Help with proof of Girsanov Theorem

Question

I'm studying this proof of Girsanov Theorem and trying the figure out the details however I need some help with this. I noticed there are, also here on stackexchange, a lot of different versions of the Theorem so I start by stating Girsanov the way I know it:

Theorem: Let $X=X_0+A+M$ be a continuous semimartingale on $(\Omega, \mathcal{F}, \mathbb{P})$ w.r.t. a filtration $\mathbb{F}$. Let $\mathbb{Q}$ be another probability measure on $(\Omega, \mathcal{F})$ such that $\mathbb{Q}_t << \mathbb{P}_t$ for all $t \geq 0$ and with density process $Z$. Then the stochastic integral $Z^{-1} \cdot \langle Z, M \rangle$ is well defined under $\mathbb{Q}$ and $X$ is a continuous martingale on $(\Omega, \mathcal{F}, \mathbb{Q})$ with local martingale part $$M^{\mathbb{Q}}=M-Z^{-1} \cdot \langle Z, M \rangle$$ Moreover the quadratic covariation $\langle X,Y\rangle$ of two semimartingales under $\mathbb{P}$ is the same under $\mathbb{P}$ and $\mathbb{Q}$.

I will now give the proof from the lecture notes with my ideas and problems.

Let $T^n = \inf\{t > 0 \;:\; Z_t < \frac{1}{n}\}$. On each $[0,T^n]$ we have that $Z^{-1} \cdot \langle Z, M \rangle$ is $\mathbb{P}$-a.s. finite and therefore also $\mathbb{Q}$-a.s. finite. Since $T^n \to \infty$ $\mathbb{Q}$-a.s. the process $Z^{-1} \cdot \langle Z, M \rangle$ is well defined under $\mathbb{Q}$.

I assume for $Z^{-1} \cdot \langle Z, M \rangle$ to be well defined means that it is $\mathbb{Q}$-a.s. finite, correct? And I see that on $[0,T^n]$ the process $Z$ stays away from zero, by why is $Z^{-1} \cdot \langle Z, M \rangle$ $\mathbb{P}$-a.s. finite?

Next the proof continuous with \begin{align*} (M^{\mathbb{Q}}Z)_t^{T^n} &= M_0Z_0 + \int_0^{T^n \wedge t} M^{\mathbb{Q}} dZ + \int_0^{T^n \wedge t}ZdM^{\mathbb{Q}} + \langle M^{\mathbb{Q}}, Z \rangle_{T^n \wedge t}\\ &=M_0Z_0 + \int_0^{T^n \wedge t} M^{\mathbb{Q}} dZ + \int_0^{T^n \wedge t}ZdM \end{align*}

I calculate that $\int_0^{T^n \wedge t}ZdM^{\mathbb{Q}} = \int_0^{T^n \wedge t}ZdM - \langle M^, Z \rangle_{T^n \wedge t}$. Now what to do with the last term? If I use linearity for quadratic covariation (I don know if this even holds) I get $\langle M^{\mathbb{Q}}, Z \rangle_{T^n \wedge t} = \langle M^, Z \rangle_{T^n \wedge t} - \langle Z, \langle Z,M \rangle\rangle_{T^n \wedge t}$. The first term cancels, but what to do with the second?

I now see how it follows that $M^{\mathbb{Q}}$ is a local martingale, but how does that show that $X$ is a continuous semimartingale w.r.t. $\mathbb{Q}$? And that it's local martingale part is $M^{\mathbb{Q}} $? Lastly is there an easy way to see the last claim with the quadratic covariation?

Thanx in advance!

Answer 1

In order of increasing technicality:

1. $\langle X,Y \rangle$ is the same under $P$ and $Q$ because $\langle X, Y \rangle_t$ is a $P_t$-limit in probability of $\sum_{i}(X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_{i}})$ for $\{t_i\}$ a partition $\Delta$ of $[0,t]$ as $|\Delta| \to 0$. Since $Q_t \ll P_t$ the $Q_t$-limit also converges, showing that it must be the $Q$-joint variation as well.

2. The martingale $Z$ is strictly positive $Q$-a.s. First suppose $Q \ll P$ and let $\tau$ be the first time $Z_t$ or $Z_{t-}$ hits $0$, and let $\tau_n$ be the first time $Z \leq 1/n$. Then for positive rational $q$ we have $1/n \geq E[Z_{\tau_n} 1_{\tau_n < \infty}] \geq E[Z_{\tau+q}1_{\tau_n < \infty}]$. Sending $n\to \infty$ give $E[Z_{\tau+q}1_{\forall n, \tau_n < \infty}] = 0$ so $Z_{\tau+q} = 0$ a.s. all $q$ on $\{\tau < \infty\} \subseteq \{\forall n, \tau_n < \infty\}$. Right continuity then implies $Z = 0$ on $[\tau,\infty)$ a.s. Thus $Z_{\infty} = 0$ on $\{\tau < \infty\}$, and hence $Q(\tau<\infty) = \int 1_{\tau < \infty} Z_\infty dP = 0$. For the case $Q_t \ll P_t$ for all $t$, just stop $Z$ at $t$ and apply the result to get $Z$ is a.s. strictly positive on $[0,t]$ for all $t$ a.s., and hence on $[0,\infty)$ a.s.

3. If $X$ is cadlag and $XZ$ is a $P$-local martingale, then $X$ is a $Q$-local martingale because a localizing sequence under $P$ is also such under $Q$. In particular $MZ - \langle M,Z \rangle = \int Z dM + \int MdZ$, so $ZM-\langle Z,M\rangle$ is a $P$-local martingale, and hence $M-\frac{1}{Z}\langle Z,M\rangle$ is a $Q$-local martingale. Since $\frac{1}{Z}$ and $\langle Z,M\rangle$ are $Q$ semimartingale, $\frac{1}{Z}\langle Z,M\rangle = \int\frac{1}{Z}d\langle Z,M\rangle + \int\langle Z,M\rangle d(\frac{1}{Z})$ under $Q$. Thus $$M-\int\frac{1}{Z}d\langle Z,M\rangle = (M-\frac{1}{Z}\langle Z,M\rangle) + (\frac{1}{Z}\langle Z,M\rangle-\int\frac{1}{Z}d\langle Z,M\rangle)$$ $$ = (M-\frac{1}{Z}\langle Z,M\rangle) + \int\langle Z,M\rangle d(\frac{1}{Z}) $$ is the sum of two $Q$ local martingales.