Worked Problems in Girsanov Drift Changes

A Finite-Horizon Change of Measure

On a finite time interval, a Girsanov calculation has a practical sequence: choose the density that produces the desired drift, verify that it defines a probability measure, identify the Brownian motion under that measure, and then compute the quantity of interest using that Brownian motion. The sign in the drift correction is essential: with the convention used here, the new Brownian motion is the old one plus the time integral of the Girsanov kernel.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix a finite horizon $T>0$. An integrand $H$ for an Ito integral $\int_0^t H_s\,dB_s$ is assumed progressively measurable and in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. For the density below, we also require that its kernel $\theta$ is progressively measurable and satisfies the stated Novikov condition.

Definition: Given such a kernel $\theta$, define $$ Z_t=\exp\left(-\int_0^t\theta_s\,dB_s-\frac12\int_0^t\theta_s^2\,ds\right),\qquad 0\le t\le T. $$ Novikov’s condition on $[0,T]$ is $$ \mathbb{E}\left[\exp\left(\frac12\int_0^T\theta_s^2\,ds\right)\right]<\infty. $$ Under this condition, the standard Girsanov theorem says that $Z_t$ is a true martingale with $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$ for $s\le t\le T$, and $\mathbb{E}[Z_T]=1$. Thus $d\mathbb{Q}=Z_T\,d\mathbb{P}$ defines a probability measure equivalent to $\mathbb{P}$ on $\mathcal{F}_T$. Under $\mathbb{Q}$, $$ B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds $$ is Brownian motion. These martingale and Brownian-motion conclusions are the standard Girsanov theorem, established in the earlier material on Girsanov’s theorem. Whenever a stochastic integral appears below, its integrand is progressively measurable and square-integrable under the measure used to define that integral.

Equivalence means that $\mathbb{P}$ and $\mathbb{Q}$ assign probability zero to the same events in $\mathcal{F}_T$. For an integrable random variable $Y$, the change-of-measure formula is $\mathbb{E}^{\mathbb{Q}}[Y]=\mathbb{E}[Z_TY]$. A useful computation often avoids this weighted expectation: rewrite $Y$ using $B^{\mathbb{Q}}$, whose increments under $\mathbb{Q}$ are centred normal variables.

Why Deterministic Kernels Are Easy to Check

Proposition: If $\theta$ is a deterministic square-integrable function on $[0,T]$, then Novikov’s condition holds and $Z_t$ is a martingale on $[0,T]$. In particular, $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$ for $0\le s\le t\le T$.

Proof. Since $\theta$ is deterministic and square-integrable, $a_t=\int_0^t\theta_u^2\,du$ is finite and non-random. Hence $\mathbb{E}[\exp(\tfrac12a_T)]=\exp(\tfrac12a_T)<\infty$, which verifies Novikov’s condition. For $s\le t$, write $$ Z_t=Z_s\exp\left(-\int_s^t\theta_u\,dB_u-\frac12\int_s^t\theta_u^2\,du\right). $$ The integral over $[s,t]$ is independent of $\mathcal{F}_s$ and is normal with mean zero and variance $v=\int_s^t\theta_u^2\,du$. The normal moment-generating function gives the conditional expectation of the exponential factor as $\exp(\tfrac12v-\tfrac12v)=1$. Therefore $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$. Also $Z_t$ is integrable: its expectation is $1$, as follows by taking $s=0$ in the same calculation and using $Z_0=1$. $\square$

This direct argument verifies both integrability and the conditional-expectation identity for deterministic kernels. For general adapted kernels, the theorem’s martingale conclusion is not obtained merely by writing down a stochastic exponential; Novikov’s condition is a sufficient condition that ensures it.

Worked Drift Changes

Worked Example: A Constant Kernel and a Tilted Moment

Take $T=2$ and $\theta_t=\tfrac12$. Find $\mathbb{E}^{\mathbb{Q}}[B_2]$, $\mathbb{E}^{\mathbb{Q}}[B_2^2]$, and $\mathbb{E}^{\mathbb{Q}}[e^{B_2}]$.

