are $X-3=(-3,-1,1,3)$ and $Y-2.5=(-1.5,-0.5,0.5,1.5)$. Thus the covariance is $(4.5+0.5+0.5+4.5)/4=2.5$, while the control variance is $(2.25+0.25+0.25+2.25)/4=1.25$. Hence $\beta^*=2.5/1.25=2$.

The adjusted outcomes $X-2(Y-2.5)$ are, in order, $0-2(-1.5)=3$, $2-2(-0.5)=3$, $4-2(0.5)=3$, and $6-2(1.5)=3$. Their mean remains $3$, and their variance is zero. Check: the original variance is $(9+1+1+9)/4=5$, and the theorem gives $5-(2.5)^2/1.25=0$, as the direct adjusted outcomes show.

Common Pitfall: Reversing the Sign

For positively correlated $X$ and $Y$, the adjustment subtracts $\beta(Y-\mu_Y)$, with $\beta$ usually positive. In the last example, the incorrect expression $X+2(Y-2.5)$ produces the values $-3,1,5,9$, whose variance is $(36+4+4+36)/4=20$, larger than the original variance $5$. The correction is to subtract the centered control: high control outcomes then lower the adjusted payoff, offsetting the payoff's tendency to be high. The optimal coefficient's sign follows the covariance; if covariance is negative, the minimizing coefficient is negative and the same subtraction formula still applies.

Check Your Understanding

Use finite second moments throughout. For option-pricing questions, take expectations under the risk-neutral measure.

For $\operatorname{Cov}(X,Y)=3$ and $\operatorname{Var}(Y)=6$, find the optimal coefficient and the variance reduction from the control.
Why does centering $Y$ at its known expectation preserve the target expectation for every fixed coefficient?
For $G$ lognormal with log-mean $m$ and log-variance $v>0$, write the formula for $\mathbb{E}[G^2\mathbf{1}_{\{G>K\}}]$.
What happens to the optimal coefficient if the target payoff is multiplied by a deterministic discount factor $d>0$ while the control is unchanged?
Explain why estimating the control mean from the same simulation sample and treating it as known can invalidate the stated unbiasedness argument.