# What is the stationary distribution for the contact process on the half line?

The contact process is a well-studied Markov process. I'm just concerned with the one-dimensional nearest-neighbor version here.

The state space is $\eta\in\{0,1\}^\mathbb Z$, and for state $\eta$ at each site $x\in\mathbb Z$ the dynamics are: $0\rightarrow 1$ at rate $\lambda\sum_{y:|x-y|=1}\eta(y)$ and $1\rightarrow0$ at rate $1$.

There is a critical rate $\lambda_c$ such that for $\lambda>\lambda_c$ we have a positve probability of survival for any initial distribution that does not place unit mass of the state of all zeros. We have convergence to a stationary $\nu_\lambda$ such that $$\rho_\lambda=\nu_\lambda\left(\left\{\eta : \eta(x)=1\right\}\right)>0$$ for $\lambda>\lambda_c$ and is independent of $x.$

What I am interested in is this process restricted to the half-line, $\mathbb Z^+_0=\{0,1,2,\ldots\}$. In particular, I want to know what the stationary distribution is. It seems like we would have $$u_\lambda(x)=\nu_\lambda\left(\left\{\eta : \eta(x)=1\right\}\right)>0$$ an increasing function of $x$ with $u_\lambda(0)>0$ and $u_\lambda(x)\rightarrow\rho_\lambda$ as $x\rightarrow\infty$.

I can't seem to find any papers that discuss this. Does anyone have any relevant references or maybe can provide some relevant calculations that help to understand $u_\lambda(x)?$

Edit: I'm specifically interested in the questions:
1) Does $u_\lambda(x)\rightarrow\rho_\lambda$ as $x\rightarrow\infty?$
2) Is $u_\lambda(x)$ strictly increasing in $x?$
3) What is the relationship between $u_\lambda(0)$ and $\displaystyle\lim_{x\rightarrow\infty}u_\lambda(x)?$

I'm fairly certain (1) and (2) are true, but I'd especially like quantitative results for (3). The more specific about $u_\lambda(x)$ the better, however I'm not asking for anything close to an exact formula as that is likely not possible.

You can assume I'm familiar with all the basics (monotonicity, coupling, graphical representation, duality, etc., whatever is in Liggett's two books at least).

## 1 Answer

For the first part of your question, you can do this by coupling: Define two processes: the first is the contact process on $\{0,1,2,\ldots\}$ and the second is the contact process on $\{1,2,3,\ldots\}$ with the same parameters. Call the processes $\eta$ and $\eta'$ respectively.

They can be monotonically coupled: that is: there is a probabaility measure on the stochastic processes $(\eta,\eta')$ such that $\eta(x)\ge \eta'(x)$ for all $x$ such that the evolution of $\eta$ and $\eta'$ follow the correct laws.

Now $u(x)=\nu(\eta(x)=1)=\nu'(\eta'(x+1)=1)\le \nu(\eta(x+1)=1)=u(x+1)$. The inequality here arises from the coupling: on the space where both processes are defined, the even $\{\eta'(x+1)=1\}$ is a subset of $\{\eta(x+1)=1\}$.

EDIT: Some more details here, giving a (in principle) concrete estimate on the difference $u(1)-u(0)$. We'll use the description of the coupling coming from the graphic method. So in the $\eta$ process, there are 'lifelines' above each $x\in\mathbb Z_0^+$ marked with "kill" symbols (a Poisson process with intensity 1), "left to right" arrows joining a pair of lifelines with intensity $\lambda$ and "right to left arrows" joining a pair of lifelines with intensity $\lambda$. The $\eta'$ process has the same symbols in the same locations, except there is no lifeline above 0. Now consider the lifeline above 1. If the system is in equilibrium, the system should be in state 1 at each kill symbol with probability $u(0)$. At each kill symbol with no kill symbol in the previous unit of time, the system should be in state 1 with probability at least $u(0)$. Call this an isolated kill symbol. If in addition, there are no arrows from the 2 lifeline to the 1 lifeline for the following 2 units of time, call the kill symbol a target. By the Markov property, the $\eta'$ process above $x=1$ is in state 1 immediately below at least $u(0)$ proportion of targets. Now if a target has the property that in the $\eta$ process, there is a right-to-left arrow (from 1 to 0) in the unit of time preceding the target, no kill symbols in the 0 lifeline for 1 unit before and after the target and a left-to-right arrow (from 0 to 1) in the time unit after the target, then $\eta(1)$ is 1 while $\eta'(1)$ is 0 for at least one unit of time. One can then do the calculation to get a concrete estimate of the difference. Since $u(1)\le \rho_\lambda$, this also gives a lower bound for $\rho_\lambda-u(0)$.

• Thanks for that! I'm familiar with monotone coupling, but can this be expanded on to show strict inequality? I.e. that $u_\lambda$ is strictly increasing in $x$? However, maybe your argument is as precise as I will get without significantly more work? – jdods Jun 3 '16 at 21:35
• One more comment: a nice way to prove things like this is the graphical method. Details can be found in the paper of Griffeath (sciencedirect.com/science/article/pii/0304414981900028). You can get the strict inequality from that, I believe. – Anthony Quas Jun 3 '16 at 21:53
• I'm familiar with the basic results and techniques. I doubt strict inequality falls out of this coupling, but I'll think about it. I was hoping to get an estimate on the difference between $u_\lambda(0)$ and $\rho_\lambda$ however I will likely have to accept your answer since I failed to be that specific. Would you be insulted if I edited the question to note that? Maybe even that is a faux pas... – jdods Jun 3 '16 at 22:23
• Feel free to edit the question to reflect what you need - I believe it's considered good style to mark the edits on the question so that the original question can be identified. No need to accept my answer as it might discourage others from thinking about your question if there are still open parts to it. I believe you could get a concrete lower bound for the difference between $u_\lambda(0)$ and $\rho_\lambda$ from the graphical method. – Anthony Quas Jun 3 '16 at 23:04
• Yes I think I like all of that. – Anthony Quas Jun 5 '16 at 5:31