text:
\section{Introduction}
\label{sec1}

In recent years limit theorems and statistical inference for high
frequency observations of stochastic processes have received a great
deal of attention. The most prominent class of high frequency
statistics are power variations that have been proved to be of immense
importance for the analysis of the fine structure of an underlying
stochastic process. The asymptotic theory for power variations and
related statistics has been intensively studied in the setting of It\^o
semimartingales, fractional Brownian motion and Brownian
semi-stationary processes, to name just a few; see for example \cite
{BCP11,BCP13,BGJPS05,C01,JP12} among many others.

In the recent work \cite{BLP17,BP17} power variations of stationary
increments L\'evy moving average processes have been investigated in
details. These are continuous-time stochastic processes $(X_t)_{t \geq
0}$, defined on a probability space $(\varOmega, \mathcal{F}, \mathbb
{P})$, that are given by
\begin{equation}
\label{Xmodel} X_t = \int_{-\infty}^{t}
\bigl(g(t - s) - g_0(-s) \bigr) \, \di L_s,
\end{equation}
where $L = (L_t)_{t \in\R}$ is a symmetric L\'evy process on $\R$ with
$L_0 = 0$ and without Gaussian component. Moreover, $g, g_0 : \R\to\R
$ are deterministic functions vanishing on $(-\infty, 0)$. The most
prominent subclasses include L\'evy moving average processes, which
correspond to the setting $g_0 = 0$, and the linear fractional stable
motion, which is obtained by taking $g(s) = g_0(s) = s^\alpha_+$ and
$L$ being a symmetric $\beta$-stable L\'evy process with $\beta\in(0,
2)$. The latter is a self-similar process with index $H=\alpha+1/\beta
$; see \cite{ST94}.

We introduce the $k$th order increments $\Delta_{i, k}^{n} X$ of $X$,
$k \in\N$, that are defined by
\begin{equation}
\label{filter} \Delta_{i, k}^{n} X := \sum
_{j = 0}^k (-1)^j \binom{k}{j}
X_{(i - j) /
n}, \quad i \geq k.
\end{equation}
For example, we have that $\Delta_{i, 1}^n X = \smash{X_{\frac{i}{n}}}
- \smash{X_{\frac{i - 1}{n}}}$ and $\Delta_{i, 2}^n X = \smash{X_{\frac
{i}{n}}} - 2 \smash{X_{\frac{i - 1}{n}}} + \smash{X_{\frac{i -
2}{n}}}$. The main statistic of interest is the power variation
computed on the basis of $k$th order increments:
\begin{equation}
\label{vn} V(X, p; k)_n := \sum_{i = k}^n
|\Delta_{i, k}^{n} X|^p, \quad p > 0.
\end{equation}
A variety of asymptotic results has been shown for the statistic
$V(p;k)_n$ in \cite{BLP17,BP17}. The mode of convergence and possible
limits heavily depend on the interplay between the power $p$, the form
of the kernel function $g$ and the Blumenthal--Getoor index of $L$. We
recall that the Blumenthal--Getoor index is defined via
\begin{equation}
\label{BG} \beta:= \inf \Biggl\{r \geq0 : \int_{-1}^1
|x|^r \, \nu(\di x) < \infty \Biggr\} \in[0, 2],
\end{equation}
where $\nu$ denotes the L\'evy measure of $L$. It is well known that
$\sum_{s \in[0,1]} |\Delta L_s|^p$ is finite when $p > \beta$, while
it is infinite for $p < \beta$. Here $\Delta L_s = L_s - L_{s-}$ where
$L_{s-} = \lim_{u \uparrow s, u < s} L_u$. To formulate the results of
\cite{BLP17,BP17}, we introduce the following set of assumptions on
$g$, $g_0$ and $\nu$:
\medbreak
\noindent\textbf{Assumption~(A):} The function $g : \R\to\R$
satisfies the condition
\begin{equation}
\label{kshs} g(t) \sim c_0 t^\alpha\quad\text{as } t
\downarrow0 \quad\text{for some } \alpha> 0\text{ and } c_0 \neq0,
\end{equation}
where $g(t) \sim f(t)$ as $t\downarrow0$ means that $\lim_{t
\downarrow0} g(t) / f(t) = 1$. For some $w\in(0, 2]$, $\limsup_{t \to
\infty} \nu(x : |x| \geq t) t^{w} < \infty$ and $g - g_0$ is a bounded
function in $L^{w}(\R_+)$. 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 $t \in(0, \delta)$,
$|g^{(k)}|$ is decreasing on $(\delta, \infty)$ and $g^{(j)} \in
L^w((\delta, \infty))$ for $j \in\{1, k\}$.
\medbreak
\noindent\textbf{Assumption~(A-log):} In addition to (A) suppose that
\[
\int_\delta^\infty|g^{(k)}(s)|^w
\big| \log\bigl(|g^{(k)}(s)|\bigr)\big| \, \di s < \infty.
\]
Intuitively speaking, Assumption~(A) says that $g^{(k)}$ may
have a singularity at $0$ when $\alpha$ is small, but it is smooth
outside of $0$. The theorem below has been proved in \cite{BLP17,BP17}. We recall that a sequence of $\R^d$-valued random variables
$(Y_n)_{n\geq1}$ is said to converge stably in law to a random
variable $Y$, defined on an extension of the original probability space
$(\varOmega, \mathcal{F}, \mathbb{P})$, whenever the joint convergence in
distribution $(Y_n, Z) \schw(Y,Z)$ holds for any $\mathcal
F$-measurable $Z$; in this case we use the notation $Y_n \stab Y$. We
refer to \cite{AE78,R63} for a detailed exposition of stable convergence.

\begin{theorem}[{\cite[Theorem~1.1(i)]{BLP17} and \cite[Theorem~1.2(i)]{BP17}}]
\label{th1}
Suppose that Assumption~(A) holds, the Blumenthal--Getoor index
satisfies $\beta< 2$ and $p > \beta$. If $w = 1$ assume that (A-log)
holds. Then we obtain the following cases:
\begin{enumerate}
\item\label{it:th1:1} When $\alpha<k-1/p$ then we have the stable convergence
\begin{equation}
\label{part1}
\begin{aligned} n^{\alpha p} V(X, p;
k)_n &\stab|c_0|^p \sum
_{m : T_m \in[0, 1]} |\Delta L_{T_m}|^p
V_m
\\
\quad\text{with} \quad V_m &= \sum_{l = 0}^{\infty}
|h_k(l + U_m)|^p, \end{aligned}
\end{equation}
where $(T_m)_{m\geq1}$ denote the jump times of $L$, $(U_m)_{m \geq
1}$ is a sequence of independent identically (i.i.)
$\mathcal{U}(0,1)$-distributed\querymark{Q2} variables independent of
$L$, and the function $h_k$ is defined by
\begin{equation}
\label{def-h} h_k(x) = \sum_{j = 0}^k
(-1)^j \binom{k}{j} (x - j)_{+}^{\alpha} \quad
\text{with} \quad y_+ = \max\{y, 0\}.
\end{equation}

\item\label{it:th1:2} When $\alpha=k-1/p$ and additionally
$1/p+1/w>1$, then we have
\begin{equation}
\label{part2}
\begin{aligned} \frac{n^{\alpha p}}{\log(n)} V(X,p;k)_n
&\toop | c_0 q_{k, \alpha} |^p \sum
_{s \in(0, 1]} |\Delta L_s|^p
\\
\quad\text{with} \quad q_{k, \alpha} &:= \prod_{j = 0}^{k - 1}
(\alpha- j). \end{aligned}
\end{equation}
\end{enumerate}
\end{theorem}
We remark that the first order asymptotic theory of \cite
[Theorem~1.1]{BLP17} includes two more regimes: an ergodic type limit
theorem in the setting $p < \beta$, $\alpha< k - 1 / \beta$ and
convergence in probability to a random integral in the setting $p
\geq1$, $\alpha> k - 1 / \max\{p, \beta\}$. However, in this paper we
concentrate ourselves on results of Theorem~\ref{th1}, which are quite
non-standard in the literature. More specifically, our aim is to extend
the theory of Theorem~\ref{th1} to kernels $g$ that exhibit multiple
singularities. We call a point $x \in\R_+$ a singularity point when
the $k$th derivative $g^{(k)}$ of $g$ explodes at $x$. Note that under
Assumption~(A) and condition $\alpha\leq k - 1 / p$ the function $g$
has only one singularity point at $x=0$. In practical applications a
singularity point $x \in\R_+$ leads to a strong feedback effect
stemming from the past jumps around the time $t-x$. Such effects has
been discussed in the context of turbulence modelling in \cite{GP14}.

We will show that the limits in Theorem~\ref{th1}\ref{it:th1:1} and~\ref
{it:th1:2} will be affected by the presence of multiple singularity
points. More precisely, we will see that the increments $\Delta
_{i,k}^{n} X$ can be heavily influenced by the jumps of $L$ that
happened in the past, and the time delay is determined by the
singularity points of $g$. The obtained result is similar in spirit to
the work \cite{GP14} that studied quadratic variation of Brownian
semi-stationary processes under multiple singularities of the kernel
$g$. Furthermore, we will prove that in general the stable convergence
in Theorem~\ref{th1}\ref{it:th1:1} only holds along a subsequence.

The paper is structured as follows. Section~\ref{sec2} presents the
main results of the article. Proofs are collected in Section~\ref{sec3}.

\section{Main results}
\label{sec2}

We consider stationary increments L\'evy moving average processes as
defined at \eqref{Xmodel} and recall that the driving motion $L$ is a
pure jump L\'evy process with L\'evy measure $\nu$. Now, we introduce
the condition on the kernel function $g$:
\medbreak
\noindent\textbf{Assumption~(B):} For some $w \in(0, 2]$, $\limsup_{t
\to\infty} \nu(x : |x| \geq t) t^w < \infty$ and $g - g_0$ is a
bounded function in $L^w(\R_+)$. Furthermore, there exist points $0 =
\theta_0 < \theta_1 < \cdots< \theta_l$ such that the following
properties hold:
\begin{enumerate}
\item\label{it:B:1} $g(t) \sim c_0 t^{\alpha_0}$ as $t \downarrow0$
for some $\alpha_0 > 0$ and $c_0 \neq0$.

\item\label{it:B:2} $g(t) \sim c_z |t - \theta_z|^{\alpha_z}$ as $t \to
\theta_z$ for some $\alpha_z > 0$ and $c_z \neq0$, and for all $z = 1,
\ldots, l$.

\item\label{it:B:3} $g \in C^k(\R_+ \setminus\{\theta_0, \ldots, \theta
_l\})$.

\item\label{it:B:4} There exist $\delta$, $K > 0$ such that
$|g^{(k)}(t)| \leq K |t - \theta_z|^{\alpha_z - k}$ for all $t \in
(\theta_z - \delta, \theta_z + \delta) \setminus\{\theta_z\}$, for any
$z = 0, \ldots, l$. Furthermore, there exists a $\delta' > 0$ such that
$|g^{(k)}|$ is decreasing on $(\theta_l + \delta', \infty)$ and
$g^{(j)} \in L^w ((\theta_l + \delta', \infty))$ for $j \in\{1, k\}$.
\end{enumerate}
Let us give some remarks on Assumption~(B). First of all,
conditions~(B)\ref{it:B:1} and~(B)\ref{it:B:2}, which are direct
extensions of \eqref{kshs}, mean that for small powers $\alpha_z > 0$
the points $\theta_z$ are singularities of $g$ in the sense that
$g^{(k)}(\theta_z)$ does not exist. On the other hand, condition~(B)\ref
{it:B:3} states that there exist no further singularities. The
parameter $w$ is by no means unique. It simultaneously describes the
tail behaviours of the L\'evy measure $\nu$ and the integrability of
the function $|g^{(k)}|$, which exhibit a trade-off. When $L$ is $\beta
$-stable we always take $w=\beta$. Furthermore, Assumption~(B)
guarantees the existence of $X_t$ for all $t \geq0$. Indeed, it
follows from \cite[Theorem~7]{RR89} that the process $X$ is
well-defined if and only if for all $t \geq0$,
\begin{equation}
\label{sdkfhsd;kf} \int_{-t}^\infty\int
_{\R} \bigl(| f_t(s)x |^2 \wedge1
\bigr) \, \nu (\di x) \, \di s < \infty,
\end{equation}
where $f_t(s) = g(t + s) - g_0(s)$. By adding and subtracting $g$ to
$f_t$ it follows by Assumption~(B) and the mean value theorem that
$f_t$ is a bounded function in $L^w(\R_+)$. For all $\epsilon> 0$,
Assumption~(B) implies that
\[
\int_{\R} \bigl(|y x|^2 \wedge1\bigr) \,\nu(
\di x) \leq K \bigl( \1_{\{|y| \leq
1\}} |y|^w + \1_{\{|y| > 1\}}|y|^{\beta+ \epsilon}
\bigr),
\]
which shows \eqref{sdkfhsd;kf} since $f_t$ is a bounded function in
$L^w(\R_+)$.

