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The Girsanov Theorem · Tutorial 302 of 600

Girsanov's Theorem and Sign Conventions

Learn how the sign in an exponential density determines the Brownian motion under the changed measure and the resulting drift of an SDE.

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What You'll Learn

  • State the one-dimensional Girsanov theorem on a finite time horizon.
  • Match the sign in an exponential density to the correct transformed Brownian motion.
  • Distinguish the original measure from the changed measure and state when they are equivalent.
  • Derive the drift adjustment for an SDE under the changed measure.
  • Check sign conventions by rewriting the original Brownian motion in terms of the transformed one.
  • Identify common errors involving the density exponent and drift coefficient.

Why the Sign Convention Matters

A change of probability measure alters which continuous process is Brownian, even though the underlying sample paths have not changed. The sign is fixed by the density: with an exponential density whose stochastic-integral term is negative, the Brownian motion under the new measure is the original Brownian motion plus the integrated kernel. Reversing either sign without changing the other gives an incorrect drift adjustment.

Fix \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) satisfying the usual conditions, carrying a Brownian motion \(B_t\) with respect to this filtration. Let \(\theta_t\) be progressively measurable and satisfy \(\mathbb{E}\int_0^T\theta_s^2\,ds<\infty\). All integrals with respect to \(B\) below are Ito integrals; their integrands are progressively measurable and square-integrable on the stated horizon. In particular, \(\theta\) belongs to \(\mathcal{L}^2\) on \([0,T]\). For an integral with respect to a Brownian motion under a changed measure, the corresponding integrand must likewise be progressively measurable and square-integrable under that measure; local versions can be used after stopping when only pathwise local integrability is available.

Definition: Novikov’s condition on \([0,T]\) is \(\mathbb{E}\left[\exp\left(\tfrac12\int_0^T\theta_s^2\,ds\right)\right]<\infty\). Under this condition, define \(Z_t=\exp\left(-\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds\right)\) for \(0\le t\le T\). The density process \(Z_t\) gives the measure \(\mathbb{Q}\) on \(\mathcal{F}_T\) by \(d\mathbb{Q}=Z_T\,d\mathbb{P}\).

Novikov’s criterion, a standard result established in the earlier material on exponential martingales, says that \(Z_t\) is a true \(\mathbb{P}\)-martingale and \(\mathbb{E}[Z_T]=1\). Thus \(\mathbb{Q}\) is a probability measure. The exponential is strictly positive whenever it is defined, so \(Z_T>0\) almost surely; consequently \(\mathbb{Q}\) and \(\mathbb{P}\) are equivalent on \(\mathcal{F}_T\). This distinction matters: \(B\) is Brownian under \(\mathbb{P}\), whereas the theorem identifies a generally different process as Brownian under \(\mathbb{Q}\).

The One-Dimensional Girsanov Theorem

Theorem: Under the assumptions above, including Novikov’s condition, the process \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\), for \(0\le t\le T\), is a Brownian motion with respect to \((\mathcal{F}_t)\) under \(\mathbb{Q}\). In particular, its increments are independent of the past under \(\mathbb{Q}\), and \(B_t^{\mathbb{Q}}-B_s^{\mathbb{Q}}\) has distribution \(N(0,t-s)\) for \(0\le s<t\le T\).

The sign can be read directly from the pair of formulas:

$$ Z_t=\exp\left(-\int_0^t\theta_s\,dB_s-\frac12\int_0^t\theta_s^2\,ds\right), \qquad B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds. $$

The theorem is the standard one-dimensional Girsanov theorem. Its proof by martingale characterization is given in the proof of Girsanov’s theorem; here the point is to state the hypotheses and make the associated signs unambiguous. The density is a true martingale because Novikov’s condition is assumed, not merely a local martingale. Under \(\mathbb{Q}\), \(B^{\mathbb{Q}}\) is a true martingale: Brownian increments are integrable and have conditional mean zero, so \(\mathbb{E}^{\mathbb{Q}}[B_t^{\mathbb{Q}}\mid\mathcal{F}_s]=B_s^{\mathbb{Q}}\) for \(s\le t\).

Sign rule: If the density contains \(-\int\theta\,dB\), add \(\int\theta\,ds\) to \(B\) to obtain the Brownian motion under \(\mathbb{Q}\). Equivalently, rewrite \(dB_t=dB_t^{\mathbb{Q}}-\theta_t\,dt\).

