NEW BRIGHTON -- Charlee Brooks channels her inner mermaid on "Loveren," a musical project that's making a splash.
"Loveren" already passed the first ballot in the Grammy Award nominations, moving on toward further consideration in the Best New Age Album and Best Engineered Non-Classical Album categories.
Brooks finds out Dec. 5 if her debut album reached the coveted final-five in either Grammy category.
"I probably won't get any sleep the night before," the New Brighton Area High School graduate said. "I never expected to make it through the preliminary round, so I'm definitely excited."
Brooks is the vocalist, co-writer and co-producer on "Loveren," an atmospheric concept collaboration with her Laguna Beach, Calif., neighbor, three-time Grammy nominee David Arkenstone, whose work has ranged from video games to televised sporting events such as the Kentucky Derby.
Brooks met Arkenstone at her first Los Angeles job, where she was an audio engineer for Blizzard Entertainment's highly popular World of Warcraft online multi-player game. Arkenstone scores music for Blizzard games.
Later, they were on a Pacific boat cruise with other musicians when the conversation somehow turned to mermaids.
"When you live an ocean life and you're looking at the ocean every day you wonder what's going on out there," Brooks said. "One thing just led to another, and the next I knew we had created this record," Brooks said.
"Loveren," sold on Amazon, iTunes and elsewhere, comes with a lavish 16-page insert that outlines the album's story of passion and romance in the magical world of mermaids.
Brooks sings in a contemporary, operatic ballad style that's been compared to New Age titans Sarah Brightman and Enya.
To augment the straight-forward lyrics, Brooks also created a special Mermaid language, "Mermish," she sometimes uses to provide extra emotional and exotic ambiance that further propels the storyline.
"It was interesting to deliver so much emotion phonetically without language limitation," Brooks said. "Mermish made the album universal in a way that it may impact each imagination differently, and that it not be limited to one country or another. Everyone can understand it because it's based on pure emotion."
Brooks wrote the first song, "Sessa Nulma," while back in New Brighton visiting her parents, Lisa and Steve Jurich, who work at the town's post office. Using Skype, she sang her lyrics to Arkenstone who finished the music.
"David has done a lot of concept albums in the past, but I definitely had to step out of my element," said Brooks, who began singing at age 3, accompanied by her father on guitar, which led to a guest appearance at age 11 on a Carnival Cruise ship.
Brooks performed in New Brighton High musicals and took dance classes with the Lincoln Park Performing Arts Center's Mancini Academy shortly before that Midland facility's charter school opened in 2006. At 17, she moved to California to pursue her dreams.
Those dreams have started to come true since "Loveren" debuted at a sold-out, 1,200-seat Los Angeles theater as the first half of a show called "David Arkenstone's Symphonic Adventure." Part two of the show, a celebration of Arkenstone's 25-year career, featured Brooks singing and dancing throughout, along with guest performances from Karen Briggs (violinist for PBS television favorite Yanni), Eric Rigler (a Celtic musician on the "Titanic" soundtrack) and members of Cirque du Soleil.
PBS has expressed interest in a "Loveren" special, and European concert venues have discussed a tour, Brooks said.
"But I don't have any details at the moment because I've been staying away from those management meetings," she said.
Everything's moving quickly for Brooks, who grew up an admirer of Brightman and Enya, as well as a fan of pop stars like Britney Spears and former Beaver Countian Christina Aguilera.
"Because who doesn't like Britney and Christina, right?" Brooks said via phone from California. "If you saw my iPod you'd be really confused. I've got Broadway right next to rap. It all depends on what day it is."
She hopes a day will come when friends and family back in Beaver County can watch "Loveren" on TV or see a live local performance.
For now, they can sample a "Loveren" concert clip, music video and a behind-the-scenes video on her website, charleebrooks.com, which also links to Amazon where you can buy the CD.
You can't miss the album cover that utilized green screen technology to make Brooks look like a mermaid.
"I was excited about that. It was one of the best moments," she said. "I was like, 'Look, I have a tail!"


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Woot! Washington DC is becoming one of my many second homes. Over the past year I have entertained at the Gaylord Convention Center on what seems a 'regular rotation' between DC, Baltimore, Philly, and New York City.
This week's event was for a great group of specialists at Somnia, Inc. Somnia is one of the largest anesthesia infrastructures for providing supplies across the country.
Their booth was super popular and even neighboring booth attendees kept leaning over to watch the constant creation of caricatures. This event was so popular I ended up drawing, literally, until the rugs started getting pulled out from under me!!! It was extremely fun and I'm honored that they had me and I can't wait until next year!


Set on the fringes of the popular East Devon village of Lympstone.
Bradleys are marketing this property at a price of £600,000, which benefits from gas central heating and double glazing. The current owners have improved the property to include a bespoke, hand-built country style kitchen. The accommodation includes entrance hall, living room with bay window, kitchen/family room, utility and ground-floor cloakroom. On the first floor are four bedrooms and family bathroom.
The attic conversion provides the master bedrooms and further en-suite.


