The subtle business of model reduction for stochastic chemical kinetics

Abstract

This paper addresses the problem of simplifying chemical reaction networks by adroitly reducing the number of reaction channels and chemical species. The analysis adopts a discrete-stochastic point of view and focuses on the model reaction set S₁⇌S₂→S₃, whose simplicity allows all the mathematics to be done exactly. The advantages and disadvantages of replacing this reaction set with a single S₃-producing reaction are analyzed quantitatively using novel criteria for measuring simulation accuracy and simulation efficiency. It is shown that in all cases in which such a model reduction can be accomplished accurately and with a significant gain in simulation efficiency, a procedure called the slow-scale stochastic simulation algorithm provides a robust and theoretically transparent way of implementing the reduction.

INTRODUCTION

Biochemical systems typically contain networks of many chemical reaction channels involving many molecular species. This circumstance encourages attempts to construct simpler but equivalent “reduced” reaction networks. A well known example of such a reduction is the Michaelis–Menten abridgment of the enzyme-substrate reactions,^{1, 2} which has been the subject of many refinements over the years^{3, 4} and which continues to play an important role in biochemistry today.⁵

Typically, an abridgment replaces the given reaction network with a network that involves fewer reaction channels and fewer chemical species. Perhaps the simplest reaction set that presents the opportunity for doing that, one that has several features in common with the enzyme-substrate reactions but is mathematically more tractable, is

$$S_1\underset{c_2}{\overset{c_1}{\rightleftarrows }}{S}_2\stackrel{c_3}{\to }{S}_3,$$ (1)

where we assume that c₁ and c₃ are both nonzero. It is tempting to cut to the chase and replace this set of three three-species reactions with one two-species reaction, such as

$$S_1\stackrel{c}{\to }{S}_3,$$ (2)

where the reaction constant c is given some “suitable” value. Our focus in this paper will be to determine the conditions under which it is advisable to make such a replacement and to show how the replacement should be implemented. Of course, if a modeler deliberately chooses to model the production of S₃ molecules from S₁ molecules by reaction 2 instead of by reactions 1, then this issue is moot. But we are assuming here that the modeler believes that reactions 1 really describe what is going on physically, and therefore wants any abridgement of Eq. 1, such as reaction 2, to mimic the salient effects of reactions 1 with reasonable accuracy. A modeler might choose to use reaction 2 instead of reactions 1 because the values of the rate constants c₁, c₂, and c₃ in Eq. 1 are not all known. But choosing an appropriate value for c in Eq. 2 inevitably makes assumptions about those three rate constants; thus, it might be better to use Eq. 1 with those assumptions made explicitly and openly, since that would not only preserve the topology of reactions 1 but also make it easy to incorporate later new information about the unknown rate constants.

The most obvious advantage in replacing reactions 1 with a single S₃-producing reaction like Eq. 2 is the reduction in the numbers of reactions and species that we have to contend with. Another advantage might be speeding up the numerical simulation of reactions 1. By simulation we mean here stochastic simulation, since stochasticity often plays a role in cellular systems. But there are two potential drawbacks to such a reduction: First, as will be elaborated on below, this is always an approximation, since it is simply not possible for any single reaction to exactly mimic reactions 1 in all respects. Second, if we want to have the option of embedding reactions 1 in a larger network of reactions, some of which may involve species that get removed in the model reduction, as S₂ has in Eq. 2, then it may be impossible to simulate those other reactions when using the reduced model.

In this paper, we will address these matters in detail for reaction set 1. We will begin by presenting some novel perspectives on simulation efficiency and simulation accuracy. We will show that these new perspectives imply that a one-reaction abridgment of Eq. 1 will be advisable in some circumstances, but not in others. We will then show that, in all cases where a model reduction can be done accurately and with a significant gain in stochastic simulation efficiency, implementing the reduction will be more involved than just swapping reactions 1 for reaction 2. Finally, we will establish a new perspective on the results of two recent papers, namely, the slow-scale stochastic simulation algorithm (ssSSA) of Cao et al.,⁶ and the stochastic quasi-steady-state approximation singular perturbation analysis (sQSPA) of Mastny et al.⁷

The reaction network 1 we are focusing on here is obviously very simple. But that simplicity allows all the mathematics, which even in this case is nontrivial, to be done exactly, and thus all issues to be explored thoroughly. We believe that a clarification of these issues in the context of reactions 1 can lead to a better understanding of how these issues play out in more complicated reaction networks.

QUANTIFYING THE GAIN IN SIMULATION EFFICIENCY

Well-stirred chemical systems with discrete molecular populations and stochastic reaction dynamics can be exactly simulated by the well-known SSA.⁸ The only downside is that the SSA is usually quite slow: The SSA simulates every reaction event, so the time required to make a SSA simulation run is proportional to the number of reaction events that occur.

Replacing reactions 1 with a single S₃-producing reaction, such as reaction 2, would evidently have the consequence that a new S₃ molecule would be produced by each reaction event. In contrast, the creation of a new S₃ molecule via reactions 1 usually requires more than one-reaction event. Therefore, a fair measure of the gain in simulation efficiency realized by such an abridgment would be the average number of reaction events that are needed by reactions 1 to produce one S₃ molecule. That number turns out to be surprisingly easy to compute.

Suppose a molecule starts out as an S₁ molecule or, as we shall say, starts in state S₁. On each visit to state S₂, the molecule has probability c₃∕(c₂+c₃) of going on to state S₃ and probability c₂∕(c₂+c₃) of going back to S₁. So in n visits to state S₂, the molecule would go on to state S₃ an average of nc₃∕(c₂+c₃) times; thus, in order to get an average of one visit to state S₃, the molecule needs to visit state S₂ a total of n₁ times, where n₁c₃∕(c₂+c₃)=1. It follows that n₁=(c₂+c₃)∕c₃. But each of these n₁ visits to state S₂ requires exactly two reaction events, namely, the reaction R₁ that brings the molecule to state S₂, and either reaction R₂ or R₃ that takes the molecule away. Therefore, the average number of reaction events required for the molecule to get from state S₁ to state S₃ via the reaction set 1 is 2n₁=2(c₂+c₃)∕c₃. If the molecule had started in state S₂ instead of state S₁, it would be exactly one-reaction event closer to state S₃, so the average number of reaction events for a molecule in state S₂ to reach state S₃ would be 2n₁−1, a difference that is not significant for our purposes here. We thus conclude that the gain in simulation efficiency achieved by replacing reactions 1 with a single S₃-producing reaction is approximately

$$G=2\left(\frac{c_2+c_3}{c_3}\right).$$ (3)

That is to say, simulating the production of S₃ molecules via some single reaction like Eq. 2 will be approximately G times faster than simulating via reactions 1. The overall gain in simulation efficiency would actually be less than this if reactions 1 were embedded in a larger reaction network that is also simulated.

Two things are noteworthy about the result 3. First, the gain depends on c₂ and c₃ but not on c₁. Second, we will have

$$G⪢1\text{if}\text{and}\text{only}\text{if}{c}_2⪢c_3.$$ (4)

The assertion of Eq. 3 that G=2 when c₂=0 is an obvious result; the assertion that G=4 when c₂=c₃ is perhaps less obvious. But both of those efficiency gains are modest compared to the gains achievable when c₂ is one or more orders of magnitude larger than c₃. It follows that one should carefully examine one’s goals in reducing reactions 1 when the strong inequality c₂⪢c₃ is not satisfied.

ACCURACY: THE IMPORTANCE OF BEING EXPONENTIAL

Let T(x₁,x₂) be the time to the next firing of reaction R₃ in Eq. 1 when there are x₁S₁ molecules and x₂S₂ molecules. Obviously, T(x₁,x₂) will be some kind of random variable. In this section, we will show that a necessary condition for reactions 1 to be accurately replaceable by a single S₃-producing reaction, for example, reaction 2, is that T(x₁,x₂) be approximately an exponential random variable. In Sec. 4, we will calculate the exact probability density function (pdf) of T(x₁,x₂), with the aim of finding out under what conditions this exponentiality requirement is satisfied.

