Nonexplosive finite continuous-time chain and reversible spectral gap

Set $\mathrm{\Lambda }>{\max }_a{q}_a$ and $K:=I+Q\mathrm{/}\mathrm{\Lambda }$. Every off-diagonal entry of $K$ is nonnegative, $K_{aa}=1-q_a\mathrm{/}\mathrm{\Lambda }>0$, and each row sums to one, so $K$ is a transition matrix. The balance relation in [F1] gives ${\pi }_a{K}_{ab}={\pi }_b{K}_{ba}$ for all $a,b$. Use [F2] with state space $E\times (0,1)$, initial state distributed as $p_0$ with an independent uniform mark, and transition kernel $K(a,b)d{u}^{\mathrm{\prime }}$ from $(a,u)$ to $(b,u^{\mathrm{\prime }})$. Write $Y_n$ for the state coordinate and $U_n$ for the uniform mark; the sequence $Y_n$ has transitions $K$, and the marks $U_n$ are independent uniforms independent of the state sequence. The holding intervals ${\tau }_n:=-\log U_n\mathrm{/}\mathrm{\Lambda }$ are independent exponentials of rate $\mathrm{\Lambda }$.

For every $n$, $\mathrm{Pr}({\tau }_n\ge 1\mathrm{/}\mathrm{\Lambda })=e^{-1}$. For each fixed $m$, independence gives probability $(1-e^{-1}{)}^N$ that none of ${\tau }_m,\dots ,{\tau }_{m+N-1}$ reaches $1\mathrm{/}\mathrm{\Lambda }$; this tends to zero. A countable union over $m$ shows that infinitely many holding intervals are at least $1\mathrm{/}\mathrm{\Lambda }$, almost surely. Therefore $S_n:={\sum }_{j=0}^{n-1}{\tau }_j\to \mathrm{\infty }$. For each $t\ge 0$, there is a unique $N(t)$ with $S_{N(t)}\le t<S_{N(t)+1}$; set $X_t:=Y_{N(t)}$. Thus the constructed path is piecewise constant and has only finitely many transition attempts on each bounded interval. This proves nonexplosion.

The sum of $k\ge 1$ independent rate-$\mathrm{\Lambda }$ exponential holding times has density ${\mathrm{\Lambda }}^k{s}^{k-1}{e}^{-\mathrm{\Lambda }s}\mathrm{/}(k-1)!$, as follows by induction using convolution. Integrating this density against the survival $e^{-\mathrm{\Lambda }(t-s)}$ of the next holding time gives $\mathrm{Pr}(N(t)=k)=e^{-\mathrm{\Lambda }t}(\mathrm{\Lambda }t{)}^k\mathrm{/}k!$; for $k=0$ it is $e^{-\mathrm{\Lambda }t}$. Starting from state $a$, each uniformized attempt stays at $a$ with probability $K_{aa}=1-q_a\mathrm{/}\mathrm{\Lambda }$. Since $N(t)$ is Poisson and independent of the $K$-chain, the probability of no state change by time $t$ is ${\sum }_{k\ge 0}{e}^{-\mathrm{\Lambda }t}(\mathrm{\Lambda }t{)}^k{K}_{aa}^k\mathrm{/}k!=e^{-q_{a}t}$. At the first attempt that changes state, the destination $b\mathrm{\ne }a$ has conditional probability $K_{ab}\mathrm{/}(1-K_{aa})=w_{ab}\mathrm{/}{q}_a$. Thus the actual holding time and jump destination have exactly the law stated in the Definition. The exponential holding times are memoryless and independent of the $K$-chain, so after any time the remaining clock and future transitions have the same laws from the current state; conditioning on $N(t)$ therefore yields the time-homogeneous transition matrix $P_t=e^{-\mathrm{\Lambda }t}{\sum }_{k=0}^{\mathrm{\infty }}(\mathrm{\Lambda }t{)}^k{K}^k\mathrm{/}k!=e^{tQ}$ because $Q=\mathrm{\Lambda }(K-I)$. The finite-matrix series converges absolutely, differentiates termwise, and satisfies $P_t^{\mathrm{\prime }}=P_{t}Q$ and $P_0=I$. Each term is stochastic, hence $P_t$ is stochastic. For each state $b$, detailed balance gives $(\pi Q{)}_b={\sum }_{a\mathrm{\ne }b}{\pi }_a{w}_{ab}-{\pi }_b{q}_b={\pi }_b{\sum }_{a\mathrm{\ne }b}{w}_{ba}-{\pi }_b{\sum }_{a\mathrm{\ne }b}{w}_{ba}=0$; therefore $\pi P_t=\pi$.

