text:
\section{Introduction}

Let $d,m\in {\mathbb N}$, $T\in(0, \infty)$, consider a $d$-dimensional system of autonomous stochastic differential equations (SDE)
\begin{equation}\label{sde0}
\begin{aligned}
dX(t) & = \mu(X(t)) \, dt + \sigma(X(t)) \, dW(t), \quad t\in [0,T],\\
X(0) & = x_0
\end{aligned}
\end{equation}
with a deterministic initial value $x_0\in{\mathbb R}^d$, a drift coefficient $\mu\colon{\mathbb R}^d\to{\mathbb R}^d$, a diffusion coefficient $\sigma\colon {\mathbb R}^d\to{\mathbb R}^{d\times m}$ and an $m$-dimensional driving Brownian motion $W$, and assume that \eqref{sde0} has a unique strong solution $(X(t))_{t\in[0, T]}$. Our computational task is to approximate $X(T)$ by means of methods that use finitely many evaluations of the driving Brownian motion $W$. In particular we are interested in the following question: under which assumptions on the coefficients $\mu$ and $\sigma$ exists a method of the latter type, which converges to
$X(T)$ in absolute mean with a polynomial rate?

It is well-known that if the coefficients $\mu$ and $\sigma$ are globally Lipschitz continuous then the classical Euler scheme achieves the rate of convergence $1/2$, see \cite{m55}. Moreover,
the recent literature on numerical approximation of SDEs contains a number
of results on approximation schemes that are specifically designed for non-Lipschitz coefficients and achieve
polynomial convergence rates for suitable classes of such SDEs,
see e.g.
\cite{h96,hms02
HutzenthalerJentzenKloeden2012, MaoSzpruch2013Rate,
WangGan2013,Sabanis2013ECP,TretyakovZhang2013,Beynetal2014,KumarSabanis2016, Beynetal2016}
for SDEs with globally monotone coefficients
and see e.g.
\cite{BerkaouiBossyDiop2008,GyoengyRasonyi2011,DereichNeuenkirchSzpruch2012,
Alfonsi2013,NeuenkirchSzpruch2014,HutzenthalerJentzen2014,
HutzenthalerJentzenNoll2014CIR,
LS15b,LS16, Tag16, HH16b}
for SDEs with possibly non-monotone coefficients.

On the other hand, it has recently been shown in \cite{JMGY15}  that for any sequence  $(a_n)_{n\in{\mathbb N}}\subset (0,\infty)$, which may converge to zero arbitrary slowly,  there exists an SDE \eqref{sde0} with $d=4$ and $m=1$ and with infinitely often differentiable and bounded coefficients $\mu$ and $\sigma$ such that no approximation of $X(T)$
based on finitely many
evaluations of the driving Brownian
motion $W$ converges in absolute mean  faster than the given sequence $(a_n)_{n\in{\mathbb N}}$.
More formally,
\begin{equation}
\label{eq:intro3}
  \inf_{ s_1, \dots, s_n \in [0,T] }
  \inf_{
    \substack{
      u \colon {\mathbb R}^n \to {\mathbb R}^4
    \\
      \text{measurable}
    }
  }
  {\mathbb E}
    \big\|
      X( T )
      -
      u\big( W( s_1 ), \dots, W( s_n )
      \big)
    \big\|
\geq  a_n.
\end{equation}
In particular, there exists an SDE \eqref{sde0} with infinitely often differentiable and bounded coefficients $\mu$ and $\sigma$ such that
its solution at the final time can not be approximated with a polynomial rate of convergence based on finitely many
evaluations of the driving Brownian
motion $W$. We add that the latter statement in the special case when the approximation is given by the Euler scheme with equidistant time steps has first been shown in \cite{hhj12}.

Note that the time points $s_1, \ldots, s_n\in[0,T]$ that are used by an approximation $u( W( s_1 ), \dots,$ $W( s_n ))$ in \eqref{eq:intro3} are fixed, and therefore
this negative result does not cover approximations that may choose the number as well as the
location of the evaluation sites of the driving Brownian motion $W$ in an
adaptive way, e.g. numerical schemes that adjust the actual step size according to a criterion that is based on the values of the driving Brownian motion $W$ observed so far, see e.g. \cite{Gaines1997,MG02_habil,m04, Moon2005, RW2006, LambaMattinglyStuart2007, Hoel2012,Hoel2014} and the references therein. See Section \ref{approximations} for the formal definition of that type of approximations.  It is
well-known that for SDEs \eqref{sde0} with (essentially) globally Lipschitz continuous coefficients $\mu$ and $\sigma$ adaptive approximations can not achieve a better rate of convergence compared to what is
best possible for non-adaptive ones, which at the same time coincides with the best possible rate of convergence that can be achieved by any approximation based on $W(\tfrac{T}{n}), W(\tfrac{2T}{n}), \ldots, W(T) $, see \cite{MG02_habil,m04}.
However, as has recently turned out, this is not necessarily the case anymore if the coefficients $\mu$ and $\sigma$ are not both globally Lipschitz continuous.  In ~\cite{HH16} it has been shown that for the one-dimensional squared Bessel process,
which is the solution of the SDE \eqref{sde0} with $d=m=\mu=1$ and $\sigma(x)=2\sqrt {|x|}$ for $x\in{\mathbb R}$ the following holds: the best possible rate of convergence that can be achieved by any approximation based on $W(\tfrac{T}{n}), W(\tfrac{2T}{n}), \ldots, W(T) $ equals $1/2$, i.e. there exist $c_1, c_2>0$ such that
\[
 c_1\cdot n^{-1/2}\leq  \inf_{
    \substack{
      u \colon {\mathbb R}^n \to {\mathbb R}
    \\
      \text{measurable}
    }
  }
  {\mathbb E}
    \big|
      X( T )
      -
      u\big( W(\tfrac{T}{n}), W(\tfrac{2T}{n}), \ldots, W(T)
      \big)
    \big|
\leq  c_2\cdot n^{-1/2},
\]
 while the best possible rate of convergence that can be achieved by  approximations based on $n$ adaptively chosen evaluations of the driving Brownian motion $W$ equals infinity. More formally, for every $\alpha>0$ there exists $c>0$ and a sequence of approximations $\widehat X_n$ based on $n$ adaptively chosen evaluations of $W$ such that
\[
{\mathbb E} |X(T)-\widehat X_n|\leq c\cdot
n^{-\alpha}.
\]

In view of the latter result one might hope that a non-polynomial lower error bound $a_n$
in \eqref{eq:intro3} could be overcome by using adaptive approximations, see also the discussion in ~\cite[p.\,2]{Gobet2016}. In the present article we prove that the pessimistic alternative is true.
We show that for any sequence  $(a_n)_{n\in{\mathbb N}}\subset (0,\infty)$, which may converge to zero arbitrary slowly,  there exists an SDE \eqref{sde0} with $d=4$ and $m=1$ and with infinitely often differentiable and bounded coefficients $\mu$ and $\sigma$ such that no approximation
based on $n$ adaptively chosen evaluations  of the driving Brownian
motion $W$ on average can achieve a smaller absolute mean error than the given number  $a_n$, i.e.
\[
  {\mathbb E}
    \big\|
      X( T )
      -
     \widehat X_n
    \big\|
\geq  a_n
\]
for any approximation $ \widehat X_n$ of the latter type.
This fact is an immediate consequence of Corollary \ref{cor3} in Section \ref{results}  together with an appropriate scaling argument.
For the proof of the latter result  we employ the same class of SDEs as in ~\cite{JMGY15}.
Thus, roughly speaking, these SDEs can not be solved approximately in the strong sense in a reasonable computational time by means of any kind of adaptive (or nonadaptive) method based on finitely many evaluations of the driving Brownian motion $W$.

We conjecture that a similar negative result does even if one  allows for adaptive  approximations based on finitely many  evaluations of arbitrary linear continuous functionals of the
driving Brownian motion $W$.
However, in this case one  can not employ the class of SDEs from \cite{JMGY15} since for every such SDE  its solution at the final time can be approximated with error zero based on the evaluation of only two linear continuous functionals of the
driving Brownian motion $W$, see \eqref{solutionSDE}

We add that negative results in the spirit of \eqref{eq:intro3} for quadrature problems for marginal distributions of SDEs have recently been established in \cite{MGY16}.

We briefly describe the content of the paper. In Section~\ref{not}
we fix some notation.
In Section~\ref{sec:setting} we briefly introduce the class of SDEs from \cite{JMGY15}, which is studied in this article as well. In Section~\ref{approximations} we formally define the class
of adaptive approximations, which are analysed in this article. Our lower error bounds are stated in Section \ref{results}. The proof of the
main result, Theorem \ref{t1}, is carried
out in Section~\ref{strong}.

\section{Notation}\label{not}

Throughout this article the following notation is used.
For a set $ A $, a vector space $ V$, a set $ B \subseteq V $,
and a function $ f \colon A \to B $
we
put
$
  \operatorname{supp}( f ) = \left\{ x \in A \colon f(x) \neq 0 \right\}
$. For sets $A$, $B$, a function $f \colon A \to B$ and a subset
$E\subseteq A$ we denote by $f|_E$ the restriction of $f$ to $E$.
Moreover, for $ d \in {\mathbb N} $ and $ v \in {\mathbb R}^d $ we write $
  \|v\|
$ for the Euclidean norm of $ v$.
For $n\in{\mathbb N}$ and $-\infty<a<b<\infty$ we denote by  $\mathfrak
B({\mathbb R}^n)$ and   $\mathfrak
B(C([a,b]))$  the Borel $\sigma$-fields on ${\mathbb R}^n$ and on $C([a,b])$, respectively, where the latter space is equipped with the supremum norm. For $S$ being a finite product of the latter two spaces we denote by  $\mathfrak
B(S)$ the Borel  $\sigma$-field on $S$ generated by the respective product topology.

\begin{comment}
Furthermore,
for $ x \in {\mathbb R} $
we put
$
  \lfloor x \rfloor = \max\!\left(
    {\mathbb Z} \cap ( - \infty, x ]
  \right)
$
and
$
  \lceil x \rceil = \min\!\left(
    {\mathbb Z} \cap [ x, \infty )
  \right)
$.
\end{comment}

\section{A family of SDEs with smooth and bounded coefficients}
\label{sec:setting}

Throughout this article we study
SDEs
provided by the following setting.

Let
$ T  \in (0,\infty) $,
let
$ ( \Omega, \mathcal{F}, {\mathbb P} ) $
be a probability space with a normal filtration
$ ( \mathcal{F}_t )_{ t \in [0,T] } $,
and let
$
  W \colon [0,T] \times \Omega \to {\mathbb R}
$
be a
standard $ ( \mathcal{F}_t )_{ t \in [0,T] } $-Brownian motion
on $ ( \Omega, \mathcal{F}, {\mathbb P} ) $.

Let $ 0<\tau_1<\tau_2<T$ and let $ f, g, h \in C^{ \infty }({\mathbb R}) $ be
bounded and satisfy $
  \operatorname{supp}( f ) \subseteq ( - \infty, \tau_1 ]
$,
$
  \inf_{ t\in [ 0, \tau_1/2  ] } | f'(t) | > 0
$,
$
  \operatorname{supp}( g ) \subseteq [ \tau_1, \tau_2 ]
$,
$g\not =0$,
$
  \operatorname{supp}( h ) \subseteq [ \tau_2, \infty )
$,
and
$
  \int_{ \tau_2 }^{ T } h(t) \, dt \neq 0
$.

For every $
  \psi \in C^{ \infty }( {\mathbb R})
$
let
$
  \mu^{\psi} \colon {\mathbb R}^4 \to {\mathbb R}^4
$
and
$
  \sigma \colon {\mathbb R}^4 \to {\mathbb R}^4
$
be given by
\begin{align*}
  \mu^{ \psi }(x) & = \bigl( 1, 0, 0, h( x_1 ) \cdot \cos( x_2 \, \psi( x_3 ) ) \bigr),\\
  \sigma(x) &=
  \bigl(
    0, f( x_1 ) , g( x_1 ), 0
  \bigr)
\end{align*}
and
 consider the following $4$-dimensional system of SDEs
\begin{equation}\label{sde}
\begin{aligned}
dX^{\psi }(t) & = \mu^\psi (X^{\psi }(t)) \, dt + \sigma(X^{\psi }(t)) \, dW(t), \quad t\in [0,T],\\
X^{\psi}(0) & = 0.
\end{aligned}
\end{equation}

\begin{rem}
Note that for  every $ \psi \in C^{ \infty }( {\mathbb R} ) $ the
functions $ \mu^{ \psi } $ and $\sigma$ are infinitely often
differentiable and bounded.
\end{rem}

\begin{rem}\label{solution}
It is easy to see that for every $ \psi \in C^{ \infty }( {\mathbb R}
) $ the SDE \eqref{sde} has a unique strong solution given by
\begin{align}\label{solutionSDE}
  X_1^{ \psi }(t) & = t ,
\qquad
  X_2^{ \psi }(t) = \int_0^{ \min( t, \tau_1 ) } f(s) \, dW(s),\notag
\\
  X_3^{ \psi }(t) & = 1_{ [ \tau_1 , \, T ] }( t ) \cdot
  \int_{ \min( t, \tau_1 ) }^{ \min( t , \tau_2 ) } g(s) \, dW(s) ,
\\
  X_4^{ \psi }(t) & =
  1_{ [ \tau_2 , \, T ] }(t) \cdot
  \cos\bigl(
    X_2^{ \psi }( \tau_1 ) \,
    \psi\big(
      X_3^{ \psi }( \tau_2 )
    \big)
  \bigr)
  \cdot
  \int_{ \tau_2 }^{ t } h(s) \, ds\notag
\end{align}
for all $ t \in [0,T] $.
\end{rem}

\begin{comment} LL
\begin{ex}
\label{ex:fgh}  Let $ c_1, c_2, c_3 \in {\mathbb R} $ and let $ f, g, h \colon
{\mathbb R} \to {\mathbb R} $ be given by
\begin{align*}
  f(x) & = 1_{(-\infty,\tau_1)}(x)\cdot
    \exp\Bigl(
      c_1 +
      \frac{
        1
      }{
        x - \tau_1
      }
    \Bigr)
  ,
\\
  g(x) & = 1_{(\tau_2,\tau_3)}(x)\cdot
    \exp\Bigl(
      c_2 +
      \frac{ 1 }{ \tau_2 - x } + \frac{ 1 }{ x - \tau_3 }
    \Bigr)
  ,
\\
  h(x) & =
    1_{(\tau_3,\infty)}(x)\cdot
    \exp\Bigl(
      c_3
      +
      \frac{ 1 }{ \tau_3 - x }
    \Bigr)
\end{align*}
for $x\in{\mathbb R}$. Then $ f, g$ and $h $ satisfy the conditions stated above.
\end{ex}
\end{comment}

\section{Adaptive strong approximations}
\label{approximations} Let $\delta\in (0,T]$. We study general
strong approximations of $X^\psi(T)$ based on $(W(t))_{t\in[\delta,
T]}$ and on finitely many sequential evaluations of $W$ in the
interval $(0, \delta)$. Every such approximation $\widehat X\colon
\Omega\to{\mathbb R}^4$ is defined by three sequences
\[
\varphi=(\varphi_n)_{n\in{\mathbb N}}, \quad \chi=(\chi_n)_{n\in{\mathbb N}},\quad
\phi=(\phi_n)_{n\in{\mathbb N}}
\]
of measurable mappings
\begin{align}\label{sequences}
\varphi_n&\colon{\mathbb R}^{n-1} \times {\mathbb R}^{[\delta, T]} \to (0, \delta),\notag\\
\chi_n&\colon{\mathbb R}^{n}\times {\mathbb R}^{[\delta, T]}\to\{0,1\},\\
\phi_n&\colon{\mathbb R}^{n}\times {\mathbb R}^{[\delta, T]}\to{\mathbb R}^4.\notag
\end{align}
The sequence $\varphi$ determines the evaluation sites of a trajectory
of $W$ in the interval $(0, \delta)$. The total number of
evaluations is determined by the sequence $\chi$ of stopping rules.
Finally, the sequence $\phi$ is used to obtain the approximation to
$X^\psi(T)$ from the observed data.

