2.1. Notation

Let $\mathbb{R}$ , $\mathbb{R}_{\geq 0}$ , and $\mathbb{R}_{>0}$ be the sets of real, non-negative, and positive numbers, respectively. Let $\mathbb{N}$ and $\mathbb{N}_0$ be the sets of positive and non-negative integers, respectively, and let $\mathbb{Z}$ denote the set of integers. For $x=(x_1,\dots, x_n)\in \mathbb{R}^n$ and $y=(y_1,\dots, y_n)\in \mathbb{R}^n$ , we say that $x\geq y$ if $x_i\geq y_i$ for all $i=1,\dots, n$ and that $x>y$ if $x\geq y$ and $x\neq y$ . In addition, we define $x\vee y= (\!\max\{x_1,y_1\},\dots, \max\{x_n, y_n\})$ and $\langle x,y\rangle=\sum_{i=1}^n x_iy_i$ for the inner product on $\mathbb{R}^n$ . For a finite set $\mathcal{I}$ , $\:|\mathcal{I}|$ denotes the cardinality of $\mathcal{I}$ .

2.2. Reaction networks

A reaction network is a triple $\mathcal{N}=(\mathcal{S},\mathcal{C},\mathcal{R})$ of three finite sets: a set $\mathcal{S}=\{S_1,\ldots,S_n\}$ of species, a set $\mathcal{C}$ of complexes, which are non-negative linear combinations of the species (complexes are identified as elements of $\mathbb{N}_0^n$ ), and a set $\mathcal{R}$ of reactions, which are elements of $\mathcal{C}\times\mathcal{C}$ . One could regard $(\mathcal{C},\mathcal{R})$ as a digraph. We assume that all species have a positive coefficient in some complex and that every complex is the source or target of some reaction. In this case, $\mathcal{C}$ and $\mathcal{S}$ can be determined from $\mathcal{R}$ . In examples, we simply draw the reactions.

For simplicity, for a reaction $r\in\mathcal{R}$ , we write $r=(y,y')$ or $r=y\ce{->}y'$ , where y is the source of r and y ^′ the target. In the chemical literature, these are called the reactant $\mathrm{reac} (r)=y$ and the product $\mathrm{prod} (r)=y'$ of r, respectively. The reaction vector of r is $\zeta(r) = y'-y\in\mathbb{Z}^n$ .

We introduce an associative binary operation [Reference Hoessly, Wiuf and Xia14]

(3) \begin{align}\oplus & \,: \:\big(\mathbb{N}_0^n \times \mathbb{N}_0^n\big) \times \big(\mathbb{N}_0^n \times \mathbb{N}_0^n\big) \to \mathbb{N}_0^n \times \mathbb{N}_0^n, \\(y_1, y_1')\oplus (y_2, y_2') & \,:\!=\, \big(y_1+0\vee (y_2-y_1'), \:y_2'+0\vee (y_1'-y_2)\big) \nonumber\end{align}

for $(y_1, y_1'), (y_2, y_2') \in \mathbb{N}_0^n \times \mathbb{N}_0^n$ . It can be interpreted as the sum of two reactions; for example,

\[\big( (1,1,0), (0,0,1) \big) \oplus \big( (0,1,1), (0,1,0) \big) = \big( (1,2,0), (0,1,0) \big)\]

corresponds to the sum (cf. [Reference Ingalls18, Subsection 3.1.2])

\[S_1 + S_2 \ce{->} S_3 \quad \oplus \quad S_2 + S_3 \ce{->} S_2 \quad = \quad S_1 + 2S_2 \ce{->} S_2. \]

The left-hand side of the sum consists of the molecules required for the two reactions to proceed in succession, and the right-hand side consists of the molecules that are produced and not consumed in the two reactions.

In the following, we assume that a reaction network $\mathcal{N}=(\mathcal{S},\mathcal{C},\mathcal{R})$ is given.

In graph terminology, a reaction network is weakly reversible if the components of the digraph $(\mathcal{C},\mathcal{R})$ are strongly connected.

A stochastic reaction network (SRN) models the stochastic dynamics of a reaction network. Specifically, the evolution of the molecule counts, $(X(t)\,\colon t\in \mathbb{R}_{\ge0})$ , is modelled as an $\mathbb{N}_0^n$ -valued CTMC satisfying the following stochastic equation:

(4) \begin{align}X(t)=X(0)+\sum_{r\in \mathcal{R}}Y_r\Bigg(\int_0^t\lambda_{r}(X(s))\,{\mathrm{d}}s\Bigg)\zeta(r),\end{align}

where $(Y_{r}\,\colon r\in \mathcal{R})$ is a collection of independent and identically distributed unit-rate Poisson processes and $\lambda=(\lambda_{r}\,\colon r \in \mathcal{R})$ is a collection of non-negative intensity functions on $\mathbb{N}_0^n$ , known as the kinetics. Stochastic mass-action kinetics is commonly assumed in the literature; however, it is not a prerequisite for the majority of this paper’s content. For a reaction $r = y_r \to y_r'$ , the intensity function under mass-action kinetics takes the form

\[\lambda_r (x) = \kappa_r \frac{x!}{(x - y_r)!} = \kappa_r \prod_{i=1}^n \frac{x_i !}{(x_i - y_{r,i})!}\]

for some positive constant $\kappa_r$ .

An SRN is denoted by $(\mathcal{N},\lambda)$ . Throughout, we assume the following compatibility condition.

If Condition 1 is fulfilled, then $x\to_{\mathcal{N}} z$ implies that there is positive probability of going from x to z and vice versa.

A probability measure $\pi$ on an irreducible component $\Gamma\subseteq \mathbb{N}_0^n$ of $(\mathcal{N},\lambda)$ is (see [Reference Cappelletti and Joshi5, Definition 4.1]):

1. (i) a stationary distribution if for all $x\in \Gamma$ ,

   (5) \begin{align}\pi(x)\sum_{r\in \mathcal{R}}\lambda_{r}(x)=\sum_{r\in \mathcal{R}}\pi \big(x - \zeta(r) \big)\lambda_{r} \big(x - \zeta(r) \big);\end{align}

2. (ii) a complex balanced distribution if for all complexes $\eta\in \mathcal{C}$ and all $x\in \Gamma$ ,

   (6) \begin{align}\pi(x)\sum_{y':\,\eta\to y'\in \mathcal{R}}\lambda_{\eta\to y'}(x)=\sum_{y:\,y\to \eta\in \mathcal{R}}\pi(x+y-\eta)\lambda_{y\to \eta}(x+y-\eta);\end{align}

3. (iii) a detailed balanced distribution if $\mathcal{N}$ is reversible and for all $y\ce{->}y'\in \mathcal{R}$ ,

   \begin{align*}\pi(x)\lambda_{y\to y'}(x)=\pi(x+y'-y)\lambda_{y'\to y}(x+y'-y).\end{align*}

We reserve the term ‘stationary distribution’ for a distribution on a single irreducible component. Hence, a stationary distribution $\pi$ on an irreducible component $\Gamma$ of $(\mathcal{N},\lambda)$ satisfies $\pi(x)>0$ for all $x\in \Gamma$ . Equation (5) is known in the chemical literature as the master equation (also the Kolmogorov forward equation) for the SRN $(\mathcal{N},\lambda)$ ; see [Reference Anderson and Kurtz3]. A detailed balanced distribution is a complex balanced distribution, and a complex balanced distribution is a stationary distribution.

2.3. Non-interacting species and reduction

Let $\mathcal{N}=(\mathcal{C},\mathcal{R},\mathcal{S})$ be a reaction network, and let $\mathcal{U} = \{U_1,\dots, U_m\}$ be a proper subset of $\mathcal{S}$ . Order the species such that

\[\mathcal{S} \setminus \mathcal{U} = \{S_1, \dots, S_{n-m}\}\quad \text{and} \quad \mathcal{S}=\{S_1, \dots, S_{n-m}, U_1,\dots, U_m\}.\]

Let $x=(z,u)\in\mathbb{N}_0^{n-m}\times \mathbb{N}_0^m=\mathbb{N}_0^n$ with $z=(z_1, \ldots, z_{n-m})$ and $u=(u_1,\ldots,u_m)$ . We define the projection $\rho\,\colon\mathbb{N}_0^n \to \mathbb{N}_0^m$ onto the space of the species in $\mathcal{U}$ by

(7) \begin{align}\rho (x) = \sum_{i=1}^m u_i U_i \in \mathbb{N}_0^m\quad \text{for } x =(z,u) \in \mathbb{N}_0^n.\end{align}

Example 1. Consider the reaction network (see [Reference Ingalls18, Subsection 3.1.2])

(8) \begin{align} E + A \ce{<=>} EA, \quad EA + B \ce{<=>} EAB, \quad EAB \ce{->} E + P + Q.\end{align}

It models the conversion of two substrates A and B into two other substrates P and Q by means of an enzyme E and several short-lived intermediate molecules, EA and EAB. Therefore, it is natural to treat $\mathcal{U} = \{EA, EAB\}$ as a set of non-interacting species, and thus $\mathcal{S} \setminus \mathcal{U} = \{E, A, P, Q\}$ is a set of core species. The choice is not unique; for example, $\mathcal{U}=\{E,EA,EAB\}$ is also a set of non-interacting species.

For a set $\mathcal{U}=\{U_1,U_2,\dots, U_m\}$ of non-interacting species, let $\{\mathcal{C}_0, \mathcal{C}_1, \dots, \mathcal{C}_m\}$ be the partition of $\mathcal{C}$ by

\begin{align*} \mathcal{C}_i \,:\!=\, \begin{cases}\displaystyle \big\{y\in \mathcal{C}\,:\,\rho (y) = 0 \big\}, & i = 0,\\[3pt] \displaystyle \big\{y \in \mathcal{C}\,:\,\rho (y) = U_i \big\}, &1\leq i\leq m,\end{cases}\end{align*}

and let $\{\mathcal{R}_{i,j}:\,i,j=0,\dots, m\}$ be the partition of $\mathcal{R}$ where

\[\mathcal{R}_{i,j} \,:\!=\, \big\{y\ce{->}y'\,\colon y\in \mathcal{C}_i, \, y'\in \mathcal{C}_j \big\},\quad i,j=0,\dots, m.\]

For the reaction network in (8) with $\mathcal{U} = \{EA, EAB\}$ ,

\begin{gather*}\mathcal{C}_0 = \{E+A, E + P + Q\},\quad \mathcal{C}_1 = \{EA, EA + B\},\quad \mathcal{C}_2 = \{EAB\},\\\mathcal{R}_{0,0} = \mathcal{R}_{0,2} = \mathcal{R}_{1,1} = \mathcal{R}_{2,2} = \varnothing, \quad \mathcal{R}_{0,1} = \{E + A \ce{->} EA\},\\ \mathcal{R}_{1,0} = \{EA \ce{->} E + A\}, \quad \mathcal{R}_{1,2} = \{EA + B \ce{->} EAB \},\\\mathcal{R}_{2,0} = \{EAB \ce{->} E + P + Q\}, \quad \mathcal{R}_{2,1} = \{EAB \ce{->} EA + B\}.\end{gather*}

We introduce a multi-digraph $(\mathcal{V},\mathcal{E})$ (see [Reference Sáez, Wiuf and Feliu27]) with

\[\mathcal{V}=\{U_0\}\cup \mathcal{U}\quad \mathrm{and} \quad \mathcal{E}=\big\{U_i\ce{->[{\textit{r}}]}U_j\,\colon r\in \mathcal{R}_{i,j},\: i,j=0,\dots, m\big\},\]

where $U_0$ is an additional node corresponding to all complexes not containing species in $\mathcal{U}$ . There is a one-to-one correspondence between the edges $U_i\ce{->[{\textit{r}}]}U_j$ and the reactions $r \in \mathcal{R}_{i,j}$ . As an example, the multi-digraph associated with (8) is

A q-step walk $\theta$ in $(\mathcal{V},\mathcal{E})$ is a sequence of $q\in\mathbb{N}$ edges

\begin{align*}\theta = U_{i_0}\ce{->[{$r_1$}]} U_{i_1} \ce{->[{$r_2$}]} \cdots \ce{->[{$r_q$}]} U_{i_q},\end{align*}

where $r_k\in \mathcal{R}_{i_{k-1},i_k}$ . A q-step walk is closed if $U_{i_0} = U_{i_q}$ . If $U_{i_0}=U_0$ , then $U_{i_q}$ is said to be produced; and if $U_{i_q}=U_0$ , then $U_{i_0}$ is said to be degraded. Let $\mathcal{U}_{\mathrm{pro}}$ and $ \mathcal{U}_{\deg}$ be the sets of produced and degraded non-interacting species, respectively.

