Conditional Monte Carlo for Diffusion Payoffs

Conditioning a Payoff Before Simulation

Monte Carlo estimation often simulates a diffusion path and evaluates a payoff $Y$ at its end. If part of the path has already been simulated, the remaining randomness can sometimes be averaged out analytically. The resulting payoff is a conditional expectation: it has the same mean as $Y$, but it cannot have greater variance. The gain can be substantial when the conditioning removes a large part of the payoff's randomness.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration. An Ito integral $\int_0^t H_s\,dB_s$ below is understood for a progressively measurable integrand $H_s$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^t H_s^2\,ds<\infty$. The diffusion examples will use constant integrands, which satisfy this condition on every finite time interval. Let $\mathcal{G}\subseteq\mathcal{F}$ represent the simulated information we retain—for example, the path up to a time $s<T$—and suppose the payoff $Y$ is square-integrable.

Definition: The conditional Monte Carlo payoff based on $\mathcal{G}$ is $C=\mathbb{E}[Y\mid\mathcal{G}]$. It is a $\mathcal{G}$-measurable random variable satisfying $\mathbb{E}[C\mathbf{1}_A]=\mathbb{E}[Y\mathbf{1}_A]$ for every $A\in\mathcal{G}$. The condition $Y\in L^2(\mathbb{P})$ ensures that both $Y$ and $C$ have finite variance.

In practice, one simulates the information in $\mathcal{G}$, computes $C$ from it, and averages independent copies of $C$. The key is to condition on information that leaves the conditional expectation tractable while averaging away meaningful payoff randomness.

Why the Variance Cannot Increase

Theorem: If $Y\in L^2(\mathbb{P})$, $\mathcal{G}\subseteq\mathcal{F}$, and $C=\mathbb{E}[Y\mid\mathcal{G}]$, then $$ \mathbb{E}[C]=\mathbb{E}[Y],\qquad \operatorname{Var}(Y)=\operatorname{Var}(C)+\mathbb{E}\!\left[\operatorname{Var}(Y\mid\mathcal{G})\right]. $$ Consequently $\operatorname{Var}(C)\le\operatorname{Var}(Y)$. Equality holds if and only if $Y=C$ almost surely; otherwise the variance reduction is strict.

Proof. The defining property of conditional expectation, applied to the whole space, gives $\mathbb{E}[C]=\mathbb{E}[Y]$. Set $R=Y-C$. Then $\mathbb{E}[R\mid\mathcal{G}]=0$, and $C$ is $\mathcal{G}$-measurable. Therefore $\mathbb{E}[CR]=\mathbb{E}[C\,\mathbb{E}[R\mid\mathcal{G}]]=0$; the product is integrable by the Cauchy–Schwarz inequality. Expanding the square gives $\mathbb{E}[Y^2]=\mathbb{E}[C^2]+\mathbb{E}[R^2]$. Subtracting $\mathbb{E}[Y]^2=\mathbb{E}[C]^2$ yields $\operatorname{Var}(Y)=\operatorname{Var}(C)+\mathbb{E}[R^2]$. By definition, $\operatorname{Var}(Y\mid\mathcal{G})=\mathbb{E}[R^2\mid\mathcal{G}]$, so taking expectations gives the displayed identity. Since $R^2\ge0$, the reduction is nonnegative. Equality holds precisely when $\mathbb{E}[R^2]=0$, which for a nonnegative random variable is equivalent to $R=0$ almost surely. $\square$

For $N$ independent simulated paths, the sample mean of the original payoffs has variance $\operatorname{Var}(Y)/N$; the sample mean of the conditional payoffs has variance $\operatorname{Var}(C)/N$. Both estimate the same expectation. Thus, for the same number of independent paths, conditioning cannot worsen the estimator's variance. It does not by itself guarantee a cheaper computation: evaluating the conditional expectation may require extra work.

Worked Example: Averaging Out One Random Component

Let $U$ and $V$ be independent, each taking the values $-1$ and $1$ with probability $1/2$. Define $Y=U+2V$ and condition on $\mathcal{G}=\sigma(U)$. Compare the variances of $Y$ and $C=\mathbb{E}[Y\mid\mathcal{G}]$.