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

Many popular measures of pairwise dependence, for example Kendall's
tau \citep{Kendall38} and Spearman's rho \citep{Spearman04}, have the
undesirable property that they may be zero even when the two
considered random variables $X$ and $Y$ are dependent.  Addressing
this weakness, \cite{BergsmaDassios14} introduced a new rank-based
correlation measure $\tau^*$, which, under mild conditions on the
joint distribution of $(X,Y)$, is zero if and only if $X$ and $Y$ are
independent.  Where Kendall's tau is defined in terms of concordance
and discordance of two independent copies of $(X,Y)$, the new $\tau^*$
is based on similar notions of concordance and discordance for four
independent copies of $(X,Y)$.  While a na\"{\i}ve computation of
$t^*$, the empirical version of $\tau^*$, thus requires $O(n^4)$ time
for a sample of size $n$, it was recently shown that this
computational burden can be reduced to $O(n^2\log(n))$
\citep{WeihsEtAl15}.  As $t^*$ is now computable for larger sample
sizes, understanding its asymptotic behavior becomes a problem of
practical interest and has the potential to yield simple tests of
independence that avoid Monte Carlo approximation of p-values.

We introduce the statistic $t^*$ in Section~\ref{section:
  preliminaries}, where we also review background on U-statistics.  In
Section~\ref{sec:degeneracy}, we clarify that $t^*$ is a degenerate
U-statistic under the null hypothesis that the sample is generated
under independence.  We also prove that in certain settings degeneracy
occurs only under independence.  In Section \ref{section: asymptotics
  under the null}, we use the asymptotic theory of degenerate
U-statistics to derive an explicit representation of the asymptotic
distribution of $t^*$ when the sample is generated under independence
and with marginals that are continuous or discrete.  The asymptotic
distribution takes the form of a Gaussian chaos; specifically, we find
a (in some cases infinite) sum of scaled and centered chi-square
distributions.  Simulations in Section \ref{section: simulations} then
demonstrate how the large-sample theory can be leveraged to perform
tests of independence and compute power.  Indeed, asymptotic
distributions are found to give accurate approximations for sample
sizes as small as $n=80$.  We end with a discussion in Section \ref{section: discussion}.

\section{Preliminaries}\label{section: preliminaries}

\subsection{The $t^*$ statistic} \label{subsection: t*} Let
$(x_1,y_1),\ldots,(x_n,y_n)$ be a sample of points in $\mathbb{R}^2$.
The empirical version of the Bergsma-Dassios sign covariance is
the statistic
\begin{align}\label{equation: test statistics}
t^* := \frac{(n-4)!}{n!} \sum_{\substack{1\leq i,j,k,l \leq n \\
    i,j,k,l~\text{distinct}}}
a(x_i,x_j,x_k,x_l) a(y_i,y_j,y_k,y_l),
\end{align}
where
\begin{multline}
  \label{eq:def-a}
a(z_1,z_2,z_3,z_4) =\\ I(z_1,z_3<z_2,z_4) + I(z_1,z_3>z_2,z_4) -
I(z_1,z_2<z_3,z_4) - I(z_1,z_2>z_3,z_4).
\end{multline}
Here we use $I(\cdot)$ to denote the indicator function and $a,b < c,d$ is shorthand for $\max(a,b) < \min(c,d)$.  As in
\cite{WeihsEtAl15}, we defined $t^*$ in the form of a U-statistic,
whereas \cite{BergsmaDassios14} introduced it as a V-statistic.
Indeed, $t^*$ from~(\ref{equation: test statistics}) is an unbiased
estimator of the sign covariance
\[
\tau^* :=
\Exp\left[a(X_1,X_2,X_3,X_4)a(Y_1,Y_2,Y_3,Y_4)\right]
\]
of \cite{BergsmaDassios14}.  Here, $(X_1,Y_1),...,(X_4,Y_4)$ are
random vectors drawn independently from a given bivariate distribution on
$\mathbb{R}^2$.

\begin{exmp}
  \label{ex:gaussian-taustar}
  Figure~\ref{fig:normal-taustar} shows the values of $\tau^*$ for
  bivariate normal distributions, which we computed by Monte Carlo
  simulation.  The sign covariance $\tau^*$ is an even function of the
  normal correlation $\rho$, and we thus only show values for
  $\rho\in[0,1]$.  For each considered correlation $\rho$ we
  averaged 200 values of $t^*$, each computed from a
  sample of size $n=300$.
\end{exmp}

\begin{figure}[t]
  \centering
  \includegraphics[scale=.4]{bivariate_normal_relation}
  \caption{Sign covariance $\tau^*$ of bivariate normal
    distributions.}
  \label{fig:normal-taustar}
\end{figure}

As we explain in the remainder of this subsection, the statistic $t^*$
is based on counting concordant and disconcordant quadruples.

\begin{defn}
Let $(x_1,y_1),...,(x_4,y_4)$ be four points relabelled so that $x_1\leq x_2\leq x_3\leq x_4$.  We say that the points are
\begin{align*}
	\begin{array}{lll}
		\text{\emph{inseparable}} & \mbox{if } \ \
			\begin{array}{@{}l@{}}
				\text{$x_2=x_3$ or there exists a permutation $\pi$ of $\{1,2,3,4\}$} \\
				\text{so that $y_{\pi(1)}\leq y_{\pi(2)}=y_{\pi(3)}\leq y_{\pi(4)}$,}
			\end{array}
	\end{array}
\end{align*}
and if they are not inseparable, then we call them
\begin{align*}
		\begin{array}{lll}
			\text{\emph{concordant}} & \mbox{if } \text{$\max(y_1,y_2) < \min(y_3,y_4)$\ \ or \ $\max(y_3,y_4) < \min(y_1,y_2)$,} \\
			\text{\emph{discordant}} & \mbox{if }  \text{$\max(y_1,y_2) > \min(y_3,y_4)$ and $\max(y_3,y_4) > \min(y_1,y_2) $.}
		\end{array}
\end{align*}
\end{defn}
The above definitions are mutually exclusive and exhaustive in that
any set of four points in $\mathbb{R}^2$ will be exactly one of inseparable,
concordant, or discordant. Moreover, if the points are drawn from a
bivariate distribution with continuous marginals then they will be
almost surely concordant or discordant. See Figure 3 of
\cite{BergsmaDassios14} for a visual depiction of concordance and
discordance.

