Why Change the Drift?
Suppose a diffusion payoff is zero on almost every simulated path and nonzero only when the process makes an unusual excursion. A direct Monte Carlo estimate is then based on very few successful paths. Importance sampling changes the probability law used to generate paths so that the relevant event occurs more often. Since this also changes the distribution of the simulated payoff, each payoff must be multiplied by a likelihood ratio that corrects back to the original law.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a scalar SDE on \([0,T]\), \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), with a specified initial value. Assume the SDE has a well-defined solution and that the payoff \(F\), a measurable function of the simulated path, is integrable under the original measure. The drift is \(\mu\); the diffusion coefficient is \(\sigma\). The aim is to estimate \(\mathbb{E}^{\mathbb{P}}[F]\).
For the stochastic integrals below, the integrands are progressively measurable and belong to \(\mathcal{L}^2\) under the measure in use: for example, \(\mathbb{E}^{\mathbb{P}}\int_0^T\theta_t^2\,dt<\infty\). The Ito integral is meant throughout. A standard result used here is Girsanov’s theorem; its proof is not given in this tutorial. A sufficient condition for the exponential density in that theorem to be a true martingale is Novikov’s condition, \(\mathbb{E}^{\mathbb{P}}\exp(\tfrac12\int_0^T\theta_t^2\,dt)<\infty\). Under this condition, define \(d\mathbb{Q}^{\theta}/d\mathbb{P}|_{\mathcal{F}_T} =\exp(\int_0^T\theta_t\,dB_t-\tfrac12\int_0^T\theta_t^2\,dt)\). Under \(\mathbb{Q}^{\theta}\), \(B_t^{\theta}=B_t-\int_0^t\theta_s\,ds\) is Brownian motion, so the SDE has drift \(\mu+\sigma\theta\) and the same diffusion coefficient \(\sigma\). This is a change of sampling measure, not necessarily a risk-neutral measure.
The Likelihood-Ratio Identity
Proof. The density defining \(\mathbb{Q}^{\theta}\) is strictly positive, so the measures are equivalent on \(\mathcal{F}_T\). By the Radon–Nikodym definition of conditional density, \(L_T=d\mathbb{P}/d\mathbb{Q}^{\theta}|_{\mathcal{F}_T}\). Therefore, for every integrable \(\mathcal{F}_T\)-measurable \(F\), integration against this density gives \(\mathbb{E}^{\mathbb{Q}^{\theta}}[L_TF]=\mathbb{E}^{\mathbb{P}}[F]\). To obtain the displayed expression for \(L_T\), substitute \(dB_t=dB_t^{\theta}+\theta_t\,dt\) into the exponent of \(d\mathbb{Q}^{\theta}/d\mathbb{P}\). Its exponent becomes \(\int_0^T\theta_t\,dB_t^{\theta}+\tfrac12\int_0^T\theta_t^2\,dt\); taking the reciprocal gives the stated formula. \(\square\)
Thus \(L_TF\), evaluated on paths simulated under \(\mathbb{Q}^{\theta}\), is an unbiased observation of the original payoff. If its second moment is finite, independent simulations give the estimator \(\widehat{V}_N=N^{-1}\sum_{j=1}^N L_T^{(j)}F^{(j)}\), with variance \(\operatorname{Var}_{\mathbb{Q}^{\theta}}(L_TF)/N\). The change of measure alone does not guarantee a variance reduction: an unsuitable shift can produce rare but very large likelihood ratios.
Worked Estimators
Worked Example: Checking a Weighted Payoff
Let \(T=1\), \(X_t=B_t\) under \(\mathbb{P}\), and \(F=X_1^2\). Choose the constant shift \(\theta_t=1\). Under \(\mathbb{Q}^{\theta}\), \(X_1=1+Z\), where \(Z=B_1^{\theta}\sim N(0,1)\), and \(L_1=\exp(-Z-\tfrac12)\). Compute the weighted expectation.
Writing \(\varphi(z)=(2\pi)^{-1/2}e^{-z^2/2}\), complete the square: \(\varphi(z)e^{-z-1/2}=\varphi(z+1)\). Hence \(\mathbb{E}^{\mathbb{Q}^{\theta}}[L_1X_1^2]=\int_{\mathbb{R}}(1+z)^2\varphi(z+1)\,dz\). Set \(y=z+1\); then \(1+z=y\), so the integral is \(\int_{\mathbb{R}}y^2\varphi(y)\,dy=1\).
Check: under the original measure, \(\mathbb{E}^{\mathbb{P}}[B_1^2]=1\), which agrees. In this example the shifted paths are not needed to make the payoff rare; the calculation checks the likelihood-ratio direction and sign.
Worked Example: A Rare Terminal-Tail Event
Estimate \(p=\mathbb{P}(B_1\ge2)\) using the shift \(\theta_t=2\). Under \(\mathbb{Q}^{\theta}\), \(B_1=2+Z\), \(Z\sim N(0,1)\), so the event is \(Z\ge0\), which has probability \(1/2\). The likelihood ratio is \(L_1=e^{-2Z-2}\). The weighted observation is \(Y=e^{-2Z-2}\mathbf{1}_{\{Z\ge0\}}\).
