Uniformization constructs every finite-state rate matrix chain

Put ${\lambda }_i={\sum }_{j\mathrm{\ne }i}{L}_{ij}$ and $q={\max }_i{\lambda }_i$. If $q=0$, all off-diagonal entries vanish and the zero row sums force $L=0$. On the finite probability space $(E,2^E,\nu )$ put $Y_0(\omega )=\omega$ and $X_t=Y_0$ for all $t$. Then $N_t=0$, paths are constant, the conditional transition is the identity, and $e^{tL}=I$, giving the conclusion in this case.

Suppose $q>0$. For $i\mathrm{\ne }j$, $R_{ij}=L_{ij}\mathrm{/}q\ge 0$, while $R_{ii}=1-{\lambda }_i\mathrm{/}q\ge 0$; each row sums to $1$. Let ${\mu }_q(dw)=q{e}^{-qw}dw$ on $(0,\mathrm{\infty })$, a probability measure because its density integrates to $1$. On $E\times (0,\mathrm{\infty })$ use initial law $\nu \otimes {\mu }_q$ and transition kernel $K((i,w),\{j\}\times B)=R_{ij}{\mu }_q(B)$ for Borel $B\subseteq (0,\mathrm{\infty })$. For any Borel subset of the product, its kernel value is the finite sum over $j$ of $R_{ij}$ times the ${\mu }_q$-measure of the corresponding section; thus it is measurable, nonnegative and has total mass $1$.

Apply [F2] to this initial law and kernel, and write the resulting coordinates as $Z_n=(Y_n,W_{n+1})$. Their finite-prefix law factors as the Markov-chain prefix law for $Y$ times ${\prod }_{k=1}^{n+1}{\mu }_q(d{W}_k)$, because every transition kernel chooses the next state with probabilities $R_{ij}$ and the next waiting time with the same factor ${\mu }_q$, independently of the current waiting time. Cylinder events generate the two coordinate sigma-algebras, so [F3] extends this factorization to independence of the full sequence $Y$ from the iid family $(W_k)$. The first coordinate is an $R$-Markov sequence with initial law $\nu$, as required by [F1].

Each $W_k$ has probability $e^{-1}$ of being at least $q^{-1}$. For any fixed $r$, independence gives $\mathbb{P}(W_k<q^{-1}\text{ for }r\le k\le n)=(1-e^{-1}{)}^{n-r+1}\to 0$. Continuity of probability for decreasing events, obtained by applying [F4] to complements, shows that the probability of no such event after $r$ is zero. A countable union over $r$ shows that $W_k\ge q^{-1}$ occurs infinitely often almost surely, and therefore ${\sum }_k{W}_k=\mathrm{\infty }$ almost surely. On every bounded time interval there are consequently finitely many event times; the convention $X_t=Y_{N_t}$ makes paths right-continuous with left limits.

For $n\ge 1$, convolution of the exponential densities gives the arrival-time density $f_{T_n}(u)=\frac{q^n{u}^{n-1}{e}^{-qu}}{(n-1)!},u>0.$ Indeed the formula holds for $n=1$, and convolving it with $q{e}^{-q(t-u)}$ gives $q^{n+1}{e}^{-qt}{\int }_0^t{u}^{n-1}du\mathrm{/}(n-1)!=q^{n+1}{t}^n{e}^{-qt}\mathrm{/}n!$. Since $W_{n+1}$ is independent of $T_n$, for $n\ge 1$, $\mathbb{P}(N_t=n)={\int }_0^t{f}_{T_n}(u)\mathbb{P}(W_{n+1}>t-u)du=e^{-qt}\frac{(qt{)}^n}{n!};$ for $n=0$ the probability is $\mathbb{P}(W_1>t)=e^{-qt}$.

The event process has independent rate-$q$ Poisson increments. To see the conditional assertion needed here, on $\{N_s=n\}$ the event-clock history gives $T_n\le s<T_{n+1}$; by the exponential memoryless identity $\mathbb{P}(W_{n+1}-(s-T_n)>u\mid W_{n+1}>s-T_n)=e^{-qu}$, the residual time after $s$ is rate-$q$ exponential independent of that history. The later waiting times are independent rate-$q$ exponentials as well. Repeating the calculation in step 3.2 shows that, conditional on the clock history up to $s$, $N_{s+h}-N_s$ has probabilities $e^{-qh}(qh{)}^k\mathrm{/}k!$, independent of that history. The clock is independent of $Y$ by step 2.1 and the uniformized convention [F1].

Let $H_s=(Y_0,\dots ,Y_{N_s})$ be the finite embedded-state prefix and put ${\mathcal{G}}_s=\sigma (N_u:0\le u\le s,H_s)$. Since $N_u\le N_s$ for $u\le s$, the construction filtration ${\mathcal{F}}_s=\sigma (N_u,X_u:0\le u\le s)$ is contained in ${\mathcal{G}}_s$. Given ${\mathcal{G}}_s$ and $X_s=i$, $k$ further clock events have probability $e^{-qh}(qh{)}^k\mathrm{/}k!$ by step 4.1, and the embedded chain has $k$-step transition matrix $R^k$. Hence $\mathbb{P}(X_{s+h}=j\mid {\mathcal{G}}_s)={\sum }_{k=0}^{\mathrm{\infty }}{e}^{-qh}\frac{(qh{)}^k}{k!}(R^k{)}_{ij}=(e^{-qh}{e}^{qhR}{)}_{ij}=(e^{hL}{)}_{ij},$ since $q(R-I)=L$. This conditional law depends on ${\mathcal{G}}_s$ only through $X_s$; [F5] applied to the integrable indicator ${\mathbf{1}}_{\{X_{s+h}=j\}}$ and ${\mathcal{F}}_s\subseteq {\mathcal{G}}_s$ gives the required identity for all $s\le t$. The matrix series converges absolutely because $E$ is finite.

The constructed $X$ is adapted to ${\mathcal{F}}_t$ by definition. Step 3.1 gives nonexplosion and càdlàg paths; step 5.1 gives its all-pairs conditional transition matrices. Thus it is a uniformized realization for every finite rate matrix, including reducible matrices and states with ${\lambda }_i=0$.