Drift Removal by Changing the Brownian Motion
A change of measure can alter the drift seen by a diffusion without changing its paths. The mechanism is a shift of the Brownian motion: under a new probability measure, the shifted process is Brownian, and substituting it into the stochastic differential equation changes the drift term. In one dimension, the required shift is determined by the ratio of drift to volatility.
Fix a finite horizon \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Suppose, on \([0,T]\), a continuous adapted process \(X_t\) satisfies \(dX_t=\mu_t\,dt+\sigma_t\,dB_t\), where \(\mu_t=\mu(t,X_t)\) is progressively measurable and \(\sigma_t=\sigma(t,X_t)\) is progressively measurable. The drift integral is assumed finite almost surely, and initially assume \(\mathbb{E}\int_0^T\sigma_t^2\,dt<\infty\), so the stochastic integral with respect to \(B\) is an Itô integral in \(\mathcal{L}^2\).
Assume that \(\sigma_t\ne0\) for almost every time, almost surely, and define the candidate Girsanov kernel \(\theta_t=\mu_t/\sigma_t\) there (its value on the exceptional set can be set to zero). The finite-horizon condition used below is Novikov’s condition under \(\mathbb{P}\): \(\mathbb{E}\exp\big(\frac12\int_0^T\theta_t^2\,dt\big)<\infty\). In particular, the integral of \(\theta^2\) is finite almost surely, and \(\theta\) is an admissible Itô integrand under \(\mathbb{P}\). Novikov’s condition is a sufficient condition, not a necessary one, for the stochastic exponential below to be a true martingale.
Proof. Novikov’s condition and Novikov’s criterion, a standard result established in the earlier material on stochastic exponentials, imply that \(Z_t\) is a true martingale and has expectation one. Since the exponential defining \(Z_T\) is strictly positive almost surely, the measure \(\mathbb{Q}\) with density \(Z_T\) is a probability measure equivalent to \(\mathbb{P}\). By the standard Girsanov theorem, \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\) is Brownian motion under \(\mathbb{Q}\). The pathwise identity \(dB_t=dB_t^{\mathbb{Q}}-\theta_t\,dt\) gives \(dX_t=(\mu_t-\sigma_t\theta_t)\,dt+\sigma_t\,dB_t^{\mathbb{Q}}\). Since \(\sigma_t\theta_t=\mu_t\) almost everywhere, the drift term vanishes. Equivalence of the measures transfers almost-sure finiteness of \(\int_0^T\sigma_t^2\,dt\) from \(\mathbb{P}\) to \(\mathbb{Q}\); thus the final stochastic integral is well-defined locally. This does not establish \(\mathbb{E}^{\mathbb{Q}}\int_0^T\sigma_t^2\,dt<\infty\), so it does not by itself establish that the integral is in \(\mathcal{L}^2\) under \(\mathbb{Q}\). \(\square\)
Why the Ratio Is Drift Divided by Volatility
The kernel is not guessed from the density formula: it is forced by the cancellation equation. The sign follows from the convention \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\). Replacing that plus sign by a minus sign without changing the density convention reverses the cancellation.
Proof. Substitution of \(dB_t=dB_t^{\mathbb{Q}}-\lambda_t\,dt\) into the original equation yields \(dX_t=(\mu_t-\sigma_t\lambda_t)\,dt+\sigma_t\,dB_t^{\mathbb{Q}}\). If \(\lambda_t=\mu_t/\sigma_t\) where \(\sigma_t\ne0\), then \(\mu_t-\sigma_t\lambda_t=0\). Conversely, if the drift is zero, then \(\mu_t-\sigma_t\lambda_t=0\); division by the nonzero \(\sigma_t\) gives \(\lambda_t=\mu_t/\sigma_t\). At times where \(\sigma_t=0\), changing the Brownian motion cannot alter the drift, because \(\sigma_t\lambda_t=0\). There, drift cancellation is possible only if \(\mu_t=0\). These statements concern the drift coefficients up to the usual almost-everywhere equality in the integral equation. \(\square\)
Worked Example: Constant Drift and Volatility
Let \(X_t=x_0+t+2B_t\) on \([0,2]\). Find the drift-removing kernel and verify the change of measure, including its Novikov bound.
Here \(\mu_t=1\) and \(\sigma_t=2\), so \(\theta_t=1/2\). The Novikov expectation is exactly \(\mathbb{E}\exp\big(\frac12\int_0^2(1/2)^2\,dt\big)=\exp(\frac12\cdot\frac14\cdot2)=e^{1/4}<\infty\). Thus the density is \(Z_2=\exp(-\frac12B_2-\frac14)\), since \(\int_0^2\theta_s\,dB_s=B_2/2\) and \(\frac12\int_0^2\theta_s^2\,ds=1/4\).
Under \(\mathbb{Q}\), \(B_t^{\mathbb{Q}}=B_t+t/2\). Therefore \(B_t=B_t^{\mathbb{Q}}-t/2\), and \(X_t=x_0+t+2(B_t^{\mathbb{Q}}-t/2)=x_0+2B_t^{\mathbb{Q}}\). The time-dependent terms cancel exactly. As a check, \(B_t^{\mathbb{Q}}\) is Brownian under \(\mathbb{Q}\), so \(\mathbb{E}^{\mathbb{Q}}[X_t]=x_0+2\mathbb{E}^{\mathbb{Q}}[B_t^{\mathbb{Q}}]=x_0\), consistent with the drift-free representation.
Worked Example: A Bounded State-Dependent Ratio
Suppose a solution on \([0,2]\) has drift \(\mu(t,X_t)=\sin(X_t)\) and diffusion coefficient \(\sigma(t,X_t)=2+\cos(X_t)\). Check nondegeneracy, compute the kernel, and verify Novikov’s condition.
