Proving Girsanov's Theorem with Density Processes

The Density Process and the Shifted Process

Girsanov’s theorem relates a change of probability measure to a change in drift. Its proof is a useful synthesis of Ito’s product rule, stochastic exponentials, localization, and the distinction between a local martingale and a true martingale. The central calculation is short: a carefully signed density process cancels the drift of a candidate shifted Brownian motion. The measure change and the localization checks are what make that calculation a theorem.

Fix a finite horizon $T>0$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{0\le t\le T},\mathbb{P})$ satisfying the usual conditions, and let $B_t$ be a Brownian motion under $\mathbb{P}$. Assume that $\theta_t$ is progressively measurable and that $\mathbb{E}\int_0^T\theta_s^2\,ds<\infty$. Thus the Ito integral $\int_0^t\theta_s\,dB_s$ is defined for $0\le t\le T$. Suppose also that Novikov’s condition holds:

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

Novikov’s condition is a standard sufficient criterion ensuring that the stochastic exponential below is a true martingale; its proof is a technical exponential-integrability result and is not repeated here. In the calculations that follow, any integrand known only to be locally square-integrable is interpreted by stopping when its accumulated square integral or its magnitude reaches a finite bound. The stopped integrands are square-integrable in the sense $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Definition: Define the density process $Z_t$ and the candidate shifted process $\widetilde B_t$ by $$ Z_t=\exp\left(-\int_0^t\theta_s\,dB_s-\frac12\int_0^t\theta_s^2\,ds\right), \qquad \widetilde B_t=B_t+\int_0^t\theta_s\,ds. $$ Under Novikov’s condition, $Z_t$ is a positive true $\mathbb{P}$-martingale with $Z_0=1$ and $\mathbb{E}[Z_T]=1$. Define $\mathbb{Q}$ on $\mathcal{F}_T$ by $d\mathbb{Q}=Z_T\,d\mathbb{P}$. Since $Z_T>0$ almost surely, $\mathbb{Q}$ and $\mathbb{P}$ are equivalent on $\mathcal{F}_T$.

Girsanov’s Theorem

Theorem: Under the assumptions above, $\widetilde B_t=B_t+\int_0^t\theta_s\,ds$ is a Brownian motion on $[0,T]$ under $\mathbb{Q}$. The density process is $Z_t=\dfrac{d\mathbb{Q}}{d\mathbb{P}}\big|_{\mathcal{F}_t}$.

Proof. We first compute the differentials. Ito’s formula applied to the exponential defining $Z_t$, with exponent $Y_t=-\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds$, gives $dY_t=-\theta_t\,dB_t-\tfrac12\theta_t^2\,dt$, $(dY_t)^2=\theta_t^2\,dt$, and hence $dZ_t=Z_t\,dY_t+\tfrac12Z_t(dY_t)^2=-Z_t\theta_t\,dB_t$. The stochastic integral is understood locally if its integrand is not globally square-integrable; after stopping, the integrand is square-integrable. Novikov’s condition supplies the true-martingale property of $Z$.

Apply Ito’s product rule to $Z_t\widetilde B_t$. Here $d\widetilde B_t=dB_t+\theta_t\,dt$, and the covariation term is $d[Z,\widetilde B]_t=-Z_t\theta_t\,dt$, because the finite-variation part of $\widetilde B$ has zero covariation with $Z$. Therefore

\begin{aligned} d(Z_t\widetilde B_t) &=\widetilde B_t\,dZ_t+Z_t\,d\widetilde B_t+d[Z,\widetilde B]_t\\ &=-Z_t\theta_t\widetilde B_t\,dB_t+Z_t(dB_t+\theta_t\,dt)-Z_t\theta_t\,dt\\ &=Z_t(1-\theta_t\widetilde B_t)\,dB_t. \end{aligned}

The two $dt$ terms cancel. This proves that $Z\widetilde B$ is a local martingale under $\mathbb{P}$, not necessarily a true martingale: its stochastic-integral integrand need not have finite expected square integral on the whole interval. The calculation alone does not yet establish a martingale under $\mathbb{Q}$.

