This is somehow lengthy, but I think you will better understand how Girsanov actually works. The theorem you stated are more an application of Girsanov. The motivation behind Girsanov is the following: You are interested in how semimartingales behave under a change of measure. Since the finite variation part do not change, the question reduces to how local martinagles behave under a change of measure. As I was taught, Girsanov answers this question:
Suppose you have $Q\approx P$ and assume for simplicity that the density process $Z$ is continuous. If you have a continuous local martingale $M$ null at zero (wrt $P$), i.e. $M\in \mathcal{M}_{0,loc}^c(P)$, then $$\bar{M}=M-\int\frac{1}{Z}d\langle Z, M\rangle = M-\langle L,M\rangle \in \mathcal{M}_{0,loc}^c(Q)$$
where we write $Z=Z_0\mathcal{E}(L)$.
This is what I would refer to Girsanov's theorem. Note that this implies that in particular $M$ is $Q$-Semimartingale. Of course there are generalizations of this theorem ($Z$ general etc.).
Both of your theorems are the same. It is a special case of Girsanov. Take $M=W$, where $W$ is $P$ Brownian Motion. As an application of Girsanov you get:
If $W$ is a $P$-Brownian Motion and $Q\approx P$ with density process of the form $Z=\mathcal{E}(\int \Theta_s dW_s)$, for a predictable process $\Theta$. Then $W$ is under $Q$ a Brownian Motion with drift, i.e.$$ W=\bar{W}+\int\Theta_s ds$$ for a $Q$-Brownian Motion $\bar{W}$.
This is an immediate consequence of Girsanov and the proof is straight forward using Lévy's characterization of Brownian Motion. However in most cases you have to go the other way around: Usually you do not have $Q\approx P$. This means you have a probability measure and want to construct an equivalent probability measure $Q$ such that the density process $Z$ is a stochastic exponential. Hence you start with $L\in\mathcal{M}_{0,loc}^c(P)$ and define $Z:=\mathcal{E}(L)$. You hope that $Z$ can be used to define an equivalent probability measure $Q$, as $\frac{dQ}{dP}=Z_\infty$. We have $\mathcal{E}(L)=Z$, hence $Z$ is a local martingale and strictly positive. Therefore it is a supermartingale on $[0,\infty)$ (use Fatou to prove that)! By the supermartingale convergence theorem $Z_t$ converges $P-a.s.$ to $Z_\infty$. The problem is, $Z_\infty$ can be $0$ or $E[Z_\infty]<1$ (or both together). As already mentioned you want do define $\frac{dQ}{dP}:=Z_\infty$. You want at least that this $Q$ is absolutely continuous w.r.t $P$, i.e. $Q\ll P$. Hence you need at least
- $Z_\infty >0$
- $E[Z_\infty]=1$.
A priori, as said before, $Z_\infty=0$ and/or $E[Z_\infty]<1$. Hence we must find some conditions, such that $1.$ and $2.$ are true. For $1.$ we must have $\langle L\rangle_\infty < \infty$ (by definition of $Z_\infty=e^{L_\infty -\frac{1}{2}\langle L \rangle_\infty}$). For $2.$ you can use: $E[Z_\infty]=1 $ if and only if $Z$ is a uniformly integrable $P$ martingale on $[0,\infty]$. Now there is a famous condtion, called Novikov's condition, which gives a sufficient condition of $Z=\mathcal{E}(L)$ to be a uniformly integrable martingale on $[0,\infty]$.
Looking at your question. The theorem from Wikipedia is exactly my second statement with $X:=\int\Theta_s dW$. Note that $[W,X]=\langle W,X\rangle = \langle W,\int \Theta_s dW \rangle = \int \Theta_s d\langle W,W\rangle = \int\Theta_s ds$. Furthermore $Z:=\mathcal{E}(X)$. The whole difference between the theorem from Wikipedia and Shreve is in specifying the process $X$ further. Shreve assumes that $X$ has a particular form, i.e. an integral w.r.t to a Brownian Motion. That is the only difference.
In finance you often work on finite time horizon, i.e. on $[0,T]$. You can easily extend everything to $[0,\infty)$ by setting everything equal zero outside $[0,T]$.
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.
Best Answer
Actually, writing $W_t=\tilde W_t-\theta t$, (the) Girsanov theorem says that $W_t$ is a Brownian motion with no drift under $\mathbb P$, where... $$\left.\frac{d\mathbb P}{d\mathbb Q}\right|_{\mathcal F_t}=\exp\left(\theta \color{red}{\tilde W_t} - \tfrac12\theta^2 t\right)=\exp\left(\theta W_t +\tfrac12\theta^2 t\right)=\frac{1}{Z_t}$$