Brownian Drift Changes Under Girsanov

Removing a Constant Brownian Drift

Changing measure can make a process with drift into Brownian motion. The drift does not disappear from the underlying paths; rather, the probability assigned to each path changes. Keeping track of the direction of the change of measure is essential: the density that removes a drift has a sign determined by the process we want to become Brownian.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$, and fix a finite horizon $T>0$. For a constant $\mu\in\mathbb{R}$, define $X_t=B_t+\mu t$ for $0\le t\le T$. Under $\mathbb{P}$, $X$ has drift $\mu$ and diffusion coefficient $1$. It is given pathwise by the Brownian motion and deterministic drift, so it is a strong solution of $dX_t=\mu\,dt+dB_t$ with $X_0=0$.

Define a process on this horizon by

Z_t=\exp\left(-\mu B_t-\frac12\mu^2t\right),\qquad 0\le t\le T.

We will use $Z_T$ as a density to define $\mathbb{Q}$ on $\mathcal{F}_T$. Before doing so, we verify that it is positive, has expectation one, and satisfies the stated integrability condition.

Definition: For this constant-drift change, define $\mathbb{Q}$ on $\mathcal{F}_T$ by $\dfrac{d\mathbb{Q}}{d\mathbb{P}}\big|_{\mathcal{F}_T}=Z_T$. If $Z_t=\mathbb{E}[Z_T\mid\mathcal{F}_t]$, then $Z$ is the density process. The measures $\mathbb{P}$ and $\mathbb{Q}$ are equivalent on $\mathcal{F}_T$ when $Z_T>0$ almost surely.

Checking the Density and Novikov’s Condition

The stochastic exponential has the integral representation $Z_t=\mathcal{E}(M)_t$, where $M_t=-\int_0^t\mu\,dB_s=-\mu B_t$. Here the integrand $H_s=-\mu$ is adapted and progressively measurable, and $\mathbb{E}\int_0^T H_s^2\,ds=\mu^2T<\infty$; thus $H$ belongs to the class $\mathcal{L}^2$. The Ito integral is defined for such integrands. The exponential is strictly positive for every finite $t$.

Novikov’s condition for a progressively measurable integrand $\theta_s$ on $[0,T]$ is $\mathbb{E}\exp\big(\tfrac12\int_0^T\theta_s^2\,ds\big)<\infty$. It is a sufficient condition for the associated stochastic exponential to be a true martingale, rather than merely a local martingale. In the present case the kernel is $\theta_s=\mu$, so

\mathbb{E}\exp\left(\frac12\int_0^T\mu^2\,ds\right) =\exp\left(\frac12\mu^2T\right)<\infty.

For this constant kernel, we can also verify the martingale property directly, including its conditional-expectation requirement.

Proposition: The process $Z_t=\exp(-\mu B_t-\tfrac12\mu^2t)$ is a true martingale under $\mathbb{P}$ on $[0,T]$, and $\mathbb{E}[Z_t]=1$ for every $t\in[0,T]$.

Proof. Fix $0\le s\le t\le T$. Factor the density at time $t$ as

Z_t=Z_s\exp\left(-\mu(B_t-B_s)-\frac12\mu^2(t-s)\right).

The increment $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. For a centered normal random variable $Y$ with variance $v$, its moment-generating function gives $\mathbb{E}[e^{aY}]=e^{a^2v/2}$. Therefore the conditional expectation of the second factor is $e^{\mu^2(t-s)/2}e^{-\mu^2(t-s)/2}=1$. It follows that $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$. Taking $s=0$, with $Z_0=1$, gives $\mathbb{E}[Z_t]=1$; in particular, $Z_T$ is integrable and defines a probability measure $\mathbb{Q}$. This direct computation proves a true-martingale property, not just a local-martingale property. $\square$

Worked Example: The Density at a Specified Horizon

Take $\mu=\tfrac12$ and $T=2$. Compute the terminal density, the Novikov quantity, and the mean of the density.

Substitution into the definition gives

Z_2=\exp\left(-\frac12 B_2-\frac12\left(\frac12\right)^2(2)\right) =\exp\left(-\frac12B_2-\frac14\right).

