 Solution

The contact process on $\mathbb{Z}$ has states ${\xi }_t(x)\in \{0,1\}$. Each infected site recovers at rate $\mu$, and infection passes across each oriented nearest-neighbour edge at rate $\lambda$. For the graphical representation of the contact process, put independent rate-$\mu$ recovery marks on each vertical time line and independent rate-$\lambda$ infection arrows on each oriented nearest-neighbour edge. Then $y\in {\xi }_t^A$ exactly when a forward path from some $(x,0)$ with $x\in A$ reaches $(y,t)$ by moving upward, following arrows, and avoiding recovery marks.

If $A\subseteq B$, every path starting in $A$ also starts in $B$, so ${\xi }_t^A\subseteq {\xi }_t^B$. A path starts in $A\cup B$ exactly when it starts in one of the two sets, which proves additivity of the contact process:The event ${\xi }_t^A\cap B\ne \varnothing$ says that a graphical path runs from $A\times \{0\}$ to $B\times \{t\}$. Reflecting the time interval about $t/2$ and reversing every arrow preserves the joint law of the independent Poisson processes, and the path now runs from $B$ to $A$. This proves duality of the contact process:

The survival probability of the contact process isand ${\lambda }_c(\mu )=\inf \{\lambda :\theta (\lambda ,\mu )>0\}$. By duality and translation invariance,The events on the right decrease with $t$ because the empty state is absorbing, and their intersection is eternal survival. Therefore