text:
\section{Introduction}

Zero biasing for the normal approximation of a random variable $W$ using Stein's method was introduced in Goldstein and Reinert \cite{GR}. One instance in which the zero bias method may be applied is for $W$ for which a Stein pair
$W,W'$ may be constructed, that is, for $W$ that may be coupled to a variable $W'$ such that $W,W'$ is exchangeable and
satisfies $E(W'|W)=(1-a)W$ for some $a \in (0,1]$. After giving a brief review of these methods in
Section \ref{review}, in Section \ref{general} we provide a general result allowing one to apply zero biasing when the statistic $W$ of interest is formed by certain growth processes and can be coupled in a Stein pair.

Section \ref{jack} studies a certain statistic $W_\alpha$ under the Jack$_{\alpha}$ measure on partitions.
We defer precise definitions to Section \ref{jack}, but for now mention that is of interest to
study statistical properties of Jack$_{\alpha}$ measure. The case $\alpha=1$ corresponds to the
actively studied Plancherel measure of the symmetric group. The surveys
\cite{AD},\cite{De}, \cite{O2} and the seminal papers \cite{BOO},\cite{J},\cite{O1} indicate how the Plancherel measure of
the symmetric group is a discrete analog of random matrix theory, and describe its importance in representation theory and geometry. Okounkov \cite{O2} notes that the study of Jack$_{\alpha}$ measure is an important open problem, about which relatively little is known. It is a discrete analog of Dyson's $\beta$ ensembles from random matrix theory \cite{BO1}.

The particular statistic $W_\alpha$ under Jack measure which we study is of interest for several reasons. When $\alpha=1$
it reduces to the character ratio of transpositions under Plancherel measure, or equivalently to the
spectrum of the random transposition walk. Also by Corollary 1 of \cite{DH}, there is a natural random
walk on perfect matchings of the complete graph on $n$ vertices, whose eigenvalues are precisely $\frac{W(\lambda)}{\sqrt{n(n-1)}}$, occurring
with multiplicity proportional to the Jack$_2$ measure of $\lambda$.
The proofs to date of central limit theorem for $W_\alpha$ range from combinatorial
ones using the method of moments in \cite{K1}, \cite{H}, \cite{Sn}, and the use Stein's method, which produces
an error term (but with no explicit constant) in the Kolomogorov metric \cite{F1}, \cite{F2}, \cite{SS}. Our contribution is to prove a central limit theorem in the $L^1$ metric, with a small explicit constant.

Section \ref{polya} applies the main result of Section \ref{general} to study a growth process arising from the P\'{o}lya-Eggenberger urn model. More precisely,
imagine an urn ${\mathcal U}_{A,B}$ containing $A$ white balls and $B$ black balls. At each time step one ball is drawn, and returned to the urn along
with $m$ balls of the same color. This is one of the simplest urn models, discussed in detail in the textbooks \cite{JK} and \cite{M}. We obtain a central limit theorem with explicit error term for the number of white balls drawn after $n$ steps. While \cite{JK} and \cite{M} contain many useful results and pointers to the literature, including some central limit theorems in more general settings, to the best of our knowledge the literature does not contain results that provide such error terms for this problem.

\section{Stein's method and zero biasing} \label{review}

Stein's lemma \cite{S1} states that a random variable $Z$ has the mean zero normal distribution ${\mathcal N}(0,\sigma^2)$
if and only if
\bea \label{Stein.ch}
\sigma^2 Ef'(Z)=E[Zf(Z)]
\ena
for all absolutely continuous functions $f$ for which these expectations exist. Motivated by this
characterization, for a mean zero, variance $\sigma^2$ random variable $W$ and a given function $h$ on which to test the difference between $Eh(W)$ and $Nh=Eh(Z)$, Stein \cite{S1} considered the differential equation
\bea \label{Stein.eq}
\sigma^2 f'(w)-wf(w) = h(w)-Nh.
\ena
For the unique bounded solution $h$ of (\ref{Stein.eq}), one can evaluate the required difference by
substituting $W$ for $w$ and taking expectation, to yield
\beas
Eh(W)-Nh=E[\sigma^2 f'(W)- Wf(W)].
\enas
Though it may not be immediately clear why the right hand side may be simpler to evaluate
than the left, a variety of techniques have been developed to handle various situations. For instance,
the exchangeable pair technique, from \cite{S2} handles the expectation of the right hand side when
the given random variable $W$ can be coupled to $W'$ so that $(W,W')$ is an $a$-Stein pair, that is, an
exchangeable pair that satisfies
\bea \label{WpW.Stein.pair}
E(W'|W)=(1-a)W \quad \mbox{for some $a \in (0,1)$.}
\ena
Other techniques for handling the Stein equation are discussed in detail in \cite{C1} and in the references therein, but of particular relevance here is the zero bias coupling, which we now review.

Though the mean zero normal is the unique distribution satisfying (\ref{Stein.ch}), one can ask whether
a given variable satisfies a like identity of it own. Indeed, it is shown in \cite{GR} that for every mean zero,
variance $\sigma^2$ random variable $X$, there exists a distribution for a random variable $X^*$, termed
the $X$-zero biased distribution, such that
\bea \label{zero.bia.char}
\sigma^2 Ef'(X^*)=E[Xf(X)]
\ena
for all absolutely continuous functions $f$ for which these expectations exist. The mapping of ${\mathcal L}(X)$, the
distribution of $X$, to ${\mathcal L}(X^*)$, is known as the zero bias transformation. In particular, Stein's lemma
(\ref{Stein.ch}) can be rephrased as the statement that the mean zero normal ${\mathcal N}(0,\sigma^2)$ is the unique fixed point of the zero bias transformation characterized by (\ref{zero.bia.char}).

Heuristically, then, if the transformation has a fixed point at the mean zero normal, then an approximate fixed point
should be approximately normal. This heuristic has been made precise for a variety of examples in \cite{GR}, \cite{G1}, \cite{G2}, \cite{G3} and \cite{G4} (see also \cite{C1}) in order to yield bounds in both the Kolmogorov and $L^1$ metric.
For the latter, the following result from \cite{G4} is often useful; we use $|| \cdot ||_1$ to denote the $L^1$ metric.
\begin{theorem}\label{zb2}
If the mean zero, variance 1 random variable $W$ can be coupled to $W^*$ having the $W$-zero bias distribution, then
\beas
||{\mathcal L}(W)-{\mathcal L}(Z)||_1 \le 2E|W^*-W|
\enas
where $Z$ is a standard normal variable.
\end{theorem}

Hence, to obtain $L^1$ bounds, the question reduces to finding a way to couple $W$ and $W^*$. Lemma \ref{GR} below of \cite{GR}, noting here that the result holds also for $a=1$, shows how the construction of a variable $W^*$ with the $W$-zero bias distribution can be achieved with
the help of the distribution $dF(w,w')$ of a Stein pair. First, it can easily be shown from (\ref{WpW.Stein.pair}) that if $W,W'$ is an $a$-Stein pair possessing second moments then
\bea \label{Stein.pair.a}
\ee W=0 \qmq{and} \ee(W'-W)^2=2a\mbox{Var}(W),
\ena
so in particular,
\bea \label{F.dagger.dist}
dF^\dagger(w,w')=\frac{(w'-w)^2}{2a}dF(w,w')
\ena
is a bivariate distribution.

\begin{lemma}
\label{GR}
If $W^\dagger, W^\ddagger$ have distribution
(\ref{F.dagger.dist}) where $F(w,w')$ is the joint distribution of an $a$-Stein pair, and $U$ is a uniformly distributed variable, independent of $W^\dagger,W^\ddagger$, then
\beas
W^*=UW^\dagger + (1-U) W^\ddagger
\enas
has the $W$-zero bias distribution.
\end{lemma}
In particular, if $W$ and $W^\dagger,W^\ddagger$ can be constructed on a common space,
then $W$ and $W^*$ can be also.

We remark that a number of results are available when $(W,W')$ is only an approximate Stein pair, that is, an exchangeable pair that satisfies the linearity condition (\ref{WpW.Stein.pair}) with a remainder, see for instance \cite{RR}, and \cite{C1}. Correspondingly, here we expect the conclusions of Theorems \ref{zb2} and \ref{T.cond.V} to hold for approximate Stein pairs by including in the bounds the additional terms that arise from such remainders.

In what follows we study processes for which the random variable $W$ of interest can be
written as the sum $V+T$, where $V$ is a function of a variable $\tau$ determined by the process run to a penultimate state, and $T$ a function
of running the process for one additional step. In our examples, given $\tau$, a Stein pair $(W,W')=(V+T,V+T')$ can be constructed by running two copies of the last step of chain, forming $T$ and $T'$ conditionally independent given $\tau$.

In such cases a pair of random variables with distribution (\ref{F.dagger.dist}) can be similarly constructed by forming $(W^\dagger,W^\ddagger)=(V^\Box+T_{\tau^\Box}^\dagger,V^\Box+T_{\tau^\Box}^\ddagger)$ for $V^\Box$ and $T_{\tau^\Box}^\dagger,T_{\tau^\Box}^\ddagger$ sampled by biasing the distributions of $V$ and $T,T'$ in a certain way. Our first application of Theorem \ref{T.cond.V}, to Jack measure, is particularly simple since the biasing factor to form the $V^\Box$ distribution from that of $V$ is unity, and we may therefore take $V=V^\Box$. For our second example, the P\'{o}lya-Eggenberger urn, we will see that biasing draws from the urn ${\mathcal U}_{A,B}$ in our process results in the urn ${\mathcal U}_{A+m,B+m}$.

\section{General Result} \label{general}

The purpose of this section is to prove the following theorem.

\begin{theorem} \label{T.cond.V}
Consider a bivariate distribution ${\mathcal L}(\tau,T)$ on a random object $\tau$ and random variable $T$, and a $\tau$ measurable random variable $V=V_\tau$ such that sampling $\tau$, and then, given $\tau$, sampling  $T$ and $T'$ independently from the conditional distribution ${\mathcal L}(T|\tau)$, the random variables
\bea \label{WWp.sum}
W=V+T \qmq{and} W'=V+T'
\ena
have variance one and are an $a$-Stein pair.
Denoting
\bea \label{ETWgV.noV}
\ee(T|\tau)=\mu_\tau \qmq{and} \ee((T-\mu_\tau)^2|\tau) = \sigma_\tau^2,
\ena
and the distribution of $\tau$ by $dF(\tau)$, the measure $F^\Box(\tau)$ specified by
\bea \label{dist:vtau}
dF^\Box(\tau)=\frac{\sigma_\tau^2}{a}dF(\tau)
\ena
is a probability measure, and for any coupling of $\tau$ to $\tau^\Box$ with distribution (\ref{dist:vtau}), we have
\bea \label{L1.T.cond.V}
\lefteqn{||{\mathcal L}(W)-{\mathcal L}(Z)||_1}\nonumber \\
&&\le 2 \ee |(V_{\tau^\Box}-V) +(\mu_{\tau^\Box}-\mu_\tau)|+ 2 \ee |T-\mu_\tau|+\frac{\ee |T-\mu_\tau|^3}{\mbox{Var}(T-\mu_\tau)}.
\ena
When $\mu_\tau$ equals zero and $\sigma_\tau^2$ is constant almost surely, then
\bea \label{L2.T.cond.V}
||{\mathcal L}(W)-{\mathcal L}(Z)||_1 \le 2 \ee|T|+\frac{\ee|T^3|}{\mbox{Var}(T)}.
\ena
\end{theorem}

\begin{proof} First consider the case where $\mu_\tau=0$ a.s..
Since conditional on $\tau$ the pair $T$ and $T'$ are independent,
we have $\ee[T'T|\tau]=\ee[T'|\tau]\ee[T|\tau]=0$, and therefore, from (\ref{WWp.sum}) and (\ref{ETWgV.noV}),
\bea \label{expT-Tp2}
\ee((W'-W)^2|\tau)=\ee((T'-T)^2|\tau)=2\sigma_\tau^2.
\ena
Taking expectation and applying (\ref{Stein.pair.a}), we have that
\bea \label{Esig2}
\ee \sigma_\tau^2=a,
\ena verifying that $dF^\Box(\tau)$ is a probability measure.

By construction, the joint distribution of $(T,T',\tau)$ is, with some abuse of notation, given by
\beas
dF(t,t',\tau)=dF(t'|\tau)dF(t|\tau)dF(\tau),
\enas
and therefore the pair $(W,W')$ has distribution
\bea \label{dFwwp}
dF(w,w')=\int_{\tau,t,t': v+t=w,v+t'=w'}dF(t'|\tau)dF(t|\tau)dF(\tau),
\ena
where $v=V_\tau$.
By Lemma \ref{GR}, with $U$ an independent uniform random variable on $[0,1]$,
$$
W^*=UW^\dagger+(1-U)W^\ddagger
$$
has the $W$-zero bias distribution when $(W^\dagger,W^\ddagger)$ has distribution given by
\beas
dF^\dagger(w,w')=\frac{(w'-w)^2}{2a}dF(w,w').
\enas

For any fixed $\tau$ let $F(t|\tau)$ denote
the conditional distribution of $T$ given $\tau$. By (\ref{expT-Tp2}), for every $\tau$ the measure
\bea \label{dFtau.tt.daggers}
dF_\tau^\dagger(t,t')=\frac{(t'-t)^2}{2\sigma_\tau^2}dF(t'|\tau)dF(t|\tau),
\ena
is a bivariate probability distribution.