Stochastic chemical kinetics, which encompasses both the SSA and the chemical master equation, assumes that the dynamics of reaction 2 are described by⁸

$$cdt=\text{the}\text{probability}\text{that}\text{a}\text{given}{S}_1\text{molecule}\text{in}\text{reaction}\left(2\right)\text{will}\text{become}\text{an}{S}_3\text{molecule}\text{in}\text{the}\text{next}\text{infinitesimal}\text{time}\text{interval}dt.$$ (5)

Verification of this critical condition is awkward to accomplish directly. A more convenient but completely equivalent condition is afforded by the following theorem, which is proved in Appendix A.

Theorem 1. Condition 5 is equivalent to saying that the time required for a given S₁ molecule to become an S₃ molecule via reaction 2 is an exponential random variable with mean 1∕c.

We recall that the exponential random variable with mean c⁻¹ is defined to be the random variable on 0≤t<∞ which has pdf c exp(−ct) and cumulative distribution function (cdf) 1−exp(−ct). It follows from Theorem 1 that reaction 2 cannot be simulated using the SSA, nor analyzed via the chemical master equation, unless the time required for a given S₁ molecule to become an S₃ molecule via reaction 2 is, at least approximately, an exponential random variable. This result has motivated some recent molecular dynamics studies of excluded volume effects in simple well-stirred one- and two-dimensional chemical systems.⁹

It might be thought that this exponential requirement, being stochastic, would not apply if we were content to describe reaction 2 in terms of traditional deterministic chemical kinetics. However, that is not true. To see why, recall that the traditional reaction rate equation for Eq. 2, written in terms of the number X_i(t) of S_i molecules at time t, and assuming X₃(0)=0, is

$$\frac{d{X}_3\left(t\right)}{dt}=c{X}_1\left(t\right)=c\left(X_1\left(0\right)-X_3\left(t\right)\right).$$ (6a)

The solution to this equation is

$$X_3\left(t\right)=X_1\left(0\right)\left(1-e^{-ct}\right).$$ (6b)

Consistency requires that X₃(t) in Eq. 6b should accurately describe the behavior of the average number of S₃ molecules in the stochastic formulation (note that we are dealing here with a linear first-order reaction). Let f(t) be the cdf for the time-to-reaction τ of any particular S₁ molecule; i.e., f(t) is the probability that τ≤t, and hence the probability that an S₁ molecule will have become an S₃ molecule by time t. Then since the S₁ molecules react independently of each other, the probability that exactly n of them will have become an S₃ molecule by time t is

$$\frac{X_1\left(0\right)!}{n!\left[X_1\left(0\right)-n\right]!}{\left[f\left(t\right)\right]}^n{\left[1-f\left(t\right)\right]}^{X_1\left(0\right)-n}.$$

This implies that the number of S₃ molecules created in time t is the binomial random variable B(f(t),X₁(0)). Since the mean of that random variable is f(t)X₁(0), then agreement with Eq. 6b requires that

$$X_1\left(0\right)\left(1-e^{-ct}\right)=f\left(t\right)X_1\left(0\right),$$

or

$$f\left(t\right)=1-e^{-ct}.$$ (7)

But this is the cdf of the exponential random variable with mean 1∕c. Thus we conclude that the time τ to reaction 2 for each individual S₁ molecule must be exponentially distributed in order for the deterministic rate equations 6 to be valid.

For an example of a nonexponential τ-distribution that is clearly inconsistent with Eq. 6b, suppose that τ were uniformly distributed in the interval [c⁻¹−ε,c⁻¹+ε) for some ε<c⁻¹. Then the mean time-to-reaction for each S₁ molecule would indeed be c⁻¹. But with this lifetime distribution, the number of S₃ molecules would obviously stay at zero until time c⁻¹−ε, and then rise roughly linearly to X₁(0) in a time 2ε. This is clearly not the behavior predicted by formula 6b.

The relevance of the foregoing result to the problem of replacing reactions 1 with some single S₃-producing reaction such as Eq. 2 can be understood as follows. If there are x₁S₁ molecules in the system, then Eq. 5 and the addition law of probability imply that the probability that reaction 2 will fire in the next dt is x₁×cdt=cx₁dt. More generally, any single reaction that produces one S₃ molecule will have the property that, for some state-dependent function a which is called the reaction’s propensity function, adt gives the probability that the reaction will fire in the next dt. This implies, by the same reasoning that led to Theorem 1, that the time to the next firing of that reaction will be exponentially distributed with mean a⁻¹. Therefore, if this reaction is to be a surrogate for reactions 1—a replacement that approximately replicates the way in which reactions 1 produce S₃ molecules—then the time T(x₁,x₂) to the next firing of reaction R₃ in Eq. 1 must be, at least approximately, exponentially distributed. If that turns out to be so, then an approximating surrogate reaction for Eq. 1 should exist. But if T(x₁,x₂) is found to be clearly nonexponential, then we must conclude that reactions 1cannot be accurately replaced by a single S₃-producing reaction.

DISTRIBUTION OF THE TIME TO THE NEXT R₃ REACTION

In Appendix B, we prove that the pdf P(t;x₁,x₂) of T(x₁,x₂), the time to the next R₃ reaction in Eq. 1 when there are x₁S₁ molecules and x₂S₂ molecules, is given by Eq. B18. In that formula, the four functions p(β,t|α,0) for α, β=1,2 are given explicitly by Eqs. B13, B14, and

$${\lambda }_{\pm }\equiv \frac{1}{2}\left[\left(c_1+c_2+c_3\right)\pm \sqrt{{\left(c_1+c_2+c_3\right)}^2-4c_1{c}_3}\right].$$ (8)

That P(t;x₁,x₂) in Eq. B18 is not generally exponential can be seen by noting that its form for T(1,0), the time for a single S₁ molecule to become an S₃ molecule via reactions 1, turns out to be¹⁰

$$P\left(t;1,0\right)=\frac{c_1{c}_3}{\left({\lambda }_{+}-{\lambda }_{-}\right)}\left[e^{-{\lambda }_{-}t}-e^{-{\lambda }_{+}t}\right].$$ (9)

This pdf is obviously not exponential; indeed, it vanishes at t=0, whereas the pdf of any exponential random variable has its maximum at t=0. It also follows from Eq. B18 that the pdf of the time for a single S₂ molecule to become an S₃ molecule via reactions 1 is

$$P\left(t;0,1\right)=\frac{c_3}{\left({\lambda }_{+}-{\lambda }_{-}\right)}\left[\left(c_1-{\lambda }_{-}\right)e^{-{\lambda }_{-}t}+\left({\lambda }_{+}-c_1\right)e^{-{\lambda }_{+}t}\right].$$ (10)

Although this pdf achieves its maximum at t=0, it still does not generally have a simple exponential form. Plots of the two pdfs 9, 10 for c₁=c₃=1 and c₂=0.1 are shown in Fig. 1 on a semilog scale, where a truly exponential pdf would appear as a downsloping straight line. The nonexponential character of P(t;1,0) is obvious; that of P(t;0,1) is evinced by a gradual change in slope around t=2.

Figure 1.

Semilog plots of P(t;x₁,x₂) for c₁=c₃=1 and c₂=0.1 for two cases: The solid curve is for x₁=1 and x₂=0, from Eq. 9. The dashed curve is for x₁=0 and x₂=1, from Eq. 10. Neither pdf has the straight-line form of an exponential pdf (there is a gradual change in slope in the dashed curve around t=2). The figure also shows that P(t;x₁,x₂) in this case depends on x₁ and x₂ individually, and not just on their sum.

