I will use $B:=\bar{W}$ for simplicity. Looking at what you want, we see that we would like to have
$$W_t=\int c(s,X_s)ds + B_t\iff B_t=W_t-\int c(s,X_s)ds$$
for a given process $X$. Let $Y_t:=c(t,X_s)$. Let $Z:=\mathcal{E}(Y\circ W)$, where $Y\circ W=\int Y_sdW_s$. Hence you need some structural assumptions on $c$ such that you are allowed to write $Y\circ W$. In general $Z$ is a local martingale and positive on $[0,\infty)$ hence a supermartingale (Fatou). Therefore $Z_t$ converges to $Z_\infty$ $P$-a.s. It may happen that $Z_\infty = 0$ and or $E[Z_\infty] < 1$. The condition $E[Z_\infty]=1$ is equivalent to the property that $Z$ is a (uniformly integrable) martingale on $[0,\infty]$. Here you can use Novikov's condition. So if
$$E[\exp{(\frac{1}{2}\langle Y\circ W\rangle_\infty)}]<\infty$$
then $Z$ is a uniformly integrable martingale on $[0,\infty]$. We must have $\langle Y\circ W\rangle_\infty <\infty$ to guarantee that $Z_\infty$ becomes not $0$ with positive probability. Suppose it is true that $Z_\infty >0$. Then you can define a probability measure $Q$ which is equivalent to $P$ by $\frac{dQ}{dP}=Z_\infty$. The general Girsanov tells you that for a continuous local martingale $M$ w.r.t $P$ and a density process $Z$ we have
$$\tilde{M}=M-\int\frac{1}{Z}d\langle Z,M\rangle=M-\langle L, M\rangle$$
is a continuous local martingale w.r.t $Q$, where $Z$ is of the form $Z=\mathcal{E}(L)$ for a continuous local martingale $L$.
Take $W=M$ and $L=Y\circ W$ to conclude
$$\tilde{M}=:B=M-\langle L,M\rangle=W-\int Y_s ds$$
It is easy to verify that $B$ is a $Q$ Brownian Motion (Lévy).
After all you are allowed to do that on some assumption on $c$:
- $Y\in L^2_{loc}(W)$
- $E[\exp{(\frac{1}{2}\langle Y\circ W\rangle_\infty)}]<\infty$
- $\langle Y\circ W\rangle_\infty <\infty$
Keep in mind that $2$ and $3$ are "just" sufficient condition in this case. So they are a good points to start with.
From Itô's lemma you have :
$Y_t=W_t^2 - t= \int_0^t W_sdW_s$
So what you have here is that $Y_t$ is a local martingale. To prove that it is indeed a martingale it suffices to show that :
$\forall t>0, E[\langle Y\rangle_t]<\infty$
as you can check in lemma 3 which is not too hard I think.
Best regards
Best Answer
If an Ito process has no drift, then we have a process of the type $I_t = \int_0^t H_u dW_u,$ (i.e the stochastic differential is $dI_u = H_u dW_u$). The previous integral is defined for integrand processes $H$ such that $H$ is $\{ \mathscr{F}_t \}$-adapted and $\int_0^t |H_u|^2 du < \infty $ a.s for every $t>0.$ But $I_t$ defined like that, is ("only") a local martingale. For $I_t$ to be a (true) martingale we have to check that $E \int_0^t |H_u|^2 du < \infty $ a.s for every $t>0.$