For $f,g\in L^2(E,\pi )$, detailed balance gives $⟨f,Qg{⟩}_{\pi }=⟨Qf,g{⟩}_{\pi }$. Pairing the two directed terms for each edge gives $-⟨f,Qf{⟩}_{\pi }=\frac{1}{2}{\sum }_{a\in E}{\sum }_{b\mathrm{\ne }a}{\pi }_a{w}_{ab}(f_b-f_a{)}^2\ge 0$. Equality holds only when $f_a=f_b$ along every positive-rate edge; connectedness in [F1] then makes $f$ constant. Thus the zero eigenspace is exactly the constants. By [F3], $Q$ has a real orthonormal eigenbasis; nonpositivity and the one-dimensional kernel imply all other eigenvalues are $-{\lambda }_j$ with ${\lambda }_j>0$. Expanding in this basis gives $P_{t}f=\pi (f)1+{\sum }_{j=1}^{r-1}{e}^{-{\lambda }_{j}t}⟨f,{\varphi }_j{⟩}_{\pi }{\varphi }_j$, proving the positive gap and exponential convergence.

Apply [F2] on $((E\times (0,1){)}^2)$ with initial state whose two $E$ coordinates are equal with law $\pi$ and whose two marks $U_0^{+},U_0^{-}$ are independent uniforms. From a state pair $(a^{+},a^{-})$, use the product kernel $K(a^{+},b^{+})K(a^{-},b^{-})d{u}^{\mathrm{\prime }+}d{u}^{\mathrm{\prime }-}$; this makes the two $K$-chains conditionally independent given their shared initial state and gives independent uniform marks at every step. Denote the state coordinates by $Y_n^{\pm }$, set ${\tau }_n^{\pm }:=-\log U_n^{\pm }\mathrm{/}\mathrm{\Lambda }$, $S_0^{\pm }:=0$, and $S_n^{\pm }:={\sum }_{j=0}^{n-1}{\tau }_j^{\pm }$. For $s\ge 0$, let $N^{\pm }(s)$ be the unique index with $S_{N^{\pm }(s)}^{\pm }\le s<S_{N^{\pm }(s)+1}^{\pm }$ and put $Z_s^{\pm }:=Y_{N^{\pm }(s)}^{\pm }$. Step 2.1 applied to each clock sequence ensures these indices exist for every finite $s$. Put $X_t=Z_t^{+}$ for $t\ge 0$ and $X_t=Z_{-t}^{-}$ for $t<0$. For times entirely on the nonnegative half, the Markov-chain law has the stated product form. For negative times $t_1<\cdots <t_k<0$, append the time $0$ with state $c=X_0$. In the forward parameter $s=-t$, the ordered states are $c,a_k,\dots ,a_1$, so their joint probability is ${\pi }_c{P}_{-t_k}(c,a_k){\prod }_{i=1}^{k-1}{P}_{t_{i+1}-t_i}(a_{i+1},a_i)$. Applying ${\pi }_a{P}_u(a,b)={\pi }_b{P}_u(b,a)$ first to the factor from $0$ to $t_k$ and then successively to the remaining factors gives ${\pi }_{a_1}{\prod }_{i=1}^{k-1}{P}_{t_{i+1}-t_i}(a_i,a_{i+1})P_{-t_k}(a_k,c)$. Summing over $c$ gives the negative-time marginal; $k=1$ is included by the empty-product convention. If a list also has nonnegative times, conditional independence of the halves given $X_0=c$ multiplies this negative-side factor by $P_{t_{k+1}}(c,a_{k+1}){\prod }_{i=k+1}^{n-1}{P}_{t_{i+1}-t_i}(a_i,a_{i+1})$. For adjacent times $t_k<0\le t_{k+1}$ around zero, summing over $c$ joins the two factors because ${\sum }_c{P}_{-t_k}(a_k,c)P_{t_{k+1}}(c,a_{k+1})=P_{t_{k+1}-t_k}(a_k,a_{k+1})$, by the semigroup law. Thus in every case the joint probability is ${\pi }_{a_1}{\prod }_{i=1}^{n-1}{P}_{t_{i+1}-t_i}(a_i,a_{i+1})$, which depends only on time differences. For each $a\in E$, the set $\{(s,\omega ):Z_s^{\pm }(\omega )=a\}$ equals ${\bigcup }_{n\ge 0}\{(s,\omega ):S_n^{\pm }(\omega )\le s<S_{n+1}^{\pm }(\omega ),Y_n^{\pm }(\omega )=a\}$; every set in this union is product-measurable because $S_n^{\pm }$ and $Y_n^{\pm }$ are measurable. Since $E$ is finite, both half-processes are jointly measurable, and composing the negative half with $t\mapsto -t$ and gluing on the measurable half-lines preserves joint measurability. Step 2.1 gives finitely many jumps on every bounded interval on both halves, so the glued process has the stated local path property.