Matrix Chapman–Kolmogorov equations

For all $m,n\ge 0$, $K^{m+n}=K^m{K}^n$. For $n=0$, the composition formula in [F2] and $K^0=I$ in [F1] give $K^m{K}^0=K^m$. If the identity holds at $n$, then [F1] and associativity [F3] give

Apply step 1.1 to the singleton $\{y\}$. By [F2] and [F6], $p^{(m+n)}(x,y)={\int }_E{K}^n(z,\{y\})K^m(x,dz)$. The integrand is measurable and between zero and one because $K^n$ is a probability kernel, so the integral is defined.

If $E$ is finite, list it without repetition as $e(0),\dots ,e(r-1)$; if it is countably infinite, fix a bijection $e:\mathbb{N}\to E$. Let $J=\{0,\dots ,r-1\}$ in the finite case and $J=\mathbb{N}$ in the infinite case, and set $g_N(z)={\sum }_{k\in J,k<N}{K}^n(e(k),\{y\}){\mathbf{1}}_{\{z=e(k)\}}$. These finite-support functions increase pointwise to $g(z)=K^n(z,\{y\})$ and are constant once $N\ge r$ in the finite case. By [F4], the singleton weights of $K^m(x,\cdot )$ are $p^{(m)}(x,e(k))$; [F5] therefore gives ${\int }_{E}gd{K}^m(x,\cdot )={\lim }_N{\sum }_{k\in J,k<N}{p}^{(m)}(x,e(k))p^{(n)}(e(k),y)={\sum }_{z\in E}{p}^{(m)}(x,z)p^{(n)}(z,y)$. Combining with step 2.1 proves the formula, with the nonnegative series interpreted by its finite partial sums.

If $m=0$, the row $p^{(0)}(x,z)={\mathbf{1}}_{\{x=z\}}$ leaves only the term $z=x$; if $n=0$, $p^{(0)}(z,y)={\mathbf{1}}_{\{z=y\}}$ leaves only $z=y$. When $m=n=0$, both sides are ${\mathbf{1}}_{\{x=y\}}$. Thus the zero-time endpoints, including the one-state and deterministic cases, agree. If $E=\mathrm{\varnothing }$, there are no $x,y$ and the assertion is vacuous. The proof uses kernel algebra and nonnegative sums only; no AC or conditional-probability version enters.