Girsanov’s Theorem

Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.

Assume \(\{\mathcal F_t\}_{t \geq 0}\) is a complete, right-continuous filtration. A change of measure changes conditional expectations, and can therefore change a martingale into a process with drift. We ask what happens when the underlying probability measure \(\mathop{\mathrm{\mathbf{P}\!\,}}\) is replaced by another probability measure \(\mathop{\mathrm{\mathbf{V}\!\,}}\). The density process describes the change of measure at each time. Once it is understood, the drift correction will be expressed by covariation.

Assume \(\mathop{\mathrm{\mathbf{V}\!\,}}\) is a probability measure on \((\Omega, \mathcal F)\). Suppose \(\mathop{\mathrm{\mathbf{V}\!\,}}\ll \mathop{\mathrm{\mathbf{P}\!\,}}\) on the \(\sigma\)-field \(\mathcal F_\infty\). For every \(t \in [0,\infty]\), define

\[\begin{aligned} D_t = \frac{\dif \mathop{\mathrm{\mathbf{V}\!\,}}}{\dif \mathop{\mathrm{\mathbf{P}\!\,}}}\bigg|_{\mathcal F_t} \end{aligned}\]

as the Radon-Nikodym derivative of \(\mathop{\mathrm{\mathbf{V}\!\,}}\) with respect to \(\mathop{\mathrm{\mathbf{P}\!\,}}\) on the \(\sigma\)-field \(\mathcal F_t\). Then, the process \(\{D_t\}_{t \geq 0}\) is a uniformly integrable martingale, and has a càdlàg modification. For this modification, writing this using \(D_t\) as above, we have that for every stopping time \(T\), \(D_T = \frac{\dif \mathop{\mathrm{\mathbf{V}\!\,}}}{\dif \mathop{\mathrm{\mathbf{P}\!\,}}}\bigg|_{\mathcal F_T}\). Then, if \(\mathop{\mathrm{\mathbf{V}\!\,}}\ll \mathop{\mathrm{\mathbf{P}\!\,}}\) and \(\mathop{\mathrm{\mathbf{P}\!\,}}\ll \mathop{\mathrm{\mathbf{V}\!\,}}\) on \(\mathcal F_\infty\), we have \(\inf_{t \geq 0} D_t > 0\), \(\mathop{\mathrm{\mathbf{P}\!\,}}\)-almost surely.

The density on the current information is obtained by conditioning the terminal density. To verify this, if \(A \in \mathcal F_t\), we have

\[\begin{aligned} \mathop{\mathrm{\mathbf{V}\!\,}}(A) = \mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} \mathbb 1_A = \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}} [\mathbb 1_A D_\infty] = \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}} [\mathbb 1_A \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}}[D_\infty \mid \mathcal F_t]] ~. \end{aligned}\]

The uniqueness of the Radon-Nikodym derivative on \(\mathcal F_t\) implies \(D_t = \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}} [D_\infty \mid \mathcal F_t]\) almost surely. As such, \(D\) is a closed, uniformly integrable martingale (with the almost sure limit \(D_\infty\)). Since \(\{\mathcal F_t\}_{t \geq 0}\) is complete and right-continuous, there exists a càdlàg modification of \(\{D_t\}_{t \geq 0}\). So, if \(T\) is a stopping time, the optional stopping theorem implies for every \(A\in\mathcal F_T\),

Finally, for any \(\varepsilon > 0\), define

\[\begin{aligned} T_\varepsilon = \inf \{t \geq 0 : D_t < \varepsilon\} ~. \end{aligned}\]

Because \(T_\varepsilon\) is the first hitting time of an open set by a càdlàg process, \(T_\varepsilon\) is a stopping time. Since \(\{T_\varepsilon < \infty\}\) is \(\mathcal F_{T_\varepsilon}\)-measurable, \(\mathop{\mathrm{\mathbf{V}\!\,}}(T_\varepsilon < \infty) = \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}} [\mathbb 1_{\{T_\varepsilon < \infty\}} D_{T_\varepsilon}] \leq \varepsilon\), since \(D_{T_\varepsilon} \leq \varepsilon\) on \(\{T_\varepsilon < \infty\}\) (by the right-continuity of sample paths). It then follows that

\[\begin{aligned} \mathop{\mathrm{\mathbf{V}\!\,}}\left( \bigcap_{n \geq 1} \{T_{\frac1n} < \infty\} \right) = 0 ~. \end{aligned}\]

