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A binomial guard prevents overdraw but changes Poisson noise

Example

For a pure-death reaction $A \to \emptyset$ on $E = {0, 1, \dots, n}$ with $n$ molecules and propensity $a (k) = c k$ , choose a critical threshold $0 \leq n_{c} < n$ , maximum leap $τ_{\max} = h$ , horizon $T = h$ , and step budget $B \geq 1$ , where $h = 1 / (2 c)$ . The death channel is noncritical at the initial state, its capacity is $m = n$ , and the guarded rule uses $p = a (n) h / m = 1 / 2$ on its first step.

Facts & Assumptions

Given: Countable Choice, Dependent Choice, integers $n \geq 1$ and $0 \leq n_{c} < n$ , $c > 0$ , $h = 1 / (2 c)$ , the Poisson leap Explicit Poisson tau-leaping with an invalid-state status, the binomial guard Critical-reaction binomial guarded tau-leaping, and the binomial definition below.

[F1]

Explicit Poisson tau-leaping with an invalid-state status: the unguarded firing count is $K_{P} \sim Pois (a (n) h)$ .

[F2]

Critical-reaction binomial guarded tau-leaping: for capacity $n$ and $p = a (n) h / n$ , the guarded count is $Binomial (n, p)$ and cannot exceed $n$ .

[F3]

Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials: a $Binomial (n, p)$ count is a sum of $n$ independent $Bernoulli (p)$ trials.

[F4]

Normalization, moments, and generating function of the Poisson count law: a $Pois (λ)$ count has mean and variance $λ$ .

[A1]

The critical-reaction algorithm assumes Countable Choice and Dependent Choice for its direct-SSA subroutine. Here its critical set is empty, so the algorithm skips that subroutine and defines the critical result to be no event. Critical-reaction binomial guarded tau-leaping, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain

Verification

Given: The input conditions in the Example.

1.1A1F2F3algebra

Since $n > n_{c}$ , the death channel is noncritical, and $τ = \min (T, τ_{\max}, m / a (n)) = h$ . The critical set is empty, so the algorithm skips the critical SSA call, records no critical event, and sets $δ = h$ . Thus $p = 1 / 2$ . The guarded count $K_{B}$ is a sum of $n$ Bernoulli variables, each in ${0, 1}$ by [F3], so $0 \leq K_{B} \leq n$ and the updated population $n - K_{B}$ is nonnegative. Its mean is $n p = n / 2$ .

2.1F1F3F4step 1.1algebra

Independence of the Bernoulli trials gives $Var (K_{B}) = \sum_{i = 1}^{n} p (1 - p) = n / 4$ . In contrast, the Poisson count has mean $a (n) h = n / 2$ and variance $n / 2$ by [F1, F4]. Thus the binomial guard retains the mean but halves the count variance in this finite model; it is not the same Poisson leap law.

3.1F1step 2.1algebra∎

The unguarded Poisson count has $P (K_{P} > n) > 0$ , since every mass at $k > n$ is positive when its parameter $n / 2$ is positive. Therefore its leap can overdraw the available molecules, while the guarded count cannot.

Depends on

Used by

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Sources

Tiejun Li, Lecture 21: Application in chemical reaction kinetics, Applied Stochastic Analysis course notes (standard reference, not scraped)