This question seems to be related to Theorem IX.7.28 in [J. Jacod and A. Shiryaev's *Limit theorems for stochastic processes* (2013)](https://www.google.co.kr/books/edition/Limit_Theorems_for_Stochastic_Processes/mSD4CAAAQBAJ?hl=ko&gbpv=1&printsec=frontcover), 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,



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


