Exact Markovian projection: strong lumpability

Proof. Necessity. Suppose the projected process is Markov for every initial distribution, with a transition law determined by $\bar\pi(0)$. Fix a block $B_a$ and $i,i'\in B_a$, and consider the two initial distributions $\delta_i$ and $\delta_{i'}$. Both project to the same initial block law, namely $\delta_a$ on $T$. Hence by hypothesis the projected chains started from $\delta_i$ and from $\delta_{i'}$ have the same law, and in particular the same instantaneous transition rate out of $a$ into any block $B_b$ with $b\neq a$.

Compute that rate from the source. Starting at $i$, for $b\neq a$, $$\Pr\bigl[p(X_h)=b \mid X_0=i\bigr]=\sum_{j\in B_b}\bigl(e^{hG}\bigr)_{ij} =\sum_{j\in B_b}\bigl(\delta_{ij}+h\,g_{ij}+O(h^2)\bigr) =h\sum_{j\in B_b}g_{ij}+O(h^2),$$ where $\delta_{ij}=0$ for $j\in B_b$ since $i\in B_a$ and $B_a\cap B_b=\varnothing$. Dividing by $h$ and letting $h\downarrow0$, the instantaneous $a\to b$ rate from $i$ is $\sum_{j\in B_b}g_{ij}$, and likewise from $i'$ it is $\sum_{j\in B_b}g_{i'j}$. Equality of the two projected laws forces $$\sum_{j\in B_b}g_{ij}=\sum_{j\in B_b}g_{i'j}\qquad(b\neq a).$$ For $b=a$ the condition follows from (31): since row sums vanish, $$\sum_{j\in B_a}g_{ij}=-\sum_{b\neq a}\sum_{j\in B_b}g_{ij}=-\sum_{b\neq a}\sum_{j\in B_b}g_{i'j}=\sum_{j\in B_a}g_{i'j}.$$ This is (33).

Sufficiency. Assume (33), so that $\bar g_{ab}$ in (34) is well defined.

$\bar G$ is a generator. For $a\neq b$ and $i\in B_a$, every $j\in B_b$ satisfies $j\neq i$, so $g_{ij}\ge0$ by (31); hence $\bar g_{ab}=\sum_{j\in B_b}g_{ij}\ge0$. And the row sums vanish: $$\sum_{b\in T}\bar g_{ab}=\sum_{b\in T}\sum_{j\in B_b}g_{ij}=\sum_{j\in S}g_{ij}=0 ,$$ using that the blocks partition $S$. So $\bar G$ satisfies (31) on $T$.

(1) Closure of the block probabilities. Let $\pi(t)$ solve (32) and set $\bar\pi_a(t):=\sum_{i\in B_a}\pi_i(t)$. Then $$\begin{aligned} \dot{\bar\pi}_b(t) &=\sum_{j\in B_b}\dot\pi_j(t) =\sum_{j\in B_b}\sum_{i\in S}\pi_i(t)\,g_{ij} =\sum_{a\in T}\sum_{i\in B_a}\pi_i(t)\Bigl(\sum_{j\in B_b}g_{ij}\Bigr)\\ &=\sum_{a\in T}\sum_{i\in B_a}\pi_i(t)\,\bar g_{ab} =\sum_{a\in T}\bar\pi_a(t)\,\bar g_{ab}, \end{aligned}$$ where the fourth equality is exactly (33): the inner sum is the same number $\bar g_{ab}$ for every $i\in B_a$, so it factors out of the sum over $i\in B_a$ regardless of how $\pi(t)$ distributes mass within the block. Thus $\bar\pi$ obeys the closed master equation $\dot{\bar\pi}=\bar\pi\bar G$, for every initial $\pi(0)$.

