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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