More precisely, let $\omega\in\Omega$, let $w=W(\omega)$ be the corresponding trajectory of $W$ and put $v=(w(t))_{t\in[\delta, T]}$. The sequential observation of $w$
starts at the knot $\varphi_1(v)$. After $n$ steps the
available information is then given by $D_n(\omega)=(y_1,
\ldots, y_n, v)$, where $y_1=w(\varphi_1(v))$, \ldots,
$y_n=w(\varphi_n( y_1, \ldots, y_{n-1}, v))$, and we decide
whether we stop or further evaluate $w$ according to the value of
$\chi_n(D_n(\omega))$. The total number of observations of $w$ in the
interval $(0, \delta)$ is thus given by
\begin{equation}\label{nu}
\nu(\omega)=\min\{n\in{\mathbb N}:\chi_n(D_n(\omega))=1\}.
\end{equation}
 If
$\nu(\omega)<\infty$, then the data $D_{\nu(\omega)}(\omega)$
is used to construct the estimate
$\phi_{\nu(\omega)}(D_{\nu(\omega)}(\omega))\in{\mathbb R}^4$.

For obvious reasons we require that $\nu<\infty$ ${\mathbb P}$-a.s. Then
the resulting approximation is given by
\[
\widehat X=\phi_{\nu}(D_{\nu}).
\]
Without loss of generality we assume that
\begin{equation}\label{15}
\varphi_k(y_1, \ldots, y_{k-1}, v)\not = \varphi_l(y_1, \ldots, y_{l-1}, v)
\end{equation}
for all $y\in{\mathbb R}^\infty$, all $k, l\in{\mathbb N}$ with $k\not=l$ and all $v\in C([\delta, T])$.
We put
\[
c(\widehat X)={\mathbb E} \nu,
\]
that is the expected number of evaluations of the driving Brownian
motion $W$ in the interval $(0, \delta)$. We denote by $\mathcal{X}^\delta$
the class of all methods of the above form and for $N\in{\mathbb N}$ we put
\[
\mathcal{X}_N^\delta=\{\widehat X\in\mathcal{X}^\delta: c(\widehat X)\leq N\}.
\]
Clearly,
$
\mathcal{X}_N^{\delta_1}\subseteq\mathcal{X}_N^{\delta_2}
$
for all $0<\delta_2\leq \delta_1\leq T$ and all $N\in{\mathbb N}$.

Let us stress that the class $\mathcal{X}_N^{\delta}$ contains in particular all methods from the literature, which use a step size control based on $N$ sequential
evaluations of $W$ on average, see e.g. \cite{Gaines1997,MG02_habil,m04, Moon2005, RW2006, LambaMattinglyStuart2007, Hoel2012,Hoel2014} and the references therein.
Moreover, $\mathcal{X}_N^{\delta}$ of course contains
all nonadaptive approximations $u\bigl(W(s_1), \ldots, W(s_N), $ $(W(t))_{t\in[\delta, T]}\bigr)$ based on  $N$
evaluations of $W$ at fixed time points $s_1, \ldots, s_N\in(0, \delta)$ and on $(W(t))_{t\in[\delta, T]}$, as studied  in \cite{JMGY15}.
In the latter case one can take any sequences $\varphi, \chi$ and $\phi$ satisfying
\[ \varphi_n=s_n \text{ for } n\leq N, \chi_1=\ldots \chi_{N-1}=0, \chi_N=1 \text{ and } \phi_N=u.
\]

\begin{comment} needed?
Moreover, for $N\in{\mathbb N}$ we put
\[
\mathcal{X}_N=\mathcal{X}_N^T,
\]
that is the class of all general strong approximations of
$X^\psi(T)$ based on $W(T)$ and on $N$ sequential evaluations of $W$
in the interval $(0, T)$ on average and we note that
\[
\mathcal{X}_N\subseteq\mathcal{X}_N^\delta
\]
for all $\delta\in(0, T]$ and $N\in{\mathbb N}$.
\end{comment}

\section{Main results}
\label{results} Assume the setting in Section~\ref{sec:setting} and
put

\begin{equation}\label{51}
\alpha
  =
  \inf_{ t \in [ 0 , \tau_1 /2  ] }
  | f'(t) |^2
  ,\quad
  \beta
   =
   \int_{\tau_1}^{\tau_2}  g^2(t) dt,\quad
   \gamma
   =
  \int_{ \tau_2 }^T
  h(t) \, dt
\end{equation}
as well as
\[
c_1=\frac{ \gamma \exp \bigl(- \tfrac{\pi^2}{4}-\tfrac{1}{\beta}\bigr)}{  8 \pi\sqrt{2\pi\beta}  },\quad c_2=\frac{ \gamma \exp \bigl(- \tfrac{\pi^2}{4}\bigr)}{  4 \pi  }.
\]

Our main result is stated in Theorem \ref{t1}. It provides a uniform lower bound for the mean absolute error of any
strong approximation of $ X^{ \psi }(T) $ that is based on $(W(t))_{t\in [\delta, T]}$ and on $N$
sequential evaluations of $ W $ in the
interval $(0, \delta)$ on average in the case that $\psi$ is positive, strictly increasing and satisfies $
  \lim_{ x \to \infty } \psi( x ) = \infty
$ as well as
$1\in\psi({\mathbb R})$. See Section \ref{strong} for the proof.
\begin{theorem}
\label{t1}  Let $
  \delta\in(0, T]
$ and let $ \psi \in C^{ \infty }( {\mathbb R}) $ be positive, strictly
increasing  with $
  \lim_{ x \to \infty } \psi( x ) = \infty
$ and
$1\in\psi({\mathbb R})$. Then for all $N\in{\mathbb N}$ and all $\widehat X\in\mathcal{X}_N^\delta$ we
have
\begin{equation}\label{61}
  {\mathbb E}
    \|
      X^{ \psi }( T ) -\widehat X
    \|\geq c_1\cdot \exp\bigl(-\tfrac{1}{\beta}\cdot \bigl(\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha (\min(\delta, \tau_1/2))^3}}\bigr)  N^3\bigr)\bigr)^2\bigr)- \frac{c_2}{N}.
\end{equation}

\end{theorem}

As a consequence of Theorem \ref{t1} we obtain a non-polynomial decay of the smallest possible mean absolute error of
strong approximation of $ X^{ \psi }(T) $ based on $(W(t))_{t\in [\delta, T]}$ and on $N$
sequential evaluations of $ W $ in the
interval $(0, \delta)$ on average if $\psi$ additionally satisfies an exponential growth condition.

\begin{cor}\label{cor1} Let $
  \delta\in(0, T]
$ and let $ \psi \in C^{ \infty }( {\mathbb R}) $ be positive, strictly
increasing  with $
  \lim_{ x \to \infty } \psi( x ) = \infty
$ and
$1\in\psi({\mathbb R})$. Moreover assume that for all $q\in (0,\infty)$
\[
\lim_{x\to\infty} \psi(x)\cdot \exp(-q x^2)=\infty.
\]
Then for all $q\in(0, \infty)$ we have
\[
\lim_{N\to\infty} \bigl(N^q\cdot \inf_{\widehat X\in \mathcal{X}^\delta_N}  {\mathbb E}
    \|
      X^{ \psi }( T ) -\widehat X
    \|\bigr) =\infty.
\]

\end{cor}
\begin{proof}
The assumptions on the function $\psi$ ensure that for all $q\in(0, \infty)$
\begin{equation}\label{78}
\lim_{N\to\infty}      \bigl(N^q\cdot \exp\bigl(-\tfrac{1}{\beta}\cdot \bigl(\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha (\min(\delta, \tau_1/2))^3}}\bigr)  N^3\bigr)\bigr)^2\bigr)\bigr)=\infty,
\end{equation}
see Lemma 4.5 in \cite{JMGY15}. This in particular implies that there exists $N_0\in{\mathbb N}$ such that for all $N\geq N_0$
\[
\frac{c_1}{2}\cdot \exp\bigl(-\tfrac{1}{\beta}\cdot \bigl(\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha (\min(\delta, \tau_1/2))^3}}\bigr)  N^3\bigr)\bigr)^2\bigr)\geq \frac{c_2}{N}.
\]
Employing Theorem \ref{t1}  we therefore conclude that for all $N\geq N_0$
\[
  \inf_{\widehat X\in \mathcal{X}^\delta_N} {\mathbb E}
    \|
      X^{ \psi }( T ) -\widehat X
    \|\geq \frac{c_1}{2}\cdot \exp\bigl(-\tfrac{1}{\beta}\cdot \bigl(\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha (\min(\delta, \tau_1/2))^3}}\bigr)  N^3\bigr)\bigr)^2\bigr).
\]
The latter estimate and  \eqref{78} imply the statement of the corollary.
\end{proof}

The following result shows that the smallest possible mean absolute error of
strong approximation of $X^\psi(T)$ based  $(W(t))_{t\in [\delta, T]}$ and on $N$
sequential evaluations of $ W $ in the
interval $(0, \delta_N)$ on average may converge to zero  arbitrarily slow even then when  the sequence $(\delta_N)_{N\in{\mathbb N}}$ tends to zero
with any given speed.

\begin{comment}

The following result shows that the smallest possible mean absolute error of
strong approximation of $X^\psi(T)$ based  $(W(t))_{t\in [\delta, T]}$ and on $N$
sequential evaluations of $ W $ in the
interval $(0, \delta)$ on average may converge to zero  arbitrarily slow. This holds even then when $\delta$ depends on $N$, i.e., $\delta=\delta_N$, and the sequence $(\delta_N)_{N\in{\mathbb N}}$ tends to zero
with any given speed.
\end{comment}

\begin{cor}\label{cor3}
Let $(a_N)_{N\in{\mathbb N}}\subset (0, \infty)$ and $(\delta_N)_{N\in{\mathbb N}}\subset (0, T]$ satisfy $\lim_{N\to\infty} a_N=0$ and $\lim_{N\to\infty} \delta_N=0$. Then there exists $\kappa>0$ and $ \psi \in C^{ \infty }( {\mathbb R}) $ such that for all $N\in{\mathbb N}$ we
have
\[
 \inf_{\widehat X\in \mathcal{X}^{\delta_N}_N} {\mathbb E}
    \|
      X^{ \psi }( T ) -\widehat X
    \|\geq \kappa\cdot a_N.
\]
\end{cor}
\begin{proof}
We proceed similar to the proof of Corollary 4.3 in \cite{JMGY15}.
Without loss of generality we may assume that the sequences $
  ( a_N )_{ N \in {\mathbb N} }
$ and $
  ( \delta_N )_{ N \in {\mathbb N} }
$ are strictly decreasing.
Let
\[
N_0=\min\bigl\{N\in{\mathbb N}\colon a_N+\frac{c_2}{N}\leq c_1\bigr\}
\]
and for $N\geq  N_0$ put
\[
  b_N =\sqrt{
  -\beta
    \ln\!\big(
      \tfrac{ 1}{c_1}\cdot\big(a_N+\tfrac{ c_2 }{N}
    \big)\big)}, \quad   d_N =
  \bigl(1+ \sqrt{\tfrac{96 }{\alpha (\min(\delta_N, \tau_1/2))^3}}\bigr) N^3
  .
\]
Note that the sequences $
  ( b_N )_{ N\geq N_0 }
$ and $
  ( d_N )_{ N\geq N_0 }
$ are strictly increasing and satisfy
\[
 \lim_{ N \to \infty } b_N =  \lim_{ N \to \infty } d_N = \infty
.
\]
Define a function $
  \psi \colon {\mathbb R} \to {\mathbb R}
$ by
\[
  \psi(x) =
  \begin{cases}
   d_{N_0}\cdot\bigl( 1 -
    \exp\!\big(
      \frac{ 1 }{
        x - b_{N_0}
      }
    \big)\bigr)
    ,
  &
    \text{if }
    x < b_{N_0}
    ,
\\[1ex]
  d_N ,
  &
  \text{if } x = b_N \text{ and } N\ge N_0,
\\[1ex]
 d_{N-1}  +
  \displaystyle{
    \frac{
       d_N  -  d_{N-1}
    }{
              1 +
        \exp\!\big(
          \frac{ 1 }{  x - b_{ N - 1 }  }
          -
          \frac{ 1 }{
             b_N - x
          }
        \big)
          }
  }
  ,
  &
  \text{if }
  x \in ( b_{ N - 1 } , b_N ) \text{ and } N > N_0
  .
\end{cases}
\]
Then $ \psi $ is positive, strictly increasing,  infinitely often
differentiable and satisfies\\  $
  \lim_{ x \to \infty } \psi( x ) = \infty
$ as well as $1\in\psi( {\mathbb R} )$.