For $q\in \mathbb{N}_0$ and $i\in\{0,\dots, m\}$ , we let $\Xi_i(q)$ be the collection of all $(q + 1)$ -step closed walks that start and end at $U_i$ without passing though $U_0$ . Then every element in $\Xi_i(q)$ is of the form

(9) \begin{align}\gamma = U_i\ce{->[{$r_0$}]}U_{i_1}\ce{->[{$r_1$}]}\cdots \ce{->[{$r_{q-1}$}]}U_{i_q}\ce{->[{$r_q$}]}U_i,\end{align}

where $i_1,\dots, i_q \in \{1,\dots, m\}$ , $r_j \in \mathcal{R}_{i_j,i_{j+1}}$ for $0\leq j\leq q$ , and by definition $i_0 = i_{q+1}= i$ . In particular, if $\gamma\in\Xi_i(0)$ , then $\gamma$ consists of one edge $U_i \ce{->[{\textit{r}}]} U_i$ only; and if $i=0$ , then $\gamma$ is a reaction in $\mathcal{R}$ between core species only. Furthermore, define

\[\Xi_i \,:\!=\, \bigcup_{q=0}^{\infty}\Xi_i(q),\quad i=0,\ldots,m. \]

For the reaction network in (8) with $\mathcal{U} = \{EA, EAB\}$ , we find that

In (8), $\mathcal{U} = \{EA, EAB\}$ is a set of non-interacting species but not a set of intermediate species, because $EA + B \in \mathcal{C}_1$ , yet $\rho(EA + B) = EA \neq EA + B$ . On the other hand, if $\mathcal{U} = \{EAB\}$ , then EAB is an intermediate species, since $\rho(EAB) = EAB$ .

In (8) we have $\mathcal{U}_{\mathrm{pro}} = \mathcal{U}_{\deg} $ , and thus $\mathcal{U}$ is eliminable. Definition 6 differs from Definition 4.1 of [Reference Hoessly, Wiuf and Xia14], with the latter encompassing more general situations. However, these two definitions align in the context of non-interacting species.

Proof. To prove $\mathcal{U}_{\mathrm{pro}} = \mathcal{U}_{\deg}$ , it suffices to prove the one-directional inclusion $\mathcal{U}_{\mathrm{pro}} \subseteq \mathcal{U}_{\deg}$ . The reverse direction follows by the same reasoning. Let $U \in \mathcal{U}_{\mathrm{pro}}$ . Then there exists $q \in \mathbb{N}_{\geq 0}$ and a walk in $(\mathcal{V}, \mathcal{E})$ ,

\[ U_0 \ce{->} U_{i_1} \ce{->} \cdots \ce{->} U_{i_q} \ce{->} U_i,\]

where the edge labels are skipped for simplification. If $q = 0$ , then $U_0 \ce{->} U_i \in \mathcal{E}$ , and thus $U_i \in \mathcal{U}_{\deg}$ by weak reversibility.

For $q \geq 1$ , we prove the lemma by induction. First, $U_{i_q} \ce{->} U_i$ implies the existence of a reaction $y \ce{->} y' \in \mathcal{R}_{i_q, i}$ . Then, owing to weak reversibility, there exists a sequence of reactions in $\mathcal{R}$ such that

\[ y' \ce{->} y_1\ce{->}\cdots \ce{->} y_{q'}\ce{->} y.\]

If there exists $j\in \{1,\dots, q'\}$ such that $y_j \in \mathcal{C}_0$ , then we are done. Otherwise, by the induction hypothesis, we have $U_{i_q} \in \mathcal{U}_{\deg}$ . This implies that there exists a walk of the form

\[ U_0 \ce{->} \cdots \ce{->} U_{i_q} \ce{->} U_i \ce{->} \cdots \ce{->} U_{i_q} \ce{->} \cdots \ce{->} U_0.\]

Therefore $U_i \in \mathcal{U}_{\deg}$ . The proof is complete.

We introduce a map $\mathfrak{R}$ defined on the collection of all walks in $(\mathcal{V},\mathcal{E})$ :

Then we have

\[\mathcal{R}_{i,i} = \mathfrak{R}(\Xi_i(0)) \subseteq \mathfrak{R} (\Xi_i), \quad i=0,\dots, m.\]

In particular, $\rho (y) = \rho (y') = 0$ for $\gamma \in \Xi_0$ with $ \mathfrak{R}(\gamma)=(y,y')$ .

Definition 7. Let $\mathcal{U}$ be an eliminable set of species. The reduced reaction network $\mathcal{N}_{\mathcal{U}}$ obtained by elimination of $\mathcal{U}$ from $\mathcal{N}$ is a triple $(\mathcal{S}_{\mathcal{U}},\mathcal{C}_{\mathcal{U}},\mathcal{R}_{\mathcal{U}})$ where

\[ \mathcal{R}_{\mathcal{U}} \,:\!=\, \mathfrak{R}(\Xi_0) \setminus \{(y,y)\,\colon y\in \mathbb{N}_0^{n_{\mathcal{U}}}\},\quad \mathcal{C}_{\mathcal{U}} \,:\!=\, \{y,y'\,\colon y\to y'\in\mathcal{R}_{\mathcal{U}}\},\]

\[\mathcal{S}_{\mathcal{U}} \,:\!=\, \mathop\bigcup_{y\in \mathcal{C}_{\mathcal{U}}} \textrm{supp}(y)\subseteq \mathcal{S}\setminus \mathcal{U}.\]

Here, $\textrm{supp} (y) = \{S_i\,\colon y_i>0\}$ is the support of $y = \sum_{j=1}^n y_j S_j$ in $\mathbb{N}_0^{\mathcal{S}}$ , and $n_{\mathcal{U}} = |\mathcal{S}_{\mathcal{U}}|$ .

By Definition 7, we find that for (8),

\[\mathfrak{R}(\Xi_0) = \left\{\begin{array}{c} E + A \ce{->} E + A, \quad E + A + B \ce{->} E + A + B, \\[3pt] E + A + B \ce{->} E + P + Q \end{array} \right\}\!,\]

and thus the reduced reaction network is the triple $(\mathcal{S}_{\mathcal{U}},\mathcal{C}_{\mathcal{U}}, \mathcal{R}_{\mathcal{U}})$ with

(10) \begin{align} \mathcal{S}_{\mathcal{U}} =\{E, A, B, P, Q\}, \quad \mathcal{C}_{\mathcal{U}} = \{E + A + B, E + P + Q\}, \\[-28pt] \nonumber \end{align}

\[\mathcal{R}_{\mathcal{U}} = \{E + A + B \ce{->} E + P + Q\}.\]

The set $\mathcal{R}_{\mathcal{U}}$ may contain infinitely many reactions, and thus the reduced reaction network may not be a reaction network.

Example 2. ([Reference Hoessly, Wiuf and Xia14, Example 2].) Consider the following reaction network with $\mathcal{U} = \{G'\}$ :

\[ G \ce{<=>} G', \quad G' \ce{->} G' + P, \quad P \ce{->} \varnothing.\]

Then $\mathcal{C}_0 = \{P, \varnothing\}$ , $\mathcal{C}_1 = \{G, G', G' + P\}$ , $\mathcal{R}_{0,0} = \{P \ce{->} \varnothing\}$ , $\mathcal{R}_{0,1} = \{G \ce{->} G'\}$ , $\mathcal{R}_{1,0} = \{G' \ce{->} G\}$ , $\mathcal{R}_{1,1} = \{G' \ce{->} G' + P\}$ , and the associated multi-digraph is

Therefore, $\Xi_0(0) = \{U_0 \ce{->[{ \rm{P}}]} U_0\}$ and

for all $k \geq 1$ . As a result,

\[ \mathcal{R}_{\mathcal{U}} = \{P \ce{->} \varnothing\} \cup \{G \ce{->} G + k P \,\colon k = 1,2,\dots\}\]

is an infinite set.

The example is further discussed in the conclusion section. Furthermore, as observed in [Reference Hoessly, Wiuf and Xia14], if $\mathcal{N}$ is weakly reversible and $\mathcal{U}$ is a set of non-interacting species, $\mathcal{N}_{\mathcal{U}}$ is not necessarily weakly reversible.

Example 3. ([Reference Hoessly, Wiuf and Xia14, Example 8].) Consider the following weakly reversible reaction network with the set of non-interacting species $\mathcal{U}=\{U_1,U_2\}$ :

\[S_1\ce{->} U_1\ce{->} S_2\ce{->} U_2\ce{->} S_1,\quad S_3+U_2\ce{<=>} S_4.\]

The reduced reaction network with $\mathcal{R}_\mathcal{U}=\{S_1\ce{<=>} S_2, S_2+S_3\ce{->} S_4\ce{->} S_3+S_1\}$ is not weakly reversible.

Condition 2 below is equivalent to $\mathcal{R}_{\mathcal{U}}$ being finite [Reference Hoessly, Wiuf and Xia14], and $\mathcal{N}_\mathcal{U}$ is thus a reaction network if Condition 2 is fulfilled. The proof is given in Subsection 6.1.

Let $(\mathcal{N},\lambda)$ be an SRN. We introduce a kinetics $\lambda_{\mathcal{U}}$ on the reduced reaction network $\mathcal{N}_{\mathcal{U}}$ (see [Reference Hoessly and Wiuf13]). For $r \in \mathcal{R}_{i,j}\subseteq \mathcal{R}$ where $i,j\in \{0,\dots, m\}$ , define the function $\beta_r \,\colon \mathbb{N}_0^{n}\to [0,1]$ by

(11) \begin{align}\beta_r (x) \,:\!=\, \lambda_r(x) \bigg/ \sum_{k = 0}^m\sum_{r'\in \mathcal{R}_{i,k}}\lambda_{r'}(x),\end{align}

with the convention $0/0 \,:\!=\, 0$ . Then

(12) \begin{align}\sum_{k=0}^m \sum_{r\in \mathcal{R}_{i,k}} \beta_r(x) = \mathbf{1}_{\mathcal{D}_i}(x),\quad i=0,\ldots,m,\end{align}

where

\[\mathcal{D}_i \,:\!=\, \bigcup_{k=0}^m \bigcup_{r\in \mathcal{R}_{i,k}} \big\{x\in \mathbb{N}_0^n\,\colon x\geq \mathrm{reac} (r)\big\}.\]

For $\gamma\in \Xi_0$ , define $\lambda^*_{\mathcal{U},\gamma}\,\colon\mathbb{N}_0^{n}\to \mathbb{R}_{\geq 0}$ by

(13) \begin{align}\lambda^*_{\mathcal{U},\gamma}(x) \,:\!=\, \lambda_{r_0}(x) \prod_{j=1}^{q}\beta_{r_j} \bigg(x+\sum_{i=1}^{j}\zeta_{i - 1}\bigg),\end{align}

where $\zeta_{i- 1} \,:\!=\, \zeta (r_{i- 1})$ is the reaction vector of $r_{i - 1}$ and $i_0 = i_{q+1} \,:\!=\, 0$ . Define the kinetics $\lambda_{\mathcal{U}} = (\lambda_{\mathcal{U},r} \,\colon r\in \mathcal{R}_{\mathcal{U}})$ by

(14) \begin{align}\lambda_{\mathcal{U},r}(x)= \sum_{\gamma\in \Xi_0 :\,\mathfrak{R}(\gamma) = r}\lambda^*_{\mathcal{U},\gamma}(x),\quad x\in \mathbb{N}_0^{n_{\mathcal{U}}},\;r\in\mathcal{R}_\mathcal{U}.\end{align}

Then $\lambda_{\mathcal{U},r}(x)> 0$ if and only if $x\geq \mathrm{reac}(r)$ ; see [Reference Hoessly and Wiuf13, Corollary 3.11]. As a consequence, $(\mathcal{N}_{\mathcal{U}},\lambda_{\mathcal{U}})$ satisfies Condition 1. Furthermore, using Definition 7, we have

$$ \sum_{r\in\mathcal{R}_\mathcal{U}} \lambda_{\mathcal{U},r}(x)\le \sum_{\gamma\in\Xi_0} \lambda^*_{\mathcal{U},\gamma}(x)=\sum_{r\in\mathcal{R}} \lambda_r(x) <\infty$$

(see [Reference Hoessly and Wiuf13, Lemma 3.10]). This implies that the evolution of the molecule counts of the core species, $(Z(t)\,\colon t\in \mathbb{R}_{\ge0})$ , satisfies the following stochastic equation, similar to (4):

\begin{align*}Z(t)=Z(0)+\sum_{r\in \mathcal{R}_\mathcal{U}}Y_r\Big(\!\int_0^t\lambda_{r}(Z(s))\,{\mathrm{d}}s\Big)\zeta (r),\end{align*}

where $(Y_r$ , $r\in\mathcal{R}_\mathcal{U})$ , is a (possibly infinite) collection of unit-rate independent and identically distributed Poisson processes.