The converse sign convention follows by applying the theorem to the kernel \(-\theta\). If \(\exp\left(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds\right)\) defines the density, then the transformed Brownian motion is \(B_t-\int_0^t\theta_s\,ds\). The sign in the density’s stochastic-integral term and the sign in the Brownian adjustment change together.

How an SDE’s Drift Changes

The drift coefficient is the coefficient of \(dt\); the diffusion coefficient is the coefficient of \(dB_t\). Suppose an adapted continuous process \(X_t\) satisfies, under \(\mathbb{P}\), \(dX_t=a_t\,dt+c_t\,dB_t\). Here \(a_t\) and \(c_t\) are progressively measurable, and the integrals are well-defined on \([0,T]\) (for example, \(c\in\mathcal{L}^2\) under \(\mathbb{P}\), with the required local integrability under \(\mathbb{Q}\)). Substitution of \(dB_t=dB_t^{\mathbb{Q}}-\theta_tdt\) gives the drift adjustment.

Proposition: With the preceding assumptions, the same process can be written under \(\mathbb{Q}\) as \(dX_t=(a_t-c_t\theta_t)\,dt+c_t\,dB_t^{\mathbb{Q}}\), using local stochastic integrals if necessary. Thus the new drift is the old drift minus the diffusion coefficient multiplied by the Girsanov kernel.

Proof. By definition, \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\), so \(B_t=B_t^{\mathbb{Q}}-\int_0^t\theta_s\,ds\). Substituting this identity into the integral form of the SDE gives

$$ X_t=X_0+\int_0^t a_s\,ds+\int_0^t c_s\,dB_s =X_0+\int_0^t(a_s-c_s\theta_s)\,ds+\int_0^t c_s\,dB_s^{\mathbb{Q}}. $$

The substitution is valid first for bounded, simple integrands by linearity of the sums defining the integrals. For progressively measurable integrands for which the displayed integrals exist, it extends by the defining limits of stochastic integration, or by localization when the integrals are only locally defined. The finite-variation term is the ordinary time integral \(\int c_s\theta_s\,ds\). This proves the stated drift and diffusion coefficients. \(\square\)

This algebra does not assert that every chosen \(\theta\) is admissible: the density and measure change require the hypotheses of the theorem. Nor does changing measure change the pathwise coefficient \(c_t\) in this representation. It changes the drift term when the original Brownian motion is expressed using the Brownian motion under \(\mathbb{Q}\).

Worked Sign Checks

Worked Example: Negative Exponent, Positive Brownian Adjustment

Take the constant kernel \(\theta_t=2\) on \([0,T]\). It is progressively measurable and \(\mathbb{E}\int_0^T\theta_s^2\,ds=4T<\infty\). Novikov’s condition holds because \(\mathbb{E}\left[\exp(\tfrac12\int_0^T4\,ds)\right]=\exp(2T)<\infty\). The density and transformed Brownian motion are therefore

$$ Z_t=\exp(-2B_t-2t),\qquad B_t^{\mathbb{Q}}=B_t+2t. $$

Solving the second equation for the original Brownian motion gives \(B_t=B_t^{\mathbb{Q}}-2t\), or \(dB_t=dB_t^{\mathbb{Q}}-2\,dt\). Under \(\mathbb{Q}\), the theorem says \(B^{\mathbb{Q}}\), not \(B\), is Brownian. A mistaken choice \(B_t-2t\) reverses the adjustment: in terms of the actual Brownian motion it equals \(B_t^{\mathbb{Q}}-4t\), which has a nonzero deterministic drift and is not the process asserted to be Brownian by the theorem.

Check: Substituting \(B_t=B_t^{\mathbb{Q}}-2t\) into \(B_t+2t\) gives \(B_t^{\mathbb{Q}}-2t+2t=B_t^{\mathbb{Q}}\), confirming the sign.

Worked Example: Adjusting the Drift of an SDE

Suppose \(dX_t=3\,dt+2\,dB_t\), and take \(\theta_t=\tfrac12\) on \([0,T]\). The kernel is in \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^T(\tfrac12)^2ds=\tfrac14T\). Novikov’s condition is satisfied: its left-hand side is \(\exp(\tfrac18T)\), which is finite. The theorem gives \(dB_t=dB_t^{\mathbb{Q}}-\tfrac12dt\). Substitution yields

$$ dX_t=3\,dt+2\left(dB_t^{\mathbb{Q}}-\frac12\,dt\right) =(3-1)\,dt+2\,dB_t^{\mathbb{Q}} =2\,dt+2\,dB_t^{\mathbb{Q}}. $$

The transformed drift is \(a-c\theta=3-2(\tfrac12)=2\); the diffusion coefficient remains \(2\). To check by reversing the substitution, \(dB_t^{\mathbb{Q}}=dB_t+\tfrac12dt\), so \(2dt+2dB_t^{\mathbb{Q}}=2dt+2dB_t+dt=3dt+2dB_t\), the original SDE.

