In the proof of the FCLT Theorem 7.4.1 of Ethier and Kurtz (1986) "Markov Processes", the last four lines on p.355 write ($\tau_n^r = \inf\{t: |X_n(t)|\geq r\}$):

"Consequently, if $X$ is a solution of the martingale problem for $(A, v)$, then $X(\cdot\wedge\tau_n^r)\Rightarrow X(\cdot\wedge\tau^r)$ for all $r$ such that $P\{lim_{s\to r} \tau^s = \tau^r\} = 1$ (here $\tau^r = \inf\{t: |X(t)|\geq r\}$). But $\tau_r\to \infty$ as $r\to\infty$ (since $X$ has sample paths in $C_{\mathbb R^d}[0,\infty)$), so $X_n\Rightarrow X$."

I understand why $X_n(\cdot\wedge\tau_n^r)\Rightarrow X(\cdot\wedge\tau^r)$ from the earlier development of the proof, but there is a gap between this and the desired weak convergence $X_n\Rightarrow X$.

My thoughts to fill this gap: since we already have $X(\cdot\wedge\tau_n^r)\Rightarrow X(\cdot\wedge\tau^r)$, the proof of $X_n\Rightarrow X$ is complete by Corollary 3.3.3 on p.110 of EK86 if we show

  1. $d\big(X_n,X_n(\cdot\wedge\tau_n^r)\big)\stackrel{p}{\to}0$ as $n,r\to\infty$
  2. $d\big(X(\cdot\wedge\tau^r),X\big)\stackrel{p}{\to}0$ as $r\to\infty$

where $d(x,y)=\sum_{j=1}^d d_{\infty}(x_j,y_j)$ is the product Skorohod metric of $D_{\mathbb R^d}[0,\infty)$ where $d_{\infty}(x_j,y_j)$ is defined as Eq. (5.2) on p.117 of EK86.

Based on a previous post (Martingale central limit theorem proof on p.341 of Ethier and Kurtz (1986)), we can estimate (WLOG, set dimension $d=1$): $$ d(X_n,X_n(\cdot\wedge\tau_n^r)) \leq e^{-\tau_n^r} $$ and $$ d(X(\cdot\wedge\tau^r),X) \leq e^{-\tau^r}. $$ The proof for (2) can be finished by using the the second estimation together with the fact $\tau_r\to \infty$ as $r\to\infty$ a.s.

Question: The problem is (1). I am not sure how to show $\tau_n^r\stackrel{p}{\to}\infty$ as $r,n\to\infty$. A fact may be relevant: $\tau_n^r\Rightarrow\tau^r$ for all but countably many $r$ noted on the 5th line on page 355. Is this enough for our purpose? Don't we need process convergence of $\tau_n^r\Rightarrow\tau^r$ in $r$ to complete the job? (But the stopping time is not a continuous mapping in the topology generated by $d(x,y)$ by Whitt (1980))

Any help would be greatly appreciated! Thank you very much.


After some thoughts, I found that showing (1) is actually not hard. I provide some details here.

Recall that $X_n(\cdot\wedge\tau_n^r)\Rightarrow X(\cdot\wedge\tau^r)$ in $D_{\mathbb R^d}[0,\infty)$ for all but countably many $r$ (pointwisely for each of them), if $T<\tau_n^r$, we have $X_n \Rightarrow X$ in $D_{\mathbb R^d}[0,T]$, and by continuous mapping theorem we also have

$$ \sup_{t\leq T}\|X_n(t)\|_2 \Rightarrow \sup_{t\leq T}\|X(t)\|_2 \quad\mbox{as }n\to\infty, $$ where $\sup_{t\leq T}\|X(t)\|_2$ is a continuous random variable (r.v.) because $X(t)$ is a Gaussian process with a.s. continuous sample path.

Therefore, for any $T>0$, for all but countably many $r$ that $X_n(\cdot\wedge\tau_n^r)\Rightarrow X(\cdot\wedge\tau^r)$ in $D_{\mathbb R^d}[0,\infty)$, $$ P(\tau_n^r>T)=P\big(\sup_{0 \leq t\leq T}\|X_n(t)\|_2<r\big) = P\big(\sup_{0 \leq t\leq T}\|X(t)\|_2<r\big) + R_n(r) $$

where $R_n(r) = P(\sup_{t\leq T}\|X_n(t)\|_2<r)-P(\sup_{t\leq T}\|X_n(t)\|_2<r)$ satisfies $\sup_{r\in\mathbb R_+}|R_n(r)|=o(1)$ as $n\to\infty$ because the convergence in distribution to continuous r.v. implies uniform convergence of CDFs (Convergence in law implies uniform convergence of cdf's).

As $n\to\infty$ and $r\to\infty$ along an increasing countable sequence, since $R_n=o(1)$ uniformly in $r$, we obtain that $P(\tau_n^r>T)\to 1$ (since $P(\sup_{0 \leq t\leq T}\|X(t)\|_2<r)\to 1$ as $r\to\infty$ along an increasing countable sequence) for any $T>0$. This verifies that $\tau_n^r\stackrel{p}{\to}\infty$ as $r$ and $n$ $\to\infty$.


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