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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]$ $$ E[f(X_t)] - f(x) = E\int_0^t (A f(Y_s^0) + Q f(Y_s^0)) ds + O(t^2) \;. $$

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.

The construction given in the question 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]$ $$ E[f(X_t)] - f(x) = E\int_0^t (A f(Y_s^0) + Q f(Y_s^0)) ds + O(t^2) \;. $$

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]$ $$ E[f(X_t)] - f(x) = E\int_0^t (A f(Y_s^0) + Q f(Y_s^0)) ds + O(t^2) \;. $$

The construction given in the question 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 $Y_0=x$ if $i=0$ and else $Y_0 \sim \alpha(Y_{\tau_i - \tau_{i-1}}^{i-1}, \cdot)$ .

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]$ $$ E[f(X_t) - f(x)] = E\int_0^t (A f(Y_s^0) + Q f(Y_s^0)) ds + O(t^2) \;. $$

The construction given in the question 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]$ $$ E[f(X_t)] - f(x) = E\int_0^t (A f(Y_s^0) + Q f(Y_s^0)) ds + O(t^2) \;. $$