The consequences of the nonexponential form of the pdf P(t;1,0) in Eq. 9 are illustrated in Fig. 2. The jagged solid curve shows a single X₃(t) trajectory obtained in an exact SSA run of reactions 1, using the parameter values c₁=c₃=1 and c₂=0.1 and the initial conditions X₁(0)=300 and X₂(0)=X₃(0)=0. The dashed curve shows the average of 10 000 such trajectories. It can be shown that the mean of the pdf in Eq. 9 is (c₁+c₂+c₃)∕c₁c₃, which in this case equals 2.1; i.e., the average time for an S₁ molecule to become an S₃ molecule via reactions 1 in this case is 2.1. If we made the usual deterministic assumption that formula 6a applies with c=(2.1)⁻¹, then Eq. 6b would give the trajectory shown as the dotted curve in Fig. 2. The mismatch between that curve and the dashed curve illustrates the inappropriateness of replacing reactions 1 with reaction 2 when the time between R₃ reactions is not exponentially distributed.

Figure 2.

The solid curve shows a single X₃(t) trajectory obtained in a SSA run of reactions 1 with c₁=c₃=1, c₂=0.1, X₁(0)=300, and X₂(0)=X₃(0)=0. The dashed curve shows the average of 10 000 such trajectories. The dotted curve plots the function 6b with c=(2.1)⁻¹, which corresponds to the same mean S₁→S₃ conversion time. The mismatch between the dashed curve and the dotted curve indicates the error that results from replacing reactions 1 with reaction 2 in this nonexponential case.

The additional revelation in Fig. 1 that P(t;1,0) is not the same curve as P(t;0,1) illustrates another potential problem for model reduction: While any acceptable single-reaction abridgment of reactions 1 will accurately replicate the time evolution of the S₃ population, and hence also the time evolution of the totalS₁ and S₂ population, the abridgment might not accurately replicate the time evolutions of the S₁ and S₂ populations separately; e.g., reaction 2 gives us no indication of the S₂ population. Therefore, if P(t;x₁,x₂) depends on x₁ and x₂individually, and not just on their sum

$$x_1+x_2\equiv x_{12},$$ (11)

as Fig. 1 shows happens when c₁=c₃=1, c₂=0.1, and x₁₂=1, then the lack of information about the individual values of x₁ and x₂ could make using the abridged reaction in a real simulation impossible, even if P(t;x₁,x₂) were exponential.

A close inspection of Eqs. B18, B13, B14 reveals that P(t;x₁,x₂) is in general a polynomial in e^−λ₋t and e^−λ₊t. From Eq. 8, it can be shown that when both c₁ and c₃ are positive, as we are assuming here, then 0<λ₋≤λ₊. Therefore, a necessary condition for P(t;x₁,x₂) to be approximately exponential is for the rate constants to be such that

$${\lambda }_{-}⪡{\lambda }_{+},$$ (12)

because then, all terms involving e^−λ₊t will be negligibly small for t⪢1∕λ₊, and we can hope that the t-dependence for t⪢1∕λ₊ will be given by some power of e^−λ₋t.

When 0<λ₋≤λ₊, the extreme inequality 12 will be satisfied if and only if

$$\frac{{\lambda }_{+}{\lambda }_{-}}{{\left({\lambda }_{+}+{\lambda }_{-}\right)}^2}⪡1.$$ (13)

Since Eq. 8 implies that λ₊λ₋=c₁c₃ and λ₊+λ₋=c₁+c₂+c₃, then condition 13 is the same as

$$\frac{c_1{c}_3}{{\left(c_1+c_2+c_3\right)}^2}\equiv \left(\frac{c_1}{c_1+c_2+c_3}\right)\left(\frac{c_3}{c_1+c_2+c_3}\right)⪡1.$$ (14)

Since each factor in the middle of Eq. 14 is less than 1, then the right inequality in Eq. 14 can be satisfied if and only if at least one of those two factors is ⪡1. The first factor will be ⪡1 if and only if c₂+c₃⪢c₁, which is the same as either c₂⪢c₁ or c₃⪢c₁. And the second factor will be ⪡1 if and only if c₁+c₂⪢c₃, which is the same as either c₁⪢c₃ or c₂⪢c₃. Thus we conclude that condition 12 will be satisfied if and only if at least one of the following four conditions holds:

$$c_2⪢c_1,$$ (15a)

$$c_3⪢c_1,$$ (15b)

$$c_1⪢c_3,$$ (15c)

$$c_2⪢c_3.$$ (15d)

Note that these four conditions are not mutually exclusive; e.g., the condition c₂⪢c₃⪢c₁ satisfies both conditions 15d, 15b. Nor are these conditions collectively exhaustive; e.g., the condition c₁=c₂=c₃ satisfies none of conditions 15. But satisfaction of at least one of conditions 15 is necessary, and as we shall see shortly sufficient, for P(t;x₁,x₂) to be exponential.

Assume now that at least one of conditions 15 is satisfied. Then the strong inequality 14 will also be satisfied, so we will have from Eq. 8 that

$${\lambda }_{\pm }=\frac{\left(c_1+c_2+c_3\right)}{2}\left[1\pm \sqrt{1-\frac{4c_1{c}_3}{{\left(c_1+c_2+c_3\right)}^2}}\right]\approx \frac{\left(c_1+c_2+c_3\right)}{2}\left[1\pm \left(1-\frac{1}{2}\frac{4c_1{c}_3}{{\left(c_1+c_2+c_3\right)}^2}\right)\right],$$

whence

$${\lambda }_{+}\approx c_1+c_2+c_3,{\lambda }_{-}\approx \frac{c_1{c}_3}{c_1+c_2+c_3}.$$ (16)

When Eq. 16 is substituted into Eqs. B13, B14, and the results are substituted into Eq. B18, we obtain since λ₋⪡λ₊,

$$P\left(t;x_1,x_2\right)\approx \frac{c_1{c}_3{x}_1}{c_1+c_2+c_3}{e}^{-{\lambda }_{-}\left(x_1+x_2\right)t}{\left(\frac{c_1+c_2}{c_1+c_2+c_3}\right)}^{x_2}+\frac{c_1{c}_3{x}_2}{c_1+c_2+c_3}{e}^{-{\lambda }_{-}\left(x_1+x_2\right)t}{\left(\frac{c_1+c_2}{c_1+c_2+c_3}\right)}^{x_2-1}\left(t⪢1∕{\lambda }_{+}\right).$$ (17)

The restriction on t here ensures that all terms involving e^−λ₊t have become negligibly small. Again, this approximation assumes that at least one of conditions 15 is satisfied.

Now let us examine 17 for the individual conditions 15. First, if either condition 15a or 15b holds, so that c₂+c₃⪢c₁, then Eq. 16 gives λ₊≈c₂+c₃ and λ₋≈c₁c₃∕(c₂+c₃). Equation 17 simplifies slightly, in that c₁ gets dropped from all denominators. Further simplification of Eq. 17 follows from the observation that the conditionc₂+c₃⪢c₁ implies that reaction R₁, which creates S₂ molecules, will occur much less frequently than reactions R₂ and R₃, which destroy S₂ molecules. The S₂ population will thus usually be very small, and a reasonable approximation would be to set x₂≈0, and hence x₁≈x₁₂. With those approximations, the second term in Eq. 17 is effectively removed and the equation finally reduces to

$$P\left(t;x_1,x_2\right)\approx \left(\frac{c_1{c}_3{x}_{12}}{c_2+c_3}\right)e^{-\left(c_1{c}_3{x}_{12}∕(c_2+c_3)\right)t}\left(t⪢{\left(c_2+c_3\right)}^{-1},\text{for}\left(15\text{a}\right)\text{or}\left(15\text{b}\right)\right).$$ (18)

Since this pdf has the exponential form, an accurate single-reaction abridgment should be possible. And the decay constant in Eq. 18 will be the propensity function of the surrogate reaction. The fact that this decay constant depends on x₁ and x₂ only through their sum x₁₂ suggests that the reduced model should be amenable to simulation.