For $N\in{\mathbb N}$ put
\[
\varepsilon_N =  \inf_{\widehat X\in\mathcal{X}_N^{\delta_N}} {\mathbb E}
    \|
      X^{ \psi }( T ) -\widehat X
    \|.
\]
Theorem \ref{t1} implies that for
all $ N\geq N_0$
\[
  \varepsilon_N  \geq c_1
  \cdot
  \exp\bigl(
    - \tfrac{ 1 }{ \beta }
    \cdot
    (
      \psi^{ - 1 }(
       d_N
      )
    )^2
  \bigr)-\frac{c_2}{N}
  = c_1  \cdot\exp\bigl( - \tfrac{ 1 }{ \beta } \cdot b_N ^2 \bigr)-\frac{c_2}{N} = a_N
  .
\]
Since the sequence $(\varepsilon_N)_{N\in{\mathbb N}}$ is decreasing hence for all $ N \in \{ 1, 2, \dots, N_0 \} $
\[
 \varepsilon_N
\ge \varepsilon_{N_0} \ge a_{ N_0 }.
\]
Using the assumption that the sequence $
  ( a_N )_{ N \in {\mathbb N} }
$ is strictly decreasing we therefore conclude that for all $N\in{\mathbb N}$
\[
\varepsilon_N \ge  \min\{1,a_{N_0}/a_N\}\cdot  a_N \ge
\frac{a_{N_0}}{a_1}\cdot a_N,
\]
which completes the proof of the corollary with $\kappa=
a_{N_0}/a_1$.
\end{proof}

\section{Proof of Theorem \ref{t1}}
\label{strong}

Let $\delta\in (0, T]$ and let $\widehat X\in\mathcal{X}^\delta$ be given
by sequences $\varphi=(\varphi_n)_{n\in{\mathbb N}}, \chi=(\chi_n)_{n\in{\mathbb N}}$ and
$\phi=(\phi_n)_{n\in{\mathbb N}}$, see \eqref{sequences}. Recall the
definition \eqref{nu} of $\nu$. We first determine the regular
conditional distribution ${\mathbb P}^{W|D_\nu}$.

For $n\in{\mathbb N}$ put
\[
S_n=\{s\in(0, \delta)^n: |\{s_1, \ldots, s_n\}|=n\}.
\] For $n\in{\mathbb N}$, $s\in S_n$, $y\in{\mathbb R}^n$ and $v\in
C([\delta, T])$  define functions
\[
m_{s,y,v}\colon [0, T]\to {\mathbb R}\,\,\text{ and }\,\, R_s\colon [0,
T]^2\to {\mathbb R}
\] as follows.
If $s_1<\ldots <s_n$  put $s_0=y_0=0$, $s_{n+1}=\delta$ and
$y_{n+1}=v(\delta)$ and let
\[
m_{s,y,v}(t)=\begin{cases}
 \frac{s_{k}-t}{s_k-s_{k-1}}\cdot y_{k-1}+\frac{t-s_{k-1}}{s_k-s_{k-1}}\cdot y_k
    ,
  &
    \text{if }
    t\in [s_{k-1}, s_k) \text{ for } k\in\{1, \ldots, n+1\},  \\
    v(t)
    ,
  &
    \text{if }
    t\in [\delta, T]
\end{cases}
\]
as well as
\[
R_s(r,t)=\begin{cases}
 \frac{(s_{k}-\max(r,t))\cdot (\min(r,t)-s_{k-1})}{s_k-s_{k-1}}
    ,
  &
    \text{if }
    r,t\in [s_{k-1}, s_k) \text{ for } k\in\{1, \ldots, n+1\},  \\
    0
    ,
  &
    \text{otherwise},
\end{cases}
\]
for $r,t\in [0,T]$. Otherwise put
\[
m_{s,y,v}=m_{(s_{\pi(1)},\ldots,s_{\pi(n)}),
(y_{\pi(1)},\ldots,y_{\pi(n)}),v}, \quad
R_s=R_{(s_{\pi(1)},\ldots,s_{\pi(n)})},
\]
where $\pi$ is the permutation of $\{1, \ldots, n\}$ such that $s_{\pi(1)}<\ldots<s_{\pi(n)}$.

 For $n\in{\mathbb N}$, $y\in{\mathbb R}^n$,
$v\in C([\delta, T])$ and $k=1,\ldots,n$ put
\begin{equation}\label{14}
s^{y,v}_k=\varphi_k( y_1, \ldots, y_{k-1},v).
\end{equation}
Note that due to the assumption \eqref{15} we have $|\{s^{y,v}_1,
\ldots, s^{y,v}_n\}|=n$. Let $Q_{y,v}$ denote the Gaussian measure
on $\mathfrak B(C([0, T]))$ with mean $m(Q_{y,v})=m_{s^{y,v},y,v}$
and covariance function $R(Q_{y,v})=R_{s^{y,v}}$. Consider the
measurable space
\begin{equation}\label{Omega1}
 (\Omega_1, \mathcal F_1)=\Bigl(\bigcup_{n=1}^\infty {\mathbb R}^n\times C([\delta, T]), \sigma\Bigl(\bigcup_{n=1}^\infty \mathfrak B({\mathbb R}^n\times C([\delta, T]))\Bigr)\Bigr).
\end{equation}
It  is easy to see that $D_\nu\colon\Omega\to\Omega_1$ is $\mathcal
F$-$\mathcal F_1$ measurable. Define the mapping
\[
K\colon \Omega_1 \times \mathfrak B(C([0, T]))\to [0,1]
\]
by
\[
K((y,v), A)=Q_{y,v}(A)
\]
for all $y\in{\displaystyle \bigcup_{n=1}^\infty {\mathbb R}^n}, v\in
C([\delta, T])$  and $A\in \mathfrak B(C([0, T]))$.

\begin{lemma}\label{regular}
$K$ is a version of the regular conditional distribution
${\mathbb P}^{W|D_\nu}$.
\end{lemma}

In the case of $\delta=T$ the statement of Lemma \ref{regular}
seems to be well-known, see, e.g., \cite{hmr01, m02, MG02_habil, m04}, but a
proof of it seems not to be available in the literature. If,
additionally, $\nu$ is constant then Lemma \ref{regular} follows
from Lemma 2.9.7 in~\cite[p.~474]{TWW88}, but measurability issues
have not been fully addressed in the proof of the latter result. For
convenience of the reader we therefore provide a proof of Lemma
\ref{regular} here.

\begin{proof}
Clearly, for all $(y,v)\in \Omega_1$ the mapping
\[
\mathfrak B(C([0, T]))\ni A\mapsto K((y,v), A)\in [0,1]
\]
is a probability measure on $\mathfrak B(C([0, T]))$.

Next, let $A\in\mathfrak B(C([0, T]))$. We show that the mapping
\begin{equation}\label{l6}
\Omega_1\ni (y,v) \mapsto K((y,v), A)\in [0,1]
\end{equation}
is $\mathcal F_1$\,-\,$\mathfrak B([0,1])$ measurable. For $n\in{\mathbb N}$,
$s\in S_n$ and $u\in C([0,\delta])$ define a function
\[
F_{s,u}\colon [0,T]\to{\mathbb R}
\]
by
\[
F_{s,u}(t)=\begin{cases}
 u(t)-m_{s, (u(s_1), \ldots, u(s_n)),0}(t)
    ,
  &
    \text{if }
    t\in [0,\delta],
    \\
    0
    ,
  &
    \text{if }
    t\in (\delta, T]
    \end{cases}
\]
for $t\in[0, T]$.
 It is
easy to see that for all $(y, v)\in \Omega_1$
\[
Q_{y,v}={\mathbb P}^{F_{s^{y,v},(W(t))_{t\in[0, \delta]}}+m_{s^{y,v},y,v}},
\]
and therefore for all $(y, v)\in \Omega_1$
\begin{equation}\label{l5}
 K((y,v), A)=\int_{C([0, \delta])} 1_A(F_{s^{y,v},u}+m_{s^{y,v},y,v})\,\,{\mathbb P}^{(W(t))_{t\in[0, \delta]}}(du).
\end{equation}
Clearly,
\[
\{((y,v),u)\in\Omega_1\times C([0, \delta]):
F_{s^{y,v},u}+m_{s^{y,v},y,v}\in A\}=\bigcup_{n=1}^\infty A_n,
\]
where
\[
A_n=\{(y,v,u)\in{\mathbb R}^n\times C([\delta,T])\times C([0, \delta]):
F_{s^{y,v},u}+m_{s^{y,v},y,v}\in A\}
\]
 for $n\in{\mathbb N}$.
\begin{comment}
By assumption,  $\varphi_k$ is $\mathfrak{B}({\mathbb R}^{k-1}\times C([\delta, T]))$\,-\,$\mathfrak{B}((0,\delta))$ measurable for every $k=1, \ldots, n$.
Hence the mapping
\[
{\mathbb R}^n\times C([\delta, T]) \ni (y,v)\mapsto s^{y,v}\in (0, \delta)^n
\]
is $\mathfrak{B}({\mathbb R}^n\times C([\delta, T]))$\,-\,$\mathfrak{B}((0,\delta)^n)$ measurable. Moreover, the mapping
\[
(0,\delta)^n\times{\mathbb R}^n\times C([\delta, T]) \ni (s,y,v)\mapsto m_{s,y,v}\in C([0, T])
\]
is continuous and thus $\mathfrak B((0,\delta)^n\times{\mathbb R}^n\times C([\delta, T]))$\,-\,$\mathfrak B(C([0, T]))$ measurable. Thus, the mapping
\[
{\mathbb R}^n\times C([\delta, T]) \ni (y,v)\mapsto m_{s^{y,v},y,v}\in C([0, T])
\]
is $\mathfrak{B}({\mathbb R}^n\times C([\delta, T]))$\,-\,$\mathfrak B(C([0, T]))$ measurable.
\end{comment}
The measurability of the functions  $\varphi_n$, $n\in{\mathbb N}$, imply that for
every $n\in{\mathbb N}$ the mapping
\[
{\mathbb R}^n\times C([\delta, T]) \ni (y,v)\mapsto s^{y,v}\in S_n
\]
is $\mathfrak{B}({\mathbb R}^n\times C([\delta, T]))$\,-\,$\mathfrak{B}(S_n)$
measurable. Thus, observing that for every $n\in{\mathbb N}$ the mappings
\[
S_n\times C([0,\delta]) \ni (s,u)\mapsto F_{s,u}\in C([0, T])
\]
and
\[
S_n\times{\mathbb R}^n\times C([\delta, T]) \ni (s,y,v)\mapsto m_{s,y,v}\in C([0, T])
\]
are continuous we conclude that for every  $n\in{\mathbb N}$ the mapping
\[
{\mathbb R}^n\times C([\delta, T])\times C([0,\delta]) \ni (y,v,u)\mapsto
F_{s^{y,v},u}+m_{s^{y,v},y,v}\in C([0, T])
\]
is $\mathfrak{B}({\mathbb R}^n\times C([\delta, T])\times
C([0,\delta])$\,-\,$\mathfrak B(C([0, T]))$ measurable. Hence for
every  $n\in{\mathbb N}$
\[
A_n\in \mathfrak{B}({\mathbb R}^n\times C([\delta, T])\times
C([0,\delta]))\subset \mathcal F_1\otimes \mathfrak
B(C([0,\delta])),
\]
 which implies that the mapping
\[
\Omega_1\times C([0, \delta]) \ni((y,v),u)\mapsto
1_A(F_{s^{y,v},u}+m_{s^{y,v},y,v})\in{\mathbb R}
\]
is $\mathcal F_1\otimes \mathfrak B(C([0,\delta]))$\,-\,$\mathfrak
B(R)$ measurable. Using \eqref{l5} and employing Fubini's theorem we
thus conclude that the mapping \eqref{l6} is $\mathcal
F_1$\,-\,$\mathfrak B([0,1])$ measurable.

Finally, let $A\in\mathfrak B(C([0, T]))$ and $E\in \mathcal F_1$.
We show that
\begin{equation}\label{l3}
{\mathbb P} (\{W\in A\}\cap\{D_\nu\in E\})=\int_E K((y,v), A)\,\,
{\mathbb P}^{D_\nu}(d(y,v)).
\end{equation}
We have
\begin{align*}
\{D_\nu\in E\}\cap \{\nu<\infty\}&=\bigcup_{n=1}^\infty (\{D_\nu\in
E\}\cap \{\nu=n\})\\
&=\bigcup_{n=1}^\infty\Bigl(\{D_n\in E\}\cap
\{\chi_n(D_n)=1\}\cap\bigcap_{k=1}^{n-1}\{\chi_k(D_k)=0\}\Bigr)\\
&=\bigcup_{n=1}^\infty \{D_n\in E\cap C_n\},
\end{align*}
where $C_n\in \mathfrak B({\mathbb R}^n\times C([\delta, T]))$ is given by
\[
C_n= \chi_n^{-1}(\{1\})\cap \bigcap_{k=1}^{n-1} \{(y,v)\in
{\mathbb R}^n\times C([\delta, T]): (y_1, \ldots, y_k,v)\in
\chi_k^{-1}(\{0\})\}
\]
for $n\in{\mathbb N}$. Since $\nu<\infty$ a.s. we thus obtain
\begin{align*}
{\mathbb P} (\{W\in A\}\cap\{D_\nu\in E\})&=\sum_{n=1}^\infty {\mathbb P} (\{W\in
A\}\cap\{D_n\in E\cap C_n\})\\
&= \sum_{n=1}^\infty \int_{E\cap C_n}{\mathbb P}^{W|D_n=(y,v)} (A)
\,\,{\mathbb P}^{D_n}(d(y,v)).
\end{align*}
Similarly to the proof of Lemma 2.9.7 on page 474 in ~\cite{TWW88}
one can show that for all $n\in{\mathbb N}$  and all $G\in\mathfrak
B({\mathbb R}^n\times C([\delta, T]))$
\[
\int_{G}{\mathbb P}^{W|D_n=(y,v)} (A) \,\,{\mathbb P}^{D_n}(d(y,v))=\int_{G}Q_{y,v}
(A) \,\,{\mathbb P}^{D_n}(d(y,v)).
\]
Hence
\begin{align*}
{\mathbb P} (\{W\in A\}\cap\{D_\nu\in E\})&= \sum_{n=1}^\infty \int_{E\cap C_n}Q_{y,v}(A)
\,\,{\mathbb P}^{D_n}(d(y,v))\\
&=\sum_{n=1}^\infty \int_{E\cap C_n}K((y,v), A)
\,\,{\mathbb P}^{D_n}(d(y,v)).
\end{align*}
Since ${\mathbb P}^{D_n}(G\cap C_n) = {\mathbb P}^{D_\nu} (G)$  for all $n\in{\mathbb N}$ and
all $G\in\mathfrak B({\mathbb R}^n\times C([\delta, T]))$ we thus conclude
\[
{\mathbb P} (\{W\in A\}\cap\{D_\nu\in E\})= \sum_{n=1}^\infty \int_{E\cap ({\mathbb R}^n\times C([\delta, T]))}K((y,v), A)
\,\,{\mathbb P}^{D_\nu}(d(y,v)),
\]
 which implies \eqref{l3} and completes the proof of the
lemma.
\end{proof}

