Forward equation and stationary laws for finite-state rate processes

If $T=0$, the only time is $t=0$ and $p(0)$ is its initial law. Suppose $T>0$. By the first-jump convention, the no-jump probability from $x$ over $[t,t+h]$ is $H_x(t,t+h)=\exp (-{\int }_t^{t+h}{\lambda }_x(s)ds)=1-{\lambda }_x(t)h+o(h)$. For $x\mathrm{\ne }y$, the probability of exactly one jump $x\to y$ and no further jump is ${\int }_t^{t+h}{H}_x(t,s){\kappa }_{xy}(s)H_y(s,t+h)ds={\kappa }_{xy}(t)h+o(h)$. Paths with at least two jumps contribute $O(h^2)$ by [F2]. Hence the conditional endpoint transition probabilities equal ${\kappa }_{xy}(t)h+o(h)$ for $y\mathrm{\ne }x$, and $1-{\lambda }_x(t)h+o(h)$ for $y=x$. Averaging this expansion over the finite states gives $\mathrm{\mid }{p}_y(t+h)-p_y(t)\mathrm{\mid }=O(h)$; the same estimate from the left shows $p$ is continuous. By the Markov property, for each $y\in E$, $p_y(t+h)=p_y(t)(1-{\lambda }_y(t)h)+{\sum }_{x\mathrm{\ne }y}{p}_x(t){\kappa }_{xy}(t)h+o(h)\mathrm{.}$ There are finitely many states, so the remainder remains $o(h)$ after summation. Dividing by $h$ and taking $h\downarrow 0$ gives the stated forward equation; applying the same expansion at $t-h$ gives the left derivative. Its right-hand side is continuous because the rate functions and each $p_x$ are continuous, hence $p$ is continuously differentiable on the interior, with the corresponding one-sided derivatives at the endpoints.

Suppose $\pi Q(t)=0$ and $p(0)=\pi$. The constant function $p(t)=\pi$ satisfies the forward equation. If $d(t)$ is the difference of any other solution and $\pi$, then $d(0)=0$ and $d^{\mathrm{\prime }}=dQ$. For $t>0$, each component of $d$ is differentiable on $[0,t]$, with one-sided endpoint derivatives, and its derivative is continuous, hence integrable. Applying the second fundamental theorem in [F4] componentwise gives $d(t)={\int }_0^{t}d(s)Q(s)ds\mathrm{.}$ For $t=0$ this identity holds by the zero-interval convention. Use the vector $1$-norm $\mathrm{\parallel }d{\mathrm{\parallel }}_1:={\sum }_x\mathrm{\mid }{d}_x\mathrm{\mid }$ and let $B:={\sup }_{s\in [0,T]}{\max }_x{\sum }_y\mathrm{\mid }{Q}_{xy}(s)\mathrm{\mid }<\mathrm{\infty }$, the maximum absolute row sum. Let $D:={\sup }_{s\in [0,T]}\mathrm{\parallel }d(s){\mathrm{\parallel }}_1<\mathrm{\infty }$. Since $d$ is a row vector, the triangle inequality gives $\mathrm{\parallel }d(s)Q(s){\mathrm{\parallel }}_1\le {\sum }_x\mathrm{\mid }{d}_x(s)\mathrm{\mid }{\sum }_y\mathrm{\mid }{Q}_{xy}(s)\mathrm{\mid }\le B\mathrm{\parallel }d(s){\mathrm{\parallel }}_1$. Iterating the integral equation yields $\mathrm{\parallel }d(t){\mathrm{\parallel }}_1\le D(Bt{)}^n\mathrm{/}n!$ for every integer $n\ge 1$. This bound tends to zero as $n\to \mathrm{\infty }$, so $d(t)=0$ for every $t$. Therefore $p(t)=\pi$.

In the constant-rate case, $\pi Q=0$ is exactly the coordinate condition ${\sum }_{x\mathrm{\ne }y}{\pi }_x{\kappa }_{xy}={\pi }_y{\lambda }_y$ for each $y$, so step 2.1 proves invariance of the stated generator-stationary law.