Proof of Girsanov's Theorem by Martingale Characterization

The Change of Measure and the Drift-Adjusted Process

Girsanov’s theorem identifies what happens to Brownian motion when probability is changed using a stochastic exponential. The central proof idea is to combine the density process with a candidate drift-adjusted process: under the original measure, their product has no drift. A localization argument then turns that cancellation into a local martingale statement under the new measure. Finally, quadratic variation identifies the resulting process as Brownian motion.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$, with the filtration satisfying the usual conditions. Fix $T>0$. Let $\theta_t$ be progressively measurable and satisfy $\mathbb{E}\int_0^T\theta_t^2\,dt<\infty$, so that its Ito integral is defined. All stochastic integrals below are Ito integrals. In particular, no Stratonovich correction is being used.

Theorem: Suppose Novikov’s condition holds: $$ \mathbb{E}\left[\exp\left(\frac12\int_0^T\theta_t^2\,dt\right)\right]<\infty. $$ Set $$ Z_t=\exp\left(-\int_0^t\theta_s\,dB_s-\frac12\int_0^t\theta_s^2\,ds\right),\qquad \widetilde B_t=B_t+\int_0^t\theta_s\,ds. $$ Then $Z_t$ is a strictly positive martingale with $\mathbb{E}[Z_T]=1$. Define $\mathbb{Q}$ on $\mathcal{F}_T$ by $d\mathbb{Q}=Z_T\,d\mathbb{P}$. Under $\mathbb{Q}$, $\widetilde B_t$, $0\le t\le T$, is a Brownian motion.

Novikov’s condition is a sufficient condition for the stochastic exponential $Z$ to be a true martingale; it is not necessary. It also implies the required square integrability of $\theta$, since $x\le e^{x/2}$ for $x\ge0$, applied to $x=\int_0^T\theta_t^2dt$. Strict positivity of the displayed exponential gives equivalence of $\mathbb{P}$ and $\mathbb{Q}$ on $\mathcal{F}_T$: they have the same null events. The sign convention here is important: the density uses $-\int\theta\,dB$, so the process that becomes Brownian is $B+\int\theta\,dt$.

Drift Cancellation Under the Original Measure

First compute the dynamics of the density. Put $U_t=-\int_0^t\theta_s\,dB_s-\frac12\int_0^t\theta_s^2ds$, so $Z_t=e^{U_t}$. The quadratic variation is $[U]_t=\int_0^t\theta_s^2ds$. For $f(u)=e^u$, $f'(u)=e^u$ and $f''(u)=e^u$; Ito’s formula therefore gives

dZ_t=Z_t\,dU_t+\frac12 Z_t\,d[U]_t =Z_t\left(-\theta_t\,dB_t-\frac12\theta_t^2dt\right) +\frac12Z_t\theta_t^2dt =-Z_t\theta_t\,dB_t.

Also $d\widetilde B_t=dB_t+\theta_tdt$. The Ito product rule for $Z_t\widetilde B_t$ has cross-variation $d[Z,\widetilde B]_t=-Z_t\theta_tdt$, because the finite-variation part $\int\theta\,dt$ has zero covariation with a continuous semimartingale. Thus

d(Z_t\widetilde B_t) =Z_t\,d\widetilde B_t+\widetilde B_t\,dZ_t+d[Z,\widetilde B]_t =Z_t(1-\widetilde B_t\theta_t)\,dB_t.

For completeness, the product rule is Ito’s formula for $f(z,y)=zy$: $f_z=y$, $f_y=z$, $f_{zz}=f_{yy}=0$, and $f_{zy}=1$. Its second-order term is precisely $d[Z,\widetilde B]$, which cancels the drift $Z_t\theta_tdt$. The displayed differential shows that $Z\widetilde B$ is a continuous local martingale under $\mathbb{P}$. It need not be a true martingale: the stochastic integrand $Z_t(1-\widetilde B_t\theta_t)$ is not asserted to be square integrable over the whole interval. The local martingale property is the statement needed for the change-of-measure argument.

The Localization Step

Lemma: Let $Z_t$ be a strictly positive, continuous, true $\mathbb{P}$-martingale on $[0,T]$ with $Z_0=1$, and define $\mathbb{Q}$ by $d\mathbb{Q}=Z_Td\mathbb{P}$. If $X_t$ is a continuous semimartingale with bounded $X_0$, and $Z_tX_t$ is a $\mathbb{P}$-local martingale, then $X_t$ is a $\mathbb{Q}$-local martingale.

Proof. Choose a deterministic constant $C$ such that $|X_0|\le C$ almost surely. For integers $n>C$, let $\tau_n=\inf\{t\ge0:|X_t|\ge n\}$, with the infimum of the empty set taken as infinity. Continuity implies $|X_{t\wedge\tau_n}|\le n$: before the first hitting time the absolute value is below $n$, at a finite hitting time it equals $n$, and at time zero it is at most $C<n$. Thus this localization does not assume that an unbounded random initial value is below a deterministic threshold.