Now using (\ref{dFwwp}), (\ref{Esig2}) and (\ref{dFtau.tt.daggers})
\bea
\lefteqn{dF^\dagger(w,w')}\nn \\
&=& \frac{(w'-w)^2}{2a}\int_{\tau,t,t':v+t=w,v+t'=w'}dF(t'|\tau)dF(t|\tau)dF(\tau)\nn \\
&=& \int_{\tau,t,t':v+t=w,v+t'=w'}\frac{(w-w')^2}{2a} dF(t'|\tau)dF(t|\tau)dF(\tau)\nn \\
&=& \int_{\tau,t,t':v+t=w,v+t'=w'}\frac{\sigma_\tau^2}{a}\frac{(t'-t)^2}{2\sigma_\tau^2} dF(t'|\tau)dF(t|\tau)dF(\tau)\nn \\
&=& \int_\tau  \left( \int_{t,t':v+t=w,v+t'=w'} \frac{(t'-t)^2}{2\sigma_\tau^2} dF(t'|\tau)dF(t|\tau) \right) \frac{\sigma_\tau^2}{a} dF(\tau) \label{mixing.measure}\\
&=& \int_\tau  \left( \int_{t,t':v+t=w,v+t'=w'} dF_\tau^\dagger(t',t) \right) dF^\Box(\tau).\nn
\ena

The factorization in the integral indicates that given $\tau^\Box$ with distribution $dF^\Box(\tau)$,
the pair $(W^\dagger, W^\ddagger)$ can be generated by sampling $T_{\tau^\Box}^\dagger,T_{\tau^\Box}^\ddagger$ from $dF_{\tau^\Box}^\dagger(t',t)$, and then
setting
\beas
W^\dagger=V_{\tau^\Box}+T_{\tau^\Box}^\dagger \qmq{and} W^\ddagger=V_{\tau^\Box}+T_{\tau^\Box}^\ddagger,
\enas
where $V_{\tau^\Box}$ is the value of $V$ on $\tau^\Box$. In particular, letting
\bea \label{TBox}
T^{\tau^\Box}=UT_{\tau^\Box}^\dagger+(1-U)T_{\tau^\Box}^\ddagger,
\ena
we have that
\beas
W^*=U(V_{\tau^\Box}+T_{\tau^\Box}^\dagger)+(1-U)(V_{\tau^\Box}+T_{\tau^\Box}^\ddagger)=
V_{\tau^\Box}+T^{\tau^\Box}
\enas
has the $W$-zero biased distribution.

For a fixed $\tau$, let $T_\tau$ and $T_\tau'$ denote independent copies of
a random variable with distribution $dF(t|\tau)$. Clearly $T_\tau$ and $T_\tau'$ are exchangeable, and as $\mu_\tau=0$, we have
$\ee(T)=\ee \left( \ee(T|\tau) \right)= \ee \mu_\tau = 0$ and therefore $\ee(T'|T)=\ee(T')= 0$.
Hence $(T,T')$ is a $1$-Stein pair. In view of (\ref{dFtau.tt.daggers}), Lemma \ref{GR} yields that when $T_\tau^\dagger,T_\tau^\ddagger$ have
distribution $F_\tau^\dagger(t,t')$ and $U$ is an independent uniform random variable,
\bea \label{Tnu}
T_\tau^*=UT_\tau^\dagger + (1-U)T_\tau^\ddagger
\ena
has the $T_\tau$-zero biased distribution.

As $\ee(T)=0$, by (\ref{Esig2}) we obtain
\beas
a=\ee \sigma_\tau^2 = \ee \left( \ee(T^2|\tau)\right) = \ee (T^2) = \mbox{Var}(T).
\enas
Comparing (\ref{TBox}) and (\ref{Tnu}), we see that the distribution
${\mathcal L}(T^{\tau^\Box})$ is the mixture of the distributions ${\mathcal L}(T_\tau^*)$ with mixing measure
 $\sigma_\tau^2/\mbox{Var}(T)$, by (\ref{mixing.measure}). Therefore, by Theorem 2.1 of \cite{G3}, $T^{\tau^\Box}$ has the $T$-zero bias distribution.
Applying the zero
bias identity (\ref{zero.bia.char}) with $f(x)=(1/2)x^2 \mbox{sign}(x)$, we have
\beas
\ee|T^{\tau^\Box}|= \frac{\ee|T^3|}{2\mbox{Var}(T)}.
\enas

Now, with $\tau$ and $\tau^\Box$ the given coupling, letting $V=V_\tau$ and $T$ be sampled from ${\mathcal L}(T|\tau)$,
 setting $(W,W^*)=(V+T,V^\Box+T^{\tau^\Box})$ yields a coupling
of $W$ and $W^*$ on the same space, satisfying
\beas
\lefteqn{\ee|W^*-W|} \\
&=&    \ee|V_{\tau^\Box}-V+T^{\tau^\Box}-T|  \\
&\le&  \ee|V_{\tau^\Box}-V|+\ee|T|+\ee|T^{\tau^\Box}|  \\
&=&    \ee|V_{\tau^\Box}-V|+ \ee|T|+\frac{\ee|T^3|}{2\mbox{Var}(T)}.
\enas

Theorem \ref{zb2} now yields
\bea \label{L1.bd.mu0}
||{\mathcal L}(W)-{\mathcal L}(Z)||_1  \le  2\ee|V_{\tau^\Box}-V|+2 \ee|T|+\frac{\ee|T^3|}{\mbox{Var}(T)}.
\ena
When $\sigma_\tau^2$ is constant we have that $dF^\Box(\tau)=dF(\tau)$, and hence may let $\tau^\Box=\tau$; taking $V_{\tau^\Box}=V$ in (\ref{L1.bd.mu0}) now yields (\ref{L2.T.cond.V}).

To obtain the result for general $\mu_\tau$, we reduce to the case $\mu_\tau=0$ by writing
\beas
(W,W')=(V+T,V+T')=((V+\mu_\tau)+(T-\mu_\tau),(V+\mu_\tau)+(T'-\mu_\tau)).
\enas
Replacing $V$ and $T$ in (\ref{L1.bd.mu0}) by $V+\mu_\tau$ and $T-\mu_\tau$, respectively, yields (\ref{L1.T.cond.V}).
\end{proof}

\section{The Jack measure} \label{jack}

In this section we apply Theorem \ref{T.cond.V}
to study a property of the Jack$_{\alpha}$ measure on the set of
partitions of size $n$. For $\alpha>0$ the Jack$_{\alpha}$
measure chooses a partition $\lambda$ of size $n$ with probability
\bea \label{def:Jackalpha}
\mbox{Jack}_\alpha(\lambda)=
\frac{\alpha^n n!}{\prod_{x \in \lambda} (\alpha a(x) + l(x) +1)
(\alpha a(x) + l(x) + \alpha)},
\ena
where in the product over all boxes $x$
in the partition $\lambda$, $a(x)$ denotes the number of boxes in the same
row of $x$ and to the right of $x$ (the ``arm'' of $x$), and $l(x)$
denotes the number of boxes in the same column of $x$ and below $x$
(the ``leg'' of $x$). For example one calculates that the partition
\[ \lambda=
\begin{array}{c c c} \framebox{}& \framebox{}& \framebox{} \\
\framebox{}& \framebox{}& \end{array} \] of 5 has Jack$_{\alpha}$
measure
\[ \mbox{Jack}_\alpha(\lambda)= \frac{60 \alpha^2}{(2 \alpha+2)(3 \alpha+1) (\alpha+2)(2
\alpha+1)(\alpha+1)}.\]

With $\lambda$ having the $\mbox{Jack}_\alpha$ distribution, we apply the theory of Section \ref{general} to prove
an explicit $L_1$ normal approximation bound for the statistic
\beas
W_{\alpha}(\lambda) = \frac{\sum_{x \in
\lambda} c_{\alpha}(x)}{\sqrt{\alpha {n \choose 2}}}
\enas
where $c_{\alpha}(x)$ denotes the
``$\alpha$-content'' of $x$, defined as
\[ c_{\alpha}(x) = \alpha(\mbox{column number of $x - 1$}) - (\mbox{row number of $x - 1$}).\]
In the diagram below representing a partition of 7, each box is filled with its $\alpha$-content: \[
\begin{array}{c c c c} \framebox{0}& \framebox{$\alpha$}& \framebox{2 $\alpha$}&
\framebox{3 $\alpha$} \\ \framebox{$-1$}& \framebox{$\alpha-1$}&& \\ \framebox{$-2$} &&&
\end{array}. \]

In the Kolmogorov metric, the paper \cite{F1} proved an $O(n^{-1/4})$ error term for the normal
approximation of $W_{\alpha}$; this rate was sharpened in \cite{F4} using martingales to $O(n^{(-1/2)+\epsilon})$ for any $\epsilon>0$ and in \cite{F3} to $O(n^{-1/2})$ using Bolthausen's inductive
approach to Stein's method, but without an explicit constant. The text \cite{HO} proves a central limit theorem, with no error term, for $W_{\alpha}$ using quantum probability. Here we give an explicit $L_1$ bound to the normal with small constants.

To obtain our bound we construct an exchangeable pair using Kerov's growth process for generating a random partition distributed according to Jack$_\alpha$ measure. Given a box $x$ in the diagram of $\lambda$, again letting $a(x)$ and $l(x)$ denote the arm and leg of $x$ respectively, set
\beas c_{\lambda}(\alpha) =
	\prod_{x \in \lambda} (\alpha a(x) + l(x) +1), \quad
	c_{\lambda}'(\alpha) = \prod_{x \in \lambda} (\alpha a(x) +
	l(x) + \alpha)
\enas
and, for $\tau$ a partition obtained from $\lambda$ by removing a single corner box,
 \[ \psi_{\lambda/\tau}'(\alpha) = \prod_{x \in
C_{\lambda/\tau}-R_{\lambda/\tau}} \frac{(\alpha a_{\lambda}(x) +
l_{\lambda}(x)+1)}{(\alpha a_{\lambda}(x) + l_{\lambda}(x)+\alpha)}
\frac{(\alpha a_{\tau}(x) + l_{\tau}(x)+\alpha)}{(\alpha a_{\tau}(x) +
l_{\tau}(x)+1)}
\]
where $C_{\lambda/\tau}$ is the union of columns of $\lambda$ that
intersect $\lambda-\tau$ and $R_{\lambda/\tau}$ is the union of rows of $\lambda$ that intersect $\lambda-\tau$.

The state of Kerov's growth process at times $n=1,2,\ldots$ is a partition of size $n$, starting at time one with the unique partition of 1. If at stage $n-1$ the state of the process is the partition $\tau$, a transition to the partition $\lambda$ occurs with probability
\[ \frac{c_{\tau}(\alpha)}{c_{\lambda}(\alpha)} \psi_{\lambda/\tau}'(\alpha).\] As shown in \cite{K2}, \cite{F4}, if $\tau$ is chosen from the Jack$_{\alpha}$ measure on partitions of size $n-1$, then transitioning according to this rule results in a partition $\lambda$ of $n$ distributed according to Jack$_{\alpha}$ measure.

We now present an $L^1$ bound for the normal approximation of $W_\alpha$.

\begin{theorem} \label{jackbound} Let
\bea
\label{def:Walpha}
W_{\alpha}(\lambda)=\frac{\sum_{x \in
\lambda} c_{\alpha}(x)}{\sqrt{\alpha {n \choose 2}}}
\ena
and let $W_\alpha$ be the value of $W_\alpha(\lambda)$ when $\lambda$
has the Jack$_{\alpha}$ measure distribution for some $\alpha>0$. Then for $Z$
a standard normal random variable,
\bea \label{L1.Plancherel}
||{\mathcal L}(W_{\alpha})-{\mathcal L}(Z)||_1 \le \sqrt{\frac{2}{n}} \left( 2 + \sqrt{2+\frac{\max(\alpha,1/\alpha)}{n-1}} \right).
\ena
\end{theorem}

\begin{proof} First we show (\ref{L1.Plancherel}) holds for all $\alpha \ge 1$.
Constructing $\tau$ from the Jack measure on partitions of size $n-1$ and then taking one
step in Kerov's growth process yields $\lambda$ with the Jack measure on partitions of size $n$,
and we may write
\beas
W_{\alpha}=V+T
\enas
where
\beas
V= \frac{\sum_{x \in
\tau} c_{\alpha}(x)}{\sqrt{\alpha {n \choose 2}}} \qmq{and} T=\frac{c_{\alpha}(\lambda/\tau)}{\sqrt{\alpha {n \choose 2}}},
\enas
and $c_{\alpha}(\lambda/\tau)$ denotes the $\alpha$-content of the box added to $\tau$ to form $\lambda$.