Let $S_4$ be the set of permutations on 4 elements, and for $\pi \in
S_4$ and $(z_1,z_2,z_3,z_4) \in \mathbb{R}^{4}$, let $z_{\pi(1,2,3,4)}
:= (z_{\pi(1)},z_{\pi(2)},z_{\pi(3)},z_{\pi(4)})$.  Introducing the
symmetric function
\begin{align}\label{eq: kernel}
h((x_1,y_1),\ldots,(x_4,y_4)) := \frac{1}{4!} \sum_{\pi \in S_4}
a(x_{\pi(1,2,3,4)}) a(y_{\pi(1,2,3,4)}),
\end{align}
we may rewrite $t^*$ as a sum of permutation invariant terms,
namely,
\begin{align}\label{eq: Bergsma-Dassios U-statistic}
t^* = \frac{1}{\binom n4} \sum_{(i,j,k,l) \in C(n,4)} h\left((x_i,y_i),(x_j,y_j),(x_k,y_k),(x_l,y_l)\right),
\end{align}
where $C(n,4)= \{(i,j,k,l) \,:\,1 \leq i < j < k < l \leq n \}$.  Lemma
1 in \cite{WeihsEtAl15} gives the following result.

\begin{lem} \label{lemma: kernel values}
Let $A = \{(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)\}\subset\mathbb{R}^2$. Then
\begin{align*}
	h((x_1,y_1),..,(x_4,y_4)) = \left\{
		\begin{array}{lll}
			2/3 & \mbox{if } \text{the points in $A$ are concordant,}\\
			-1/3 & \mbox{if }  \text{the points in $A$ are discordant,} \\
			0 & \mbox{if }  \text{the points in $A$ are inseparable.}
		\end{array}
	\right.
\end{align*}
\end{lem}

Equation~(\ref{eq: Bergsma-Dassios U-statistic}) expresses $t^*$ in
the familiar form of a U-statistic with symmetric kernel $h$, and we
proceed to review some of the tools available for the study of
U-statistics.

\subsection{Theory of U-statistics} \label{subsection: theory of
  u-statistics} Let $Z_1,Z_2,\dots$ be i.i.d.\ random variables taking
their values in $\mathbb{R}^d$ with $d\geq 1$.  Let $k:\left(\mathbb{R}^d\right)^m
\to \mathbb{R}$ be a \emph{kernel function} invariant to permutation of its
$m$
arguments. For $n\geq m$, the \emph{U-statistic with kernel $k$} is
the statistic
\begin{align}
  \label{eq:Un}
	U_n := \frac{1}{{n \choose m}} \sum_{(i_1,...,i_m)\in C(n,m)} k(Z_{i_1},...,Z_{i_m}),
\end{align}
where $C(n,m) = \{(i_1,...,i_m)\in \{1,...,n\}^m: i_1 < i_2 < ... <
i_m\}$.  Note that $\Exp[U_n] = \Exp[k(Z_1,...,Z_m)]$ so that $U_n$ is
an unbiased estimator of $\theta:=\Exp[k(Z_1,...,Z_m)]$.

Of central importance in determining the asymptotics of U-statistics
are the functions
\begin{equation}
  \label{eq:k_i}
  k_i(z_1,...,z_i) =  \Exp[k(z_1,...,z_i, Z_{i+1},\dots,Z_{m})], \quad i
=1,\dots,m,
\end{equation}
 and their variances
\begin{equation}
  \label{eq:sigma_i^2}
  \sigma_i^2 = \Var[k_i(Z_1,...,Z_i)], \quad i=1,...,m.
\end{equation}
It is well known that $\sigma_1^2 \leq \sigma_2^2 \leq ...\leq
\sigma_m^2$.  In particular, if $\sigma_m^2$ is finite then so are all
other $\sigma_i^2$.  We now recall two theorems on the large-sample
distribution of the U-statistic $U_n$ \cite[Chapter 5]{Serfling:1980}.

\begin{thm} \label{theorem: normal asymptotics}

  If the kernel $k$ of the statistic $U_n$ from~(\ref{eq:Un})
  has variance $\sigma_m^2 < \infty$, then
\begin{align*}
	\sqrt{n}(U_n-\theta) \overset{d}{\to} N(0, m^2 \sigma_1^2).
\end{align*}
\end{thm}

If $\sigma_1^2 = 0$, then the Gaussian limit is degenerate, and we
have $\sqrt{n}(U_n-\theta) \overset{p}{\to} 0$.  Indeed, if
$\sigma_1^2 = 0$ and $\sigma_2^2 >0$, then scaling $U_n$ by a factor
of $n$ results in a non-Gaussian asymptotic distribution.  To present
this result, we write $\chi^2_1$ for the chi-square distribution with
one degree of freedom and define $A_k$ to be the operator that acts
via $g(\cdot) \mapsto \Exp[(k_2(\cdot, Z_1) - \theta)g(Z_1)]$  on
square-integrable functions $g$ (that is,  $\Exp[g(Z_1)^2]<\infty$).

