Non-explosivity of Stochastically Modeled Reaction Networks that are Complex Balanced

Abstract

We consider stochastically modeled reaction networks and prove that if a constant solution to the Kolmogorov forward equation decays fast enough relatively to the transition rates, then the model is non-explosive. In particular, complex-balanced reaction networks are non-explosive.

Appendix

Let $E\subset {{\mathbb {Z}}}^d$ and for $x,y\in E$, $x\ne y$, let $q(x,y)\ge 0$ and assume Condition 1 is satisfied. As already discussed in the Introduction, a continuous-time Markov chain on E with transition intensities $\{q(x,y)\}$ is intuitively a stochastic process $(X_t)_{t\ge 0}$ in E such that

$$\begin{aligned} P(X_{t+h}=y|{{\mathcal {F}}}_t^X)=q(X_t,y)h+o(h),\quad t,h\ge 0, \end{aligned}$$

where $\{{{\mathcal {F}}}_t^X\}$ is the filtration generated by $(X_t)_{t\ge 0}$. There are several ways of making this intuition precise.

Definition 5

• (a) The one-dimensional distributions of $(X_t)_{t\ge 0}$ satisfy the forward or master equation if for every $x\in E$

  $$\begin{aligned} \frac{\text {d}}{\text {d}t}P_t(x)=\sum _{y\in E\setminus \{x\}} P_t(y)q(y,x)-\sum _{y\in E\setminus \{x\}} P_t(x)q(x,y), \end{aligned}$$

  (18)

  where the derivative at $t=0$ is interpreted as a right derivative.

• (b) Consider

  $$\begin{aligned} {{\mathcal {D}}}(E)=\{f:E\rightarrow {{\mathbb {R}}}: \#\{x\in E:f(x)\ne 0\}<\infty \} \end{aligned}$$

  and define the linear operator A on ${{\mathcal {D}}}(E)$ by

  $$\begin{aligned} Af(x)=\sum _{y\in E\setminus \{x\}}q(x,y)(f(y)-f(x)). \end{aligned}$$

  Then $(X_t)_{t\ge 0}$ is a solution of the martingale problem for A if

  $$\begin{aligned} f(X_t)-f(X_0)-\int _0^tAf(X_s)\hbox {d}s \end{aligned}$$

  (19)

  is a $\{{{\mathcal {F}}}_t^X\}$-martingale for each $f\in {{\mathcal {D}}}(E)$.

• (c) For $x,y\in E$, $x\ne y$, let $(N^{xy}_t)_{t\ge 0}$ be a Poisson process with intensity q(x, y). The stochastic equation for $(X_t)_{t\ge 0}$ is given by

  $$\begin{aligned} f(X_t)=f(X_0)+\sum _{x,y\in E :x\ne y}\int _0^t(f(y)-f(X_{s-}))\mathbf{1}_{\{X_{s-}=x\}}\hbox {d}N^{xy}_s. \end{aligned}$$

  (20)

Remark 2

Note that by Condition 1, for $f\in {{\mathcal {D}}}(E)$, $\sup _{x\in E}|Af(x)|<\infty$.

Furthermore, we introduce an additional state $\partial$ as done in the main text. Solutions of (18) need to satisfy $P_t(x)\ge 0$, $\sum _{x\in E}P_t(x)\le 1$ and $P_t(\partial )=1-\sum _{x\in E}P_t(x)$. Moreover, for solutions of the martingale problem and the stochastic equation, we extend $f\in {{\mathcal {D}}}(E)$ to $E\cup \{\partial \}$ by defining $f(\partial )=0$ and define $X_t=\partial$, if $X_t\notin E$.

With this understanding of solutions, it follows that any solution of the stochastic equation (20) is a solution of the martingale problem. Indeed, we can write

$$\begin{aligned} f(X_t)-f(X_0)-\int _0^tAf(X_s)\hbox {d}s=\sum _{x,y\in E :x\ne y}\int _0^t(f(y)-f(X_{s-}))\mathbf{1}_{\{X_{s-}=x\}}\hbox {d}\tilde{N}^{xy}_s, \end{aligned}$$

where

$$\begin{aligned} \tilde{N}^{xy}_s = N^{xy}_s-q(x,y)s \end{aligned}$$

is a martingale. Moreover, for any solution of the martingale problem, $P_t(x)=P(X_t=x)$ is a solution of (18): this follows by considering $f=\mathbf{1}_{\{x\}}$ for $x\in E$ and by taking expectation (Anderson and Kurtz 2015). The converse of these observations also holds.

Theorem 4

If $(P_t)_{t\ge 0}$ is a solution of (18), then there exists a solution of the martingale problem such that for $x\in E\cup \partial$, $P(X_t=x)=P_t(x)$. If $(X_t)_{t\ge 0}$ is a solution of the martingale problem, then there exists a solution $(\tilde{X}_t)_{t\ge 0}$ of the stochastic equation such that $(X_t)_{t\ge 0}$ and $(\tilde{X}_t)_{t\ge 0}$ have the same distribution.

Proof

The first statement follows from Corollary 3.2 of Kurtz and Stockbridge (1998). The second statement can be proved by arguments similar to the proof of (17) in Kurtz (2011). $\square$

It follows from Theorem 4 that weak uniqueness for any of the three characterizations implies weak uniqueness for the other two, for a fixed initial distribution. Moreover, the law of a solution to the stochastic equation is uniquely determined up to time $T_{\infty }$, where $T_{\infty }$ is defined as in Definition 1. This observation provides the proof of the next result.

Theorem 5

If for a fixed initial distribution $P_0$ we have $P(T_{\infty }=\infty )=1$, then there exists a unique process $(X_t)_{t\ge 0}$ with initial distribution $P_0$ satisfying any of the three characterization of Definition 5.

We conclude with the following result, which can be read from Ethier and Kurtz (1986, Theorem 9.17 of Chapter 4) and is due to Echeverría (1982).

Theorem 6

If $\pi$ is a constant solution to the forward Kolmogorov equation, namely if

$$\begin{aligned} \sum _{y\in E\setminus \{x\}} \pi (x)q(x,y)=\sum _{y\in E\setminus \{x\}} \pi (y)q(y,x), \end{aligned}$$

then there exists a solution of the martingale problem which is a stationary process with stationary distribution $\pi$.

Note that under the hypotheses of Theorem 6, Theorem 4 implies the existence of a solution to the martingale problem with $P_t=\pi$ for all $t\ge 0$, and a stationary solution to the stochastic equation (20).