It is shown in \cite{F1} that constructing $\lambda'$ by taking another step in Kerov's growth
process from $\tau$, independently of $\lambda/\tau$ given $\tau$, and then forming $W_{\alpha}'$ from $\lambda'$ as $W$
is formed from $\lambda$, results in exchangeable variables $W_{\alpha},W_{\alpha}'$ that satisfy (\ref{WpW.Stein.pair}) with $a=2/n$.
Hence, (\ref{WWp.sum}) of Theorem \ref{T.cond.V} is satisfied. Corollary 5.3 of \cite{F1} gives that $Var(W)=1$.

From Section 3 of \cite{F3}, one recalls the following three facts:
\begin{enumerate}
\item \label{Jack.1} $E[T|\tau]=0$ for all $\tau$.
\item \label{Jack.2} $E[T^2|\tau]=\frac{2}{n}$ for all $\tau$.
\item \label{Jack.3} $E[T^4]= \frac{\alpha^2 {n \choose 2} + \alpha(\alpha-1)^2(n-1) + 3 \alpha^2 {n-1 \choose 2}}{\alpha^2 {n \choose 2}^2}$
\end{enumerate}

As $V$ is measurable with respect to the $\sigma$-algebra generated by $\tau$, condition (\ref{ETWgV.noV})
is satisfied. From properties (\ref{Jack.1}) and (\ref{Jack.2}) above we have, respectively, that $\mu_\tau=0$ and $\sigma_\tau^2$ is a constant, almost surely. Hence the bound (\ref{L2.T.cond.V}) of Theorem \ref{T.cond.V} holds.

Applying the Cauchy-Schwarz inequality gives that $E|T| \leq \sqrt{ET^2} = \sqrt{2/n}$, accounting
for the first term in the bound. From property (\ref{Jack.3}), now applying $\alpha \geq 1$, we have
\[ E[T^4] \leq \left[ \frac{{n \choose 2} + 3 {n-1 \choose 2}}{{n \choose 2}^2} \right] + \frac{\alpha(n-1)}{{n \choose 2}^2} \leq \frac{8}{n^2} + \frac{4 \alpha}{n^2(n-1)}. \] The Cauchy-Schwarz
inequality gives that $E|T^3| \leq \sqrt{E[T^2] E[T^4]}$, and properties (\ref{Jack.1}) and (\ref{Jack.2}) give
$\mbox{Var}(T)=2/n$, yielding the final term in the bound (\ref{L1.Plancherel}). Thus the result is shown
when $\alpha \ge 1$.

To obtain a bound for all $\alpha>0$ note first that when taking the transpose $\lambda^t$ of a partition $\lambda$ the roles of the arms $a(x)$ and legs $l(x)$ become interchanged; hence, letting $\lambda_\alpha$ be a partition with the $\mbox{Jack}_\alpha$ distribution,
from (\ref{def:Jackalpha}), for all $\alpha>0$ we have
\beas
{\mathcal L}(\lambda_\alpha)={\mathcal L}(\lambda_{1/\alpha}^t).
\enas
Next, as $W_\alpha(\lambda)=-W_{1/\alpha}(\lambda^t)$ for all $\lambda$, and ${\mathcal L}(Z)={\mathcal L}(-Z)$,
\beas
\lefteqn{||{\mathcal L}(W_\alpha(\lambda_\alpha))-{\mathcal L}(Z)||_1}\\
&=&||{\mathcal L}(-W_{1/\alpha}(\lambda_\alpha^t))-{\mathcal L}(Z)||_1\\
&=&||{\mathcal L}(-W_{1/\alpha}(\lambda_{1/\alpha}))-{\mathcal L}(-Z)||_1\\
&=&||{\mathcal L}(W_{1/\alpha}(\lambda_{1/\alpha}))-{\mathcal L}(Z)||_1.
\enas
Hence, as the bound (\ref{L1.Plancherel}) holds for all $\alpha \ge 1$, it holds for all $\alpha>0$.
\end{proof}

\section{P\'{o}lya-Eggenberger urn model} \label{polya}

For $m,n,A,B>0$ fixed integers, we define a probability distribution on the
set $\{0,1,\cdots,n\}$ by
\bea \label{def:Mnk}
M_{n,A,B}(k) = {n \choose k} \frac{(A/m)_k (B/m)_{n-k}}{(A/m+B/m)_n}.
\ena
Unless clarity demands it, we will simply write $M_n(k)$ for $M_{n,A,B}(k)$.
Here $x_r=x(x+1) \cdots (x+r-1)$, the rising factorial,
where we set $x_0$=1.

It is well known \cite{K3}, \cite{M}, \cite{JK} that the distribution $M_n(k)$ can be achieved
in the following way. Imagine an urn ${\mathcal U}_{A,B}$ that initially has $A$ white and $B$ black balls. At each time step,
one ball is drawn uniformly from the urn and then returned back along with $m$ balls of the same color. If $S_n$ is the number of white balls drawn in the first $n$ draws, then
\beas
P(S_n=k)=M_n(k) \quad \mbox{for $k=0,1, \ldots, n$.}
\enas
We note that when $S_n=k$ the urn ${\mathcal U}_{A,B}$ contains $A+km$ white balls.

In this section we prove the following $L^1$ normal approximation to the distribution of $S_n$, properly standardized.

\begin{theorem}
\label{thm:mainAB}
For $n \in \mathbb{N}$ let $S_n$ be the number of white balls added to ${\mathcal U}_{A,B}$ after $n$ time steps, and set
\bea \label{def:WnAB}
W_n= \sqrt{\frac{(A+B+m)n}{AB(A+B+nm)}} \left[ A - \frac{(A+B)S_n}{n} \right].
\ena
Then $W_n$ has mean zero and variance 1, and for $Z$ a standard normal random variable, for $n \ge (A+B+m)/2m$
\beas
\lefteqn{||{\mathcal L}(W_{n+1})-{\mathcal L}(Z)||_1}\\
&\le&  \left(\frac{4mn}{A+B+m}+  \frac{A^2+6AB+B^2}{AB} \right)\sqrt{\frac{(A+B+m)^3}{AB(A+B+nm+m)(n+1)}}
\enas
while for $n<(A+B+m)/2m$,
\beas
\lefteqn{||{\mathcal L}(W_{n+1})-{\mathcal L}(Z)||_1}\\
&\le& \left( \frac{A^2+8AB+B^2}{AB} \right)\sqrt{\frac{(A+B+m)^3}{AB(A+B+nm+m)(n+1)}}.
\enas
\end{theorem}

From Theorem 3.2 of \cite{M}, we know with $A,B,m$ fixed and $n \rightarrow \infty$,
\beas
S_n/n \rightarrow_d {\mathcal B}(A/m,B/m), \quad \mbox{}
\enas
that is, the fraction of white balls drawn converges to the Beta distribution with parameters
$A/m,B/m$. In particular, the limiting value of the bound as $n \rightarrow \infty$, giving
an $L^1$ bound between the standardized Beta distribution and the normal, is
$4\sqrt{m(A+B+m)/AB}$; for, say $A=B$, the bound specializes to $4\sqrt{m(2A+m)}/A$, which tends
to zero at rate $1/\sqrt{A}$ if $m$ is fixed and $A$ grows.

For what follows it is useful to relate the distribution $M_n(k)$ to up and
down chains. On the set $\Gamma_n= \{(n,k): 0 \leq k \leq n \}$,
placing directed edges from $(n-1,k)$ to $(n,k)$ and to $(n,k+1)$ results in what is known
as known as Pascal's lattice \cite{K3}. It is convenient to define $d((n,k))={n \choose k}$,
the number of paths from $(0,0)$ to $(n,k)$.
More generally, one defines $d((n,k) / (m,j))$ to be the number of
paths from $(m,j)$ to $(n,k)$; this is ${n-m \choose k-j}$.

We define an ``up'' chain that transitions from $(n,k)$ to $(n+1,k)$ with probability
$(B+nm-km)/(A+B+nm)$ and to $(n+1,k+1)$ with probability $(A+km)/(A+B+nm)$. We also define
a ``down'' chain that transitions from $(n,k)$ to $(n-1,k-1)$ with probability $k/n$
and to $(n-1,k)$ with probability $1-(k/n)$. One easily checks that if $(n-1,k)$ is distributed
according to $M_{n-1}$, then applying the up chain gives an element of $\Gamma_n$ distributed according to
$M_n$. Similarly, if $(n,k)$ is distributed according to $M_n$, one checks that applying
the down chain gives an element of $\Gamma_{n-1}$ distributed according to $M_{n-1}$.

We denote the up chain from $\Gamma_{n-1}$ to $\Gamma_n$ by $U_{n-1}$ and the down chain
from $\Gamma_n$ to $\Gamma_{n-1}$ by $D_n$. A straightforward computation yields that
\begin{equation} \label{commut} D_{n+1}U_n = c_n U_{n-1}{D_n} + (1-c_n) I_n \end{equation}
with $c_n=\frac{n(A+B+nm-m)}{(n+1)(A+B+nm)}$, so that the tools of \cite{F5} are in force.

The following lemma shows how to use the up and down chains to construct a Stein pair, that is, a pair of
exchangeable random variables satisfying (\ref{WpW.Stein.pair}).
\begin{lemma} \label{expair:AB}
Let $W_n$ be given by (\ref{def:WnAB}) with
$S_n$ the number of white balls added to ${\mathcal U}_{A,B}$ after $n$ time steps. Now construct $S_n'$ by transitioning
down using $D_n$ and then up using $U_{n-1}$, and let $W_n'$ be given by (\ref{def:WnAB})
with $S_n$ replaced by $S_n'$. Then $W_n,W_n'$ is an $a_n$-Stein pair with
\beas
a_n = \frac{A+B}{n(A+B+nm-m)}.
\enas
\end{lemma}

\begin{proof} By Theorem 4.3 of \cite{F5} and equation \eqref{commut}, a left eigenvector with eigenvalue $1-a_n$ is obtained by applying the operator $U^{n-1}$ to $(1,0)-(1,1)$. From the general theory of down-up chains (see \cite{F5}), one has that
\begin{eqnarray*} U^{n-1}(1,0) & = & \sum_{k=0}^n \frac{M_n(k) d((n,k)/(1,0))}{M_1(0) d(n,k)} \cdot (n,k) \\
& = & \sum_{k=0}^n M_n(k) \frac{{n-1 \choose k} (A+B)}{{n \choose k} B} \cdot (n,k). \end{eqnarray*} Similarly,
\begin{eqnarray*} U^{n-1}(1,1) & = & \sum_{k=0}^n \frac{M_n(k) d((n,k)/(1,1))}{M_1(1) d(n,k)} \cdot (n,k) \\
& = & \sum_{k=0}^n M_n(k) \frac{{n-1 \choose k-1} (A+B)}{{n \choose k} A} \cdot (n,k). \end{eqnarray*}

Since $U_{n-1}D_n$ is a reversible Markov chain with stationary distribution $M_n$, its
right eigenvectors are obtained from its left eigenvectors by dividing by $M_n$. Thus \[ \frac{{n-1 \choose k} (A+B)}{{n \choose k} B}
 - \frac{{n-1 \choose k-1} (A+B)}{{n \choose k} A} = \frac{A+B}{AB} \left[ A - \frac{k(A+B)}{n} \right] \]
is a right eigenvector of $U_{n-1}D_n$ with eigenvalue $\left(1 - \frac{A+B}{n(A+B+nm-m)} \right)$. Since $W_n(k)$ is
a scalar multiple of $\frac{A+B}{AB} \left[ A - \frac{k(A+B)}{n} \right]$, the result follows. \end{proof}

The next goal is to compute the mean and variance of $W_n$ given by (\ref{def:WnAB})
with $S_n$ the number of white balls drawn in the first $n$ draws. Clearly for all $n \ge 1$ one may write
\beas
S_n = {\bf 1}_0 + \cdots + {\bf 1}_{n-1}
\enas
where ${\bf 1}_j=1$ if a white ball is drawn at time $j$, and ${\bf 1}_j=0$ otherwise.
The next lemma computes the mean and covariance of the indicators ${\bf 1}_j$.

\begin{lemma} \label{term}
For $j=0,\ldots,n-1,$ let ${\bf 1}_j$ denote the indicator that a white ball is drawn from ${\mathcal U}_{A,B}$ at time $j$. Then
\begin{enumerate}
\item $E[{\bf 1}_j] = \frac{A}{A+B}$ for all $j \in \{0,\ldots,n-1\}$.
\item $E[{\bf 1}_h {\bf 1}_j]= \frac{A(A+m)}{(A+B)(A+B+m)}$ for all $0 \le h < j \le n-1$
\item $E[S_n]=\frac{nA}{A+B}$.
\end{enumerate}
\end{lemma}

\begin{proof} It is classical and elementary that the indicators ${\bf 1}_j, j=0,\ldots,n-1$ are an exchangeable sequence (see
\cite{JK} or \cite{M} for a proof). Thus $E[{\bf 1}_j]$ is the probability that the first ball drawn is white, and
$E[{\bf 1}_h {\bf 1}_j]$ is the probability that the first two balls drawn are white. These observations, and linearity of expectation, yields the lemma.
\end{proof}

With the help of Lemma \ref{term}, we now compute the mean and variance of $W_n$.
\begin{lemma}
\label{mean.var.W.AB}
If $W_n$ is given by (\ref{def:WnAB}) where $S_n$ is the number of white balls added to ${\mathcal U}_{A,B}$ after
$n$ time steps, then
\beas
E[S_n^2]=\frac{nA}{A+B} + 2 {n \choose 2} \frac{A(A+m)}{(A+B)(A+B+m)},
\enas
\beas
EW_n=0 \qmq{and} \mbox{Var}(W_n) = 1.
\enas
\end{lemma}

\begin{proof}  Since $W_n,W_n'$ is a Stein pair we have that $EW_n=0$ by (\ref{Stein.pair.a}). Now, using the fact that ${\bf 1}_i^2={\bf 1}_i$, and both parts of Lemma \ref{term}, we obtain
\beas
E[S_n^2] & = & E[({\bf 1}_0+\cdots+{\bf 1}_{n-1})^2] \\
& = & E \left[ \sum_{i=0}^{n-1} {\bf 1}_i + 2 \sum_{0 \leq h<j \leq n-1} {\bf 1}_h{\bf 1}_j \right]\\
& = & \frac{nA}{A+B} + 2 {n \choose 2} \frac{A(A+m)}{(A+B)(A+B+m)},
\enas
yielding the first claim.