First, $\int_0^2\theta_s^2\,ds=\tfrac14\cdot2=\tfrac12$, so the Novikov expectation is $\exp(\tfrac12\cdot\tfrac12)=e^{1/4}<\infty$. The density is $Z_2=\exp(-\tfrac12B_2-\tfrac14)$. Under $\mathbb{Q}$, $B_t^{\mathbb{Q}}=B_t+\tfrac12t$, so $B_2=B_2^{\mathbb{Q}}-1$. Since $B_2^{\mathbb{Q}}\sim N(0,2)$ under $\mathbb{Q}$, its mean is $0$ and its second moment is $2$. Consequently, $$ \mathbb{E}^{\mathbb{Q}}[B_2]=-1,\qquad \mathbb{E}^{\mathbb{Q}}[B_2^2]=2+(-1)^2=3. $$ For the exponential moment, the normal moment-generating function gives $$ \mathbb{E}^{\mathbb{Q}}[e^{B_2}] =e^{-1}\mathbb{E}^{\mathbb{Q}}[e^{B_2^{\mathbb{Q}}}] =e^{-1}e^{2/2}=1. $$ Check: directly, $\mathbb{E}[Z_2]=e^{-1/4}\mathbb{E}[e^{-B_2/2}]=e^{-1/4}e^{(1/4)\cdot2/2}=1$, confirming density normalisation. Also $B_2$ under $\mathbb{Q}$ has mean $-1$ and variance $2$, so its second moment is $2+1=3$, as computed.

Worked Example: Choosing a Kernel to Change an SDE Drift

Under $\mathbb{P}$, let $X_0=0$ and $dX_t=\tfrac32\,dt+2\,dB_t$ for $0\le t\le2$. Choose a constant kernel so that the drift of $X$ under the new measure is $\tfrac12$, and compute its mean and second moment at time $2$.

If $\theta$ is constant, then $dB_t=dB_t^{\mathbb{Q}}-\theta\,dt$. Substitution gives $$ dX_t=\left(\frac32-2\theta\right)dt+2\,dB_t^{\mathbb{Q}}. $$ Setting the drift equal to $\tfrac12$ requires $\tfrac32-2\theta=\tfrac12$, so $2\theta=1$ and $\theta=\tfrac12$. For this choice, Novikov’s expectation on $[0,2]$ is $e^{1/4}$, finite. The changed equation is $dX_t=\tfrac12\,dt+2\,dB_t^{\mathbb{Q}}$, and therefore $X_2=1+2B_2^{\mathbb{Q}}$. It follows that $$ \mathbb{E}^{\mathbb{Q}}[X_2]=1,\qquad \mathbb{E}^{\mathbb{Q}}[X_2^2]=1^2+2^2\cdot2=9. $$ Check: using $B_2^{\mathbb{Q}}=B_2+1$, the original solution gives $X_2=3+2B_2=1+2B_2^{\mathbb{Q}}$, the same expression. Its variance under $\mathbb{Q}$ is $2^2\cdot2=8$, and $8+1^2=9$, agreeing with the second moment.

Worked Example: A Piecewise Kernel and a Cross Moment

On $[0,2]$, let $\theta_t=1$ for $0\le t\le1$ and $\theta_t=-\tfrac12$ for $1<t\le2$. Compute $\mathbb{E}^{\mathbb{Q}}[B_1]$, $\mathbb{E}^{\mathbb{Q}}[B_2]$, $\mathbb{E}^{\mathbb{Q}}[B_1B_2]$, and $\mathbb{E}^{\mathbb{Q}}[B_2-B_1]$.

The squared-kernel integral is $\int_0^2\theta_s^2\,ds=1^2\cdot1+(\tfrac12)^2\cdot1=\tfrac54$. Thus the Novikov expectation is $e^{5/8}<\infty$. The accumulated kernel is $\int_0^1\theta_s\,ds=1$ and $\int_0^2\theta_s\,ds=1-\tfrac12=\tfrac12$. Writing $B_t=B_t^{\mathbb{Q}}-\int_0^t\theta_s\,ds$ yields $$ B_1=B_1^{\mathbb{Q}}-1,\qquad B_2=B_2^{\mathbb{Q}}-\frac12. $$ The means are therefore $\mathbb{E}^{\mathbb{Q}}[B_1]=-1$ and $\mathbb{E}^{\mathbb{Q}}[B_2]=-\tfrac12$. Brownian covariance under $\mathbb{Q}$ gives $\mathbb{E}^{\mathbb{Q}}[B_1^{\mathbb{Q}}B_2^{\mathbb{Q}}]=1$, since the covariance is $\min(1,2)=1$ and both means are zero. Expanding the product, $$ \mathbb{E}^{\mathbb{Q}}[B_1B_2] =1-\frac12\mathbb{E}^{\mathbb{Q}}[B_1^{\mathbb{Q}}] -\mathbb{E}^{\mathbb{Q}}[B_2^{\mathbb{Q}}]+\frac12 =\frac32. $$ Finally, $B_2-B_1=(B_2^{\mathbb{Q}}-B_1^{\mathbb{Q}})+\tfrac12$, so $\mathbb{E}^{\mathbb{Q}}[B_2-B_1]=\tfrac12$. Check: the mean difference is also $-\tfrac12-(-1)=\tfrac12$. The covariance of $B_1$ and $B_2$ remains $1$, since subtracting deterministic constants does not change covariance; their mean product is $( -1)(-\tfrac12)=\tfrac12$, giving the cross moment $1+\tfrac12=\tfrac32$.

