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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 $(\kappa_t)_{t\ge0}$
• $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$(Af)(x):=\int f(y)-f(x)\:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E_b$$

Question: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$

The idea is that the local behavior between jumps of $(X_t)_{t\ge0}$ is described by $(Y_t)_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.

The process should be described by something like $$X_t=\sum_{n\in\mathbb N_0}1_{[\tau_n,\:\tau_{n+1})}(t)Y^{(n)}_{t-\tau_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.

However, how do we need to define the $\tau_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?

I'm aware of the following simpler result: If $(W_n)_{n\in\mathbb N_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $r>0$ and $W$ is independent of $N$, then $$Z_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-r)}\right)_{t\ge0}$ and generator $r\left(\kappa-\operatorname{id}_{\mathcal E_b}\right)$.

In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W_n=\sum_{i=1}^n\xi_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$

Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.

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 $(\kappa_t)_{t\ge0}$
• $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$Af(x):=\int_E ( f(y)-f(x)) \:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E_b$$

Question: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$

The idea is that the local behavior between jumps of $(X_t)_{t\ge0}$ is described by $(Y_t)_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.

The process should be described by something like $$X_t=\sum_{n\in\mathbb N_0}1_{[\tau_n,\:\tau_{n+1})}(t)Y^{(n)}_{t-\tau_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.

However, how do we need to define the $\tau_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?

I'm aware of the following simpler result: If $(W_n)_{n\in\mathbb N_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $r>0$ and $W$ is independent of $N$, then $$Z_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-r)}\right)_{t\ge0}$ and generator $r\left(\kappa-\operatorname{id}_{\mathcal E_b}\right)$.

In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W_n=\sum_{i=1}^n\xi_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$

Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.

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 $(\kappa_t)_{t\ge0}$
• $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$(Af)(x):=\int f(y)-f(x)\:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E_b$$

Question: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$

The idea is that the local behavior between jumps of $(X_t)_{t\ge0}$ is described by $(Y_t)_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.

The process should be described by something like $$X_t=\sum_{n\in\mathbb N_0}1_{[\tau_n,\:\tau_{n+1})}(t)Y^{(n)}_{t-\tau_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.

However, how do we need to define the $\tau_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?

I'm aware of the following simpler result: If $(W_n)_{n\in\mathbb N_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\alpha>0$ and $W$ is independent of $N$, then $$Z_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-\alpha)}\right)_{t\ge0}$ and generator $\alpha\left(\kappa-\operatorname{id}_{\mathcal E_b}\right)$.

In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W_n=\sum_{i=1}^n\xi_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$

Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.

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 $(\kappa_t)_{t\ge0}$
• $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$(Af)(x):=\int f(y)-f(x)\:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E_b$$

Question: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$

The idea is that the local behavior between jumps of $(X_t)_{t\ge0}$ is described by $(Y_t)_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.

The process should be described by something like $$X_t=\sum_{n\in\mathbb N_0}1_{[\tau_n,\:\tau_{n+1})}(t)Y^{(n)}_{t-\tau_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.

However, how do we need to define the $\tau_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?

I'm aware of the following simpler result: If $(W_n)_{n\in\mathbb N_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $r>0$ and $W$ is independent of $N$, then $$Z_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-r)}\right)_{t\ge0}$ and generator $r\left(\kappa-\operatorname{id}_{\mathcal E_b}\right)$.

In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W_n=\sum_{i=1}^n\xi_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$

Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.

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 $(\kappa_t)_{t\ge0}$
• $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$(Af)(x):=\int f(y)-f(x)\:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E_b$$

Question: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$

The idea is that the local behavior between jumps of $(X_t)_{t\ge0}$ is described by $(Y_t)_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.

The process should be described by something like $$X_t=\sum_{n\in\mathbb N_0}1_{[\tau_n,\:\tau_{n+1})}(t)Y^{(n)}_{t-\tau_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.

However, how do we need to define the $\tau_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?

I'm aware of the following simpler result: If $(W_n)_{n\in\mathbb N_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\alpha>0$ and $W$ is independent of $N$, then $$Z_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-\alpha)\right)_{t\ge0}$ and generator $\alpha\left(\kappa-\operatorname{id}_{\mathcal E_b}\right)$.

In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W_n=\sum_{i=1}^n\xi_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$

Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.

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 $(\kappa_t)_{t\ge0}$
• $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$(Af)(x):=\int f(y)-f(x)\:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E_b$$

Question: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$

The idea is that the local behavior between jumps of $(X_t)_{t\ge0}$ is described by $(Y_t)_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.

The process should be described by something like $$X_t=\sum_{n\in\mathbb N_0}1_{[\tau_n,\:\tau_{n+1})}(t)Y^{(n)}_{t-\tau_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.

However, how do we need to define the $\tau_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?

I'm aware of the following simpler result: If $(W_n)_{n\in\mathbb N_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N_t)_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\alpha>0$ and $W$ is independent of $N$, then $$Z_t:=W_{N_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-\alpha)}\right)_{t\ge0}$ and generator $\alpha\left(\kappa-\operatorname{id}_{\mathcal E_b}\right)$.

In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W_n=\sum_{i=1}^n\xi_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$

Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.