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The graphical representation of the contact process preserves coordinatewise order. Its law from the all-infected state decreases in stochastic domination with time, and its limit is invariant by the Feller semigroup property. The all-healthy state is absorbing. Any invariant probability law lies between these two laws. Duality of the contact process gives $\overline{\nu }(\eta (x)=1)=\theta (\lambda )$ when the recovery rate is one. Consequently the two extremal laws coincide exactly when the survival probability of the contact process is zero.

Use the contact process convention that each infected vertex recovers at rate one and sends infection arrows to each neighbour at rate $\lambda$. Starting from a finite set, the process stays finite and is nonexplosive: its total birth rate is at most $\lambda (d+1)$ times its current size, so a linear-rate branching process dominates the number of births. The bound involving $(d-1{)}^{-1}$ presupposes the branching-tree case $d\ge 2$.

For a nonempty finite infected set $A$, put $a=|A|$, let $e(A)$ count its internal edges, and let $b(A)$ count edges with exactly one endpoint in $A$. Each such boundary edge contributes infection rate $\lambda$, even when several lead to the same vacant vertex. Since the induced graph is a forest, $e(A)\le a-1$, and the edge boundary of a finite forest in a regular tree givesA recovery multiplies ${\nu }_{\rho }(A)={\rho }^a$ by ${\rho }^{-1}$, while an infection multiplies it by $\rho$. Therefore its infinitesimal generator isSubstituting the boundary lower bound givesAt the absorbing empty set the generator is zero, rather than this nonempty-set expression. If $\lambda \rho (d-1)\ge 1$, the generator is nonpositive in every state. Both ${\nu }_{\rho }$ and its generator are bounded, since $a{\rho }^a$ is bounded and $b(A)\le (d+1)a$. Applying Dynkin's formula over successive time intervals with the Markov property proves the exponential cardinality supermartingale for the contact process on a tree. In particular,If $\tau$ is the extinction time, $1_{\{\tau \le t\}}\le {\nu }_{\rho }({\xi }_t^A)$, so ${\mathbb{P}}_{\lambda }(\tau <\infty )\le {\rho }^{|A|}<1$. Choosing $\rho \in [1/(\lambda (d-1)),1)$ proves global survival whenever $\lambda >1/(d-1)$. Obtaining a strict improvement of the threshold requires an additional comparison.

Here is a two-vertex branching comparison for contact-process survival. Delete one edge incident to a starting vertex and retain its forward rooted subtree, so every vertex has $d$ children. Partition it into cells as follows: pair each cell root $u$ with one distinguished child $v$; the other $d-1$ children of $u$ and all $d$ children of $v$ become roots of child cells, and repeat recursively. Each cell thus has $2d-1$ child cells.

Suppress arrows back from a child cell into its parent, allow each child cell to be activated only once, and let infection inside each two-vertex cell evolve with the usual recoveries and reinfections until both vertices are healthy. A child cell is activated at its root by the first outgoing arrow to it while the corresponding parent vertex is infected. This process is contained in the original infection process under the graphical representation of the contact process: only infection paths have been removed. The Poisson clocks inside a child cell and on its outgoing edges have not previously been used. The Strong Markov property and independence of the clocks therefore make different cells' offspring counts independent and identically distributed. Dependence among different children of the same cell does not matter. The cell genealogy is a Galton-Watson process.

Let $a$ be the probability that a cell, initially infected only at $u$, sends an arrow along one specified outgoing edge from $u$ before internal extinction. Let $b$ be the corresponding probability for one specified outgoing edge from $v$, and let $c$ be the probability of success for the first target when both cell vertices are initially infected. First-step conditioning on the three nonempty pair states givesFor example, from the state with only $u$ infected, recovery at rate one fails, an internal arrow at rate $\lambda$ leads to the two-infected state, and the target arrow at rate $\lambda$ succeeds. From the state with only $v$ infected, the target cannot be reached until $u$ is infected. Symmetry identifies this latter success probability with $b$. Solving these equations yieldsBy linearity of expectation, the offspring mean is $M(\lambda )=(d-1)a+db$. At the comparison rate ${\lambda }_0=1/(d-1)$,These finite-state probabilities are continuous in $\lambda$. Hence some $0<{\lambda }^{\prime }<{\lambda }_0$ still has $M({\lambda }^{\prime })>1$. The branching-process extinction criterion gives positive probability of infinitely many activated cells. Nonexplosion prevents infinitely many activations in bounded time, and each activation is connected to its ancestor by an infection path. Thus infection survives for all time on that event. The global survival threshold of the contact process consequently satisfiesRetaining reinfection inside pairs is what supplies the strict improvement; a comparison that permits only one infection per vertex would give merely the non-strict bound.

