## Lecture 25. Girsanov theorem

In this section, we describe a theorem which has far reaching consequences in mathematical finance: The Girsanov theorem. It describes the impact of a probability change on stochastic calculus.

Let $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ be a filtered probability space. We assume that $(\mathcal{F}_t)_{t \ge 0}$ is the usual completion of the filtration of a Brownian motion $(B_t)_{t \ge 0}$. Let $\mathbb{Q}$ be a probability measure on $\mathcal{F}_\infty$ which is equivalent to $\mathbb{P}$. We denote by $D$ the density of $\mathbb{Q}$ with respect to $\mathbb{P}$.

Theorem (Girsanov theorem) There exists a progressively measurable process $\left( \Theta_t \right)_{t \geq 0}$ such that for every $t \ge 0$, $\mathbb{P} \left( \int_0^t \Theta_s^2ds < +\infty \right)=1$ and $\mathbb{E} \left( D \mid \mathcal{F}_t \right)=\exp \left( \int_0^t \Theta_s dB_s - \frac{1}{2} \int_0^t \Theta_s ^2 ds \right).$ Moreover, the process $B_t - \int_0^t \Theta_s ds$ is a Brownian motion on the filtered probability space $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{Q})$. As a consequence, a continuous and adapted process $(X_t)_{t \ge 0}$ is a $\mathbb{P}$-semimartingale if and only if it is a $\mathbb{Q}$-semimartingale.

Proof. Since $\mathbb{P}$ and $\mathbb{Q}$ are equivalent on $\mathcal{F}_\infty$, there are of course also equivalent on $\mathcal{F}_t$ for every $t \ge 0$. The density of $\mathbb{Q}_{/ \mathcal{F}_t}$ with respect to $\mathbb{P}_{/ \mathcal{F}_t}$ is given by $D_t=\mathbb{E}^{\mathbb{P}} \left( D \mid \mathcal{F}_t \right)$. As a consequence, the process $D_t$ is a positive martingale. From Itō’s representation theorem, we therefore deduce that there exists a progressively measurable process $(u_t)_{t \ge 0}$ such that $D_t=1+\int_0^t u_sdB_s.$ Let now $\Theta_t=\frac{u_t}{D_t}$. We have then,
$D_t=1+\int_0^t \Theta_s D_s dB_s.$
By using Itō’s formula to the process $D_t \exp \left( -\int_0^t \Theta_s dB_s +\frac{1}{2} \int_0^t \Theta_s ^2 ds \right)$, we see that it implies
$D_t=\exp \left( \int_0^t \Theta_s dB_s - \frac{1}{2} \int_0^t \Theta_s ^2 ds \right).$
We now want to prove that the process $B_t - \int_0^t \Theta_s ds$ is a $\mathbb{Q}$-Brownian motion. It is clear the $\mathbb{Q}$-quadratic variation of this process is $t$. From the Levy’s characterization result, we therefore just need to prove that it is a $\mathbb{Q}$ local martingale. For this, we are going to prove that that the process
$N_t= \left( B_t - \int_0^t \Theta_s ds\right) \exp \left( \int_0^t \Theta_s dB_s - \frac{1}{2} \int_0^t \Theta_s ^2 ds \right)$
is a $\mathbb{P}$-local martingale. Indeed, from the integration by parts formula, it is immediate that
$dN_t= D_t dB_t +\left( B_t - \int_0^t \Theta_s ds\right) dD_t.$
Since $D_t$ is the density of $\mathbb{Q}_{\mathcal{F}_t}$ with respect to $\mathbb{P}_{\mathcal{F}_t}$, it is then easy to deduce that $N_t$ is a $\mathbb{P}$-local martingale and thus a $\mathbb{P}$ Brownian motion $\square$

Exercise. Let $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathbb{P})$ be a filtered probability space that satisfies the usual conditions. As before, let $\mathbb{Q}$ be a probability measure on $\mathcal{F}_\infty$ which is equivalent to $\mathbb{P}$. We denote by $D$ the density of $\mathbb{Q}$ with respect to $\mathbb{P}$ and $D_t=\mathbb{E}^{\mathbb{P}} (D \mid \mathcal{F}_t)$. Let $(M_t)_{t \ge 0}$ be a $\mathbb{P}$ local martingale. Show that the process
$N_t=M_t-\int_0^t\frac{d \langle M, D \rangle_s}{D_s}$
is a $\mathbb{Q}$ local martingale. As a consequence, a continuous and adapted process $(X_t)_{t \ge 0}$ is a $\mathbb{P}$-semimartingale if and only if it is a $\mathbb{Q}$-semimartingale.