For example, consider the reduced reaction network in (8) and (10). We have

\[\mathfrak{R} (\gamma)= ( E + A + B \ce{->} E + P + Q),\]

where

\[ \gamma = U_0 \ce{->[{$E + A \to EA$}]} U_1 \ce{->[{$EA + B \to EAB$}]} U_2 \ce{->[{$EAB \to E + P + Q$}]} U_0.\]

Therefore, by (11), (13), and (14), the kinetics for the reduced reaction network is

\begin{align*} \lambda_{\mathcal{U}, E + A + B \to E + P + Q} (x) & = \lambda_{\mathcal{U},\gamma}^* (x) = \lambda_{E + A \to EA} (x) \beta_{1} (x) \beta_{2} (x),\end{align*}

where

\begin{align*} \beta_1 (x) \,:\!& =\beta_{EA + B \to EAB} (x + EA - E - A) \\ & =\frac{\lambda_{EA + B \to EAB}}{\lambda_{EA + B \to EAB} + \lambda_{EA \to E + A}}\,(x + EA - E - A),\\ \beta_2 (x) \,:\!& =\beta_{EAB \to E + P + Q} (x - E - A - B + EAB)\\ &= \frac{\lambda_{EAB \to E + P + Q}}{\lambda_{EAB \to EA + B} + \lambda_{EAB \to E + P + Q}}\,(x - E - A - B + EAB).\end{align*}

3.1. A remark on Theorem 1

The proof of Theorem 1(ii) presented in Subsection 6.4 is based on the recurrence of two irreducible discrete-time Markov chains (DTMCs). Recurrence is deduced from the finiteness of the state spaces (of the DTMCs), as a result of Condition 3. The state space being finite is not a necessary condition for recurrence, and nor is recurrence necessary for the validity of Theorem 1. However, we have not been able to identify any other simple conditions that imply recurrence of these DTMCs. Condition 3 is not necessary, as demonstrated by the following (artificial) example, although Theorem 1(ii) remains valid in this particular case.

Example 5. Consider the reaction network in Example (4),

(16)

where for all $x=(x_A,x_U)\in \mathbb{N}_0^2$ ,

\begin{equation*}\begin{aligned}\lambda_1(x)=&\:(x_A!)^2\mathbf{1}_{\mathbb{N}}(x_A),\\\lambda_2(x)=&\:[(x_A+1)!]^2\mathbf{1}_{\mathbb{N}}(x_U),\end{aligned}\qquad\begin{aligned}\lambda_3(x)=&\:2[(x_A+1)!]^2\mathbf{1}_{\mathbb{N}^2}(x_A,x_U),\\ \lambda_4(x)=&\:[(x_A+2)!]^2\mathbf{1}_{\mathbb{N}}(x_U).\end{aligned}\end{equation*}

Note the unusual squared factorial rates that are instrumental for showing that Condition 3 is not necessary. The compatibility condition (Condition 1) is fulfilled, even though all rates depend on $x_A$ . Then $\Gamma=\mathbb{N}_0^2\setminus \{(0,0)\}$ is an irreducible component of (16). One can show that

\[\pi(x) \,:\!=\, \frac{1}{3\times 2^{x_A+x_U}}\quad \,for \,\,\, x = (x_A,x_U)\in \Gamma\]

is a detailed balanced distribution of (16) on $\Gamma$ . Therefore, Proposition 2 ensures that the reduced SRN

\[ \Big\{n A \ce{<=>[{$\lambda_{n,m}$}][{$\lambda_{m,n}$}]} m A\,\colon n,m \in \mathbb{N}, \: n < m\Big\},\]

where $\lambda_{n,m}$ is defined in Subsection 2.3, has detailed balanced stationary distribution $\pi_0(x_A) = C \pi (x_A,0)$ , $x_A \in \mathbb{N}$ , for some constant $C > 0$ . In other words, Condition 3 is not necessary in Theorem 1.

Following the strategy in Step 1 of the proof of Theorem 1 (see Subsection 6.4), we construct a DTMC on $\mathcal{X} = \{x\,\colon x \geq 0\}\cup \{\partial\}$ (where $\partial$ is an ceremony state) with transition probabilities

\begin{align*}P_{\partial,\partial} = 0,\quad P_{\partial, 0} = 1,\quad P_{x,\partial} = \frac{\lambda_2(x,1)}{(\lambda_2+\lambda_3+\lambda_4)(x,1)},\\[3pt] P_{x,x + 1} = \frac{\lambda_4(x,1)}{(\lambda_2+\lambda_3+\lambda_4)(x,1)},\quad P_{x,x - 1} = \frac{\lambda_3(x,1)}{(\lambda_2+\lambda_3+\lambda_4)(x,1)},\end{align*}

and $P_{x,x'} = 0$ otherwise. Then Theorem 1(ii) is valid if $\mathbb{P}_{x} (\tau < \infty) = 1$ , where $\tau \,:\!=\, \min\{ n\geq 1\,\colon X_n = \partial\}$ . However, this is unfortunately not true. In fact, for any $n\geq 2$ ,

\begin{align*}\mathbb{P}_{x}(\tau > n) &\geq\mathbb{P}_{x}\big(X_k=x + k\ \forall k=1,\dots,n\big)\\& =\prod_{k=1}^{n} P_{x + k -1,x + k } = \prod_{k = 1}^{n} \frac{\lambda_4(x + k-1,1)}{(\lambda_2+\lambda_3+\lambda_4)(x + k - 1,1)}\\&=\prod_{k=1}^{n}\frac{\prod_{j=1}^{x + k+1}j^2}{ \prod_{j=1}^{x + k }j^2 + 2\prod_{j=1}^{x + k }j^2 + \prod_{j=1}^{x + k+1} j^2 }\\ &=\prod_{k=1}^{n}\big(3 (x + k+1)^{-2}+1\big)^{-1}\ge \exp\!\left(-3\sum_{k=1}^\infty x^{-2} \right)>0,\end{align*}

using that $-\log (3 x^{-2} + 1) \geq - 3x^{-2}.$ Therefore, $\mathbb{P}_{x}(\tau = \infty)>0$ . This implies that replicating the arguments in the proof of Theorem 1 does not suffice to show that $\pi_0$ is a stationary distribution for the reduced SRN. This is due to the failure of Condition 3 in (16).

3.2. Two special cases

Assume that $\mathcal{U}$ is a set of intermediate species in a weakly reversible SRN. Then it can be demonstrated that Conditions 2 and 3 are both satisfied. Therefore Theorem 1 applies. For a set of intermediate species $\mathcal{U}$ , we have that for $i,j \in\{1,\dots, m\}$ :

1. (i) $\mathcal{C}_i = \{U_i\}$ ;

2. (ii) $\mathcal{R}_{i,i} = \varnothing$ and $\mathcal{R}_{i,j} \subseteq \{U_i\ce{->}U_j\}$ for $i\neq j$ .

As noted in Subsection 2.2, a detailed balanced distribution is a complex balanced distribution, and a complex balanced distribution is a stationary distribution. Therefore, the conclusions of Propositions 1 and 2 are much stronger than those in the general case presented in Theorem 1. This is because, in the case of intermediate species or detailed balance, the reduced SRN takes a simpler form while preserving essential properties of the original SRN. The proofs of Propositions 1 and 2 are given in Subsections 6.2 and 6.3, respectively.

3.3. Positive recurrence of the reduced SRN

Proof. By assumption, $\pi_{0}$ is a stationary distribution on an irreducible component $\tilde{\Gamma}\subseteq\Gamma_0$ . To show positive recurrence on $\tilde{\Gamma}$ , we apply the sufficient condition for positive recurrence from [Reference Asmussen4, Corollary 4.4]. The condition holds if the following is true for $\tilde{\Gamma}$ :

$$\sum_{z\in\tilde{\Gamma}}\sum_{r\in\mathcal{R}_{\mathcal{U}}}\lambda_{\mathcal{U},r}(z)\pi_0(z)<\infty.$$

Using that complex balanced distributions for mass-action SRNs take the form $M\frac{c^x}{x!}$ (see [Reference Anderson, Craciun and Kurtz2]), we can replace $\pi_0(z)=\pi(z)/\pi(\Gamma_{0})$ by $M_0\frac{c^x}{x!}$ with $M_0=M/\pi(\Gamma_0)$ and $c^x \,:\!=\, \prod_{i=1}^n c_i^{x_i}$ , for any $c \in \mathbb{R}_+^n$ and $x\in \mathbb{R}^n$ . Furthermore, by Lemma 3.10 in [Reference Hoessly and Wiuf13],

$$\sum_{r\in\mathcal{R}_\mathcal{U}}\lambda_{\mathcal{U},r}(z) \leq \sum_{r\in\mathcal{R}}\lambda_{r}(z,0).$$

Since $\tilde{\Gamma}\subseteq\Gamma_0\subseteq \mathbb{N}^{n_\mathcal{U}}_0$ and $\mathcal{R}$ is finite, it is enough to show that

$$M_0\sum_{z\in\mathbb{Z}^{n_\mathcal{U}}_{\geq 0}}\lambda_{r}(z,0)\frac{c^{(z,0)}}{z!}<\infty. $$

This final inequality follows as $\lambda_{r}(z,0)$ is a polynomial of fixed degree (in z) for stochastic mass-action kinetics.

As a corollary of Theorem 1, Propositions 1 and 2, and Theorem 2, we get the following result for reductions of complex balanced SRNs with stochastic mass-action kinetics.

Then $(\mathcal{N}_{\mathcal{U}},\lambda_{\mathcal{U}})$ is positive recurrent on each irreducible component of $\Gamma_0$ .

6.1. Proof of Lemma 2

( $\Rightarrow$ ) Suppose that Condition 2 holds. Given a walk $\theta\subseteq (\mathcal{V}, \mathcal{E})$ , define $\mathcal{U}_{\theta} \,:\!=\, \{U_i\in \mathcal{U} \,\colon U_i \in \theta\}$ . Consider a closed walk $\gamma \in \Xi_{i_0}$ for any $i_0 \in \{1,\dots, m_{\mathrm{pro}}\}$ . Then $U_{i_0} \in \mathcal{U}_{\gamma}$ . In addition, $\mathcal{R}_{i_0,i_0}=\varnothing$ , because otherwise $r = y + U_{i_0} \ce{->} y' + U_{i_0} \in \mathcal{R}_{i_0,i_0}$ with $y \neq y'$ contradicts Condition 2. This implies $|\mathcal{U}_{\gamma}|\geq 2$ , and

\[\mathfrak{R}(\Xi_{i_0})=\bigcup_{k=2}^{m_{\mathrm{pro}}}A_{i_0,k}, \quad A_{i_0, k} \,:\!=\, \{\mathfrak{R}(\gamma)\,\colon |\mathcal{U}_{\gamma}|=k\}.\]

We will show that $|\mathfrak{R}(\Xi_0)| < \infty$ , which is the key to proving $|\mathcal{R}_{\mathcal{U}}| < \infty$ .