Applying the expression for $E[S_n]$ given by Lemma \ref{term}, it follows that
\begin{eqnarray*}
\mbox{Var}(S_n) & = & \left[ \frac{nA}{A+B} + 2 {n \choose 2} \frac{A(A+m)}{(A+B)(A+B+m)} - \frac{n^2A^2}{(A+B)^2} \right] \\
& = & \frac{AB(A+B+nm)n}{(A+B+m)(A+B)^2}. \end{eqnarray*} Hence, from the definition (\ref{def:WnAB}) of $W_n$ we conclude
that \[ \mbox{Var}(W_n) = \frac{(A+B+m)(A+B)^2}{AB(A+B+nm)n} \mbox{Var}(S_n) = 1.\]
\end{proof}

We will apply Theorem \ref{T.cond.V} by writing $W_{n+1}=V+T$ where
\bea \label{def:V}
V = \sqrt{\frac{(A+B+m)(n+1)}{AB(A+B+nm+m)}} \cdot \left[A - \frac{(A+B)S_n}{n+1} \right]
\ena
and
\bea \label{def:urnT}
T = - \sqrt{\frac{(A+B+m)}{AB(A+B+nm+m)(n+1)}} \cdot (A+B) \cdot {\bf 1}_n,
\ena
and letting $\tau=S_n$. We note that the condition in Theorem \ref{T.cond.V} that $V$ be $\tau$ measurable is here clearly satisfied.
The following lemma gives the properties of $T$ needed for computing an $L^1$ bound using Theorem \ref{T.cond.V}.

\begin{lemma} \label{t1}
Let $T$ be given by (\ref{def:urnT}) and $\tau=S_n$.
\begin{enumerate}

\item \label{condmeanAB} The conditional mean $\mu_\tau=E(T|\tau)$ is given by
\beas
\mu_\tau=- \sqrt{\frac{(A+B+m)}{AB(A+B+nm+m)(n+1)}} \cdot \frac{(A+B)  (A+mS_n)}{A+B+mn}.
\enas

\item \label{cond.var.T.urn} The conditional variance $\sigma_\tau^2=E((T-\mu_\tau)^2|\tau)$ is given by
\beas
\sigma_\tau^2 = \frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)}  \frac{(A+mS_n)(nm-mS_n+B)}{(A+B+mn)^2}.
\enas

\item \label{urn.std.2rd.t} The variance $\mbox{Var}(T-\mu_\tau)$ satisfies
\beas
\mbox{Var}(T-\mu_\tau)=  \frac{(A+B)}{(n+1)\left( A+B+nm\right)}.
\enas

\item \label{EcondT-mutau} The absolute deviation of $T$ about $\mu_\tau$ satisfies
\beas
E|T-\mu_\tau| \leq \sqrt{\frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)}}.
\enas

\item \label{urn.std.3rd.t.-mu} The third order deviation of $T$ about $\mu_\tau$, standardized by $\mbox{Var}(T-\mu_\tau)$,
satisfies
\beas
\frac{E|T-\mu_{\tau}|^3}{\mbox{Var}(T-\mu_\tau)} \leq  \sqrt{\frac{(A+B+m)^3}{AB(A+B+nm+m)(n+1)}} \cdot \frac{(A+B)^2}{AB}.
\enas

\end{enumerate}
\end{lemma}

\begin{proof} Parts \ref{condmeanAB} and \ref{cond.var.T.urn} follow immediately from (\ref{def:urnT}) and that
\bea \label{P1jgivenk}
P({\bf 1}_j=1|S_j=k) = \frac{A+mk}{A+B+mj}
\ena
for all $j=0,\ldots,n-1,k=0,\ldots,j$.

For part (\ref{urn.std.2rd.t}), first note that as $E(T-\mu_\tau|\tau)=0$ we have $\mbox{Var}(T-\mu_\tau)=E(T-\mu_\tau)^2$.  Now again using (\ref{P1jgivenk}), we have that $E\left( (T-\mu_\tau)^2 |S_n \right)$ equals
\beas
\lefteqn{\left( \frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)} \right)  E\left[ \left( {\bf 1}_n - \frac{A+mS_n}{A+B+nm}\right)^2| S_n \right]}\\
&=& \left( \frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)} \right) \left( \frac{(A+mS_n)(B+m(n-S_n))}{\left( A+B+nm\right)^2} \right).
\enas
Expanding the product $(A+mS_n)(B+m(n-S_n))$, taking expectation using the expressions for $E[S_n]$ and $E[S_n^2]$ provided by Lemmas \ref{term} and \ref{mean.var.W.AB}, respectively, the claim follows after some simplification.

For part \ref{EcondT-mutau}, one has that
\begin{eqnarray*}
& & E|T-\mu_\tau| \\
& = & E[E|T-\mu_\tau||S_n] \\
& = & \sqrt{\frac{A+B+m}{AB(A+B+nm+m)(n+1)}} \cdot (A+B) \\
& & \cdot E \left[ \left| {\bf 1_n}- \frac{A+mS_n}{A+B+nm} \right| | S_n \right] \\
& \leq & \sqrt{\frac{A+B+m}{AB(A+B+nm+m)(n+1)}} \cdot (A+B). \end{eqnarray*} The second equality used (\ref{P1jgivenk}),
and the inequality that
\bea \label{1n-E1n}
E \left[ \left| {\bf 1_n}- \frac{A+mS_n}{A+B+nm} \right|^p | S_n \right] \leq 1 \quad \mbox{for all $p \ge 0$}
\ena
with $p=1$.

Now, for part \ref{urn.std.3rd.t.-mu}, similarly, applying (\ref{1n-E1n}) with $p=3$ we obtain
\begin{eqnarray*}
E|T-\mu_\tau|^3 & = & E[E|T-\mu_\tau|^3|S_n] \\
& = & \left[ \frac{A+B+m}{AB(A+B+nm+m)(n+1)} \right]^{3/2} (A+B)^3 \\
 & & \cdot E \left[ \left| {\bf 1_n}- \frac{A+mS_n}{A+B+nm} \right|^3 | S_n \right] \\
& \leq & \left[ \frac{A+B+m}{AB(A+B+nm+m)(n+1)} \right]^{3/2} (A+B)^3.
\end{eqnarray*}
Part \ref{urn.std.3rd.t.-mu} now follows from part \ref{urn.std.2rd.t} by division.

\end{proof}

Specializing (\ref{dist:vtau}) to the case at hand, with $M_{n,A,B}(k)$ the distribution of $S_n$ given by (\ref{def:Mnk}), we now consider constructing a coupling of $S_n$ to a random variable $S_n^\Box$ with distribution
\bea \label{def:Mnbox}
M_{n,A,B}^\Box(k) = \frac{\sigma_k^2}{a_{n+1}}M_{n,A,B}(k)
\ena
where $a_{n+1}$ is given by Lemma \ref{expair:AB}. The next result shows that one can achieve a variable with distribution $S_n^\Box$ by adding
 $2m$ additional balls to the urn at time zero, $m$ white and $m$ black, that is, by using the urn ${\mathcal U}_{A+m,B+m}$.

\begin{lemma} \label{MnMnBox}
\beas
M_{n,A,B}^\Box = M_{n,A+m,B+m}.
\enas
\end{lemma}

\begin{proof}
From (\ref{cond.var.T.urn}) of Lemma \ref{t1} and Lemma \ref{expair:AB}, we have
\beas
\frac{\sigma_k^2}{a_{n+1}}= \frac{(A/m+B/m)(A/m+B/m+1)(A/m+k)(B/m+n-k)}{(A/m)(B/m)(A/m+B/m+n)(A/m+B/m+n+1)}.
\enas
Hence, for all $k \in \{0,\ldots,n\}$,
\beas
\lefteqn{M_{n,A,B}^\Box (k)}\\
&=& \frac{(A/m+B/m) (A/m+B/m+1) (A/m+k)(B/m+n-k)}{(A/m)(B/m)(A/m+B/m+n)(A/m+B/m+n+1)} \\
& & \cdot {n \choose k}\frac{(A/m)_k(B/m)_{n-k}}{(A/m+B/m)_n}\\
&=&{n \choose k}\frac{(A/m+1)_k(B/m+1)_{n-k}}{(A/m+B/m+2)_n}\\
&=&M_{n,A+m,B+m}(k).
\enas
\end{proof}
Lemma  \ref{MnMnBox} shows that for the process $S_n$ on the urn ${\mathcal U}_{A,B}$, the process $S_n^\Box$ is for the urn ${\mathcal U}_{A+m,B+m}$. As for both processes no additional balls have been added at time zero, we have that
\bea \label{S0SBox0}
S_0=0 \qmq{and} S_0^\Box=0.
\ena
As at times $n \ge 1$ both of these chains increase by $1$ when a white ball has been selected, if
$S_n=k$ and $S_n^\Box=j$, then $S_{n+1}=k+1$ and $S_{n+1}^\Box=j+1$ with respective probabilities
\bea \label{def:snksnkbox}
s_n(k)= \frac{A+km}{A+B+mn} \qmq{and} s_n^\Box(j)=\frac{A+m+jm}{A+B+2m+mn}.
\ena

We now couple $S_n$ and $S_n^\Box$ by coupling, at each stage, the two Bernoulli variables that indicate the drawing of a white ball in each urn. In particular, we couple these two Bernoulli variables so that the chance they are not equal is minimized.

\begin{theorem} \label{thm:ABcoupling}
Let $s_n(k)$ and $s_n^\Box(j)$ be given by (\ref{def:snksnkbox}) for $n,j,k \in \mathbb{N}$. Then
 the bivariate chain taking values in $\mathbb{N} \times \mathbb{N}$ characterized by
the initial condition $(S_0,S_0^\Box)=(0,0)$ and transitions
\beas
p_{n+1,n}(u,v|k,j)=P(S_{n+1}=u,S_{n+1}^\Box=v|S_n=k,S_n^\Box=j)
\enas
at times $n \ge 0$ according to
\beas
p_{n+1,n}(u,v|k,j)=\left\{
\begin{array}{cl}
\min(s_n(k),s_n^\Box(j))                  & (u,v)=(k+1,j+1)\\
(s_n^\Box(j)-s_n(k))^+                    & (u,v)=(k,j+1)\\
(s_n(k)-s_n^\Box(j))^+                    & (u,v)=(k+1,j)\\
1 - \max(s_n(k),s_n^\Box(j))         & (u,v)=(k,j)
\end{array}
\right.
\enas
is a coupling on a joint space of the urn models ${\mathcal U}_{A,B}$ and ${\mathcal U}_{A+m,B+m}$, respectively.

In addition, letting
\beas
N=\inf\{n \ge 1: S_n \not = S_n^\Box \}
\enas
we have
\bea \label{absSNSNBoxis1}
|S_N - S_N^\Box|=1
\ena
and
\beas
\mbox{if $S_N^\Box=S_N+1$ then $S_n^\Box \ge S_n$ for all $n \ge 0$,}
\enas
while, otherwise,
\beas
\mbox{if $S_N=S_N^\Box+1$ then $S_n \ge S_n^\Box$ for all $n \ge 0$}.
\enas

\end{theorem}

\begin{proof} That we must have $(S_0,S_0^\Box)=(0,0)$ is clear by (\ref{S0SBox0}).
As marginally for $S_n$ we have
\beas
P(S_{n+1}=k+1|S_n=k)= \min(s_n(k),s_n^\Box(j))  + (s_n(k)-s_n^\Box(j))^+   = s_n(k),
\enas
and similarly for $S_n^\Box$, both marginal transition functions
agree with those specified by (\ref{def:snksnkbox}), hence the joint chain is a coupling
of the two urn models in question.
Further, since $S_0=S_0^\Box$, and at most one white ball is drawn from either of the two urns at each time $n \ge 0$, (\ref{absSNSNBoxis1}) holds.