\begin{remark}[Toy example]
\label{rem1}
Recall the following well-known results about the power variation of a
pure jump L\'evy process $L$:
\[
V(L,p;k)_n \toop\sum_{s \in[0,1]} |\Delta
L_s|^p < \infty
\]
for any $k \geq1$ and any $p > \beta$. Let us now consider a simple
stationary increments L\'evy moving average process $X$ with $g_0 = 0$
and $g(x) =
\1_{[0, 1]}(x)$. In this case we may call the points $\theta_0 = 0$ and
$\theta_1 = 1$ the singularities of $g$, although they do not precisely
correspond to conditions~(B)\ref{it:B:1} and~(B)\ref{it:B:2}, and we
observe that $X_t = L_t - L_{t - 1}$. Hence, we obtain the convergence
in probability
\[
V(X,p;k)_n \toop\sum_{s \in[0, 1]} |\Delta
L_s|^p + \sum_{s \in[-1,
0]} |\Delta
L_s|^p
\]
for any $k \geq1$ and any $p > \beta$. This result demonstrates that
even in the simplest setting multiple singularities lead to a different limit.
\end{remark}
It turns out that only the minimal powers among $\{\alpha_0,
\ldots, \alpha_l\}$ determine the asymptotic behaviour of the statistic
$V(X, p; k)_n$. Thus, we define
\begin{equation}
\label{Aalpha} \alpha:= \min\{\alpha_0, \ldots,
\alpha_l\} \qquad\text{and} \qquad \mathcal{A} := \{z :
\alpha_z = \alpha\}.
\end{equation}
Furthermore, we introduce the notation $h_{k, 0} := h_k$ and
\begin{equation}
\label{hkz} h_{k, z}(x) = \sum_{j = 0}^k
(-1)^j \binom{k}{j} |x - j|^{\alpha_z} \quad\text{for } z =
1, \ldots, l.
\end{equation}
In the main result below we consider a subsequence $(n_j)_{j \in\N}$
such that the following condition holds:
\begin{equation}
\label{nj} \lim_{j \to\infty} \{n_j
\theta_z\} = \eta_z \in[0, 1] \quad\text {for all } z
\in\mathcal{A},
\end{equation}
where $\{x\}$ denotes the fractional part of $x \in\R$. Obviously,
such a subsequence always exists since $\{n \theta_z\}$ is a bounded
sequence. Sometimes we will require a stronger condition, which is
analogous to Assumption~(A-log):
\medbreak
\noindent\textbf{Assumption~(B-log):} Condition~(B) holds and we have that
\[
\int_{\theta_l+\delta'}^{\infty} |g^{(k)}
(t)|^{w} \big|\log\bigl(|g^{(k)} (t)|\bigr)\big| \, \di t < \infty.
\]
The main result of the paper is the following theorem.

\begin{theorem}
\label{th2}
Suppose that Assumption~(B) holds, the Blumenthal--Getoor index
satisfies $\beta< 2$ and $p > \beta$. If $w = 1$ assume that (B-log)
holds. Recall the notations \eqref{Aalpha} and~\eqref{hkz}. Then we
obtain the following cases:
\begin{enumerate}
\item\label{it:th2:1} When $\max_{0\leq z \leq l} \alpha_z <k-1/p$ and
condition~\eqref{nj} holds, then we have the stable convergence
\begin{align}
\label{part1.2} &n_j^{\alpha p}V(X,p;k)_{n_j} \stab\sum
_{z \in\mathcal{A}} |c_z|^p \sum
_{m : T_m \in[-\theta_z, 1 - \theta_z]} |\Delta L_{T_m}|^p
V_m^z
\\
&\text{with} \quad V_m^z = \sum
_{r\in\Z} \big|h_{k,z}\bigl(r+1-\{U_m+
\eta_z\} \bigr)\big|^p.
\nonumber
\end{align}
as $j \to\infty$, where $(U_m)_{m \geq1}$ is a sequence of i.i. $\mathcal
{U}(0,1)$-distributed variables independent of $L$.

\item\label{it:th2:2} Let $\alpha= \alpha_0 = \cdots= \alpha_l = k -
1/p$. Assume that the functions $f_z : \R_+ \to\R$ defined by $f_z(x)
= g(x) / |x - \theta_z|^{\alpha}$ are in $ C^k((\theta_z - \delta,
\theta_ z + \delta))$ for all $\delta< \max_{1 \leq j \leq l}(\theta_j
- \theta_{j - 1})$. If $1/p+1/w>1$, then we have
\begin{equation}
\label{part2.2} \frac{n^{\alpha p}}{\log(n)} V(X, p; k)_n \toop|q_{k, \alpha}
|^p \sum_{z = 0}^{l}
|c_z|^p (1 + \1_{\{z \geq1\}} ) \sum
_{m : T_m \in[-\theta
_z, 1 - \theta_z]} |\Delta L_{T_m}|^p,
\end{equation}
where the constant $q_{k,\alpha}$ has been introduced in Theorem~\ref
{th1}\ref{it:th1:2}.
\end{enumerate}
\end{theorem}
We remark that the stable convergence in Theorem~\ref{th2}\ref
{it:th2:1} only holds along the subsequence $(n_j)_{j \geq1}$, which
is seen from the form of the limit in \eqref{part1.2} that depends on
$(\eta_z)$. The original statistic $n^{\alpha p} V(X, p; k)_{n}$ is
tight, but does not converge except when $\theta_z \in\N$ for all $z
\in\mathcal{A}$. On the other hand, in Theorem~\ref{th2}\ref{it:th2:2}
we do not require to consider a subsequence.

Notice that the interval $ [-\theta_z, 1 - \theta_z]$, which appears in
Theorem~\ref{th2}, is the set $[0, 1]$ shifted by $\theta_z$ to the
left. Given the discussion of Remark~\ref{rem1}, such a shift in the
limit is not really surprising. We recall that a similar phenomenon has
been discovered in \cite{GP14} in the context of Brownian
semi-stationary processes. These are stochastic processes $(Y_t)_{t
\geq0}$ defined by
\[
Y_t = \int_{-\infty}^{t} g(t - s)
\sigma_s \, \di W_s,
\]
where $W$ is a two-sided Brownian motion and $(\sigma_t)_{t \in\R}$ is
a c\'adl\'ag process. When the kernel function $g$ satisfies
conditions~(B)\ref{it:B:1} and~(B)\ref{it:B:2} along with some further
assumptions, which in particular ensure the existence of $Y_t$, the
authors have shown the following convergence in probability (see \cite
[Theorem 3.2]{GP14}):
\begin{equation}
\frac{1}{n \tau_n^2} V(Y, 2; k)_n \toop\sum
_{z \in\mathcal{A}} \pi_z \int_{-\theta_z}^{1 - \theta_z}
\sigma_s^2 \, \di s
\end{equation}
where $\tau_n^2 = \E[(\Delta_{k, k}^n G)^2]$ with $G_t = \int_{-\infty
}^{t} g(t - s) \, \di W_s$, and the probability weights $(\pi_z)_{z \in
\mathcal{A}}$ are given by
\[
\pi_z = \frac{c_z^2 \| h_{k, z}\|^2_{L^2(\R)}}{\sum_{z \in\mathcal{A}}
c_z^2 \| h_{k, z} \|^2_{L^2(\R)}}.
\]
Hence, we observe the same shift phenomenon in the integration region
as in Theorem~\ref{th2}.

\section{Proofs}
\label{sec3}

Throughout this section all positive constants are denoted by $C$
although they may change from line to line. We will divide the proof of
Theorem~\ref{th2} into several steps. First, we will show the
statements \eqref{part1.2} and~\eqref{part2.2} for a compound Poisson
process. In the second step we will decompose the jump measure of $L$
into jumps that are bigger than $\epsilon$ and jumps that are smaller
than $\epsilon$. The big jumps form a compound Poisson process and
hence the claim follows from the first step. Finally, we prove
negligibility of small jumps when $\epsilon\to0$.

We start with an important proposition.

\begin{proposition}
\label{prop1}
Let $T = (T_1, \ldots, T_d)$ be a stochastic vector with a density $v :
\R^d \to\R_+$. Suppose there exists an open convex set $A \subseteq\R
^d$ such that $v$ is continuously differentiable on $A$ and vanishes
outside of $A$. Then, under condition~\eqref{nj}, it holds that
\begin{equation}
\label{mainprop} \bigl(\{n_j T + n_j \theta_z
\}\bigr)_{z \in\mathcal{A}} \stab\bigl(\{U + \eta_z\}
\bigr)_{z \in\mathcal{A}} \quad\text{as } j \to\infty,
\end{equation}
where $\{x\}$ denotes the fractional parts of the vector $x \in\R^d$
and $x + a$, $a \in\R$, is componentwise addition. Here $U = (U_1,
\ldots, U_d)$ consists of i.i. $\mathcal{U}(0, 1)$-distributed random
variables defined on an extension of the space $(\varOmega, \mathcal{F},
\mathbb{P})$ and being independent of $\mathcal{F}$.
\end{proposition}

\begin{proof}
We first show the stable convergence
\begin{equation}
\label{stabconv} \{ nT \} \stab U.
\end{equation}
This statement has already been shown in \cite[Lemma 4.1]{BLP17}, but
we demonstrate its proof for completeness. Let $f:\R^d \times\R^d\to\R
$ be a $C^1$-function, which vanishes outside some closed ball in
$A\times\R^d$. We claim that there exists a finite constant $K > 0$
such that for all $\rho> 0$
\begin{equation}
\label{eq-est-f-g} D_\rho:= \Bigg| \int_{\R^d} f\bigl(x, \{
x/\rho\}\bigr) v(x) \, \di x - \int_{\R
^k} \biggl(\int
_{[0, 1]^d} f(x,u) \, \di u \biggr) v(x) \, \di x \Bigg| \leq K \rho.
\end{equation}
By \eqref{eq-est-f-g} used for $\rho= 1/n$ we obtain that
\begin{equation}
\label{sdkfh} \E\bigl[f\bigl(T, \{n T\}\bigr)\bigr] \longrightarrow\E\bigl[f(T,
U)\bigr]\quad\text{as } n \to \infty.
\end{equation}
Moreover, due to \cite[Proposition~2(D'')]{AE78}, \eqref{sdkfh} implies
the stable convergence\break $\{n T\} \stab U$ as $n\to\infty$. Thus, we need
to prove the inequality \eqref{eq-est-f-g}. Define $\phi(x,u) := f(x,
u) v(x)$. Then it holds by substitution that
\begin{align*}
\int_{\R^d} f\bigl(x, \{ x/\rho\}\bigr) v(x) \, \di x &= \sum
_{j \in\Z^d} \int_{(0, 1]^d}
\rho^d \phi(\rho j + \rho u, u) \, \di u
\end{align*}
and
\begin{align*}
\int_{\R^d} \biggl(\int_{[0, 1]^d}
f(x,u) \, \di u \biggr) v(x) \, \di x &= \sum_{j \in\Z^d}
\int_{[0, 1]^d} \biggl(\int_{(\rho j, \rho(j +
1)]} \phi(x, u) \,
\di x \biggr) \, \di u.
\end{align*}
Hence, we conclude that
\begin{align*}
D_\rho&\leq\sum_{j \in\Z^d} \int
_{(0, 1]^d} \Bigg|\int_{(\rho j,
\rho(j + 1)]} \phi(x, u) \, \di x -
\rho^d \phi(\rho j + \rho u , u) \Bigg| \, \di u
\\
&\leq\sum_{j \in\Z^d} \int_{(0, 1]^d} \int
_{(\rho j,\rho(j + 1)]} \big|\phi(x, u)- \phi(\rho j + \rho u, u) \big| \, \di x \, \di
u.
\end{align*}
Using that $A$ is convex and open, we deduce by the mean value theorem
that there exists a positive constant $K$ and a compact set $B\subseteq
\R^d\times\R^d$ such that for all $j\in\Z^d$, $x \in(\rho j, \rho(j
+ 1)]$ and $u \in(0,1]^d$ we have
\[
\big|\phi(x, u) - \phi(\rho j + \rho u, u) \big| \leq K \rho\1_B(x, u).
\]
Thus, $D_\rho\leq K \rho\int_{(0, 1]^d} \int_{\R^d} \1_{B} (x, u) \,
\di x \, \di u$, which shows \eqref{stabconv}.

Now, we are ready to prove the statement \eqref{mainprop}. By~\eqref
{stabconv} and condition~\eqref{nj} we conclude that
\[
\bigl(\{n_j T\}, \{n_j \theta_z \}
\bigr)_{z \in\mathcal{A}} \stab(U, \eta _z)_{z \in\mathcal{A}} \quad
\text{as } j \to\infty.
\]
Next, consider the map $f : \R^d \times\R^{l'} \to\R^{d \times l'}$,
where $l'$ denotes the cardinality of $\mathcal A$, given by
\[
f(x, y_1, \ldots, y_{l'}) = \bigl(\{x + y_1
\}, \ldots, \{x + y_{l'}\}\bigr).
\]
This map is discontinuous exactly in those points $x, y_1, \ldots,
y_{l'}$ for which $x_j + y_i \in\Z$ for some $i \in\{1, \ldots, l'\}$
and some $j \in\{1, \ldots, d\}$. Note that the probability of the
limiting variable $(U, \eta_z )_{z \in\mathcal{A}}$ lying in the
latter set is $0$. Hence, it follows from the continuous mapping
theorem for stable convergence that
\[
f\bigl(\{n_j T\}, \bigl(\{n_j \theta_z\}
\bigr)_{z \in\mathcal{A}}\bigr) \stab f\bigl(U,(\eta_z)_{z \in\mathcal{A}}
\bigr) = \bigl(\{U + \eta_z\}\bigr)_{z \in
\mathcal{A}}
\]
as $j \to\infty$. Since $x = \{x\} + \lfloor x \rfloor$ we have the
identity $\{x + y\} = \{\{x\} + \{y\}\}$ and the left-hand side becomes
\[
f\bigl(\{n_j T\}, \bigl(\{n_j \theta_z\}
\bigr)_{z \in\mathcal{A}} \bigr) = \bigl(\{n_jT + n_j
\theta_z \}\bigr)_{z \in\mathcal{A}},
\]
which concludes the proof of Proposition~\ref{prop1}.
\end{proof}
Now, we introduce the notation
\begin{equation}
\label{gin} g_{i,n}(x) = \sum_{j = 0}^{k}
(-1)^j \binom{k}{j} g\bigl((i - j)/n - x\bigr),
\end{equation}
and observe the identity
\[
\Delta_{i,k}^n X= \int_{\R}
g_{i, n}(s) \, \di L_s.
\]
The next lemma presents some estimates for the function $g_{i, n}$. Its
proof is a straightforward consequence of Assumption~(B) and the Taylor
expansion.

\begin{lemma}
\label{lem1}
Suppose that Assumption~(B) holds and let $z=1,\ldots, l$. Then there
exists an $N \in\N$ such that for all $n\geq N$ and $i \in\{k, \ldots
, n\}$ the following hold:
\begin{enumerate}
\item\label{it:lem1:1} $|g_{i,n}(x)| \leq C (|i/n - x - \theta
_z|^{\alpha_z} + n^{-\alpha_z} )$ for all $x\in[\frac{i-2k}{n} - \theta
_z, \frac{i + 2k}{n} - \theta_z]$.

