1 What is Girsanov’s Theorem?
The Girsanov theorem states that if we change the drift coefficient of a given Itô Process (with a non-degenerate diffusion coefficient), then the law of the process will not change dramatically. In fact, the law of the new process will be absolutely continuous with respect to the law of the original process and we can compute explicitly the Radon-Nikodym derivative.
Let \(B(t) = (B_1(t), \ldots, B_n(t))\) be an \(n\)-dimensional Brownian motion on a filtered probability space \((\Omega, \mathcal{F}, \{\mathcal{F}_t\}, \mathbb{P})\). Let \(\theta(t) = (\theta_1(t), \ldots, \theta_n(t))\) be an adapted process such that the Novikov condition holds: \[\mathbb{E}_{\mathbb{P}}\left[\exp\left(\frac{1}{2}\int_0^T \|\theta(t)\|^2 dt\right)\right] < \infty.\]
Define the Radon-Nikodym derivative: \[\begin{align}\frac{d\mathbb{Q}}{d\mathbb{P}}\bigg|_{\mathcal{F}_T} & = Z(T) \\ Z(t) & = \exp\left(-\int_0^t \theta(s) \cdot dB(s) - \frac{1}{2}\int_0^t \|\theta(s)\|^2 ds\right).\end{align}\] Then under the measure \(\mathbb{Q}\), the process \[ \widetilde{B}(t) = B(t) + \int_0^t \theta(s) ds \] is an \(n\)-dimensional Brownian motion.
Before sketching the proof we first introduce two crucial building blocks, the Lévy characterization of BM and Bayes’ Rule.
2 Lévy Characterization of BM
The Lévy characterization provides the key verification tool in the proof. When we want to show that \(\tilde{B}(t)\) is a Brownian motion under the new measure \(\mathbb{Q}\), we could try to verify all the original properties of a BM directly (continuity, independent increments with the right distribution, etc.). That would be very difficult.
Instead, the Lévy characterization gives us a much easier path! We only need to verify two things:
- \(\tilde{B}(t)\) is a martingale under \(\mathbb{Q}\), and
- Its quadratic variation is \(\left< \tilde{B}_{i}, \tilde{B}_{j} \right>_{t} = \delta_{ij}t\).
Let \(X(t)\) be an \(n\)-dimensional SP on probability space \((\Omega, \mathcal{H}, \mathbb{Q})\) taking values in \(\mathbb{R}^n\). Then the following are equivalent: 1. \(X(t)\) is BM w.r.t. \(\mathbb{Q}\) i.e. the law of \(X(t)\) w.r.t. \(\mathbb{Q}\) is the same as the law of an \(n\)-dimensional BM. 2. Both: 1. \(X(t)\) is a martingale w.r.t. \(\mathbb{Q}\) (and w.r.t. its own filtration); and 2. \(X_{i}(t)X_{j}(t)-\delta_{ij}t\) is a martingale w.r.t. \(\mathbb{Q}\) (and w.r.t. its own filtration) for all \(i,j\in\{1,2,\dots,n\}\).
One may replace condition 2.2 with the following:
The cross-variation processes \(\left< X_{i}, X_{j} \right>_{t}\) satisfy the identity \[ \left< X_{i},X_{j} \right> _{t}(\omega) = \delta_{ij}(t)\quad\text{a.s.}\quad 1\leq i,j\leq n \] where \[ \left< X_{i},X_{j} \right> _{t} = \frac{1}{4}[\left< X_{i}+X_{j}, X_{i}+X_{j} \right> _{t}-\left< X_{i}-X_{j}, X_{i}-X_{j} \right> _{t}], \] \(\left< Y,Y \right>_{t}\) being the quadratic variation process.
3 Bayes’ Rule
The Bayes’ rule handles the mechanism of changing probability measures. When we switch from \(\mathbb{P}\) to \(\mathbb{Q}\), conditional expectations change. Girsanov’s proof needs to show that certain properties hold under the new measure \(\mathbb{Q}\), but we have more information about what happens under the original measure \(\mathbb{P}\).
Bayes’ rule is the bridge, it tells us how to relate \(\mathbb{E}_{\mathbb{Q}}[\cdot|\mathcal{F}_{t}]\) in terms of \(\mathbb{E}_{\mathbb{P}}[\cdot|\mathcal{F}_{t}]\) and the Radon-Nikodym derivative.