We give the localization argument explicitly. For each positive integer $n$, stop $\widetilde B$ when its absolute value first reaches $n$, or at $T$ if it does not; call this time $\tau_n$. The process $Z_t\widetilde B_{t\wedge\tau_n}$ is a local $\mathbb{P}$-martingale: before $\tau_n$ the preceding product calculation applies, and after $\tau_n$ the stopped factor is constant, leaving only a stochastic integral against $Z$. It is in fact a true martingale. Indeed, $|\widetilde B_{t\wedge\tau_n}|\le n$, so its values at stopping times are bounded in absolute value by $nZ_\rho$, where $\rho\le T$ is the relevant stopping time. The family $\{Z_\rho:\rho\le T\}$ is uniformly integrable because $Z_t=\mathbb{E}[Z_T\mid\mathcal{F}_t]$ and $Z_T$ is integrable. Thus the stopped local martingale is uniformly integrable and hence a true martingale.

For $s\le t$, Bayes’ formula for conditional expectation gives $\mathbb{E}^{\mathbb{Q}}[\widetilde B_{t\wedge\tau_n}\mid\mathcal{F}_s] =Z_s^{-1}\mathbb{E}^{\mathbb{P}}[Z_T\widetilde B_{t\wedge\tau_n}\mid\mathcal{F}_s]$. Since the stopped variable is $\mathcal{F}_t$-measurable and $\mathbb{E}^{\mathbb{P}}[Z_T\mid\mathcal{F}_t]=Z_t$, the numerator equals $\mathbb{E}^{\mathbb{P}}[Z_t\widetilde B_{t\wedge\tau_n}\mid\mathcal{F}_s]$. The true-martingale property just proved makes this $Z_s\widetilde B_{s\wedge\tau_n}$. Dividing by $Z_s>0$ shows that $\widetilde B_{t\wedge\tau_n}$ is a $\mathbb{Q}$-martingale. Since $\tau_n$ increases to $T$ almost surely by continuity of $\widetilde B$, $\widetilde B$ is a continuous local martingale under $\mathbb{Q}$.

Finally, adding a continuous finite-variation process does not change Brownian quadratic variation, so $[\widetilde B]_t=[B]_t=t$ almost surely. Levy’s characterization states that a continuous local martingale starting at zero with quadratic variation $t$ is Brownian motion. This is a standard result; its proof is not repeated here. Its hypotheses hold under $\mathbb{Q}$, and $\widetilde B_0=0$. Therefore $\widetilde B$ is Brownian motion under $\mathbb{Q}$. $\square$

Worked Checks of the Density and Sign

Worked Example: A Constant Kernel

Take $T=1$ and $\theta_t=2$. Novikov’s condition holds because $\mathbb{E}[\exp(\tfrac12\int_0^1 2^2\,ds)]=e^2<\infty$. The terminal density is $Z_1=e^{-2B_1-2}$. Since $B_1\sim N(0,1)$, its moment-generating function gives $\mathbb{E}[Z_1]=e^{-2}\mathbb{E}[e^{-2B_1}]=e^{-2}e^{(-2)^2/2}=1$.

To check the direction of the shift, the density tilts the standard normal density of $B_1$ by $e^{-2x-2}$. Completing the square gives $-x^2/2-2x-2=-(x+2)^2/2$. Thus under $\mathbb{Q}$, $B_1\sim N(-2,1)$, and $\widetilde B_1=B_1+2\sim N(0,1)$, as required.

Check: $\mathbb{E}^{\mathbb{Q}}[\widetilde B_1]=(-2)+2=0$ and $\operatorname{Var}^{\mathbb{Q}}(\widetilde B_1)=1$. The shifted process has the Brownian endpoint mean and variance; the theorem supplies its full Brownian law.

Worked Example: A Piecewise-Constant Kernel

On $[0,1]$, set $\theta_t=1$ for $0\le t\le\tfrac12$ and $\theta_t=-1$ for $\tfrac12<t\le1$. The value at the single time $\tfrac12$ does not affect either integral. Novikov’s expectation is $e^{1/2}$, since $\tfrac12\int_0^1\theta_s^2\,ds=\tfrac12$. Write $U=B_{1/2}$ and $V=B_1-B_{1/2}$; under $\mathbb{P}$, these are independent $N(0,\tfrac12)$ variables. Direct substitution gives $Z_1=\exp(-U+V-\tfrac12)$.