\item\label{it:lem1:2} $|g_{i,n}(x)| \leq Cn^{-k}|(i-k)/n-x - \theta
_z|^{\alpha_z-k}$ for all $x \in(\frac{i}{n}-\delta- \theta_z,\tfrac
{i-k}{n} - \theta_z)$ if $\alpha_z - k < 0$.

\item\label{it:lem1:3} $|g_{i,n}(x)| \leq C n^{-k} |(i-k)/n - x - \theta
_z|^{\alpha_z - k}$ for all $x \in(\frac{i + k}{n} - \theta_z, \frac{i
- k}{n} + \delta- \theta_z)$ if $\alpha_z - k < 0$.

\item\label{it:lem1:4} $|h_{k,z}(x)| \leq|x - k|^{\alpha- k}$ for all
$x\geq k+1$ and $|h_{k,z}(x)| \leq|x + k|^{\alpha- k}$ for all $x
\leq-k - 1$, if $\alpha_z - k < 0$.

\item\label{it:lem1:5} For each $\varepsilon>0$ it holds that
\begin{align*}
n^k|g_{i,n}(s)| \1_{(-\infty, \frac{i}{n} - \varepsilon- \theta_l]}(s) &\leq
C_\varepsilon \bigl(\1_{[-\theta_l - \delta', 1 - \theta_l]}(s)
\\
&\quad+ \1_{(-\infty, -\theta_l - \delta')}(s) |g^{(k)}(-s)| \bigr).
\end{align*}
\end{enumerate}
Furthermore, similar estimates hold for $z = 0$ with obvious
adjustments that account for the fact that $g$ and $h_{k,0}$ are both
vanishing on $(- \infty, 0)$.
\end{lemma}

\subsection{Proof of Theorem~\ref{th2} in the compound Poisson case}
\label{sec3.1}
In this subsection we assume that $L$ is a compound Poisson process.
Recall that $(T_m)_{m \geq1}$ denotes the jump times of $L$. Let
$\varepsilon> 0$ and consider $n_j \in\N$ such that $\varepsilon n_j
> 4k$. Define the set\querymark{Q3}
\begin{align*}
\varOmega_\varepsilon= \bigl\{&\omega\in\varOmega: \text{for all $m \in\N $
with $T_m(\omega) \in[-\theta_l, 1]$ it holds}
\\
&|T_{m}(\omega) - T_i(\omega)| > 2\varepsilon,
T_m(\omega) + \theta_{z} - \theta_{z'} \notin
\bigl[T_i(\omega) - 2 \varepsilon, T_i(\omega) + 2
\varepsilon\bigr]
\\
&\forall i \neq m \ \forall z, z' \in\{0, \ldots, l\} \text{ and $
\Delta L_s(\omega) = 0$}
\\
&\text{for all $s \in[- \varepsilon- \theta_z, -
\theta_z + \varepsilon ] \cup[1 - \varepsilon- \theta_z,
1 -\theta_z + \varepsilon]$} \ \forall z\in\{0, \ldots, l\} \bigr\}.
\end{align*}
Roughly speaking, on the set $\varOmega_\varepsilon$ the jump times in
$[-\theta_l, 1]$ are well separated, their increments are outside a
small neighbourhood of $\theta_{z} - \theta_{z'}$, and there are no
jumps around the fixed points $-\theta_z$ and $1 - \theta_z$. In
particular, it obviously holds that $\mathbb{P}(\varOmega_\varepsilon) \to
1$ as $\varepsilon\to0$.

Throughout the proof we assume without loss of generality that $0 \in
\mathcal A$. Now, we introduce a decomposition, which is central for
the proof. Recalling the definition of $g_{i, n}$ at \eqref{gin}, we
observe the identity
\begin{equation}
\label{maindec} \Delta_{i, k}^n X = \sum
_{z \in\mathcal{A}} M_{i, n, \varepsilon, z} + \sum_{z \in\mathcal{A}^c}
M_{i, n, \varepsilon, z} + R_{i, n,
\varepsilon},
\end{equation}
where for $z = 1, \ldots, l$
\begin{align*}
M_{i, n, \varepsilon, 0} &= \int_{\frac{i}{n} - \varepsilon}^{\frac{i}{n}}g_{i, n}(s) \, \di L_s,
\qquad
M_{i, n, \varepsilon, z} = \int_{\frac{i}{n} - \theta_z - \varepsilon}^{\frac{i}{n} - \theta_z + \frac{\lfloor n\varepsilon\rfloor}{n}} g_{i,n}(s) \, \di L_s\\
R_{i ,n, \varepsilon} &= \int_{-\infty}^{\frac{i}{n} - \theta_l - \varepsilon}
g_{i, n}(s) \, \di L_s + \sum_{z = 1}^{l}\int_{\frac
{i}{n} - \theta_z + \frac{\lfloor n\varepsilon\rfloor}{n}}^{\frac
{i}{n} - \theta_{z - 1} - \varepsilon} g_{i, n}(s) \, \di
L_s.
\end{align*}
It turns out that the first term $\sum_{z \in\mathcal{A}} M_{i, n,
\varepsilon, z}$ is dominating, while the other two are negligible.

\subsubsection{Main terms in Theorem~\ref{th2}\ref{it:th2:1}}
\label{sec3.1.2}

In this subsection we consider the dominating term in the decomposition
\eqref{maindec}. We want to prove that, on $\varOmega_\varepsilon$, then
for $j \to\infty$
\begin{equation}
\label{toshow} n_j^{\alpha p} \sum_{i = k}^{n_j}
\Bigg| \sum_{z \in\mathcal{A}} M_{i, n_j, \varepsilon, z} \Bigg|^p \stab
\sum_{z \in\mathcal A} |c_z|^p \sum
_{m : T_m \in[-\theta_z, 1 - \theta_z]} |\Delta L_{T_m}|^p
V_m^z,
\end{equation}
where the limit has been introduced in \eqref{part1.2}. Let us fix an
index $z \in\mathcal{A}$. Then, on $\varOmega_\varepsilon$, for each jump
time $T_m \in(-\theta_z, 1 - \theta_z]$ there exists a unique random
variable $i_{m, z} \in\N$ such that
\[
T_m \in \biggr(\frac{i_{m, z} - 1}{n} - \theta_z,
\frac{i_{m, z}}{n} - \theta_z \biggl].
\]
We also observe the following implication, which follows directly from
the definition of the set $\varOmega_\varepsilon$:\querymark{Q4}
\[
\text{On } \varOmega_\varepsilon, \text{ if } M_{i,n,\varepsilon,z} \neq 0
\text{ for some } z \in\mathcal{A} \implies M_{i,n,\varepsilon,z'} = 0 \text{ for any }
z'\neq z \text{ in } \mathcal{A}.
\]
Indeed, this is the consequence of the definition of the term $M_{i, n,
\varepsilon, z}$ and the statement
\[
T_m(\omega) + \theta_z - \theta_{z'} \notin
\bigl[T_{m'}(\omega) - 2\varepsilon, T_{m'}(\omega) + 2
\varepsilon\bigr] \ \forall m'\neq m \ \forall z, z'
\in\{0, \ldots, l\},
\]
which holds on $\varOmega_\varepsilon$. Hence, we conclude that
\[
n^{\alpha p}\sum_{i = k}^{n} \Bigg|\sum
_{z \in\mathcal{A}} M_{i, n,
\varepsilon, z} \Bigg|^p =
n^{\alpha p} \sum_{z \in\mathcal{A}} \sum
_{i = k}^{n} | M_{i, n, \varepsilon, z} |^p
\]
on $\varOmega_\varepsilon$, and we obtain the representation
\begin{align}
&n^{\alpha p}\sum_{i = k}^{n}
|M_{i, n, \varepsilon, z}|^p = V_{n,
\varepsilon, z} \quad\text{with}
\nonumber
\\
\label{Vnz} & V_{n, \varepsilon, z} = n^{\alpha p} \sum
_{m : T_m \in
(-\theta_z, 1 - \theta_z]} |\Delta L_{T_m}|^p \sum
_{u = -\lfloor
n\varepsilon\rfloor}^{\lfloor n\varepsilon\rfloor+ v_m^z} |g_{i_{m,
z} + u, n}(T_m)|^p,
\end{align}
where $v_m^z$ are random variables taking values in $\{-2, -1, 0\}$
that are measurable with respect to $T_m$. If $z = 0$ then the sum
above is one-sided, i.e. from $u = 0$ to $\lfloor n\varepsilon\rfloor
$, cf. \cite[Eq.~(4.2)]{BLP17}. Next, we observe the identity
\[
\{nT_m + n \theta_z\} = nT_m + n
\theta_z - \lfloor nT_m + n \theta_z
\rfloor= nT_m + n \theta_z - (i_{m, z} -1).
\]
Due to Assumption~(B), we can write $g(x)= c_z|x - \theta_z|^{\alpha}
f(x)$ with $f(x) \to1$ as $x \to\theta_z$, for any $z\in\mathcal{A}$
(for $\theta_0 = 0$ we need to replace $|x|^{\alpha}$ by $x_+^{\alpha
}$). This allows us to decompose
\begin{align}
&n^{\alpha} g \biggl(\frac{i_{m, z} + u - r}{n} - T_m \biggr)\nonumber\\
&\quad = c_z n^{\alpha} \Big|\frac{i_{m, z} + u -r}{n} - T_m - \theta_z \Big|^{\alpha} f \biggl(\frac{i_{m, z} + u -r}{n} - T_m \biggr)\nonumber\\
&\quad = c_z | u - r + i_{m,z} - nT_m - n\theta_z |^{\alpha} f \biggl(\frac{u-r}{n} +n^{-1} (i_{m,z} - n T_m) \biggr)\nonumber\\
&\quad = c_z | u - r + 1 - \{nT_m + n\theta_z\} |^{\alpha} f \biggl(\frac{u-r}{n} +n^{-1} \bigl(n\theta_z + 1 - \{n T_m + n\theta_z\}\bigr) \biggr)\nonumber\\
&\quad = c_z | u - r + 1 - \{nT_m + n \theta_z\}|^{\alpha} f \biggl(\frac{u - r + 1 - \{n T_m + n\theta_z\}}{n} +\theta_z \biggr),\label{gdecomp}
\end{align}
for any $m \in\N$, $0 \leq r \leq k$ and $z \in\mathcal{A}$. Since
$f(x) \to1$ as $x \to\theta_z$, we find that for any $d \in\N$
\begin{align*}
& \biggl(n_j^{\alpha} g \biggl(\frac{i_{m, z} + u - r}{n_j} -
T_m \biggr) \biggr)_{|u|, m \leq d,\, 0\leq r \leq k,\, z \in\mathcal A}
\\
&\stab\bigl(c_z |u-r + 1 - \{U_m +
\eta_z\}|^{\alpha}\bigr)_{|u|, m \leq d,\, 0
\leq r \leq k,\, z \in\mathcal A},
\end{align*}
which holds due to condition~\eqref{nj}, decomposition~\eqref{gdecomp}
and Proposition~\ref{prop1} (for $\theta_0 = 0 $ we again need to
replace $|x|^{\alpha}$ by $x_+^{\alpha}$). Hence, by continuous mapping
theorem for stable convergence we deduce that
\begin{equation}
\label{gstab} \bigl(n_j^{\alpha} g_{i_{m,z} +u, n_j}(T_m)
\bigr)_{|u|, m \leq d,\, z \in
\mathcal A} \stab\bigl(c_z h_{k, z} \bigl(1 + u -
\{U_m + \eta_z\}\bigr) \bigr)_{|u|, m
\leq d,\, z \in\mathcal A}
\end{equation}
as $j \to\infty$,
which is a key result of the proof. We now define a truncated version
of $V_{n, \varepsilon, z}$ introduced in \eqref{Vnz}:
\[
V_{n,\varepsilon,z,d} := n^{\alpha p} \sum_{\substack{m \leq d: \\ T_m
\in(-\theta_z, 1 - \theta_z]}} |
\Delta L_{T_m}|^p \Biggl(\sum_{u =
-\lfloor\varepsilon d \rfloor}^{\lfloor\varepsilon d \rfloor+ v_m^z}
|g_{i_{m, z} + u, n}(T_m)|^p \Biggr).
\]
From \eqref{gstab} and properties of stable convergence we conclude that
\begin{equation}
\label{vndconv} (V_{n_j, \varepsilon, z, d} )_{z \in\mathcal{A}} \stab(V_{\varepsilon
, z, d})_{z \in\mathcal{A}}
\quad\text{as } j \to\infty,
\end{equation}
where
\[
V_{\varepsilon, z, d} = |c_z|^p \sum
_{\substack{m \leq d : \\ T_m \in
(-\theta_z, 1 - \theta_z]}} |\Delta L_{T_m}|^p \Biggl(\sum
_{u =
-\lfloor\varepsilon d \rfloor}^{\lfloor\varepsilon d \rfloor+ v_m^z} \big|h_{k,z} \bigl(1+u -
\{U_m +\eta_z\}\bigr)\big|^p \Biggr).
\]
Applying a monotone convergence argument, we deduce the almost sure convergence
\begin{equation}
\label{domconv} V_{\varepsilon, z, d} \uparrow V_z =
|c_z|^p \sum_{T_m \in(-\theta_z,
1 - \theta_z]} |\Delta
L_{T_m}|^p \biggl(\sum_{u \in\Z}
\big|h_{k,z} \bigl(1 + u - \{U_m + \eta_z\}
\bigr)\big|^p \biggr)
\end{equation}
as $d \to\infty$, where the second sum on the right-hand side is
finite, since $|h_{k, z}(x)| \leq C|x|^{\alpha- k}$ for large enough
$|x|$ and all $z \in\mathcal A$, and $\alpha< k - 1/p$. In view of
\eqref{vndconv} and \eqref{domconv}, we are left to prove the convergence
\[
\lim_{d \to\infty} \limsup_{n \to\infty} |
V_{n, \varepsilon, z, d} - V_{n, \varepsilon, z} | = 0
\]
on $\varOmega_{\varepsilon}$. Set $K_d = \sum_{m > d : T_m \in(-\theta_z,
1 - \theta_z]} |\Delta L_{T_m}|^p$ and observe that $K_d \to0$ as $d
\to\infty$, since $L$ is a compound Poisson process. Due to Lemma~\ref
{lem1} we conclude that $|n^{\alpha} g_{i, n}(x)| \leq C \min\{1, |i/n
- x|^{\alpha- k}\}$ and thus
\[
|V_{n, \varepsilon, z, d} - V_{n, \varepsilon, z} | \leq C \biggl(K_d + \sum
_{|u| > \lfloor\varepsilon d \rfloor} |u|^{p(\alpha- k)} \biggr) \quad\text{for all
} z \in\mathcal{A},
\]
and the latter converges to $0$ almost surely as $d \to\infty$,
because $\alpha< k - 1/p$. Consequently, we have shown \eqref{toshow}.
\qed