Now suppose that either condition 15c or condition 15d holds. Then c₁+c₂⪢c₃, and Eq. 16 gives λ₊≈c₁+c₂ and λ₋≈c₁c₃∕(c₁+c₂). The relation c₁+c₂⪢c₃ implies that c₃ can be dropped from all denominators in Eq. 17. That equation then reduces to

$$P\left(t;x_1,x_2\right)\approx \left(\frac{c_1{c}_3{x}_{12}}{c_1+c_2}\right)e^{-\left(c_1{c}_3{x}_{12}∕(c_1+c_2)\right)t}\left(t⪢{\left(c_1+c_2\right)}^{-1},\text{for}\left(15\text{c}\right)\text{\hspace{0.17em}or}\left(15\text{d}\right)\right).$$ (19)

Again this pdf has the exponential form, with the decay constant depending on x₁ and x₂ only through their sum x₁₂. Therefore, replacing reactions 1 with a single S₃-producing reaction, whose propensity function is the decay constant in Eq. 19, should be feasible. Note that Eq. 19 does not assume, as Eq. 18 does, that x₂≈0.

IMPLEMENTING THE REDUCED MODEL

We showed in Sec. 4 that an accurate replacement of reactions 1 by a single S₃-producing reaction should be possible under conditions 15a, 15b, 15c, 15d. More specifically, the result in Eq. 18 shows that under conditions 15a, 15b the S₃-producing reaction should have the propensity function

$$a=\frac{c_1{c}_3{x}_{12}}{c_2+c_3}\left(\text{for}{c}_2⪢c_1\text{or}{c}_3⪢c_1\right),$$ (20)

with the understanding that x₂≈0, and that we are not interested in phenomena occurring on timescales of order (c₂+c₃)⁻¹ or smaller. And the result in Eq. 19 shows that under conditions 15c, 15d the S₃-producing reaction should have propensity function

$$a=\frac{c_1{c}_3{x}_{12}}{c_1+c_2}\left(\text{for}{c}_1⪢c_3\text{or}{c}_2⪢c_3\right),$$ (21)

with no restrictions on x₂, but with the understanding that we are not interested in phenomena occurring on timescales of order (c₁+c₂)⁻¹ or smaller. But exactly how should the replacement reaction be framed in these cases?

First let us dispose of two “obvious” cases in whichc₂≈0, and reaction R₂ practically never fires. The first case couples that condition with condition 15b, c₃⪢c₁: In a short time of order $c_3^{-1}$, all S₂ molecules become S₃ molecules via reaction R₃; thereafter, the S₁ molecules convert to S₃ molecules essentially via reaction 2 with the approximate propensity function a=c₁x₁, since each R₁ firing will practically always be followed immediately by an R₃ firing. This result also follows by setting c₂≈0 in Eq. 20, and remembering that x₁₂≈x₁ since Eq. 20 assumes thatx₂≈0. The other obvious case couples condition c₂≈0 with condition 15c, c₁⪢c₃: In a short time of order $c_1^{-1}$, all S₁ molecules become S₂ molecules via reaction R₁; thereafter, the S₂ molecules convert to S₃ molecules via reaction R₃ with the approximate propensity function a=c₃x₂=c₃x₁₂. This result also follows by putting c₂≈0 in Eq. 21. In both of these obvious cases, the simulation speedup factor realized by the abridgment is about 2, which is rather modest.

A more interesting case arises by conjoining conditions 15a, 15b, and requiring that bothc₂ and c₃ be ⪢c₁, a condition that we will write c₂, c₃⪢c₁. This condition has been analyzed in detail by Mastny et al.⁷ using a reduction method which they call the stochastic sQSPA. The conclusion of their analysis (see Ref. 7, Table II) expressed in our notation here is that reactions 1 can be replaced by reaction 2 with propensity function a=c₁c₂x₁∕(c₂+c₃). This is the same as our result in Eq. 20, since the assumption c₂, c₃⪢c₁ in Ref. 7 implies x₂≈0, and hence x₁≈x₁₂. Our first-passage-time analysis thus confirms the result of Mastny et al.,⁷ including the proviso which is implicit in their derivation that this approximation is valid only on timescales larger than (c₂+c₃)⁻¹. The resulting gain in simulation efficiency 3 will be large or small according to whether c₂∕c₃ is large or small. But we note that the condition x₂≈0 would appear to pose a problem if we wanted to embed the abridged reaction in a network of other reactions, some of which create or consume S₂ molecules.

Another interesting case is Eq. 15d, c₂⪢c₃. We showed in Sec. 2 that this is the condition for a truly substantial speedup in stochastic simulation. But it turns out that simply replacing reactions 1 with reaction 2 using the propensity function 21 has some limitations. To illustrate, we have used the exact SSA to simulate each of reactions 1, 2 for parameter values

$$c_1=3,c_2=2,c_3={10}^{-4}$$ (22a)

and the initial conditions

$$X_1\left(0\right)=300,X_2\left(0\right)=X_3\left(0\right)=0.$$ (22b)

Figure 3 shows the results of the SSA simulation of reactions 1. In this figure, the species populations have been plotted out immediately after the occurrence of each R₃ reaction, so only 300 points get plotted in the conversion of the 300 S₁ molecules. But approximately 1.2×10⁷ reaction events had to be simulated in order to get those 300 R₃ reactions, so there were on average 4×10⁴R₁ and R₂ reaction events between successive R₃ reaction events, a figure that agrees with formula 3.

Figure 3.

A “true” picture of reactions 1 for the parameter settings 22 is provided by this SSA run of those reactions. Here the species populations have been plotted out only after each R₃ event. Since the S₃ population remains constant between successive R₃ reactions, this plotting strategy reveals the full trajectory of X₃(t). But of course, the populations of species S₁ and S₂ are not constant between successive R₃ reactions.

Figure 4 shows the results of the SSA simulation of the surrogate reaction 2 using the propensity function in Eq. 21 and the same parameter values 22 as used in Fig. 3. Here the populations have been plotted out after each simulated reaction. Since only 300 reaction events were simulated in this run, compared to the 1.2×10⁷ events that were simulated to produce Fig. 3, the gain in simulation efficiency achieved by using the surrogate reaction 2 is truly large. Comparing Fig. 4 with Fig. 3 shows that the surrogate reaction 2 does give a satisfactory representation of the X₃(t) trajectory, just as we expect on the basis of our analysis. But reaction 2 evidently does not provide a satisfactory representation of the X₁(t) trajectory; furthermore, it gives us no information at all about the X₂(t) trajectory. The explanation for these shortcomings is not hard to fathom: When we stop simulating reactions R₁ and R₂, as we do when we substitute reaction 2 for reactions 1, we lose the ability to accurately track the populations of species S₁ and S₂.

Figure 4.

A SSA simulation of the surrogate reaction 2 with propensity function 21, using the same settings 22 as in Fig. 3. Only 300 reaction events were simulated here, as compared to 1.2×10⁷ reaction events in Fig. 3, so the gain in computational speed over reactions 1 is truly enormous. The X₃(t) trajectory has been accurately rendered. But the X₁(t) trajectory has not, and the X₂(t) trajectory has been completely lost.

If we were interested in only X₃(t), and if reactions 1 were the only reactions in the system that involve species S₁ and S₂, then we might be satisfied with this state of affairs. But we are often concerned with situations in which reactions 1 take place concurrently with other reactions, some of which have one or both of species S₁ and S₂ as reactants. With no reliable information about the instantaneous populations of S₁ and S₂ when using reaction 2, how are we to evaluate the propensity functions of those other reactions in order to simulate their firings along with the firings of reaction 2? Evidently, simply replacing reactions 1 with reaction 2 when c₂⪢c₃ will not be satisfactory if there are other reactions in the system that have S₁ and S₂ as reactants, or if we want to see how species S₁ and S₂ behave on the timescale of reaction R₃.

THE SLOW-SCALE SSA: A ROBUST RECIPE FOR CONDITION 15d

We will show in this section that, under condition 15d,

$$c_2⪢c_3,$$ (23)

replacing reactions 1 with a single S₃-producing reaction can be accurately and robustly accomplished using a procedure called the ssSSA. Designed more generally for “stiff” stochastic systems (systems with a wide separation of timescales with the fastest mode being stable), the ssSSA was introduced in Ref. 6 by some of the present authors, and is basically a refinement of ideas introduced earlier by Haseltine and Rawlings¹¹ and Rao and Arkin.¹² Instead of replacing reactions 1 with a single new reaction like Eq. 2, the ssSSA eliminates reactions R₁ and R₂, and then uses a modified propensity function for reaction R₃.