\begin{thm} \label{theorem: degenerate asymptotics} If the kernel $k$
  of the statistic $U_n$ from~(\ref{eq:Un}) has variance
  $\sigma_m^2 < \infty$ and $\sigma_1^2 = 0$, then
\begin{align*}
	n(U_n - \theta) \overset{d}{\to} {m \choose 2}\sum_{i=1}^\infty \lambda_i (\chi_{1i}^2 - 1)
\end{align*}
where $\chi_{11}^2,\chi_{12}^2,\dots$ are i.i.d.~$\chi^2_1$ random
variables, and the $\lambda_i$'s are the eigenvalues, taken with
multiplicity, associated to a system of orthonormal eigenfunctions of
  the operator $A_k$.
\end{thm}

We will use Theorem \ref{theorem: degenerate asymptotics} in Section
\ref{section: asymptotics under the null} to find the asymptotic
distribution of $t^*$ under the null hypothesis of independence.

\section{Degeneracy of the sign covariance}
\label{sec:degeneracy}

Let $Z_i=(X_i,Y_i)$ for $i = 1,2,\dots,n$ be an i.i.d.\ sequence
comprising copies of a random vector $(X,Y)$ with values in $\mathbb{R}^2$.
Let $t^*$ be the (empirical) Bergsma-Dassios sign covariance for this
sample.  We begin our study of the asymptotic properties of the
U-statistic $t^*$ by studying its degeneracy.  Our first observation
is that $t^*$ is degenerate when $X$ and $Y$ are independent, denoted
$X\perp\!\!\!\perp Y$.  Next, in a particular setting that has $(X,Y)$
continuously distributed, we are able to show that $t^*$ is degenerate
only if $X\perp\!\!\!\perp Y$.

The statistic $t^*$ has the kernel $h$ from \eqref{eq: kernel}, which
has $m=4$ arguments.  Specializing the definitions from~(\ref{eq:k_i})
and~(\ref{eq:sigma_i^2}) to the present setting, we may define
functions $h_1,...,h_4$ with variances $\sigma_1^2,...,\sigma_4^2$.
The kernel $h$ is a bounded function and thus $\sigma_4^2<\infty$.
Hence, Theorem~\ref{theorem: normal asymptotics} applies and yields
the following result.

\begin{cor}\label{cor: asymptotics under dependence}
  As $n\to\infty$, the sign covariance converges to a normal limit,
  namely,
  \begin{align*}
    \sqrt{n}(t^* - \tau^*) \overset{d}{\to} N(0, 16\sigma_1^2).
  \end{align*}
\end{cor}

The result just stated provides a non-trivial distributional
approximation to $t^*$ only if $\sigma_1^2>0$.  The following lemma
observes that this fails to be the case under the null hypothesis of
independence, under which $t^*$ is a degenerate U-statistic.  The
proof of the lemma as well as the proofs of all other results in this
section are deferred to Appendix~\ref{section: proofs-degen}.

\begin{lem}\label{lemma: degeneracy of h_1}
  If $X\perp\!\!\!\perp Y$ then $\sigma_1^2 = \Var[h_1(X_1,Y_1)] =0$ so that
  $h_1(X_1,Y_1)$ is a degenerate random variable.
\end{lem}

According to Lemma~\ref{lemma: degeneracy of h_1} and Theorem
\ref{theorem: normal asymptotics}, if $X\perp\!\!\!\perp Y$ we have
$\sqrt{n}t^*\overset{p}{\to} 0$ because $\Exp[t^*] =\tau^*= 0$ under
independence \citep{BergsmaDassios14}.  We thus need to appeal to
Theorem \ref{theorem: degenerate asymptotics} to find a non-degenerate
asymptotic distribution for $t^*$ when $X\perp\!\!\!\perp Y$.  This is the topic
of Section~\ref{section: asymptotics under the null}.

\begin{rem} \label{remark: sigma values} In the continuous case with
  $X\perp\!\!\!\perp Y$, it is possible to compute all of the variances
  $\sigma_1^2,...,\sigma_4^2$ exactly.  We report these values to be
  \begin{align*}
    \sigma_1^2 = 0,\ \ \ \sigma_2^2 = \frac{1}{225}, \ \ \ \sigma_3^2 = \frac{8}{225}, \ \ \ \sigma_4^2 = \frac{50}{225}.
  \end{align*}
  The fact that $\sigma_1^2=0$ was shown in generality in
  Lemma~\ref{lemma: degeneracy of h_1}.  The value of $\sigma_2^2$ can
  be computed as the sum of the squared eigenvalues of $h_2$ which are
  derived in the proof of Theorem \ref{theorem: continuous asymptotic
    distribution}; in particular, we have that
  $\sigma_2^2 = \sum_{i=1}^\infty\sum_{j=1}^\infty \frac{6^2}{\pi^8
    i^4j^4} = \frac{1}{225}$. Finally, $\sigma_3^2$ and $\sigma_4^2$
  can be computed explicitly from the representation of $h$ in Lemma
  \ref{lemma: kernel values}, this computation is trivial for
  $\sigma_4^2$ but quite lengthy for $\sigma_3^2$ and thus is omitted.
\end{rem}