Since \(-1\le\cos(X_t)\le1\), \(1\le2+\cos(X_t)\le3\), so the volatility is never zero. The kernel is \(\theta_t=\sin(X_t)/(2+\cos(X_t))\). The inequalities \(|\sin(X_t)|\le1\) and \(2+\cos(X_t)\ge1\) give \(|\theta_t|\le1\). Hence, path by path, \(\frac12\int_0^2\theta_t^2\,dt\le\frac12\cdot2=1\), and consequently \(\mathbb{E}\exp\big(\frac12\int_0^2\theta_t^2\,dt\big)\le e<\infty\).
Girsanov therefore gives \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\) as Brownian motion under \(\mathbb{Q}\). Substitution yields \(dX_t=(\sin(X_t)-(2+\cos(X_t))\theta_t)\,dt+(2+\cos(X_t))\,dB_t^{\mathbb{Q}}\). The drift is \(\sin(X_t)-(2+\cos(X_t))\sin(X_t)/(2+\cos(X_t))=0\). The volatility factor is unchanged. This computation checks cancellation along the process; it does not require solving the state-dependent SDE explicitly.
Worked Example: Why Nonzero Volatility Matters
Consider \(X_t=t\), written as \(dX_t=1\,dt+0\,dB_t\), on any finite interval \([0,T]\) with \(T>0\). Can an equivalent change of measure remove its drift?
Here \(\mu_t=1\) and \(\sigma_t=0\), so \(\mu_t/\sigma_t\) is undefined. More fundamentally, any Brownian shift contributes \(\sigma_t\lambda_t=0\) to the drift adjustment. The drift after the shift remains \(1-0\cdot\lambda_t=1\), not zero. Equivalence of measures preserves the almost-sure identity \(X_t=t\); it cannot turn this deterministic path into a process with zero drift. Thus nonzero volatility is essential wherever the original drift is nonzero.
Check: if instead both coefficients were zero, then the drift would already be zero. The ratio \(0/0\) would still not define a kernel, but no drift-removing change would be needed.
Checking the Finite-Horizon Condition
The practical test is performed under \(\mathbb{P}\), using the ratio along the diffusion. A useful sufficient bound is a deterministic one: if \(|\theta_t|\le C\) almost everywhere almost surely on \([0,T]\), then \(\frac12\int_0^T\theta_t^2\,dt\le C^2T/2\). It follows that the Novikov expectation is at most \(e^{C^2T/2}\), which is finite. The bound must hold on the whole horizon under consideration; a check on each short interval does not, without further argument, automatically establish a condition on a longer horizon.
More generally, a deterministic function \(g\) with \(\theta_t^2\le g(t)\) almost everywhere and \(\int_0^Tg(t)\,dt<\infty\) gives the bound \(\mathbb{E}\exp(\frac12\int_0^T\theta_t^2\,dt)\le\exp(\frac12\int_0^Tg(t)\,dt)<\infty\). If the ratio is random and unbounded, its pointwise finiteness is not enough: the exponential expectation itself must be checked or another applicable martingale criterion used. Failure to verify Novikov’s condition means this sufficient test has not established the measure change; it does not, by itself, prove that the stochastic exponential is not a martingale.
Common Sign and Integrability Pitfalls
A common sign error is to set \(B_t^{\mathbb{Q}}=B_t-\int_0^t\theta_s\,ds\) while retaining the kernel \(\theta_t=\mu_t/\sigma_t\). With that convention, \(dB_t=dB_t^{\mathbb{Q}}+\theta_t\,dt\), and the transformed drift becomes \(\mu_t+\sigma_t\theta_t\), which is \(2\mu_t\), not zero. The correction is to use the paired convention in this tutorial: density exponent \(-\int\theta\,dB-\frac12\int\theta^2dt\), Brownian motion \(B^{\mathbb{Q}}=B+\int\theta\,dt\), and kernel \(\theta=\mu/\sigma\).
Another mistake is to infer square-integrability under \(\mathbb{Q}\) merely because the diffusion integral is finite almost surely. Equivalence of \(\mathbb{P}\) and \(\mathbb{Q}\) transfers null events, not necessarily integrability of expectations. Unless the additional \(\mathbb{Q}\)-expectation bound is known, the accurate conclusion is that the integral is locally defined.
Finally, removing drift by Girsanov does not assert strong or weak uniqueness for the SDE. A strong solution is constructed on a specified probability space using its Brownian motion; a weak solution includes the probability space and Brownian motion as part of the solution. Pathwise uniqueness compares solutions driven by the same Brownian motion, whereas uniqueness in law compares their distributions. A change of measure provides a law under which the drift form changes; it does not establish any of these uniqueness properties.
Check Your Understanding
For each question, state the measure and finite-horizon assumptions relevant to your answer.
- For drift \(\mu_t=3\) and volatility \(\sigma_t=2\) on \([0,1]\), find \(\theta_t\) and evaluate the Novikov expectation exactly.
- If \(|\mu(t,X_t)/\sigma(t,X_t)|\le C\) on \([0,T]\), derive a deterministic upper bound for the Novikov expectation.
- Explain why \(\sigma_t=0\) and \(\mu_t\ne0\) prevent drift removal by a Brownian shift.
- With the convention \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\), derive the transformed drift and its cancellation condition.
- What additional assumption is needed to call \(\int_0^T\sigma_t\,dB_t^{\mathbb{Q}}\) square-integrable under \(\mathbb{Q}\), rather than merely locally defined?