Next, assume that $\delta\in(0, \tau_1]$, let $n\in{\mathbb N}$,  $y\in{\mathbb R}^n$ and $v\in C([\delta,T])$. Recall the definition \eqref{14} of the time points
$s^{y,v}_1, \ldots, s^{y,v}_n$ and put
\begin{equation}\label{36}
s^{y,v}_0=0,\quad s^{y,v}_{n+1}=\delta.
\end{equation}
Let $\pi$ be the permutation of $\{0, \ldots, n+1\}$ such that $s^{y,v}_{\pi(0)}<\ldots <s^{y,v}_{\pi(n+1)}$.
  Let $i^*\in\{0, \ldots, n\}$ and put
\begin{equation}\label{t0t1}
t_0=s^{y,v}_{\pi(i^*)}, \quad t_1=s^{y,v}_{\pi(i^*+1)}.
\end{equation}
Define mappings
\[
\widetilde W\colon C([0,T])\to C([0, t_0]\cup [t_1, \tau_1]), \,\, B\colon C([0,T])\to C([t_0, t_1])
\]
by
\begin{equation}\label{18}
\widetilde W(w)=w|_{[0, t_0]\cup [t_1, \tau_1]}
\end{equation}
for $w\in C([0,T])$ and
\begin{equation}\label{20}
B(w)(t)=w(t)-\frac{t_1-t}{t_1-t_0}w(t_0)-\frac{t-t_0}{t_1-t_0}w(t_1)
\end{equation}
for $w\in C([0,T])$ and $t\in [t_0, t_1]$. In the following lemma we present properties of the measures $Q_{y,v}^{\widetilde W}$, $Q_{y,v}^{ B}$ and $Q_{y,v}^{(\widetilde W, B)}$, which will be used in the proof of Theorem~\ref{t1}.
\begin{lemma}\label{induced}
We have
\begin{itemize}
\item[(i)] $Q_{y,v}^{\widetilde W}$ is the Gaussian measure on $\mathfrak B(C([0, t_0]\cup [t_1, \tau_1]))$ with  mean\\ $m(Q_{y,v}^{\widetilde W})=m_{s^{y,v},y,v}|_{[0, t_0]\cup [t_1, \tau_1]}$ and covariance function
$R(Q_{y,v}^{\widetilde W})=R_{s^{y,v}}|_{([0, t_0]\cup [t_1,
\tau_1])^2}$.
\item[(ii)]$Q_{y,v}^{B}$ is the Gaussian measure on $\mathfrak B(C([t_0, t_1]))$ with  mean $m(Q_{y,v}^{B})=0$ and covariance function
\[
R(Q_{y,v}^{B})(r,t)=\frac{(t_1-\max(r,t))\cdot (\min(r,t)-t_0)}{t_1-t_0}, \quad r,t\in[t_0, t_1].
\]
\item[(iii)] $Q_{y,v}^B=Q_{y,v}^{-B}$,
\item[(iv)] $Q_{y,v}^{(\widetilde W, B)}=Q_{y,v}^{\widetilde W}\times Q_{y,v}^{B}$.
\end{itemize}
\end{lemma}
\begin{proof}
The property (i) is obvious. Since $B$ is a linear continuous mapping hence $Q_{y,v}^{B}$ is a
Gaussian measure. Next, observe that for all $t\in [t_0, t_1]$ we have
\begin{align*}
m(Q_{y,v})(t)&=\frac{t_1-t}{t_1-t_0}y_{\pi(i^*)}+\frac{t-t_0}{t_1-t_0}y_{\pi(i^*+1)}.
\end{align*}
In particular, for all $t\in[t_0, t_1]$,
\begin{equation}\label{19}
m(Q_{y,v})(t)-\frac{t_1-t}{t_1-t_0}m(Q_{y,v})(t_0)
-\frac{t-t_0}{t_1-t_0}m(Q_{y,v})(t_1)=0.
\end{equation}
Hence for all
$t\in [t_0, t_1]$,
\begin{align*}
m(Q_{y,v}^{B})(t)&=\int_{C([t_0, t_1])}b(t)Q_{y,v}^{B}(db)\\
&= \int_{C([0, T])}\Bigl(w(t)-\frac{t_1-t}{t_1-t_0}w(t_0)-\frac{t-t_0}{t_1-t_0}w(t_1)\Bigr)Q_{y,v}(dw)\\
&=m(Q_{y,v})(t)-\frac{t_1-t}{t_1-t_0}m(Q_{y,v})(t_0) -\frac{t-t_0}{t_1-t_0}m(Q_{y,v})(t_1)=0
\end{align*}
and for all $r,t\in [t_0, t_1]$,
\begin{align*}
R(Q_{y,v}^{B})(r,t)
&=\int_{C([t_0, t_1])}\bigl(b(r)-m(Q_{y,v}^{B})(r)\bigr)\cdot \bigl(b(t)-m(Q_{y,v}^{B})(t)\bigr) Q_{y,v}^{B}(db)\\
&= \int_{C([0, T])}\Bigl(w(r)-\frac{t_1-r}{t_1-t_0}w(t_0)-\frac{r-t_0}{t_1-t_0}w(t_1)\Bigr)\\
&\hspace{3cm}\cdot \Bigl(w(t)-\frac{t_1-t}{t_1-t_0}w(t_0)-\frac{t-t_0}{t_1-t_0}w(t_1)\Bigr)Q_{y,v}(dw),
\end{align*}
and hence
\begin{align*}
R(Q_{y,v}^{B})(r,t)&= \int_{C([0, T])}\Bigl(w(r)-m(Q_{y,v})(r)-\frac{t_1-r}{t_1-t_0}(w(t_0)-m(Q_{y,v})(t_0))\\
&\hspace{6cm}-\frac{r-t_0}{t_1-t_0}(w(t_1)-m(Q_{y,v})(t_1))\Bigr)\\
&\hspace{3cm}\cdot \Bigl(w(t)-m(Q_{y,v})(t)-\frac{t_1-t}{t_1-t_0}(w(t_0)-m(Q_{y,v})(t_0))\\
&\hspace{6cm}-\frac{t-t_0}{t_1-t_0}(w(t_1)-m(Q_{y,v})(t_1))\Bigr)Q_{y,v}(dw)\\
&=R(Q_{y,v})(r,t)-\frac{t_1-t}{t_1-t_0}R(Q_{y,v})(r,t_0)-\frac{t-t_0}{t_1-t_0}R(Q_{y,v})(r,t_1)\\
&\quad-\frac{t_1-r}{t_1-t_0}\Bigl(R(Q_{y,v})(t_0,t)-\frac{t_1-t}{t_1-t_0}R(Q_{y,v})(t_0,t_0)-\frac{t-t_0}{t_1-t_0}R(Q_{y,v})(t_0,t_1)\Bigr)\\
&\quad-\frac{r-t_0}{t_1-t_0}\Bigl(R(Q_{y,v})(t_1,t)-\frac{t_1-t}{t_1-t_0}R(Q_{y,v})(t_1,t_0)-\frac{t-t_0}{t_1-t_0}R(Q_{y,v})(t_1,t_1)\Bigr).
\end{align*}
Observing that $R(Q_{y,v})(t_0,t)=R(Q_{y,v})(t_1,t)=0$ for all $t\in [t_0, t_1]$ we thus obtain
\[
R(Q_{y,v}^{B})(r,t)=R(Q_{y,v})(r,t)
\]
for all $r,t\in [t_0, t_1]$, which completes the proof of (ii). The property (ii) implies that
$Q_{y,v}^{-B}$ is the Gaussian measure on $\mathfrak B(C([t_0, t_1]))$ with  mean $m(Q_{y,v}^{-B})=0$ and
covariance function $R(Q_{y,v}^{-B})=R(Q_{y,v}^{B})$, which yields
the property (iii). Next, we prove (iv). Using the properties (i),
(ii) as well as \eqref{19} and the fact that $R(Q_{y,v})(r,t)=0$ for all $r\in [0,t_0]\cup [t_1,
\tau_1]$ and $t\in [t_0, t_1]$ we obtain
\begin{align*}
&\int_{C([0,T])}\bigl(\widetilde W(w)(r)-m(Q_{y,v}^{\widetilde
W})(r)\bigr)\cdot\bigl(B(w)(t)-m(Q_{y,v}^{B})(t)\bigr)
Q_{y,v}(dw)\\
&=\int_{C([0,T])}\bigl(w(r)-m(Q_{y,v})(r)\bigr)\cdot\Bigl(w(t)-\frac{t_1-t}{t_1-t_0}w(t_0)-\frac{t-t_0}{t_1-t_0}w(t_1)\Bigr)
Q_{y,v}(dw)\\
&=\int_{C([0,T])}\bigl(w(r)-m(Q_{y,v})(r)\bigr)\cdot\Bigl(w(t)-m(Q_{y,v})(t)-\frac{t_1-t}{t_1-t_0}(w(t_0)-m(Q_{y,v})(t_0))\\
&\hspace{8cm}-\frac{t-t_0}{t_1-t_0}(w(t_1)-m(Q_{y,v})(t_1))\Bigr)
Q_{y,v}(dw)\\
&=R(Q_{y,v})(r,t)-\frac{t_1-t}{t_1-t_0}R(Q_{y,v})(r,t_0)-\frac{t-t_0}{t_1-t_0}
R(Q_{y,v})(r,t_1)=0
\end{align*} for all $r\in [0,t_0]\cup [t_1,
\tau_1]$ and $t\in [t_0, t_1]$, which means that $\widetilde W(r)$ and $B(t)$ are uncorrelated.
This and the fact that $Q_{y,v}^{\widetilde W}$ and $Q_{y,v}^{B}$
are Gaussian measures implies (iv).

\end{proof}

In the proof of Theorem~\ref{t1} we employ the following lower bound
for the first absolute moment of the sine of a normally distributed
random variable, which is a generalization of Lemma 4.2 from
~\cite{JMGY15}, where a centered normally distributed random
variable has been considered.

\begin{lemma}
\label{l2}
Let $a\in{\mathbb R}$, $ \tau \in [1,\infty) $,
and let $ Y \colon \Omega \to {\mathbb R} $ be a
$ \mathcal{N}( a, \tau^2 ) $-distributed random variable.
Then
\[
  {\mathbb E}\big[
    | \sin(Y) |
  \big]
  \ge
  \frac{ \exp\bigl(- \tfrac{\pi^2}{8}\bigr)}{ \sqrt{ 8 \pi } }
  .
\]
\end{lemma}
The proof of Lemma \ref{l2} is a straighforward generalization of the proof of Lemma 4.2 from
~\cite{JMGY15}.

\begin{comment}
\begin{proof}
   We have
\begin{align*}
  {\mathbb E}\big[
    | \sin(Y) |
  \big]
  &=
  \frac{ 1 }{ \sqrt{ 2 \pi } }
  \int_{\mathbb R}
  | \sin( \tau (z+a) ) |
  \exp(
    -
    z^2/ 2 )
  dz
\\
  &
  \ge
  \frac{ \exp\bigl(- \tfrac{\pi^2}{8}\bigr)
  }{
    \sqrt{ 2 \pi }
  }
  \int_0^{ \frac{ \pi }{ 2 } }
  \left| \sin( \tau (z+a) ) \right|
  dz\\
  &=
  \frac{
     \exp\bigl(- \tfrac{\pi^2}{8}\bigr)
  }{
    \tau \sqrt{ 2 \pi }
  }
  \int_0^{ \frac{ \tau \pi }{ 2 } }
  \left| \sin(x+a\tau) \right|
  dx
  .
\end{align*}
This and the fact that
\[
  \int_0^{
    \frac{ \tau \pi }{ 2 }
  }
  | \sin(x+a\tau) | \, dx
\ge
  \int_0^{
    \lfloor \tau \rfloor \cdot \frac{ \pi }{2 }
  }
  |\sin(x+a\tau)| \, dx
=
  \lfloor \tau \rfloor \cdot
  \int_0^{ \frac{ \pi }{ 2 } }
  \sin(x+a\tau)\, dx
=
  \lfloor \tau \rfloor
\ge
  \frac{ \tau }{ 2 }
\]
complete the proof.
\end{proof}
\end{comment}

We now proceed with the proof of Theorem \ref{t1}.

