Convergence of discretized process when its predictable part converges to infinite variation process

Question

This question seems to be related to Theorem IX.7.28 in J. Jacod and A. Shiryaev's Limit theorems for stochastic processes (2013), and it is very important to prove asymptotic properties of my statistical estimator. The situation can be much like simplified as follows.

Here is a question.

Let our discretized process be \begin{align} X^n_t = \sum_{i=1}^{[nt]} \left(W_{\frac{i}{n}} - W_{\frac{i-2}{n}}\right), \end{align} where $W$ is a standard Wiener process. My intuition says that it would converge stably in law to $2W$ (in the Skorokhod space $\mathbb{D}([0,1])$), i.e., \begin{align} X^n \xrightarrow{\mathcal{L}-s} 2W. \end{align} But I'm not sure it is ok to apply Theorem IX.7.28 to show this, because the predictable part (denote it as $B^n_t$) converges to Wiener process itself which has "infinite variation": \begin{align} B^n_t &:= \sum_{i=1}^{[nt]} \mathbb{E}_{\frac{i-1}{n}} \left[ W_{\frac{i}{n}} - W_{\frac{i-2}{n}} \right] \\ &= \sum_{i=1}^{[nt]} \left(W_{\frac{i-1}{n}} - W_{\frac{i-2}{n}}\right) \\ &= W_{\frac{[(n-1)t]}{n}} \xrightarrow{\mathbb{P}} W_t, \end{align} while Jacod and Shiryaev's theorem says that $B^n_t$ should converge to a predictable "finite variation" process $B_t$. So how can I show this?

Here is a theorem.

Without considering the truncation function or jumps of $X$, the theorem says:

IX.7.28 Theorem. For every càdlàg process $X$, we use the following notation: \begin{align*} X_t^n = X_{[nt]/n},\qquad\qquad \Delta_i^n X = X_{i/n} - X_{(i-1)/n} = \Delta X^n_{i/n}, \end{align*} We also consider the discretized process of the form \begin{align*} X_t^n = \sum_{i=1}^{[nt]} \chi_i^n, \end{align*} where each $\chi_i^n$ is $\mathcal{F}_{i/n}$-measurable. Assume that each $\chi_i^n$ is square-integrable, and $X$ is a continuous and $\mathbb{E}[|X_t|^2] < \infty$ for all $t$ with the canonical decomposition $X_t = B_t + M_t$ where $B_t$ is predictable finite variation process and $M_t$ is square-integrable local martingale.

Suppose also that for all $t>0$ and all uniformly integrable martingale $N$ which are orthogonal to $X$ we have \begin{align*} \sup_t \left| \sum_{i=1}^{[nt]} \mathbb{E} _{\frac{i-1}{n}}[\chi_i^n] - B_t \right| &\xrightarrow{\mathbb{P}} 0, \\ \sum_{i=1}^{[nt]} \mathbb{V}_{\frac{i-1}{n}}[\chi_i^n] &\xrightarrow{\mathbb{P}} \langle M, M \rangle_t + \langle w \cdot W', w \cdot W' \rangle_t \\ \sum_{i=1}^{[nt]} \mathbb{E}_{\frac{i-1}{n}}[\chi_i^n \Delta_i^n M] &\xrightarrow{\mathbb{P}} \langle M, M \rangle_t, \\ \sum_{i=1}^{[nt]} \mathbb{E}_{\frac{i-1}{n}}[\chi_i^n \Delta_i^n N] &\xrightarrow{\mathbb{P}} 0. \end{align*} Then there is a very good canonical Wiener extension of $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}, \mathbb{P})$ with Wiener process $W'$ and a continuous $X$-biased $\mathcal{F}$-progressive conditional martingale PII $X'$ on this extension such that \begin{align*} X^n \xrightarrow{\mathcal{L}-s} X' = X + w \cdot W', \end{align*} where $w \cdot W' = \int w\,dW'$ and $w$ is a predictable process.

If I have any misunderstanding please let me know. Any help will be appreciated.

Thanks,

Answer 1

Let $h_i:=W_{i/n}-W_{(i-1)/n}$ and $n_t:=\lfloor nt\rfloor$, so that $t-1/n\le n_t/n\le t$. Then $$X^n_t=\sum_{i=1}^{n_t}(h_{i-1}+h_i) =\sum_{i=1}^{n_t}h_{i-1}+\sum_{i=1}^{n_t}h_i=W_{n_t/n-1/n}-W_{-1/n}+W_{n_t/n},$$ whence $$|X^n_t-2W_t|\le|W_{n_t/n-1/n}-W_t|+|W_{-1/n}|+|W_{n_t/n}-W_t|$$ and, for each real $u>0$, $$P(\sup_{0\le t\le 1}|X^n_t-2W_t|>3u)\le2nP(\max_{0\le s\le2/n}|W_s|>u)+P(|W_{-1/n}|>u).$$ Next, $P(|W_{-1/n}|>u)=P(|W_1|>u\sqrt n)\to0$ (as $n\to\infty$) and $$ \begin{aligned} 2nP(\max_{0\le s\le2/n}|W_s|>u) &=2nP(\max_{0\le t\le1}|W_t|>u\sqrt{n/2}) \\ &\le2nP(\max_{0\le t\le1}W_t>u\sqrt{n/2}) \\ &=4nP(W_1>u\sqrt{n/2}) \\ &\le2n\exp\{-u^2n/4)\to0. \end{aligned} $$ So, $X^n\to2W$ in probability in $C[0,1]$ and hence in probability in $D[0,1]$ and hence stably in $D[0,1]$, as desired.