Suppose that $k=2$ . Then $\gamma$ has the form

\[\gamma = U_{i_0}\ce{->[{$r_0$}]}U_{j}\ce{->[{$r_1$}]}U_{i_0}\ce{->[{$r_{2}$}]}\cdots\ce{->[{$r_{2q}$}]}U_j\ce{->[{$r_{2q+1}$}]}U_{i_0}\]

for some $j \in \{1,\dots, i_0 - 1, i_0 + 1,\dots, m_{\mathrm{pro}}\}$ and $q\in\mathbb{N}_0$ . As a result of Proposition 1 from [Reference Hoessly, Wiuf and Xia14] and Condition 2,

(23) \begin{align}\mathfrak{R}(\gamma) = \bigoplus_{i=0}^q \big(r_{2i}\oplus r_{2i+1}\big) = \bigoplus_{i=0}^q (y_i+U_{i_0},y_i+U_{i_0})\end{align}

for some $y_i \in \mathbb{N}_0^{\mathcal{S}}$ with $i=0,\dots, q$ . Note that $\{r\oplus r' \,\colon r \in \mathcal{R}_{i_0,j},\, r' \in \mathcal{R}_{j,i_0}\}$ is a finite set. This implies that

\begin{align*} \big\{ y \in \mathbb{N}_{0}^{\mathcal{S}} \,\colon (y + U_{i_0}, y + U_{i_0}) = r \oplus r', \: r \in \mathcal{R}_{i_0,j}, \: r' \in \mathcal{R}_{j,i_0}\big\}\end{align*}

is a finite set. Combining this observation with Proposition 2(iii) from [Reference Hoessly, Wiuf and Xia14], we conclude that $A_{i_0,2}$ is finite.

We prove finiteness of $A_{i_0,k}$ for general $k\geq 3$ by induction. Let $\gamma \in A_{i_0,k}$ be of the form (9) with $i=i_0$ . It suffices to prove the claim that

(*) \begin{equation} \text{the collection of all possible $\mathfrak{R}(\gamma)$ is finite, assuming $i_0 \notin \{i_1,\dots, i_q \}.$} \end{equation}

Otherwise, we can decompose $\gamma$ into a ‘summation’ of closed walks, where $U_{i_0}$ only appears as the initial and terminal nodes in each closed walk. Here, we say that a walk $\theta$ is the summation of two walks $\theta_1$ and $\theta_2$ , written as $\theta = \theta_1 + \theta_2$ , if

\[ \theta = U_{i_1}\ce{->[{$r_1$}]}U_{i_2}\ce{->[{$r_2$}]}\cdots \ce{->[{$r_{q''-1}$}]}U_{i_{q''}}, \\[-10pt] \nonumber \]

\[\theta_1 = \{U_{i_1}\ce{->[{$r_1$}]}\cdots \ce{->[{$r_{q'-1}$}]} U_{i_{q'}}\},\quad \mathrm{and}\quad \theta_2 =\{U_{i_{q'}}\ce{->[{$r_{q'}$}]}\cdots \ce{->[{$r_{q''-1}$}]}U_{i_{q''}}\}.\]

Therefore, $\mathfrak{R}(\gamma)$ can be written in the form of (23) and finiteness of $A_{i_0,k}$ follows from [Reference Hoessly, Wiuf and Xia14, Proposition 2(iii)], and the finiteness of the collection of all possible $y_i$ is finite, where the latter is a consequence of claim (*).

Suppose that $\gamma$ of the form (9) with $i=i_0$ and $i_0 \notin \{i_1,\dots, i_q\}$ . Then the collection of all possible $\gamma$ can be divided into two sets,

\begin{equation*} B_1 \,:\!=\, \{\gamma \,\colon i_{j}\neq i_{j'} \text{ for all } 1\leq j < j' \leq q\},\quad B_2 \,:\!=\, \{\gamma \,\colon \gamma \text{ is not in } B_1\}.\end{equation*}

It is trivial that $B_1$ is a finite set. It suffices to show that $B_2$ is finite, too. Let $\gamma \in B_2$ . We decompose $\gamma$ into the summation of closed walks connected by paths

\[\gamma = \theta_{1} + \gamma_{1} + \theta_{2} + \dots + \gamma_{q} + \theta_{q+1},\]

where for all i, $\gamma_{i}\in \Xi_{j_i'}$ and $\theta_{i}$ is a path in the sense that no repeated nodes are allowed.

Without loss of generality, we assume that $j_i'\neq j_{i'}'$ for all $1\leq i< i'\leq q'$ , where $j_i'$ is such that $\gamma_i \in \Xi_{j_i'}$ . Thus, $q\leq m_{\mathrm{pro}}$ . This is because, otherwise, if we let $\gamma'_{i} = \gamma_{i} + \theta_{i+1} + \cdots + \theta_{i'} + \gamma_{i'}\in \Xi_{j_i'}$ and write

\[\gamma = \theta_{1} + \gamma_{1} + \theta_{2} + \dots + \theta_{i} + \gamma_i' + \theta_{i' + 1} + \gamma_{i' + 1} + \dots + \gamma_{q} + \theta_{q+1},\]

this process can be iterated until no repeated $j_i'$ occurs. Note that the collection of all paths in $(\mathcal{V},\mathcal{E})$ is finite. This implies that the collection of all possible $\mathfrak{R}(\theta_{i})$ for $1\leq i\leq q'+1$ is finite. In addition, for any $i\in\{1,\dots, q\}$ , $U_{i_0}$ is not included in $\gamma_{i}$ , and thus $\mathfrak{R}(\gamma_{i})\in A_{j_i', k'}$ with some $k'<k$ . Thanks to the induction hypothesis $|A_{j_i', k'}| < \infty$ , the collection of all possible $\mathfrak{R}(\gamma_i)$ is finite for all $i = 1,\dots, q$ . Hence, $B_2$ (the collection of all possible $\mathfrak{R}(\gamma) = \mathfrak{R} (\theta_{1})\oplus \mathfrak{R}(\gamma_1) \oplus \cdots \oplus \mathfrak{R}(\theta_{q+1})$ ) is finite owing to [Reference Hoessly, Wiuf and Xia14, Proposition 2(iii)] and the fact that $q\leq m_{\mathrm{pro}}$ . This completes the proof of claim (*) for $i_0$ and thus for all $i = 1,\dots, m_{\mathrm{pro}}$ and $k = 2, \dots, m_{\mathrm{pro}}$ .

The rest of the proof is straightforward. For any $\gamma \in \Xi_0$ , we can decompose it into a summation of at most $m_{\mathrm{pro}}$ closed walks connected by paths. Since the ranges of the function $\mathfrak{R}$ on paths and closed walks are all finite, the collection of all possible $\mathfrak{R}(\gamma)$ is also finite. As a result, $|\mathfrak{R}(\Xi_0)| < \infty$ .

( $\Leftarrow$ ) Suppose that $\mathcal{R}_{\mathcal{U}}$ consists of finitely many reactions but Condition 2 fails. Then there exists a closed walk $\gamma_0 \in \Xi_{i_0}$ for some $i_0\in \{1,\dots, m_{\mathrm{pro}}\}$ such that $\mathfrak{R}(\gamma_0) = y_0 + U_{i_0} \ce{->} y_0' + U_{i_0}$ with $y_0\neq y_0'$ . Since $\mathcal{U}_{\mathrm{pro}}\subseteq\mathcal{U}_{\deg}$ (the set $\mathcal{U}$ can be eliminated), there are paths $\theta_1$ and $\theta_2$ directed from $U_0$ to $U_{i_0}$ and from $U_{i_0}$ to $U_{0}$ , respectively. It follows that for any $q\in \mathbb{N}$ ,

\[\gamma_{q} \,:\!=\, \theta_1 + \gamma_{q}^* +\theta_2 \in \Xi_0 \quad \text{where } \gamma_{q}^* \,:\!=\, \underbrace{\gamma + \dots + \gamma}_{{\textit{q} {\textrm{ instances of }}\gamma}}\!.\]

Let $(x_q,x_q') \,:\!=\, \mathfrak{R}(\gamma_{q})$ . We will show that $\{ (x_q,x_q') \,\colon q\in \mathbb{N}\}\cap \mathcal{R}_\mathcal{U}$ is an infinite set. Recall that $y_0\neq y_0'$ . Without loss of generality, assume that $y_{0,1} \neq y_{0,1}'$ . Here, we only prove the case of $y_{0,1}>y_{0,1}'$ ; the other case, $y_{0,1} < y_{0,1}'$ , can be proved similarly. Let

\[(z_q+U_{i_0},z_q'+U_{i_0}) \,:\!=\, \mathfrak{R}(\gamma_{q}^*)=\bigoplus_{i=1}^q(y+U_{i_0},y'+U_{i_0}).\]

Then, by Definition 3 and $y_{0,1}>y_{0,1}'$ , we have

\[(z_{q,1},z_{q,1}')=\bigl(y_{0,1}+q(y_{0,1} - y_{0,1}'),\, y_{0,1}'\bigr).\]

This yields that $\{z_{q,1}\,\colon q \in \mathbb{N}\}$ is a strictly increasing sequence. Let $\mathfrak{R}(\theta_1)=(y_1,y_1'+U_{1})$ and $\mathfrak{R}(\theta_2)=(y_2+U_{1},y_2')$ . Then, for all $q > M$ with M large enough, $z_{q,1}\geq y_{1,1}'$ holds. As a result,

\begin{align*}x_{q,1} &=y_{1,1}+z_{q,1}-y_{0,1}'+(y_{2,1}-z_{q,1}')\vee 0\\[3pt] \nonumber &=y_{1,1}+y_{0,1}+q(y_{0,1}-y_{0,1}')-y_{1,1}'+(y_{2,1}-y_{0,1}')\vee 0.\end{align*}

Thus, $\{x_{q,1}, q > M\}$ is a strictly increasing sequence. This contradicts $\{\mathfrak{R}(\gamma_{q}) = (x_{q},x_{q}') \,\colon q>M\} \subseteq \mathcal{R}_\mathcal{U}$ being a finite set. The proof is complete.

