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A zero-total-hazard state is absorbing

Statement

Assume Countable Choice and Dependent Choice. Let $E = {0, 1}$ with one death channel of increment $- 1$ , propensity $a (1) = 1$ , and $a (0) = 0$ . Starting from $1$ , the process waits an $Exp (1)$ time $W$ , moves to $0$ , and remains there forever. In particular, state $0$ has total hazard zero and is absorbing.

Facts & Assumptions

Given: Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain), the finite-state reaction system above and the exact Gillespie method with unlimited event budget.

[F1]

Positive propensity requires the poststate $x + ν$ to lie in the state set; the reaction event and generator use the stated increment and rate. (Finite-channel reaction propensity systems and their generator)

[F2]

Under Countable Choice and Dependent Choice, at positive total rate the finite-state Gillespie process with a fixed initial state and unlimited event budget has exponential holding time and the channel probabilities are propensity divided by total rate; zero total rate is absorbing. (Finite-state Gillespie and next-reaction methods have the specified CTMC law)

[F3]

If the direct method reaches its zero-propensity test before the horizon, it returns the constant continuation with status absorbed; a proposed event at or beyond the horizon instead gives status horizon without recording that event. (Gillespie direct method with horizon and budget statuses)

Proof

technique · direct

1.1F1F2givenalgebra

The only positive rate is $a (1) = 1$ , and its poststate $1 + (- 1) = 0$ lies in $E$ , so the system is admissible. At $1$ , [F2] gives $W \sim Exp (1)$ ; at $0$ , the total rate is $0$ .

2.1F2F3step 1.1algebra

Once the first event occurs, the state is $0$ and every propensity is zero. By [F2] the process remains at $0$ at all later times. In an unlimited-budget call with finite horizon $T$ , if $W < T$ , the event is recorded and the next zero-propensity test returns status absorbed; if $W \geq T$ , [F3] returns status horizon without recording the event. The underlying process therefore satisfies $P (X_{t} = 1) = e^{- t}$ and $P (X_{t} = 0) = 1 - e^{- t}$ for $t \geq 0$ , independently of the call's horizon censoring.

3.1F1step 1.1algebra∎

The generator is $(Q f) (1) = f (0) - f (1)$ and $(Q f) (0) = 0$ for every bounded $f : E \to R$ , so the zero-hazard state has no outgoing transition and the example satisfies the operator convention.

Depends on

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Sources

Daniel T. Gillespie, Exact Stochastic Simulation of Coupled Chemical Reactions (1977) (standard reference, not scraped)