This is a (probably basic) question about the generator of a Markov process.

Let $(E,d)$ be a locally compact metric space. We consider a Feller process $X=(\{X_t\}_{t \ge 0},\{P_x\}_{x \in E})$ on $E$. That is, for any $t>0$, the semigroup $P_t$ of $X$ maps $C_{\infty}(E)$ into itself. Here, $C_\infty(E)$ denotes the space of continuous functions on $E$ vanishing at infinity.

We assume that the generator $L$ of $X$ is described as follows: for $f \in C_{\infty}(E)$ and $x \in E$, \begin{align*} Lf(x)=\int_{X \setminus \{x\}}(f(y)-f(x))c(x,y)\,\mu(dy). \end{align*} Here, $\mu$ denotes a Radon measure on $E$, and $c(x,y)$ a nonnegative bounded function on $E \times E \setminus \text{diag}$ with compact support. Therefore, the Feller processs $X$ is a pure jump process.

For a bounded open set $U$, we define $\tau_U=\inf\{t>0 \mid X_t \notin U\}$. The part process $X^U$ of $X$ on $U$ is defined as \begin{align*} X^U_t=\begin{cases} X_t,&\quad t<\tau_U,\\ \partial ,&\quad t \ge \tau_U. \end{cases} \end{align*} We assume also that $X^U$ is also Feller process. Then, can we describe the generator $L^U$ of $X^U$?

We write $E_x$ for the expectation under $P_x$, the law of $X$ starting from $x$. For $f \in C_{\infty}(U)$ and $x \in U$, we have \begin{align*} L^Uf(x)&=\lim_{t \to 0}\frac{E_{x}[f(X^U_t)]-f(x)}{t}\\ &=\lim_{t \to 0}\left(\frac{E_{x}[f(X_t)]-f(x)}{t}-\frac{E_{x}[f(X_t):t \ge \tau_U]}{t} \right). \end{align*} Can wa characterized the quantity $\lim_{t \to 0}E_{x}[f(X_t):t \ge \tau_U]/t$ in terms of $c(x,y)$ and $\mu$?


1 Answer 1


My second advice for those who work with generators of Markov processes is: Use Dynkin's characteristic operator! [*]

If $f$ is in the domain of the generator $L$ of $X$, then $$ L f(x) = \lim_{B \to \{x\}} \frac{E_x f(X(\tau_B)) - f(x)}{E_x \tau_B} \, , $$ where the expression under the limit means, for example, that $B = B(x, r)$ with $r \to 0^+$. The same formula holds for $f$ in the domain of the generator $L^U$ of $X^U$, as long as we restrict the class of sets $B$ to those that are contained in $U$.

So if $f$ is in the domain of both $L$ and $L^U$, then we get $L f(x) = L^U f(x)$. Therefore, if the intersection of the domains of $L$ and $L^U$ is sufficiently rich (for example, if for every $x \in U$ and every $f$ in the domain of $L^U$, there is $g$ in the domain of $L$ such that $f = g$ in a neighbourhood of $x$), then you get the result that you are looking for.

In the general case, things can get more complicated. However, your expression for the generator suggests that you know in advance that the domains of the generators are rich enough, so I will stop here.

Edit: For the record, let me add a short description of a peculiar example where there are too few functions in the intersection of the domains of $L$ and $L^U$.

Let $X_t$ be the standard 1-D Brownian motion with added extra jumps from $2$ to $0$, according to the following rule. We let $X_t$ run as the usual Brownian motion until the local time at $2$ hits an independently chosen, exponentially distributed threshold. At that time, we artificially move $X_t$ to zero, and restart the above procedure. Lather, rinse, repeat.

A function $f$ belongs to the domain of $L$ if and only if it is $C^2_\infty$ in $\mathbb R \setminus \{2\}$, with $f''_+(2) = f''_-(2)$ and $$ (f'_+(2) - f'_-(2)) + (f(0) - f(2)) = 0 . $$ Suppose that $U = (-1, 1)$. If $f$ is in the domain of both $L$ and $L^U$, then the latter condition implies that $f = 0$ in the complement of $U$), so that the former condition leads to simply $f(0) = 0$.

It follows that every function $f$ in the intersection of the domains of $L$ and $L^U$ satisfies $f(0) = 0$. This is already bad, but still not too bad.

Consider, however, a similarly constructed process which jumps from $1 + n$ to $q_n$, where $q_n$ is the enumeration of rational numbers in $U = (-1, 1)$. By the same argument, every function in the intersection of the domains of $L$ and $L^U$ is zero at every $q_n$, and hence identically zero. This is bad.

Of course these two processes are seemingly quite different from the ones mentioned in the question, but in fact I am rather sure one can cook up similar examples using only jumps governed by sufficiently irregular functions $c(x,y)$.

[*] The first advice reads: Don't do that unless you really have to. Most of the time there is a way to avoid the use of the generator whatsoever.

  • $\begingroup$ Thank you for your answer! I recognized that the generator of the part process is identified with that of the original process under a relatively general setting. $\endgroup$
    – sharpe
    Sep 21, 2021 at 8:39
  • $\begingroup$ This is indeed rather general, but not entirely general. I added a comment to my answer to show possible limitations. Generators are often very counter-intuitive creatures. $\endgroup$ Sep 21, 2021 at 10:55
  • $\begingroup$ Thank you for writing a concrete example. Dealing with generators is not easy for me... $\endgroup$
    – sharpe
    Sep 21, 2021 at 11:42
  • $\begingroup$ @MateuszKwaśnicki Could you elaborate on your advice in [*] not to use the generator if possible? Why is it undesirable to work with it, and how can one avoid it? $\endgroup$
    – Nate River
    Sep 22, 2021 at 5:14
  • $\begingroup$ Why? — It's a wild animal! How? — Get a domesticated one instead, like the resolvent or the semigroup. This is a partially humorous advice that I actually got from prof. Byczkowski. The main reason to avoid the use of generators is related to problems with description of its domain. The question here is a good example: the relationship between the original process and its part ("killed process") is absolutely clear, but at the level of generators it is extremely difficult to describe $L^U$ in terms of $L$, at least in the general case. $\endgroup$ Sep 22, 2021 at 9:12

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