6.2. Proof of Proposition 1

If $\mathcal{U}$ consists of intermediate species, then $\Gamma_0$ is an irreducible component of $\mathcal{N}_{\mathcal{U}}$ ; see [Reference Hoessly, Wiuf and Xia14, Theorem 5.6]. In addition, if $\gamma\in \Xi_0$ , then $\mathfrak{R}(\gamma)=y\ce{->}y'$ , where $y=\mathrm{reac} (r_{0})$ and $y'=\mathrm{prod} (r_{q})$ . Thus, $\mathcal{C}_{\mathcal{U}}\subseteq \mathcal{C}_0=\{y\in \mathcal{C}\,\colon \rho (y) = 0\}$ . To prove the proposition, it suffices to show that the following equation holds for all $\eta\in \mathcal{C}_{\mathcal{U}}$ and $x\in\Gamma_0$ :

(24) \begin{align}\sum_{r\in \mathrm{out}_\mathcal{U}(\eta)} \pi(x)\lambda_{\mathcal{U},r}(x)=\sum_{r\in \mathrm{inc}_\mathcal{U}(\eta)} \pi \big(x-\zeta(r) \big)\lambda_{\mathcal{U},r} \big(x-\zeta (r) \big),\end{align}

where $\mathrm{out}_\mathcal{U} (\eta)$ and $\mathrm{inc}_\mathcal{U}(\eta)$ denote the sets of outgoing and incoming reactions of $\eta$ in $\mathcal{R}_{\mathcal{U}}$ , respectively. If $r\in \mathfrak{R}(\Xi_0)\setminus\mathcal{R}_\mathcal{U}$ , then $r=(\eta,\eta)$ for some $\eta$ , and r is both an outgoing ‘reaction’ of $\eta$ and an incoming ‘reaction’ of $\eta$ , with $\zeta(r) = \eta - \eta = 0$ . It follows that (24) is equivalent to

(25) \begin{align}\sum_{r\in \mathrm{out}_\mathcal{U}^* (\eta)} \pi(x)\lambda_{\mathcal{U},r}(x) = \sum_{r\in \mathrm{inc}_\mathcal{U}^* (\eta)} \pi \big(x - \zeta(r) \big) \lambda_{\mathcal{U},r} \big(x-\zeta (r) \big),\end{align}

where $\mathrm{out}_\mathcal{U}^* (\eta)$ and $\mathrm{inc}_\mathcal{U}^*(\eta)$ denote the sets of outgoing and incoming ‘reactions’ of $\eta$ in $\mathfrak{R}(\Xi_0)$ and $\lambda_{\mathcal{U}}$ is extended to $\mathfrak{R}(\Xi_0)$ by (14).

Recall that $\pi$ is a complex balanced distribution for $(\mathcal{N},\lambda)$ and that $\eta$ as in (25) is in $\mathcal{C}_{\mathcal{U}}\subseteq \mathcal{C}_0\subseteq \mathcal{C}$ . It follows that

\begin{align*}\sum_{r\in \mathrm{out} (\eta)} \pi(x)\lambda_r(x)=\sum_{r\in \mathrm{inc} (\eta)} \pi \big(x - \zeta (r) \big) \lambda_r \big(x - \zeta(r) \big),\end{align*}

where $\mathrm{out} (\eta)$ and $\mathrm{inc} (\eta)$ denote the sets of outgoing and incoming reactions of $\eta$ in $\mathcal{R}$ .

Denote by $L_0 (\eta)$ and $R_0 (\eta)$ the left- and right-hand sides of (25), respectively, and by $L (\eta)$ and $R (\eta)$ the left- and right-hand sides of (24), respectively. If $x\not\geq \eta$ , then $L_0(\eta)=R_0(\eta)=0$ . Thus, it suffices to show (25) assuming $x\geq \eta$ .

Step 1: $L(\eta)=L_0(\eta)$ . We have

\[L(\eta)=\pi(x)\left(\sum_{r\in \mathrm{out}(\eta)\cap \mathcal{R}_{0,0}}\lambda_r(x)+\sum_{i = 1}^{m_{\mathrm{pro}}} \sum_{r\in \mathrm{out}(\eta)\cap \mathcal{R}_{0,i}}\lambda_r(x)\right)\]

and

\begin{align*}L_0(\eta) = & \pi(x)\sum_{\gamma\in \Xi_0 :\,g(\gamma)=\eta}\lambda^*_{\gamma}(x) \\ = & \pi(x) \left(\sum_{r\in \mathrm{out}(\eta)\cap \mathcal{R}_{0,0}}\lambda_r(x) + \sum_{k = 1}^{\infty} \;\sum_{\gamma\in \Xi_0 (k) :\,g(\gamma)=\eta}\lambda^*_{\gamma}(x)\right)\!,\end{align*}

where $m_{\mathrm{pro}}$ is given in (22) and $g(\gamma) \,:\!=\, \mathrm{reac}(\mathrm{init}(\gamma))$ is the reactant of the reaction of the initial edge of $\gamma$ . We claim that for any reaction $r_* = \eta$ $\overset{r_*}{\longrightarrow} U_{\iota}$ with $U_\iota\in \mathcal{U}_{\mathrm{pro}}$ ,

(26) \begin{align}\lambda_{r_*} (x) = \sum_{k = 1}^{\infty} \sum_{\gamma\in \Xi_0(k):\, \mathrm{init}(\gamma)=U_0\overset{r_*}{\longrightarrow} U_{\iota}}\lambda^*_{\gamma}(x).\end{align}

Then, by summing both sides of (26) over $r_* \in \mathrm{out} (\eta)\setminus \mathcal{R}_{0,0}$ , we get

\begin{align*}\sum_{r_*\in \mathrm{out}(\eta)\setminus \mathcal{R}_{0,0}}\lambda_{r_*}(x)=\sum_{\gamma\in \Xi_0:\, g(\gamma)=\eta}\lambda^*_{\gamma}(x)\end{align*}

and thus $L(\eta)=L_{0}(\eta)$ .

To prove (26), we construct an auxiliary DTMC X on $\mathcal{X}=\{U_0\} \cup \mathcal{U}_{\mathrm{pro}}$ . By (12), the transition probabilities can be defined as

\[\mathbb{P}_{U_i} (U_j) = P_{i,j} = \displaystyle \sum_{r \in \mathcal{R}_{i,j}} \beta_{r} \big(x - \eta + \mathrm{reac} (r)\big).\]

Thus, for any $i,j\in\{0,\dots, m_{\mathrm{pro}}\}$ , $P_{i,j}>0$ if and only if $U_i \ce{->} U_j \in \mathcal{R}$ . In addition, since $\mathcal{U}$ is eliminable, it follows that X is irreducible. Let $\tau=\min\{k\geq 1\,\colon X_k=U_0\}$ . Note that

\begin{align*} \mathbb{P}_{U_{\iota}} (\tau = 1) = P_{\iota, 0} &= \sum_{r \in \mathcal{R}_{\iota, 0}} \beta_r (x - \eta + U_{\iota}) = \lambda_{r_*}(x)^{-1} \sum_{r \in \mathcal{R}_{\iota, 0}} \lambda_{r_*}(x) \beta_r (x - \eta + U_{\iota})\\ &= \lambda_{r_*}(x)^{-1} \sum_{\gamma \in \Xi_0(1):\, \mathrm{init} (\gamma) = U_0\overset{r_*}{\longrightarrow} U_{\iota} } \lambda_{\gamma}^*(x).\end{align*}

By iteration, one can show that for every $k \in \mathbb{N}$ ,

\begin{align*} \mathbb{P}_{U_{\iota}} (\tau = k) = \lambda_{r_*}(x)^{-1} \sum_{\gamma \in \Xi_0(k):\,\mathrm{init} (\gamma) = U_0\overset{r_*}{\longrightarrow} U_{\iota} } \lambda_{\gamma}^*(x).\end{align*}

Taking (13) into account and using that the Markov chain is irreducible (cf. [Reference Norris23, Theorems 1.5.6 and 1.5.7]), we have

\begin{align*}1 = \mathbb{P}_{U_{\iota}}(\tau<\infty) = \sum_{k=1}^{\infty}\mathbb{P}_{U_{\iota}}(\tau = k) = \lambda_{r_*} (x)^{-1} \sum_{k=1}^{\infty} \sum_{\gamma\in \Xi_0 (k):\, \mathrm{init}(\gamma) = U_0\overset{r_*}{\longrightarrow} U_{\iota}}\lambda_{\gamma}^*(x).\end{align*}

This proves (26) and thus $L(\eta)=L_0(\eta)$ .

Step 2: $R(\eta)=R_0(\eta)$ . Similarly, in order to show that $R(\eta)=R_0(\eta)$ , it suffices to verify that for any $r^{*} = U_{\iota} $ $\overset{r^*}{\longrightarrow} \eta$ with some $U_{\iota} \in \mathcal{U}_{\deg} = \mathcal{U}_{\mathrm{pro}}$ (see Lemma 1), the following equation holds:

(27) \begin{align}&\pi(x-\eta+U_{\iota}) \lambda_{r^*} (x-\eta+U_{\iota}) \nonumber \\[4pt]& \quad = \sum_{k=1}^{\infty} \sum_{\gamma\in \Xi_0(k) :\, \mathrm{ter}(\gamma) = U_{\iota} \overset{r^*}{\longrightarrow} U_0}\pi \big(x - \zeta (\mathfrak{R}(\gamma)) \big) \lambda_{\gamma}^* \big(x - \zeta (\mathfrak{R}(\gamma)) \big).\end{align}

For any $i,j \in \{0,\dots, m_{\mathrm{pro}}\}$ , let

\[P_{i,j}=\begin{cases}\displaystyle \frac{\sum_{y :\, y \to \eta \in \mathcal{R}_{j,0}}\pi(x - \eta + y) \lambda_{y \to \eta}(x - \eta + y) }{\pi (x) \sum_{y' :\, \eta \to y'\in \mathcal{R} } \lambda_{\eta \to y'}(x) }, & i = 0,\\[12pt]\displaystyle \frac{ \sum_{r\in \mathcal{R}_{j,i}} \pi(x - \eta + \mathrm{reac}(r)) \lambda_{r}(x - \eta + \mathrm{reac}(r)) }{\pi (x - \eta + U_i)\sum_{y':\, U_i \to y' \in \mathcal{R}} \lambda_{U_i \to y' }(x - \eta + U_i) }, & i \neq 0. \\[12pt]\end{cases}\]

Recall that $\pi$ is a complex balanced distribution for $(\mathcal{N},\lambda)$ on $\mathcal{N}$ ; then, from (6) and Condition 1, we see that $\mathbb{P}_{U_i} (U_j) = P_{i,j}$ for $i,j \in \{0,\dots, m_{\mathrm{pro}}\}$ forms a transition probability for a DTMC, say Y, on $\mathcal{X}$ . Let $\tau \,:\!=\, \min\{q\geq 1:\, Y_q = U_0\}$ , and for any $k\in \mathbb{N}$ let

\begin{align*}R_{1, k}(\eta) \,:\!=\, \sum_{\gamma\in \Xi_0(k) :\, \mathrm{ter}(\gamma)= U_{\iota} \overset{r^*}{\longrightarrow} U_0}\pi \big(x-\zeta(\mathfrak{R}(\gamma)) \big) \lambda_{\gamma}^* \big(x - \zeta (\mathfrak{R}(\gamma)) \big).\end{align*}

We aim to show that

(28) \begin{align} \mathbb{P}_{U_{\iota}} (\tau = k) = R_{1,k}(\eta) / L_1(\eta)\end{align}

for all $k \geq 1$ , where $L_1(\eta)$ is the left-hand side of (27). Suppose that $k = 1$ . Then

(29) \begin{align} \mathbb{P}_{U_{\iota}} (\tau = 1) = P_{\iota,0} = & \frac{ \sum_{r\in \mathcal{R}_{0,\iota}} \pi(x - \eta + \mathrm{reac}(r)) \lambda_{r}(x- \eta + \mathrm{reac}(r)) }{\pi (x - \eta + U_{\iota}) \sum_{y':\, U_i \to y' \in \mathcal{R}} \lambda_{U_i \to y' }(x - \eta + U_i) }.\end{align}

On the other hand, for any $r' = y\ce{->} y'\in \mathcal{R}_{i,j}$ with some $i,j\in \{0,\dots, m_{\mathrm{pro}}\}$ and $x'\geq y$ , it follows from (11) and (12) that

\begin{align*}\lambda_{r'}(x') = \lambda_{r'}(x')\sum_{j'=0}^{m}\sum_{r''\in \mathcal{R}_{i,j'}}\beta_{r''}(x') = \beta_{r'}(x') \sum_{j'=0}^{m}\sum_{r'' \in \mathcal{R}_{i,j'}}\lambda_{r''}(x') > 0\end{align*}

and therefore

(30) \begin{align}\beta_{r'}(x') = \lambda_{r'}(x') \bigg/ \sum_{j'=0}^{m}\sum_{r'' \in \mathcal{R}_{i,j'}}\lambda_{r''}(x').\end{align}

As a consequence, we have

(31) \begin{align}R_{1,1}(\eta) = & \sum_{y :\, y\to U_{\iota} \in \mathcal{R}_{0,\iota}} \pi(x - \eta + y) \lambda_{y \to U_{\iota}} (x-\eta + y) \beta_{r^{\,*}} (x-\eta + y - y + U_{\iota}) \nonumber\\[3pt]= & \frac{ \lambda_{r^*} (x-\eta + U_{\iota})\sum_{y :\, y\to U_{\iota} \in \mathcal{R}_{0,\iota}} \pi(x-\eta + y) \lambda_{y \to U_{\iota}} (x-\eta + y) }{ \sum_{y' :\, U_{\iota} \to y' \in \mathcal{R}}\lambda_{U_{\iota} \to y'}(x-\eta + U_{\iota})}.\end{align}

By comparing (27), (29), and (31), one obtains equation (28) with $k = 1$ .

