= Solution
Because $A$ is a real <symmetric matrix>, the <spectral theorem> gives $\|A\|_2=\rho(A)$. For $\beta\rho<1$, the <Neumann series> satisfies
$$
R=(I-\beta A)^{-1}=\sum_{k=0}^\infty(\beta A)^k,
\qquad \|R\|_2\leq\frac1{1-\beta\rho}.
$$
Let $m=\sum_iX_i(0)$ be the initial infected count. Since its entries are <indicator random variables>, $\|X(0)\|_2=\sqrt m$, while $\|\mathbf1\|_2=\sqrt n$. The <Cauchy-Schwarz inequality> and the <operator norm> bound yield
$$
\boxed{\mathbb E Z\leq
\frac{\sqrt{nm}}{1-\beta\rho}}.
$$
Thus a sufficient asymptotic condition for a small outbreak is
$$
\boxed{\frac{\sqrt{m/n}}{1-\beta\rho}\longrightarrow0}.
$$
For example, it holds when $m=o(n)$ and $\beta\rho\leq1-\eta$ for a fixed $\eta>0$. For any fixed $\delta>0$, the <Markov inequality> then gives
$$
\mathbb P(Z\geq\delta n)\leq
\frac{\sqrt{m/n}}{\delta(1-\beta\rho)}\longrightarrow0.
$$
This 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.
Back to article page