Taking the difference between the probabilities of drawing a white ball from either of the two urns yields
\bea \nn
\lefteqn{s^\Box_n(j)-s_n(k)}\\
&=&m \left(\frac{(A+mn) (j-k-1)+B (j-k+1) +2m(n -k)}{(A+B+mn) (A+B+2m+mn)}\right).\label{sdifference}
\ena
Suppose now that $S_N^\Box = S_N + 1$. We show by induction that $S_n^\Box \ge S_n+1$ for all $n \ge N$. Clearly
the claim is true for $n=N$. Assume that $S_n^\Box \ge S_n+1$ for some $n \ge N$, say $(S_n,S_n^\Box)=(k,j)$ with $j-k \ge 1$. Then, by (\ref{sdifference}) we see that
$s_n^\Box(j) \ge s_n(k)$, and
hence $(S_{n+1},S_{n+1}^\Box)$ equals $(k+1,j+1),(k,j+1)$ or $(k,j)$ with respective
probabilities $s_n(k),s_n^\Box(j)-s_n(k)$ and $1-s_n^\Box(j)$. In particular, $S_{n+1}^\Box \ge S_{n+1}+1$.

As the same argument applies in the case $S_n \ge S_n^\Box+1$, and since $S_n^\Box=S_n$ for all $0 \le n < N$
by the definition of $N$, the final claim of the lemma is shown.
\end{proof}

We now compute a bound on $E|S_n^\Box-S_n|$ for the coupling provided by  Theorem \ref{thm:ABcoupling}.

\begin{lemma}
The joint chain $(S_n,S_n^\Box)$ as specified in Theorem \ref{thm:ABcoupling} satisfies
\beas
E|S_n^\Box-S_n| \le \frac{2mn}{A+B+m}{\bf 1}\left(n \ge \frac{A+B+m}{2m}\right) +{\bf 1}\left(n < \frac{A+B+m}{2m}\right).
\enas
\end{lemma}

\begin{proof} By Theorem \ref{thm:ABcoupling}, with $N$ as defined there, we have
\beas
|S_n^\Box-S_n| = (S_n^\Box-S_n){\bf 1}_{\{n \ge N, S_N^\Box = S_N+1\}}+(S_n-S_n^\Box){\bf 1}_{\{n \ge N, S_N = S_N^\Box+1\}}.
\enas

For the first expectation,
\beas
\lefteqn{E \left[(S_n^\Box-S_n){\bf 1}_{\{n \ge N, S_N^\Box = S_N+1\}} \right]}\\
&=&\sum_{t=1}^{n-1} E \left[ (S_n^\Box-S_n){\bf 1}_{\{N=t, S_N^\Box = S_N+1\}} \right] +P(N=n, S_N^\Box = S_N+1)\\
&=&\sum_{t=1}^{n-1} \sum_{u \ge 0} E \left[ (S_n^\Box-S_n){\bf 1}_{\{N=t, S_N^\Box = S_N+1, S_N=u\}} \right]+P(N=n, S_N^\Box = S_N+1)\\
&=& \sum_{t=1}^{n-1} \sum_{u  \ge 0} E\left( S_n^\Box-S_n |N=t,S_N^\Box = S_N+1, S_N=u \right)\\
&& \cdot P(N=t,S_N^\Box = S_N+1, S_N=u)+P(N=n, S_N^\Box = S_N+1).
\enas

For $1 \le t \le n-1$, on the conditioning event, urn ${\mathcal U}_{A,B}$ has $A+mu$ white balls and $B+mt-mu$ black balls
at time $t$, and then has been run for time $n-t$. At each of these time steps, by Lemma \ref{term},
there is probability $(A+mu)/(A+B+mt)$ that a white ball will be selected from urn ${\mathcal U}_{A,B}$.

Similarly, for
$1 \le t \le n-1$, on the conditioning event, urn ${\mathcal U}_{A+m,B+m}$ has $A+m+(mu+m)=A+mu+2m$ white balls and
$B+m+mt-(mu+m)=B+mt-mu$ black balls
at time $t$, and then has been run for time $n-t$. At each of these time steps, by Lemma \ref{term},
the probability is $(A+mu+2m)/(A+B+mt+2m)$ that a white ball is selected from urn ${\mathcal U}_{A+m,B+m}$.

Hence, as it may be that all the balls chosen from ${\mathcal U}_{A,B}$ before time $N$ are black,
that is, we may have $S_N=u$ for $u=0$, we have
\beas
\lefteqn{E\left( S_n^\Box-S_n |N=t,S_N^\Box = S_N+1, S_N=u \right)}\\
&=& (n-t) \left( \frac{A+mu+2m}{A+B+mt+2m}-\frac{A+mu}{A+B+mt} \right)\\
&=& (n-t)\left(\frac{2m(B+mt-mu)}{(A+B+mt) (A+B+mt+2m)}\right)\\
&\le& 2m(n-t)\left(\frac{(B+mt)}{(A+B+mt) (A+B+mt+2m)}\right)\\
&\le& \frac{2m(n-t) (B+mt)}{(A+B+mt)^2}\\
&\le& \frac{2m(n-t)}{A+B+mt}.
\enas

Therefore
\beas
\lefteqn{E \left[ (S_n^\Box-S_n){\bf 1}_{\{n \ge N, S_N^\Box = S_N+1\}} \right]}\\
&\le& \sum_{t=1}^{n-1} \frac{2m(n-t)}{A+B+mt} \sum_{u  \ge 0} P(N=t,S_N^\Box = S_N+1, S_N=u)\\
&&+ P(N=n, S_N^\Box = S_N+1) \\
&=& \sum_{t=1}^{n-1} \frac{2m(n-t)}{A+B+mt} P(N=t,S_N^\Box = S_N+1) + P(N=n, S_N^\Box = S_N+1).
\enas

Reversing the roles of $S_n$ and $S_n^\Box$, though here noting that
it is necessary that $u \le t-1$
for the event $\{N=t,S_N = S_N^\Box+1, S_N^\Box=u\}$ to have positive probability, we similarly obtain
\beas
\lefteqn{E \left[(S_n-S_n^\Box){\bf 1}_{\{n \ge N, S_N = S_N^\Box+1\}} \right]}\\
&\le& \sum_{t=1}^{n-1} \frac{2m(n-t)}{A+B+mt} P(N=t,S_N = S_N^\Box+1) + P(N=n, S_N = S_N^\Box+1).
\enas
Now using that $(n-t)/(A+B+mt)$ is a decreasing
function of of $t \ge 0$, summing yields
\beas
\lefteqn{E|S_n^\Box-S_n|}\\
&\le& 2m\sum_{t=1}^{n-1} \frac{n-t}{A+B+mt} P(N=t) + P(N=n)\\
&\le& \frac{2mn}{A+B+m}P(N \le n-1) + P(N=n)\\
&\le& \frac{2mn}{A+B+m} {\bf 1}\left(n \ge \frac{A+B+m}{2m}\right) +{\bf 1}\left(n < \frac{A+B+m}{2m}\right),
\enas
as claimed, where in the final inequality we have used the fact that since $\alpha + \beta \le 1$
for $\alpha=P(N \le n-1)$ and  $\beta=P(N=n)$, then for any nonnegative numbers $a$ and $b$ we have
$\alpha a + \beta b \le \max(a,b)$.

\end{proof}

{\em Proof of Theorem \ref{thm:mainAB}}. That $EW_n=0$ and $\mbox{Var}(W_n)=1$ is
the content of Lemma \ref{mean.var.W.AB}.

We now compute the $L^1$ bound using Theorem \ref{T.cond.V}.
Applying (\ref{condmeanAB}) of Lemma \ref{t1} with $\tau^\Box$ and $\tau$ we obtain
\beas
\lefteqn{|\mu_{\tau^\Box}-\mu_\tau|}\\
&=&\sqrt{\frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)}}  \cdot \frac{(A+m|S_n^\Box-S_n|)}{A+B+mn}\\
&\le& \sqrt{\frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)}},
\enas
since both $S_n$ and $S_n^\Box$ take values between $0$ and $n$.

Applying the definition (\ref{def:V}) of $V$ on $\tau^\Box$ and $\tau$,
\beas
|V_{\tau^\Box} - V| = \sqrt{\frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)}} \cdot |S_n^\Box-S_n|,
\enas
so that for $n \ge (A+B+m)/2m$ we have
\beas
E|V_{\tau^\Box} - V| = \frac{2mn}{A+B+m}\sqrt{\frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)}},
\enas
while for $n < (A+B+m)/2m$,
\beas
E|V_{\tau^\Box} - V| = \sqrt{\frac{(A+B+m)(A+B)^2}{AB(A+B+nm+m)(n+1)}}.
\enas
The calculation is completed by using (\ref{EcondT-mutau}) and (\ref{urn.std.3rd.t.-mu}) of Lemma \ref{t1} for the final two terms, and then applying the inequality $(A+B+m)(A+B)^2 \le (A+B+m)^3$.
\bbox

redpajama_set_name: RedPajamaArXiv


FTP, or File Transfer Protocol, is among the most widely used means for exchanging web files between a computer and a web server. Using an FTP client program like FileZilla, you can set up a connection to your web hosting account and drag & drop files or entire folders in both directions and the software will handle all the rest. There are a number of pluses of using File Transfer Protocol, among them the option to restart a download/upload if there is some problem with the connection or the possibility to set up different FTP accounts. Thanks to the latter option, you can create FTP users that can access only particular folders in your web hosting account, but no other folders, email accounts or other info, which makes it the perfect solution if you have to grant a web designer access, for instance. You can also use an FTP account with famous web design applications like Dreamweaver or FrontPage and upload the web content that you've created straight from them, without the need to resort to 3rd-party software.
You can set up as many FTP accounts as you like with any of the cloud hosting plans offered by us, since we have decided not to restrict this functionality. Thus, you can maintain multiple sites simultaneously or give other individuals access to separate folders. As soon as a given job has been fulfilled and you wouldn't like that individual to be able to connect to your files anymore, you can update the password for the FTP account or you can simply remove that account with a couple of clicks of the mouse and stop worrying about someone accessing it once again. The Hepsia Control Panel, which is included with all shared packages, will allow you to check the complete list of the existing FTP accounts listed in alphabetical order. It takes only one single click of the mouse to manage different options for any of them – to edit the password or the access path, to download an auto-configuration file, etc.


text:
kleiner-Slider_1
MMI RWTH Aachen > Research > Environment > Storm Simulation
MMI RWTH Aachen › Forschung › Storm Simulation
Storm Simulation
In 2007 storm Kyrill unleashed its wrath on Europe. In the German state of North Rhine-Westphalia forest damage of an estimated 1.5 billion euros was incurred. In order to minimize such damages in the future, storm simulations can be consulted as a suitable aid to support forestry planning. This video shows a realistic visualization of a computational fluid dynamics (CFD) data set, which has been calculated for an area near Schmallenberg using the parameters for Kyrill. Through intuitive visualization by particle tracing it immediately becomes apparent why exactly windfalls had come about. Through the deployment of the demonstrated simulation technology such risk areas can be identified and taken into consideration accordingly in forestry planning.
Please activate JavaScript to view this video.
Video-Link: https://www.youtube.com/watch?v=lG677VTn_WY
More Research Subjects
AllEnvironmenteRoboticsIndustrySpace
Rigid Body Dynamics
When it comes to simulation algorithms, the simulation of the dynamic behavior of the mechanical...
The development of increasingly sophisticated Advanced Driver Assistance Systems (ADAS) and highly complex Autonomous Vehicles...
Virtual Testbeds
Virtual test beds provide the foundations for the realization of new crystallization points for the...
Projective Virtual Reality
Intelligent man and machine interfaces represent the key technologies for applications in service robotics, in...
The objective of innovations in eRobotics is to provide a comprehensive software environment for the...
Environment Modelling
Models are the basis of every simulation and therefore the basis of almost every eRobotic...
3D Simulation Technology
Almost all eRobotics techniques are based on the implementation of 3D simulation technology. The basis...
Forest Machine Simulation
3D simulation of forestry machines facilitates training that is efficient and free from danger, e.g....
Virtual Forest
With the Virtual Forest, the era of the proverbial "you can't see the wood for...
When it comes to the construction machine simulator, the realistic simulations of the machines themselves,...
Satellite Servicing
Life without satellites has become inconceivable in this day and age. Whether it is used...
Exploration Robots
Robots are indispensable to planetary exploration and in the exploration of our solar system.
With its virtual International Space Station, MMI has developed a comprehensive, accessible and interactive...
3D Control Technology
"Simulation-based control" techniques in the context of 3D simulation technology enable the design and testing...
Motion Simulator
In the case of virtual reality applications the (audio)-visual feedback of the simulation takes often...
The development and operation of autonomous underwater vehicles (AUV) is a complex, dangerous and expensive...
In 2007 storm Kyrill unleashed its wrath on Europe. In the German state of North...
Combine Harvester Simulation
In the case of the combine harvester simulation, aside from realistic simulation of the machines,...
Simulation-based Reasoning
The so-called Sense-Think-Act paradigm is considered as the operative definition of a (mobile) robot. A...
Virtual Commissioning
"Virtual Commissioning" aims at the setup and initial programming of complex automation systems based on...
MMI | Institute for Man-Machine Interaction
RWTH Aachen University | Ahornstraße 55 | 52074 Aachen | Germany
Director: Univ.-Prof. Dr.-Ing. Jürgen Roßmann
Contact | Team | Site Credits | Disclaimer | Sitemap
© 2022 MMI RWTH AACHEN UNIVERSITY. All rights reserved.