Its mean is \(\mathbb{E}^{\mathbb{Q}^{\theta}}[Y]=\int_0^\infty e^{-2z-2}\varphi(z)\,dz\). Since \(e^{-2z-2}\varphi(z)=\varphi(z+2)\), this equals \(\int_2^\infty\varphi(y)\,dy=1-\Phi(2)\), the desired probability. Numerically, \(\Phi(2)\approx0.9772499\), so \(p\approx0.0227501\).
The second moment is also explicit: \(\mathbb{E}^{\mathbb{Q}^{\theta}}[Y^2]=e^{-4}\int_0^\infty e^{-4z}\varphi(z)\,dz\). Completing the square gives \(e^{-4z}\varphi(z)=e^8\varphi(z+4)\), so the second moment is \(e^4\Phi(-4)\). Using \(e^4\approx54.59815\) and \(\Phi(-4)\approx0.00003167124\), this is approximately \(0.0017292\). The variance is therefore \(0.0017292-(0.0227501)^2\approx0.0012116\).
Check: direct simulation under \(\mathbb{P}\) has Bernoulli variance \(p(1-p)\approx0.0222325\), substantially larger here. This comparison is specific to this shift and payoff; it is not a guarantee for every importance-sampling choice.
Worked Example: Estimating a Barrier-Crossing Probability
Let \(X_t=B_t\) under \(\mathbb{P}\), for \(0\le t\le1\), and estimate \(p=\mathbb{P}(\sup_{0\le t\le1}X_t\ge2)\). Use the constant shift \(\theta_t=2\). Under \(\mathbb{Q}^{\theta}\), \(X_t=2t+B_t^{\theta}\), and the exact continuous-path estimator is \(Y=\exp(-2B_1^{\theta}-2)\mathbf{1}_{\{\sup_{0\le t\le1}(2t+B_t^{\theta})\ge2\}}\).
The likelihood-ratio identity proves \(\mathbb{E}^{\mathbb{Q}^{\theta}}[Y]=p\). The shift makes the event more frequent: if \(X_1\ge2\), continuity and \(X_0=0\) imply that the path crossed level \(2\); and under \(\mathbb{Q}^{\theta}\), \(X_1=2+Z\), so \(\mathbb{Q}^{\theta}(X_1\ge2)=1/2\). Thus the barrier event has probability at least \(1/2\) under the sampling measure. Under \(\mathbb{P}\), the reflection principle gives \(p=2(1-\Phi(2))\approx2(1-0.9772499)=0.0455002\). The reflection principle is used here as a standard Brownian-motion result.
For an illustrative simulated path, suppose the two Brownian increments under \(\mathbb{Q}^{\theta}\), over intervals of length \(1/2\), are \(1/2\) and \(-1/4\). Then \(B_1^{\theta}=1/2-1/4=1/4\), and at time \(1\), \(X_1=2+1/4=9/4\). Since \(X_0=0<2\) and \(X_1=9/4>2\), this path crosses the barrier. Its likelihood ratio is \(e^{-2(1/4)-2}=e^{-5/2}\), so this realization contributes \(e^{-5/2}\) to the estimator.
Check: the correction uses the terminal Brownian value \(B_1^{\theta}=1/4\), not the terminal diffusion value \(X_1=9/4\). Also, a grid that never observes a crossing does not by itself establish that a continuous path did not cross between grid points. The estimator above concerns continuous crossing; a discrete-monitoring approximation is a different payoff.
Choosing and Checking a Drift Shift
The shift changes the drift only in directions supplied by the diffusion coefficient: in one dimension the new drift is \(\mu(t,X_t)+\sigma(t,X_t)\theta_t\). Consequently, where \(\sigma=0\), this measure change cannot alter the instantaneous drift. In multiple dimensions, a proposed drift change must lie in the range of the diffusion matrix. The chosen \(\theta\) must also satisfy conditions that make the measures equivalent and the likelihood ratio usable; Novikov’s condition is a convenient sufficient condition, not a necessary one.
In practice, a useful shift moves typical simulated paths toward the payoff event without making the weights excessively variable. For a terminal upper-tail event, a positive constant shift is a natural first choice. For a path-dependent event, a state-dependent shift may be more effective, but it must remain progressively measurable and meet the required integrability conditions. Variance should be assessed from the weighted observations \(L_TF\), not from the unweighted event frequency alone.
Check Your Understanding
For each question, distinguish the original probability measure from the auxiliary sampling measure.
- Under a constant shift \(\theta_t=c\), write the simulated drift and the likelihood ratio in terms of \(B_T^{\theta}\).
- Why does the likelihood-ratio identity require weighting the payoff, even though the event is more frequent under the sampling measure?
- For \(B_1\ge a\) with \(a>0\), what constant shift makes the event under the auxiliary measure equivalent to a standard normal variable being nonnegative?
- State one reason a shift that increases event frequency might nevertheless increase estimator variance.
- Explain why observing no barrier crossing on a finite time grid is not conclusive for a continuously monitored barrier.