Monte Carlo Estimation of SDE Functionals

From a Diffusion Expectation to a Monte Carlo Estimate

Many questions about a diffusion ask for the expectation of a quantity computed from its terminal value or its whole trajectory. If the distribution of that quantity is difficult to integrate analytically, independent simulated paths provide a direct estimator: compute the quantity on each path and average. The estimator’s mean and variance follow from the corresponding moments of a single path payoff.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. For specificity, take a scalar diffusion on $[0,T]$ with deterministic initial value $X_0=x_0$ and integral representation

X_t=x_0+\int_0^t\mu(s,X_s)\,ds+\int_0^t\sigma(s,X_s)\,dB_s,\qquad 0\le t\le T.

Here $\mu$ is the drift and $\sigma$ is the diffusion coefficient. The Ito integral is defined for progressively measurable integrands $H_s$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; in this equation we assume $\sigma(s,X_s)$ satisfies this condition. Assume also that $\int_0^T|\mu(s,X_s)|\,ds$ is finite almost surely. We take the solution’s law as given and suppose the sampling procedure can draw independent paths with exactly that law. The results below concern sampling error, not the existence of such a procedure.

Definition: For a terminal-value payoff, choose a measurable function $g:\mathbb{R}\to\mathbb{R}$ and define $Y=g(X_T)$. For a path-dependent payoff, choose a measurable function $\varphi:C([0,T])\to\mathbb{R}$ and define $Y=\varphi(X)$, where $C([0,T])$ is the space of continuous real-valued paths. The target is $\theta=\mathbb{E}[Y]$, provided $\mathbb{E}|Y|<\infty$. If $\mathbb{E}[Y^2]<\infty$, the payoff variance $\operatorname{Var}(Y)$ is also finite.

A terminal payoff is a special case of a path payoff: it depends on the path only through $X_T$. A path payoff can instead use several times, an ordinary time integral, or another measurable feature of the trajectory. Measurability makes the expectation meaningful; finite first moment makes the target finite; finite second moment is the additional condition needed for the variance formula.

The Sample-Average Estimator

Let $X^{(1)},\ldots,X^{(N)}$ be independent paths, each with exactly the same law as $X$, where $N$ is a positive integer. The copies may be regarded as coordinate processes on a product probability space. Evaluate the same payoff on every path, setting $Y_j=g(X_T^{(j)})$ or $Y_j=\varphi(X^{(j)})$, and define

\widehat{\theta}_N=\frac{1}{N}\sum_{j=1}^{N}Y_j.

Independence refers to the paths, not to the observations within a single path: values $X_t^{(j)}$ at different times $t$ on one path are generally dependent. Every $Y_j$ has the same distribution as $Y$, so $\mathbb{E}|Y|<\infty$ gives a well-defined target and an unbiased estimator.

Theorem: If $\mathbb{E}|Y|<\infty$, then $\mathbb{E}[\widehat{\theta}_N]=\theta$. If, in addition, $\mathbb{E}[Y^2]<\infty$, then $$ \operatorname{Var}(\widehat{\theta}_N)=\frac{\operatorname{Var}(Y)}{N}. $$

Proof. Linearity of expectation and equality of the payoff distributions give $\mathbb{E}[\widehat{\theta}_N]=N^{-1}\sum_{j=1}^N\mathbb{E}[Y_j]=N^{-1}N\theta=\theta$. For the variance statement, the second-moment assumption ensures every $Y_j$ has finite variance. Expanding the variance of the sum gives the sum of the individual variances plus pairwise covariance terms. For $j\ne k$, independence implies $\mathbb{E}[Y_jY_k]=\mathbb{E}[Y_j]\mathbb{E}[Y_k]$, so $\operatorname{Cov}(Y_j,Y_k)=0$. Thus $\operatorname{Var}(\widehat{\theta}_N)=N^{-2}\sum_{j=1}^N\operatorname{Var}(Y_j)=N^{-2}N\operatorname{Var}(Y)=\operatorname{Var}(Y)/N$. $\square$

The theorem separates two quantities that are easy to confuse. The variance of one path payoff is $\operatorname{Var}(Y)$; the variance of the average of $N$ independent payoffs is $\operatorname{Var}(Y)/N$. Independence and equal distribution are essential to this calculation. Without independence, the covariance terms need not vanish.

Worked Calculations

Worked Example: A Squared Brownian Terminal Value

Take $X_t=B_t$, $T=2$, and $Y=X_T^2$. Find the target and the variance of the sample average for $N=4$ independent exact paths.

Since $B_2$ is normal with mean zero and variance $2$, its second and fourth moments are $\mathbb{E}[B_2^2]=2$ and $\mathbb{E}[B_2^4]=3(2^2)=12$. Therefore the target is $\theta=\mathbb{E}[Y]=2$, and the variance of one payoff is $\operatorname{Var}(Y)=\mathbb{E}[B_2^4]-(\mathbb{E}[B_2^2])^2=12-2^2=8$. The estimator is the average of four such payoffs, so $\operatorname{Var}(\widehat{\theta}_4)=8/4=2$.