Let \(\mu\) and \(\nu\) be two probability measures on a measurable space \((\Omega, \mathcal{G})\) such that \[d\nu(\omega)=f(\omega)d\mu(\omega),\]for some \(f \in L^1(\mu)\). Let \(X\) be a random variable on \((\Omega,\mathcal{G})\) such that \[\mathbb{E}_{\nu}[\lvert X \rvert ] = \int_{\Omega}\lvert X(\omega) \rvert f(\omega)~d\mu(\omega)<\infty.\]Let \(\mathcal{H}\) be a \(\sigma\)-algebra, \(\mathcal{H}\subset \mathcal{G}\). Then \[\mathbb{E}_{\nu}[X|\mathcal{H}]\cdot\mathbb{E}[f|\mathcal{H}]=\mathbb{E}_{\mu}[fX|\mathcal{H}]\quad\text{a.s.}.\]
Proof: From the definition of Conditional Expectation we have that if \(H \in \mathcal{H}\) then \[ \begin{align} \int_{H}\mathbb{E}_{\nu}[X|\mathcal{H}] ~fd\mu & = \int_{H}\mathbb{E}_{\nu}[X|\mathcal{H}] ~d\nu \\ & = \int_{H}X~d\nu \\ & = \int_{H}X ~fd\mu \\ & = \int_{H}\mathbb{E}_{\mu}[fX|\mathcal{H}]~d\mu. \end{align} \] Now, since both \(\mathbb{E}_{\nu}[X|\mathcal{H}] \cdot \mathbb{E}_{\mu}[f|\mathcal{H}]\) and \(\mathbb{E}_{\mu}[fX|\mathcal{H}]\) are \(\mathcal{H}\)-measurable (as products/conditional expectations of \(\mathcal{H}\)-measurable random variables), and for any \(H \in \mathcal{H}\):
\[\int_{H} \mathbb{E}_{\nu}[X|\mathcal{H}] \cdot \mathbb{E}_{\mu}[f|\mathcal{H}] ~d\mu = \int_{H}\mathbb{E}_{\nu}[X|\mathcal{H}] ~fd\mu = \int_{H}\mathbb{E}_{\mu}[fX|\mathcal{H}]~d\mu,\]
by the definition of conditional expectation, we have \[\mathbb{E}_{\nu}[X|\mathcal{H}]\cdot\mathbb{E}_{\mu}[f|\mathcal{H}]=\mathbb{E}_{\mu}[fX|\mathcal{H}]\quad\text{a.s.}.\] \(\square\)
4 Girsanov’s Theorem Proof Sketch
By the Novikov Condition, \(Z(t)\) is a positive martingale with \(\mathbb{E}[Z(T)] = 1\), so we can define a probability measure \(\mathbb{Q}\) by \(d\mathbb{Q} = Z(T) d\mathbb{P}\) on \(\mathcal{F}_T\).
For \(t \leq T\), define \(\widetilde{\mathcal{F}}_t = \sigma(B(s) : s \leq t)\) and consider \(\widetilde{B}(t) = B(t) + \int_0^t \theta(s) ds\).
By Itô’s lemma applied to \(Z(t)\): \[ dZ(t) = -Z(t)\theta(t) \cdot dB(t), \] so \(Z(t)\) is a local martingale under \(\mathbb{P}\).
To verify that \(\widetilde{B}(t)\) is a Brownian motion under \(\mathbb{Q}\), we use the Lévy characterization. It suffices to show:
- \(\widetilde{B}(t)\) is a martingale under \(\mathbb{Q}\) with respect to \(\{\widetilde{\mathcal{F}}_t\}\)
- \(\widetilde{B}_i(t)\widetilde{B}_j(t) - \delta_{ij}t\) is a martingale under \(\mathbb{Q}\)
For the first property, by Bayes’ rule, we need to show that for any bounded \(\widetilde{\mathcal{F}}_t\)-measurable \(X\): \[\mathbb{E}_{\mathbb{Q}}[\widetilde{B}(t) | \widetilde{\mathcal{F}}_0] = \widetilde{B}(0) = 0.\]
This follows because \(\widetilde{B}(t)\) is a martingale under \(\mathbb{P}\) plus a drift term that gets absorbed by the change of measure. The second property follows similarly by noting that the quadratic variation is preserved under measure change.
Thus \(\widetilde{B}(t)\) is a Brownian motion under \(\mathbb{Q}\). \(\square\)
5 Key Consequences and Applications
Change of drift in SDEs: If \(X(t)\) satisfies \[ dX(t) = \mu(t) dt + \sigma(t) dB(t), \] and we set \(\theta(t) = \sigma(t)^{-1}\mu(t)\), then under \(\mathbb{Q}\) the process becomes \[ dX(t) = \sigma(t) d\widetilde{B}(t), \] i.e., a martingale. This is why Girsanov’s theorem is essential for removing drift in stochastic models.
Measure-theoretic interpretation: The theorem shows that changing the drift of a diffusion corresponds to a change of measure via an absolutely continuous transformation. This is crucial in mathematical finance for arbitrage-free pricing and risk-neutral measure change.