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 be a filtered probability space. We assume that is the usual completion of the filtration of a Brownian motion . Let be a probability measure on which is equivalent to . We denote by the density of with respect to .

**Theorem*** (Girsanov theorem) There exists a progressively measurable process such that for every , and Moreover, the process is a Brownian motion on the filtered probability space . As a consequence, a continuous and adapted process is a -semimartingale if and only if it is a -semimartingale. *

**Proof.** Since and are equivalent on , there are of course also equivalent on for every . The density of with respect to is given by . As a consequence, the process is a positive martingale. From Itō’s representation theorem, we therefore deduce that there exists a progressively measurable process such that Let now . We have then,

By using Itō’s formula to the process , we see that it implies

We now want to prove that the process is a -Brownian motion. It is clear the -quadratic variation of this process is . From the Levy’s characterization result, we therefore just need to prove that it is a local martingale. For this, we are going to prove that that the process

is a -local martingale. Indeed, from the integration by parts formula, it is immediate that

Since is the density of with respect to , it is then easy to deduce that is a -local martingale and thus a Brownian motion

**Exercise.*** Let be a filtered probability space that satisfies the usual conditions. As before, let be a probability measure on which is equivalent to . We denote by the density of with respect to and . Let be a local martingale. Show that the process
is a local martingale. As a consequence, a continuous and adapted process is a -semimartingale if and only if it is a -semimartingale.*

**Exercise** * Let be a Brownian motion. We denote by the Wiener measure, by the coordinate process and by its natural filtration.*

- Let and be the distribution of the process . Show that for every , and that

- Is it true that
- For , we denote Compute the density function of (You may use the previous question).
- More generally, let be a measurable function such that for every , . We denote by the distribution of the process . Show that for every ,

and that

Let be a filtered probability space that satisfies the usual conditions and let be a Brownian motion on it. Let now be a progressively measurable process such that for every , . We denote

As a consequence of Itō's formula, it is clear that is a local martingale. In general is not a martingale, but the following two lemmas provide simple sufficient conditions that it is.

**Lemma.** *If for every , then is a uniformly integrable martingale.*

**Proof.** The process is a non negative local martingale and thus a super martingale

**Lemma.** *(Novikov’s condition) If , then is a uniformly integrable martingale.*

**Proof.** We denote . As a consequence of , the random variable has moments of all order. So from Burkholder-Davis-Gundy inequalities, has moments of all orders, which implies that is a uniformly integrable martingale. We have then

The Cauchy-Schwarz inequality implies then that .

We deduce from the Doob's convergence theorem that the process is a uniformly integrable submartingale. Let now . We have

Holder's inequality shows then that

If we can prove that , then by letting in the above inequality, we would get

and thus .

Let such that . Consider and so that . Using

and then Holder's inequality, shows that there is a constant (depending only on ) such that for any stopping time

By the Doob's maximal inequality, it implies that the local martingale is actually a true martingale. This implies and the desired conclusion

We now assume that is a uniformly integrable martingale. In that case, it is easy to see that on the -field , there is a unique probability measure equivalent to such that for every , The same argument as before shows then that with respect to , the process

is a Brownian motion.