redpajama_set_name: RedPajamaCommonCrawl


text:
<?php
if(isset($_GET['model'])){
	$model = $_GET['model'];
}
else{
	$model = "judge";
}
switch($model){
	case "judge":include("template/judge.php");break;
	case "rank":include("template/rank.php");break;
	case "about":include("template/about.php");break;
}
?>

redpajama_set_name: RedPajamaGithub


text:
<?php

namespace PUGX\MultiUserBundle\Tests\Stub;

class AnotherUserProfileForm
{
    public function getName()
    {
        return "form_name";
    }
}


Position: Legal Services of Central New York, Inc. will hire a staff attorney for our Utica office.
Background: Legal Services of Central New York (LSCNY) is committed to achieving justice for low-income people and those who have difficulty accessing the justice system. For more than 50 years, LSCNY has represented clients in housing, income security, consumer, education, employment, and access to health care matters. We seek to make a difference in clients' lives and their communities by eliminating the barriers that cause poverty or reduce our clients' ability to escape from poverty. We continually search for ways to improve the lives of our clients and their communities by creating new projects such as the Faith Seidenberg Racial Justice Initiative, the Community Counsel Project, and a re-entry project in collaboration with the Onondaga County Assigned Counsel Program.
Responsibilities: Staff attorneys engage in comprehensive legal representation including advice, negotiation, litigation, and administrative and legislative advocacy. Attorneys represent individuals and engage in broad-based impact advocacy. Attorneys collaborate in teams led by an experienced supervisor.
Admitted or eligible for admission to the bar of the State of New York preferred.
Experience or interest in housing work is a plus.
A demonstrated commitment to providing high quality legal assistance to low-income and under-served populations.
Excellent analytical, legal writing, and advocacy skills.
Demonstrated ability to work effectively with individuals from diverse backgrounds.
Closing date: Closing date for applications Friday, February 1, 2019 or when filled.
Legal Services of Central New York, Inc. is committed to hiring, retaining, and promoting diversity and culturally appropriate inclusion so we can be more creative, effective, and just, bringing more varied perspectives, experiences, backgrounds, talents, and interests to the practice of law and the administration of justice. We accept, respect, and value differences that include attributes such as age, race, gender, ethnicity, religion, sexual orientation, gender identity and expression, disabilities, language, socioeconomic status, marital and parental status, and cultural background.


Domestic terrorism is visited upon a Hamburg family with tragic effect in the German offering, Into the Fade (Aus dem Nicht). Diane Kruger stars as a wife and mother who seeks revenge against the neo-Nazi sympathizers who kill her Kurdish husband and their young son in a bomb blast.
Into the Fade has a running time of 1 hour 46 minutes and is rated R.


Too often it happens that unfamiliar compositions beautifully played are accompanied by the realization that conservative program making may well keep us from ever hearing the pieces again.
The Chamber Music Society of Lincoln Center once more prompted such a thought with Monday night's concert at the Harris Theater for Music and Dance.
The second half of the program offered a chestnut, Franz Schubert's "Trout" Quintet. But all six works before it were unfamiliar and performed so tellingly that those who braved the cold should have felt warmed for having heard the selections in concert at least once in their lives.
The longest piece was Ernest Chausson's 1881 Trio, a composition written under the influence of his teachers Jules Massenet and Cesar Franck. Mystery, brooding and melancholy characterized many works of art, visual and musical, from the period. This one has a hypersensitive atmosphere that requires special sympathy, which was most clearly displayed by pianist Inon Barnatan, violinist Areta Zhulla and cellist Nicholas Cannelakis in passages the composer directed to be held back.
The more withdrawn moments of the Chausson were effectively prepared for by Antonin Dvorak's Opus 40 Nocturne and four selections from his "Cypresses." The Nocturne, in particular, (played by Zhulla, Cannelakis, Ani Kavafian, Matthew Lipman and David Grossman) had ideal shimmering gentleness, with especially smooth and dreamy tone. The first played of the "Cypresses" – No. 5 in Dvorak's numbering – also caught the atmosphere of soft moonlight in the prominent viola playing of Lipman, who replaced the ailing Lawrence Power.
Schubert's "Trout" Quintet is nowadays so often performed with hard-edged clarity that it shines brightly where it really should only glimmer. This time sharp corners were rounded but not overly polished. So the performance was nicely responsive to shifts in mood, having variety of attack and tone throughout, which made for particularly well-characterized variations in the fourth movement.
Much of this was owing to Barnatan's pianism. As indicated at the Harris a fortnight ago, he has affinity for Schubert's inwardness as well as lightness. Nothing sounded pressured on Monday; on the contrary, as in the Chausson, there was a sense of withdrawing from tension to occupy spaces of serene release.
Sharing in the delicately dappled approach were Kavafian, Lipman, Grossman and Gary Hoffman, with the latter's cello achieving hushed perfection in the fifth variation of the Andantino.


text:
How Costume Designer Heidi Bivens Secretly Created Your Wardrobe
Heidi Bivens is the woman behind The Beach Bum's sleazy jubilation and Mid90s' skate realness.
By Rachel Tashjian
TBB_0004.dngAtsushi Nishijima - Courtesy of NEON and VICE
Back in November 2017, when director Harmony Korine started filming The Beach Bum, his costume designer, Heidi Bivens, asked the director if she should call in some flame-print Ugg boots designed by their friend Jeremy Scott. The aesthetic was almost eerily perfect for Korine's universe—high fashion meets bad taste—and after all, Korine had told Bivens, "I just want [Matthew McConaughey] to look like he's on fire." But they decided against it, Bivens said. "Harmony was like, 'We can't use any fashion, because whatever we do is going to end up becoming fashion.' "
That seems to have happened even faster than Korine and Bivens thought. As soon as filming began, paparazzi images began circulating from the set, a year and a half before the movie's March 29 release date. The images went viral: By now you've surely seen countless photographs of McConaughey, who plays the wandering Florida stoner Moondog, with his hair a stringy, bleach-blond straggle, wearing a burnt-orange-and-blue palm-tree-print shirt with matching shorts, and white tube socks and Velcro dad sneakers.
Bivens out and about.
Or in a grandma's-Florida-condo-chintz-print blouse and shorts. Or skateboarding in silver Uggs appliquéd with purple sequined hearts. Every look was more ridiculous than the last: In mid-December, McConaughey was in a kind of cheesy '90s mall version of a Memphis Design pattern, riding a hoverboard while eating a lobster with his hands. His shirt was unbuttoned, and he was wearing a lobster bib. It was righteous.
As images kept arriving over the next few months, it was as if there were two performances happening: the film itself and the making of the film, distinct from the movie's plot, which unfolded with its own cinematic twists and surprises. Here was Isla Fisher, playing Moondog's girlfriend, Minnie, in skintight jersey with a giant pink margarita glass. There was Snoop Dogg, wearing purple silk paisley. Then Zac Efron showed up in a denim vest covered in evangelical Christian patches, his beard carved into stripes (Korine recently said this was inspired by "a panini") and his badly bleached hair moussed into a freakish wedge. Then McConaughey was reading a book that matched his shirt and wearing a madras blazer while wearing two pairs of glasses. Tom and Lorenzo called it "the role he was born to play."
Korine has a pretty bizarre conception of celebrity; he doesn't so much wink as he does look at you, deadpan, and blink slowly. It seemed like McConaughey was playing a caricature of the most extreme version of public persona—"Alright alright alriiiiiiight" but with a wardrobe to express the joyful chaos of his inner life—and Efron was toying with his inherent cheesiness, and Snoop Dogg was...getting paid to play Snoop Dogg, the ultimate Snoop Dogg finesse. The film's costumes, and the moments that seemed to erupt in tandem with their exuberance, were so extreme, so potent, that if the film never came out, we'd still have felt its influence.
And then a weird thing started happening. Uggs came back into style. Prada showed a flame-print suit last summer. Footwear News wrote that Justin Bieber appeared at Coachella "dressed up like Matthew McConaughey's character in the actor's upcoming film, The Beach Bum," and in June, Vanity Fair anointed Bieber and Jonah Hill and Pete Davidson as kings of "scumbro" style—"the R.E.I.-clad trustafarian co-ed meets Supreme. It's pizza for breakfast and caviar pizza for dinner." The film hasn't even been released yet, but its fingerprints are all over fashion already. How did Bivens and Korine tap into the unlikely way that a ton of guys are dressing before this movie has even come out yet?
"The costume designer's incredible, Heidi Bivens," says Hayes. "She did Spring Breakers, and she also did Jonah Hill's movie, Mid 90s. She's kind of like an auteur costume designer. It's incredible what she does."
It isn't that Bivens is some kind of sleazecore whisperer; by all appearances, she seems to be a cool blonde who wears a lot of black clothing by The Row and Christopher Kane. Instead, Bivens shares Korine's bizarre Gen X vibe: an obsession with subculture and brands and ideas and people the mainstream might overlook or discount, whether that's Uggs or Disney stars. But The Beach Bum isn't the first time she's filtered those interests into film. Bivens, who is 42, started her career working in magazines, styling for W, Purple, and Vogue, and has worked as a costume designer since the early 2000s—she's styled music videos for Kanye West and Beck, for example, and did costumes for David Lynch's Inland Empire (2006), and that great Spike Jonze Kenzo perfume ad from 2016 in which Margaret Quallay dances like a maniac, and short fashion films directed by the likes of Tim Barber, Cass Bird, and Daphne Guinness.
More recently, her approach has bled into—or straight-up helped inspire—the current '90s revival. In addition to Korine's upcoming film, Bivens also worked last year as the costume designer for Jonah Hill's directorial debut Mid90s. "I hired Heidi three years before we even started shooting," Hill told GQ around the time of the film's release. "I knew very early on who I wanted to do costumes." In other words, she's sort of a legend in that universe. For that film, she scoured her stash of contemporaneous skate magazines and recreated graphic T-shirts by brands like Blind and Chocolate; she even called in Spike Jonze to look at a screen test to make sure the baggy fit of the jeans was just right.
Jonze is a telling reference: Bivens has that Gen X energy, though hers leans distinctly more toward that era's cool girls. She has the same "girls have to stick together" outlook, protective without mothering, that helped make Kim Gordon, Natasha Lyonne, and Chloe Sevigny icons. That led her to Korine, who she'd never actually met despite moving in the same skateboarding, fashion, and art circles throughout the '90s. "We have a lot of mutual friends in common, like more than I can count, but I just never knew him when he was in New York," she said. So when Spring Breakers was moving into production, she asked her then-agent set up a meeting, for which she went down to see Korine in Nashville. (He now lives in Florida.) "And I think it was clear to him from the start that we could have a shorthand," Bivens said.
Her background makes her a unique quantity in Korine's universe. When we talked about her first working with Korine on Spring Breakers, for example, which stars three young women opposite a savage James Franco and an extremely hard Gucci Mane, she said that she and Korine both felt "that it would be beneficial to have a female costume designer working with the girls. There were a lot of times during that shoot where the girls could have felt vulnerable, and I think it was good for them to have me to lean on and turn to in terms of building the wardrobe for the film."
But something feels different this time around. I told her that, while lots of people are thinking about The Beach Bum as Korine's magnum opus—his most fully realized and coherent film—it feels like her magnum opus, too. Something about the insane coherence of its chaotic look feels inevitable, like Korine and Bivens have been waiting to articulate this for a long time.
As she did in Mid90s, for example, she called on her long term brain trust for The Beach Bum. Efron's vest, for example, is a piece Bivens made in collaboration with New York artist Chris Habib, another friend and fan of both Korine and Bivens. "I told him about Zac's character—that he was sort of a nihilist, but also a Jesus freak. And he just took the ball and ran with it." The patches are earnestly boneheaded, and therefore hilarious: "These ARE my church clothes," reads one.
Korine and Bivens are sort of like the Astaire and Rogers of bad taste: he dictates the broad strokes, and she fills them in with fancy and totally absurd footwork. "He writes a lot of detail in his script regarding how he sees the clothing, and how he sees the costume," she said. Like the Uggs: Korine put them in the script, but Bivens painted them silver and added the purple sequin hearts. Or the marabou feather pink robe: "It's written into the script that he would be wearing Minnie's robe, and then it was up to me to interpret what that robe would be. I just wanted to make it ridiculous and over the top."
And maybe that's the secret of The Beach Bum's style: there's an appealing zeal that's made the look so viral. Being totally dedicated to A Look has historically meant setting boundaries—that's the appeal of men's suiting—but the more dedicated you become to looking wild, the freer you are. Who wouldn't want to dress like that? Like when Snoop Dogg wears a Snoop Dogg robe: "He just showed up on set that day to shoot a scene in that robe," she said. "You don't really tell Snoop Dogg what to do." She and Harmony decided to just let him wear it. "And in the end, it kind of creates another comedic moment, if people even notice that it says 'Snoop Dogg' on it," she said. "It's obvious that Snoop Dogg is basically playing Snoop Dogg."
Or when Korine told Bivens that McConaughey should look like he was on fire. "I don't even think he was specifically thinking of literal flames—we ended up using this photorealistic print of flames—but he just wanted to convey visually that he was on fire," Bivens said. "I just wanted to make it ridiculous and over the top. Because we were really going for comedy. That was stressed throughout the visual direction: if we had a choice, go for comedy." Good fashion is so serious; doesn't it feel good to be able to laugh at it for once?
I asked Bivens how she accounts for the strange coincidences—the Uggs, the flames, the scumbros. "We chose [McConaughey]'s shoes before we realized that there was a dad shoe trend happening," she said, and "Harmony said this thing about how he wanted it to look like he was on fire before that season came out for Prada. And Prada has used flames in their archive, so it's not that Miuccia saw the paparazzi photos and then was inspired."
"But I like to think that there is a creative ether that exists, and we're all influenced by stuff even when we don't realize it, even if we're not conscious of it." As for whether the film's release will mean even more guys dressing a la McConaughey and Snoop, Bivens said, "We'll see what happens this summer. I think it's fun because it's not taking itself too seriously. And I'm happy to see that, just because it feels refreshing." Catch you guys on my hoverboard.
Related Stories for GQMatthew McConaugheySnoop Dogg
Harmony Korine, Glorious Weirdo
By Zach Baron
Photography by Bruce Gilden/Magnum Photos


