Do Girsanov transformations constitute a group?

Question

Just to briefly sum up, I have a process $\phi \in \Lambda^2_{loc}$ , a Brownian motion $B_t$ and the Doleans-Dade exponential defined as usual:

$Z_t:=\exp \left(\int_0^t \phi_s d B_s - \frac{1}{2} \int_0^t \phi^2_s ds \right).$

If we assume that Novikov's condition is fulfilled

$ \mathbb{E} \left[\exp \left(\frac{1}{2} \int_0^T \phi^2_s ds \right) \right]<+\infty $,

then the process

$\tilde{B}_t := B_t - \int_0^t \phi_s ds$

is still a Brownian motion, but under the equivalent probability measure defined via the Radon-Nikodym derivative

$ \frac{d \mathbb{Q}}{d \mathbb{P}}=Z_T$.

My question is: do Girsanov's transformations constitute a group (with respect to the sum) ? That is, given a Girsanov transformation, is its inverse a Girsanov transformation as well? And what about the composition, that is the sum of two Girsanov transformations?

For the moment, I reasoned as follows. The $0$ is clearly represented by $\phi \equiv 0$. Given a transformation, its inverse is

$\tilde{B}_t := B_t + \int_0^t \phi_s ds$.

Obviously, $-\phi \in \Lambda^2_{loc}$ and Novikov's condition is still fulfilled. However, I am not sure about the DD exponential. Can we still define a Girsanov change of measure with

$Z^*_t:=\exp \left(-\int_0^t \phi_s d B_s - \frac{1}{2} \int_0^t \phi^2_s ds \right)?$

Regarding the product, I just considered

$d \tilde{B}_1(t) = d B_1(t) + \phi_1(t) dt $

$d \tilde{B}_2(t) = d \tilde{B}_1(t) + \phi_2(t) dt $,

which leads to

$d \tilde{B}_2(t) = d B_1(t) + (\phi_1(t) + \phi_2(t)) dt$?

How could I proceed? Thanks a lot for any help!!

Answer 1

I might have an answer to this.

As far as I can tell it is sufficient that $\tilde{\mathbb{P}}=\mathcal{E(A)}\mathbb{P}$ is a probability to preform a transformation. This in turn is equivalent to that $\mathcal{E(A)}$ is a martingale.

One can show that $X$ is a martingale under a measure $P$ iif $XM$ is martingale under $Q$ given that $M=\frac{P}{Q}$, in turn implying that $M^{-1}$ is a martingale.

Update

Also Novikov actually says that $\mathcal{E(A)}$ is a martingale given that the $\mathcal{A}$ satisfies the condition which is what you asked right?