We should note that condition 23 differs from, and thus corrects, the condition advertised in Ref. 6 for applying the ssSSA to reactions 1.¹³ Also, as will be explained in the next paragraph, condition 23 does not need to be supplemented by c₁⪢c₃.

When condition 23 is satisfied, an S₂ molecule is much more likely to become an S₁ molecule than an S₃ molecule. Thus, successive occurrences of reaction R₃ will usually be separated by very many occurrences of reactions R₁ and R₂; indeed, as we showed in Sec. 2, there will be on average 2(c₂+c₃)∕c₃R₁ and R₂ reactions occurring between successive R₃ reactions, and that number will be ⪢1 when condition 23 holds. Since R₁ and R₂ will be firing much more frequently than R₃, we call R₁ and R₂ “fast reactions,” and R₃ a “slow reaction.” Notice that the designation of R₁ as a fast reaction under condition 23 is justified regardless of the size of c₁>0, because between two successive R₃ reactions, there will inevitably be as many R₁ firings as R₂ firings. And it is the averagefrequency of firing, not the size of the reaction rate constant, that determines whether a reaction is “fast” or “slow” for the ssSSA. Species S₁ and S₂ are then designated as “fast species” because their populations get changed by a fast reaction, and S₃ is called a “slow species” because its population does not.

The fast species populations evolving under only the fast reactions, i.e., $S_1\underset{c_2}{\overset{c_1}{\rightleftarrows }}{S}_2$, constitute what is called the virtual fast process. We denote it by $\left({\widehat{X}}_1\left(t\right),{\widehat{X}}_2\left(t\right)\right)$, using the hat to distinguish it from the real fast process (X₁(t),X₂(t)) which evolves under all three reactions in Eq. 1. For the virtual fast process (but not for the real fast process), we have the conservation relation

$${\widehat{X}}_1\left(t\right)+{\widehat{X}}_2\left(t\right)=x_{12}\left(\text{a}\text{constant}\right);$$ (24)

therefore, the virtual fast process has only one independent variable. We choose it to be ${\widehat{X}}_2\left(t\right)$, and then take ${\widehat{X}}_1\left(t\right)=x_{12}-{\widehat{X}}_2\left(t\right)$. The process ${\widehat{X}}_2\left(t\right)$ thus evolves according to the propensity functions

$${\widehat{a}}_1\left(x_2\right)=c_1\left(x_{12}-x_2\right),{\widehat{a}}_2\left(x_2\right)=c_2{x}_2,$$

with ${\widehat{X}}_2\left(t\right)$ increasing by 1 each time R₁ fires, and decreasing by 1 each time R₂ fires. This simple stochastic process has been well studied.¹⁴ It can be shown that its t→∞ limit ${\widehat{X}}_2\left(\infty \right)$ is the binomial random variable with parameters c₁∕(c₁+c₂) and x₁₂:

$${\widehat{X}}_2\left(\infty \right)=\mathcal{B}\left(c_1∕\left(c_1+c_2\right),x_{12}\right).$$ (25)

Since B(p,N) has mean Np and variance Np(1−p), then

$$⟨{\widehat{X}}_2\left(\infty \right)⟩=\frac{c_1{x}_{12}}{c_1+c_2}\text{and}\text{var}\left\{{\widehat{X}}_2\left(\infty \right)\right\}=\frac{c_1{c}_2{x}_{12}}{{\left(c_1+c_2\right)}^2}.$$ (26a)

It then follows from Eq. 24 (or by symmetry) that

$$⟨{\widehat{X}}_1\left(\infty \right)⟩=\frac{c_2{x}_{12}}{c_1+c_2}\text{and}\text{var}\left\{{\widehat{X}}_1\left(\infty \right)\right\}=\frac{c_1{c}_2{x}_{12}}{{\left(c_1+c_2\right)}^2}.$$ (26b)

Notice that the asymptotic distribution of the virtual fast process depends on the current state (x₁,x₂,x₃)≡x only through the quantity x₁+x₂=x₁₂. That these few facts about ${\widehat{X}}_2\left(t\right)$ are all that is needed to construct a computationally viable abridgment of reactions 1 when c₂⪢c₃ is a consequence of the following theorem.

Theorem 2. Given condition 23, let the system be in state (x₁,x₂,x₃)≡x at time t. Then for any δt that is large compared to the expected time to the next R₁ or R₂ reaction, but small compared to the expected time to the next R₃ reaction, the probability that reaction R₃ will fire in [t,t+δt) is approximately ${\overline{a}}_3\left(\mathbf{x}\right)\delta t$, where

$${\overline{a}}_3\left(\mathbf{x}\right)\equiv c_3⟨{\widehat{X}}_2\left(\infty \right)⟩.$$ (27)

Furthermore, ${\widehat{X}}_2\left(\infty \right)$ and $x_{12}-{\widehat{X}}_2\left(\infty \right)$ provide good estimates of the populations of species S₂ and S₁ at any time aftert+δt but before the next R₃ reaction occurs.

This theorem is proved in Appendix C. It says, first of all, that ${\overline{a}}_3\left(\mathbf{x}\right)$ as defined in Eq. 27 is the “effective propensity function” of reaction R₃ on the timescale of that (slow) reaction. This is so because the defining attribute of a propensity function is that its product with an “effectively infinitesimal” time span gives the probability that the reaction will occur in that time span. With Eq. 26a, Eq. 27 takes the explicit form

$${\overline{a}}_3\left(\mathbf{x}\right)=\frac{c_3{c}_1{x}_{12}}{c_1+c_2}.$$ (28)

Note that this is the same as the propensity function 21 that our first-passage-time analysis gave for condition 23. Theorem 2 also tells us that the S₂ and S₁ populations at any time greater than δt after the last R₃ reaction can be estimated by drawing a sample x₂ of the random variable ${\widehat{X}}_2\left(\infty \right)$ in Eq. 25, and then taking X₂=x₂ and X₁=x₁₂−x₂.

The critical assumption used in proving Theorem 2 (see Appendix C) is that between successive firings of reaction R₃ there will typically be many firings of reactions R₁ and R₂. We showed in Sec. 2 that this will always be so if condition 23 holds. To see that the result 28 is consistent with this fact, we reason as follows: Since ${\overline{a}}_3\left(\mathbf{x}\right)\delta t$ is (approximately) the probability that R₃ will fire in the next δt, then the mean time to the next firing of R₃ will be (approximately)

$$\frac{1}{{\overline{a}}_3\left(\mathbf{x}\right)}=\frac{c_1+c_2}{c_3{c}_1{x}_{12}}.$$ (29a)

And since the average probability that either R₁ or R₂ will fire in the next dt is $c_1⟨{\widehat{X}}_1\left(\infty \right)⟩dt+c_2⟨{\widehat{X}}_2\left(\infty \right)⟩dt$, then the average mean time to the next firing of either R₁ or R₂ will be (approximately)

$$\frac{1}{c_1⟨{\widehat{X}}_1\left(\infty \right)⟩+c_2⟨{\widehat{X}}_2\left(\infty \right)⟩}=\frac{c_1+c_2}{2c_2{c}_1{x}_{12}},$$ (29b)

where the last step follows upon substituting from Eq. 26. Now observe that, under condition c₂⪢c₃, the time 29b will indeed be very much smaller than the time 29a; moreover, no other condition is required to ensure this.

The strategy of the ssSSA is to use the standard SSA to simulate only reaction R₃, but taking the propensity function for that reaction to be the function 28 instead of c₃x₂. At each firing of R₃, the ssSSA increases the S₃ population by 1 and decreases x₁₂ by 1. The ssSSA then “waits” for a time of order δt, which is very small on the timescale of reaction R₃ but nevertheless large enough for the fast species populations to “relax” to their t=∞ values, and it then estimates the populations of the fast species by sampling the binomial random variable 25. The full ssSSA simulation procedure for reactions 1 thus proceeds as follows.

