0
$\begingroup$

Background:

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 $f \in C_b$

Problem:

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.

$\endgroup$
2
  • $\begingroup$ 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, 2016 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. $\endgroup$ Jan 28, 2016 at 17:42

1 Answer 1

0
$\begingroup$

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 $X_t > N$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.