 We first consider the case  $\delta\in (0, \tau_1/2]$.
Let $N\in{\mathbb N}$ be such that $\widehat X\in \mathcal{X}_{N}^\delta$. We may then assume that $\widehat X\in \mathcal{X}_{N+1}^\delta$ and that $\widehat X$ uses  the
evaluation site $\delta/2$.
  Let
$ \psi \in C^{ \infty }( {\mathbb R}) $ be positive, strictly increasing
with $
  \lim_{ x \to \infty } \psi( x ) = \infty
$ and $
  1\in \psi({\mathbb R}).
$
Using Remark \ref{solution}, the assumptions on the functions $f$
and $g$ and It\^o's formula we obtain that ${\mathbb P}$-a.s.
\begin{equation}\label{44}
X_2^\psi(\tau_1)=-\int_0^{\tau_1} f'(t)W(t)dt, \quad X_3^\psi(\tau_2)=-\int_{\tau_1}^{\tau_2} g'(t)W(t)dt.
\end{equation}
Thus, ${\mathbb P}$-a.s.
\begin{equation}\label{11}
X_4^{ \psi }( T )=F(W),
\end{equation}
where $F\colon C([0,T])\to {\mathbb R}$ is given by
\[
F(w)=\gamma \cdot \cos\Bigl(\int_0^{\tau_1} f'(t)w(t)dt \cdot \psi\Bigl(-\int_{\tau_1}^{\tau_2} g'(t)w(t)dt\Bigr)\Bigr)
\]
and $\gamma$ is defined in \eqref{51}.
Recall the definition \eqref{Omega1} of the measurable space $(\Omega_1, \mathcal F_1)$ and define a function $G\colon \Omega_1\to{\mathbb R}^4$ by
\[
G(y,v)=\phi_n(y,v)
\]
for $y\in{\mathbb R}^n$ and $v\in C([\delta, T])$. Due to the measurability of the functions $\phi_n$, $n\in{\mathbb N}$, the function $G$ is $\mathcal F_1$-$\mathfrak B({\mathbb R}^4)$ measurable. Moreover,
\begin{equation}\label{12}
\widehat X=\phi_{\nu}(D_{\nu})=G(D_{\nu}).
\end{equation}
Let $\text{pr}_4\colon {\mathbb R}^4\to{\mathbb R}$ denote the projection to the
fourth component. Due to \eqref{11} and \eqref{12} we have
\begin{equation}\label{42}
{\mathbb E}| X_4^{ \psi }( T ) -\widehat X_4|={\mathbb E}
    |
      F(W) -\text{pr}_4(G(D_{\nu}))
    |={\mathbb E}\,\bigl({\mathbb E}\bigl(
    |
      F(W) -\text{pr}_4(G(D_{\nu}))
    |\,\bigr|\,D_\nu\bigr)\bigr).
\end{equation}
Lemma \ref{regular} implies that for ${\mathbb P}^{D_\nu}$-a.a. $(y,v)\in\Omega_1$
\begin{align}\label{13}
{\mathbb E}\bigl(
    |
      F(W) -\text{pr}_4(G(D_{\nu}))
    |\,\bigr|\,D_\nu=(y,v)\bigr)
    &=\int_{C([0,T])}  |
      F(w) -\text{pr}_4(G(y,v))
    |\, Q_{y,v}(dw).
\end{align}

Fix $n\in{\mathbb N}$,  $y\in{\mathbb R}^n$ and $v\in C([\delta,T])$.
We show that
\begin{align}\label{41}
&\int_{C([0,T])}  |
      F(w) -\text{pr}_4(G(y,v))
    |\, Q_{y,v}(dw)\notag\\
    &\qquad\qquad\qquad\geq \frac{ \gamma \exp \bigl( - \tfrac{\pi^2}{ 4}\bigr) }{  8 \pi  }
  \cdot 1_{[1, \infty)}\Bigl(\sqrt{\tfrac{\alpha\delta^3}{96 n^3}}\cdot\psi\Bigl(-\int_{\tau_1}^{\tau_2} g'(t)\, v(t)dt\Bigr)\Bigr),
\end{align}
where $\alpha$ is given by \eqref{51}.
Recall the definitions \eqref{14} and \eqref{36} of the time points
$s^{y,v}_0, \ldots, s^{y,v}_{n+1}$, let $\pi$ be the permutation of $\{0, \ldots, n+1\}$ such that $s^{y,v}_{\pi(0)}<\ldots <s^{y,v}_{\pi(n+1)}$,  and
let $t_0$ and  $t_1$ be given by \eqref{t0t1} with
\begin{equation}\label{34}
i^*=\min\{i\in\{0, \ldots, n\}:|s^{y,v}_{\pi(i+1)}-s^{y,v}_{\pi(i)}|\geq\delta/(n+1)\}.
\end{equation}
Put
\[ C_1=C([0, t_0]\cup [t_1, \tau_1]),\quad C_2=C([t_0,
t_1]),\quad C_3=C([\tau_1, \tau_2]),
\] let $ \widetilde W\colon
C([0,T])\to C_1$ and $B\colon C([0,T])\to C_2$ be given by
\eqref{18} and \eqref{20}. respectively, and define  $\overline W\colon C([0,T])\to
C_3$ by
\[
\overline W(w)=w|_{[\tau_1,\tau_2]}
\]
for $w\in C([0,T])$. Moreover, define mappings $H_1\colon C_1
\to{\mathbb R}$, $H_2\colon C_2\to{\mathbb R}$, $H_3\colon C_3\to{\mathbb R}$ and $J\colon
C_1\times C_2\times C_3\to{\mathbb R}$ by
\begin{align*}
&H_1( \widetilde w)=\int_{0}^{t_0} f'(t)\, \widetilde w(t)dt+\int_{t_0}^{t_1} f'(t)\,\Bigl(\frac{t-t_0}{t_1-t_0} \widetilde w(t_1)+\frac{t_1-t}{t_1-t_0} \widetilde
w(t_0)\Bigr)dt+\int_{t_1}^{\tau_1} f'(t)\, \widetilde w(t)dt,\\
&H_2(b)=\int_{t_0}^{t_1} f'(t)\,b(t)dt, \quad H_3(\overline w)=-\int_{\tau_1}^{\tau_2} g'(t)\, \overline w(t)dt
\end{align*}
as well as
\[
J( \widetilde w,b, \overline w)=\gamma\cdot
\cos\bigl((H_1(\widetilde w)+H_2(b))\cdot \psi(H_3( \overline
w))\bigr)
\]
 for $\widetilde w\in C_1$, $b\in C_2$ and $\overline w\in C_3$. We
 then have
\[
F=J(\widetilde W, B, \overline W).
\]
Clearly, $ Q^{\overline W}_{y,v}$ is the dirac measure concentrated at $v|_{[\tau_1, \tau_2]}$.
 Using Lemma \ref{induced}(iii),(iv) and the triangle inequality we
thus obtain
\begin{align*}
&\int_{C([0,T])}  |
      F(w) -\text{pr}_4(G(y,v))
    |\, Q_{y,v}(dw)\\
    &\qquad\quad =\int_{C_1}\int_{C_2} |
      J(\widetilde w, b, v|_{[\tau_1, \tau_2]}) -\text{pr}_4(G(y,v))
    |\, Q^{B}_{y,v}(db)\,
    Q^{\widetilde W}_{y,v}(d\widetilde w)\\
    &\qquad\quad  =\int_{C_1}\int_{C_2} \frac{1}{2}\, \bigl(|
      J(\widetilde w, b, v|_{[\tau_1, \tau_2]})  -\text{pr}_4(G(y,v))
    |\\
    &\hspace{5cm}+|
      J(\widetilde w, -b, v|_{[\tau_1, \tau_2]})  -\text{pr}_4(G(y,v))
    |\bigr)\, Q^{B}_{y,v}(db)\,
    Q^{\widetilde W}_{y,v}(d\widetilde w)\\
    &\qquad\quad  \geq\int_{C_1}\int_{C_2}   \frac{1}{2}\,|
      J(\widetilde w, b, v|_{[\tau_1, \tau_2]})-
      J(\widetilde w, -b, v|_{[\tau_1, \tau_2]})
    |\, Q^{B}_{y,v}(db)\,
 Q^{\widetilde W}_{y,v}(d\widetilde w).
\end{align*}
Put $z=H_3(v|_{[\tau_1, \tau_2]})$. The fact that for all
$x,y\in{\mathbb R}$
\[
\cos(x)-\cos(y)=2\sin(\tfrac{y-x}{2}) \sin(\tfrac{y+x}{2})
\]
thus implies
\begin{align*}
&\int_{C([0,T])}  |
      F(w) -\text{pr}_4(G(y,v))
    |\, Q_{y,v}(dw)\\
    &\qquad\qquad \geq\gamma \int_{C_1}|
     \sin\bigl(H_1( \widetilde w)\cdot \psi(z)\bigr)|\,  Q^{\widetilde W}_{y,v}(d\widetilde w)\cdot \int_{C_2}
    |\sin\bigl(H_2(b)\cdot \psi(z)\bigr)
    |\, Q^{B}_{y,v}(db)\\
    &\qquad\qquad =\gamma \int_{{\mathbb R}}|
     \sin(y\cdot \psi(z))|\,  Q^{H_1(\widetilde W)}_{y,v}(dy)\cdot \int_{{\mathbb R}}
    |\sin(y\cdot \psi(z))
    |\, Q^{H_2(B)}_{y,v}(dy).
\end{align*}
Due to Lemma \ref{induced}(i),(ii), $Q^{\widetilde W}_{y,v}$ and
$Q^{B}_{y,v}$ are Gaussian measures on $\mathfrak B(C_1)$ and
$\mathfrak B(C_2)$, respectively. Since the mappings $H_1$ and $H_2$
are linear and continuous we conclude that $Q^{H_1(\widetilde
W)}_{y,v}$ and $Q^{H_2(B)}_{y,v}$ are Gaussian measures on
$\mathfrak B({\mathbb R})$. Let $m_1, \sigma_1^2$ and  $m_2, \sigma_2^2$ denote the mean and the
variance of $Q^{H_1(\widetilde W)}_{y,v}$ and $Q^{H_2(B)}_{y,v}$,
respectively. Applying Lemma \ref{l2} we obtain
\[
\int_{{\mathbb R}}|
     \sin(y\cdot \psi(z))|\,  Q^{H_1(\widetilde W)}_{y,v}(dy)\geq \frac{ \exp \bigl(- \tfrac{\pi^2}{8}\bigr) }{ \sqrt{ 8 \pi } }
  \cdot 1_{[1, \infty)}(\sigma_1\cdot\psi(z))
\]
as well as
\[
\int_{{\mathbb R}}
      |\sin(y\cdot \psi(z))
    |\, Q^{H_2(B)}_{y,v}(dy)\geq \frac{ \exp \bigl(- \tfrac{\pi^2}{8}\bigr) }{ \sqrt{ 8 \pi } } \cdot 1_{[1, \infty)}(\sigma_2\cdot\psi(z)).
\]
Hence
\begin{equation}\label{40}
\int_{C([0,T])}  |
      F(w) -\text{pr}_4(G(y,v))
    |\, Q_{y,v}(dw)\geq \frac{ \gamma  \exp\bigl(- \tfrac{\pi^2}{4}\bigr) }{  8 \pi  }
 \cdot 1_{[1, \infty)}\bigl(\min(\sigma_1,\sigma_2)\cdot\psi(z)\bigr).
\end{equation}
Next we derive lower bounds for $\sigma_1^2$ and $\sigma_2^2$.  Due to Lemma \ref{induced}(ii) we have
\[
m_2=\int_{C_2} H_2(b)\, Q^B_{y,v}(db)=\int_{t_0}^{t_1}f'(t)\cdot m(Q^{B}_{y,v})(t)\,dt=0
\]
as well as
\begin{align*}
\sigma_2^2&=\int_{C_2} (H_2(b)-m_2)^2\, Q^B_{y,v}(db)=\int_{t_0}^{t_1}\int_{t_0}^{t_1} f'(r)\cdot f'(t)\cdot R(Q^{B}_{y,v})(r,t)dr dt\\
&=\int_{t_0}^{t_1}\int_{t_0}^{t_1} f'(r)\cdot f'(t)\cdot \frac{(t_1-\max(r,t))\cdot (\min(r,t)-t_0)}{t_1-t_0}dr dt.
\end{align*}
It is easy to see that for all $a,b\in{\mathbb R}$ with $a<b$
\begin{equation}\label{30}
\int_{a}^{b}\int_{a}^{b} \frac{(b-\max(r,t))\cdot (\min(r,t)-a)}{b-a}dr dt=\frac{(b-a)^3}{12}.
\end{equation}
Moreover, the assumption $ \inf_{ t\in [ 0, \tau_1/2  ] } | f'(t) | > 0$ implies that for all $r,t\in[0, \tau_1/2]$
\begin{equation}\label{31}
f'(r)\cdot f'(t)=|f'(r)\cdot f'(t)|\geq \alpha.
\end{equation}
Observing \eqref{34} we thus obtain
\begin{equation}\label{35}
\sigma_2^2\geq \frac{\alpha(t_1-t_0)^3}{12}\geq  \frac{\alpha\delta^3}{12(n+1)^3}.
\end{equation}
Put $s_i=s^{y,v}_{\pi(i)}$ for $i=0, \ldots, n+1$ as well as  $s_{n+2}=\tau_1$. Clearly,
\begin{align*}
m_1&=\int_{C_1} H_1(\widetilde w)\, Q^{\widetilde W}_{y,v}(d\widetilde w )=\sum_{i\in \{0, \ldots, n+1\}\setminus \{i^*\}}\int_{s_i}^{s_{i+1}} f'(t) m(Q^{\widetilde W}_{y,v})(t)dt\\
&\hspace{5.3cm}+\int_{t_0}^{t_1} f'(t)\Bigl(\frac{t-t_0}{t_1-t_0}  m(Q^{\widetilde W}_{y,v})(t_1)+\frac{t_1-t}{t_1-t_0}  m(Q^{\widetilde W}_{y,v})(t_0)\Bigr)dt.
\end{align*}
Hence
\begin{align*}
\sigma_1^2&=\int_{C_1} (H_1(\widetilde w)-m_1)^2\, Q^{\widetilde W}_{y,v}(d\widetilde w )\\
&= \int_{C_1} \Bigl(\sum_{i\in \{0, \ldots, n+1\}\setminus \{i^*\}}\int_{s_i}^{s_{i+1}} f'(t)\cdot (\widetilde w(t)-m(Q^{\widetilde W}_{y,v})(t))dt\\
&\hspace{2cm} +\int_{t_0}^{t_1} f'(t)\cdot\frac{t-t_0}{t_1-t_0} \cdot (\widetilde w(t_1)-m(Q^{\widetilde W}_{y,v})(t_1))dt\\
&\hspace{4cm}+\int_{t_0}^{t_1}  f'(t)\cdot \frac{t_1-t}{t_1-t_0}\cdot (\widetilde w(t_0)-  m(Q^{\widetilde W}_{y,v})(t_0))dt\Bigr)^2 Q^{\widetilde W}_{y,v}(d\widetilde w ).
\end{align*}
Lemma \ref{induced}(i)  implies that $R(Q^{\widetilde W}_{y,v})(r, t)=0$ for all  $r\in[s_i, s_{i+1}]$ and $t\in [s_j, s_{j+1}]$ and all $i, j\in\{0, \ldots, n\}$ with $i\not =j$
as well as  $R(Q^{\widetilde W}_{y,v})(r, t)=0$ for all $r\in [0, \tau_1]$ and $t\in \{t_0, t_1\}\cup [s_{n+1}, s_{n+2}]$. Thus
\begin{align*}
\sigma_1^2
&=\sum_{i\in \{0, \ldots, n\}\setminus \{i^*\}}\int_{s_i}^{s_{i+1}}\int_{s_i}^{s_{i+1}} f'(r)\cdot f'(t)\cdot R(Q^{\widetilde W}_{y,v})(r,t)dr dt\\
&= \sum_{i\in \{0, \ldots, n\}\setminus \{i^*\}}\int_{s_i}^{s_{i+1}}\int_{s_i}^{s_{i+1}} f'(r)\cdot f'(t)\cdot \frac{(s_{i+1}-\max(r,t))\cdot (\min(r,t)-s_i)}{s_{i+1}-s_i}dr dt.
\end{align*}
Using \eqref{31}, \eqref{30} and the H\"older inequality we therefore obtain
\begin{align*}
\sigma_1^2
&\geq \sum_{i\in \{0, \ldots, n\}\setminus \{i^*\}}\int_{s_i}^{s_{i+1}}\int_{s_i}^{s_{i+1}} |f'(r)\cdot f'(t)|\cdot \frac{(s_{i+1}-\max(r,t))\cdot (\min(r,t)-s_i)}{s_{i+1}-s_i}dr dt\\
&\geq \frac{\alpha}{12}\sum_{i\in \{0, \ldots, n\}\setminus \{i^*\}} (s_{i+1}-s_i)^3\\
&\geq \frac{\alpha}{12 n^2}\cdot\bigl(\sum_{i\in \{0, \ldots, n\}\setminus \{i^*\}} (s_{i+1}-s_i) \bigr)^3= \frac{\alpha}{12 n^2}\cdot (\delta-(t_1-t_0))^3.
\end{align*}
The assumption that $\widehat X$ uses the evaluation site $\delta/2$ implies that $t_1-t_0\leq\delta/2$. Hence
\begin{equation}\label{360}
\sigma_1^2\geq  \frac{\alpha\delta^3}{96 n^2}.
\end{equation}
The desired lower bound \eqref{41} follows from \eqref{40}, \eqref{35} and \eqref{360}.

