Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 36 / 4 / a / Solution

Take the initial removed set to be empty, as required by the displayed equality in this discrete SIR epidemic on a graph. If initially removed individuals are allowed, add their indicators $Y_i(0)$ to the final-removal count; an isolated initially removed vertex already shows why this correction is necessary. Condition on a fixed initial infection vector.

Given the history through time $k$, the union bound over currently infected neighbours givesWrite $u(k)=\mathbb{E}X(k)$. Taking expectations and iterating this componentwise inequality for the nonnegative adjacency matrix of a graph gives $u(k)\le (\beta A{)}^{k}X(0)$.

A vertex is infected at most once, and an infection lasts exactly one step before permanent removal. Hence $Y_i(\infty )={\sum }_{k\ge 0}{X}_i(k)$ under the stated initial convention. The Tonelli theorem now givesThe walk count from powers of an adjacency matrix explains the terms: all possible transmission walks are counted, including walks that overcount because removal prevents reinfection. This is the adjacency-matrix bound for a discrete SIR epidemic.