A Drift-Selection Rule and Its Justification

Proposition: Suppose under $\mathbb{P}$ a process satisfies $dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$, and select a progressively measurable kernel $\theta_t$ for which Novikov’s condition holds on $[0,T]$. Under $\mathbb{Q}$, its drift and diffusion representation is $$ dX_t=\big(\mu(t,X_t)-\sigma(t,X_t)\theta_t\big)\,dt+\sigma(t,X_t)\,dB_t^{\mathbb{Q}}. $$ Where $\sigma(t,X_t)\ne0$, a desired drift $b(t,X_t)$ is obtained by choosing $\theta_t=(\mu(t,X_t)-b(t,X_t))/\sigma(t,X_t)$, provided this kernel satisfies the required measurability, integrability, and Novikov conditions. Where $\sigma=0$, the drift cannot be altered by this formula.

Proof. By the definition of the transformed Brownian motion, $B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds$, so $dB_t=dB_t^{\mathbb{Q}}-\theta_t\,dt$. Substitute this identity into the given equation: $$ dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\big(dB_t^{\mathbb{Q}}-\theta_t\,dt\big) =\big(\mu(t,X_t)-\sigma(t,X_t)\theta_t\big)\,dt+\sigma(t,X_t)\,dB_t^{\mathbb{Q}}. $$ If $\sigma(t,X_t)\ne0$, solving $\mu-\sigma\theta=b$ gives the stated choice of $\theta$. If $\sigma(t,X_t)=0$, the correction $-\sigma\theta$ is zero, so this substitution cannot change the drift at that point. The Girsanov theorem applies only when the selected kernel meets its hypotheses, including Novikov’s condition here. $\square$

Common Sign and Validity Checks

A frequent error is to use the wrong sign when identifying the transformed Brownian motion. For the constant-kernel example with $\theta=\tfrac12$, the correct relation is $B_t^{\mathbb{Q}}=B_t+\tfrac12t$, hence $B_t=B_t^{\mathbb{Q}}-\tfrac12t$. The incorrect substitution $dB_t=dB_t^{\mathbb{Q}}+\tfrac12dt$ would give drift $\tfrac32+2(\tfrac12)=\tfrac52$ for the SDE in the second example, not the desired $\tfrac12$. Substitution with the correct sign gives $\tfrac32-2(\tfrac12)=\tfrac12$.

A density also has to be a true martingale on the horizon, not merely a positive stochastic exponential. In these examples the kernels are deterministic and square-integrable, so the proposition above verifies the conditional-expectation identity directly; Novikov’s condition also holds. For a general adapted kernel, check Novikov’s condition or another applicable martingale criterion before treating $Z_T$ as a probability density. Finally, distinguish the drift from the diffusion coefficient: this change of measure adjusts the drift by $-\sigma\theta$, while the coefficient of $dB_t^{\mathbb{Q}}$ remains $\sigma$.

Calculation	What to verify
Selecting $\theta$	Substitute into $\mu-\sigma\theta$ and obtain the target drift.
Defining $\mathbb{Q}$	Check Novikov’s condition, or another valid condition ensuring a true density martingale.
Computing moments	Rewrite using $B_t^{\mathbb{Q}}$ and use its Brownian law under $\mathbb{Q}$.
Checking a cross moment	Separate covariance from the product of the changed means.

Check Your Understanding

For each problem, state the measure and the horizon to which your calculation applies.

For the constant kernel $\theta_t=c$ on $[0,T]$, write $Z_T$ and compute the Novikov expectation.
If $dX_t=\alpha\,dt+\sigma\,dB_t$ with $\sigma\ne0$, solve for a constant kernel that changes the drift to $\beta$, and state the sign convention used.
For a deterministic kernel, explain how independence of Brownian increments verifies $\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s$.
Under a changed measure, if $B_T=B_T^{\mathbb{Q}}-a$ and $B_T^{\mathbb{Q}}\sim N(0,T)$, compute $\mathbb{E}^{\mathbb{Q}}[B_T^2]$.
Give one reason a proposed adapted kernel may fail to define a probability measure through the stated exponential density.

Calculation	What to verify
Selecting \(\theta\)	Substitute into \(\mu-\sigma\theta\) and obtain the target drift.
Defining \(\mathbb{Q}\)	Check Novikov’s condition, or another valid condition ensuring a true density martingale.
Computing moments	Rewrite using \(B_t^{\mathbb{Q}}\) and use its Brownian law under \(\mathbb{Q}\).
Checking a cross moment	Separate covariance from the product of the changed means.