Since \(\mathop{\mathrm{\mathbf{P}\!\,}}\ll \mathop{\mathrm{\mathbf{V}\!\,}}\), the same conclusion follows. This implies \(\mathop{\mathrm{\mathbf{P}\!\,}}\)-almost surely, there exists an integer \(n \geq 1\) such that \(T_{\frac1n} = \infty\). \(\blacksquare\)

The stochastic logarithm gives a convenient representation of a strictly positive continuous local martingale \(D\). Its initial value need not be one. Define the zero-started continuous local martingale

\[L_t=\int_0^tD_s^{-1}\,\mathrm dD_s,\qquad L_0=0.\]

On each finite interval, positivity and continuity allow localization away from zero, so the reciprocal integrand is locally bounded. Itō’s formula gives

\[\log D_t-\log D_0 =L_t-\frac12\int_0^tD_s^{-2}\,\mathrm d\langle D\rangle_s =L_t-\frac12\langle L\rangle_t.\]

Thus the representation is

\[D_t=D_0\exp\left(L_t-\frac12\langle L\rangle_t\right)=D_0\mathcal E(L)_t.\]

Conversely, Itō’s formula applied to the exponential gives \(\mathrm dD_t=D_t\,\mathrm dL_t\), which recovers the displayed integral for \(L\). This proves uniqueness among continuous local martingales starting at zero. The factor \(D_0\) disappears only when the density is normalized to one at time zero. ◻

Now, we can state and prove Girsanov’s theorem. Assume probability measures \(\mathop{\mathrm{\mathbf{P}\!\,}}\) and \(\mathop{\mathrm{\mathbf{V}\!\,}}\) defined on \((\Omega, \mathcal F)\) are mutually absolutely continuous on \(\mathcal F_\infty\). Let \(\{D_t\}_{t \geq 0}\) be the martingale with càdlàg sample paths such that, for all \(t \geq 0\), \(D_t = \frac{\dif \mathop{\mathrm{\mathbf{V}\!\,}}}{\dif \mathop{\mathrm{\mathbf{P}\!\,}}}\big|_{\mathcal F_t}\). Suppose \(D\) has continuous sample paths, and \(L\) is the unique continuous local martingale such that \(D_t = D_0\mathcal E(L)_t\). Then, if \(M\) is a continuous local martingale starting at zero under \(\mathop{\mathrm{\mathbf{P}\!\,}}\), the process \(\widetilde M = M - \langle M, L \rangle\) is a continuous local martingale under \(\mathop{\mathrm{\mathbf{V}\!\,}}\).

Clearly, we’ve already shown that \(D_t = D_0\mathcal E(L)_t\). Furthermore, that fact shows that \(D_t > 0\) (because the infimum is strictly positive) almost surely. Therefore, we can let \(T\) be a stopping time and \(X\) be an adapted process with continuous sample paths.

The measure-change step is the following conditional-expectation identity: if \((XD)^T\) is a martingale with respect to \(\mathop{\mathrm{\mathbf{P}\!\,}}\), then \(X^T\) is a martingale under \(\mathop{\mathrm{\mathbf{V}\!\,}}\). The density formula at a stopping time allows both integrability and the martingale identity to be checked under the original measure. Since \(\mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} |X_{T \wedge t}| = \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}} |X_{T \wedge t} D_{T \wedge t}| < \infty\), we have \(X^T_t \in L^1(\mathop{\mathrm{\mathbf{V}\!\,}})\) for all \(t\). Then, for \(s < t\), pick \(A \in \mathcal F_s\). Because \(A \cap \{T > s\} \in \mathcal F_s\), we have, because \((XD)^T\) is a martingale under \(\mathop{\mathrm{\mathbf{P}\!\,}}\),

\[\begin{aligned} \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}} [\mathbb 1_{A \cap \{T > s\}} X_{T \wedge t} D_{T \wedge t}] = \mathbb{E}_{\mathop{\mathrm{\mathbf{P}\!\,}}} [\mathbb 1_{A \cap \{T > s\}} X_{T \wedge s} D_{T \wedge s}] ~. \end{aligned}\]

