# Change of probability measure and a continuous-time Markov chain

Let $(\Omega,\mathcal{F},\mathbb{P},\mathbb{F})$ be a complete filtered probability space, with $W$ a Wiener process and $\alpha$ a continuous-time Markov chain (taking values in $\{1,...,M\}$). We assume that $W$ and $\alpha$ are independent processes and that the filtration $\mathbb{F}$ is such that $$\mathcal{F}_t = \sigma ( \mathcal{F}_t^W \cup \mathcal{F}_t^\alpha),$$ for all $t \in [0;T]$, satisfying the usual hypothesis. I also have a stochastic process $\lambda$ such that $\lambda$ is progressively measurable and $$\int_0^T E(|\lambda_t|^2) dt < + \infty.$$ If $\lambda$ satisfies that Novikov condition, by applaying Grisanov's theorem, I get a probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ defined by $$\frac{d\mathbb{Q}}{d\mathbb{P}} = Z_T^{\lambda} = \exp \left( -\int_0^T \lambda_s dW_s - \frac{1}{2} \int_0^T |\lambda_s|^2 ds \right),$$ and under which the process $W^\lambda$ defined by $$W_t^\lambda = W_t + \int_0^t \lambda_s ds,$$ for all $t \in [0;T]$, is a Wiener process on $(\Omega,\mathcal{F},\mathbb{Q},\mathbb{F})$. To be more precise, the stochastic process $\lambda$ depends on (and only on) $\alpha$; i.e. $\lambda_t = f(\alpha_t)$, for all $t \in [0;T]$ and a function $f : \mathbb{R} \rightarrow \mathbb{R}$.

My questions are:

• how can I prove that $W^\lambda$ and $\alpha$ are independent processes on $(\Omega,\mathcal{F},\mathbb{Q},\mathbb{F})$?

• how can I prove that $\alpha$ is always a continuous-time Markov chain on $(\Omega,\mathcal{F},\mathbb{Q},\mathbb{F})$?

For the first question, my idea is to prove that $$\int_\Omega 1_{F_1} 1_{F_2} \frac{d\mathbb{Q}}{d\mathbb{P}} d\mathbb{P} = \int_\Omega 1_{F_1} \frac{d\mathbb{Q}}{d\mathbb{P}} d\mathbb{P}\int_\Omega 1_{F_2} \frac{d\mathbb{Q}}{d\mathbb{P}} d\mathbb{P},$$ for all $F_1 \in \mathcal{F}^W$ and $F_2 \in \mathcal{F}^\alpha$, with $\mathcal{F}^W$ and $\mathcal{F}^\alpha$ the smallest $\sigma$-algebras for which the processes $W$ and $\alpha$ are measurable, respectively. But I don't know how I could prove that, if it's possible... However, for the second question, a hint would also be welcome... –  user73191 Apr 27 '13 at 12:15
That's exactly why I'm here, to ask for some help... It's not because I don't want to try by myself, but I'm here to ask for hints, if someone has any idea. I just remarked that the first question is not right, the good one is: "is $W^\lambda$ (not $W$) independent of $\alpha$?" –  user73191 Apr 27 '13 at 12:54