For Novikov’s condition, $\tfrac12\mu^2T=\tfrac12\cdot\tfrac14\cdot2=\tfrac14$, so its left-hand side is $e^{1/4}<\infty$. Independently, since $B_2\sim N(0,2)$,

\mathbb{E}[Z_2] =e^{-1/4}\mathbb{E}[e^{-B_2/2}] =e^{-1/4}\exp\left(\frac{(-1/2)^2\cdot2}{2}\right) =e^{-1/4}e^{1/4}=1.

Check: the Gaussian moment-generating-function calculation gives exactly the normalization required for $Z_2$ to be a probability density. The conditional-expectation calculation above verifies the stronger statement that the densities at all intermediate times form a martingale.

Which Process Becomes Brownian?

Apply the standard Girsanov theorem on the finite horizon $[0,T]$. In the version used here, if a progressively measurable kernel $\theta$ satisfies Novikov’s condition and $d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_T}=\mathcal{E}(-\int_0^T\theta_s\,dB_s)_T$, then $B_t+\int_0^t\theta_s\,ds$ is Brownian motion under $\mathbb{Q}$. The proof of this general change-of-measure theorem is beyond this tutorial and is treated later in the course. Here $\theta_s=\mu$, the integral is well-defined since the constant integrand is in $\mathcal{L}^2$, and Novikov’s condition was checked above. Consequently,

B_t^{\mathbb{Q}}:=B_t+\mu t $$ $$ \text{is Brownian motion under }\mathbb{Q}\text{ for }0\le t\le T.

The filtration is unchanged; the probability measure is not. By construction $X_t=B_t+\mu t=B_t^{\mathbb{Q}}$, so the drifted process $X$ is Brownian under $\mathbb{Q}$. This is a change in law, not a claim that the same paths have become drift-free under $\mathbb{P}$. No issue of pathwise uniqueness or uniqueness in law is needed for this explicit process: $X$ is defined directly from $B$, and the stated Brownian law is under $\mathbb{Q}$.

Worked Example: Checking the Drift Sign

Take $\mu=-\tfrac13$ and $T=3$. Find the density and the Brownian motion under the changed measure.

In this case, $X_t=B_t-\tfrac13t$. Substituting $\mu=-\tfrac13$ into the density gives

Z_t=\exp\left(\frac13B_t-\frac12\left(\frac19\right)t\right) =\exp\left(\frac13B_t-\frac{t}{18}\right).

Novikov’s exponent is $\tfrac12\cdot\tfrac19\cdot3=\tfrac16$, so the required expectation is $e^{1/6}<\infty$. Girsanov’s theorem gives $B_t^{\mathbb{Q}}=B_t-\tfrac13t$, which is exactly $X_t$. At the terminal time, $Z_3=\exp(\tfrac13B_3-\tfrac16)$. Since $B_3\sim N(0,3)$ under $\mathbb{P}$, its mean is

\mathbb{E}[Z_3] =e^{-1/6}\exp\left(\frac{(1/3)^2\cdot3}{2}\right) =e^{-1/6}e^{1/6}=1.

Check: the drift-adjusted process has the same sign as the drift in $X_t=B_t-\tfrac13t$. The density has a positive coefficient on $B_t$, as required by $Z_t=\exp(-\mu B_t-\tfrac12\mu^2t)$ when $\mu=-\tfrac13$.

Transforming a Terminal-Event Probability

For any event $A\in\mathcal{F}_T$, the definition of the Radon–Nikodym derivative gives $\mathbb{P}(A)=\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A\,d\mathbb{P}/d\mathbb{Q}]$. Since $Z_T>0$, the inverse density is $d\mathbb{P}/d\mathbb{Q}|_{\mathcal{F}_T}=1/Z_T$. Substituting $B_T=B_T^{\mathbb{Q}}-\mu T$ into this inverse density yields the following formula.