Because

\[\begin{aligned} D_{T \wedge t} = \frac{\dif \mathop{\mathrm{\mathbf{V}\!\,}}}{\dif \mathop{\mathrm{\mathbf{P}\!\,}}}\bigg|_{\mathcal F_{T \wedge t}} ~~,~~ D_{T \wedge s} = \frac{\dif \mathop{\mathrm{\mathbf{V}\!\,}}}{\dif \mathop{\mathrm{\mathbf{P}\!\,}}}\bigg|_{\mathcal F_{T \wedge s}} ~, \end{aligned}\]

and \(A \cap \{T > s\} \in \mathcal F_{T \wedge s} \subset \mathcal F_{T \wedge t}\), it follows that

\[\begin{aligned} \mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} [\mathbb 1_{A \cap \{T > s\}} X_{T \wedge t}] = \mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} [\mathbb 1_{A \cap \{T > s\}} X_{T \wedge s}] ~. \end{aligned}\]

This, combined with the straightforward fact that \(\mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} [\mathbb 1_{A \cap \{T \leq s\}} X_{T \wedge t}] = \mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} [\mathbb 1_{A \cap \{T \leq s\}} X_{T \wedge s}]\), implies \(\mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} [\mathbb 1_A X_{T \wedge t}] = \mathbb{E}_{\mathop{\mathrm{\mathbf{V}\!\,}}} [\mathbb 1_A X_{T \wedge s}]\). This shows the claim is true. This immediately implies that if \(XD\) is a continuous local martingale under \(\mathop{\mathrm{\mathbf{P}\!\,}}\), \(X\) is a continuous local martingale under \(\mathop{\mathrm{\mathbf{V}\!\,}}\).

It remains to identify the correction that makes the product with the density a local martingale. Let \(M\) be a continuous local martingale under \(\mathbf P\), starting at zero, and set \(\widetilde M=M-\langle M,L\rangle\). The product formula uses covariation of the two factors:

\[\widetilde M_tD_t=\int_0^t\widetilde M_s\,\mathrm dD_s+ \int_0^tD_s\,\mathrm d\widetilde M_s+\langle\widetilde M,D\rangle_t.\]

A continuous finite-variation process has zero covariation with a continuous semimartingale, so \(\langle\widetilde M,D\rangle=\langle M,D\rangle\). The integral definition of \(L\) gives

\[\mathrm d\langle M,L\rangle_s=D_s^{-1}\,\mathrm d\langle M,D\rangle_s.\]

Substitute these two identities into the product formula:

\[\begin{aligned} \widetilde M_tD_t &=\int_0^t\widetilde M_s\,\mathrm dD_s+\int_0^tD_s\,\mathrm dM_s -\int_0^tD_s\,\mathrm d\langle M,L\rangle_s+\langle M,D\rangle_t\\ &=\int_0^t\widetilde M_s\,\mathrm dD_s+\int_0^tD_s\,\mathrm dM_s. \end{aligned}\]

The remaining integrals are continuous local martingales under \(\mathbf P\), since their continuous integrands are locally bounded. Their sum starts at zero. The measure-change argument above therefore makes \(\widetilde M\) a continuous local martingale under \(\mathbf V\). Starting at zero avoids an additional initial-integrability hypothesis; one may add an \(\mathcal F_0\)-measurable initial variable if it is integrable under both measures. ◻

Reading: Le Gall (2016), Brownian Motion, Martingales, and Stochastic Calculus.