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

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.

Put $Z={\sum }_i{Y}_i(\infty )$ and $1=(1,\dots ,1{)}^T$. Sum the bounds in part (a) over the vertices, using linearity of expectation and the Tonelli theorem for the nonnegative series:The bound remains valid if the series diverges, although it then gives no finite estimate. When the spectral radius of $\beta A$ is less than one, the series is a Neumann series, giving the resolvent bound for a discrete SIR epidemic.

Because $A$ is a real symmetric matrix, the spectral theorem gives $\Vert A{\Vert }_2=\rho (A)$. For $\beta \rho <1$, the Neumann series satisfiesLet $m={\sum }_i{X}_i(0)$ be the initial infected count. Since its entries are indicator random variables, $\Vert X(0){\Vert }_2=\sqrt{m}$, while $\Vert 1{\Vert }_2=\sqrt{n}$. The Cauchy-Schwarz inequality and the operator norm bound yieldThus a sufficient asymptotic condition for a small outbreak isFor example, it holds when $m=o(n)$ and $\beta \rho \le 1-\eta$ for a fixed $\eta >0$. For any fixed $\delta >0$, the Markov inequality then givesThis is the spectral condition for a small discrete SIR outbreak. A uniform gap and few initial infections are needed for this conclusion; the finite-population condition $\beta \rho <1$ alone does not control an asymptotically vanishing gap or an initially macroscopic infected set.