• 1.In state (x₁,x₂,x₃) at time t, and with x₁₂=x₁+x₂, evaluate ${\overline{a}}_3$ in Eq. 28.
• 2.Draw a unit-interval uniform random number r and compute the time to the next R₃ reaction,$\tau =\left(1∕{\overline{a}}_3\right)\text{ln}\left(1∕r\right)$.
• 3.Advance time to the next R₃ reaction by replacingt←t+τ. Then actualize that reaction by replacingx₃←x₃+1 and x₁₂←x₁₂−1.
• 4.Generate the “relaxed” populations of the fast species by taking x₂ to be a sample of the binomial random variable 25, and x₁=x₁₂−x₂.
• 5.Record (t,x₁,x₂,x₃) if desired. Then return to step 1, or else stop.

Figure 5 shows the results of a ssSSA run made in this way for the parameter values 22. The results are seen to be practically indistinguishable (in a statistical sense) from the exact SSA results in Fig. 3. But whereas the SSA run took about 6 min to execute, the ssSSA run took only a fraction of a second. Notice that the ssSSA remedies the deficiencies of the reaction 2 simulation in Fig. 4 as regards species S₁ and S₂.

Figure 5.

A ssSSA simulation of reactions 1 using the same settings 22 as used in Figs. 34. Here the fast reactions R₁ and R₂ have been skipped over, and only firings of the slow reaction R₃ have been simulated, using, however, the modified propensity function 28 or 21. As in Fig. 4, only 300 reaction events were simulated in this run (but here those were modified R₃ reactions), and the population of the slow species S₃ has been accurately rendered. But this ssSSA run evidently gives a much more accurate picture of the behavior of the fast species S₁ and S₂ than does the run in Fig. 4. Notice also that the initial rapid relaxation in Fig. 3 of X₁ (from 300) and X₂ (from 0) is accurately replicated in this ssSSA run.

What happens if reactions 1 are embedded in a network of other reactions, some of which involve the fast species S₁ and S₂ as reactants? The answer to this question depends on whether the other reactions are fast or slow. If any of the other reactions are as fast or faster than reactions R₁ and R₂, then we must start the analysis all over by finding, if possible, a new virtual fast process that is asymptotically stable. But if all of the new reactions are slow—i.e., they occur infrequently relative to reactions R₁ and R₂—then they can easily be accommodated in the above simulation procedure. For example, the additional slow reaction R₄, $S_1+S_4\stackrel{c_4}{\to }{S}_5$, which has propensity function a₄(x)=c₄x₁x₄, would be assigned the effective propensity function

$${\overline{a}}_4\left(\mathbf{x}\right)=c_4⟨{\widehat{X}}_1\left(\infty \right)⟩x_4=\frac{c_4{c}_2{x}_{12}{x}_4}{c_1+c_2},$$

where the last equality follows from Eq. 26b. And the additional slow reaction $S_1+S_2\stackrel{c_5}{\to }{S}_6$ with propensity function a₅(x)=c₅x₁x₂ would be assigned the effective propensity function

$${\overline{a}}_5\left(\mathbf{x}\right)=c_5⟨{\widehat{X}}_1\left(\infty \right){\widehat{X}}_2\left(\infty \right)⟩=\frac{c_5{c}_1{c}_2}{{\left(c_1+c_2\right)}^2}{x}_{12}\left(x_{12}-1\right).$$

The last step here follows by first writing

$$⟨{\widehat{X}}_1\left(\infty \right){\widehat{X}}_2\left(\infty \right)⟩=⟨\left(x_{12}-{\widehat{X}}_2\left(\infty \right)\right){\widehat{X}}_2\left(\infty \right)⟩=x_{12}⟨{\widehat{X}}_2\left(\infty \right)⟩-⟨{\widehat{X}}_2^2\left(\infty \right)⟩,$$

then using $⟨{\widehat{X}}_2^2\left(\infty \right)⟩={⟨{\widehat{X}}_2\left(\infty \right)⟩}^2+\text{var}⟨{\widehat{X}}_2\left(\infty \right)⟩$, and finally invoking Eq. 26. Thus, any new slow reactions involving the fast species S₁ and S₂ can be accommodated by the ssSSA, despite the fact that we have no sure knowledge of the instantaneous populations of those fast species.

The status of the fast species populations in the ssSSA merits further comment: Although the values for x₁ and x₂ generated in step 4 get plotted in step 5, those values are not used in the computations that drive the simulation; therefore, if plots of the fast species populations are not needed, step 4 can be omitted without any impairment to simulation accuracy. The fact is that x₁ and x₂ are not individually “tracked” by the ssSSA, because the ssSSA does not simulate reactions R₁ and R₂. Step 4 merely estimates how the values of x₁ and x₂ might appear on the slow timescale. But the sum x₁+x₂=x₁₂ is accurately tracked, and that sum is all that we need to implement reaction R₃, or any other slow reaction that involves one or both of S₁ and S₂ as reactants.

SUMMARY AND CONCLUSIONS

In this paper, we have shown that replacing reactions 1 with a single reaction that produces S₃ cannot be done accurately unless the time to the next creation of an S₃ molecule via reactions 1 can be well approximated by an exponential random variable. We showed that this applies even to the associated deterministic reaction rate equations. The specific requirement for accuracy is that P(t;x₁,x₂), the pdf of the time to the next R₃ event in Eq. 1 given x₁S₁ molecules and x₂S₂ molecules, should be well approximated by the canonical exponential form ae^−at. If that is so, then a surrogate reaction that accurately mimics the production of S₃ molecules should exist, and a will be its propensity function. If, however, the surrogate reaction is unable to accurately track the S₁ and S₂ populations individually, then even if the exponential approximation obtains, a model reduction will be feasible only if a depends on x₁ and x₂ only through their sum x₁₂.

Against this background, we derived using first-passage-time theory an exact formula for P(t;x₁,x₂). We then showed that there are only four situations in which that function satisfies the foregoing conditions: c₂⪢c₁ [Eq. 15a], c₃⪢c₁ [Eq. 15b], c₁⪢c₃ [Eq. 15c], and c₂⪢c₃ [Eq. 15d]. We found that if either of conditions 15a or 15b holds, then, under the reasonable assumption that the S₂ population is practically always zero, the propensity function of the surrogate reaction will have the form a=c₁c₃x₁₂∕(c₂+c₃). And we found that if either of conditions 15c or 15d holds, then the propensity function of the surrogate reaction will have the form a=c₁c₃x₁₂∕(c₁+c₂), with no assumptions being made regarding the S₂ population. Note that conditions 15a, 15b, 15c, 15d are not mutually exclusive; e.g., if c₂⪢c₃⪢c₁, then conditions 15d, 15b are both satisfied, and each of the two different formulas for a in those two cases reduces to the same result, c₁c₃x₁₂∕c₂.

We pointed out that abridgment solely for the sake of reducing the size of the model is not always prudent. Abridging a set of reactions is always an approximation, so there is always some loss of accuracy. In particular, although we can be confident that in the scenarios 15a, 15b, 15c, 15d the true behavior of the S₃ population in reactions 1 will be accurately replicated by the surrogate reaction, that might not be so for the S₁ and S₂ populations, since most model reductions will eliminate or severely constrain those two species. That might not matter if reactions 1 occur in isolation, in which case it would be a clear benefit of the abridgment. But it could give rise to a serious problem if reactions 1 are embedded in a larger network of reactions, some of which have S₁ and S₂ as reactants or products.

