Killed and absorbed transition kernels

Let $K$ be a probability kernel on $(E,\mathcal{E})$ and $D\in \mathcal{E}$. The kernel absorbed on $D$ is the kernel candidate on $E$ defined by $K_D^{\mathrm{a}\mathrm{b}\mathrm{s}}(x,A)=1_D(x)1_A(x)+1_{D^c}(x)K(x,A),A\in \mathcal{E}\mathrm{.}$ Thus every $x\in D$ has the Dirac transition ${\delta }_x$, while transitions from $D^c$ are unchanged. For killing, adjoin a point $\mathrm{\Delta }\mathrm{\notin }E$ and equip $E_{\mathrm{\Delta }}=E\bigsqcup \{\mathrm{\Delta }\}$ with ${\mathcal{E}}_{\mathrm{\Delta }}=\{A:A\in \mathcal{E}\}\cup \{A\cup \{\mathrm{\Delta }\}:A\in \mathcal{E}\}\mathrm{.}$ The kernel killed upon exiting $D$ is $K_D^{\mathrm{k}\mathrm{i}\mathrm{l}\mathrm{l}}(x,B)=\{\begin{array}{ll}K(x,B\cap D)+K(x,D^c)1_B(\mathrm{\Delta }),& x\in D,\\ 1_B(\mathrm{\Delta }),& x\in D^c\cup \{\mathrm{\Delta }\},\end{array}B\in {\mathcal{E}}_{\mathrm{\Delta }}\mathrm{.}$ In particular, a start outside $D$ is killed immediately and the cemetery is absorbing. If $D=E$, killing never occurs from $E$; if $D=\mathrm{\varnothing }$, every start is sent to $\mathrm{\Delta }$. These formulas are choice-free.

The following schematics compare the two modifications. An arrow to a set denotes a transition law on that set, not a transition to a single state. For absorption, the loop at $x\in D$ has total mass one; from $y\in D^c$ the entire original law is retained.

For killing, the upper arrow retains the measure $A\mapsto K(x,A\cap D)$ on $D$; the other arrow redirects all exit mass to $\mathrm{\Delta }$. Every outside start goes to $\mathrm{\Delta }$ with mass one, and $\mathrm{\Delta }$ then stays fixed.

The diagrams illustrate the formulas in the Definition and do not impose a finite state space or any additional assumptions. In the boundary case $D=E$, there are no outside starts and $K(x,D^c)=0$ for $x\in E$; for $D=\mathrm{\varnothing }$, the killing diagram has only the transitions to the cemetery. The definition is independent of these illustrations.