\subsubsection{Main terms in Theorem~\ref{th2}\ref{it:th2:2}}
\label{sec3.1.3}
We start with a simple lemma.

\begin{lemma}
\label{lem2}
Let $(a_i)_{i \in\N}$ be a sequence of positive real numbers such that\break
$\lim_{i \to\infty} ia_i =1$. Then it holds that
\[
\lim_{n \to\infty} \frac{1}{\log(n)} \sum
_{i = 1}^{cn} a_i =1
\]
for any fixed $c \in\N$.
\end{lemma}

\begin{proof}
Due to the assumption of the lemma, we have that $(a_i)_{i \in\N}$ is
a bounded sequence and for each $\epsilon> 0$ there exists an $N =
N(\epsilon)$ with
\[
|a_i - i^{-1}| \leq\epsilon i^{-1} \quad
\text{for all } i \geq N.
\]
It obviously holds that $\lim_{n \to\infty} \sum_{i = 1}^{cn} i^{-1} /
\log(n) = 1$. On the other hand, we obtain that
\[
\limsup_{n \to\infty} \frac{1}{\log(n)} \sum
_{i = N}^{cn} |a_i - i^{-1}|
\leq\epsilon\limsup_{n \to\infty} \frac{1}{\log(n)} \sum
_{i
= 1}^{cn} i^{-1} = \epsilon.
\]
Since $\epsilon> 0$ is arbitrary, we conclude the statement of
Lemma~\ref{lem2}.
\end{proof}
Now, we will again use the decomposition \eqref{Vnz}, which
holds on $\varOmega_{\varepsilon}$, and treat each term $V_{n, \varepsilon
, z}$ separately. We consider $z \geq1$ and we will show that
\begin{equation}
\label{apprbyh} \frac{1}{\log(n)}\sum_{u = -\lfloor n\varepsilon\rfloor}^{\lfloor
n\varepsilon\rfloor+ v_m^z}
\big|n^{\alpha}g_{i_{m, z} + u, n}(T_m) - c_z
h_{k, z}\bigl(u + 1 - \{nT_m + n \theta_z\}
\bigr) \big|^p \to0
\end{equation}
as $n \to\infty$, for any $m \in\N$. Let us first consider the case
$|u| \geq k$. Recall that we have assumed that $f_z(x) = g(x) / |x -
\theta_z|^{\alpha}$ is in $C^k((\theta_z - \delta, \theta_z + \delta))$
for any $\delta< \max_{1 \leq j \leq l}(\theta_j - \theta_{j- 1})$.
Now, due to identity \eqref{gdecomp} and Taylor expansion of order $k$,
we obtain the bound (cf. \cite[Eqs.~(4.8) and~(4.9)]{BP17})
\[
\sum_{u = -\lfloor n\varepsilon\rfloor}^{\lfloor n\varepsilon\rfloor
+ v_m^z} \big|n^{\alpha}g_{i_{m,z} + u,n}(T_m)
- c_z h_{k,z}\bigl(u+1-\{ nT_m + n
\theta_z\}\bigr) \big|^p \1_{\{|u| \geq k \}} \leq C,
\]
for any $\varepsilon< \max_{1 \leq j \leq l}(\theta_j - \theta_{j -
1})$. Since $|n^{\alpha} g_{i_{m, z} + u, n}(T_m)|$ is bounded for any
$|u| < k$ due to Lemma~\ref{lem1}, we deduce the convergence in \eqref{apprbyh}.

Next, for large enough $|u|$ we observe the bounds
\[
|q_{k, \alpha}|^p a_u \leq\big|h_{k, z}
\bigl(u + 1 - \{nT_m + n \theta_z\}\bigr)\big|^p
\leq|q_{k, \alpha}|^p a_{u - k - 1} \quad\text{where }
a_u = |u|^{-1}.
\]
Hence, by Lemma~\ref{lem2}, we conclude the convergence
\begin{equation}
\label{convforh} \frac{1}{\log(n)}\sum_{u = -\lfloor n\varepsilon\rfloor}^{\lfloor
n\varepsilon\rfloor+ v_m^z}
\big| h_{k, z}\bigl(u + 1 - \{nT_m + n \theta_z\}
\bigr) \big|^p \to2|q_{k, \alpha}|^p \quad\text{as } n\to
\infty.
\end{equation}
The same statement holds for $z=0$, but the limit becomes $|q_{k,\alpha
}|^p$, since in this setting the sum is one-sided.
We set $\| x\|_p^{p} = \sum_{i=1}^m |x_i|^p$ for any $x \in\R^m$ and
$p>0$, and recall that $\| x\|_p$ is a norm for $p \geq1$. It holds that
\begin{equation}
\label{ineq}
\begin{aligned} |\| x\|_p^p -
\| y\|_p^p| &\leq\| x - y \|_p^p
\quad\text{ when } p \in(0, 1],
\\
|\| x\|_p - \| y\|_p| &\leq\| x - y\|_p
\quad\text{ when } p > 1. \end{aligned}
\end{equation}
By \eqref{apprbyh}, \eqref{convforh} and~\eqref{ineq}, and taking into
account the definition of $V_{n, \varepsilon, z}$ at \eqref{Vnz}, we
readily deduce the convergence
\[
\frac{V_{n, \varepsilon,z}}{\log(n)} \toop|q_{k, \alpha} c_z|^p (1 +
\1 _{\{z\geq1\}}) \sum_{m : T_m \in[-\theta_z, 1 - \theta_z]} |\Delta
L_{T_m}|^p
\]
as $n \to\infty$, and hence
\[
n^{\alpha p}\sum_{i = k}^{n} \Bigg|\sum
_{z = 0}^{l} M_{i, n,
\varepsilon, z}
\Bigg|^p \toop|q_{k,\alpha} |^p \sum
_{z = 0}^l |c_z|^p (1 +
\1_{\{z \geq1\}} ) \sum_{m : T_m \in[-\theta_z, 1- \theta
_z]} |\Delta
L_{T_m}|^p
\]
as $n \to\infty$, on $\varOmega_{\varepsilon}$.\qed

\subsubsection{Negligible terms}
\label{sec3.1.1}
Due to inequalities at \eqref{ineq}, it suffices to show that on $\varOmega
_{\varepsilon}$
\begin{equation}
\label{neglconv} a_n \sum_{i = k}^{n}
|R_{i, n, \varepsilon}|^p \toop0 \quad\text {and} \quad a_n
\sum_{i = k}^{n} |M_{i, n, \varepsilon,z}|^p
\toop0 \quad\text{for } z \in \mathcal{A}^c,
\end{equation}
as $n \to\infty$, where $a_n = n^{\alpha p}$ in Theorem~\ref{th2}\ref
{it:th2:1} and $a_n = n^{\alpha p}/ \log(n)$ in Theorem~\ref{th2}\ref
{it:th2:2},
and this will prove that these terms do not affect the limits in
Theorem~\ref{th2}. At this stage we notice that outside the singularity
points the kernel function $g$ satisfies the same properties under
Assumption~(B) (resp. Assumption~(B-log)) as under Assumption~(A)
(resp. Assumption~(A-log)). Consequently, we can apply the estimates
for the term $R_{i, n, \varepsilon}$ derived in \cite[Eqs.~(4.8)
and~(4.12)]{BLP17} and \cite[Section 4]{BP17} under conditions~(A) and~(A-log)
\begin{align*}
\sup_{n \in\N, i = k, \ldots, n} n^k |R_{i, n, \varepsilon}| &< \infty
\text{ almost surely if } w \in(0, 1],
\\
\sup_{n \in\N, i = k, \ldots, n} \frac{n^k |R_{i, n, \varepsilon
}|}{(\log(n))^q} &< \infty\text{ almost
surely if } w \in(1,2],
\end{align*}
where $q$ is determined via $1 / q + 1 / w = 1$, since $R_{i, n,
\varepsilon}$ is only affected by the function $g$ outside the
singularity points $\theta_z$. We readily conclude the first
convergence at \eqref{neglconv} in the setting of Theorem~\ref{th2}\ref
{it:th2:1}, because $\alpha<k-1/p$. It also holds in the setting of
Theorem~\ref{th2}\ref{it:th2:2}, where for $w \in(1,2]$ we use the
assumption that $1 / p + 1 / w > 1$.

Now, we show the second statement of \eqref{neglconv}, which is only
relevant in the setting of Theorem~\ref{th2}\ref{it:th2:1}. Since
$\alpha_z < k -1/p$ for all $z$, we can apply to $\sum_{i = k}^{n}
|M_{i, n, \varepsilon, z}|^p$, $z \in\mathcal{A}^c$, the same
techniques as for $\sum_{i = k}^{n} |M_{i, n, \varepsilon, z}|^p$, $z
\in\mathcal{A}$. Hence, using the same methods as in Section~\ref
{sec3.1.2}, we conclude that on $\varOmega_{\varepsilon}$
\[
n^{\alpha p} \sum_{i = k}^{n}
|M_{i, n, \varepsilon, z}|^p = O_{\mathbb
{P}} \bigl(n^{p(\alpha- \alpha_z)}
\bigr) \quad\text{for all } z \in \mathcal{A}^c,
\]
where the notation $Y_n = O_{\mathbb{P}} (a_n)$ means that the sequence
$a_n^{-1} Y_n$ is tight. Since $\alpha_z > \alpha$ for all $z \in
\mathcal{A}^c$, we obtain the second statement of \eqref{neglconv}. The
results of Sections~\ref{sec3.1.2}--\ref{sec3.1.1} and the fact that
$\varOmega_{\varepsilon} \uparrow\varOmega$ as $\varepsilon\to0$ imply the
assertion of Theorem~\ref{th2} in the compound Poisson case.\qed