We conclude from \eqref{42},\eqref{13} \eqref{41} and \eqref{44} that
\begin{equation}\label{71}
{\mathbb E}| X_4^{ \psi }( T ) -\widehat X_4|\geq \frac{ \gamma \exp \bigl(- \tfrac{\pi^2}{4}\bigr) }{  8 \pi  }\cdot
   {\mathbb P} \Bigl(\Bigl\{\psi(X_3^\psi(\tau_2))\geq \sqrt{\tfrac{96 \nu^3}{\alpha \delta^3}}\Bigr\}\Bigr).
\end{equation}
Put
\[
A=\Bigl\{\psi(X_3^\psi(\tau_2))\geq \bigl(1+ \sqrt{\tfrac{96 }{\alpha \delta^3}}\bigr) N^3\Bigr\}, \quad B=\{\nu\leq N^2\}.
\]
Clearly,
\begin{align}\label{70}
{\mathbb P} \Bigl(\Bigl\{\psi(X_3^\psi(\tau_2))\geq \sqrt{\tfrac{96 \nu^3}{\alpha \delta^3}}\Bigr\}\Bigr)&\geq {\mathbb P} \Bigl(\Bigl\{\psi(X_3^\psi(\tau_2))\geq \sqrt{\tfrac{96 \nu^3}{\alpha \delta^3}}\Bigr\}\cap \{\nu\leq N^2\}\Bigr)\notag\\
&\geq {\mathbb P}(A\cap B)\geq{\mathbb P}(A)-{\mathbb P}(B^c).
\end{align}
Note that $X_3^\psi(\tau_2)\sim  \mathcal{N}( 0, \beta ) $, where $\beta$ is given by \eqref{51}. Moreover, the assumption that $\psi$ is continuous with $\lim_{x\to\infty} \psi(x)=\infty$ and  $1\in\psi({\mathbb R})$ ensures that
\[
\bigl(1+ \sqrt{\tfrac{96 }{\alpha \delta^3}}\bigr) N^3\in\psi({\mathbb R}).
\]
Hence
\begin{align}\label{72}
{\mathbb P} (A)&={\mathbb P}\Bigl(\Bigl\{X_3^\psi(\tau_2)\geq \psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha \delta^3}}\bigr)  N^3\bigr)\Bigr\}\Bigr)\notag\\
&\geq \frac{1}{\sqrt{2\pi\beta}}\int_{\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha \delta^3}}\bigr)  N^3\bigr)}^{1+\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha \delta^3}}  \bigr)N^3\bigr)} \exp\bigl(-\tfrac{x^2}{2\beta}\bigr) dx\notag\\
&\geq \frac{1}{\sqrt{2\pi\beta}}\cdot\exp\bigl(-\tfrac{1}{2\beta}\cdot \bigl(1+\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha \delta^3}}\bigr)  N^3\bigr)\bigr)^2\bigr)\notag\\
&\geq \frac{\exp\bigl(-\tfrac{1}{\beta}\bigr)}{\sqrt{2\pi\beta}}\cdot\exp\bigl(-\tfrac{1}{\beta}\cdot \bigl(\psi^{-1}\bigl(\bigl(1+ \sqrt{\tfrac{96 }{\alpha \delta^3}}\bigr)  N^3\bigr)\bigr)^2\bigr).
\end{align}
By the Markov inequality and the fact that $\widehat X\in\mathcal{X}^\delta_{N+1}$,
\begin{equation}\label{73}
{\mathbb P}(B^c)\leq \frac{{\mathbb E}\nu}{N^2}\leq \frac{N+1}{N^2}\leq  \frac{2}{N}.
\end{equation}
Estimates \eqref{71}, \eqref{70}, \eqref{72} and \eqref{73} imply \eqref{61}, which completes the proof of the theorem in the case $\delta\in(0, \tau_1/2]$.

If $\delta\in (\tau_1/2,T]$ then the lower bound \eqref{61} follows from the  fact that $\mathcal{X}_N^\delta\subseteq \mathcal{X}_N^{\tau_1/2}$ and the lower bound \eqref{61} in the case $\delta=\tau_1/2$.

\section*{Acknowledgement}
I am
grateful to Thomas M\"uller-Gronbach for stimulating discussions
on the topic of this article.

\bibliographystyle{acm}

redpajama_set_name: RedPajamaArXiv


text:
Breed Specific Healthcare
Ear Cropping Surgery and Care
Hospice and Euthanasia Services
Lantana Animal Hospital
Providing pampering for pets & people from Lantana, Bartonville, Copper Canyon, Double Oak, Highland Village, Flower Mound & surrounding areas. Drop by for a treat for you & your pet & take a tour of our hospital!
4110 Justin Road
Flower Mound,Texas 75077
M, T, W, F: 7:00am-6:00pm
Th: 7:00am-8:00pm
EMERGENCY (469) 464-2964 Flower Mound Emergency Clinic
Dr. Leigh Anne Robertson, DVM
Dr. Robertson graduated from Texas A&M College of Veterinary Medicine with honors in May of 2000. Although animals have been her lifelong passion, veterinary medicine is her second career. After completing her BS from Texas A&M in 1990, she spent 5 years working for Andersen Consulting as a Technology Integration Services Consultant. She resigned this position to attend veterinary school, pursuing her passion for animal care and to truly make a difference for people too.
She is a native of Bryan-College Station and was raised on a family ranch active in raising exotic animals and birds. She was a member of 4-H, showing rabbits, lambs, steers and horses in numerous fairs, livestock shows, rodeos and barrel racing events. Dr. Robertson shows Doberman Pinschers in conformation and performance events and is an active member of the Doberman Pinscher Club of Dallas and the Doberman Pinscher Club of America.
For the past 4 years, Lantana has been her home along with her dogs Derby, Jezebel and Blondie. She has two awesome adopted clinic cats, SK & Ranger, who will be greeting you at the hospital. SK loves dogs and may volunteer to perform a a CAT scan for your pet! When Dr. Robertson is not at work pampering your pets, she will most likely be found riding or racing her mountain bike. She also is actively involved with Valley Creek Church and especially enjoys Soul Sisters.
Dr. Katherine Hall, DVM
Dr. Hall graduated from Texas A&M College of Veterinary Medicine with honors in May of 2018. While she loves all pets, cats are her passion. During school, she was actively involved in the Student Chapter of the American Association of Feline Practitioners. Before vet school, she worked with alpacas and llamas.
Dr. Hall was born and raised in Farmers Branch and is thrilled to be back in North Texas. She has just recently joined our practice making a wonderful new addition to our growing team! Growing up, she rode horses and practiced karate. Outside of the clinic, she enjoys knitting, spinning yarn, and playing computer games.
She shares her home with two cats- Gabby and Kale'a. She target trained both of her cats and taught one to sit on command. She is still working on teaching the more stubborn one to sit.
Gail Dunlap - Client Services Representative
Gail Dunlap has been with Lantana Animal Hospital since opening in 2013 and has four wonderful Doberman Pinschers, Vendetta, Brooklyn, Archie and Chicago. Gail belongs to the Doberman Pinscher Club of Dallas and enjoys showing her Dobermans and additionally runs Agility.
Deana Orlandi - Client Services Representative
Deana recently joined the practice in August of 2018. She was born and raised in here in Lubbock,Texas. She received her BBA from Letourneau University in and has 13 years experience in the Luxury Hospitality Industry. She enjoys interacting with people and providing personalized customer service. She has traveled extensively, and especially loves Hawaii and Australia!
Outside of work she likes to spend time camping and hiking with her family and two Shetland sheep dogs Kalea and Kai.
Andrea Terranova- Client Services Representative
Andrea has been with Lantana Animal Hospital since September 2018. Born and raised in Southern California she moved to North Texas in 2006. Andrea has worked in the veterinary field as a Customer Service Representative and Veterinary Assistant since 2006. Prior to working in the veterinary field Andrea worked for 2 professional drag racing teams in the NHRA both in the office and working events trackside.
In her free time she enjoys cooking / baking, catching up on her DVR and spending time with her 2 children, and Bulldog "Dozer" and Siberian Husky "Skyla".
Kacie Packer - Practice Manager
Kacie joined the practice in 2014 as our Technician Director. She grew up in Los Angeles, California and moved to Dallas in 2010. Kacie began her career in veterinary medicine in 2008 as a receptionist and shortly thereafter moved into the role of Veterinary Technician, where she became the Director of Nursing at her previous practice. Kacie spent 6 years with a corporate veterinary practice, where she was a Practice Mananger for 3 of those 6 years, but decided that private practice was where she preferred to be long term.
Kacie studied at California State University Channel Islands, and is currently pursing her Licensing as a Veterinary Technician with an Associate of Science in Veterinary Medicine from Penn Foster College. Kacie is a Certified Veterinary Assistant, Level III through the Texas Veterinary Medical Association.
When she isn't seeing patients, she enjoys spending time with her husband, Warren, traveling the world, and her dog, Bandie, running Agility. She also has 5 cats - Boots, Blush, Meiki, Chatters and Maeby.
Amanda Knappenberger - Veterinary Technician
Amanda has been with us, as a Veterinary Technician, since 2014.
Amanda has been in the veterinary industry since 2008, starting as a kennel technician and working her way up to an animal surgical technician within 2 months! She was promoted to a veterinary assistant shortly thereafter and earned her level 1 Texas Veterinary Medical Association certification in veterinary assisting in 2012. Amanda has always had a love for animals and now has a lifelong passion for taking care of them.
Amanda has had animals all of her life and currently has three dogs, Sam, Eleanor and Cruz, and three cats, Dusty, Larry and Curly. She is also currently studying to attain her Associates of Science in Veterinary Technology and then become licensed as a veterinary technician. In her spare time Amanda enjoys working on sewing projects!
CJ Johnson- Veterinary Technician
CJ joined our practice in May of 2018. She has been in the veterinary industry off and on since 2013. CJ started out as a receptionist and quickly moved her way into a Veterinary Technician position when she realized her love for taking care of animals. She currently has her Associates of Arts from Tarrant County College and is pursuing her certification from Texas Veterinary Medical Association. She has always had a love for animals both big and small. She tends to love them in twos as she currently has two horses, Decker and Shadow, two dogs, Dexter and Deus, and two cats, Luna and Grayson.
In her spare time,CJ loves trying new adventures with her husband Luke, traveling, and spending time with her fur babies!
Kendall Byrd- Veterinary Technician
Kendall joined our practice in August of 2018. She has been in the veterinary field for 3 years. She graduated from Lone Star College Veterinary Technician Program in December of 2017 receiving her certification as a Veterinary Assistant, Associate Degree of Applied Science in Veterinary Technology, and is working towards becoming licensed as a Veterinary Technician.
She moved to Dallas from a small town in Montgomery, Texas, where she lived on an exotic deer ranch with her parents. She has always had a love for animals as she grew up surrounded by them because of her sweet mother, which led her to this career path.
When She is not working, she loves to spend time and watch movies with her boyfriend Troy and cat Lieutenant Dan. She also loves being with her family out on the ranch in Montgomery.
Alexa Suarez - Animal Care Technician
Alexa was born and raised in Denton, Texas and is attending the University of North Texas and is currently seeking a bachelors degree in biology and a minor in chemistry. Her hope, after graduating from UNT, is to attend Texas A&M University to pursue a degree in veterinary medicine. She has wanted to be a vet ever since she was little and she couldn't imagine doing anything else with her life. She has always had a love for animals and ever since she was born she's had at least one pet in her household. She currently has four dogs and two rabbits. When she is not in school, she likes to draw and paint or do anything art related!
Jordyn Kadura- Animal Care Technician
Jordyn is a pre-veterinary student at the University of North Texas, pursuing a Bachelor of Arts in Biology. While veterinary medicine has not always been her plan she has always loved animals and cannot remember a time she didn't have at least one. Right now, she has 3 dogs, Charlie, Daizy and Martin, but has had rabbits, turtles, and sugar gliders in the past. While she is unsure of what field of veterinary medicine she would like to go into, small animal, large animal, and exotics all strike her interest. When she is not in school she enjoys anything outdoors, working out and spending time with her boyfriend, family and dogs.
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Food Recall: Bravo Packing Inc
Welcome Dr. Hall!
Flower Mound, Texas 75077

redpajama_set_name: RedPajamaCommonCrawl


text:
Samantha Ronson
Full Name: Samantha Ronson
Hometown: London, UK
Birthday: August 7, 1977
Born in 1977 in London, England to socialite Ann Dexter-Jones and band manager Laurence Ronson, Samantha Ronson is a DJ and musical talent who is a staple in the Hollywood club scene. Her career was jumpstarted when she spun live for MTV's New Year's Eve bash in 2001.
Ronson was the first rock act to sign with Jay-Z's Roc-A-Fella Records, and she released four songs and wrote and recorded her first album, Red, under this label, though it was never officially released. Ronson has since left the label to focus exclusively on DJing.
She and ex Lindsay Lohan called it quits in April 2009 after a high-profile on-again, off-again two years of dating.