At ENG Automotive Group our priority is on providing complete customer satisfaction. Our team is dedicated to meeting your automotive needs with a complete line of exceptionally reconditioned and quality pre-owned and off lease vehicles.
Visit our location and take advantage of our great deals on quality vehicles! An experienced salesperson will be happy to assist you in selecting a vehicle that best suits your needs and desires.
If you are looking for a Luxury or an Exotic Sports Car, you are in the right place, we have a standard to be the best in the business of fine cars second to none, extraordinary customer service if you are wanting that ultimate car of your dreams. ENG Automotive Group can locate that special car for you. Whether it be a domestic, import or an exotic sports machine let ENG put you in the right vehicle for the best price.
The honesty and friendly service will have you coming back time after time.
ENG Automotive Group is a member of the Used Car Dealers Association (UCDA).
Copyright © 2015 Engautos.com. All Rights Reserved.


For each visitor, we collect your IP address and standard web log information, such as your browser type and the pages you accessed on our website.
We may collect the e-mail address and phone numbers of those who communicate with us and any other information volunteered by the consumer. Persons who supply their telephone numbers or email address will only receive contact from us to reply to the contact form submitted.
For registered users of our website we store a username, password and email address. If provided by the users we also store names, website address and a biography.
We store contact information of neighborhood watch members so CPD "beat officers" may inform them of active crimes within their neighborhoods or their immediate block. The most prevalent contact by the officers will be E-mails within a subdivision or on particular streets, and many times they want to contact only Captains. The following member information is stored securely and only available to authorized users: names, address/subdivision/beat/block description, phone number(s), email address, notes and training dates.
Upon request we provide site visitors with access to all personally identifiable information that we maintain about them.


text:
Latest Voting News
TUNIS, Tunisia (AP) — Tunisia's president and its shaky, decade-long experiment with democracy are facing an important test Sunday as voters cast ballots in the second round of parliamentary elections. Turnout was just 11% in the first round of voting last month, as many...
PRAGUE (AP) — Petr Pavel, a retired army general, decisively defeated populist billionaire Andrej Babis in a runoff vote Saturday to become the Czech Republic's new president. Pavel, 61, will succeed controversy-courting Milos Zeman in the largely ceremonial but prestigious post....
HARRISBURG, Pa. (AP) — Republican lawmakers who have spread election conspiracy theories and falsely claimed that the 2020 presidential outcome was rigged are overseeing legislative committees charged with setting election policy in two major political battleground states. Divided...
Constitutional amendments stalled with Pa. House stalemate
HARRISBURG, Pa. (AP) — Pennsylvania voters almost certainly will not see a package of proposed constitutional amendments on the statewide ballot in May's primary election after a deadline set by Democratic Gov. Josh Shapiro's administration came and went Friday for lawmakers to pass the...
Republican National Committee Chair Ronna McDaniel won her bid Friday to lead the GOP for two more years, prevailing in an election that highlighted fierce internal divisions that threaten to plague the party into the next presidential season. McDaniel, whom Donald Trump tapped as...
Czechs pick successor to Milos Zeman in presidential runoff
PRAGUE (AP) — Czechs started voting Friday for a new president, choosing between a retired army general and a populist billionaire in a two-day runoff election to succeed controversy-courting Milos Zeman in the largely ceremonial post. Gen. Petr Pavel and Andrej Babis, who is a...
Poland's opposition criticizes amendments to electoral code
WARSAW, Poland (AP) — Poland's opposition leader on Friday criticized a new electoral reform package introduced by the governing coalition and said his party was planning to strongly monitor the voting process during general elections set for the fall. Donald Tusk, of the Civic...
Greek government survives no-confidence vote over wiretaps
ATHENS, Greece (AP) — Greece's government on Friday survived, as expected, a no-confidence vote following a motion by the country's opposition leader over a wiretapping scandal that has stung the center-right governing party in the run-up to elections. Of parliament's 300...
RICHMOND, Va. (AP) — In a series of key votes Thursday, Virginia Senate Democrats defeated several bills that would have restricted abortion access in the state, including a proposed 15-week ban with exceptions that was a priority for Republican Gov. Glenn Youngkin. They are the...
Gov. Walz proposes $3.3B for infrastructure across Minnesota
ST. PAUL, Minn. (AP) — Gov. Tim Walz proposed a $3.3 billion public infrastructure package Thursday that would use a combination of borrowing and cash to finance improvements to roads, bridges, water systems, housing and the environment across Minnesota. "The investments in our...
Grants Pass Daily Courier


text:
'Real Housewives of New Jersey' recap: The return of Dina Manzo and the wrath of Kim Granatell
Updated: Apr. 01, 2019, 1:36 p.m. | Published: Aug. 03, 2010, 4:50 a.m.
BRAVOThe "Real Housewives of New Jersey" stars, sans Danielle Staub, gather for Audriana Giudice's christening party.
Let's kick off this "Real Housewives of New Jersey" recap by stating for the record that I haven't seen acting this bad since Gia Giudice auditioned for that Christian Slater film. And I'm not even talking about Gia's acting -- I'm talking about Teresa Giudice denying she's a stage mom. It's just so obvious that every single scene here is a carefully (or not so carefully) orchestrated set piece, from Christine Staub's oh-so-offhand revelation that Teresa has been talking about Danielle Staub's search for her birth mother to Caroline Manzo's casual visit to the Brownstone, where son Chris tries to set her up on a lunch date with Kim Granatell, to Kim G.'s excrutiating blow-out with Danielle at an upscale Italian trattoria. (Why do these things always have to be in restaurants? Are producers hoping for an overturned table, perhaps some flying cutlery? Answer: Yes.)
BRAVODina Manzo, baby Audriana's godmother, prepares for the baby's christening on tonight's episode of "Real Housewives of New Jersey."
A party to end all parties (we hope): Teresa is now saying that husband Joe kept their money woes a secret, that she always lives within her means, and that she's still not clear on how much money she and Joe owe. Let's revisit the timeline once again: She and Joe filed for bankruptcy in October. Her signature is on the papers, which say she and Joe owe about $11 million. The party the Giudices held to celebrate baby Audriana's christening was held sometime in November or December. It featured a cocktail hour and a sit-down dinner for 200. Ice sculptures. A Marie Antoinette lookalike emerging out of a table to serve sushi. This is living within your means if you're a hedge fund boss, not if you're slinging pizzas in Hillside and flipping apartments in the Oranges. I actually do feel a little bad for Joe. He clearly knows how bad this looks, even if he's at a loss to reign in his wife, who seems to think that her free-spending ways are utterly endearing. "Make it nice, not too nice, you know what I mean?" he helplessly tries to direct Chris, who is managing the event. On the way to the party, Joe points out a $5 shoe store in beautiful downtown Paterson, and Teresa snips: "Before you used to never know how much I used to pay for parties. I used to just write the checks." Joe: "Now you don't have a checkbook anymore, so now I know." Dina Manzo, Audriana's godmother, returns to the show for the party, which she helped plan and decorate, and the event is hailed as a huge triumph, unless, of course, you're the Giudices' bankruptcy trustee.
OB-GYN OMG: Danielle takes Christine, 16, to her first visit to the gynecologist. Oh lord, will this assault on common decency never end? Thankfully, the doctor informs her (and us) that no, there will be no exam (though I wouldn't put that past Bravo). He just wants to talk to her about getting the HPV vaccine, which is actually not a bad idea. Instead of using this as an opportunity to educate viewers about a vaccine that can prevent cervical cancer and genital warts, Danielle uses this as an opportunity to humiliate her daughter. After asking an incisive question about contracting disease through oral sex, she turns to her daughter. "Can you just please tell me you're not going to have sex?" she asks Christine. "I'm just sitting here thinking, why would you want to after this conversation?" Why would anyone want to, after watching this scene? "Real Housewives of New Jersey" should be prescribed as birth control.
Double date: Chris is hanging out with his best friend, John Granatell, Kim G.'s son, at Kim's house. Kim comes home and tells Chris that she really would like to be friends with his mother, but Caroline draws the line at anyone who is friends with Danielle. "I tried to break her," Kim G. says. "She's tough to break," Chris shoots back. She pumps Chris for tips on how to get close to Caroline -- she doesn't say why, though it's perfectly clear that her main goal is to consolidate her position on the show, not because she wants to trade recipes with Caroline and give each other manicures. Chris says he'll talk to his mother. When Caroline JUST SO HAPPENS to stop by the Brownstone later, Chris broaches the idea of a mother-and-son double-date, and Caroline, to no one's surprise, says no.
Mamma Mia: Danielle confides in Kim G., her "dearest friend," that she wants to track down her biological mother, who gave her up for adoption when she was 15. ("I just want to smell her. I just want to smell my mom." Hate to break this to you, but she probably smells like BenGay.) Then we see Christine telling Danielle that one of her friends heard from her aesthetician that Teresa was gossiping about Danielle's search for her birth mom. Danielle puts two and two together and "realizes" that Kim G. must be in touch with Teresa, Caroline and Jacqueline Laurita, because Kim G. is the only one she's confided in about this. Well, Kim G., and a roomful of Bravo production staffers. She tells the girls she needs to step outside to make a phone call. She calls Danny Provenzano, of course, who wants to know how they're going to handle it. Subtext: He wants to take Kim G. on a one-way getaway to the Pine Barrens.
The sound and the fury: Kim G. busts in on Jacqueline and starts cussing up a storm about Danielle in front of Jacqueline's infant son. Nice. She's angry because she found out that Danielle told another friend to stay away from her. (Which is not exactly how Kim G. described their falling-out to me when I interviewed her in June: She told me she had invited Danielle to a fancy party, and Danielle turned her down, and later she found out that Danielle went on her own. Not quite "Et tu Brute?" territory.) Anyway, Kim G. tells Jacqueline that she was right about Danielle all along. Danielle, meanwhile, is still venting to Danny and to the camera about Kim G's treachery, that Kim G. just wanted to gather information from her so she could get in good with Teresa, Caroline and Jacqueline. In perhaps the oddest moment of the evening, Danielle adopts a bizarre hillbilly/baby accent to imitate Kim G.: "If I attack Danielle, maybe, just maybe they'll be my bes' fwends toooo." Danielle and Kim G. meet up at Portobello in Oakland, where Danielle immediately goes on the offensive, accusing her of talking with Teresa, Jacqueline and Caroline. "Absolutely," Kim G. says. I will say this about her, she has never denied playing both sides. She just calls it "being stuck in the middle." Danielle tells her that she never should have passed along intimate details about Danielle to "those animals." Kim G. insists she's been a good friend to Danielle, despite Danielle putting her in some very uncomfortable situations. And then the scenery-chewing commences. Danielle gets up to walk out of the restaurant, and Kim G. follows her out, yelling that she's had enough of Danielle's crap. "Kim G. showed her true colors, and it was a huge display of disgusting," Danielle says in voiceover. As Danielle scurries into her SUV piloted, I believe, by Danny, Kim G. screams -- and I can't vouch entirely for the accuracy of this screed because of all the bleeps -- "You get the (bleep) out of here, you piece of (bleep). Look at you, you're going to be a (bleeping) old lady with fake and square (bleeps)." Fake and square bleeps? Them's fightin' words!