Check: the estimator’s expectation is $2$, because it averages four variables each with expectation $2$. Its variance $2$ is smaller than the single-payoff variance $8$ by exactly the factor $4$, as the theorem requires.

Worked Example: A Geometric Brownian Terminal Payoff

Let $a=\sqrt{\ln 2}$ and define $X_t=\exp(-a^2t/2+aB_t)$ for $0\le t\le1$. Use the terminal payoff $Y=X_1$ and find the target and the variance of its sample average for $N=16$ independent exact paths.

This process is a geometric Brownian motion with drift $0$ and diffusion coefficient $aX_t$. To verify the SDE, apply Ito’s formula to $f(t,b)=\exp(-a^2t/2+ab)$. Its derivatives are $f_t=-a^2f/2$, $f_b=af$, and $f_{bb}=a^2f$. The formula, including its second-derivative correction, gives $dX_t=(f_t+\tfrac12 f_{bb})dt+f_b\,dB_t=0\,dt+aX_t\,dB_t$. The integrand is square-integrable on this time interval because $\mathbb{E}\int_0^1(aX_t)^2dt=a^2\int_0^1e^{a^2t}dt=e^{a^2}-1=1$.

Since $B_1$ is standard normal, its moment-generating function gives $\mathbb{E}[e^{cB_1}]=e^{c^2/2}$. Consequently, $\mathbb{E}[Y]=e^{-a^2/2}e^{a^2/2}=1$, while $\mathbb{E}[Y^2]=e^{-a^2}e^{(2a)^2/2}=e^{a^2}=2$. Thus $\operatorname{Var}(Y)=2-1^2=1$, and $\operatorname{Var}(\widehat{\theta}_{16})=1/16$.

Check: substituting $a^2=\ln 2$ into the second moment gives $e^{a^2}=e^{\ln 2}=2$. Subtracting the squared mean $1$ gives payoff variance $1$; dividing by $16$ gives estimator variance $1/16$.

Worked Example: A Time-Integrated Brownian Path

For $X_t=B_t$ on $[0,3]$, take the path payoff $Y=\int_0^3 B_s\,ds$, with the integral an ordinary time integral along the continuous path. Find its target and the variance of the sample average for $N=9$ independent exact paths.

The mean is zero because $\mathbb{E}[B_s]=0$ for every $s$. Brownian covariance is $\mathbb{E}[B_sB_t]=\min(s,t)$, so $\mathbb{E}[Y^2]=\int_0^3\int_0^3\min(s,t)\,ds\,dt$. The integrals are justified since $\mathbb{E}|B_sB_t|\le\sqrt{st}$, which is integrable over the square. By symmetry of the integrand, the double integral is $2\int_0^3\int_0^t s\,ds\,dt =2\int_0^3(t^2/2)\,dt =\int_0^3t^2\,dt =27/3=9$. Hence $\operatorname{Var}(Y)=9$, and $\operatorname{Var}(\widehat{\theta}_9)=9/9=1$.

Check: the target is zero, so the payoff variance equals its second moment, $9$. Averaging nine independent path payoffs divides that value by nine, giving estimator variance $1$.

Why Exact Sampling Matters

The theorem describes an estimator built from independent draws from the exact law of the diffusion path. If a numerical method instead produces approximate paths $X^{h,(j)}$, applying the payoff and averaging estimates the expectation under the approximate path law. That average may have its own sampling variation, but its expected value is generally $\mathbb{E}[\varphi(X^h)]$, not $\mathbb{E}[\varphi(X)]$. Exact-path unbiasedness therefore does not automatically apply to a discretized scheme.

In practice, “exact paths” means that the simulated payoff has the desired exact distribution, rather than merely that the simulation uses many time steps. For some models and payoffs, an exact sampling method is available; for others, time discretization is needed. Keeping these cases distinct identifies what the sample-average variance formula does—and does not—describe.

A Common Mistake

A frequent error is to report the variance of one payoff as the variance of the Monte Carlo estimate. In the Brownian example with $T=2$, the payoff variance is $8$. The incorrect calculation for $N=4$ is $\operatorname{Var}(\widehat{\theta}_4)=8$, which forgets that the estimator averages four independent values. The corrected calculation is $\operatorname{Var}(\widehat{\theta}_4)=8/4=2$. The target remains $2$; increasing $N$ changes the estimator variance, not the target expectation.

Check Your Understanding

For each question, distinguish the single-path payoff from the sample average and check that the stated moment assumptions hold.

If $Y$ has finite mean but infinite second moment, which conclusion of the theorem still applies, and which variance formula is unavailable?
For $X_t=B_t$ on $[0,1]$ and $Y=X_1^2$, compute the target, payoff variance, and sample-average variance for $N=2$.
For independent paths with payoff variance $6$, what is the variance of the average of $N=3$ payoffs?
Explain why observations at different times on the same Brownian path cannot be treated as independent copies in the theorem.
If a numerical scheme generates approximate paths, which expectation does the average estimate in general: the exact-path target or the expectation under the approximate path law?