Because $\mathbb{E}[U]=\mathbb{E}[V]=0$, $\mathbb{E}[Y]=0$. Independence gives $\mathbb{E}[UV]=\mathbb{E}[U]\mathbb{E}[V]=0$, and $U^2=V^2=1$. Hence $$ \operatorname{Var}(Y)=\mathbb{E}[(U+2V)^2] =1+4\cdot 0+4=5. $$ Given $U$, the independent variable $V$ still has mean zero, so $C=U+2\mathbb{E}[V\mid U]=U$. Thus $\mathbb{E}[C]=0$ and $\operatorname{Var}(C)=\mathbb{E}[U^2]=1$.

Check: the conditional residual is $Y-C=2V$, so $\mathbb{E}[(Y-C)^2]=4$. The variance decomposition gives $5=1+4$, as computed directly. Here conditioning removes the variability contributed by $V$, but the remaining variability in $U$ means the conditional payoff is not constant.

Conditionally Gaussian Terminal States

Consider the arithmetic Brownian diffusion $$ X_t=x_0+\mu t+\sigma B_t,\qquad 0\le t\le T, $$ where $x_0,\mu,\sigma$ are constants. Equivalently, $dX_t=\mu\,dt+\sigma\,dB_t$. The drift is $\mu$, the diffusion coefficient is $\sigma$, and the stochastic integral uses the progressively measurable constant integrand $H_s=\sigma$, with $\mathbb{E}\int_0^T H_s^2\,ds=\sigma^2T<\infty$. Since Brownian increments after $s$ are independent of $\mathcal{F}_s$, for $0\le s<T$, $$ X_T=X_s+\mu(T-s)+\sigma(B_T-B_s). $$ Consequently the conditional distribution of $X_T$ given $\mathcal{F}_s$ is Gaussian with mean $m_s=X_s+\mu(T-s)$ and variance $v=\sigma^2(T-s)$.

For a call payoff $(X_T-K)^+=\max(X_T-K,0)$, suppose $v>0$. Write $\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}$ for the standard normal density and $\Phi(z)=\int_{-\infty}^z\phi(u)\,du$ for its CDF. Standardizing the conditional Gaussian gives the conditional payoff formula below.

Formula: With $d_s=(m_s-K)/\sqrt{v}$, $$ \mathbb{E}[(X_T-K)^+\mid\mathcal{F}_s] =(m_s-K)\Phi(d_s)+\sqrt{v}\,\phi(d_s). $$ If $v=0$, then $X_T=m_s$ given $\mathcal{F}_s$, and the conditional payoff is instead $(m_s-K)^+$.

To verify the formula, condition on $\mathcal{F}_s$ and write $X_T=m_s+\sqrt{v}Z$, where $Z$ has the standard normal distribution. The payoff is positive precisely when $Z>(K-m_s)/\sqrt{v}=-d_s$. Its expectation is therefore the integral of $(m_s-K)+\sqrt{v}z$ over $z>-d_s$. The constant term contributes $(m_s-K)\Phi(d_s)$, using the symmetry of the normal distribution. Since $\phi'(z)=-z\phi(z)$, the integral of $z\phi(z)$ from $-d_s$ to infinity is $\phi(d_s)$. This gives the stated expression. The formula supplies a version of the conditional expectation as a function of $X_s$; it does not require the event $X_s=x$ to have positive probability.

Worked Example: Conditioning a Gaussian Call Payoff

Let $x_0=0$, $\mu=0$, $\sigma=2$, $s=1$, $T=2$, and $K=1$. Find the conditional Monte Carlo payoff as a function of $X_1$, then evaluate that function at $X_1=1$.

Here $T-s=1$, so $m_1=X_1$ and $v=\sigma^2(T-s)=2^2\cdot1=4$. Thus $d_1=(X_1-1)/2$, and the conditional estimator is $$ C=\mathbb{E}[(X_2-1)^+\mid\mathcal{F}_1] =(X_1-1)\Phi\!\left(\frac{X_1-1}{2}\right) +2\phi\!\left(\frac{X_1-1}{2}\right). $$ At state value $X_1=1$, $d_1=0$, so $C=2\phi(0)=2/\sqrt{2\pi}=\sqrt{2/\pi}$.