text:
Christian convert brutally murdered
Star Report
Wed Mar 23, 2016 12:00 AM Last update on: Wed Mar 23, 2016 03:42 AM
Unidentified assailants slit the throat of a Christian convert at Garialpara in Kurigram town yesterday.
The attack came a week after a Shia homeopathic doctor was killed in Jhenidah and a month after a Hindu priest was murdered in Panchagarh.
Hossain Ali, 68, a freedom fighter and a retired inspector of Family Planning Department in Kurigram, was on his way home in Garialpara from his routine morning walk around 7:00am when the attackers swooped on him, said police and locals.
Three men riding a motorcycle intercepted Hossain, who had converted to Christianity from Islam I the 90's. Two of them got down from the bike and attacked him with sharp weapons, leaving him dead on the spot, said locals.
As a pedestrian witnessed the incident, the killers chased him and hurled a homemade bomb at him, they added.
After the explosion, locals rushed to the scene and tried to stop the killers from fleeing.
The attackers threw two more bombs while fleeing, as some people tried to block a road by felling trees. The bombs, however, did not explode.
Witnesses said two of the killers were wearing helmets.
Mohammad Tabarak Ullah, superintendent of police in Kurigram, said police reached the spot after being informed and recovered the body.
The body was sent to Kurigram Sadar Hospital morgue for an autopsy.
Nazrul Islam, resident doctor of the hospital, said Hossain's throat was cut and there were hacking marks in his neck and back.
He died of excessive bleeding, said the doctor.
SP Tabarak, who visited the spot following the incident, said police collected evidence from the spot.
Filing of a case was underway, police said.
Sirajul Islam, sub-inspector of Kurigram Sadar Police Station, said police picked up three people for interrogation in connection with the murder. He, however, refused to disclose their identities.
A guard of honour was given to Hossain in front of his home in the town by the police and the local administration around 5:45pm. Later, his body was sent to his village home at Sitaijhar for burial.
Father of two daughters and a son, Hossain converted to Christianity in 1999, said his family members.
It could not be immediately known whether Hossain had changed his family name after converting to Christianity. He was known to the locals by his Muslim name.
He had been living at his Garialpara house for the last two decades.
The victim's son Ruhul Amin Azad said he was clueless about who killed his father, as Hossain had no enemies.
Azad, however, said the family has disputes with locals over ownership of a piece of land.
He also said a person identifying himself as Abul Bashir rented a room of their house in the first week of this month.
Azad said Bashir told them that he was from Rangpur' Mahiganj. Although he initially said he would live with his family, he lived alone.
Bashir left the house a few days ago saying he would bring his family but had not returned, the victim's son said, adding that they found Bashir's mobile phone switched off.
On February 14, unidentified criminals hacked Abdur Razzak, 48, a Shia homeopathy doctor, to death in Kaliganj upazila of Jhenidah.
Razzak used to give religious sermons to some 20 followers at his pharmacy on Ganna Road in Kaliganj municipality, said police.
Just after dawn on February 21, criminals slit the throat of priest Jogeswar Dasadhikari of Sri Sri Shonto Gaurio temple in Debiganj of Panchagarh and fled the scene firing shots and exploding homemade bombs, which injured two devotees.
The attackers fled the scene on a motorbike.
Global terror outfit Islamic State (IS) claimed responsibility for the two incidents, according to US-based SITE Intelligence Group.
However, The Daily Star could not independently verify both the claims.
Our Kurigram correspondent contributed to this report.
brutal murder / christian convert / garialpara / Kurigram
3 die in Bogra road crash
Indian govt asked to pay Rs 5 lakh to Felani's family
Ctg couple to walk gallows for murder
Machetes strike again: This time, a tailor killed in Tangail
Felani killing: Justice denied, border killings continue
১৯ মিনিট আগে|নারায়ণগঞ্জ সিটি নির্বাচন
অনুকরণীয় রাজনীতিবিদ ডা. সেলিনা হায়াৎ আইভী


Can't make it to our Business Showcase but want to be part of this great event? No problem! Your logo can be displayed prominently on our Greenwich Chamber of Commerce Walk of Fame banner at our April 25th Business Showcase. Let everyone see your name as they walk into our Business Showcase. A great advertising opportunity. Send us your logo now!
Fee: $500. Your name will also be recognized in our Chamber's special Business Showcase website page. Plus, two free tickets to the event for your friends. Maximum ten logos so reserve yours.


Now that I've got a Lucky Sheep sleeping bag, I don't have to sleep in a synthetic bag anymore when I'm outside, and my sleep quality is better than ever. And when I sleep inside on the floor, the Lucky Sheep bag is perfect for that as well. It's so cozy, it makes minimalist sleeping feel like a complete luxury. Goodbye plush mattresses and cumbersome box springs! A Lucky Sheep bag and a foam pad on the floor, and you'll get a perfect night of sleep.


text:
Justia Dockets & Filings Ninth Circuit California Northern District Apple Inc. v. Samsung Electronics Co. Ltd. et al Filing 606 Attachment 1
Apple Inc. v. Samsung Electronics Co. Ltd. et al
EXHIBITS re #602 Administrative Motion to File Under Seal re Samsung's January 10, 2012 Filings Exhibit 3 - [Proposed] Public Redacted Version of Samsung's Motion for Clarification filed bySamsung Electronics Co. Ltd.. (Attachments: #1 Declaration, #2 Exhibit A-J, #3 Exhibit K, #4 Exhibit L, #5 Proposed Order)(Related document(s) #602 ) (Maroulis, Victoria) (Filed on 1/11/2012)
1 QUINN EMANUEL URQUHART & SULLIVAN, LLP Charles K. Verhoeven (Cal. Bar No. 170151)  charlesverhoeven@quinnemanuel.com 50 California Street, 22nd Floor  San Francisco, California 94111 Telephone: (415) 875-6600  Facsimile: (415) 875-6700  Kevin P.B. Johnson (Cal. Bar No. 177129) kevinjohnson@quinnemanuel.com  Victoria F. Maroulis (Cal. Bar No. 202603) victoriamaroulis@quinnemanuel.com  555 Twin Dolphin Drive 5th Floor Redwood Shores, California 94065  Telephone: (650) 801-5000 Facsimile: (650) 801-5100 Michael T. Zeller (Cal. Bar No. 196417)  michaelzeller@quinnemanuel.com 865 S. Figueroa St., 10th Floor  Los Angeles, California 90017 Telephone: (213) 443-3000  Facsimile: (213) 443-3100  Attorneys for SAMSUNG ELECTRONICS CO., LTD., SAMSUNG ELECTRONICS  AMERICA, INC. and SAMSUNG TELECOMMUNICATIONS AMERICA, LLC   UNITED STATES DISTRICT COURT  NORTHERN DISTRICT OF CALIFORNIA, SAN JOSE DIVISION  APPLE INC., a California corporation, CASE NO. 11-cv-01846-LHK DECLARATION OF BRETT ARNOLD IN SUPPORT OF SAMSUNG'S MOTION FOR CLARIFICATION REGARDING THE COURT'S DECEMBER 22, 2011 ORDER (DKT NO. 535)  Plaintiff, vs.  SAMSUNG ELECTRONICS CO., LTD., a Korean business entity; SAMSUNG  ELECTRONICS AMERICA, INC., a New York corporation; SAMSUNG  TELECOMMUNICATIONS AMERICA, LLC, a Delaware limited liability company,  Defendants.   Date: January 18, 2012 Time: 2:00 pm Place: Courtroom 5, 4th Floor Magistrate Judge Paul S. Grewal [PROPOSED] PUBLIC REDACTED VERSION   02198.51855/4543853.1 Case No. 11-cv-01846-LHK ARNOLD DECLARATION IN SUPPORT OF SAMSUNG'S MOTION FOR CLARIFICATION 1 I, Brett Arnold, declare: 2 1. I am an associate in the law firm of Quinn Emanuel Urquhart & Sullivan, LLP, 3 counsel for Samsung Electronics Co., Ltd., Samsung Electronics America, Inc. and Samsung 4 Telecommunications America, LLC (collectively, "Samsung"). I have personal knowledge of 5 the facts set forth in this declaration and, if called upon as a witness, I could and would testify as 6 follows. 7 2. On September 20, 2011, during preliminary injunction discovery, counsel for the 8 parties conducted a meet and confer in which Samsung offered a compromise that would allow 9 Mr. Sherman to see CAD files, inventor notebooks, the deposition transcript of Christopher 10 Stringer (who was the only design inventor to have been deposed by that time), and Apple's 11 presentations showing that certain design features are functional. Apple rejected this 12 compromise. 13 3. On January 5, 2012, lead trial counsel met and conferred. I am informed that at 14 the meeting, counsel for Samsung indicated that it would like to show Mr. Sherman several 15 additional categories of design documents. Counsel for Apple indicated that it would consider 16 the categories that Samsung would provide. On January 10, 2012, counsel for Apple indicated 17 without explanation that Apple would limit Mr. Sherman's access to the categories identified in 18 Magistrate Judge Grewal's order. I immediately inquired as to Apple's reasons for this decision. 19 As of the time of filing, counsel for Apple had not responded. 20 4. On January 5, 2012, Apple stated in a letter that it had "thousands" of models and 21 parts relating to its products that it would make available for inspection. 22 5. On December 24, 2011, Apple confirmed in a letter from its counsel to counsel for 23 Samsung that it was running a number of basic design search terms through its designers' emails 24 and files, and was beginning to produce the results on a rolling basis. By agreement between the 25 parties, these documents were due to be produced prior to the depositions of Apple's designers 26 back in October 2011. Even now, the parties continue to negotiate over the terms and delimiters 27 Apple is using on many of these design search terms. 28 02198.51855/4543853.1 Case No. 11-cv-01846-LHK ARNOLD DECLARATION IN SUPPORT OF SAMSUNG'S MOTION FOR CLARIFICATION -2- 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 I declare under penalty of perjury that the foregoing is true and correct. Executed in 25 Redwood Shores, California on January 10, 2012. 26 27 /s/ Brett Arnold 28 02198.51855/4543853.1 Case No. 11-cv-01846-LHK ARNOLD DECLARATION IN SUPPORT OF SAMSUNG'S MOTION FOR CLARIFICATION -3- GENERAL ORDER ATTESTATION 1 2 I, Victoria Maroulis, am the ECF user whose ID and password are being used to file the 3 foregoing document. I hereby attest pursuant to General Order 45.X.B. that concurrence in the 4 electronic filing of this document has been obtained from Brett Arnold. 5 /s/ Victoria Maroulis 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 02198.51855/4543853.1 Case No. 11-cv-01846-LHK ARNOLD DECLARATION IN SUPPORT OF SAMSUNG'S MOTION FOR CLARIFICATION -4-


text:
Nomada furvoides är en biart som beskrevs av Stoeckhert 1944. Nomada furvoides ingår i släktet gökbin, och familjen långtungebin. Inga underarter finns listade i Catalogue of Life.

Källor

Externa länkar

Gökbin
furvoides

redpajama_set_name: RedPajamaWikipedia


I ordered some checks to the dealer. The clutch hose of 340R was no care. It means the rumor of official counter clutch hose is not true.
I changed tires new. All cost of Yokohama A038LTS is about EURO 610. It's cheaper, isn't it?
The fenders hit the tires at gaps of road, so I reinforced them with a repair set of epoxy resin. It includes glass sheets & sandpapers.
You can see white color, can't you? 150g is heavy weight, because the fenders are only 800g.


text:
Q: Why the deference of pointer to a const is changed while the const is still const? const int c= 7 ;
const int * b= &c ;
int *a = (int *)b ;
*a = 6;
cout<<*a<<endl;
cout<<*b<<endl;
cout<<c<<endl;

I run it by g++ complier , and the output is :
6
6
7
I know it's inadvisable, but I am confused with it. I think the pointer b points to c , which is const , but why *b is changed to 6 while c is still equal to 7 ?  they refer to the same stuff , why different?
could you explain how this code works internally ?

A: §7.1.6.1 [dcl.type.cv]/p4:

Except that any class member declared mutable (7.1.1) can be modified,
  any attempt to modify a const object during its lifetime (3.8) results
  in undefined behavior.

Undefined behavior means anything can happen. As far as the language is concerned, your program may appear to run correctly, may crash, may produce incorrect results, may chomp up your hard drive, or may blow up the planet.
In this particular case, it is likely that the compiler simply generated code that prints out 7 directly, assuming that c is 7, which it is allowed to do since c is const.

A: The C-style cast (int *) performs a const_cast on b. Why this is bad has been explained plenty of times on SO.

redpajama_set_name: RedPajamaStackExchange


It's getting cold out there! Grab our Adina Black and Buffalo Plaid Tunic to keep you warm! ..And so you'll actually get out of bed in the morning. We know you never want to. So we decided to serve up some motivation in the form of a tunic.