Next, we turn our attention to the case that $X\not\perp\!\!\!\perp Y$ and
$(X,Y)$ are generated from a continuous distribution on
$\mathbb{R}^2$.  In this case we find $t^*$ to be non-degenerate.

\begin{thm}\label{theorem: nondegeneracy of h_1}
  Suppose $(X,Y)$ has a bivariate continuous distribution with a
  continuous density function $f$ with support
  $\ol{f^{-1}((0, \infty))} = [a,b]\times [c,d]$, where
  $-\infty\le a<b\le \infty$ and $-\infty\le c<d\le \infty$.  If $X$
  and $Y$ are dependent, then $\sigma_1^2 = \Var[h_1(X_1,Y_1)] > 0$.
\end{thm}

In the setting of Theorem~\ref{theorem: nondegeneracy of h_1}, we thus
have that $t^* = N(\tau^*, 16 \sigma_1^2/n) + o_p(n^{-1/2})$.

\begin{exmp} \label{example: variance of h1} To gain intuition for the
  magnitude of the asymptotic variance $16\sigma_1^2$, we use Monte
  Carlo integration to compute $16\sigma_1^2$ in the case that $(X,Y)$
  follow a bivariate normal distribution.  Since
  $\sigma_1^2$ is an even function of the correlation $\rho$ of a
  bivariate normal distribution, we consider $\rho \in [0,1]$.  In
  particular, we perform this computation letting $\rho$ take on 20,
  evenly spaced, values between $0$ and $1$.  The results of this
  computation are shown in Figure \ref{figure: variance of h1}. The
  figure shows that the asymptotic variance $16 \sigma_1^2$ gradually
  increases with larger values of $\rho$, peaking at $16 \sigma_1^2
  \approx 0.14$ when $\rho \approx 0.74$, and then decreases to 0 as
  the correlation further approaches 1. Note that a value of
  $\sigma_1^2 = 0$ when $\rho=1$ does not contradict
  Theorem~\ref{theorem: nondegeneracy of h_1} as, in this case, the
  joint distribution of $(X,Y)$ is not continuous. The shape of the curve in Figure
  \ref{figure: variance of h1} can be partially explained by the fact that
  $\sigma_1^2 \leq \sigma_4^2 = \Var(h(Z_1,...,Z_4)) = (\tau^* + 1/3)(1 - (\tau^* + 1/3))$.
  For instance, note that $(\tau^* + 1/3)(1 - (\tau^* + 1/3))$ equals 0 when $\tau^* = 2/3$
  (in which case the correlation can be seen to be 1 or -1), and is maximized at
  $\tau^* = 1/6 \approx .167$ which corresponds a correlation of approximately 0.7 (see
  Figure \ref{fig:normal-taustar}).
\end{exmp}

\begin{figure}
        \centering
        \includegraphics[width=.75\textwidth]{h1_variance.pdf}
        \caption{Monte Carlo approximations to the values of
          $16\sigma_1^2$ for bivariate normal distribution with
          different correlations.}
        \label{figure: variance of h1}
\end{figure}

\section{Asymptotics under the null hypothesis of
  independence} \label{section: asymptotics under the null}

As in the previous section, let $t^*$ be the empirical sign covariance
for an i.i.d.~sample $(X_i,Y_i)$, $i = 1,2,\dots,n$, with values in
$\mathbb{R}^2$.  Throughout this section, we assume the $(X_i,Y_i)$ to be
independent copies of a random vector $(X,Y)$ with $X\perp\!\!\!\perp Y$, so
that $t^*$ is degenerate (Lemma~\ref{lemma: degeneracy of h_1}).  We
thus need to appeal to Theorem \ref{theorem: degenerate asymptotics}
to find a non-degenerate asymptotic distribution for $t^*$.  Since
$\Exp[t^*] =\tau^*= 0$ under independence, we are led to the problem
of determining the eigenvalues of the operator
$A_h:g(\cdot) \mapsto \Exp[h_2(\cdot, Z_1)g(Z_1)]$.

A key observation is that under independence $A_h$ is a tensor product
of operators because the function $h_2$ admits the following
factorization, which along with all other results in this section is
proved in Appendix~\ref{section: proofs}.

\begin{lem}\label{lemma: product formula for h_2}
  If $X\perp\!\!\!\perp Y$ then
  \[
    h_2((x_1,y_1),(x_2,y_2)) = \frac{2}{3}\,g_{X}(x_1,x_2)
    \,g_{Y}(y_1,y_2),
  \]
  where $g_{X}(x_1,x_2) = \Exp[a(x_1,x_2,X_3,X_4)]$
  and $g_{Y}(y_1,y_2) = \Exp[a(y_1,y_2,Y_3,Y_4)]$.
\end{lem}

The function $g_X$ (and similarly $g_Y$) takes the form
\begin{align}
  \label{eq:prob-form-of-g}
	 g_{X}(x_1,x_2)
 	&\ = P(x_1, X_3 < x_2, X_4) + P(x_1, X_3 > x_2, X_4) \\
  \nonumber
	&\hspace{6mm} - P( x_1, x_2 < X_3, X_4) - P(x_1, x_2 > X_3, X_4).
\end{align}

