Finite-state killed projection has an exact lost-mass certificate

For each source state $x\in F$, the off-diagonal entries of ${\tilde{Q}}_F$ are the internal rates $q(x,y)$ and the exit rate to $†$ is ${\sum }_{y\mathrm{\notin }F}q(x,y)$. Their sum is $\lambda (x)$, so the diagonal $-\lambda (x)$ makes the column sum zero. The cemetery column is zero. Therefore ${\tilde{Q}}_F$ is a finite column generator.

By the definition of $Y$, for each $x\in F$ the event $\{Y_t=x\}$ is exactly $\{X_t=x,{\tau }_F>t\}$; the complementary event $\{Y_t=†\}$ is $\{{\tau }_F\le t\}$. Therefore $u_F(t,x)=\mathbb{P}(X_t=x,{\tau }_F>t)\le p_t(x)$ and $c_F(t)=\mathbb{P}({\tau }_F\le t)$. Returning paths are included in $p_t$ but excluded from $u_F$ by the strict first-exit condition.

Let $M_F={\max }_{x\in F}\lambda (x)$, finite because $F$ is finite. For $0\le h\le 1$, [F6] bounds the exponential series remainder by $\mathrm{\mid }{e}^{-\lambda (x)h}-(1-\lambda (x)h)\mathrm{\mid }\le h^2{\sum }_{k\ge 2}\lambda (x{)}^k\mathrm{/}k!$, so no-jump probability is $1-\lambda (x)h+o(h)$ uniformly on finite $F$. If $\lambda (x)>0$, the first-jump probability $1-e^{-\lambda (x)h}=\lambda (x)h+o(h)$ multiplied by the destination probability $q(x,y)\mathrm{/}\lambda (x)$ from [F1] gives rate $q(x,y)h+o(h)$ for each $y\mathrm{\ne }x$; if $\lambda (x)=0$, all rates from $x$ are zero and these probabilities are zero. For the aggregate exit event, its conditional destination probability is $a_x\mathrm{/}\lambda (x)$ when $\lambda (x)>0$, so its probability is $a_{x}h+o(h)$; when $\lambda (x)=0$, $a_x=0$ and this probability is zero. Here $a_x=\lambda (x)-{\sum }_{y\in F,y\mathrm{\ne }x}q(x,y)\ge 0$. After reaching $†$ there are no further changes. By [F5] applied to $-u$, $1-e^{-u}\le u$ for $u\ge 0$, so the probability of a first change within $h$ is at most $M_{F}h$; conditional on a first change that stays in $F$, a second change within the remaining time has probability at most $M_{F}h$. Hence the probability of two or more changes is at most $M_F^2{h}^2$. Thus the finite transition matrix of $Y$ is $I+h{\tilde{Q}}_F+o(h)$ as $h\downarrow 0$, uniformly over its finite state space.

Extend $u_F$ by zero outside $F$. For $x\in F$, $p_t(x)-u_F(t,x)\ge 0$ by step 1.2; for $x\mathrm{\notin }F$, the difference is $p_t(x)\ge 0$. Summing these nonnegative differences over $S$ gives $1-{\sum }_{x\in F}{u}_F(t,x)=1-\mathbb{P}({\tau }_F>t)=c_F(t)$. By [F4], this sum is exactly $\mathrm{\parallel }{p}_t-{\bar{u}}_F(t){\mathrm{\parallel }}_{{\mathrm{\ell }}^1(S)}$.

The process $Y$ is Markov: before killing this follows from the Markov property of $X$, and after killing its state $†$ is absorbing. Hence its finite column transition matrix $T_Y(h)$ satisfies $T_Y(h)=I+h{\tilde{Q}}_F+o(h)$ entrywise by step 2.1, and its law satisfies $\tilde{p}(t+h)=T_Y(h)\tilde{p}(t)$. This gives the right derivative ${\tilde{Q}}_F\tilde{p}(t)$. For $t>0$, $\tilde{p}(t)=T_Y(h)\tilde{p}(t-h)$. Since the state space has finitely many coordinates, the entrywise expansion and ${\sum }_j{\tilde{p}}_j(t-h)=1$ imply each coordinate of $\tilde{p}(t)-\tilde{p}(t-h)$ is $O(h)$; hence $\tilde{p}(t-h)\to \tilde{p}(t)$. Dividing the identity by $h$ and using the entrywise limit $(T_Y(h)-I)\mathrm{/}h\to {\tilde{Q}}_F$ gives the left derivative ${\tilde{Q}}_F\tilde{p}(t)$ as well, because every coordinate is a finite sum. Thus ${\tilde{p}}^{\mathrm{\prime }}={\tilde{Q}}_F\tilde{p}$ with initial law $\tilde{p}(0)$. By [F3], $\tilde{p}(t)=e^{t{\tilde{Q}}_F}\tilde{p}(0)$.

Let $a=\inf f$, $b=\sup f$, and $c=(a+b)\mathrm{/}2=f(†)$. The coupling uses the same path $X$: on $\{{\tau }_F>t\}$, $Y_t=X_t$, and on $\{{\tau }_F\le t\}$, $Y_t=†$. Thus $\mathrm{\mid }\mathbb{E}f(X_t)-\mathbb{E}f(Y_t)\mathrm{\mid }=\mathrm{\mid }\mathbb{E}[{\mathbf{1}}_{\{{\tau }_F\le t\}}(f(X_t)-c)]\mathrm{\mid }\le \frac{1}{2}(b-a)\mathbb{P}({\tau }_F\le t)=\frac{1}{2}\mathrm{osc}(f)c_F(t)$, as asserted.