Exercise Let $(B_t)_{t\ge 0}$ be a Brownian motion. We denote by $\mathbb{P}$ the Wiener measure, by $(\pi_t)_{t \ge 0}$ the coordinate process and by $(\mathcal{F}_t)_{t \ge 0}$ its natural filtration.

• Let $\mu \in \mathbb{R}$ and $\mathbb{P}^\mu$ be the distribution of the process $(B_t+\mu t)_{t\ge 0}$. Show that for every $t \ge 0$, $\mathbb{P}^\mu_{/ \mathcal{F}_t} \ll \mathbb{P}_{/ \mathcal{F}_t},$ and that
$\frac{d \mathbb{P}^\mu_{/ \mathcal{F}_t} }{d\mathbb{P}_{/ \mathcal{F}_t} }=e^{\mu \pi_t -\frac{\mu^2}{2} t}.$
• Is it true that $\mathbb{P}^\mu_{/ \mathcal{F}_\infty} \ll \mathbb{P}_{/ \mathcal{F}_\infty}$
• For $a \in \mathbb{R}_{\ge 0}$, we denote $T_a=\inf \{ t \ge 0, B_t+\mu t =a \}.$ Compute the density function of $T_a$ (You may use the previous question).
• More generally, let $f :\mathbb{R}_{\ge 0} \rightarrow \mathbb{R}$ be a measurable function such that for every $t \ge 0$, $\int_0^t f^2(s)ds < +\infty$. We denote by $\mathbb{P}^f$ the distribution of the process $\left(B_t+\int_0^t f(s)ds\right)_{t\ge 0}$. Show that for every $t \ge 0$,
$\mathbb{P}^f_{/ \mathcal{F}_t} \ll \mathbb{P}_{/ \mathcal{F}_t},$
and that
$\frac{d \mathbb{P}^f_{/ \mathcal{F}_t} }{d\mathbb{P}_{/ \mathcal{F}_t} }=e^{\int_0^t f(s)d\pi_s \frac{1}{2} \int_0^t f^2(s)ds}.$

Let $(\Omega, (\mathcal{F}_t)_{t \ge 0},\mathcal{F},\mathbb{P})$ be a filtered probability space that satisfies the usual conditions and let $(B_t)_{t \ge 0}$ be a Brownian motion on it. Let now $\left( \Theta_t \right)_{t \geq 0}$ be a progressively measurable process such that for every $t \ge 0$, $\mathbb{P} \left( \int_0^t \Theta_s^2ds <+\infty \right)=1$. We denote
$Z_t=\exp \left( \int_0^t \Theta_s dX_s - \frac{1}{2} \int_0^t \Theta_s l^2 ds \right), \text{ }t \geq 0.$
As a consequence of Itō's formula, it is clear that $(Z_t)_{t \ge 0}$ is a local martingale. In general $(Z_t)_{t \ge 0}$ is not a martingale, but the following two lemmas provide simple sufficient conditions that it is.

Lemma. If for every $t \geq 0$, $\mathbb{E} (Z_t)=1,$ then $(Z_t)_{t \ge 0}$ is a uniformly integrable martingale.

Proof. The process $Z$ is a non negative local martingale and thus a super martingale $\square$

Lemma. (Novikov’s condition) If $\mathbb{E} \left(\exp \left( \frac{1}{2} \int_0^\infty \Theta_s^2 ds \right) \right) < +\infty$, then $(Z_t)_{t \ge 0}$ is a uniformly integrable martingale.