\subsection{Proof of Theorem~\ref{th2} in the general case}
\label{sec3.2}
Let now $(L_t)_{t\in\R}$ be a general symmetric pure jump L\'evy
process with Blumenthal--Getoor index $\beta$. We denote by $N$ the
corresponding Poisson random measure defined by $N(A) := \#\{t \in\R:
(t, \Delta L_t) \in A\}$ for all measurable $A \subseteq\R\times(\R
\setminus\{0\})$. Next, we introduce the process
\[
X_t(m) = \int_{(-\infty, t] \times[-\frac{1}{m}, \frac{1}{m}]} x \bigl(g(t - s) -
g_0(-s) \bigr) \, N(\di s, \di x),
\]
which only involves small jumps of $L$. We will prove that
\begin{equation}
\label{smalljumps} \lim_{m \to\infty} \limsup_{n \to\infty}
\mathbb{P} \bigl( a_n V\bigl(X(m), p; k\bigr)_n > \epsilon
\bigr) = 0 \quad\text{for any } \epsilon>0,
\end{equation}
where $a_n = n^{\alpha p}$ in Theorem~\ref{th2}\ref{it:th2:1} and $a_n
= n^{\alpha p}/ \log(n)$ in Theorem~\ref{th2}\ref{it:th2:2}. First, due
to Markov's~inequality and the stationary increments of $X_t(m)$, it
follows that
\[
\mathbb{P}\bigl(a_n V\bigl(X(m), p; k\bigr)_n > \epsilon
\bigr) \leq\epsilon^{-1} a_n \sum
_{i = k}^{n} \E\bigl[| \Delta_{i, k}^n
X(m) |^p\bigr] \leq\epsilon^{-1} b_n \E\bigl[|
\Delta_{k, k}^n X(m) |^p\bigr],
\]
where $b_n = n a_n$. Hence it is enough to prove that
\begin{equation}
\label{eq:Ynm} \lim_{m \to\infty} \limsup_{n \to\infty} \E
\bigl[| Y_{n, m} |^p\bigr] = 0 \quad \text{where} \quad
Y_{n, m} = b_n^{1 / p} \Delta_{k, k}^n
X(m).
\end{equation}
Notice the representation
\[
Y_{n, m} = \int_{(-\infty, \frac{k}{n}] \times[-\frac{1}{m}, \frac
{1}{m}]} \bigl(b_n^{1/p}
g_{k, n}(s)\bigr) x \, N(\di s, \di x).
\]
Using this together with \cite[Theorem~3.3]{RR89}, \eqref{eq:Ynm} will
follow if
\begin{align*}
&\lim_{m \to\infty} \limsup_{n \to\infty}
\xi_{n, m} = 0 \quad\text {where} \quad\xi_{n, m} = \int
_{| x | \leq\frac{1}{m}} \chi_n(x) \, \nu(\di x) \quad\text{and}
\\
&\chi_n(x) = \int_{-\infty}^{\frac{k}{n}} \bigl( |
b_n^{1 / p} g_{k,
n}(s) x |^p
\1_{\{| b_n^{1 / p} g_{k,n}(s) x | \geq1 \}}
\\
&\quad\ \qquad+ | b_n^{1 / p} g_{k, n}(s) x
|^2 \1_{\{| b_n^{1 /
p} g_{k,n}(s) x | < 1 \}} \bigr) \, \di s.
\end{align*}
Suppose there exists a constant $K \geq0$ such that for all large $n
\in\N$
\begin{equation}
\label{eq:chi_leq} \chi_n(x) \leq K \bigl(| x |^p + | x
|^2\bigr) \quad\text{for all $x \in[-1, 1]$},
\end{equation}
then the dominated convergence theorem implies that
\[
\limsup_{m \to\infty} \Bigl[\limsup_{n \to\infty}
\xi_{n, m} \Bigr] \leq K \limsup_{m \to\infty} \int
_{| x | \leq\frac{1}{m}} \bigl(| x |^p + | x |^2\bigr)
\, \nu(\di x) = 0,
\]
using the assumption that $p > \beta$. We consider only \eqref
{eq:chi_leq} in the case of Theorem~\ref{th2}\ref{it:th2:1} as~\ref
{it:th2:2} is very similar, see \cite{BP17}. In the case of~\ref
{it:th2:1} then $\smash{b_n^{1/p}} = n^{\alpha+ 1 / p}$. For short
notation define $\varPhi_p : \R\to\R_+$ as the function
\[
\varPhi_p(y) = | y |^2 \1_{\{| y | \leq1\}} + | y
|^p \1_{\{| y | > 1\}}, \quad y \in\R.
\]
Note that $\varPhi_p$ is of modular growth, i.e. there exists a constant
$K_p > 0$ depending only on $p$ such that
$\varPhi_p(x + y) \leq K_p(\varPhi_p(x) + \varPhi_p(y))$
for any $x,y \in\R$. We consider the
following decomposition
\begin{align*}
\chi_n(x) &= \int_{\frac{k}{n} - \frac{1}{n}}^{\frac{k}{n}} \varPhi
_p\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, \di s + \sum
_{z = 1}^{l} \int_{\frac{k}{n} - \theta_z - \frac{1}{n}}^{\frac{k}{n} - \theta_z + \frac
{1}{n}}
\varPhi_p\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, \di s
\\
&\quad+ \sum_{z = 1}^{l} \int
_{\frac{k}{n} - \theta_z + \frac
{1}{n}}^{\frac{k}{n} - \theta_{z-1} - \frac{1}{n}} \varPhi_p
\bigl(n^{\alpha+
1/p} g_{k, n}(s) x\bigr) \, \di s
\\
&\quad+ \int_{\frac{k}{n} - \theta_l - \delta}^{\frac{k}{n} - \theta_l
- \frac{1}{n}} \varPhi_p
\bigl(n^{\alpha+ 1/p} g_{k, n}(s)x\bigr) \, \di s
\\
&\quad+ \int_{-\infty}^{\frac{k}{n} - \theta_l - \delta} \varPhi _p
\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, \di s
\\
&=: I_0(x) + \sum_{z = 1}^{l}
I_{1, z}(x) + \sum_{z = 1}^{l}
I_{2,
z}(x) + I_3(x) + I_4(x).
\end{align*}
We treat the five types of terms separately.

\paragraph{Estimation of $I_0$} By Lemma~\ref{lem1}
\[
| g_{k, n}(x) | \leq K \bigl(| \tfrac{k}{n} - s |^{\alpha_0}
\bigr) \quad\text {for all $s \in\bigl[\tfrac{k}{n} - \tfrac{1}{n},
\tfrac{k}{n}\bigr]$}.
\]
Since $\varPhi_p$ is increasing on $\R_+$ and $\alpha\leq\alpha_0$ it
follows that
\[
I_0(x) \leq K \int_{0}^{\frac{1}{n}}
\varPhi_p\bigl(x n^{\alpha+ 1/p} s^{\alpha
_0} \bigr) \, \di s
\leq K \int_{0}^{\frac{1}{n}} \varPhi_p\bigl(x
n^{\alpha+ 1/p} s^{\alpha} \bigr) \, \di s.
\]
By elementary integration it follows that
\begin{align*}
&\int_{0}^{\frac{1}{n}} | x n^{\alpha+ 1/p}
s^{\alpha} |^2 \1_{\{ | x
n^{\alpha+ 1/p} s^\alpha| \leq1\}} \, \di s
\\
&\quad\leq K \bigl(x^2 \1_{\{ | x | \leq n^{-1/p} \}} n^{2/p - 1} +
\1_{\{
| x | > n^{-1 / p} \}} | x |^{-1/\alpha} n^{-1 - 1/(\alpha p)}\bigr)
\\
&\quad\leq K\bigl(x^2 + | x |^p\bigr).
\end{align*}
The second term in $\varPhi_p$ is dealt with as follows:
\[
\int_{0}^{\frac{1}{n}} | x n^{\alpha+ 1/p}
s^{\alpha} |^p \1_{\{| x
n^{\alpha+ 1/p} s^{\alpha} | > 1\}} \, \di s \leq| x
|^p n^{\alpha p
+ 1} \int_{0}^{\frac{1}{n}}
s^{\alpha p} \, \di s = \frac{| x
|^p}{\alpha p + 1}.
\]
Combining the two estimates above it follows that $I_0(x) \leq K(|x|^2
+ |x|^p)$.

\paragraph{Estimation of $I_{1, z}$} Similarly as for $I_0$, we have,
using arguments as in part~\ref{it:lem1:1} of Lemma~\ref{lem1}, that
\[
| g_{k,n}(s) | \leq K \sum_{j = 0}^{k}
| \tfrac{k - j}{n} - s - \theta _z |^{\alpha_z} \quad\text{for
all $s \in\bigl[\tfrac{k}{n} - \theta_z - \tfrac{1}{n},
\tfrac{k}{n} - \theta_z + \tfrac{1}{n}\bigr]$}.
\]
Using the modular growth of $\varPhi_p$ it follows that
\begin{align*}
&\int_{\frac{k}{n} - \theta_z - \frac{1}{n}}^{\frac{k}{n} - \theta_z +
\frac{1}{n}} \varPhi_p
\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, d s
\\
&\quad\leq K_p \sum_{j = 0}^{k}
\int_{\frac{k}{n} - \theta_z - \frac
{1}{n}}^{\frac{k}{n} - \theta_z + \frac{1}{n}} \varPhi_p
\bigl(n^{\alpha+ 1/p} |\tfrac{k - j}{n} - s - \theta_z|^{\alpha_z}
x\bigr) \, \di s
\\
&\quad= K_p \sum_{j = 0}^{k}
\int_{-\frac{j}{n} - \frac{1}{n}}^{-\frac
{j}{n} + \frac{1}{n}} \varPhi_p
\bigl(n^{\alpha+ 1/p} |s|^{\alpha_z} x\bigr) \, \di s
\\
&\quad\leq K_p \int_{-\frac{k + 1}{n}}^{\frac{k + 1}{n}} \varPhi
_p\bigl(n^{\alpha+ 1/p} |s|^{\alpha} x\bigr) \, \di s
\\
&\quad= K_p \int_{0}^{\frac{k + 1}{n}}
\varPhi_p\bigl(n^{\alpha+ 1/p} |s|^{\alpha} x\bigr) \, \di s.
\end{align*}
As for $I_0$, we get $I_{1, z}(x) \leq K(|x|^2 + |x|^p)$.

\paragraph{Estimation of $I_{2, z}$} We decompose $I_{2, z}$ into three
terms corresponding to whether we are close to the singularity $\theta
_z$ from the right or close to the singularity $\theta_{z-1}$ from the
left or in between them, but bounded away from both. More specifically,
we decompose as
\begin{align*}
I_{2,z}(x) &= \int_{\frac{k}{n} - \theta_z + \frac{1}{n}}^{\frac{k}{n}
- \theta_z + \delta}
\varPhi_p\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, \di s
\\
&\quad+ \int_{\frac{k}{n} - \theta_z + \delta}^{\frac{k}{n} - \theta_{z
- 1} - \delta} \varPhi_p
\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, \di s
\\
&\quad+ \int_{\frac{k}{n} - \theta_{z-1} - \delta}^{\frac{k}{n} - \theta
_{z - 1} - \frac{1}{n}} \varPhi_p
\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, \di s =:
I_{2,z}^{l}(x) + I_{2,z}^{b}(x) +
I_{2,z}^{r}(x).
\end{align*}
First we note that arguments similar to Lemma~\ref{lem1}\ref{it:lem1:3}
imply that
\[
| g_{k, n}(s) | \leq K n^{-k} |\tfrac{k}{n} - s -
\theta_z|^{\alpha_z -
k} \quad\text{for all $s \in\bigl[
\tfrac{k}{n} - \theta_z + \tfrac{1}{n}, \tfrac{k}{n} -
\theta_z + \delta\bigr]$.}
\]
Using again that $\varPhi_p$ is decreasing on $\R_+$ it follows that
\begin{align*}
I_{2, z}^{l}(x) &\leq K \int_{\frac{k}{n} - \theta_z + \frac
{1}{n}}^{\frac{k}{n} - \theta_z + \delta}
\varPhi_p\bigl(n^{\alpha+ 1/p - k} | \tfrac{k}{n} - s -
\theta_z |^{\alpha_z - k} x\bigr) \, \di s
\\
&\leq K \int_{\frac{1}{n}}^{\delta} \varPhi_p
\bigl(n^{\alpha+ 1/p - k} |s|^{\alpha_z - k} x\bigr) \, \di s.
\end{align*}
If $\alpha_z = k - 1/2$ then
\begin{align*}
\int_{\frac{1}{n}}^{\delta} | x n^{\alpha+ 1/p - k}
s^{\alpha_z -
k}|^2 \1_{\{ | x^2 n^{\alpha+ 1/p - k} s^{\alpha_z - k} | \leq1 \}} \, \di s &\leq
x^2 n^{2(\alpha+ 1/p - k)} \int_{\frac{1}{n}}^{\delta}
s^{-1} \, \di s
\\
&\leq K x^2,
\end{align*}
where we used that $\alpha< k - 1/p$. For $\alpha_z \neq k - 1 / 2$ we
have that
\begin{align*}
&\int_{\frac{1}{n}}^{\delta} | x n^{\alpha+ 1/p - k}s^{\alpha_z -k}|^2 \1_{\{| x n^{\alpha+ 1/p - k} s^{\alpha_z - k}| \leq1\}} \, \di s\\
&\quad \leq K\bigl(|x|^2 n^{2(\alpha+ 1/p - k)} + |x|^2 n^{2(\alpha- \alpha_z) + 2/p - 1} \1_{\{| x | \leq n^{-1/p} \}}\\
&\qquad+ |x|^{\frac{1}{k - \alpha_z}} n^{\frac{\alpha+ 1/p - k}{k - \alpha_z}} \1_{\{|x| > n^{-1/p}\}}\bigr)\\
&\quad \leq K\bigl(x^2 + |x|^p\bigr),
\end{align*}
where we used that $\alpha\leq\alpha_z < k - 1/p$. Moreover,
\[
\int_{\frac{1}{n}}^{\delta} |x n^{\alpha+ 1/p - k}
s^{\alpha_z - k}|^p \1_{\{| x n^{\alpha+ 1/p - k} s^{\alpha_z - k}| > 1\}} \, \di s \leq K
|x|^p.
\]
The term $I_{2,z}^{r}$ is handled similarly. For the last term $I_{2,
z}^b$ we note that, since we are bounded away from both $\theta_{z-1}$
and $\theta_z$, there exists a constant $K > 0$ such that
\begin{equation*}
|g_{k,n}(s)| \leq K n^{-k} \quad\text{for all $s \in\bigl[
\tfrac{k}{n} - \theta_z + \delta, \tfrac{k}{n} -
\theta_{z-1} - \delta\bigr]$.}
\end{equation*}
This readily implies the bound $I_{2,z}^{b}(x) \leq K(x^2 + |x|^p)$.

