I am trying to compute the weak limit of the following model from mathematical biology that is supposed to exist:

Let $$L(f)(\eta)= \sum_{x \in \mathbb{Z}}\frac{1}{2}\left(1_{\eta(x+1) \neq \eta(x)}+ 1_{\eta(x-1)\neq \eta(x)} \right)(f(\eta_x)-f(\eta)), $$ where $\eta \in S:=\{0,1\}^{\mathbb{Z}}$ and $f \in D(L):=\{f \in C(S);\sum_{x \in \mathbb{Z}} \sup_{\eta \in S} |f(\eta_x)-f(\eta)|< \infty\}$ be a probability generator and $\eta_x(y):=\eta(y)$ for $y \neq x$ and $\eta_x(x):=1-\eta(x).$

Then if we start in the initial state $\eta_0(x):=1$ for $x \ge 1$ and $\eta_0(x)=0$ for $x \le 0$ and denote the probability semigroup by $T_t:C(S) \rightarrow C(S)$ we should get a weak limit

$\int T(t)f(x) d(\delta_{\eta_0})(x) \rightarrow \int f(x) d\mu(x)$ as $t \rightarrow \infty$for some measure $\mu$ and $f \in C_b$


To check this, I noticed that what this process does is to translate the distribution $\eta_0$ in the sense that the point where this distribution has its jumps (initially at zero) does a symmetric random walk on $\mathbb{Z}$ in continuous time with jump rates $\frac{1}{2}$(this number comes from the probability generator).

So let $Y:[0,\infty) \rightarrow \{0,1\}^{\mathbb{Z}}$ be the process from the original problem and $X:[0,\infty) \rightarrow \mathbb{Z}$ be the random walk of the jump point of the distribution,

then this means that we can rewrite

$$\int T(t)f(x) d(\delta_{\eta_0}(x)) = T(t)f(\eta_0) = \mathbb{E}^{\eta_0}f(Y_t) = \mathbb{E}^0 f(\eta_{X_t})= \int f(\eta_z) dP_{X_t}^0(z).$$

So I reduced the problem to the question whether there is a weak limit of the last integral, but I am pretty sure that the continuous-time symmetric random walk with jump rates $\frac{1}{2}$ does not have a weak limit, so there should be some error here if the weak-limit exists indeed.

  • $\begingroup$ You get a 50-50 mixture of all zeros and all ones, and the relevant feature of the random walk is that it is large. $\endgroup$ – user83457 Jan 28 '16 at 8:21
  • $\begingroup$ @michael could you give me a hint how to get this limit? In particular, I would like to invite you to turn your comment into an answer, so that I can accept it. Thank you very much. $\endgroup$ – Acuriousmind Jan 28 '16 at 17:42

Functions that depend on only finitely many co-ordinates are dense in continuous functions on that function space, as is plausible but also follows from Stone-Weierstrass. If f is such a function that depends only on the co-ordinates in the interval $[-N, N]$ then $$ \mathbb{E}^0 f(\eta_{X_t}) = \mathbb{E}^0 (\mathbb{E}^{\delta_1}f 1_{\lbrace X_t < -N \rbrace } + \mathbb{E}^{\delta_0}f 1_{\lbrace X_t > N \rbrace } + f(\eta_{X_t} )1_{ \lbrace -N \leq X_t \leq N \rbrace } )$$ $$= (\mathbb{E}^{\delta_1}f \mathbb{P}{\lbrace X_t < -N \rbrace } + \mathbb{E}^{\delta_0}f \mathbb{P}{\lbrace X_t > N \rbrace } + O(\mathbb{P}{\lbrace -N < X_t < N \rbrace })$$ $$ \rightarrow(\frac 12 \mathbb{E}^{\delta_1}f + \frac 12 \mathbb{E}^{\delta_0}f )$$ because, e.g., on the set $\lbrace X_t < -N \rbrace $ f sees only ones in $[-N,N]$, and the second lines is by conditioning on the value of $X_t$, and where $\delta_1$ is the measure putting all its mass on the identically 1 string. $$ $$ Let me repeat the main point: if f depends only on the coordinates $-N,....,N$ then $ \mathbb{E}^0 f(\eta_{X_t}) = \mathbb{E}^{\delta_0}f $ on the set where $X_t > N$.


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