= Solution
In the <SI model>, one infection reduces the susceptible count by one, and the total event rate is $\lambda X(t)Y(t)/n=n\,b(X_n(t))$, where $b(u)=\lambda u(1-u)$. The <Poisson time-change representation of a Markov chain> therefore gives
$$
X_n(t)=a-\frac1nN_n\left(n\int_0^t b(X_n(s))\,ds\right),
$$
where $N_n$ is a unit-rate <Poisson process>. Subtracting its clock from its count gives
$$
\boxed{\varepsilon_n(t)=\frac1n\left[
N_n\left(n\int_0^t b(X_n(s))\,ds\right)
-n\int_0^t b(X_n(s))\,ds\right]}.
$$
Thus $X_n(t)=a-\int_0^t b(X_n(s))\,ds-\varepsilon_n(t)$. The clock stops when the susceptible count reaches zero, so the representation also includes the absorbing state.
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