Let
- $(E,\mathcal E)$ be a measurable space
- $\mathcal E_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$
- $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$
- $Q$ denote the weak generator of $(\kappa_t)_{t\ge0}$; i.e. $$\mathcal D(Q):=\left\{f\in\mathcal E_b\mid\forall x\in E:[0,\infty)\ni t\mapsto(\kappa_tf)(x)\text{ is right-differentiable at }0\right\}$$ and $$(Qf)(x):=\left.\frac{\rm d}{{\rm d}t}(\kappa_tf)(x)\right|_{t=0+}\;\;\;\text{for }x\in E\text{ and }f\in\mathcal D(Q)$$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(Y_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued time-homogeneous Markov process on $(\Omega,\mathcal A,\operatorname P)$ with transition semigroup $(\kappa_t)_{t\ge0}$
Question: Let $$X_t:=\left.\begin{cases}Y^{(1)}_t&\text{, if }t<\tau_1;\\Y^{(2)}_{t-\tau_1}&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }t\ge0,$$ where $\tau_1$ is a finite stopping time on $(\Omega,\mathcal A)$ with respect to the filtration $(\mathcal F^Y_t)_{t\ge0}$ generated by $(Y_t)_{t\ge0}$ and $(Y^{(1)}_t)_{t\ge0}$ and $(Y^{(2)}_t)_{t\ge0}$ are independent $(E,\mathcal E)$-valued processes on $(\Omega,\mathcal A,\operatorname P)$ with the same distribution as $(Y_t)_{t\ge0}$.
How can we determine the transition semigroup $(T_t)_{t\ge0}$ and weak generator $L$ of $(X_t)_{t\ge0}$?
In order to determine $(T_t)_{t\ge0}$ I've tried to assume that $\tau_1(\Omega)$ is countable in a first step. This allows us to write $$\operatorname P\left[X_{s+t}\in B\mid\mathcal F^X_s\right]=\sum_{\substack{r\in\tau_1(\Omega)\\r<s+t}}\operatorname P\left[Y^{(1)}_{s+t}\in B;\tau_1=r\mid\mathcal F^X_s\right]+\sum_{\substack{r\in\tau_1(\Omega)\\r\ge s+t}}\operatorname P\left[Y^{(2)}_{s+t-r}\in B;\tau_1=r\mid\mathcal F^X_s\right]\tag1$$ for all $B\in\mathcal E$ and $s,t\ge0$. But this is not very helpful for me ...