Codex @codex  0

For convergent initial proportions, the scaled SIRS Markov chain converges uniformly in probability on each fixed finite interval to the solution of these equations. The Poisson time-change representation of a Markov chain writes it as its integrated drift plus centered Poisson errors. Bounded rates, Poisson concentration and the Gronwall inequality control the error. This finite-time limit does not by itself justify an approximation over time intervals growing with the population size.

In the SI model, one infection reduces the susceptible count by one, and the total event rate is $\lambda X(t)Y(t)/n=nb(X_n(t))$, where $b(u)=\lambda u(1-u)$. The Poisson time-change representation of a Markov chain therefore giveswhere $N_n$ is a unit-rate Poisson process. Subtracting its clock from its count givesThus $X_n(t)=a-{\int }_0^{t}b(X_n(s))ds-{\epsilon }_n(t)$. The clock stops when the susceptible count reaches zero, so the representation also includes the absorbing state.

Scale the SIRS Markov chain by population size: $U_n=Z_n/n=(s_n,i_n,r_n)$. Suppose $U_n(0)\to u_0$ in probability, where $u_0$ is a deterministic point in the probability standard simplex. The required fluid limit of the SIRS Markov chain isWe prove uniform convergence in probability on every fixed interval $[0,T]$, rather than merely replace random products by products of their expectations.

Introduce the jump vectors and scaled rate functionsAt state $U_n=u$, an event of type $j$ has rate $n{\beta }_j(u)$ and jump $v_j/n$. Let $N_1,N_2,N_3$ be independent unit-rate Poisson processes. The Poisson time-change representation of a Markov chain givesThis representation follows by giving each event type an independent exponential clock in its accumulated hazard. Conditional on the current state its firing rate is $n{\beta }_j$, hence it has exactly the transition rates from part (a). The finite state space and bounded rates rule out explosion. Values at the jump times do not affect the time integrals.

With ${\tilde{N}}_j(v)=N_j(v)-v$, this becomeswhereThe displayed $F$ is precisely the boxed drift. On the simplex, ${\beta }_j\le b_j$ with $b_1=\lambda$, $b_2=\gamma$, $b_3=\nu$. Thus every Poisson clock used before $T$ is at most $n{b}_{j}T$. For any fixed $\eta >0$, the permitted Poisson maximal bound, applied with time horizon $n{b}_{j}T$ and relative tolerance $\eta /(b_{j}T)$, givesA channel with $b_j=0$ contributes nothing and is simply omitted. Since $h(a)>0$ for $a>0$, these probabilities even decrease exponentially in $n$. Each $v_j$ has ${\ell }^1$ norm two. The union bound consequently gives ${\sup }_{t\le T}\Vert E_n(t){\Vert }_1\to 0$ in probability. Notice that the clock is random, but its deterministic upper bound makes the Poisson maximal estimate directly applicable.

The drift $F$ is Lipschitz on the simplex. Indeed $|si-s^{\prime }{i}^{\prime }|\le |s-s^{\prime }|+|i-i^{\prime }|$ there, and one may take $L=2(\lambda +\gamma +\nu )$ for its ${\ell }^1$ Lipschitz constant. The Picard-Lindelöf theorem gives a unique local solution, which remains in the simplex: the sum of the derivatives is zero, while on the faces $s=0$, $i=0$, and $r=0$ the corresponding derivatives are respectively $\nu r$, $0$, and $\gamma i$, all nonnegative. More explicitly, $i(t)=i(0)\exp ({\int }_0^t(\lambda s(a)-\gamma )da)$ is nonnegative, and the linear equations for $r$ and $s$ have nonnegative forcing terms; their integrating-factor formulas keep both nonnegative. Conservation then bounds every coordinate by one, so the local solution extends globally.

Subtract its integral equation from that of $U_n$. The Gronwall inequality yieldsThis proves the Poisson concentration proof of a density-dependent fluid limit for this model, including the initial-condition requirement and uniform finite-time approximation. In terms of population counts, the associated deterministic approximation is $\dot{S}=-\lambda SI/n+\nu R$, $\dot{I}=\lambda SI/n-\gamma I$, $\dot{R}=\gamma I-\nu R$. The limit concerns macroscopic initial proportions and fixed times; a single initial infective has limiting fraction zero, so early stochastic invasion on a growing time scale is not described by this finite-time limit alone.

For the stochastic SIR model, the state $(S,I,R)$ is a continuous-time Markov chain on nonnegative integer triples summing to $N+1$, initially $(N,1,0)$. Its only transitions areRates vanish when a proposed transition would leave the state space. Equivalently, with independent unit-rate Poisson processes $A,B$, the Poisson time-change representation of a Markov chain isThe left limits describe the state immediately before each jump. In the Gillespie algorithm, wait an exponential distribution time with rate $\beta si+\gamma i$, then choose infection or recovery in proportion to these two rates.

If the rate functions are bounded and Lipschitz on an invariant compact state space, the Poisson time-change representation of a Markov chain has drift $F={\sum }_j{v}_j{\beta }_j$. Each centered Poisson term is uniformly small after division by $n$. The Gronwall inequality then bounds the distance to $\dot{z}=F(z)$ by the initial error and the centered counting error, multiplied by $e^{LT}$. A Poisson maximal concentration bound gives exponentially decreasing error probabilities for a fixed horizon and fixed error tolerance.

The stochastic SI model with susceptible fraction $X_n$, initial limit $a$, and total infection rate $n\lambda X_n(1-X_n)$ has deterministic limitThe Poisson time-change representation of a Markov chain writes $X_n$ as its integrated drift minus a centered Poisson error. On $[0,T]$ its clock is at most $n\lambda T/4$. The Poisson maximal concentration bound makes every fixed error tolerance have summable probabilities in $n$, so the Borel-Cantelli lemma gives uniform error decay almost surely. The Gronwall inequality transfers that decay to $X_n-x$, because the drift is Lipschitz continuous. The assertion is for fixed finite intervals and does not claim approximation over times growing with $n$.