Worked Example: Choosing the Kernel to Cancel a Drift

Consider \(dX_t=-3\,dt+2\,dB_t\). To make the transformed drift zero, solve \(a-c\theta=0\): here \(-3-2\theta=0\), so \(\theta=-\tfrac32\). This kernel is in \(\mathcal{L}^2\), with \(\mathbb{E}\int_0^T\theta_s^2ds=\tfrac94T\). Novikov’s condition holds because its expectation is \(\exp(\tfrac98T)<\infty\). The density has the corresponding sign:

$$ Z_t=\exp\left(\frac32B_t-\frac98t\right),\qquad B_t^{\mathbb{Q}}=B_t-\frac32t. $$

Since \(dB_t=dB_t^{\mathbb{Q}}+\tfrac32dt\), substitution gives

$$ dX_t=-3\,dt+2\left(dB_t^{\mathbb{Q}}+\frac32\,dt\right) =(-3+3)\,dt+2\,dB_t^{\mathbb{Q}} =2\,dB_t^{\mathbb{Q}}. $$

Check: the drift computed from the proposition is \(-3-2(-\tfrac32)=-3+3=0\), as required. The positive stochastic-integral term in \(Z_t\) pairs with the negative adjustment \(B_t-\tfrac32t\); changing just one of those signs would not give this cancellation.

Common Mistakes and a Compact Reference

A frequent error is to use the sign from one convention in the density and the sign from another convention in the transformed Brownian motion. For example, with density \(\exp(-\int\theta\,dB-\tfrac12\int\theta^2ds)\), writing \(B^{\mathbb{Q}}=B-\int\theta ds\) is incorrect. It fails the defining relation \(B=B^{\mathbb{Q}}-\int\theta ds\) used to rewrite the original SDE. The correction is to use \(B^{\mathbb{Q}}=B+\int\theta ds\), and then \(dB=dB^{\mathbb{Q}}-\theta dt\).

Also distinguish a true martingale from a local martingale. Novikov’s condition is a sufficient condition ensuring that the exponential density is a true martingale with expectation one; without a condition that guarantees this, a positive stochastic exponential may be only a local martingale and may fail to have expectation one. In that case it cannot simply be used as a probability density. The theorem’s hypotheses are therefore part of the sign convention’s application, not optional technical decoration.

Density exponentBrownian motion under the changed measureRewriting \(dB_t\)New SDE drift
\(-\int\theta\,dB-\tfrac12\int\theta^2dt\)\(B_t+\int_0^t\theta_sds\)\(dB_t=dB_t^{\mathbb{Q}}-\theta_tdt\)\(a_t-c_t\theta_t\)
\(+\int\theta\,dB-\tfrac12\int\theta^2dt\)\(B_t-\int_0^t\theta_sds\)\(dB_t=dB_t^{\mathbb{Q}}+\theta_tdt\)\(a_t+c_t\theta_t\)

The table uses \(\mathbb{Q}\) for the measure defined by the indicated density, provided the relevant Novikov condition holds. Its second row follows from applying the first convention with kernel \(-\theta\). The general check is mechanical: identify the kernel from the density, write \(B\) in terms of the new Brownian motion, and only then substitute into the SDE.

Check Your Understanding

For each question, state which measure is being used and check the sign by rewriting the original Brownian motion.

  1. If the density is \(\exp(-\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2ds)\), what process is Brownian under the changed measure?
  2. For \(dX_t=a_tdt+c_tdB_t\) under \(\mathbb{P}\), what are the drift and diffusion coefficients under the measure in the first row of the table?
  3. With constant \(\theta=-1\), write \(B_t\) in terms of \(B_t^{\mathbb{Q}}\) when the density has a negative stochastic-integral term.
  4. State Novikov’s condition and explain why it matters for defining a probability measure using the exponential density.
  5. For a density with \(+\int\theta\,dB\), determine the Brownian adjustment and the transformed drift of \(dX_t=a_tdt+c_tdB_t\).