The process $Z_tX_{t\wedge\tau_n}$ is a $\mathbb{P}$-local martingale. Up to $\tau_n$ it agrees with $ZX$; after $\tau_n$ it is $X_{\tau_n}Z_t$, a bounded $\mathcal{F}_{\tau_n}$-measurable multiplier of the martingale $Z$. Moreover, its absolute value is bounded by $nZ_t$. The family $\{Z_t:0\le t\le T\}$ is uniformly integrable, since $Z_t=\mathbb{E}[Z_T\mid\mathcal{F}_t]$ and $Z_T$ is integrable. Hence the stopped local martingale $Z_tX_{t\wedge\tau_n}$, dominated by this uniformly integrable family times $n$, is a true martingale.

For $0\le s\le t\le T$, Bayes’ conditional-expectation formula and the fact that $X_{t\wedge\tau_n}$ is $\mathcal{F}_t$-measurable give $$ \mathbb{E}^{\mathbb{Q}}[X_{t\wedge\tau_n}\mid\mathcal{F}_s] =\frac{\mathbb{E}[Z_T X_{t\wedge\tau_n}\mid\mathcal{F}_s]}{Z_s} =\frac{\mathbb{E}[Z_t X_{t\wedge\tau_n}\mid\mathcal{F}_s]}{Z_s} =X_{s\wedge\tau_n}. $$ The last equality uses the true-martingale property just established. Thus each stopped process is a $\mathbb{Q}$-martingale. Since continuous paths are bounded on every compact time interval, $\tau_n$ tends to infinity almost surely as $n$ increases. The stopped martingales therefore show that $X$ is a $\mathbb{Q}$-local martingale. $\square$

Apply the lemma with $X=\widetilde B$. Its initial value is $0$, so it meets the bounded-initial-value hypothesis. The product calculation established that $Z\widetilde B$ is a $\mathbb{P}$-local martingale. Consequently $\widetilde B$ is a continuous local martingale under $\mathbb{Q}$, starting at zero.

Quadratic Variation and Brownian Identification

The added process $\int_0^t\theta_sds$ has continuous paths of finite variation on $[0,T]$, almost surely. A continuous finite-variation process has zero quadratic variation and zero covariation with a continuous semimartingale. Therefore, pathwise,

[\widetilde B]_t =\left[B+\int_0^\cdot\theta_sds\right]_t =[B]_t=t,\qquad 0\le t\le T.

The equality $[B]_t=t$ holds $\mathbb{P}$-almost surely by Brownian quadratic variation. Since $\mathbb{Q}$ is equivalent to $\mathbb{P}$, the same pathwise equality holds $\mathbb{Q}$-almost surely. This quadratic-variation statement is distinct from an ordinary chain rule: in Ito multiplication notation $dB_t\,dB_t=dt$, while $dB_t\,dt=0$ and $dt\,dt=0$.

Standard result (Lévy’s characterization): A continuous local martingale starting at zero whose quadratic variation is $t$ is a Brownian motion with respect to the given filtration. This is the standard martingale characterization of Brownian motion; its proof is not repeated here.

All hypotheses of this characterization have now been checked under $\mathbb{Q}$: $\widetilde B_0=0$, its paths are continuous, it is a $\mathbb{Q}$-local martingale, and $[\widetilde B]_t=t$. Hence it is Brownian motion under $\mathbb{Q}$. The proof does not claim that $\widetilde B$ is a true martingale merely from being a local martingale; Brownian identification follows from the stated characterization.

Worked Applications of the Proof

Worked Example: A Bounded State-Dependent Kernel

Take $T=1$ and $\theta_t=\sin(B_t)$. This process is progressively measurable, and $\theta_t^2\le1$, so $$ \mathbb{E}\int_0^1\theta_t^2dt\le1,\qquad \mathbb{E}\exp\left(\frac12\int_0^1\theta_t^2dt\right)\le e^{1/2}. $$ Thus the Ito integral is defined and Novikov’s condition holds. The density and adjusted process are $$ Z_t=\exp\left(-\int_0^t\sin(B_s)\,dB_s-\frac12\int_0^t\sin^2(B_s)\,ds\right),\qquad \widetilde B_t=B_t+\int_0^t\sin(B_s)\,ds. $$ For the drift check, $dZ_t=-Z_t\sin(B_t)dB_t$ and $d\widetilde B_t=dB_t+\sin(B_t)dt$; their product’s drift is $Z_t\sin(B_t)dt-Z_t\sin(B_t)dt=0$. Also $[\widetilde B]_1=[B]_1=1$. The theorem therefore says $\widetilde B$ is Brownian under the measure with density $Z_1$. Check: the bound on the Novikov expectation is $e^{1/2}$, not an equality in general, because $\sin^2(B_t)$ need not equal $1$.