\paragraph{Estimation of $I_3$} Arguments as in Lemma~\ref{lem1} imply that
\begin{equation*}
|g_{k, n}(s)| \leq K n^{-k}|\tfrac{k}{n} - s -
\theta_z|^{\alpha_l - k} \quad\text{for all $s \in\bigl[
\tfrac{k}{n} - \theta_l - \delta, \tfrac
{k}{n} -
\theta_l - \tfrac{1}{n}\bigr]$.}
\end{equation*}
One may then proceed as for the term $I_{2,z}^l$ above to conclude that
$I_3(x) \leq K(x^2 + |x|^p)$.

\paragraph{Estimation of $I_4$:} First we decompose the integral\querymark{Q5} into two sub-integrals:
\begin{align*}
\int_{-\infty}^{\frac{k}{n} - \theta_l - \delta} \varPhi_p
\bigl(n^{\alpha+
1/p} g_{k, n}(s) x\bigr) \, \di s &= \int
_{-\delta' - \theta_l}^{\frac{k}{n}
- \delta- \theta_l} \varPhi_p
\bigl(n^{\alpha+ 1/p} g_{k, n}(s)x \bigr) \, \di s
\\
&\quad+ \int_{-\infty}^{-\delta' - \theta_l} \varPhi_p
\bigl(n^{\alpha+ 1/p} g_{k, n}(s) x\bigr) \, \di s.
\end{align*}
In the first integral we are bounded away from $\theta_l$, hence
$|g_{k,n}(s)| \leq K n^{-k}$ for all $s$ in the interval $[-\delta' -
\theta_l, \frac{k}{n} - \delta- \theta_l]$.
For the latter integral note first that by Lemma~\ref{lem1}\ref{it:lem1:5}
\begin{equation*}
\int_{-\infty}^{-\delta' - \theta_l} \varPhi_p
\bigl(n^{\alpha+ 1/p} g_{k,
n}(s) x\bigr) \, \di s \leq\int
_{-\infty}^{-\delta' - \theta_l} \varPhi _p
\bigl(n^{\alpha+ 1/p - k} |g^{(k)}(-s)| x\bigr) \, \di s.
\end{equation*}
Now
\begin{align*}
&\int_{\delta' + \theta_l}^{\infty} | x n^{\alpha+ 1/p - k} g^{(k)}(s) |^2 \1_{\{ |x n^{\alpha+ 1/p - k} g^{(k)}(s) | \leq1 \}} \, \di s \\
&\quad \leq|x n^{\alpha+ 1/p - k}|^2 \int_{\delta' + \theta_l}^{\infty}|g^{(k)}(s)|^2 \, \di s.
\end{align*}
Since $|g^{(k)}|$ is decreasing on $(\theta_l + \delta', \infty)$ and
$g^{(k)} \in L^w((\theta_l + \delta', \infty))$ for some $w \leq2$ it
follows that the last integral is finite. Lastly, we find for $x \in
[-1 , 1]$ that
\begin{align*}
&\int_{\theta_l + \delta'}^{\infty} |x n^{\alpha+ 1/p - k}
g^{(k)}(s)|^p \1_{\{|x n^{\alpha+ 1/p - k} g^{(k)}(s)| > 1\}} \, \di s
\\
&\quad \leq|x|^p n^{p(\alpha+ 1/p - k)} \int_{\delta' + \theta_l}^{\infty}
|g^{(k)}(s)|^p \1_{\{|g^{(k)}(s)| > 1 \}} \, \di s.
\end{align*}
By our assumptions the last integral is finite, indeed
\begin{equation*}
\int_{\delta' + \theta_l}^{\infty} |g^{(k)}(s)|^p
\1_{\{|g^{(k)}(s)| >
1\}} \, \di s \leq K_p \| g^{(k)}
\|_{L^w((\delta' + \theta, \infty
))}^w < \infty.
\end{equation*}

\subsubsection{Negligibility of small jumps}

Now, we note that $X_t - X_t(m)$ is the integral \eqref{Xmodel}, where
the integrator is a compound Poisson process that corresponds to big
jumps of $L$. Hence, we obtain the results of Theorem~\ref{th2} for the
process $X-X(m)$ as in Section~\ref{sec3.1}. More specifically, under
assumptions of Theorem~\ref{th2}\ref{it:th2:1} it holds that
\[
n_j^{\alpha p}V\bigl(X - X(m), p; k\bigr)_{n_j} \stab
\sum_{z \in\mathcal A} |c_z|^p \sum
_{r : T_r \in[-\theta_z, 1 - \theta_z]} |\Delta L_{T_r}|^p
\1_{\{|\Delta L_{T_r}| > \frac{1}{m}\}} V_r^z
\]
where $V_r^z$ has been defined at \eqref{part1.2}. The term on the
right-hand side converges to the limit of Theorem~\ref{th2}\ref
{it:th2:1} as $m \to\infty$, since
\begin{equation*}
\sum_{r : T_r \in[-\theta_z, 1 - \theta_z]} |\Delta L_{T_r}|^p
< \infty\quad\text{for any $p > \beta$.}
\end{equation*}
Finally, using the decomposition $X=(X-X(m)) + X(m)$ and letting first
$n_j \to\infty$ and then $m\to\infty$, we deduce the statement of
Theorem~\ref{th2} by \eqref{smalljumps} and the inequalities \eqref
{ineq}. This completes the proof.\qed

\begin{acknowledgement}[title={Acknowledgments}]
The authors acknowledge financial support from the project
``Ambit fields: probabilistic properties and statistical inference''
funded by Villum Fonden.
\end{acknowledgement}

redpajama_set_name: RedPajamaArXiv


text:
Dr. Elemir Simko is the winner of this year's Provost's Outstanding Teaching Award.
The tops in teachers
Dr. Elemir Simko (DVM, DVSc,) believes that if you do what you love, you'll never work a day in your life.
Jeanette Neufeld
The board-certified veterinary anatomic pathologist has been part of the WCVM faculty for nearly 23 years. He is a professor in the college's Department of Veterinary Pathology and leads the college's thriving honey bee health research group.
Simko, who is Ukrainian/Rusyn and was born in the former Yugoslavia, began beekeeping in his teenage years. His hobby eventually turned into an integral part of his decades-long research career.
During his time at the WCVM, Simko has become a beloved teacher of veterinary pathology for both undergraduate and graduate students and is well known for his rigorous and encouraging teaching style.
In recognition of his dedication to his students, Simko will receive the 2022 Provost's Outstanding Teaching Award on June 1.
This annual award celebrates exceptional faculty members who demonstrate excellence in teaching and dedication to their students.
Simko says that part of what motivates him as a teacher is to teach as he would like to be taught.
"At our college, we have the best students from three western provinces and the northern territories. They are hungry for knowledge, eager to learn and ready to be challenged," says Simko, who also received the USask Provost's College Award for Outstanding Teaching in 2014.
He believes his pupils often develop a "love-hate relationship" with him as a teacher. No matter how they feel about him, Simko's goal is to provide all of his students with a strong foundation in veterinary pathology for their medical education and professional career.
"Dr. Simko has found a way to teach veterinary pathology in a manner that ensures students are exceptionally prepared to enter the world of veterinary practice," says Dr. Susantha Gomis, head of the WCVM's Department of Veterinary Pathology.
In addition to the positive feedback and success of his students, Gomis says Simko's honey bee work plays an important role in sustainable agriculture — pollinators such as honey bees have an integral role in crop production.
In response to the need for veterinarians to oversee antimicrobial use in honey bees, Simko created the first honey bee health program for veterinary students at any North American veterinary college.
Q & A with Dr. Elemir Simko
What motivates, inspires or excites you about your work?
A sparkle in a student's eyes when they learn something new, make a first discovery or publish a first paper. There is no greater professional satisfaction than helping students to achieve their goals.
Tell us about a mentor or supporter who made an impact on your career.
I was blessed to have many great teachers and mentors; those who encouraged or even demanded my maximal efforts and performance have "a special place" in my memory, and I try to imitate and emulate them. Dr. Brian Wilcock and Dr. Tony Hayes — both professors emeriti at the Ontario Veterinary College — had a profound impact on my teaching and research, respectively.
What is one piece of advice that you would give to students, colleagues or others in your field?
"Docendo discimus" (Seneca, 4 BC-AD 65). "By teaching, we learn."

redpajama_set_name: RedPajamaCommonCrawl


Now the fun part begins. It's time to find the perfect home!
Together, we will sit down to discuss your list of needs, as well as your list of wants. I will then begin a search in the MLS® system to send you all of the listings that fit your criteria. I will also continue to send you new listings as they come on the market. These listings will be sent to you in a custom Client Portal where you'll have the ability to Favorite or Trash listings, you'll also be able to mark them as Possibilities or leave comments or questions for me regarding specific listings.


text:
Doc's Computin' Tips: Recent CPU advances
We're at kind of a juncture right now as different CPU configurations make market inroads. Traditionally, we've just had one big CPU chip and every effort by the manufacturers was thrown into making it faster and faster.
However, because an increase in speed means an accompanying increase in heat, the industry has hit something of a wall lately and now you're hearing about "dual-core" and "quad-core" CPU chips as the manufacturers take a fresh approach.
In brief, a dual-core CPU is basically two CPU chips in one, and certain applications will correspondingly run twice as fast. Quad-core CPUs are another doubling up of CPU power, although in general they only increase certain functions 25% over a dual-core system and are considered something of a 'marketing gimmick' by us geeks.
To note is that it's up to the programs, themselves, as to whether or not they can utilize the multiple CPUs.
Here are two video compression programs that do the exact same thing:
And, as expected, the first one does the chore in half the time, ten minutes compared to twenty. Run it six times and that's an hour saved.
As to identifying your own system, simply right-click on the Task Bar, open Task Manager, click on the 'Performance' tab and look. If you have two windows, like the above, it's a dual-core system.
As such, if you do any kind of routine CPU-intensive process, you might want to take a peek at ye olde Task Manager and make sure the program's up to speed � literally. In the case up above, I had used DVD2One for years, but as soon as I bought the new dual-core rig and saw how DVD2One performed in Task Manager, it went straight to the scrap heap. If it had just been one of those 'percentage stories', where one program outperforms another by a blistering 2.38%, I wouldn't have bothered.
But twice just can't be argued with.
Posted by Dr. Mercury in Dr. Mercury's Computer Corner, Our Essays at 12:55 | Comments (2) | Trackbacks (0)
Doc's Computin' Tips: Index
Here's an index of my Maggie's Computin' Tips. I can't guarantee all of these will work on every Windows operating system, but most should. Programs AutoSizer � This has two great uses. It'll pop open the browser (or any program) in full-s
Weblog: Maggie's Farm
Tracked: Jul 10, 11:22
Here's an index of my Maggie's Computin' Tips. I can't guarantee all of these will work on every Windows operating system, but most should. Programs AutoSizer � This has two great uses. It'll pop open the browser (or any program) in full-scr
Tracked: Aug 28, 08:31
Here's an index of my Maggie's Computin' Tips. I can't guarantee all of these will work on every Windows operating system, but most should. Programs AutoSizer — This has two great uses. It'll pop open the browser (or any program) in full
Tracked: Jan 08, 20:00
Not all computers with a task manager that shows two processor windows actually have two processors. There were some Intel P4's from about 5 years ago that did Hyperthreading, which supposedly used unused parts of the processor to mimic a second processor. It only increased speed about 5%.
The speed increase for 2 processors is more likely to be between 60 and 80% rather then a doubling of speed if the program can handle two processors, because there is contention for other resources, like on die shared memory and computer resources like main memory and hard drive access which run at the same old speed no matter how many procs you may have.
#1 Mark on 2010-04-22 17:50 (Reply)
Mark - That Hyperthread bullshit actually dared to use two windows in Task Manager? Pretty cheeky of them. I have a good buddy who got suckered into it when they came out. I try to be a friend and not remind him constantly of this black mark on his record. That's what being friends is all about.
In reply to [ Top level ]#1: Mark on 2010-04-22 17:50 #2: Dr. Mercury on 2010-04-22 17:57


text:
CME survey points to bottom of recession, slow recovery
End of the recession?
Not yet, but amidst a sense of continued stability, Canadian manufacturers and exporters are optimistic that this economic warming effect will continue into the fall, according to the data in Canadian Manufacturers & Exporters' July Business Conditions Survey.
"The good news is that it appears the recession is bottoming out, with a number of firms seeing orders remain the same or increase over the next three months, but it's premature to say the sector is on the way to a strong recovery," says CME president and CEO Jayson Myers.
The July data reinforces the signs of stability that appeared in the last three monthly surveys.
This month, 583 companies participated in the survey conducted the first three weeks of July. While most manufacturers and exporters surveyed this month report the value of their orders is lower than it was three months ago, there is a strong indication that the downward slide has stabilized.
Over this period, 53 per cent of companies report that orders have fallen in value compared to three months ago, a 12 per cent improvement since April.
The positive trend identified for new orders the past two months continues. A majority – 71 per cent – of firms report that they expect the value of new orders to stay the same or increase in value over the next three months, a slight increase of two per cent from June.
"According to this data, there may be increased gains in production levels in the final quarter," adds Myers.
But employment prospects, while stabilized, are not as bullish. Only 12 per cent of companies expect to increase employment the next three months, which is on par with the results from the previous four surveys." The number of firms who are planning layoffs hovers around 21 per cent."
And the credit crunch is still a major obstacle for manufacturers and exporters. Still, 73 per cent of manufacturers and exporters report that they are experiencing difficulties in accessing or are unable to access financing; with the hurdles being financing for working capital purposes, operating a line of credit, capital investment purposes and investments in new technology. That's a slight improvement over last month's 77 per cent, but on par with the previous five months' data.
"The difficulties in accessing financing may thwart the pace of recovery," says Myers. "If this trend continues, my fear is that we will see a very long, drawn-out recovery."
China's manufacturing improves in October
Mid-market companies can't afford to stand still in 2013: Ernst & Young
Global motion control shipments contract in first quarter of 2013
Global motion control shipments contract in first half 2013
Robot orders down sharply in first half of 2009
Nanotechnology events attract big names to McMaster


Mann Movie : Mann is an Indian Hindi Romantic Darma film starring Aamir Khan, Manisha Koirala, Neeraj Vora, Sharmila Tagore and Anil Kapoor. The movie is directed Indra Kumar and produced by Indra Kumar & Ashok Thakeria. It was edited by Hussain Burmawala and written by Indra Kumar . Its music was given by Sanjeev Darshan and it's music was loved by audience. The movie was released on 9 July 1999. Its got the good response from critics and audience. It was one of the top movies of 1999. It did good business in india. Some more details about Mann's cast and crew.
What do you think about the Mann Movie?


