Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 36 / 1

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.

Fix $T>0$ and put $K=\lambda T/4$. Since $0\le X_n\le 1$, the clock in part (a) is at most $nK$ on $[0,T]$. The Poisson maximal concentration bound gives, for every $\delta >0$,Here $h(u)>0$ for $u>0$, so the probabilities are summable in $n$. The Borel-Cantelli lemma, first for positive rational $\delta$ and then for all $\delta$, yields ${\sup }_{t\le T}|{\epsilon }_n(t)|\to 0$ almost surely. No independence between the processes for different population sizes is required.

The deterministic SI model solves the logistic equation $x^{\prime }=-\lambda x(1-x)$ with $x(0)=a$, givingOn $[0,1]$, $b$ is Lipschitz continuous with constant $\lambda$. Subtracting the two integral equations givesThe Gronwall inequality impliesThis is the uniform almost-sure fluid limit of the SI model. One can take the intersection of the probability-one events for integer $T$ to obtain the conclusion on every finite interval. Use population sizes with $an$ integral; if the initial susceptible count is rounded, the extra initial error is at most $1/n$ and the same Gronwall inequality proof applies.