Since stochastic simulation is usually the tool of choice for analyzing complex cellular reaction networks, one reasonable goal of model reduction is to make stochastic simulation run faster. We showed that the maximum speedup factor in any single-reaction abridgment of reactions 1 is 2(c₂+c₃)∕c₃. This implies that, of the four cases 15a, 15b, 15c, 15d, the only one for which abridgment has a chance of producing a significant gain in simulation speed is case 15d, c₂⪢c₃. If that condition is satisfied, the speedup factor will be ⪢1. If it is not satisfied, the speedup factor will typically be rather small, and possibly not large enough to compensate for the loss of accuracy and robustness that invariably attends model reduction.

We showed that condition c₂⪢c₃ is the sole requirement for accurately applying the ssSSA procedure of Cao et al.⁶ to reactions 1, contrary to earlier assertions.¹³ We emphasized that the ssSSA implements a single-reaction abridgment of reactions 1 in a way that overcomes several shortcomings that arise if reactions 1 are simply replaced by reaction 2: In the ssSSA, the S₁ and S₂ populations are accurately represented on the timescale of reaction R₃, and additional slow reactions that involve S₁ and S₂ as reactants can easily be accommodated.

Finally, we showed that our first-passage-time analysis provides a framework which unites the abridgment under condition c₂⪢c₃ given by the ssSSA of Cao et al.,⁶ and the abridgment under condition c₂, c₃⪢c₁ given by the sQSPA procedure of Mastny et al.⁷ Furthermore, our first-passage-time analysis enables us to identify all of the conditions under which a single-reaction abridgment of reactions 1 is possible.

APPENDIX A: PROOF OF THEOREM 1

(Necessity) Given Eq. 5, let P₀(τ) denote the probability that the S₁ molecule will not react during the next time span τ. By the laws of probability, this function must satisfy P₀(τ+dτ)=P₀(τ)×[1−cdτ], where the last factor is the probability that the S₁ molecule, having not reacted in time τ, will not react in [τ,τ+dτ). This implies the differential equation dP₀(τ)∕dτ=−cP₀(τ). The solution of that equation for the initial condition P₀(0)=1 is P₀(τ)=exp(−cτ). Therefore, the probability that a given S₁ molecule will react in the infinitesimal time interval [τ,τ+dτ) is {the probability that it will not react in [0,τ)} times {the probability that it will react in the next dτ}: P₀(τ)×cdτ=c exp(−cτ)dτ. This implies that the pdf of the time for the S₁ molecule to react is c exp(−cτ), which is precisely the pdf of the exponential random variable with mean c⁻¹.

(Sufficiency) Given that the pdf of the time until the S₁ molecule reacts is c exp(−cτ), it follows that probability that the molecule will react in the time interval [τ,τ+dτ) is c exp(−cτ)dτ. Therefore, the probability that the molecule will react in the nextdτ, i.e., in the time interval [0,dτ), is c exp(0)dτ=cdτ, as asserted in Eq. 5.

APPENDIX B: PDF OF THE TIME TO THE NEXT R₃ EVENT

Regard Eq. 1 as depicting the transitions of a single “random walker” among three “states” S₁, S₂, and S₃. We will first derive formulas for the pdfs P_α→3(t) of the times T_α→3 (α=1 or 2) required for the walker, starting in state S_α, to first reach state S₃. To that end, let p(n,t|α,0) be the probability that the walker, having started at time 0 in state S_α (α=1 or 2), will be found at time t≥0 to be in state S_n (n=1,2,3). Since according to Eq. 1 the walker will remain in state S₃ after it arrives there, then

$$p\left(3,t|\alpha ,0\right)=\text{Prob}\left\{T_{\alpha \to 3}\le t\right\}.$$ (B1)

The probability on the right is, by definition, the cdf of the random variable T_α→3. Since the derivative of the cdf with respect to τ gives the corresponding pdf, then

$$P_{\alpha \to 3}\left(t\right)=\frac{dp\left(3,t|\alpha ,0\right)}{dt}.$$ (B2)

The laws of probability give us the following relations among the functions p(n,t|α,0) at any time t and any infinitesimally later time t+dt:

$$p\left(1,t+dt|\alpha ,0\right)=p\left(1,t|\alpha ,0\right)\times \left[1-c_{1}dt\right]+p\left(2,t|\alpha ,0\right)\times c_{2}dt,$$ (B3a)

$$p\left(2,t+dt|\alpha ,0\right)=p\left(1,t|\alpha ,0\right)\times c_{1}dt+p\left(2,t|\alpha ,0\right)\times \left[1-c_{2}dt-c_{3}dt\right],$$ (B3b)

$$p\left(3,t+dt|\alpha ,0\right)=p\left(3,t|\alpha ,0\right)+p\left(2,t|\alpha ,0\right)\times c_{3}dt.$$ (B3c)

For example, Eq. B3a is the statement that {the probability of being in state S₁ at time t+dt} is equal to the sum of {the probability of being in state S₁ at time t and then not jumping away in the next dt} plus {the probability of being in state S₂ at time t and then jumping to state S₁ in the next dt}. This logic ignores routes to state S₁ at time t+dt that involve more than one jump in time [t,t+dt), but that is permissible here since the probabilities for those paths will be of higher order than 1 in dt. Analogous reasoning gives Eqs. B3b, B3c. By algebraically rearranging each of these equations, dividing through by dt, and then taking the limit dt→0, we obtain the following set of coupled ordinary differential equations, which constitute the “master equation” for this stochastic process:

$$\frac{dp\left(1,t|\alpha ,0\right)}{dt}=-c_{1}p\left(1,t|\alpha ,0\right)+c_{2}p\left(2,t|\alpha ,0\right),$$ (B4a)

$$\frac{dp\left(2,t|\alpha ,0\right)}{dt}=c_{1}p\left(1,t|\alpha ,0\right)-\left(c_2+c_3\right)p\left(2,t|\alpha ,0\right),$$ (B4b)

$$\frac{dp\left(3,t|\alpha ,0\right)}{dt}=c_{3}p\left(2,t|\alpha ,0\right).$$ (B5)

Note that Eqs. B4a, B4b constitute a closed pair of coupled differential equations for p(1,t|α,0) and p(2,t|α,0). Once that pair of equations has been solved, p(3,t|α,0) can be obtained either by solving Eq. B5 or more simply from the fact that

$$p\left(3,t|\alpha ,0\right)=1-p\left(1,t|\alpha ,0\right)-p\left(2,t|\alpha ,0\right).$$ (B6)

Combining Eqs. B2, B5, we see that the function P_α→3(t) can be computed as

$$P_{\alpha \to 3}\left(t\right)=c_{3}p\left(2,t|\alpha ,0\right).$$ (B7)

Equations B4 can be solved in a standard way that begins by writing them as

$$\frac{d{\mathbf{p}}_{\alpha }\left(t\right)}{dt}=-\mathbf{A}•{\mathbf{p}}_{\alpha }\left(t\right),$$ (B8)

where

$${\mathbf{p}}_{\alpha }\left(t\right)\equiv \left(\begin{array}{c}p\left(1,t|\alpha ,0\right)\\ p\left(2,t|\alpha ,0\right)\end{array}\right)\text{and}\mathbf{A}\equiv \left(\begin{array}{rr}\hfill c_1& \hfill -c_2\\ \hfill -c_1& \hfill c_2+c_3\end{array}\right).$$ (B9)

The solution to Eq. B8 turns out to involve the eigenvalues λ₊ and λ₋ of A. These are evidently the solutions of the quadratic equation

$$\left(c_1-{\lambda }_{\pm }\right)\left(c_2+c_3-{\lambda }_{\pm }\right)-c_1{c}_2=0,$$ (B10)

and are easily found to be

$${\lambda }_{\pm }\equiv \frac{1}{2}\left[\left(c_1+c_2+c_3\right)\pm \sqrt{{\left(c_1+c_2+c_3\right)}^2-4c_1{c}_3}\right].$$ (B11)

A little algebra shows that the quantity under the radical here is never negative, so

$$0\le {\lambda }_{-}\le {\lambda }_{+}.$$ (B12)

We shall not belabor the process by which the solutions to Eq. B4 for α=1 and 2 are obtained, because it can be verified by simple differentiation that the functions below satisfy Eq. B4. And it is also easy to verify that they satisfy the required initial conditions.