text:
/ first team
/ 319193 alisson liverpool napoli exclusive interview
NEXT ARTICLE Media Watch: Read the latest LFC transfer rumours
Alisson out to create more 'good memories' at Napoli
Sam Williams @SamWilIiams
Alisson Becker knows how to win at Stadio San Paolo.
But Liverpool's goalkeeper is also well aware of the difficulties involved in doing so.
Back in March, Alisson was part of an AS Roma side that defeated Napoli on their own turf in Serie A and, on Wednesday, he returns as a Liverpool player in the Champions League.
Different team, different time, different competition. Same outcome?
Since his last visit to Naples, Alisson has played in Champions League quarter and semi-finals, represented his country at the World Cup and sealed a high-profile transfer to Anfield.
That, then, the finer details of a 4-2 win at the San Paolo seven months ago remain instantly-accessible in the Brazilian's mind is perhaps indicative of the scale of the Reds' task this week.
"They had 24 attempts on goal, 13 on target. I made 11 saves and conceded two goals," Alisson told Liverpoolfc.com.
"It's always tough playing away against a great club. In Napoli, we'll definitely find really passionate fans, and a team that comes onto the pitch with a huge desire to win. The last and only time I played there, we happily got a win, but it was one of the games I had to work hardest in during the season.
"It was one of the toughest games of the season for us in Serie A, so it's definitely going to be a tough game. They've had the same basis of the team for three to four years.
"I hope we have a great game and achieve the victory, but it'll definitely be a big challenge for us."
Watch Alisson's crucial saves in Chelsea 1-1 LFC on Saturday
Eleven saves and three points.
Little wonder, then, that while acknowledging the toughness of Wednesday's assignment, Alisson also speaks of 'good memories' of the San Paolo.
"I've only played once there, the season before that I was on the bench - I think we lost, I can't remember! But actually playing there, just good memories, a victory," the 'keeper, who turns 26 on Tuesday, said with a smile.
"It's great to win there because it's a great footballing atmosphere, an atmosphere that drives you and stokes your desire to win."
Highlights of LFC 3-2 PSG
Liverpool travel to Italy atop of Champions League Group C having beaten Paris Saint-Germain 3-2 at Anfield on matchday one a fortnight ago.
And Alisson thinks the manner of that victory - sealed by Roberto Firmino in injury time after PSG had fought back from 2-0 down to equalise - illustrates the fortitude and spirit of togetherness within Jürgen Klopp's squad.
"We're confident in our potential. To beat a great team like PSG fills you with confidence, but it'll be a different game so we have to look at it differently as well, bringing our desire onto the pitch, doing everything that we've been doing up until now at the start of the season," he continued.
"On the pitch, we play as a team. Playing as a team not as individuals, that's our main quality. Our strength isn't in one player - it's in all the players running together and being dedicated together on the pitch to always represent and lift up the name of Liverpool."
Alisson Becker: We're focusing only on our own performances
Trent, Thiago and Alisson react to FIFA FIFPro World11 inclusion
Four Liverpool players named in FIFA FIFPro Men's World11
Jürgen Klopp on Leipzig draw: 'It's exciting, but as tough as it gets'
Form, familiar faces and a first meeting - the lowdown on RB Leipzig
Liverpool v Manchester United: Team news
Sadio Mane interview | United clash, Thiago and Lee's goal celebration


We went from a one man show to a whole team of seven, and we're looking forward to having more engineering power in the spring as 1 or 2 University of Michigan students join our forces.
We earned our first dollar (and a few more) thanks to the support of eight early customers – eat catering company, Grange Kitchen & Bar, Harvest Kitchen, Jolly Pumpkin Cafe & Brewery, Logan – An American Restaurant, The Royal Park Hotel, Selma Cafe, and Zingerman's Roadhouse.
We've given a number of presentations and visited a lot of great farmers markets, farms, expos, and tasting events.
We were the guest chefs at SELMA Cafe, served breakfast to 182 people, and spread the word about the great restaurants and businesses highlighted on Real Time Farms.
Photos (over 10,000!) and information have been added to the site from farms and farmers markets across the country…and abroad!
An entire University of Michigan Ross School of Business class will be focused on Real Time Farms next semester working on strategies for expansion and marketing.
We've received local and national press, and Food Inc. called us their favorite new website!
A launch redesign happening soon – stay tuned!!
We are preparing to add restaurants in major metropolitan areas in the Spring!
We are launching our Food Warriors internship program for university students to become local food ambassadors for their region, help populate Real Time Farms with data, and engage people in their area. Interested? Applications available soon!
We have also partnered with 2 musicians, Lafe Dutton and Coco Kallis who make a Get Fresh! tour of 100 farmers markets every year. We'll be sponsoring them next year and they will be collecting data for Real Time Farms at every market and sharing stories on our blog.
Thanks for all of the support and encouragement this year and we look forward to growing with you in the years to come!
This entry was posted in Aw Snaps! (our best photos) and tagged photos, Real Time Farms, year in review. Bookmark the permalink.


Kingsize Galway is located on the Tuam Road, exactly 1 mile from Eyre Square(opposite the travelodge hotel). It is Irelands largest Big and Tall menswear shop at 2500ft, with 30 free car parking spaces outside the front door. Our full Big,Tall and Footwear range is available here.
We stock well-known brands such as McIvor Shirts, Wrangler, Columbia, Skopes, Brax, Mustang, Baileys as well as leading footwear brands such as Vans, Rockport, Timberland, Lloyd and Merrell. We run our online business from Galway, so if you purchase anything from www.kingsize.ie it will be dispatched from here.
So, if you are fortunate enough to live nearby or have any reason to visit one of Ireland's great cities you should call out to see us.
Shane has been working with us since 2011 and has been running our Galway store since May 2012.
A native of Longford, Shane is a keen swimmer, five a side guy and an avid Man Utd fan.
You'll know Shane when you go to our Galway store as he will be the smallest man in it. He is by no means Ireland's smallest man, but he is the smallest man selling big men's clothes in Ireland.


In my most recent blog I recounted an interview with BASF Corporation's General Counsel, Matt Lepore.
In that interview Mr. Lepore described how he went about ditching the arbitrary taxi meter of hourly billing.
And he described how he secured lower costs and client-friendly incentives through alternative fee arrangements.
As its title states, this blog is about controlling corporate legal costs through effective management and (proactive) prevention of legal trouble.


text:
You asked: Do Wheel of Fortune contestants pay their travel expenses?
Do Wheel of Fortune contestants keep their winnings?
Can contestants on Wheel of Fortune take cash instead of a trip?
Do contestants pay their own way to be on Wheel of Fortune?
How much does Pat Sajak make a year?
When you win a trip on Wheel of Fortune is it for two?
What is the most money ever won on Wheel of Fortune?
How Much Does Alex Trebek make per show?
How much does it cost to be on Wheel of Fortune?
How much does it cost to buy a vowel on Wheel of Fortune?
What happens when you win a car on a game show?
Wheel of Fortune does not pay travel expenses. If a contestant is unable to get to the site where the show is recorded, their name goes back into the selection process.
In the case of Wheel of Fortune, when you win trips, the show allows you to find less expensive versions of the trips you win, thereby decreasing your overall tax bill. But, if you've won any significant amount of cash and/or prizes, that will still leave you with a hefty tax payment at the end of the show.
IT IS INTERESTING: How does bet behind work in blackjack?
Cannonie shared some details about appearing on the show: Wheel of Fortune does not pay for contestants to get to a show taping. … Contestants are given audience passes for five people. Contestants receive two sets of tickets because they are called to appear on one of two possible tapings.
Pat Sajak's net worth sits at around $70 million currently, according to Celebrity Net Worth. What's more, Sajak earns roughly $15 million annually.
Yes, trips are for two.
As a result, her grand total of winnings came to $1,030,340, making her the biggest winner in the show's history. Years earlier, on October 14, 2008, Michelle Loewenstein became the first million-dollar winner on the show when she won a total of $1,026,080.
Alex Trebek Jeopardy Salary
Trebek's annual salary from "Jeopardy!" was $18 million. The show taped 46 days a year, and he shot five episodes per day. In other words, he earned around $391,000 every tape day, $78,000 per actual episode.
You have to be at least 18 to be on the regular shows. How much does it cost to be a contestant on Wheel of Fortune? Applying for the show is free, but you must pay for your travel expenses.
IT IS INTERESTING: Frequent question: Is Jack Black Lip balm for men only?
The price of a vowel hasn't been adjusted for inflation over the course of 30 years. It's still only $250.
If you win money or prizes on a game show, the winnings are taxable. If you win at least $600, you'll probably get a 1099-MISC tax form from the entity that awarded you the cash prize, and they'll also send a copy to the IRS. Even if you don't get a 1099, you still have to report the value of your winnings.
Where is the beach in Casino Royale?
In 2006's "Casino Royale," it's beefy Brit Daniel Craig who rises out of the
What happens if you bet money and it's a draw?
If you bet on a boxer to win and the fight is a draw,
You asked: What is a double bet in horse racing?
The double bet is one of the most simple forms of a multiple that
Best answer: How many devices can stream bet plus?
Best answer: What was the highest amount of money won on Wheel of Fortune?
Did the author use foreshadowing as a device in the story the lottery?
Quick Answer: What percentage of health lottery goes to charity?
Who should you hire when you win the lottery?
Question: Is it true that casinos don't have clocks?
Question: Where can I gamble in Miami?


text:
package com.fedevela.core.avaluo.pojos;

/**
 * Created by fvelazquez on 27/03/14.
 */
import java.io.Serializable;
import java.math.BigInteger;
import java.util.Date;
import java.util.List;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.EmbeddedId;
import javax.persistence.Entity;
import javax.persistence.JoinColumn;
import javax.persistence.JoinColumns;
import javax.persistence.ManyToOne;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.OneToMany;
import javax.persistence.Table;
import javax.persistence.Temporal;
import javax.persistence.TemporalType;

@Entity
@Table(name = "AV_CAJA_LOTE", catalog = "", schema = "PROD")
@NamedQueries({
        @NamedQuery(name = "AvCajaLote.findAll", query = "SELECT a FROM AvCajaLote a"),
        @NamedQuery(name = "AvCajaLote.findByLote", query = "SELECT a FROM AvCajaLote a WHERE a.avCajaLotePK.lote = :lote"),
        @NamedQuery(name = "AvCajaLote.findByCaja", query = "SELECT a FROM AvCajaLote a WHERE a.avCajaLotePK.caja = :caja"),
        @NamedQuery(name = "AvCajaLote.findByStatus", query = "SELECT a FROM AvCajaLote a WHERE a.status = :status"),
        @NamedQuery(name = "AvCajaLote.findByFechaCierre", query = "SELECT a FROM AvCajaLote a WHERE a.fechaCierre = :fechaCierre"),
        @NamedQuery(name = "AvCajaLote.findByUsuarioCierre", query = "SELECT a FROM AvCajaLote a WHERE a.usuarioCierre = :usuarioCierre"),
        @NamedQuery(name = "AvCajaLote.findByObservaciones", query = "SELECT a FROM AvCajaLote a WHERE a.observaciones = :observaciones"),
        @NamedQuery(name = "AvCajaLote.findByCliente", query = "SELECT a FROM AvCajaLote a WHERE a.avCajaLotePK.cliente = :cliente")})
public class AvCajaLote implements Serializable {
    private static final long serialVersionUID = 1L;
    @EmbeddedId
    protected AvCajaLotePK avCajaLotePK;
    @Column(name = "STATUS")
    private Character status;
    @Column(name = "FECHA_CIERRE")
    @Temporal(TemporalType.DATE)
    private Date fechaCierre;
    @Column(name = "USUARIO_CIERRE")
    private String usuarioCierre;
    @Column(name = "OBSERVACIONES")
    private String observaciones;
    @JoinColumns({
            @JoinColumn(name = "LOTE", referencedColumnName = "LOTE", insertable = false, updatable = false),
            @JoinColumn(name = "CLIENTE", referencedColumnName = "CLIENTE", insertable = false, updatable = false)})
    @ManyToOne(optional = false)
    private AvLoteAvaluo avLoteAvaluo;
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "avCajaLote")
    private List<AvAvaluoCaja> avAvaluoCajaList;

    public AvCajaLote() {
    }

    public AvCajaLote(AvCajaLotePK avCajaLotePK) {
        this.avCajaLotePK = avCajaLotePK;
    }

    public AvCajaLote(int lote, BigInteger caja, short cliente) {
        this.avCajaLotePK = new AvCajaLotePK(lote, caja, cliente);
    }

    public AvCajaLotePK getAvCajaLotePK() {
        return avCajaLotePK;
    }

    public void setAvCajaLotePK(AvCajaLotePK avCajaLotePK) {
        this.avCajaLotePK = avCajaLotePK;
    }

    public Character getStatus() {
        return status;
    }

    public void setStatus(Character status) {
        this.status = status;
    }

    public Date getFechaCierre() {
        return fechaCierre;
    }

    public void setFechaCierre(Date fechaCierre) {
        this.fechaCierre = fechaCierre;
    }

    public String getUsuarioCierre() {
        return usuarioCierre;
    }

    public void setUsuarioCierre(String usuarioCierre) {
        this.usuarioCierre = usuarioCierre;
    }

    public String getObservaciones() {
        return observaciones;
    }

    public void setObservaciones(String observaciones) {
        this.observaciones = observaciones;
    }

    public AvLoteAvaluo getAvLoteAvaluo() {
        return avLoteAvaluo;
    }

    public void setAvLoteAvaluo(AvLoteAvaluo avLoteAvaluo) {
        this.avLoteAvaluo = avLoteAvaluo;
    }

    public List<AvAvaluoCaja> getAvAvaluoCajaList() {
        return avAvaluoCajaList;
    }

    public void setAvAvaluoCajaList(List<AvAvaluoCaja> avAvaluoCajaList) {
        this.avAvaluoCajaList = avAvaluoCajaList;
    }

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (avCajaLotePK != null ? avCajaLotePK.hashCode() : 0);
        return hash;
    }

    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof AvCajaLote)) {
            return false;
        }
        AvCajaLote other = (AvCajaLote) object;
        if ((this.avCajaLotePK == null && other.avCajaLotePK != null) || (this.avCajaLotePK != null && !this.avCajaLotePK.equals(other.avCajaLotePK))) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "com.adeadms.core.transfer.pojos.AvCajaLote[avCajaLotePK=" + avCajaLotePK + "]";
    }