By Lemma~\ref{lemma: product formula for h_2}, $A_h=A_{g_X}\otimes
A_{g_Y}$ and thus the spectrum of $A_h$ is the product of the spectra
of $A_{g_X}$ and $A_{g_Y}$.  We record the general version of this
fact in the next lemma.  Here, eigenvalues are always repeated
according to their multiplicity, and we let
$\mathbb{N}_+=\{1,2,\dots\}$.

\begin{lem} \label{lemma: h_values} Let $g_1$ and $g_2$ be symmetric
  real-valued functions with $\Exp[g_1(X_1,X_2)] = \Exp [g_2(Y_1,Y_2)]
  = 0$ and $\Exp[g_1(X_1,X_2)^2], \Exp [g_2(Y_1,Y_2)^2] <\infty$.  For
  $i=1,2$, let $\lambda_{i,j}$, $j\in\mathbb{N}_+$, be the nonzero eigenvalues
  of $A_{g_i}$.

  Then the products $\lambda_{1,j_1}\lambda_{2,j_2}$, $j_1,j_2\in\mathbb{N}_+$,
  are the nonzero eigenvalues of $A_k$ for $k((x_1,y_1),(x_2,y_2)) :=
  g_1(x_1,x_2)g_2(y_1,y_2)$.
\end{lem}

In the sequel, we use the factorization results from
Lemmas~\ref{lemma: product formula for h_2} and~\ref{lemma: h_values}
to obtain the asymptotic distribution of $t^*$ when $X$ and $Y$ are
continuous (Section~\ref{subsection: continuos}), and when $X$ and $Y$
are discrete with finite support (Section~\ref{subsection: discrete}).
A straightforward extension covers the mixed continuous and discrete
case (Section~\ref{subsection: discrete}).

\subsection{Continuous variables}\label{subsection: continuos}

Suppose now that $X\perp\!\!\!\perp Y$ with $X$ and $Y$ following continuous
marginal distributions.  Since $h((X_1,Y_1),...,(X_4,Y_4))$ depends
only on the joint ranks of $(X_1,Y_1),...,(X_4,Y_4)$, it follows that
$\tau^*$ (and $t^*$) are invariant to monotonically increasing
transformations of the marginals of $(X,Y)$.  As such we may, and
will, assume that $X$ and $Y$ are i.i.d.~$\text{Uniform}(0,1)$.  Then
$(X,Y)$ is uniform on the unit square $(0,1)\times (0,1)$.  In this
case the factorization described in Lemma \ref{lemma: product formula
  for h_2} has a particularly nice form.

\begin{lem} \label{lemma: continuous factorization}
  If $X,Y\overset{i.i.d.}{\sim}\text{Uniform}(0,1)$, then for $(x_1,y_1),(x_2,y_2) \in (0,1)^2$,
  \begin{align*}
    h_2((x_1, y_1), (x_2, y_2)) &= 6\, c(x_1,x_2)\,c(y_1,y_2)
\end{align*}
where
\[
  c(x_1,x_2)=\frac{1}{2} x_1^2+ \frac{1}{2}x_2^2- x_1\vee
  x_2+\frac{1}{3}
\]
and $x_1\vee x_2:=\max\{x_1,x_2\}$.
\end{lem}

Somewhat surprisingly, the function $c$ corresponds to the kernel of
the well studied Cram\'{e}r-von Mises statistic.  Leveraging the fact
that the eigenvalues of $A_c$ are already known, we are now able to
derive the asymptotic distribution of $t^*$.

\begin{thm} \label{theorem: continuous asymptotic distribution} If $X$
  and $Y$ are independent continuous random variables, then
\begin{align*}
	nt^* &\overset{d}{\to} \frac{36}{\pi^4}\sum_{i=1}^\infty\sum_{j=1}^\infty \frac{1}{i^2j^2}(\chi_{1,ij}^2-1)

\end{align*}
where $\{\chi^2_{1,ij} : i,j \in \mathbb{N}_+ \}$ is a collection of i.i.d.\
$\chi^2_1$ random variables.
\end{thm}

Remarkably the asymptotic distribution just given is simply a scale
multiple of the asymptotic distribution of the U-statistic for
Hoeffding's $D$ where
\[
  D=\iint (F_{X,Y}(x,y)-F_X(x)F_Y(y))^2 \ \text{d}F_{X,Y}(x,y);
\]
see \cite{Hoeffding48}.  When $(X,Y)$ has a continuous joint
distribution, it is readily seen that $D=0$ if and only if
$X\perp\!\!\!\perp Y$.  However, this may fail in non-continuous cases.

\subsection{Discrete variables} \label{subsection: discrete}

We now treat the case where $X$ and $Y$ are independent discrete
random variables with finite supports.  Unlike in the continuous case,
the asymptotic distribution of $t^*$ then depends on how $X$ and $Y$
distribute their probability mass marginally. In practical applications these
marginal probabilities must be estimated before using our limit theorem.