Meet Rekha Mishra from Mumbai, India, a RPF officer who rescued 434 children in last one year alone at Chatrapati Shivaji Terminus (CST) railway station.
Most of children she rescued were from 13 to 16 aged groups, most of them were from UP and Bihar districts from India.
When we asked Mishra about how you figure out who needs to rescued, she said the children who are running away from their home will mostly get down in the final destination of that train.
She also added most of the children she rescued are missing kids those who took wrong trains and kids who are running away from their home.
She also added "I can easily figure out the peoples who are all on those categories".
Whenever is on the duty, she always looking for the children "who are vulnerable, and looking for people to assist them."
She said that these children are not criminals they are victims who need someone to protect them.
She also added, most of the kids who are all in these situations were mostly sexually abused, we got so many complaints filed like this.
So, it's our duty to take care of Children who are vulnerable, standing lonely in railway stations, bus stations.
However, they could trace the parents of only 28 children out of 434 kids. The Remaining kids were handed over to the Child welfare Organisations.
Rekha Mishra now works with a team of members who are trained to take of these children keeping their best interest in their mind.
One of the team member from this Community Committed Development Trust said that "After medical tests, and following procedures like making entries with and filling up documents at station manager's office, we take the child for further investigation.
We asked those kids why you ran away from your home like that most of the kids with eyes full of tears they said that their parents don't take care of them, some kids said their parents don't even provide foods and all.
Parents must take care of their children, they are precious.
Since the beginning of this year, Rekha Mishra rescued more than 100 children. Senior officials also mentioned that they are in the process of recommending her for awards for this wonderful duty of saving 434 kids.


The property consists of a single-storey house having about 80 sq.m. of living are area, a garage, farm buildings a good-sized plot of land spreading over 1500 sq.m.
It is in a big and well developed village 5 km from the town of General Toshevo, 25 km from Dobrich, 40 km from the beach of Kavarna and 72 km from Varna International Airport.
The house consists of a porch, a spacious corridor, 4 rooms and a cellar. Inside bathroom and toilet are not available and they have to be installed additionally.
The village is supplied with grocery stores, pub, Post Office and a regular bus transport connects it to the town of General Toshevo where a railway transport is available.
Near the town, Near sea, Near village, Near airport, Near bus station, Near railway station.


On Wednesday, December 21st, after 6:00 pm, all of the computers in the library will be unavailable as they undergo upgrades. We will have laptops available as temporary replacements. Ask at any library service desk for assistance.
Our new Teen Space is now open! There are a few little construction details still to be finished but, if you are in grades 6-12, please feel free to use the Teen Space to study (finals week) and socialize. Tell your friends.
Over the weekend of December 3rd and 4th, our catalog will be upgraded to improve your experience when searching for books, movies, music, or any other library materials.
During this weekend the library catalog will be down. You will be unable to search, place holds, or check your library account. In-library loans will be working as normal, however we ask that you bring your library card with you when checking out items.
We'll be closed on Thursday, November 24th for Thanksgiving and closing at 5:00 pm the day before. Even though we won't be open, you'll still be able to use many library resources.
Search the catalog and place holds; download an eBook with Media on Demand; stream movies, audiobooks, and more with Hoopla; or expand your technical and creative skills with Lynda.com.
We need your help! Please answer a few questions about how we're doing. Complete the survey online, or look for a paper copy at any library service desk. Thank you for your honest, anonymous answers.
October 21st and 22nd, from 10:00 am-4:00 pm is Annual Book Sale Time! This sale, hosted by the Friends of the Eisenhower Library, features thousands of books, movies and more. Every book sold helps to support the Friends group in its mission to enhance the library and its services.
We're excited to offer you a new learning resource at Eisenhower.
Tutor.com provides access to real time, online tutoring in over 25 subjects for kindergartners through adult learners. It also offers a dropoff service for overnight reviews of essays, book reports, resumes, cover letters, and more. Assessment quizzes help you practice your skills in over a dozen subjects. You can also access thousands of resources like flashcards, videos, ans sample resumes. Personalized features such as favorite tutors, previous session reviews, and virtual lockers make studying much easier. All of this is available to Norridge and Harwood Heights residents at the library, at school, at home, and even on your mobile device.
Visit tutor.com/eisenhower to get all this help.
We'll be closed on Sunday, September 4th and Monday, September 5th for Labor Day Weekend. Even though we won't be open, you'll still be able to use many library resources.
We're entering a new chapter at Eisenhower. This week, construction has officially begun on our new teen space and study rooms. Keep up with the progress on our construction blog.
There's still time to "Read for the Win" this summer! All you have to do to get in the game is read! Kids win prizes each week and adults/teens earn the chance to win Cubs, Second City. or Muvico tickets. Get all the details by clicking here.


The Coalition for Affordable Housing, formed in 2009, in which Bimkom is a member, seeks to develop and promote the concept of affordable housing in Israel and to assist in creating a renewed and vibrant urban model. The Coalition promotes innovative affordable housing solutions including social housing, public housing and other forms of subsidized housing which address both the practical need for affordable housing and the social need, namely to integrate populations of varying levels of income in the same residential areas in ways that encourage social diversity and avoid social exclusion.The Coalition is composed of NGOs and university-affiliated associates from the fields of civil rights law, spatial planning, and community organizing. The members of the coalition include the Association for Civil Rights in Israel (ACRI), Bimkom–Planners for Planning Rights, the Association for Distributive Justice (ADJ), Shatil, the Clinic for Social Planning at the Hebrew University, the Haifa University Clinic for Human Rights, the Movement for Life with Dignity, Itach-Ma'aki: Women Lawyers for Social Justice, the Arab Center for Alternative Planning (ACAP) and Sikkuy – The Association for the Advancement of Civic Equality in Israel.


text:
Sunday Review|Wanted: a Tea Party Speaker
Ross Douthat
Wanted: a Tea Party Speaker
IN an earlier, cozier Washington, D.C., John Boehner could have been the kind of House speaker whose memory is held dear by high-minded chin strokers on Sunday morning television programs: An icon of sadly bygone bipartisanship, a cutter of the grandest bargains, a man who, by God, made legislation move.
In this Washington, D.C., alas for him, Boehner was a humble bomb defuser, and the only grand bargains he cut were between his more intransigent backbenchers and the demands of political and constitutional reality. And now D.C. looks at his record, his resignation and his possible successors and asks: Can anyone do better?
Probably not anyone with Boehner's precise profile. Probably not Kevin McCarthy, another genial dealcutter distrusted on the right, who would have recapitulated Boehner's struggles had his candidacy not been doomed by gaffes and whiffs of scandal.
But maybe the lesson of those struggles is that the speakership simply isn't a job for a professional dealmaker and institutionalist at the moment. Instead, maybe it's a job for a conviction politician, an ideologue (in the best way!) who's also interested in governing.
Maybe, in other words, House Republicans need a speaker who's an ambassador from the Tea Party to the G.O.P.'s K Street/Chamber of Commerce wing, rather than the other way around.
The reality is this: The only way the Republican House majority can become less dysfunctional and chaotic in the short run is if the next speaker wins the trust of enough conservative backbenchers to quell or crush revolts from the rest. And the best way to win that trust is to be seen as fundamentally on the insurgents' side, which is a feat that Boehner, given his background and priorities, could never hope to manage.
Hence the recent appeal of drafting Paul Ryan to replace him. Ryan is the party's leading policy entrepreneur, his blueprints have plainly pulled the G.O.P.'s center of gravity rightward, and he has stronger movement-conservative bona fides than anyone else in the House leadership.
The suspicions that the right always had about Boehner, and would have had about McCarthy — that they care more about the deal than about the outcome, more about the party's donors than any defined small-government principle — does not attach to Ryan in the same way. So he would enter the job with a deposit of ideological credibility that might do more than all of Boehner's backroom skills to keep (some) of the caucus's rightward flank in line.
But Ryan is not really of the Tea Party. In the Bush era he voted for bills like No Child Left Behind, Medicare Part D, and TARP, all of which today's conservative insurgents despise. And he's a dove on immigration, the issue where the party's base always expects — with good reason! — their leadership is poised to sell them out.
A more ideal speaker would share Ryan's conservative credentials and wonkish spirit, but he wouldn't have that baggage. The ideal speaker, in fact, would probably have led Tea Party-driven brinksmanship at some point in the Obama-era past, the better to channel it more productively in the future.
Such a figure exists. Unfortunately, he's in the other chamber: He's Utah's junior senator, Mike Lee.
Lee has an insurgent's résumé: He was elected with the Tea Party wave in 2010, defeating an incumbent Republican, Bob Bennett, along the way. He was Ted Cruz's partner in crime during the government shutdown debates. His scorecard with Heritage Action, often the scourge of G.O.P. leaders, currently stands at 100 percent. And unlike almost every member of the House and Senate leadership, he's a genuine foe of comprehensive immigration reform.
At the same time, like Ryan (and unlike Cruz), Lee been a real policy entrepreneur. He authored a pro-family tax plan that breaks with some (if perhaps not enough) of the G.O.P. donor class's orthodoxies. He has offered serious proposals on transportation, higher education and religious liberty. And just this week he was part of a bipartisan breakthrough on criminal justice reform, one of the rare issues where the late Obama years still offer hope for compromise.
In Lee's ambitions, you can see what the House insurgents want to be — a force that moves conservative policy making away from donor service and toward genuine reform — rather than the purely nihilistic force they often threaten to become. You can see the outlines of the kind of agenda that might satisfy (some) intransigents and also provide some (very) modest ground for bipartisanship.
And then in his record and persona, you can see a — let's be frank — tribal identification with insurgency that might make easier for him to persuade the G.O.P.'s right flank to accept the real limits on the House's power.
Unfortunately the House insurgents do not appear to have a Mike Lee in their ranks.
But there is also no rule preventing the House from electing a senator as its speaker.
Maureen Dowd is off today.
A version of this article appears in print on , Section SR, Page 9 of the New York edition with the headline: A Tea Party Speaker. Order Reprints | Today's Paper | Subscribe


Just recieved these little monsters in the mail. Not only did I get 5 good sized babies but an additional 3 smaller babies for a total of 8 they were packaged extremely well and with care and wrapped extremely well. Thanks guys:) well worth the money!
I give this 5 pack a 5 star review. These were given as a birthday gift during dormancy, but since Spring is here the five plantlets have exploded with growth. All plants arrived in excellent condition when received 4 months ago. The neat rosette fashion in which these grow add to the charm of the "Big Mouth".