(2) Process-level Markovianity. Closure of the one-time marginals is not by itself the Markov property, and we now establish the latter. Let $\mathcal F:=\{f\colon S\to\mathbb R\;:\;f\ \text{is constant on each block}\}$, and for $f\in\mathcal F$ write $f_b$ for its value on $B_b$. For $i\in B_a$, $$(Gf)(i)=\sum_{j\in S}g_{ij}f(j)=\sum_{b\in T}\sum_{j\in B_b}g_{ij}f_b=\sum_{b\in T}f_b\Bigl(\sum_{j\in B_b}g_{ij}\Bigr)=\sum_{b\in T}\bar g_{ab}f_b ,\tag{35}$$ using (33) in the last step. The right-hand side depends on $i$ only through its block $a$. Hence $Gf\in\mathcal F$: the subspace $\mathcal F$ of block-constant functions is invariant under $G$. Being finite-dimensional and $G$-invariant, $\mathcal F$ is invariant under $e^{tG}$ for all $t\ge0$ (expand the exponential; each partial sum preserves $\mathcal F$, and $\mathcal F$ is closed).

Apply this with $f=\mathbf 1_{B_b}\in\mathcal F$. Invariance gives that $i\mapsto\bigl(e^{tG}\mathbf 1_{B_b}\bigr)(i)=\Pr[p(X_t)=b\mid X_0=i]$ is constant on blocks: writing $\bar P_{ab}(t)$ for its common value on $B_a$, $$\Pr\bigl[p(X_t)=b\mid X_0=i\bigr]=\bar P_{ab}(t)\qquad\text{for every }i\in B_a .\tag{36}$$ Moreover (35) identifies the generator of $\bar P(t)$: reading (35) in $\mathcal F\cong\mathbb R^{T}$ shows $G|_{\mathcal F}$ acts as $\bar G$, so $\bar P(t)=e^{t\bar G}$.

Now take any times $0\le t_1<\dots<t_n<t_{n+1}$, any blocks $b_1,\dots,b_{n+1}$, and any initial law $\pi(0)$. Write $$H:=\{p(X_{t_1})=b_1,\dots,p(X_{t_n})=b_n\}$$ for the projected past, and assume $\Pr[H]>0$; histories of zero probability impose no condition, since the Markov property is a statement about conditional probabilities, which are defined only on positive-probability events. (Equivalently, one may phrase the argument with regular conditional distributions, which exist automatically here because $S$ is finite; nothing changes.)

Let $\nu$ denote the conditional law of $X_{t_n}$ given $H$, i.e. $\nu(i):=\Pr[X_{t_n}=i\mid H]$. It is a probability measure supported in $B_{b_n}$, and it may depend intricately on the whole past $H$ and on $\pi(0)$. The Markov property of $X$ at time $t_n$, together with (36), gives $$\begin{aligned} \Pr\bigl[p(X_{t_{n+1}})=b_{n+1}\,\big|\,H\bigr] &=\sum_{i\in B_{b_n}}\nu(i)\;\Pr\bigl[p(X_{t_{n+1}-t_n})=b_{n+1}\,\big|\,X_0=i\bigr]\\[2pt] &=\sum_{i\in B_{b_n}}\nu(i)\;\bar P_{b_nb_{n+1}}(t_{n+1}-t_n)\\[2pt] &=\bar P_{b_nb_{n+1}}(t_{n+1}-t_n), \end{aligned}\tag{37}$$ where the second equality is (36) , the summand is independent of $i$ , and the third follows because $\nu$ is a probability measure on $B_{b_n}$, so its masses sum to $1$. The right-hand side of (37) depends on the past only through $b_n$, and on the times only through $t_{n+1}-t_n$. This is exactly the Markov property with time-homogeneous transition semigroup $\bar P(t)=e^{t\bar G}$, and it holds for every $\pi(0)$ , which is the assertion of strong lumpability.

The crux, and the reason marginal closure alone would not suffice, is visible in (37): the unknown within-block conditional law $\nu$ , which is precisely the information the readout discards, and which does depend on the whole past , drops out only because $\bar P_{b_n b_{n+1}}$ does not depend on which $i\in B_{b_n}$ carries the mass. That independence is (36), i.e. (33). ◻

Proof. The instantaneous rates were computed in the necessity half of Theorem 9.3 to be $\sum_{j\in B_b}g_{ij}$ and $\sum_{j\in B_b}g_{i'j}$, which differ by hypothesis. A time-homogeneous Markov generator on $T$ would assign a single rate $\bar g_{ab}$ to the projected state $a$, contradiction. ◻