Stochastic Processes – Problem 24: Prove that there is positive probability that the system never returns to the all-healthy state

Consider the contact process on the infinite 100-regular tree, described as follows. Each vertex has two states: healthy or infected. Each vertex is equipped with two independent Poisson clocks: a "recovery" clock with rate 1, and an "infection" clock with rate 100. When the recovery clock rings, if the vertex is infected, it recovers to the healthy state. When the infection clock rings, the vertex independently and uniformly selects a neighbor and infects it (i.e., sets the neighbor's state to infected). Assume that initially exactly one vertex is infected. Prove that there is positive probability that the system never returns to the all-healthy state.

Step 1. Starting from the initially infected vertex (root), consider the "effective offspring" count of infection spread. Each infected vertex: its infection clock rings at rate 100, each time selecting a random neighbor. In the early stages (few infected), almost all neighbors are healthy due to the infinite tree and high degree.

Step 2. A single infected vertex: recovery time $R \sim \mathrm{Exp}(1)$. Given $R = r$, the number of infection events is $N \sim \mathrm{Poisson}(100r)$. In the early phase, each event infects a new vertex with probability close to 1. Mean offspring: $E[\text{new infections}] = E[E[\mathrm{Poisson}(100R) | R]] = E[100R] = 100$.

Step 3. This is a continuous-time branching process with lifetime $\sim \mathrm{Exp}(1)$ and birth rate 100. Mean offspring $m = 100 \gg 1$. By branching process theory, when $m > 1$ (supercritical), the extinction probability $q < 1$, i.e., survival probability $1 - q > 0$.

Step 4. "Never returning to the all-healthy state" is equivalent to the contact process never going extinct. The generating function of the offspring distribution is: $G(s) = E[s^N] = E[e^{100R(s-1)}] = \frac{1}{1 - 100(s-1)} = \frac{1}{101 - 100s}$. Solving $q = G(q)$: $q(101 - 100q) = 1$, i.e., $100q^2 - 101q + 1 = 0$. Solutions: $q = 1$ or $q = 1/100 = 0.01$. The extinction probability is the smaller root $q = 0.01$. Therefore the survival probability is $1 - 0.01 = 0.99 > 0$.

QED.