}

redpajama_set_name: RedPajamaGithub


Originally hosted by Mass-Lift, the 3 Day Raw Powerlifting Program is a no frills powerlifting program that has the athlete squatting 3x per week, benching 3x per week, and deadlifting 1x per week.
Developed by Stephan Korte, the 3×3 program is a two-phased powerlifting program: one high volume, one heavy weight. While only calling for three workouts per week, each session includes the squat, bench, and deadlift.
The program can be run as a 4 week or 8 week program, depending upon whether one or two phases are run.


text:
US Judge says Nikulin 'competent to stand trial', wants to appoint Public Defender
Yevgeniy Nikulin, Russian citizen detained in the US and accused of cybercrimes, has been transferred to Maguire Correctional Facility in California and may be deprived of his attorney's assistance, reported Russian news agency Izvestia.
Nikulin's lawyer Arkady Bukh claimed that American federal judge suggested he would restrict Nikulin from seeking assistance from private attorney, instead he would be assigned a Public Defender. Currently, Nikulin's lawyers insist that he cannot be tried due to his mental health problems.
Another hearing on Nikulin's case is scheduled to take place on August 6. Although Nikulin has been evaluated by a psychiatrist, who stated that the man has "many mental problems," it is expected that next Tuesday the date of the trial will be set. The judge ignores the psychiatrist's statement, explaining that psychologists say Nikulin is "competent to stand trial."
"The government has proven by a preponderance of the evidence that defendant's mental disorder does not render him incompetent and that any refusal to participate in his own defense is due to his own choice," San Francisco-based judge William Alsup wrote in the order.
"The difference between a psychologist and a psychiatrist is huge," said Bukh. "It is not in a field of psychologist's responsibility to make such conclusions. It is obvious that due to his mental health problems Yevgeniy would not be able to fully protect himself at a trial. Moreover, we told the judge that at this conjuncture we were not able to even get ready to the trial properly, as our client behaves apathetically and does not want to cooperate."
"Yet the judge intends to assign a public Defender to Nikulin and put his team of lawyers aside," said Bukh.
"It is unfortunate that American justice makes it possible to take such actions."
Nikulin, who pleaded not guilty, was arrested in Prague, Czech Republic, in 2016 and later extradited to the US. Russia's Foreign Ministry then called the extradition "a conscious, politically motivated step by the Czech side aimed at undermining the constructive basis of bilateral cooperation."
Nikulin previously claimed that the FBI tried to force him to take responsibility for the hacking of the DNC servers.
Criminal Justice The US Yevgeniy Nikulin


text:
Q: Python OptionMenus keep disappearing and reappearing - how can I make them "stay"? I have a simple school assignment to do - a converter between F, C and K.
I'm having problems with OptionMenus, because they keep disappearing and reappearing when I move over with the mouse. For example if I choose from OptionMenu2, OptionMenu1 will disappear. How can I make them "stay"?
Code includes 2 pictures. If you want to run the code, you'll need one for button. You can probably just delete the one on top :)
I'll appreciate any help on this!

# -*- coding: cp1250 -*-

from Tkinter import *
import tkMessageBox
import tkFont
from array import *
from decimal import Decimal

class MojGUI(Frame):
    def __init__(self, master=None):
        Frame.__init__(self, master)

        ##NASLOV
        image1 = PhotoImage(file="naslov.gif")
        panel1 = Label(root, image=image1,background="#FFFFFF",height=50,width=400)
        panel1.pack(side=TOP)
        panel1.image = image1

        panel2 = Label(root,background="#FFFFFF")
        panel2.pack(side=TOP)
        self.vnesibesedilofont = tkFont.Font(family="Verdana",size=16,weight="bold")

        ##VPIS CIFRE
        self.entryfont = tkFont.Font(family="Verdana",size=14,weight="bold")
        self.entryWidget = Entry(panel2, width="4",font=self.entryfont,foreground="#FFFFFF", background="#bb0000")
        self.entryWidget.pack(side=LEFT)

        ##ENOTA1
        self.text1 = StringVar(master)
        self.text1.set("C") # default value
        self.enota1 = OptionMenu(panel2, self.text1, "C", "F", "K")
        self.enota1.pack(side=LEFT)

        ##ENACAJ
        self.enacaj = tkFont.Font(family="Verdana",size=16,weight="bold")
        self.znak = StringVar()
        self.znak.set(" = ")
        self.entryLabel = Label(panel2,textvariable=self.znak, background='#FFFFFF',font=self.enacaj)
        self.entryLabel.pack(side=LEFT)

        ##VREDNOST
        self.textvrednost = StringVar()
        self.vredno = tkFont.Font(family="Verdana",size=14,weight="bold")
        self.vrednost = Label(panel2,textvariable=self.textvrednost, width="9", foreground='#000000',background='#FFFFFF',font=self.vredno)
        self.vrednost.pack(side=LEFT)
        self.textvrednost.set("")

        ##ENOTA2
        self.text2 = StringVar(panel2)
        self.text2.set("C") # default value
        self.enota2 = OptionMenu(panel2, self.text2, "C", "F", "K")
        self.enota2.pack(side=LEFT)

        ##GUMB
        image2 = PhotoImage(file="pretvori.gif")
        entryButton = Button(panel2,text="",bd="0",cursor="hand2",background='#FFFFFF',activebackground="#FFFFFF",command=self.pretvori,image=image2)
        entryButton.pack(side=LEFT)
        entryButton.image = image2

        self.pack()

    def pretvori(self):

        enota1=self.text1.get()
        enota2=self.text2.get()
        original=Decimal(self.entryWidget.get())
        rezultat= StringVar()
        rezultat.set(str(original))

        if (enota1 == "C") &(enota2 == "K"):
            rezultat.set(str(round(original+273,2)))

        if (enota2 == "C") &(enota1 == "K"):
            rezultat.set(str(round(original-273,2)))

        if (enota1 == "K") &(enota2 == "F"):
            rezultat.set(str(round( Decimal(original-273) * Decimal(1.8)+32 ,2)))

        if (enota2 == "K") &(enota1 == "F"):
            rezultat.set(str(round( Decimal(original-32) / Decimal(1.8)+273 ,2)))

        if (enota1 == "C") &(enota2 == "F"):
            rezultat.set(str(round(original*Decimal(1.8)+32,2)))

        if (enota2 == "C") &(enota1 == "F"):
            rezultat.set(str(round((original-32)/Decimal(1.8),2)))

        self.textvrednost.set(rezultat.get())
        self.znak.set(" = ")

root = Tk()

root.title('Pretvornik')
root.wm_minsize(500, 200)
root.wm_resizable(0, 0)
w = root.winfo_screenwidth()
h = root.winfo_screenheight()
rootsize = tuple(int(_) for _ in root.geometry().split('+')[0].split('x'))
x = (w - 500)/2
y = (h - 200)/2
root.geometry("%dx%d+%d+%d" % (rootsize + (x, y)))
root.config(background="#FFFFFF")
app = MojGUI(master=root)
root.mainloop()

A: You are doing something very unusual when you use a label as a container for other widgets. The label is not an appropriate widget to use as a container for other widgets. While this should be allowable, it apparently is causing the behavior that you are seeing. If you change panel1 and panel2 to be frames rather than labels, your problem goes away.

redpajama_set_name: RedPajamaStackExchange


text:
The Last Hurdle: El Camino Real de Tierra Adentro
At the end of an arduous 6 month, 1600 mile trek northward to Spain's new colonial capital of Santa Fe, Spanish colonists faced one last hurdle – the La Bajada Mesa. Noted archeologist Michael Marshall walks us through the trail as it travels up the mesa.
Buffalo Soldiers in New Mexico
The history of the Buffalo soldiers and how they helped New Mexico reach statehood.
Reliving the perilous La Bajada Mesa trail, with archaeologist Michael Marshall.
In Her Own Voice - Doña Teresa Aguilera y Roche
The story of the only woman in NM arrested by the inquisition.
The Estancia Press
Go back in time to learn how the Estancia Press contributed to the life of the small town.
Rough Riders in New Mexico
A history of Teddy Roosevelt's Rough Riders in New Mexico
USS New Mexico BB40: The Drinan Diary
The history of the USS New Mexico, including firsthand accounts of a kamikaze attack.
Tesoros De Devoción
History of the traditional santero carvings and paintings.
Fashioning New Mexico: Victorian Secrets
A history of how the arrival of the railroad changed fashion in New Mexico forever.
Palace of the Governors
Examining the history of New Mexico's Palace of the Governors.
Tracing the history of Santa Fe versus today through photographs.
Your Obedient Servant W.H. Bonney
A look at some of Billy the Kid's letters to then Governor of New Mexico, Lew Wallace.
Segresser Paintings
Examining the history of the New Mexico Segresser Paintings.


Located two blocks from the MIT/Kendall Square station, Watermark Kendall West provides residences with convenience inside and outside of their condo. With fine architecture, and amenities such as a top of the line fitness center, spa, shops, and a Zen Garden; this building makes you feel at home. Conveniences include a concierge, dry cleaning drop off, and parking. All apartments are fully pre-wired with Local Area Network lines. Watermark is a cat and small dog friendly building.


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Q: Phpexcel crypted output data My PHPExcel Export works and it is proper if i save the file, but when i want also make an output to the browser than i get the same file totaly crypted.
See hardcopy:

I work on PHPExcel 1.80
Here my code:
    /** PHPExcel */
    require_once 'inc/phpExcel/Classes/PHPExcel.php';

    $objPHPExcel = new PHPExcel();

    .....

    // Save Excel 2003 file
    // redirect output to client browser
    $objWriter = PHPExcel_IOFactory::createWriter($objPHPExcel, 'Excel5');
    header('Content-Type: application/vnd.ms-excel');
    header('Content-Disposition: attachement;filename="synt_ch_fct_export.xls"');
    //header('Cache-Control: max-age=0');
    $objWriter->save('php://output');
    $objWriter->save('img/test.xls');

I have searched the web, but i couldn't find any solution.
If you have an idea ???
THX in advance


Chapel Cottage Studio's colourful, artistic, eCommerce website offers a beautifully simple online experience for students booking art classes in the Black Mountains, South Wales.
Welsh artist & teacher wanted to revamp their existing website with a fresh design and the ability to take bookings for their numerous art courses. With this in mind, I put together a clean and fresh web design that provides a professional online appearance with an easy to manage online course booking facility.
Full course management, easy to update content and a fully mobile responsive design was key to providing the best user experience.
Working closely with Jantien and her daughter, Tiz, the new web design was developed to ensure the site reflected the look and feel of their promotional literature. Colourful, artistic and clean were the key components and the new web design certainly encapsulates this.
If you're an artist looking for a beautiful website to display and sell your artwork, please get in touch to discuss your specific requirements and get the website your work deserves!


SHOCK VIDEO: Instagram Model ADMITS ON YOUTUBE . . . That She's 'MOLESTED CHILDREN'!!!
A couple of years back, video model Montana DeLeon was one of the most popular instagram models online. She's since gotten older and decided to retire from the modeling game. She now makes videos documenting her life for YouTube.
Anyway, the girl dropped a BOMB yesterday. Montana was giving advice to parents on how to protect your children from child predators. During the video she made a startling revelation. Montana says that she's "molested kids." In fact the video is entitled, "I molested kids."
She admits that she victimized children in the video. WOW!!


Our first employee ownership spotlight is on Larry Litke, Quality and Continuous Improvement Manager. For 5 years, Larry has strived to improve processes at Grand River Rubber and ensure that the company is meeting and exceeding customers' expectations. He is a regular speaker at the ASQ Geneva chapter and will be presenting at the National Quality Auditors Convention in Dallas, Texas next week about transitioning to ISO 9001-2015. We are honored to have him kick off our employee ownership series.
When Larry started working at Grand River Rubber 5 years ago, he was amazed by the company's commitment to quality. The company had an initiative in place that is still used today called "IDR". The company-wide initiative encourages employee owners at all levels of the company to submit ideas for more efficient or better quality ways of doing day-to-day operations. Employees submit an IDR form and the form gets reviewed by the Executive Staff, VP of Operations, Human Resources and the Maintenance Manager.
"What initially surprised me about the IDR initiative is how engrained it is Grand River Rubber's daily operations. Everyone from our newest employees to our oldest are encouraged to share their ideas. I was impressed by how carefully the IDR team reviews each request. Whether they accept it or not, they give details into what went into their decision. If the IDR is accepted, the employee gets an in-depth timeline of when action will take place", says Larry.
Larry brings 32 years of quality management expertise to the table and previously worked in the automotive industry. Since joining Grand River Rubber Rubber's staff, he has used his automotive background to improve the corrective action process for problems both inside and outside of the plant.
"I implemented an automotive 8D corrective action process into Grand River Rubber. The process digs into what really happened to make an error and what do we need to do to fix it. Once a week, I meet with the Quality Specialist, VP of Operations, Area Supervisors, Materials Manager, Sales Manager, and managers from our 3 product line departments: Lathe Cut, Drum & Pail, and Sweeper Belt", explains Larry.
The automotive 8D corrective action process encourages creative thinking and new solutions for problems. Since people are hearing about issues outside of their own department, they are able to view the problem with a fresh set of eyes and come up with new solutions. Since implementing the process, customer complaints have decreased dramatically. The quality process team are also better able to anticipate potential problems before they occur and take preventative action.
Coming from an automotive background, he was amazed how strikingly different the culture is at Grand River Rubber. The employee engagement at Grand River Rubber is something that makes Larry excited to come to work every day. The employees have an all-in-this-together mentality and have a vested interest in the company's success, unmatched by any company he had worked at previously.
"Employee Owners at Grand River Rubber go above the standard and see the big picture of their job and the impact they can have in the company's success. When they retire, they want to be able to say, 'I helped build that to where it is today'. Our retirees regularly come back to visit and see how the company is doing. That is something that is so rare and just goes to show, once you're part of the Grand River Rubber team, you're part of it for life", says Larry.
One of the biggest differentiators about working at Grand River Rubber is the 'work hard, play hard' mentality. The company has many initiatives that encourage employee success including milestone celebrations and retirement acknowledgements but the most exciting part of the year is the annual company picnic.
"Every year, we have a function to celebrate our company's hard work and success during the year. For several years, the company has held their company picnic at Waldemeer Park in Pennsylvania. All the employees and their immediate families are able to participate in the event free of charge. Some of my best memories are at Waldemeer with my family and co-workers", says Larry.
When asked what employee ownership means to him, Larry stated the following: "We have pride in what we do because we understand that when we work, it benefits ourselves and our families. We are all bosses who work together for the betterment of the company. We know that with hard work, our stock will increase and we will have even more security when we retire. Employee ownership is empowering", states Larry.


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