Similarly, assume $k = 2$ . Then we can write

\begin{align*}\mathbb{P}_{U_{\iota}} (\tau = 2) = \sum_{i=1}^{m_{\mathrm{pro}}} \sum_{y :\, y \to U_{i}\in \mathcal{R}_{0,i} } \bigg[ &\frac{\pi(x+\eta-y) \lambda_{y\to U_{i}}(x - \eta + y) }{\sum_{y' :\, U_i \to y'\in\mathcal{R}} \lambda_{U_i \to y'}(x) }\\&\times \frac{ \lambda_{U_{i} \to U_{\iota}}(x- \eta + U_i) }{\pi(x-\eta+U_{\iota})\sum_{y' :\, U_{\iota} \to y' \in \mathcal{R}}\lambda_{U_{\iota} \to y'}(x-\eta + U_{\iota})}\bigg],\end{align*}

and using (30) we have

\begin{align*}R_{1,2}(\eta)& =\sum_{i=1}^{m_{\mathrm{pro}}} \sum_{y :\, y \to U_{i}\in \mathcal{R}_{0,i} } \bigg[ \pi(x-\eta + y) \lambda_{y \to U_{i}} (x-\eta + y) \\ & \hspace{7.5em}\times \,\beta_{U_i\to U_{\iota}} (x-\eta + U_i) \beta_{r^{\,*}} (x-\eta + U_{\iota}) \bigg]\\ &= \bigg[\sum_{i=1}^{m_{\mathrm{pro}}} \sum_{y :\, y \to U_{i}\in \mathcal{R}_{0,i} } \frac{ \lambda_{y \to U_{i}} (x-\eta + y) \lambda_{U_i\to U_{\iota}} (x-\eta + U_i) }{ \sum_{y' :\, U_{i} \to y' \in \mathcal{R}}\lambda_{U_i \to y'}(x-\eta + U_i)}\bigg]\\[3pt] \nonumber & \quad \times \frac{ \lambda_{U_{\iota}\to \eta}(x-\eta + U_{\iota}) }{ \sum_{y' :\, U_{\iota} \to y' \in \mathcal{R}}\lambda_{U_{\iota} \to y'}(x-\eta + U_{\iota}) }.\end{align*}

This implies that equation (28) holds with $k = 2$ as well. By using an iteration argument, one should be convinced that (28) is true for all $k \in \mathbb{N}$ . This leads to (27) and hence $R_0(\eta) = R(\eta)$ . The proof of this proposition is complete.

6.3. Proof of Proposition 2

The reduced reaction network $\mathcal{N}_{\mathcal{U}}$ is reversible; see [Reference Hoessly, Wiuf and Xia14, Theorem 5.1]. For $\gamma\in \Xi_0$ , by reversibility, there exists the walk in the opposite direction,

\[\gamma'=U_0\ce{->[{$r_q'$}]}U_{i_q}\ce{->}\cdots \ce{->[{$r_1'$}]}U_1\ce{->[{$r_0'$}]}U_0\ \ \in \Xi_0,\]

where $r_j' =y'\ce{->}y$ if $r_j = y\ce{->}y'$ for $0\leq j\leq q$ with $i_0=i_{q+1}=0$ . This implies that $\zeta (r_j) = -\zeta (r_j')$ for all $0\leq j\leq q$ , and thus $\zeta (\gamma) = - \zeta(\gamma')$ with $\zeta(\gamma) \,:\!=\, \sum_{j=0}^q \zeta (r_j)$ . For any $x\in \Gamma_0$ , define

\begin{align*}x' \,:\!=\, x+\zeta(\gamma) = x + \sum_{j=0}^q \zeta(r_j) = x - \sum_{j=0}^q \zeta(r_j') = x - \zeta(\gamma').\end{align*}

By (14), it is enough to verify the following equation:

(32) \begin{align}\pi(x)\lambda_{\gamma}^* (x) = \pi (x') \lambda_{\gamma'}^* (x'),\end{align}

provided that $x\geq \mathrm{reac}(\mathfrak{R}(\gamma))$ . Recall that $\pi$ is a detailed balanced distribution for $(\mathcal{N},\lambda)$ . By definition,

\[\pi(x)\lambda_{r_j}(x) = \pi \big(x - \zeta (r_j') \big)\lambda_{r_j'} \big(x - \zeta(r_j') \big) = \pi \big(x + \zeta(r_j) \big) \lambda_{r_j'} \big(x + \zeta(r_j) \big).\]

Therefore,

\begin{align*}\pi(x)\lambda^*_{\gamma}(x)&=\pi(x)\lambda_{r_0}(x)\prod_{j=1}^{q}\frac{\lambda_{r_j}\big(x + \zeta(r_0) + \dots + \zeta (r_{j-1}) \big)}{\sum_{j'=0}^m\sum_{r \in \mathcal{R}_{i_j,j'}} \lambda_{r} \big(x + \zeta(r_0) + \dots + \zeta(r_{j-1})\big)}\\[3pt]&=\lambda_{r_0'} \big(x+\zeta(r_0)\big) \pi\big(x + \zeta(r_0)\big) \frac{\lambda_{r_1}\big(x+\zeta(r_0) \big)}{\sum_{j'=0}^m\sum_{r'\in \mathcal{R}_{i_1,j'}}\lambda_{r'}\big(x+\zeta(r_0)\big)}\\[3pt]&\quad\times\prod_{j=2}^{q}\frac{\lambda_{r_j} \big(x + \zeta(r_0) + \dots + \zeta(r_{j-1}) \big)}{\sum_{j'=0}^m \sum_{r''\in \mathcal{R}_{i_j,j'}} \lambda_{r''}\big(x + \zeta (r_0) + \dots + \zeta(r_{j-1}) \big)}\\[3pt]&=\beta_{r_0'}\big(x'+\zeta(r_q') + \dots + \zeta(r_1') \big) \pi\big(x+\zeta(r_0)\big) \lambda_{r_1} \big(x + \zeta(r_0)\big)\\[3pt]&\quad \times\prod_{j=2}^{q}\frac{\lambda_{r_j}\big(x+\zeta(r_0) + \dots + \zeta(r_{j-1}) \big)}{\sum_{j'=0}^m\sum_{r''\in\mathcal{R}_{i_j,j'}}\lambda_{r''}\big(x+\zeta(r_0)+\dots+\zeta(r_{j-1})\big)}.\end{align*}

Similarly, we deduce that

\begin{align*}& \pi \big(x+\zeta(r_0)\big)\lambda_{r_1} \big(x+\zeta(r_0)\big) \prod_{j=2}^{q}\frac{\lambda_{r_j}\big(x+\zeta(r_0)+\dots+\zeta(r_{j-1})\big)}{\sum_{j'=0}^m\sum_{r \in \mathcal{R}_{i_j,j'}}\lambda_{r}\big(x+\zeta(r_0)+\dots+\zeta(r_{j-1})\big)}\\[3pt]&\quad = \beta_{r_1'}\big(x' + \zeta(r_q')+\dots+\zeta(r_2')\big) \pi \big(x+\zeta(r_0)+\zeta(r_1)\big) \lambda_{r_2} \big(x+\zeta(r_0)+\zeta(r_1)\big)\\[3pt]&\qquad\times\prod_{j=3}^{q}\frac{\lambda_{r_j}\big(x+\zeta(r_0)+\dots+\zeta(r_{j-1})\big)}{\sum_{j'=0}^m\sum_{r'\in \mathcal{R}_{i_j,j'}}\lambda_{r'}\big(x+\zeta(r_0)+\dots+\zeta(r_{j-1})\big)}.\end{align*}

Upon repeating the above arguments, one ends up with the equation

\begin{align*}\pi(x)\lambda^*_{\gamma}(x) = & \pi(x')\lambda_{\gamma'}^*(x')\prod_{j=1}^q \beta_{r_{j-1}'} \left(x' + \sum_{j'=j}^{q} \zeta(r_{j'}') \right) = \pi(x')\lambda_{\gamma'}^* (x').\end{align*}

This proves (32), and thus $\pi_0$ is a detailed balanced distribution (or linear combination of detailed balanced distributions) for $(\mathcal{N}_{\mathcal{U}},\lambda_{\mathcal{U}})$ on $\Gamma_0$ . The proof is complete.

6.4. Proof of Theorem 1

Proof of (i). It suffices to show that for any $x\in \Gamma_0$ and $x'\in\mathbb{N}_0^{n_{\mathcal{U}}}$ , $x\to_{\mathcal{N}_{\mathcal{U}}} x'$ implies (a) $x'\in\Gamma_0$ and (b) $x'\to_{\mathcal{N}_{\mathcal{U}}} x$ .

By definition, $x\to_{\mathcal{N}_{\mathcal{U}}} x'$ if and only if there exist reactions $r_1,\dots, r_k \in \mathcal{R}_{\mathcal{U}}$ such that $r_1\oplus \cdots \oplus r_k = (y,y')$ with $x\geq y$ and $x' - x = y' - y$ . Using an iteration argument, we can assume that $k = 1$ . By the definition of the reduction of reaction networks, either $r_1 \in \mathcal{R}$ or there exists a $\gamma \in \Xi_0$ such that $r_1 = \mathfrak{R} (\gamma)$ . In either case, we have $x\to_{\mathcal{N}} x'$ with $\rho(x') = 0$ and thus $x' \in \Gamma_0$ .

By the weak reversibility of $\mathcal{N}$ , if $r_1 = y_1\ce{->}y' \in \mathcal{R}$ , then there exists a sequence of reactions such that $y'\ce{->}y_1\ce{->}y_2\ce{->}\cdots \ce{->}y_{k'}\ce{->} y \subseteq \mathcal{R}$ . We further assume that $y_i \neq y_j$ for all $1 \leq i < j \leq k'$ . Otherwise, if $y_i = y_j$ , one can work with the sequence $y'\ce{->}y_1\ce{->} \cdots \ce{->}y_{i} \ce{->}y_{j + 1} \ce{->} \cdots \ce{->} y$ , which is also included in $\mathcal{R}$ . If $y_i \in \mathcal{C}_{0}$ for all $i=1,\dots, k'$ , then each reaction is also an element of $\mathcal{R}_{\mathcal{U}}$ . This proves $x'\to_{\mathcal{N}_{\mathcal{U}}} x$ . Otherwise, there are $1 \leq i_1 < i_2 < \dots < i_{k''} \leq k'$ for some $k''<k'$ such that $y_{i}\in \mathcal{C}_0$ if and only if $i \in \{i_1,\dots, i_{k''}\}$ . For each $q \in \{0,\dots, k''\}$ , let $\gamma_q$ be the closed walk in $(\mathcal{V},\mathcal{E})$ given by the reactions $y_{i_q} \ce{->} y_{i_q+1} \ce{->}\cdots \ce{->} y_{i_{q+1}}$ , where by convention $y_{i_0} \,:\!=\, y'$ and $y_{i_{k'+1}} \,:\!=\, y$ . Then $\gamma_q \in \Xi_0$ for all q. By removing all $\gamma_q$ such that $\mathfrak{R}(\gamma_q) = (y,y)$ for some $y\in\mathbb{N}_0^n$ and recalling the assumption that $y_i \neq y_j$ for all $1 \leq i < j \leq k'$ , we know that $r_{(q)} \,:\!=\, \mathfrak{R}(\gamma_q) = y_{i_q} \ce{->} y_{i_{q + 1}} \in \mathcal{R}_{\mathcal{U}}$ . Therefore $\bigoplus_{q=0}^{k''} r_{(q)} = (y',y)$ . This proves $x'\to_{\mathcal{N}_{\mathcal{U}}} x$ . The proof of Theorem 1(i) is thus complete.