For the local-survival estimate, set $h(x)={\rho }^{g(x)}$ and $w_{\rho }(A)={\sum }_{x\in A}h(x)$. Each vertex has one neighbour of weight ${\rho }^{-1}h(x)$ and $d$ neighbours of weight $\rho h(x)$. Recoveries remove their vertex weights. Ignoring the restriction that an infection target must be vacant only increases the birth contribution, soThese weights have finite expectations on finite time intervals: a birth-only branching process with the same spatial arrows has expected weight at most $w_{\rho }(A)e^{\lambda ({\rho }^{-1}+d\rho )t}$. This justifies localization and the expectation identity for the generator. If $c=1-\lambda ({\rho }^{-1}+d\rho )>0$, Dynkin's formula and the integrating-factor form of the Gronwall inequality giveFor $d\ge 2$, take $\rho =d^{-1/2}$, which minimizes ${\rho }^{-1}+d\rho$ and makes it $2\sqrt{d}$. Starting from $\{x\}$ then givesIn particular the expected total occupation time of $x$ is finite. Its recovery clock has rate one, so the expected number of recovery marks that occur while $x$ is infected equals this occupation-time integral and is finite. Each infection episode ends almost surely at a recovery. Infection at unbounded times would therefore require infinitely many such recoveries, an event of probability zero. This height-weighted local extinction bound for the contact process provesThis is a bound for the local survival threshold of the contact process, not just a bound on single-time infection probabilities. For the degree-two tree $d=1$, the same local conclusion follows by choosing $\rho <1$ sufficiently close to one for each $\lambda <1/2$; the displayed expression $1/(d-1)$ for the global threshold is not a finite numerical assertion in that case.

Use recovery rate one and infection rate $\lambda$ along each directed nearest-neighbour edge. The graphical representation of the contact process places independent rate-one recovery marks at graph vertices and rate-$\lambda$ arrows on directed edges. Infection travels along time-directed paths following arrows and avoiding recovery marks. The same clocks couple all initial configurations and preserve coordinatewise order. Denote its semigroup by $P_t$.

The empty configuration is absorbing, so the lower invariant measure is $\underline{\nu }={\delta }_0$. To construct the upper one, put ${\mu }_t={\delta }_1{P}_t$, where $1$ is the all-infected configuration. Since ${\delta }_1{P}_s{\le }_{\mathrm{st}}{\delta }_1$, order preservation and the semigroup property implyThus these laws decrease in stochastic order. More concretely, use clocks on the entire time axis and start from all infected at time $-t$, looking at the configuration at time zero. These configurations decrease pointwise as $t$ increases: an infection arriving from an earlier start is also obtainable from the all-infected later start. Each coordinate therefore has a limit. Their joint law $\overline{\nu }$ satisfies ${\mu }_t\to \overline{\nu }$ in weak convergence of probability measures on the compact product space.

This limit is invariant. The graphical construction is Feller: for finitely many sites and a bounded time interval, the backward ancestor exploration is almost surely finite, since its size is dominated by a finite-rate branching process. Hence a cylinder function evolved for a fixed time is continuous in the initial configuration, by approximation with finite ancestor sets. Cylinder functions uniformly approximate continuous functions on the product space. We may therefore pass to the limit into get $\overline{\nu }{P}_s=\overline{\nu }$. These are indeed the extremal invariant measures of the contact process: if $\nu$ is any invariant probability law of a Markov process, apply $P_t$ to ${\delta }_0{\le }_{\mathrm{st}}\nu {\le }_{\mathrm{st}}{\delta }_1$ and take limits to obtain $\underline{\nu }{\le }_{\mathrm{st}}\nu {\le }_{\mathrm{st}}\overline{\nu }$.

For the relation with survival, reverse the graphical paths in the interval $[0,t]$, also reversing arrows. The law is unchanged because the rates are symmetric between neighbours. The set of ancestors of a site $x$ has the law of the contact process started from $\{x\}$. ConsequentlyThis is duality of the contact process. The events on the right decrease with time because the empty state is absorbing, and their limiting probability is $\theta (\lambda )$. It follows thatIf $\theta (\lambda )=0$, every coordinate is zero almost surely under $\overline{\nu }$; a countable union over the sites gives $\overline{\nu }={\delta }_0=\underline{\nu }$. Conversely, if the two invariant measures coincide, this one-site marginal is zero, so $\theta (\lambda )=0$. HenceA convention that divides the infection rate by the degree only rescales the parameter; the argument and equivalence are unchanged.

The previous part and the permitted assumption $p_c(\mathrm{bond})<1$ imply $p_c(\mathrm{site})<1$. We build an independent oriented site percolation process whose open paths carry contact-process infections, and whose site probability tends to one as $\lambda$ increases.