Check: at $X_1=1$, the future change $X_2-X_1$ is a centered normal variable with variance $4$. By symmetry, its expected positive part is half its expected absolute value. A centered normal variable with standard deviation $2$ has expected absolute value $2\sqrt{2/\pi}$, giving half of that, $\sqrt{2/\pi}$, as above. The conditional formula is not asserting that the event $X_1=1$ has positive probability; it evaluates a version of the conditional expectation at that state.

Another Payoff: The Conditional Expectation of a Square

Worked Example: A Square Payoff and the Variance Decomposition

For $X_t=2+t+2B_t$ on $0\le t\le2$, use $\mathcal{F}_1$ to replace the payoff $Y=X_2^2$ by its conditional expectation. Compute both variances in the variance decomposition.

Given $\mathcal{F}_1$, $X_2=X_1+1+2(B_2-B_1)$. The increment has conditional mean zero and conditional variance $2^2(2-1)=4$. If $Z$ has conditional mean $a=X_1+1$ and conditional variance $4$, then $\mathbb{E}[Z^2\mid\mathcal{F}_1]=a^2+4$, by expanding $Z^2=(Z-a)^2+2a(Z-a)+a^2$. Therefore $$ C=\mathbb{E}[X_2^2\mid\mathcal{F}_1]=(X_1+1)^2+4. $$ Since $X_1$ is normal with mean $3$ and variance $4$, $X_1+1$ has mean $4$ and variance $4$. For a normal variable with mean $a$ and variance $q$, its square has variance $2q^2+4a^2q$, obtained by expanding the square around its mean and using the centered normal moments $\mathbb{E}[Z^2]=q$, $\mathbb{E}[Z^3]=0$, and $\mathbb{E}[Z^4]=3q^2$. Hence $$ \operatorname{Var}(C)=2\cdot4^2+4\cdot4^2\cdot4=32+256=288. $$

Unconditionally, $X_2$ is normal with mean $2+2=4$ and variance $2^2\cdot2=8$. Thus $$ \operatorname{Var}(Y)=2\cdot8^2+4\cdot4^2\cdot8=128+512=640. $$ The conditional variance of $X_2^2$, given $\mathcal{F}_1$, is $2\cdot4^2+4(X_1+1)^2\cdot4=32+16(X_1+1)^2$. Taking expectations, $\mathbb{E}[(X_1+1)^2]=4^2+4=20$, so $\mathbb{E}[\operatorname{Var}(Y\mid\mathcal{F}_1)]=32+16\cdot20=352$.

Check: the variance identity gives $288+352=640$, matching the direct unconditional calculation. The conditional payoff preserves the mean, too: $\mathbb{E}[C]=20+4=24=\mathbb{E}[X_2^2]$.

Common Pitfall and Practical Scope

Do not replace a random future value by its conditional mean inside a nonlinear payoff. In the call example with $X_1=K=1$, the conditional mean of $X_2$ is $1$. Plugging that mean into the payoff gives $(1-1)^+=0$, which is not the conditional expected payoff. The correct value is $\sqrt{2/\pi}$, because the future Gaussian increment sometimes puts $X_2$ above the strike. In general, $\mathbb{E}[g(X_T)\mid\mathcal{F}_s]$ is not $g(\mathbb{E}[X_T\mid\mathcal{F}_s])$ for nonlinear $g$.

For a diffusion whose conditional terminal distribution or conditional payoff expectation is not available in a useful form, this particular conditioning step may not be practical. Conditioning can also reduce variance without changing any discretization bias already present: if the payoff is computed from an approximate diffusion path, the conditional estimator targets the expectation of that approximate payoff unless the conditional calculation accounts for the exact target. The method is most useful when a substantial portion of the path can be simulated and the remaining payoff randomness can be integrated out accurately at modest cost.

Check Your Understanding

Use the conditional variance identity and the Gaussian calculations above to organize your reasoning.

For $Y\in L^2(\mathbb{P})$, explain why the conditional Monte Carlo payoff has the same expectation as $Y$.
State the necessary and sufficient condition for conditioning on $\mathcal{G}$ to leave the variance unchanged.
For $X_t=x_0+\mu t+\sigma B_t$, identify the conditional mean and variance of $X_T$ given $\mathcal{F}_s$.
Explain why substituting the conditional mean of a Gaussian terminal state into a call payoff does not generally produce its conditional expected payoff.
In the square-payoff example, verify that the conditional variance contribution and the variance of the conditional payoff add to the unconditional payoff variance.