Proof of (ii). For any $\eta \in \mathcal{C}_0$ , let

\begin{align*}L_0 (\eta) = &\sum_{r \in \mathrm{out}_{\mathcal{U}}^* (\eta)} \pi(x) \lambda_{\mathcal{U},r}(x) \\= & \sum_{r\in \mathrm{out} (\eta)\cap \mathcal{R}_{0,0}} \pi(x)\lambda_r(x)+\sum_{\gamma\in \Xi_0:\, \mathrm{init}(\gamma)\in \mathrm{out}(\eta)} \pi(x)\lambda^*_{\gamma}(x),\end{align*}

\begin{align*}R_0 (\eta) = &\sum_{r\in \mathrm{inc} (\eta)\cap \mathcal{R}_{0,0}} \pi \big(x - \zeta(r) \big) \lambda_r \big(x - \zeta (r) \big)\\&+ \sum_{\gamma\in \Xi_0 :\, \mathrm{ter}(\gamma)\in \mathrm{inc}(\eta)} \pi \big(x - \zeta (\mathfrak{R}(\gamma))\big)\lambda^*_{\gamma} \big(x - \zeta (\mathfrak{R}(\gamma)\big),\end{align*}

where $x\in \Gamma_0$ satisfies $x\geq \eta$ . We use $L (\eta)$ and $R (\eta)$ to denote the left- and right-hand sides, respectively, of the following complex balanced equation:

\begin{align*}\pi(x)\sum_{r\in \mathrm{out} (\eta)} \lambda_r(x)=\sum_{r\in \mathrm{inc} (\eta)} \pi \big((x-\zeta(r) \big)\lambda_r \big(x - \zeta(r) \big).\end{align*}

Then we prove $L_0 (\eta) = R_0 (\eta)$ by verifying that $L (\eta) = L_0(\eta)$ and $R (\eta)=R_0 (\eta)$ . Once this has been done, because $L_0(\eta) = R_0(\eta) = 0$ if $x\not\geq \eta$ , we have

\[\pi(x)\sum_{r\in \mathfrak{R}(\Xi_0)} \lambda_{\mathcal{U},r}(x) = \sum_{\eta \in \mathcal{C}_0} L_0(\eta) = \sum_{\eta \in \mathcal{C}_0} R_0(\eta) =\sum_{r\in \mathfrak{R}(\Xi_0)} \pi \big(x - \zeta(r)\big)\lambda_{\mathcal{U},r} \big(x - \zeta(r) \big).\]

This implies that

\[\pi(x)\sum_{r\in \mathcal{R}_{\mathcal{U}}} \lambda_{\mathcal{U},r}(x)=\sum_{r\in \mathcal{R}_{\mathcal{U}}} \pi \big(x - \zeta (r) \big)\lambda_{\mathcal{U},r} \big(x - \zeta (r)\big),\]

and thus $\pi$ is a stationary distribution for $(\mathcal{N}_{\mathcal{U}},\lambda_{\mathcal{U}})$ .

To achieve this goal, we introduce two DTMCs, which are used in the proofs of $L (\eta) = L_0(\eta)$ and $R (\eta)=R_0 (\eta)$ . Denote by $\Theta_0$ the set of all open (not closed) walks in $(\mathcal{V},\mathcal{E})$ of the form

\[\theta=\left\{U_0\ce{->[{$r_0$}]}U_{i_1}\ce{->[{$r_1$}]}\cdots \ce{->[{$r_{q-1}$}]}U_{i_q}\right\},\quad q\in \mathbb{N}_0,\; i_1,\dots, i_q \in \{1,\dots, m_{\mathrm{pro}}\}.\]

Define $\mathcal{X} \,:\!=\, \mathcal{X}_0\cup \{\partial\}$ , where $\partial$ is a ceremony state, and

(33) \begin{align}\mathcal{X}_0 \,:\!=\, \big\{x + \zeta (\mathfrak{R}(\theta)) :\, \theta \in \Theta_0,\: x\geq \mathrm{reac}(\mathfrak{R}(\theta))\big\}.\end{align}

Analogously, define $\mathcal{Y} \,:\!=\, \mathcal{Y}_0 \cup \{\partial\}$ , where

(34) \begin{align}\mathcal{Y}_0 \,:\!=\, \big\{y\in \mathbb{N}_0^n :\, y + \zeta (\mathfrak{R}(\theta)) = x, \:\theta \in \Theta_0', \: y\geq \mathrm{reac}(\mathfrak{R}(\theta)) \big\},\end{align}

with $\Theta_0'$ consisting of all open walks in $(\mathcal{V},\mathcal{E})$ of the form

\[\theta=\{U_{i_1}\ce{->[{$r_1$}]}\cdots \ce{->[{$r_{q-1}$}]}U_{i_q}\ce{->[{$r_q$}]} U_0\},\quad q\in \mathbb{N}_0,\; i_1,\dots, i_q \in \{1,\dots, m_{\mathrm{pro}}\}.\]

The proof of Lemma 4 is in Subsection 6.5.

Step 1: $L(\eta) = L_0(\eta)$ . As in Step 1 of Subsection 6.2, to prove $L(\eta) = L_0(\eta)$ it suffices to show that for every reaction $r_* = \eta\ce{->}y'+U_{\iota} \in \mathcal{R}_{0,\iota}$ ,

(35) \begin{align}\sum_{k=1}^{\infty}\:\sum_{\gamma\in \Xi_0(k) :\, \mathrm{init}(\gamma) = r_*} \lambda^*_{\gamma}(x)= \lambda_{r_*}(x).\end{align}

To prove (35), we introduce a DTMC X and relate (35) to the recurrence of X. For a clearer understanding, we refer readers to Subsection 6.2, where a simpler instance is detailed to illustrate how the recurrence of X leads to the validity of (35). Because of the non-interacting structure, for any $z\in \mathcal{X}_0$ we have $\rho(z) = U_i$ for some $U_i\in \mathcal{U}_{\mathrm{pro}}$ . This allows us to define the transition probabilities of the Markov chain X as follows:

\begin{align*}\mathbb{P}_{z} (z') = P_{z,z'} \,:\!=\, \begin{cases}\displaystyle \sum_{r\in \mathcal{R}_{0,0}}\beta_{r}(x), & \quad z = z' = \partial,\\[12pt]\displaystyle \sum_{r \in \mathcal{R}_{i,0}} \beta_{r}(z), & \quad \rho (z) = U_i, \,z' = \partial,\\[12pt]\displaystyle \sum_{r \in \mathcal{R}_{0,j}:\, \zeta (r) = z' - x }\beta_{r}(x), & \quad z = \partial, \,\rho (z') = U_j, \\[12pt]\displaystyle \sum_{r\in \mathcal{R}_{i,j}:\, \zeta (r) = z' - z} \beta_{r} (z), & \quad \rho (z) = U_i, \,\rho (z') = U_j,\end{cases}\end{align*}

where $i,j\in \{1,\dots,m_{\mathrm{pro}}\}$ . Moreover, (35)

(36) \begin{align}\mathbb{P}_{x + \zeta(r_0)}(\tau < \infty)=1,\quad \text{where } \tau = \min \{k\geq 1\,\colon X_k = \partial\}.\end{align}

By weak reversibility of $\mathcal{N}$ , X is irreducible, and thus (36) is true because $\mathcal{X}_0$ is a finite set; see Lemma 4(i).

Step 2: $R(\eta) = R_0(\eta)$ . Using the same idea as in Step 2 of Subsection 6.2, we can reduce this task to showing that for every reaction $r^* = y' + U_{\iota} \ce{->} \eta \in \mathcal{R}_{\iota,0}$ ,

(37) \begin{align}\sum_{\gamma\in \Xi_0,\,\mathfrak{R}(\mathrm{ter}(\gamma))=r^*} \pi \big(x - \zeta (\mathfrak{R}(\gamma)) \big) \lambda^*_{\gamma} \big(x - \zeta (\mathfrak{R}(\gamma) )\big) = \pi \big(x - \zeta(r^*) \big) \lambda_{r^*} \big(x - \zeta (r^*) \big)\end{align}

for all $x\in \Gamma_0$ such that $x - \zeta (r^*) \geq \mathrm{reac} (r^*)$ .

Let Y be a DTMC on $\mathcal{Y}$ with transition probabilities

\[P_{z,z'}=\begin{cases}\displaystyle \sum_{r\in \mathcal{R}_{0,0}}\frac{ \pi \big(x - \zeta (r) \big) \lambda_{y\to z} \big(x - \zeta(r) \big) }{\pi(x)\sum_{j = 0}^m \sum_{r'\in \mathcal{R}_{0,j}} \lambda_{r'}(x) }, & \quad z = z' = \partial,\\[12pt]\displaystyle \sum_{r\in \mathcal{R}_{i,0}:\, \zeta(r) = z' - x}\frac{\pi(z') \lambda_{r}(z') }{\pi(x)\sum_{j = 0}^m \sum_{r'\in \mathcal{R}_{0,j}} \lambda_{r'}(x) }, &\quad z = \partial, \,\rho (z') = U_i, \\[12pt]\displaystyle \sum_{r\in \mathcal{R}_{0, j}}\frac{\pi \big(z - \zeta(r) \big) \lambda_{r} \big(z - \zeta(r) \big) }{\pi (z)\sum_{j'=0}^{m}\sum_{r'\in \mathcal{R}_{j,j'}} \lambda_{r'}(z) }, &\quad \rho (z) = U_j , \,z' = \partial,\\[12pt]\displaystyle \sum_{r\in \mathcal{R}_{i, j}:\, \zeta (r) = z' - z}\frac{\pi(z') \lambda_{r}(z' ) }{\pi (z)\sum_{j'=0}^{m}\sum_{r'\in \mathcal{R}_{j,j'}} \lambda_{r'}(z) }, &\quad \rho (z) = U_j , \, z' = U_i,\end{cases}\]

where $i,j\in \{1,\dots, m_{\mathrm{pro}}\}$ . Following the same line of argument as in Step 2 of Subsection 6.2, we can show that (37) is equivalent to $\mathbb{P}_{y' + U_i} (\tau < \infty) = 1$ , where $\tau \,:\!=\, \min \{k \geq 1 :\, Y_k = \partial\}$ . This is due to the recurrence of an irreducible DTMC in a finite state space. The proof of Theorem 1(ii) is thus complete.

6.5. Proof of Lemma 4

It suffices to show Lemma 4(i) by proving the following statements.

1. (A) If Condition 3(i) fails, then $\mathcal{X}_0$ in (33) is infinite for some $x\in \mathbb{N}_0^n$ .

2. (B) If Condition 3(i) holds, then $\mathcal{X}_0$ in (33) is finite for all $x\in \mathbb{N}_0^n$ .