Use the graphical representation of the contact process with recovery rate one and per-neighbour arrow rate $\lambda$. Fix $\delta >0$. On the parity lattice $(x,n)$ with $n\ge 0$ and $x+n$ even, declare $(x,n)$ good if there is no recovery mark at $x$ during $[(n-1)\delta ,(n+1)\delta ]$, and there is at least one arrow from $x$ to each of $x-1,x+1$ during $[n\delta ,(n+1)\delta ]$. Extend the Poisson clocks to negative times solely to define the root event; the process itself still starts at time zero. Goodness has probability

The independence here is important. At a fixed spatial site $x$, the allowed time indices differ by two. Its two-unit recovery intervals therefore have disjoint interiors, and its outgoing-arrow intervals are disjoint. At different spatial sites, recovery clocks are different and outgoing arrows have different tails, hence use different directed-edge clocks. Every defining random input belongs to just one parity site's event, up to deterministic interval endpoints at which a Poisson mark has probability zero. Thus the good-site indicators really are independent, with common probability $q$; no theorem about dependent percolation is needed.

Suppose $(x,n)$ and $(x\pm 1,n+1)$ are both good and $x$ is infected at time $n\delta$. The source has no recovery before $(n+1)\delta$, and it sends an arrow to the chosen neighbour during that interval. The target's good event excludes recovery throughout $[n\delta ,(n+2)\delta ]$, so the target is infected at time $(n+1)\delta$. Induction shows that any infinite oriented path of good sites starting at $(0,0)$ sustains infection at every time layer. Along each transmitting interval there is always an infected site, so the contact process never reaches its empty absorbing state.

Take $\delta ={\lambda }^{-1/2}$. ThenFor some finite $\lambda$ it exceeds $p_c(\mathrm{site})$, and the good-site process has positive survival probability. The contact process begun from the origin therefore also survives with positive probability. HenceThis is the independent oriented-percolation comparison for the contact process. If the infection convention assigns total rate $\lambda$ rather than rate $\lambda$ to each of the two neighbours, replace the arrow rate by $\lambda /2$; the same limit proves the same finiteness conclusion.

For the graphical representation of the contact process, put independent death marks on each vertical time-line $\{x\}\times [0,\infty )$ at rate $\delta$, and independent infection arrows $x\to y$ at rate $\lambda$ for every ordered nearest-neighbour pair. A time-directed path moves vertically upward, cannot pass a death mark, and may follow an infection arrow. From an initial set $A$, the infected set ${\xi }_t^A$ consists of sites reachable at time $t$ from $A\times \{0\}$. Thus infected sites recover at rate $\delta$, and a healthy site with $k$ infected neighbours becomes infected at rate $k\lambda$. Rates here are per ordered edge, fixing the normalization of $\lambda$.

With $\delta =1$, writeCouple two infection rates by adding independent arrows of rate ${\lambda }_2-{\lambda }_1$ to the rate-${\lambda }_1$ construction, retaining the same deaths. Every path available at the lower rate remains available at the higher rate. Hence ${\theta }_d(\lambda )$ is nondecreasing.

For the lower bound, dominate the infection population by a linear birth-death branching process $Z_t$, started from one particle, with per-particle birth rate $2d\lambda$ and death rate $1$. Assign one branching particle to every infection; allow an offspring at every attempted transmission, including attempts into an already infected site, and let the extra particles continue independently. Matched deaths remove their infection. This constructs $|{\xi }_t|\le Z_t$. The linear birth rate is nonexplosive, so in particular the contact process from a finite initial set is well-defined with finite population at finite times. Its branching mean solvesConsequentlyIf $2d\lambda <1$, let $t\to \infty$ to obtain ${\theta }_d(\lambda )=0$. This proves the branching bound for contact-process survival

For the upper bound use the dimension comparison for the contact process. Define the height $h(x)=x_1+\cdots +x_d$. Construct an auxiliary infected set ${\eta }_t\subseteq \mathbb{Z}$ together with an infected representative $r_n(t)\in {\xi }_t$ of height $n$ for every $n\in {\eta }_t$. Initially ${\eta }_0=\{0\}$ and $r_0=0$. Use the original graphical marks as follows.

This rule maintains a live representative for every auxiliary infection. Conditional on the full past, each active representative has death rate $1$ and outgoing-arrow rate $d\lambda$ in each of the two height directions. Representatives at distinct heights are different sites, using distinct death clocks and distinct outgoing ordered-edge clocks. Independence and memorylessness of the underlying Poisson processes give exactly the one-dimensional contact-process generatorThus $\eta$ has the law of the one-dimensional contact process with infection rate $d\lambda$, even though the representatives themselves depend on the higher-dimensional history. Since ${\eta }_t\ne \varnothing$ implies ${\xi }_t\ne \varnothing$ at every time,Every $\lambda$ for which $d\lambda >{\lambda }_c(1)$ is therefore supercritical whenever the one-dimensional survival set is nonempty; if that critical value were infinite the inequality would be trivial. Taking infima gives

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