Construction of a Markov process with prescribed local behavior and state-dependent jump distribution

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.

Answer 1

The construction given by the OP is almost correct. Here is a slight correction: $$ X_t = \sum_{n=0}^{\infty} 1_{[\tau_n,\tau_{n+1})}(t) Y_{t - \tau_n}^{n} \;, \tag{1} $$ where we have introduced

• $\{\tau_i\}$ are a sequence of jump times defined via $\tau_{i+1}=\tau_i+\xi_i$, $\tau_0=0$, and $\{\xi_i \} \overset{i.i.d.}{\sim} \operatorname{Exp}(1)$ ; and,
• $\{Y^{i}\}$ are independent realizations of $Y$ with $Y_0^i=x$ if $i=0$ and else sample $Y_0^i \mid (Y^0, \dots, Y^{i-1}, \xi_0, \dots, \xi_i) \sim \alpha(Y_{\tau_i - \tau_{i-1}}^{i-1}, \cdot)$ .

In other words, $$ X_t = \begin{cases} Y^0_t & t < \tau_1 \;, \\ Y^1_{t-\tau_1} & \tau_1 \le t < \tau_2 \;, \\ Y^2_{t-\tau_2} & \tau_2 \le t < \tau_3 \;, \\ \vdots \end{cases} $$

To see that the weak generator of (1) is indeed $L=Q+A$, write $f (X_t) - f(x) = \rm{I} + \rm{II} + \rm{III}$ where \begin{align*} \rm{I} &:= (f(X_t) - f(x)) 1_{\{t < \tau_1 \}} \;, \\ \rm{II} &:= (f(Y_0^1) - f(Y_{\tau_1}^0)) 1_{\{t \ge \tau_1 \}} \;, \\ \rm{III} &:= (f(Y_{\tau_1}^0) - f(x) + f(X_t) - f(Y_0^1)) 1_{\{t \ge \tau_1 \}} \;. \end{align*} Then \begin{align*} E[\rm{I}] &= e^{-t} ( \kappa_t f(x) - f(x) ) = e^{-t} E \int_0^t Qf (Y_s^0) ds \;, \\ E[{\rm II} \mid \tau_1 = s] &= E[f(Y_0^1) - f(Y_{s}^0)] 1_{\{ t \ge s \}} = E[ A f(Y_s^0) ] 1_{\{ t \ge s \}} \;, \\ E[ \rm{II} ] &= E \int_0^{\infty} E[ {\rm II} \mid \tau_1 = s] e^{-s} ds = E \int_0^t e^{-s} A f(Y_s^0) ds \;. \end{align*} One can similarly show that $E( \rm{III} )$ is $O(t^2)$ for $t \in [0,1]$. Therefore, combining the above and using $(e^{-s} - e^{-t}) \le (t-s)$ for $t \ge s$, one obtains that for all $ t \in [0,1]$ $$ E[f(X_t)] - f(x) = E\int_0^t (A f(Y_s^0) + Q f(Y_s^0)) ds + O(t^2) \;. $$ While this construction/analysis covers the case of constant jump rates, the case of state-dependent jump rates can be treated similarly as discussed in the comments below.