In order to present the result, we associate a matrix to a discrete
random variable as follows.  Let $U$ be a random variable with finite
support $\{u_1,\dots,u_r\}$, cumulative distribution function $F_U$
and probability mass function $p_U$.  We then define $R^U$
to be the $r\times r$ symmetric matrix whose $(i,j)$-th entry is
\begin{align} \label{equation: R matrix form}
	R^U_{ij} =&
	\ \sqrt{p_U(u_i)p_U(u_j)} \Bigg\{ \bigg[(F_U(u_i\wedge u_j) - p_U(u_i\wedge u_j))^2 + (1-F_U(u_i\vee u_j))^2\bigg] \\
	&- I(u_i\not=u_j)\bigg[F_U(u_i\wedge u_j)(1-F_U(u_i\wedge u_j))
	+ \sum_{u_i\wedge u_j < u_\ell  < u_i\vee u_j} p_U(u_\ell)(1-F_U(u_\ell))\bigg] \nonumber
	 \Bigg\}.
\end{align}

\begin{thm}\label{theorem: discrete asymptotic distribution}
  Let $X$ and $Y$ be independent discrete random variables with
  finite supports of size $r$ and $s$, respectively.  Let
  $\lambda_1^X,\ldots,\lambda_r^X$ be the eigenvalues of $R^X$, and
  let $\lambda^Y_1,\ldots,\lambda^Y_s$ be the eigenvalues of $R^Y$.
  Then
 \begin{align*}
	nt^* &\overset{d}{\to} 4\sum_{i=1}^r\sum_{j=1}^s \lambda_i^X \lambda_j^Y (\chi_{1,ij}^2-1)
\end{align*}
where $\{\chi^2_{1,ij} : i \leq r,~j\leq s \}$ is a collection of $rs$ i.i.d.\ $\chi_1^2$ random variables.
\end{thm}

In the special case that $X$ and $Y$ are Bernoulli random variables,
the asymptotic distribution can be presented in simple form.

\begin{exmp}
  If $X\sim\text{Bernoulli}(p)$ for $p\in(0,1)$, then
  \begin{align*}
    R^X = \begin{pmatrix}
      p^2(1-p) & -(p(1-p))^{3/2} \\
      -(p(1-p))^{3/2} & p(1-p)^2
    \end{pmatrix}
  \end{align*}
  has rank one and its nonzero eigenvalue is $p(1-p)$.  It follows
  that if $Y$ is a second independent random variable with
  $Y\sim\text{Bernoulli}(q)$ for $q\in(0,1)$, then
  \[
    nt^* \overset{d}{\to} 4pq(1-p)(1-q)(\chi^2_1 - 1).
  \]
  So, $t^*$ can be centered and scaled to become asymptotically
  chi-square.
\end{exmp}

\begin{exmp}
  For a ternary random variable $X$ with $P(X=1)=p_1$, $P(X=2)=p_2$
  and $P(X=3)=p_3=1-p_1-p_2$, we have
  \begin{align*}
    R^X =  \begin{pmatrix}
      p_1(1-p_1)^2
      & -\sqrt{p_1p_2}\left[ p_1(1-p_1)-p_3^2\right]
      & -\sqrt{p_1p_3}\left[ p_3(1-p_3)+p_1p_2\right]\\
      .
      & p_2\left( p_1^2+p_3^2\right)
      & -\sqrt{p_2p_3}\left[p_3 (1 -p_3)- p_1^2 \right]\\
      .
      & .
      &p_3(1-p_3)^2\\
    \end{pmatrix},
  \end{align*}
  where we show only the upper half of the symmetric matrix.  No
  simple formula seems to be available to determine the eigenvalues of
  $R^X$ in this case, but the eigenvalues can readily be computed
  numerically for any (possibly estimated) values of $p_1$ and $p_2$.
\end{exmp}

Finally,  if $X$ is discrete with finite support and $Y$ is
continuous, then a simple extension of Theorems \ref{theorem:
  continuous asymptotic distribution} and \ref{theorem: discrete
  asymptotic distribution} gives the following result.

\begin{cor}\label{theorem: mixed asymptotic distribution}
  Let $X$ and $Y$ be independent random variables, where $X$ has
  finite support of size $r$ and $Y$ is continuous. Let $\lambda_1,...,\lambda_r$
  be the eigenvalues of $R^X$. Then
 \begin{align*}
   nt^* &\overset{d}{\to} \frac{12}{\pi^2}\sum_{i=1}^r\sum_{j=1}^\infty \frac{\lambda_i}{j^2} (\chi_{1,ij}^2-1)
\end{align*}
where $\{\chi^2_{1,ij} : i \leq r,~j \in\mathbb{N}_+ \}$ is a collection of
i.i.d.\ $\chi^2_1$ random variables.
\end{cor}

\section{Simulations}\label{section: simulations}

The results from Section~\ref{section: asymptotics under the null} can
be used to form asymptotic tests of independence, and we now explore
which sample sizes are needed for the asymptotic approximations to be
accurate.  As a test based on $t^*$ has asymptotic power against all
alternatives to independence, it is also of interest to make
comparisons against other tests known to be (most) powerful for
particular settings and alternatives.  Finally, we demonstrate how the
results of Section~\ref{sec:degeneracy} can be used for sample size
computations.  Code for performing asymptotic tests of independence
has been
incorporated in the TauStar\footnote{See
 \url{https://cran.r-project.org/web/packages/TauStar/index.html}} R
package available on CRAN, the Comprehensive R Archive Network
\citep{R, taustar-package}.