Proof of statement (A). Suppose that Condition 3(i) fails; then there exists $\gamma \in \Xi_i$ such that $\mathfrak{R} (\gamma) = (y + U_i, y' + U_i) \in \Xi_i$ for some $U_i\in \mathcal{U}_{\mathrm{pro}}$ and $y < y'$ . Since $U_i\in \mathcal{U}_{\mathrm{pro}}$ , there exists a walk $\theta$ in $(\mathcal{V},\mathcal{E})$ with $\mathrm{init} (\theta) = U_0$ and $\mathrm{ter} (\theta) = U_i$ . Consider the sequence of paths

\[\theta_q \,:\!=\, \theta + \underbrace{\gamma + \cdots + \gamma}_{{{q} {\textrm { instances of}} \gamma}} \in \Theta_0,\quad q\geq 1.\]

Let $\mathfrak{R}(\theta) = (y_0, y_0')$ . Then

\begin{align*}(z_q,z_q') \,:\!=\, \mathfrak{R}(\theta_q) = \mathfrak{R}(\theta) \oplus \underbrace{\mathfrak{R}(\gamma) \oplus \cdots \oplus \mathfrak{R}(\gamma)}_{{\textit{q} \ { \textrm{instances of }\ } \mathfrak{R}(\gamma)}} = (y_0,y_0') \oplus (y, y+q(y'-y)).\end{align*}

Using Proposition 2(4) from [Reference Hoessly, Wiuf and Xia14], we have $z_q\leq y_0+y_1$ and $z_q' \geq y+q(y'-y)$ . It follows that the set

\begin{align*}\big\{x + \zeta\big(\mathfrak{R}(\theta_q)\big) = y_0 + z_q' - z_q:\, q\in \mathbb{N}\big\}\end{align*}

is an infinite subset of $\mathcal{X}_0$ . This completes the proof of statement (A).

The proof of statement (B) is based on the next two lemmas.

Lemma 6. Let $x\in \mathbb{N}_0^n$ , let q be a positive integer, and let $\mathcal{Z} \,:\!=\, \{\zeta_1,\dots, \zeta_q\}\subseteq \mathbb{Z}^n\setminus \mathbb{N}_0^n$ . Let $\mathcal{H}$ consist of all linear combinations of elements in $\mathcal{Z}$ with non-negative integer coefficients,

\[\mathcal{H} \,:\!=\, \bigg\{\sum_{i=1}^q k_i \zeta_i:\, k=(k_1,\dots,k_q)\in \mathbb{N}_0^q\bigg\}.\]

Then (i) $\mathcal{H}_x \,:\!=\, \{\eta\in \mathcal{H}\,:\, x+\eta\in \mathbb{N}_0^n\}$ is a finite set if and only if (ii) $\mathcal{H}\cap \mathbb{N}_0^n\setminus\{0\}=\varnothing$ .

Proof of statement (B). Suppose that Condition 3(i) holds. For any $\theta\in \Theta_0$ , there exists a collection of closed walks $\{\gamma_i\}_{i=1}^q$ connected by paths $\{\theta_i\}_{i=1}^{q+1}$ such that

\[\theta = \theta_1 + \gamma_1 + \theta_2 + \dots + \gamma_{q} + \theta_{q+1},\]

where $\gamma_i\in \Xi_{j_i}$ with $j_i\in \{1,\dots, m_{\mathrm{pro}}\}$ for $i=1,\dots, q$ such that $j_i\neq j_{i'}$ for all $i\neq i'$ . Then $q\leq m_{\mathrm{pro}}$ . As the number of paths in $(\mathcal{V},\mathcal{E})$ is finite, it suffices to show that the set

\begin{align*}A_i(x) \,:\!=\, \big\{x + \zeta(\mathfrak{R}(\gamma)) \,:\, \gamma \in \Xi_i, \, \mathrm{reac} (\mathfrak{R}(\gamma)) \leq x\big\}\end{align*}

is finite for every $x\in \mathbb{N}_0^n$ and $i\in \{1,\dots, m_{\mathrm{pro}}\}$ . For any $i\in\mathcal{C}$ , let

\[A_{i,k}(x) = \big\{ x + \zeta(\mathfrak{R}(\gamma)) \,:\, \gamma \in \Xi_i, \, x\geq \mathrm{reac} (\mathfrak{R}(\gamma)),\, |\mathcal{C}_{\gamma}| = k \big\},\]

with $k\in\{1,\dots, m_{\mathrm{pro}}\}$ and $\mathcal{C}_{\gamma} = \{U_i\in \mathcal{U} \,:\,U_i\in\gamma\}$ . Then $A_i(x) = \bigcup_{k=1}^{m_{\mathrm{pro}}} A_{i,k} (x)$ for $i\in \{1,\dots, m_{\mathrm{pro}}\}$ . We aim to show that $A_{i,k} (x)$ is a finite set for all $i\in \{1,\dots, m_{\mathrm{pro}}\}$ , $1\leq k\leq m_{\mathrm{pro}}$ , and $x\in \mathbb{N}_0^n$ by induction on k.

Fix any $i_0\in \{1,\dots, m_{\mathrm{pro}}\}$ . Note that $\{\Gamma\in \Xi_{i_0}:\, |\mathcal{C}_{\gamma}| = 1\} = \mathcal{R}_{i_0,i_0}$ is a finite set. Thus, by Condition 3(i) and Lemma 6, $A_{i_0,1}(x)$ is finite. For $k\geq 2$ , we show that $A_{i_0,k}(x)$ is finite by contradiction. Suppose that $A_{i_0,k}(x)$ is not a finite set. Then there exists a sequence of closed walks $\{\gamma_j\}_{j\geq 1} \subseteq \Xi_{i}$ such that the following hold:

1. (a) $|\mathcal{C}_{\gamma_j}| = k$ and $x\geq \mathrm{reac} (\mathfrak{R} (\gamma_j))$ for all $i\in \mathbb{N}$ ;

2. (b) $\{x + \zeta(\mathfrak{R}(\gamma_j))\}_{j\geq 1} \subseteq A_{i,k}$ is an infinite subset of $\mathbb{N}_0^n$ ;

3. (c) $ \zeta(\mathfrak{R} (\gamma_j)) \neq \zeta(\mathfrak{R} (\gamma_{j'}))$ for all indexes $j\neq j'$ .

Using Lemma 5, we can assume that $\{\zeta(\mathfrak{R}(\gamma_j))\}_{j\geq 1}$ is a strictly increasing sequence. Next, for every $j\in \mathbb{N}$ , $\gamma_j$ can be decomposed as $\gamma_j = \gamma_{j,1} + \dots + \gamma_{j,q_j}$ such that for $j'=1,\dots, q_j$ , $\gamma_{j,j'}\in \Xi_{i_0}$ , and $U_{i_0}$ only appears as the initial and terminal nodes of $\gamma_{j,j'}$ . Assume that

(38) \begin{align}\sum_{j'=k_1}^{k_2} \zeta \big(\mathfrak{R}(\gamma_{j,j'})\big) \neq 0\quad \text{for all }1\leq k_1\leq k_2\leq q_i.\end{align}

Otherwise, we can replace $\gamma_j$ by $\gamma_j' = \gamma_{j,1} + \dots + \gamma_{j,k_1 - 1} + \gamma_{j,k_2 + 1}$ . Then $\mathrm{reac} (\gamma_j') \leq \mathrm{reac}(\gamma_j) \leq x$ and $\zeta(\mathfrak{R}(\gamma_j')) = \zeta(\mathfrak{R}(\gamma_j))$ . Thus, with this substitution, properties (a)–(c) still hold.

For any $i\in \mathbb{N}$ , $\gamma_{i,1}$ can be decomposed into at most $k-1$ closed walks connected by paths. By the induction hypothesis, for any $x\in\mathbb{N}_0^n$ , the set $\{x + \zeta(\mathfrak{R}(\gamma_{i,1}))\}_{i\geq 1}$ is finite. Thus, for any positive integer q and any index j, $\{x + \zeta(\mathfrak{R}(\gamma_{j,1})) + \dots + \zeta(\mathfrak{R}(\gamma_{j,j'}))\}_{j'=1}^{q_j\wedge q}$ is a finite set. Therefore, if $\{q_j\}_{j\geq 1}$ is finite, then so is $\{x + \zeta(\mathfrak{R}(\gamma_j))\}_{j\geq 1}$ . This contradicts (b). Thus, by taking subsequences if necessary, we can assume that $\{q_j\}_{j\geq 1}$ is a strictly increasing sequence of positive integers. Let

\[\eta_{j,j'} \,:\!=\, x + \zeta\big(\mathfrak{R}(\gamma_{j,1})\big) + \dots + \zeta \big(\mathfrak{R}(\gamma_{j,j'})\big)\]

for all $j\in \mathbb{N}$ and $j' \in \{1,\dots, q_j\}$ . It follows from (a) that $\eta_{j,j'} \in \mathbb{N}_0^n$ . Recall that $\{\eta_{j,1} \}_{j\geq 1} = \{x + \zeta(\mathfrak{R}(\gamma_{j,1}))\}_{j\geq 1}$ is a finite set. There exists a subsequence $\{\eta_{n_{j}^1,1}\}_{j\geq 1}$ of $\{\eta_{j,1}\}_{j\geq 1}$ such that $\eta_{n_j^1,1} = y_1\in\mathbb{N}_0^n$ for all indexes j ^′. Iteratively, for any $k\geq 2$ , there exists a subsequence $\{\eta_{n_j^k,k} \}_{j\geq 1}$ of $\{\eta_{n_j^{k-1},k} \}_{j\geq 1}$ such that $\eta_{n_j^k,k} = y_{k}\in\mathbb{N}_0^n$ for all $j\in \mathbb{N}$ . Concerning Condition 3(i), the sequence $\{y_k\}_{k\geq 1}\subseteq \mathbb{N}_0^n$ satisfies the property that

(39) \begin{align}y_{k} - y_{k'} = \eta_{n_{k}^{k}, k} - \eta_{n_{k}^{k}, k'} = \sum_{j=k'+1}^{k} \zeta\big(\mathfrak{R}(\gamma_{n_{k}^k, j})\big) \notin \mathbb{N}_0^n\setminus\{0\}\quad \text{for all } k > k' \geq 1.\end{align}

On the other hand, as a result of assumption (38), we have $y_{k} \neq y_{k'}$ for all $k\neq k'$ . Thus, according to Lemma 5, there exists a strictly increasing subsequence of $\{y_k\}_{k\geq 1}$ , which contradicts (39). This proves that $A_{i,k}(x)$ is a finite set for all $i\in \{1,\dots, m_{\mathrm{pro}}\}$ and $1\leq k\leq m_{\mathrm{pro}}$ . Therefore, $A_i(x) = \bigcup_{k=1}^{m_{\mathrm{pro}}} A_{i,k}(x)$ is also finite, and so is $\mathcal{X}_0$ . The proof of statement (B) is complete.

Now it suffices to prove Lemmas 5 and 6. Note that Lemma 5 is elementary, so we only provide the proof of Lemma 6.

Conversely, suppose that (ii) holds. Let $V\subseteq \mathbb{N}_0^q$ be a set such that $\mathcal{H}_{x,V} \,:\!=\, \{x+v(\mathcal{Z}) \,:\, v\in V\}=\mathcal{H}_x$ . By Zorn’s lemma, we can assume that V is minimal, namely for any $v\in V$ , with $V'=V\setminus \{v\}$ one has that $\mathcal{H}_{x,V'}\neq \mathcal{H}_x$ . To prove that $H_x$ is finite, it suffices to show that V is a finite set. Suppose that V is an infinite set. Then there is a sequence $\{v_i\}_{i \geq 1} \subseteq V$ such that $v_i\neq v_j$ for any $i\neq j$ . Referring to Lemma 5, by taking subsequences, we can assume that $\{v_i\}_{i\geq 1}$ is strictly increasing.

For any $\eta\in \mathcal{H}_x$ , we have $\eta \geq -x$ . Thus, $\{v_i(\mathcal{Z})\}_{i\geq 1}\subseteq \mathcal{H}_x$ is bounded from below. Owing to the minimality of V, we have $v_i(\mathcal{Z})\neq v_j(\mathcal{Z})$ whenever $i\neq j$ , and thus by Lemma 5 we assume that $\{v_i(\mathcal{Z})\}_{i\geq 1}$ is strictly increasing. Because $\{v_i\}_{i\geq 1}$ is strictly increasing, it follows that $v_0 \,:\!=\, v_{2}-v_{1} \in \mathbb{N}_0^{q,*}$ and $ v_0(\mathcal{Z})=v_{2}(\mathcal{Z})-v_{1}(\mathcal{Z}) \in \mathcal{H}\cap \mathbb{N}_0^n\setminus\{0\}$ . This contradicts (ii). The proof is complete.