Worked Example: A Predictable Kernel on a Random Event

Let $s=\tfrac14$, $v=\tfrac34$, $A=\{B_{1/4}\ge0\}$, and $\theta_t=2\mathbf{1}_A\mathbf{1}_{(s,v]}(t)$ on $[0,1]$. The event $A$ belongs to $\mathcal{F}_s$, so this is a progressively measurable simple integrand. Since $B_{1/4}$ is a centered nondegenerate normal variable, $\mathbb{P}(A)=\tfrac12$. Its squared-integral and Novikov expectation are $$ \int_0^1\theta_t^2dt=4\left(\frac34-\frac14\right)\mathbf{1}_A=2\mathbf{1}_A,\qquad \mathbb{E}e^{\frac12\int_0^1\theta_t^2dt} =\frac12+\frac12e=\frac{1+e}{2}<\infty. $$ For $g_t=(t\wedge v-s)^+$ and $D_t=B_{(t\wedge v)\vee s}-B_s$, direct integration gives $$ Z_t=\exp(-2\mathbf{1}_A D_t-2\mathbf{1}_A g_t),\qquad \widetilde B_t=B_t+2\mathbf{1}_A g_t. $$ Here $g_t=0$ before $s$, increases at rate $1$ from $s$ to $v$, and equals $\tfrac12$ thereafter. On $A^c$, the kernel vanishes and $Z_t=1$; on $A$, its stochastic-integral term is $2(B_{t\wedge v}-B_s)$ when $s\le t$. Check: the quadratic-integral value on $A$ is $4\cdot\tfrac12=2$, so the exponential inside Novikov is $e^{2/2}=e$; averaging its values $1$ and $e$ over events of probability $\tfrac12$ yields $(1+e)/2$.

Worked Example: An Occupation-Time Adjustment

On $[0,2]$, take $\theta_t=\mathbf{1}_{\{B_t\ge0\}}$. This is progressively measurable and bounded by $1$. Consequently $$ \int_0^2\theta_t^2dt\le2,\qquad \mathbb{E}\exp\left(\frac12\int_0^2\theta_t^2dt\right)\le e, $$ so Novikov’s condition holds. The transformed process is $$ \widetilde B_t=B_t+\int_0^t\mathbf{1}_{\{B_s\ge0\}}\,ds. $$ Writing $A_t=\int_0^t\mathbf{1}_{\{B_s\ge0\}}ds$, the product calculation gives $d(Z_t\widetilde B_t)=Z_t(1-\widetilde B_t\mathbf{1}_{\{B_t\ge0\}})dB_t$, with no $dt$ term. Since $A_t$ is continuous and of finite variation, $[\widetilde B]_2=[B]_2=2$. Check: the Novikov exponent is at most $\tfrac12\cdot2=1$, giving the stated upper bound $e^1=e$. Thus $\widetilde B$ is Brownian under the measure with density $Z_2$, even though its added drift depends on the path.

Common Mistake: Confusing Local and True Martingales

A frequent error is to stop a process $X$ at the first time $|X_t|\ge n$ and assert $|X_{t\wedge\tau_n}|\le n$ without controlling $X_0$. If $X_0$ is random and can exceed $n$, the asserted bound fails at time zero. The lemma above assumes bounded $X_0$ and chooses $n>C$; in this theorem the process being localized is $\widetilde B$, with $\widetilde B_0=0$. This is the needed justification, not an implicit assumption that a random initial value is smaller than every chosen threshold.

A second error is to infer that a local martingale must be a martingale. The product $Z\widetilde B$ was only shown to be a local martingale at first. The proof used bounded stopping of $\widetilde B$ and uniform integrability of the density martingale to establish the stopped martingale identities needed under $\mathbb{Q}$. It then concluded only that $\widetilde B$ is a local martingale, before invoking Lévy’s characterization.

The roles of the ingredients are separate: the density process cancels the drift, localization justifies the measure-change step, and quadratic variation supplies the Brownian variance structure. A drift adjustment alone does not prove Brownian motion, and quadratic variation alone does not prove a martingale property.

Check Your Understanding

Use the theorem’s hypotheses and proof steps to answer the following.

Why does Novikov’s condition imply that the stochastic integral with integrand $\theta$ is square integrable on $[0,T]$?
In the product calculation, which two $dt$ terms cancel, and what is the remaining stochastic-integral coefficient?
Why is the bound $|X_{t\wedge\tau_n}|\le n$ valid in the localization lemma as stated?
Explain why equivalence of $\mathbb{P}$ and $\mathbb{Q}$ preserves the almost-sure quadratic-variation identity.
What additional property, beyond continuity and quadratic variation $t$, is required before Lévy’s characterization applies?