Does weak convergence of stopped Markov chains implies weak convergence of the corresponding unstopped Markov chains in some particular case?
In my case:
Assumptions on the sequence of Markov chains:
$\forall n \in \mathbb{N}$
- $\{X^n(t)\}_{t\geq 0 }\subset \frac{1}{n}\mathbb{N}^d\setminus\{\mathbf{0}\}$ Markov chain
- $\tau_n=\inf\{t:X_i^n(t)= \frac{1}{n} \text{ for some } i\}$ is a finite stopping time
- if $s\geq t$ then $||X^n(s)||_1\leq ||X^n(t)||_1$
- $X^n(0)\to \mathbf{x}_0\in \mathbb{R}_{>0}^d$
Assumptions on the limiting process:
- $\{X(t)\}_{t\geq 0 } \subset \mathbb{R}_{\geq 0}^d$ strongly continuous Markov process
- $X(0)=\mathbf{x}_0$
- $\forall t\geq ||\mathbf{x}_0||_1, X(t)=\mathbf{0}$.
- $\inf\{t:X_i(t)= 0 \text{ for some } i\}=||\mathbf{x}_0||_1$
Weak convergence:
- We know that $\{X^n(t\wedge \tau_n)\}_{t\geq 0}$ converges weakly to $X$.
I am trying to prove that $X^n$ should also converge weakly to $X$, because, heuristically, it seems that this is implied by $\tau_n\to ||\mathbf{x}_0||_1$, $X^n(\tau_n)\to X(||\mathbf{x}_0||_1)=\mathbf{0}$ and for $t\geq \tau_n$, $||X^n(t) ||_1\leq ||X^n(\tau_n)||_1\to 0$.
Is my intuition correct? Is there any classical result that could help me here?