been introduced in [@BNBV].

To describe our main results we need to introduce some notation and a set of assumptions. In this work we consider the $k$th order increments $\Delta_{i,k}^{n} X$ of $X$, $k\in {\mathbb{N}}$, that are defined by $$\begin{aligned}
 \label{filter}
\Delta_{i,k}^{n} X:= \sum_{j=0}^k (-1)^j \binom{k}{j} X_{(i-j)/n}, \qquad i\geq k.\end{aligned}$$ For instance, we have that $\Delta_{i,1}^n X = X_{\frac{i}{n}}-X_{\frac{i-1}{n}}$ and $\Delta_{i,2}^n X = X_{\frac{i}{n}}-
2X_{\frac{i-1}{n}}+X_{\frac{i-2}{n}}$. Our main functional is the power variation computed on the basis of $k$th order filters: $$\begin{aligned}
 \label{vn}
V(p;k)_n := \sum_{i=k}^n |\Delta_{i,k}^{n} X|^p, \qquad p>0.\end{aligned}$$ Now, we introduce the following set of assumptions on $g$ and $\nu$:

**Assumption (A):** The function $g\!:{\mathbb{R}}\to{\mathbb{R}}$ satisfies $$\begin{aligned}
\label{kshs}
g(t)\sim c_0 t^\alpha\qquad \text{as } t\downarrow 0\quad \text{for some }\alpha>0\text{ and }c_0\neq  0, \end{aligned}$$ where $g(t)\sim f(t)$ as $t\downarrow 0$ means that $\lim_{t\downarrow 0}g(t)/f(t)= 1$. For some $\theta\in (0,2]$, $\limsup_{t\to \infty} \nu(x\!:|x|\geq t) t^{\theta}<\infty$ and $g - g_0$ is a bounded function in $L^{\theta}({\mathbb{R}}_+)$. Furthermore, $g$ is $k$-times continuous differentiable on $(0,\infty)$ and there exists a $\delta>0$ such that $|g^{(k)}(t)|\leq K t^{\alpha-k}$ for all $t\in (0,\delta)$, $g^{(k)}\in L^\theta((\delta,\infty))$ and $|g^{(k)}|$ is decreasing on $(\delta,\infty)$.

**Assumption (A-log):** In addition to (A) suppose that $\int_\delta^\infty |g^{(k)}(s)|^\theta \log(1/|g^{(k)}(s)|)\,ds<\infty$.

Assumption (A) ensures that the process 