• $(\Omega,\mathcal A)$ be a measurable space;
  • $(E,\mathcal E)$ be a measurable space with $\{x\}\in\mathcal E$;
  • $(Y_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued time-homogeneous Markov process on $(\Omega,\mathcal A)$;
  • $\operatorname P_x$ be a probability measure on $(\Omega,\mathcal A)$ with $$\operatorname P_x[Y_0=x]=1$$ for $x\in E$;
  • $\rho:E\to[0,\infty)$ be $\mathcal E$-measurable and $$M_t:=\exp\left(-\int_0^t\rho(Y_s)\:{\rm d}s\right)\;\;\;\text{for }t\ge0;$$
  • $$(Q_tf)(x):=\operatorname E_x\left[M_tf(Y_t)\right]$$ for $x\in E$, bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $t\ge0$.

Let $A$ denote the generator of $(Y_t)_{t\ge0}$. We easily see that the generator $B$ of $(Q_t)_{t\ge0}$ is given by $$(Bf)(x)=\lim_{t\to0+}\frac{(Q_tf)(x)-f(x)}t=(Af)(x)-f(x)\rho(x)\tag1$$ for all $x\in E$ and bounded $\mathcal E$-measurable $f:E\to\mathbb R$.

Now let $\xi$ be an exponentially ditributed (with parameter $1$) random variable on $(\Omega,\mathcal A)$, independent of $Y$, and $$\tau:=\inf\left\{t\ge0:\int_0^t\rho(Y_s)\:{\rm d}s\ge\xi\right\}.$$ Furthermore, assume that $\pi(x,\;\cdot\;):=\operatorname P_x$ for $x\in E$ is a Markov kernel and let

  • $(Y^0,\tau^0)$ be a realization of $(Y,\tau)$ with $Y_0=x$;
  • $\mu$ be a probability measure on $(E,\mathcal E)$;
  • $(Y^n,\tau^n)$, $n\in\mathbb N$, be independent identically distributed realizations of $(Y,\tau)$ with $Y_0\sim\mu$ under $\operatorname P_\mu:=\mu\pi$.

How can we show that the generator $C$ of $$X_t:=\sum_{n=0}^\infty1_{[T^n,\:T^{n+1})}(t)Y^{n}_{t-T_n}\;\;\;\text{for }t\ge0,$$ where $T^n:=\sum_{i=0}^{n-1}\tau^i$ for $n\in\mathbb N_0$, is given by $$(Cf)(x)=(Af)(x)+\rho(x)\int (f(y)-f(x))\:\mu({\rm d}y)\tag2$$ for all $x\in E$ and bounded $\mathcal E$-measurable $f:E\to\mathbb R$?

Most probably (but I don't know) the domain of $C$ (or of $A$) is not the whole space of bounded $\mathcal E$-measurable functions (equipped with the supremum norm) and maybe we need to impose some regularity assumptions. But even with that in mind, what is the formal (not necessarily rigorous) way to obtain $(2)$?

  • 1
    $\begingroup$ A good start would be to take $A=0$, so that $X$ is a continuous-time Markov chain, and convince yourself that the formula gives the correct generator. Letting $E$ be discrete may make it even more clear. $\endgroup$ May 28, 2022 at 0:59
  • $\begingroup$ @NateEldredge Thank you very much for your helpful comment. Do you sugget to directly calculate $(Cf)(x)=\lim_{t\to0+}\frac{(T_tf)(x)-f(x)}t$, where $(T_tf)(x)=\operatorname E_x[f(X_t)]$, in these cases? $\endgroup$
    – 0xbadf00d
    May 28, 2022 at 9:10
  • $\begingroup$ @NateEldredge Assuming $E$ is discrete, do you mean to directly calculate $\operatorname P_\mu[X_{s+t}=x\mid\mathcal F_s]=\sum_{k=1}^\infty\operatorname P_\mu\left[s+t\in\left[\sum_{i=1}^{k-1}\tau_i,\sum_{i=1}^k\tau_i\right),Y^{(k)}_{s+t-\sum_{i=1}^{k-1}\tau_i}=x\mid\mathcal F_s\right]$? Seems still to be quite cumbersome. $\endgroup$
    – 0xbadf00d
    May 28, 2022 at 9:27
  • $\begingroup$ @NawafBou-Rabee Thanks for mentioning. I've fixed the definition. However, meanwhile I think the question boils down to the following easier to describte question: math.stackexchange.com/q/4464715/47771. Maybe you can take a look. $\endgroup$
    – 0xbadf00d
    Jun 3, 2022 at 12:59

1 Answer 1


By conditioning on $\tau_0$: $$ T_tf(x) = Q_tf(x) +E_x\left(\int_0^t \mu [T_{t-s}f]\rho(Y_s)\,ds\right). $$ Differentiate at $t=0$ to get $$ Cf(x)=Bf(x)+\mu(f)\cdot \rho(x)=Af(x)+\rho(x)\cdot \int_E [f(y)-f(x)]\mu(dy). $$ A more rigorous treatment could be made using resolvents (Laplace transform in $t$).

  • 1
    $\begingroup$ Thank you for your answer. What do you mean "by conditioning on $\tau_0$"? And I guess you're defining $(T_tf)(x):=\operatorname E_x[f(X_t)]$ as I did in my comment below the question, right? $\endgroup$
    – 0xbadf00d
    May 28, 2022 at 19:22
  • $\begingroup$ Since I infer from your answer that you could clarify my question, I would be really thankful for a response. $\endgroup$
    – 0xbadf00d
    May 31, 2022 at 15:39
  • $\begingroup$ You have $E_x[\tau_0\le t]=E_x[\int_0^t M_s\rho(Y_s)\, ds]$. On the event $\{\tau_0\le t\}$, given that $\tau_0=s$, the conditional distribution of $X_t$ is the same as that of $X$ started at time $t-s$ in a state chosen according to $\mu$ (and independent of what has gone before). $\endgroup$ May 31, 2022 at 23:53
  • $\begingroup$ Thank you for your comment. Well, okay, now I see where my problem is: How did you obtain $E_x[\tau_0\le t]=E_x[\int_0^t M_s\rho(Y_s)\, ds]$? $\endgroup$
    – 0xbadf00d
    Jun 1, 2022 at 6:04
  • 1
    $\begingroup$ $$ P_x[\tau_0>t]=P_x\left[\int_0^t\rho(Y_s)\,ds<\xi\right]=E_x\left[\exp\left(-\int_0^t\rho(Y_s)\,ds\right)\right]=E_x[M_t]. $$ And $$ \int_0^t M_s\rho(Y_s)\,ds =-\int_0^t dM_s=1-M_t. $$ $\endgroup$ Jun 1, 2022 at 14:45

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