Proposition: For every $A\in\mathcal{F}_T$, $$ \mathbb{P}(A)=\mathbb{E}^{\mathbb{Q}}\left[\mathbf{1}_A\exp\left(\mu B_T^{\mathbb{Q}}-\frac12\mu^2T\right)\right]. $$ In particular, for a Borel set $C\subseteq\mathbb{R}$, take $A=\{X_T\in C\}$ to obtain $$ \mathbb{P}(X_T\in C)=\mathbb{E}^{\mathbb{Q}}\left[\mathbf{1}_{\{B_T^{\mathbb{Q}}\in C\}}\exp\left(\mu B_T^{\mathbb{Q}}-\frac12\mu^2T\right)\right]. $$

Proof. By the definition of $\mathbb{Q}$, $d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_T}=Z_T$. Strict positivity implies $d\mathbb{P}/d\mathbb{Q}|_{\mathcal{F}_T}=Z_T^{-1}$. Hence $\mathbb{P}(A)=\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_A Z_T^{-1}]$. Since $B_T=B_T^{\mathbb{Q}}-\mu T$, direct substitution gives $$ Z_T^{-1}=\exp\left(\mu B_T+\frac12\mu^2T\right) =\exp\left(\mu B_T^{\mathbb{Q}}-\mu^2T+\frac12\mu^2T\right) =\exp\left(\mu B_T^{\mathbb{Q}}-\frac12\mu^2T\right). $$ This proves the first identity. For $A=\{X_T\in C\}$, the pathwise identity $X_T=B_T^{\mathbb{Q}}$ proves the second. $\square$

Worked Example: Reweighting an Interval Event

Let $\mu=1$, $T=1$, and $A=\{X_1\in[0,1]\}$. Express $\mathbb{P}(A)$ under $\mathbb{Q}$, simplify the integral, and check it from the distribution under $\mathbb{P}$.

Under $\mathbb{Q}$, $B_1^{\mathbb{Q}}\sim N(0,1)$ and $X_1=B_1^{\mathbb{Q}}$. The event formula gives

\mathbb{P}(A) =\mathbb{E}^{\mathbb{Q}}\left[\mathbf{1}_{\{0\le B_1^{\mathbb{Q}}\le1\}}e^{B_1^{\mathbb{Q}}-1/2}\right] =\int_0^1 e^{x-1/2}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx.

For every $x$, $e^{x-1/2}e^{-x^2/2}=e^{-(x-1)^2/2}$, because $x-\tfrac12-\tfrac12x^2=-\tfrac12(x-1)^2$. Thus

\mathbb{P}(A)=\int_0^1\frac{1}{\sqrt{2\pi}}e^{-(x-1)^2/2}\,dx =\Phi(0)-\Phi(-1).

Check: under $\mathbb{P}$, $X_1=B_1+1$ has distribution $N(1,1)$. Therefore $\mathbb{P}(0\le X_1\le1)=\Phi((1-1)/1)-\Phi((0-1)/1)=\Phi(0)-\Phi(-1)$, agreeing with the reweighted calculation. The interval includes both endpoints; for this continuous distribution, changing endpoint inclusion does not change the probability.

Common Sign Error

A frequent mistake is to use $Z_T=\exp(+\mu B_T-\tfrac12\mu^2T)$ while intending to make $B_t+\mu t$ Brownian under $\mathbb{Q}$. That density corresponds to the kernel $\theta=-\mu$ in Girsanov’s theorem, so it makes $B_t-\mu t$, not $B_t+\mu t$, Brownian under the changed measure. For the intended process $X_t=B_t+\mu t$, the correct density is $Z_T=\exp(-\mu B_T-\tfrac12\mu^2T)$.

The two measures also have distinct roles in the probability identity: $\mathbb{P}$ is the original measure and $\mathbb{Q}$ is defined by the density $Z_T=d\mathbb{Q}/d\mathbb{P}$. To express an original probability as an expectation under $\mathbb{Q}$, use the inverse density $1/Z_T$, not $Z_T$. The exponential factor in the terminal-event formula is precisely this inverse density rewritten in terms of the Brownian motion under $\mathbb{Q}$.

Check Your Understanding

Use the density convention and event identity above to test the sign, normalization, and measure in each step.

For constant $\mu$, compute $\mathbb{E}[Z_T]$ directly from the moment-generating function of $B_T$.
If $X_t=B_t-\nu t$ for a constant $\nu$, what density makes $X$ Brownian under the changed measure?
State Novikov’s condition for a deterministic constant kernel and verify it on a finite horizon.
For $A\in\mathcal{F}_T$, explain why transforming $\mathbb{P}(A)$ into an expectation under $\mathbb{Q}$ uses $d\mathbb{P}/d\mathbb{Q}$.
In the interval example, verify the event formula by standardizing the $N(1,1)$ terminal distribution under $\mathbb{P}$.