$$p\left(1,t|1,0\right)=\frac{1}{\left({\lambda }_{+}-{\lambda }_{-}\right)}\left[\left({\lambda }_{+}-c_1\right)e^{-{\lambda }_{-}t}+\left(c_1-{\lambda }_{-}\right)e^{-{\lambda }_{+}t}\right],$$ (B13a)

$$p\left(2,t|1,0\right)=\frac{c_1}{\left({\lambda }_{+}-{\lambda }_{-}\right)}\left[e^{-{\lambda }_{-}t}-e^{-{\lambda }_{+}t}\right];$$ (B13b)

$$p\left(1,t|2,0\right)=\frac{c_2}{\left({\lambda }_{+}-{\lambda }_{-}\right)}\left[e^{-{\lambda }_{-}t}-e^{-{\lambda }_{+}t}\right],$$ (B14a)

$$p\left(2,t|2,0\right)=\frac{1}{\left({\lambda }_{+}-{\lambda }_{-}\right)}\left[\left(c_1-{\lambda }_{-}\right)e^{-{\lambda }_{-}t}+\left({\lambda }_{+}-c_1\right)e^{-{\lambda }_{+}t}\right].$$ (B14b)

The pdfs of the first passage times T_1→3 and T_2→3 can now be obtained simply by substituting Eqs. B13b, B14b into formula B7. However, our main concern here is with the more general case in which x₁ random walkers are initially in state S₁ and x₂ are initially in state S₂. The pdf P(t;x₁,x₂) of the time T(x₁,x₂) required for the first of those walkers to reach state S₃ can be computed by reasoning as follows: Since the walkers evolve independently of each other, the probability that none of them will reach state S₃earlier than time t is

$$\text{Prob}\left\{T\left(x_1,x_2\right)>t\right\}={\left(\text{Prob}\left\{T_{1\to 3}>t\right\}\right)}^{x_1}{\left(\text{Prob}\left\{T_{2\to 3}>t\right\}\right)}^{x_2}.$$ (B15)

This is equivalent to

$$\text{Prob}\left\{T\left(x_1,x_2\right)\le t\right\}=1-{\left(1-\text{Prob}\left\{T_{1\to 3}\le t\right\}\right)}^{x_1}{\left(1-\text{Prob}\left\{T_{2\to 3}\le t\right\}\right)}^{x_2}.$$

Using Eq. B1, this last equation can be written

$$\text{Prob}\left\{T\left(x_1,x_2\right)\le t\right\}=1-{\left(1-p\left(3,t|1,0\right)\right)}^{x_1}{\left(1-p\left(3,t|2,0\right)\right)}^{x_2}.$$ (B16)

But the left side of Eq. B16 is, by definition, the cdf of the random variable T(x₁,x₂). Therefore, the derivative of Eq. B16 with respect to t gives the pdf of T(x₁,x₂):

$$P\left(t;x_1,x_2\right)=\frac{d}{dt}\left[1-{\left(1-p\left(3,t|1,0\right)\right)}^{x_1}{\left(1-p\left(3,t|2,0\right)\right)}^{x_2}\right].$$ (B17)

Upon evaluating this derivative with the help of Eqs. B5, B6, we get

$$P\left(t;x_1,x_2\right)=x_1{c}_{3}p\left(2,t|1,0\right){\left(p\left(1,t|1,0\right)+p\left(2,t|1,0\right)\right)}^{x_1-1}{\left(p\left(1,t|2,0\right)+p\left(2,t|2,0\right)\right)}^{x_2}+x_2{c}_{3}p\left(2,t|2,0\right){\left(p\left(1,t|1,0\right)+p\left(2,t|1,0\right)\right)}^{x_1}{\left(p\left(1,t|2,0\right)+p\left(2,t|2,0\right)\right)}^{x_2-1}.$$ (B18)

Since all the p-functions on the right side of Eq. B18 are given explicitly by Eqs. B13, B14, we have in Eq. B18 an exact, explicit formula for the pdf of the first-passage time T(x₁,x₂).

APPENDIX C: PROOF OF THEOREM 2

That it is possible, when c₂∕c₃⪢1, to choose a time span δt that contains very many R₁ and R₂ events but practically no R₃ events, follows from the fact established in Sec. 2 that successive R₃ reactions will, on average, be separated by (c₂+c₃)∕c₃ pairs of R₁ and R₂ reactions. Let [t^′,t^′+dt^′) be an infinitesimal subinterval of the interval [t,t+δt). The probability that R₃ will fire in [t^′,t^′+dt^′) is c₃X₂(t^′)dt^′. But

$$c_3{X}_2\left(t^{\prime }\right)d{t}^{\prime }\approx c_3{\widehat{X}}_2\left(t^{\prime }\right)d{t}^{\prime },$$ (C1)

because the dearth of R₃ events in [t,t+δt) implies that the real fast process X₂(t^′) can be well approximated there by the virtual fast process ${\widehat{X}}_2\left(t^{\prime }\right)$. The probability that R₃ will fire anywhere in the interval [t,t+δt) can now be computed by summing the probabilities C1 over all the dt^′ subintervals of [t,t+δt):

$${\int }_t^{t+\delta t}{c}_3{\widehat{X}}_2\left(t^{\prime }\right)d{t}^{\prime }\equiv c_3\left\{\frac{1}{\delta t}{\int }_t^{t+\delta t}{\widehat{X}}_2\left(t^{\prime }\right)d{t}^{\prime }\right\}\delta t.$$ (C2)

This invocation of the addition law of probability for mutually exclusive events is justified since the probability for more than one R₃ firing in [t,t+δt) is practically zero. Now let K be an integer that is roughly equal to the expected number of firings of R₁ and R₂ in [t,t+δt), a number that will be ⪢1. Subdividing [t,t+δt) into K subintervals of equal length δt∕K, we can approximate the integral in braces in Eq. C2 as

$$\frac{1}{\delta t}{\int }_t^{t+\delta t}{\widehat{X}}_2\left(t^{\prime }\right)d{t}^{\prime }\approx \frac{1}{\delta t}\sum _{k=1}^K{\widehat{X}}_2\left(t_k\right)\left(\frac{\delta t}{K}\right)=\frac{1}{K}\sum _{k=1}^K{\widehat{X}}_2\left(t_k\right),$$ (C3)

where t_k (k=1,…,K) locates the center of the kth subinterval. After the first few R₁ and R₂ firings, the process ${\widehat{X}}_2\left(t\right)$ will effectively “decorrelate” and “relax” to its time-independent form ${\widehat{X}}_2\left(\infty \right)$; thus, the K values ${\widehat{X}}_2\left(t_1\right),\dots ,{\widehat{X}}_2\left(t_K\right)$ in Eq. C3 can collectively be approximated by Ksample values ${\widehat{X}}_2{\left(\infty \right)}^{\left(1\right)},\dots ,{\widehat{X}}_2{\left(\infty \right)}^{\left(K\right)}$ of the random variable ${\widehat{X}}_2\left(\infty \right)$. Equation C3 then becomes

$$\frac{1}{\delta t}{\int }_t^{t+\delta t}{\widehat{X}}_2\left(t^{\prime }\right)d{t}^{\prime }\approx \frac{1}{K}\sum _{k=1}^K{\widehat{X}}_2{\left(\infty \right)}^{\left(k\right)}\approx ⟨{\widehat{X}}_2\left(\infty \right)⟩.$$ (C4)

Substituting Eq. C4 into Eq. C2, we conclude that the probability that reaction R₃ will fire in (t,t+δt) is approximately equal to $c_3⟨{\widehat{X}}_2\left(\infty \right)⟩\delta t$. That is the first assertion of Theorem 2. The second assertion follows from the fact that, for any t^′>t+δt prior to the next R₃ event, ${\widehat{X}}_2\left(t^{\prime }\right)$ can be approximated by ${\widehat{X}}_2\left(\infty \right)$.