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A Poisson leap can make a pure-death population negative

Example

Take one species with physical states $E = {0, 1}$ and one death reaction $ν = - 1$ , $a (1) = c > 0$ , $a (0) = 0$ . Start at $x_{0} = 1$ and take one Poisson leap of length $h > 0$ with draw budget $B = 1$ . The exact reaction chain stays in $E$ .

Facts & Assumptions

Given: The pure-death propensity system above and the leap Explicit Poisson tau-leaping with an invalid-state status.

[F1]

Finite-channel reaction propensity systems and their generator: one firing at state $1$ changes the state by $- 1$ , and the rate at state $0$ is zero.

[F2]

Explicit Poisson tau-leaping with an invalid-state status: the number $K$ of firings in one step from state $1$ has law $Pois (c h)$ and the proposed state is $1 - K$ .

[F3]

The scalar Poisson count law: the Poisson mass at $k$ is $e^{- c h} (c h)^{k} / k!$ .

Verification

Given: $c > 0$ and $h > 0$ .

1.1F2F3algebra

The leap state is negative exactly when $1 - K < 0$ , equivalently $K \geq 2$ . Using the masses in [F3], $P (K \geq 2) = 1 - P (K = 0) - P (K = 1) = 1 - e^{- c h} (1 + c h) > 0.$ The strict inequality follows from $e^{c h} > 1 + c h$ for $c h > 0$ .

2.1F1step 1.1algebra∎

For small $h$ , Taylor expansion of $e^{- c h} (1 + c h)$ gives $P (K \geq 2) = \frac{1}{2} c^{2} h^{2} + O (h^{3})$ . The exact chain cannot become negative because its only positive-rate event sends $1$ to $0$ , after which the rate is zero by [F1]. Thus the invalid tau-leap event is a distinct numerical exit from the physical state set.

Depends on

Used by

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Sources

Daniel T. Gillespie, Approximate accelerated stochastic simulation of chemically reacting systems (standard reference, not scraped)