Proof. We denote $M_t=\int_0^t \Theta_s dB_s$. As a consequence of $\mathbb{E} \left(\exp \left( \frac{1}{2} \langle M \rangle_\infty \right) \right)<+\infty$, the random variable $\langle M \rangle_\infty$ has moments of all order. So from Burkholder-Davis-Gundy inequalities, $\sup_{t \ge 0} |M_t |$ has moments of all orders, which implies that $M$ is a uniformly integrable martingale. We have then
$\exp\left( \frac{1}{2} M_\infty \right) = \exp\left( \frac{1}{2}M_\infty-\frac{1}{4} \langle M\rangle_\infty \right)\exp\left( \frac{1}{4} \langle M \rangle_\infty \right).$
The Cauchy-Schwarz inequality implies then that $\mathbb{E} \left( \exp\left( \frac{1}{2} M_\infty \right) \right) < +\infty$.
We deduce from the Doob's convergence theorem that the process $\exp\left( \frac{1}{2} M \right)$ is a uniformly integrable submartingale. Let now $\eta < 1$. We have
$\exp \left( \eta M_t -\frac{\eta^2}{2} \langle M \rangle_t \right)=\left( \exp \left( M_t -\frac{1}{2} \langle M \rangle_t \right)\right)^{\eta^2}\exp\left( \frac{\eta M_t}{1+\eta} \right)^{1-\eta^2}.$
Holder's inequality shows then that
$\mathbb{E} \left( \exp \left( \eta M_t -\frac{\eta^2}{2} \langle M \rangle_t \right) \right)$
$\le \mathbb{E} \left( \exp \left( M_t -\frac{1}{2} \langle M \rangle_t \right) \right)^{\eta^2} \mathbb{E} \left( \exp\left( \frac{\eta M_t}{1+\eta} \right) \right)^{1-\eta^2}$
$\le \mathbb{E} \left( \exp \left( M_t -\frac{1}{2} \langle M \rangle_t \right) \right)^{\eta^2} \mathbb{E} \left( \exp\left( \frac{M_t}{2} \right) \right)^{2\eta(1-\eta)}$
$\le \mathbb{E} \left( \exp \left( M_t -\frac{1}{2} \langle M \rangle_t \right) \right)^{\eta^2} \mathbb{E} \left( \exp\left( \frac{M_\infty}{2} \right) \right)^{2\eta(1-\eta)}$
If we can prove that $\mathbb{E} \left( \exp \left( \eta M_t -\frac{\eta^2}{2} \langle M \rangle_t \right) \right)=1$, then by letting $\eta \to 1$ in the above inequality, we would get
$\mathbb{E} \left( \exp \left( M_t -\frac{1}{2} \langle M \rangle_t \right) \right) \ge 1$
and thus $\mathbb{E} \left( \exp \left( M_t -\frac{1}{2} \langle M \rangle_t \right) \right) = 1$.

Let $p > 1$ such that $\frac{\eta \sqrt{p}}{\sqrt{p}-1} \le 1$. Consider $r=\frac{\sqrt{p}+1}{\sqrt{p}-1}$ and $s =\frac{\sqrt{p}+1}{2}$ so that $1/r+1/s=1$. Using
$\exp \left( \eta M_t -\frac{\eta^2}{2} \langle M \rangle_t \right)^p = \exp \left( \sqrt{\frac{p}{r}} \eta M_t -\frac{p}{2} \eta^2 \langle M \rangle_t \right) \exp \left( \left( p \eta -\sqrt{\frac{p}{r} } \right)M_t \right)$
and then Holder's inequality, shows that there is a constant $C$ (depending only on $M$) such that for any stopping time $T$
$\mathbb{E} \left( \exp \left( \eta M_T -\frac{\eta^2}{2} \langle M \rangle_T \right)^p \right) \le C.$
By the Doob's maximal inequality, it implies that the local martingale $\exp \left( \eta M_t -\frac{\eta^2}{2} \langle M \rangle_t \right)$ is actually a true martingale. This implies $\mathbb{E} \left( \exp \left( \eta M_t -\frac{\eta^2}{2} \langle M \rangle_t \right) \right)=1$ and the desired conclusion $\square$

We now assume that $(Z_t)_{t \ge 0}$ is a uniformly integrable martingale. In that case, it is easy to see that on the $\sigma$-field $\mathcal{F}_\infty$, there is a unique probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ such that for every $t \ge 0$, $\frac{d\mathbb{Q}_{/\mathcal{F}_t}}{d\mathbb{P}_{/\mathcal{F}_t}}=Z_t,\text{ }\mathbb{P}-a.s.$ The same argument as before shows then that with respect to $\mathbb{Q}$, the process
$B_t - \int_0^t \Theta_s ds$ is a Brownian motion.

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