Continuous time Markov chains and invariance principle

MathOverflow Asked by sharpe on December 6, 2020

This question may be elementary for experts

Let ${xi_n}_{n=1}^{infty}$ be an i.i.d random variables on a probability space $(Omega,mathcal{F},P)$. We assume that the mean of $xi_n$ is zero, and the variance is $1$. For $n in mathbb{N}$ and $t ge 0$, we set
X_n&=sum_{k=1}^{n} xi_k,quad

Here, $[cdot]$ denotes the floor function. By the definition, $Y={Y_t}_{t ge 0}$ is the linear interpolation of the simple random walk ${X_n}_{n=1}^{infty}$.
We define $Z_t^{(n)}=Y_{nt}/sqrt{n}$, Then, each ${Z_{t}^{(n)}}_{t ge 0}$ induces a probability measure on $C([0,infty))$, the space of continuous functions on $[0,infty).$ We denote by $P_n$ the probability measure. Donsker’s invariance principle states that ${P_n}_{n=1}^{infty}$ converges to the Wiener measure.

My question

We write $n^{-1}mathbb{Z}={cdots,-2/n,-1/n,0,1/n,2/n,cdots}$. Let $S^{(n)}={S_t^{(n)}}_{t ge 0}$ be a (continuous time) Markov chain on $n^{-1}mathbb{Z}$. My question is the following:

Are there "Donsker’s invariance principles" for ${S^{(n)}}_{n=1}^{infty}$ (or scaled ${S^{(n)}}_{n=1}^{infty}$) ?

Because ${S^{(n)}}_{n=1}^{infty}$ are continuous time Markov chains, there is no interpolated process like $Y$.
Although the state space of each $S^{(n)}$ is $n^{-1}mathbb{Z}$, this should be regarded as a jump process on $mathbb{R}$. Then, each $S^{(n)}$ induces an probability measure on $D([0,infty))$, the space of right continuous functions on $[0,infty)$ with finite left limits. Wiener measures are also regarded as a probability measure on $D([0,infty))$.

Please let me know if you have any preceding results.

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