The Gaussian moment-generating function gives $\mathbb{E}[e^{-U+V}]=e^{(1/2)(1/2)+(1/2)(1/2)}=e^{1/2}$, so $\mathbb{E}[Z_1]=e^{-1/2}e^{1/2}=1$. Under the density tilt, the $U$-factor is proportional to $e^{-U}$ and the $V$-factor to $e^V$. Completing the square in each normal density shows that under $\mathbb{Q}$, $U\sim N(-\tfrac12,\tfrac12)$ and $V\sim N(\tfrac12,\tfrac12)$; they remain independent because the density factors into a function of $U$ times a function of $V$.

The integrated kernel over the first half is $\tfrac12$, and over the second half it is $-\tfrac12$. Consequently the two increments of $\widetilde B$ are $U+\tfrac12$ and $V-\tfrac12$. Each has mean zero and variance $\tfrac12$, and they are independent under $\mathbb{Q}$. Their sum has mean zero and variance $1$, consistent with the Brownian endpoint law.

Check: The endpoint shift is zero, so $\widetilde B_1=B_1$. Under $\mathbb{Q}$, $B_1=U+V$ has mean $-\tfrac12+\tfrac12=0$ and variance $\tfrac12+\tfrac12=1$.

Worked Example: A Brownian Motion with Constant Drift

Let $T=2$, choose $\theta_t=-\tfrac12$, and define $\mathbb{Q}$ by the theorem. Then $Z_2=\exp(\tfrac12B_2-\tfrac14)$, because $\tfrac12\int_0^2\theta_s^2\,ds=\tfrac14$. Novikov’s expectation is $e^{1/4}$, which is finite. Normal tilting gives $\mathbb{E}[Z_2]=e^{-1/4}e^{(\frac12)^2(2)/2}=e^{-1/4}e^{1/4}=1$. Completing the square shows $B_2\sim N(1,2)$ under $\mathbb{Q}$, so $B_t$ has acquired drift $\tfrac12$ under the new measure.

Here $\widetilde B_t=B_t-\tfrac12t$. At $t=2$, its mean under $\mathbb{Q}$ is $1-\tfrac12(2)=0$, and its variance is $2$, since subtracting a deterministic constant does not change variance. Thus $\widetilde B_2\sim N(0,2)$, matching the Brownian endpoint distribution.

Check: If one instead used $\theta=+\tfrac12$, the theorem’s density would tilt $B_2$ to have mean $-1$, and the shifted process would be $B_t+\tfrac12t$. The sign in the density and the sign in the shift must be chosen together.

Common Sign Error and Why the Proof Matters

A frequent error is to use $Z_t=\exp(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds)$ while retaining the shift $B_t+\int_0^t\theta_s\,ds$. In the product calculation this incorrect density gives $dZ_t=Z_t\theta_t\,dB_t$, so the covariation term is $+Z_t\theta_t\,dt$. The product drift is then $Z_t\theta_t\,dt+Z_t\theta_t\,dt=2Z_t\theta_t\,dt$, not zero. The correction is to reverse the sign in the stochastic integral of the density, or to reverse the shift, so that the drift terms cancel.

The density process has two jobs: it defines a probability measure, and it converts the product-rule calculation into a local-martingale statement under that measure. Novikov’s condition ensures the first job by making $Z$ a true martingale; localization and Bayes’ formula justify the second. The final quadratic-variation check is indispensable: a continuous local martingale is not automatically Brownian, while a continuous local martingale with quadratic variation $t$ is Brownian by Levy’s characterization. The measure $\mathbb{Q}$ is the changed probability measure here; in finance it may be called risk-neutral only when additional model assumptions justify that interpretation.

Check Your Understanding

For each question, state which measure is being used and verify any density or integrability condition before applying the theorem.

For constant $\theta=c$ on $[0,T]$, write $Z_T$ and evaluate $\mathbb{E}[Z_T]$ using the normal moment-generating function.
In the product-rule proof, identify the covariation term $d[Z,\widetilde B]_t$ and show where it cancels the drift.
Explain why the local-martingale conclusion under $\mathbb{P}$ does not, by itself, imply that the candidate process is Brownian under $\mathbb{Q}$.
If $\theta=-\mu$ is constant, write the density and shifted process, and determine the mean of $B_T$ under $\mathbb{Q}$.
State the two properties of $\widetilde B$ needed to apply Levy’s characterization.

Proving Girsanov's Theorem with Density Processes

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