Farar & Lewis LLP is a top rated law firm, with over 40 years of combined experience. Our attorneys have won over $100 million, in combined verdicts and settlements for our clients. You deserve compensation for your pain and suffering, and it's our job to get it for you.
At Farar & Lewis LLP, we pride ourselves on our ability to be there for our clients, 24/7 when tragedy or questions arise. Every single one of our clients has access to their assigned lawyer.
We're proud to say that our personal injury lawyers in Los Angeles, are recognized as leaders in the state of California. Our firm has won millions, in compensation for our clients.
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If you have been injured in an auto accident or another type of accident that you did not cause, it is important that you seek the advice of a qualified California personal injury attorney who knows the law and has the ability to secure the compensation you deserve. At the law firm of Farar & Lewis LLP, we represent Southern California clients through some of the most difficult times of their lives.
We know how overwhelming it can be for injury victims to deal with overzealous insurance adjusters who want to settle their claims for as little as possible. We take the burden of the legal issues off your shoulders and work directly with the insurance company to negotiate a settlement that provides the medical care and financial support necessary to facilitate your recovery. If a fair settlement cannot be negotiated, we will take your case to court.
We are committed to you and your right to no-nonsense legal representation, timely communication and full compensation. We pride ourselves on providing punctual and honest case assessments.
A personal injury lawyer is an attorney who represents someone who claims to have suffered physical or emotional injury as a result of wrongdoing or neglect by another individual or entity such as a company or government agency. The term, "personal injury" encompasses anything from a slip or a fall to brain injury or wrongful death.
What may seem like a minor injury from slipping or tripping can actually trigger a surprising amount of physical damage. You fall, you curse, you pause to gain your composure and slowly get back up to your feet with nothing more than a bruised ego. Three weeks later, you wake up with searing back pain. You slowly inch your way out of bed and find it painful to walk on one side. You take the day off work and stay in bed, gorging on daytime television and comfort food.
Days later, you finally drag yourself to see the doctor. He diagnoses a sprained pelvis. If you had gone to see him immediately after the fall, he could have done something to avoid the seven-out-of-ten on a scale of 1 to 10 pain you are experiencing now. You leave his office, clutching a prescription for painkillers and instructions to stay off your feet for the next four to six weeks.
You decide you have a legitimate claim against the owner of the property where you took the fall that caused you to lose nearly two months' wages. There are medical bills, physiotherapy costs, pharmacy expenses. Your next telephone call is to your lawyer.
The first thing he is likely to tell you is that nine out of ten personal injury claims fail. The law appears to favor property owners. The biggest obstacle you are going to face is proving negligence. This isn't because the courts like property owners better than everyone else; it is to protect them from spurious claims. There are people out there who sniff out money and have no scruples about throwing themselves down a stair or two for the sake of claiming thousands of dollars in compensation.
In order for a property owner to be held responsible for an injury resulting from someone having an accident on his premises, there has to be a hazardous or dangerous condition and there has to have been a sufficient time interval for the owner to correct it. If a waiter drops a pot of tea and you happen to slip on the spill while he is reaching for a traffic cone and a cloth, the restaurant owner will not be held responsible. If, on the other hand, the waiter wanders off to serve another customer and forgets about the spill for any length of time, you may have a case.
If you do decide to take action, make sure you have a skilled and experienced Palmdale personal injury lawyer on your team.
If I am the victim of a trucking accident, who can I sue?
What can I do if the driver doesn't have insurance?


text:
'Michelle Will Not Run For Office': Obama's Daytime TV Confessional
TALK SOUP
Morning talk shows should be a doddle for a President: softball questions easily answered with a platitude or three. But despite his folksiness, President Obama dug himself deeper into a rut on this morning's 'Live With Kelly & Michael'
Lloyd Grove
Editor at Large
Updated Jul. 12, 2017 4:24PM ET / Published May. 30, 2014 12:10PM ET
"I'm not too intimidating, I don't think."
President Obama was telling Kelly Ripa and Michael Strahan that Malia's high school prom date had nothing to fear. But it will hardly be surprising if detractors and kibitzers from both political parties use that bit of self-reflection to diagnose one of the fundamental problems afflicting the Obama White House.
Nobody seems to be afraid of the guy. That, according to conventional Washington wisdom, is a major reason why his political and policy agendas are stuck in a ditch on Capitol Hill, and why Vladimir Putin and Iran's Supreme Ayatolla, Ali Khamenei, seem to view the Leader of the Free World with something less than shock and awe.
When Kelly told the president "We're terrified!" and Michael claimed "We're scared to death!", they were simply being the big kidders we all know them to be.
The president has been having a tough time of it lately—a blossoming scandal at the Veterans Administration, foreign policy troubles that have only gotten worse, and his own job approval ratings down in the basement—so it's understandable that he'd seek sweet relief by appearing on Friday's installment of Live With Kelly & Michael. After all, with the notable exception of Jerry Springer, daytime television, with its largely female viewership, is a combat-free environment, potentially allowing an embattled politician to communicate his message unchallenged.
Yet, in a sit-down that was taped at the White House on Thursday, Obama—his legs crossed in relaxed mode—might have been charming, self-deprecating, earnest, and appealing, but he did little to help himself out of his political rut and probably even dug in deeper.
Asked straight-up by Strahan whether he would ask for Eric Shinseki's resignation as Secretary of Veterans' Affairs, Obama promised to "have a serious conversation with him about whether he is prepared and has the capacity to take on the job of fixing it," but not before spending two minutes trying to contextualize the scandal with a bowlful of warm verbal porridge that didn't conceal the rat's head popping up in the middle.
Hearing the president ladle praise on the quality of the VA's medical services, and claim credit for reducing veterans homelessness by 25 percent, while distancing himself from historical "backlogs" and "inefficiencies," many viewers had to wonder what planet he is on. Obama might have been factually correct in many of his assertions, but he was tonally discordant—and politically out of this world. (Later on Friday morning Obama accepted Shinseki's resignation.)
Indeed, he brought more obvious passion to a discussion of the perils of concussions in high school and college sports—the subject of an all-day White House summit—than to the infuriating incompetence and possible criminality occurring (at least, organizationally) right under his nose.
In another, possibly problematic, point in the session, Obama stopped just short of endorsing Hillary Clinton for president, and seemed to convey that he considers Vice President Biden—who has openly mused about a 2016 candidacy—an also-ran.
"We're buddies," he said of his former secretary of state, who ran against him 2008 in an often-acrid primary battle. "I always admired her. As soon as she got here, she couldn't have been more effective and more loyal, and since that time is a really, really good friend."
And here's the money quote that is apt to give Joe Biden fits: "I don't know what she's gonna decide to do, but I know if she were to run for president, I think she'd be very effective at that." He relegated his vice president, meanwhile, to a group of "some hardworking, effective…people around me…and I love them to death."
As required on daytime television chatfests, the president also revealed some personal details: He generally goes to bed at 2 a.m., after reading through his briefing papers, and gets up at 7 a.m. for a daily workout. He has no plans to dye his graying hair. "I see Michelle in the hair salon, and it's just too much work. It takes my barber 20 minutes to cut my hair."
He told Ripa that if Michelle Obama ever told him she'd like to run for office herself, "I would say, 'Where did you take my wife?' Had there been an alien body-snatching going on? One thing I can promise you: Michelle will not run for office."
He indicated he wasn't mad at an aggressive player in a pickup basketball game a couple of years ago who inadvertently split Obama's lip with an errant elbow, requiring the president to get 12 stitches. "I'm not sure he got invited back," Obama confided, "but I did send him a picture [taken at the moment of the injury by a White House photographer] and wrote him, 'You're the only man who ever got away with cold-cocking the president.' "
He also noted how "jarring" it was to see Malia, his daughter in high heels for the first time when she went to that first prom. He would not say if she had a male date.
At one point, Strahan asked him what he'd most like to do if he could just be a normal person.
"I'd take a walk," the president answered, and there was a dreamy look in his eyes. "I would walk through this gate…I might walk out to the Lincoln Memorial and sit on there. Maybe I'd wander around and find myself at a little outdoor café or something and just sit and watch people go by. The thing you miss most about being president is just anonymity."
There was more genuine heartfelt feeling in that particular answer than any of the others Obama gave.


text:
Lighting up the "Open Sign" for waterfront business – a festival
September 26, 2014 By Carlanne
In a collaborative effort, City of Seattle entities, non-profit organizations and Seattle Waterfront businesses are presenting The New Waterfront Festival on Oct. 4th, 2014 to increase awareness and gain continued consumer support for businesses during Seawall construction throughout the fall and winter of 2014/15. On October 1st, 2014, seawall construction is scheduled to start up… Read More
More Delta flights scheduled for Seattle-Tacoma International Airport
November 19, 2013 By Carlanne
Celebrations abound as Delta Air Lines expands their presence in the Pacific Northwest and Seattle in particular. Delta Air Lines teamed up with Michael Bublé to celebrate the airline's expansion into the Seattle region. Delta hosted the event, which included a VIP Receiption and intimate Q+A with Bublé for its elite SkyMiles members and key partners, in recognition… Read More
Taste your way through ethnic Chicago
February 25, 2012 By Jodie Jacobs
Picture eating octopus ceviche from a martini glass while deep-sea fish watch from above. You are not near an ocean but in a Chicago neighborhood restaurant.
Brides say 'I Do' to bridal showcase at Pan Pacific Hotel
February 25, 2011 By Carlanne
Last week in Seattle, 175 brides said "I do" to an invitation to a bridal show case. More than 100 of those brides-to-be showed up on February 17th at the Pan Pacific Hotel to watch a mock wedding and attend a reception for the "bride" and "groom", along with the "parents" of the lucky couple and other "friends" and "relatives". It was a fascinating experience that entranced many brides-to-be.
Olympic Sculpture Park offers-visitors & locals-art & exercise
February 5, 2011 By Carlanne
Seattle's Olympic Sculpture Park offers visitors and locals both art and exercise, no matter what the season. Pictures in this article were taken in late January at the park as multitudes of people walked the trails, admired the sculpture and watched the trains as they rolled through the park.
Copperleaf named to 2011 Gayot top 10 best new restaurants
January 27, 2011 By Carlanne
Copperleaf Restaurant, located at the Cedarbrook Lodge in Seattle, has been named as best new restaurant in the US by Gayot for its farm-to-table cuisine and commitment to locally sourced ingredients. The Copperleaf Restaurant seats only 30 people per sitting, so call for reservations in advance. It is worth the extra effort.


text:
Q: jsoup: How to select the parent nodes, which have children satisfying a condition Here's the part of the HTML (simplified for the question):

What I want to get is the vector containing the ids of the auctions, for which the 2.png image is displayed (id=4672 in this case). How to construct the Selector query in order to obtain this?
http://jsoup.org/apidocs/org/jsoup/select/Selector.html - Here I can only find how to select the children, not the parents...
Any help appreciated, including the usage of other libraries. I've tried Jsoup because it seemed to be the most popular.

A: You can use parent() method:
final String html = " \n"
        + "
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        + "
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        + "    \n"
        + "
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        + "
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        + "  \n"
        + "
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        + "
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        + "    \n"
        + "
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        + "
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Document doc = Jsoup.parse(html);

for( Element element : doc.select("img") ) // Select all 'img' tags
{
    Element divGuarantee = element.parent(); // Get parent element of 'img'
    Element divProgress = divGuarantee.parent(); // Get parent of parent etc.

    // ...
}

redpajama_set_name: RedPajamaStackExchange


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text:
Will Your Self-Driving Car Turn on You? and Other Important Questions
KUT 90.5
Stanford Center for Internet and Society/flickr
Researchers, scientists, professors and engineers from around the U.S. and the world are in Austin for the 29th annual Conference on Artificial Intelligence. They're here to talk about the latest developments in the field of artificial intelligence and how those developments are affecting human lives.
Some of the field's prominent names are speaking about AI's potential impact on the human race. Will robots steal everyone's jobs? Will machines render humans irrelevant? Or will they rise up and dominate the human species with their superior intelligence?
Not exactly, says Berkeley professor Stuart Russell. He says people – the media, in particular – have blown AI doomsday scenarios out of proportion.
"Usually they will bring out pictures of Terminator robots, or they'll talk about the Matrix, or Transcendence, other sci-fi movies about AI," Russell says. "I think take that with about three buckets full of salt."
Russell says one key aspect those in AI are working on is "value alignment." It's important to develop artificial intelligence that has the same goals and motivations as human intelligence, he says.
"How do we make sure the machines want exactly what we want? That is, are they our perfect servants in the way, in PG Wodehouse's novels, that Jeeves is the perfect servant to Wooster?" he says. "Wooster is not the sharpest knife in the drawer, and Jeeves anticipates his master's wishes, cleans everything up when his master makes a mess, and smooths things over."
Driving is one of the things that's becoming increasingly automated with the development of AI technology, and Russell believes that cars, in the near future, will essentially be horizontal elevators.
"You get in, press a button to say where you want to go, you get out. And you don't think about that there's a whole bunch of engineering and algorithms in the background making sure you get to the right floor. And you wouldn't dream of going back to the days where you had to have a driver drive your elevator, and try to get it to stop at the right place, and pull back a bit to get out of the door," he says.
Both Russell and Rice University professor Moshe Vardi are speaking at this year's conference. Vardi is speaking about how the development of robotics will result in the elimination of the need for human labor in certain job sectors. Many jobs that fall into the middle class – positions that don't require complex physical or cognitive skills – are becoming automated, he says.
"Information technology is pushing on the middle class jobs, so there's talk that the middle class is declining, while the high end is thriving," Vardi says. "There is a lot of fearmongering, and a lot of optimism, and somewhere between these two – technooptimism and technopessimism – what's the right way to walk between these two ends?"
Russell believes it's possible to align machine values with those of humans, and the alignment would provide some protection from the doomsday scenarios of science fiction. Perhaps then, value-aligned machines wouldn't want to intentionally "steal" a human's job, if the machine were aware that the human in question had a family to take care of.
Both Vardi and Russell believe that those questions are pretty far out on the horizon at this point, but they know the conversation needs to be happening now to guide technology down the right path.
The Conference on Artificial Intelligence runs through Friday. Follow the conference on Twitter here.
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China: Alipay Health Code app controls movement in China
Software on smartphones dictates whether an individual should be quarantined. Chinese citizens in 200 cities, beginning with Hangzhou, are required to install the Alipay Health Code app, developed by Hangzhou's local government with the help of Alipay owner Ant Financial, on their smartphones. After users fill in a form with personal details, the software generates a QR code in one of three colors. Green enables its holder to move about unrestricted. Those with yellow codes may be asked to stay home for seven days. Red means a two-week quarantine. The app also appears to share information with the police. The Times's analysis found that as soon as a user grants the software access to personal data, a piece of the program labeled "reportInfoAndLocationToPolice" sends the person's location, city name and an identifying code number to a server. The software does not make clear to users its connection to the police. But according to China's state-run Xinhua news agency and an official police social media account, law enforcement authorities were a crucial partner in the system's development. Reporter Paul Mozur, who worked on this story, has posted video clips showing the police and app in action.
https://www.nytimes.com/2020/03/01/business/china-coronavirus-surveillance.html
https://twitter.com/paulmozur/status/1234310754452434945
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