\subsection{Empirical convergence to the asymptotic
  distribution} \label{subsection: empirical convergence}

Let $t^*$ be computed from a sample of size $n$ drawn from the joint
distribution of a bivariate random vector $(X,Y)$ with $X\perp\!\!\!\perp Y$.
Since $t^*$ only depends on ranks, its distribution does not change
when applying monotonically increasing marginal transformations to $X$
and $Y$.  When $X$ and $Y$ both have continuous distributions, we may
thus transform their distributions to $N(0,1)$ without changing the
distribution of $t^*$.  When one or both of $X$ and $Y$ are discrete
however, the distribution of $t^*$ depends on how $X$ and $Y$
distribute their probability mass making it impossible to provide an
exhaustive empirical study of convergence properties.  Instead we will
consider selected examples.  Specifically, we consider the following cases:
\begin{enumerate}[(i)]
	\item The continuous case with $X,Y\sim N(0,1)$.
	\item A discrete case with $P(X=i) = 1/10$ for $1\leq i\leq
          10$, and $P(Y=i) \propto 2^{-i}$ for $1\leq i\leq 12$.
	\item A mixed case with $X\sim N(0,1)$ and $P(Y = i) = 1/5$
          for $1\leq i\leq 5$.
\end{enumerate}
In each setting we compute, for different sample sizes $n$, a kernel
density estimate for the distribution of $t^*$ and plot it alongside
the asymptotic density.  The resulting plots are shown in Figure
\ref{figure: density convergence}, which demonstrates that the
asymptotic and finite-sample distributions are in close agreement
already when $n=80$.  While we present only one example each for the
discrete and mixed cases we found similar results when simulating with
many other choices of distributions.

\begin{figure}
        \centering
         \subfloat[Continuous case\label{figure: continuous density}]{%
		\includegraphics[width=.33\textwidth]{cont_dens_conv}
	}
	\subfloat[Discrete case\label{figure: discrete density}]{%
		\includegraphics[width=.33\textwidth]{dis_dens_conv}
	}
	\subfloat[Mixed case\label{figure: mixed density}]{%
		\includegraphics[width=.33\textwidth]{mixed_dens_conv}
	}
        \caption{Kernel density estimates from 10,000 simulated values
          of $nt^*$ at sample sizes
          $n \in \{10, 15, 20, 25, 30, 40, 50, 60, 70, 80\}$; smaller
          values of $n$ are shown in lighter color.  The plots also
          show the density of the asymptotic distributions from
          Section \ref{section: asymptotics under the null} in dashed
          blue line.}
        \label{figure: density convergence}
\end{figure}

\begin{rem}
  Computing the asymptotic densities shown in Figure \ref{figure:
    density convergence} is non-trivial and requires the numerical
  inversion of the characteristic function for the asymptotic
  distributions.  To perform this numerical inversion we use the
  techniques described in Section 7 of \citet*{BlumEtAl61}; these
  computations are done automatically in the aforementioned TauStar
  package for R.
\end{rem}

\subsection{Power comparisons} \label{subsection: power comparisons}

We explore the power of an asymptotic test based on $t^*$ in six
cases:
\begin{enumerate}[(i)]
\item First, we take $(X,Y)$ as bivariate normal with correlation
  $\rho \in\{0, .1, .2, \dots, 1\}$; the distribution of $t^*$ then does
  not depend on the means and variances which may thus be set to zero
  and one, respectively.  We compare the test based on $t^*$ to the
  two-sided test based on the standard Pearson
  correlation $\hat{\rho}$.  We implement the latter test using the
  fact that $\hat{\rho}\sqrt{(n-2)/(1-\hat{\rho}^2)}$ has a
  $t$-distribution with $n-2$ degrees of freedom.

\item Next, we consider three discrete cases all of which have $(X,Y)$
  taking values in the grid $\{1,2,\dots,5\}^2$.  In each of these cases
  we compare our test to the chi-square test of independence.
  \begin{enumerate}[(a)]
  \item In the first discrete case, $(X,Y)$ follows a mixture between
    the uniform distribution on $\{(1,1),(2,2),\dots,(5,5)\}$ and the
    uniform distribution on $\{1,\dots,5\}^2$, illustrated in Figures
    \ref{figure: dg ordinal} and \ref{figure: dg uniform},
    respectively.  We let the mixture weight $p$ for the former
    component range through the set $\{0,.1,\dots,1\}$.
  \item The second discrete case is analogous but a mixture between
    the distributions from Figures \ref{figure: dg nonordinal} and
    \ref{figure: dg uniform}.
  \item The third discrete case is as the previous two but mixes the
    distributions from Figures \ref{figure: dg L shape} and
    \ref{figure: dg uniform}.
  \end{enumerate}

\begin{figure}
        \centering
         \subfloat[\label{figure: dg ordinal}]{%
		\begin{tikzpicture}
			\fill[black!20!white] (0,2) rectangle (.5,2.5);
			\fill[black!20!white] (.5,1.5) rectangle (1,2);
			\fill[black!20!white] (1,1) rectangle (1.5,1.5);
			\fill[black!20!white] (1.5,.5) rectangle (2,1);
			\fill[black!20!white] (2,0) rectangle (2.5,.5);
			\draw[step=0.5cm,black,very thin] (0,0) grid (2.5,2.5);
		\end{tikzpicture}
	}
	\hspace{10mm}
	\subfloat[\label{figure: dg nonordinal}]{%
		\begin{tikzpicture}
			\fill[black!20!white] (0,2) rectangle (.5,2.5);
			\fill[black!20!white] (2,1.5) rectangle (2.5,2);
			\fill[black!20!white] (1,1) rectangle (1.5,1.5);
			\fill[black!20!white] (1.5,.5) rectangle (2,1);
			\fill[black!20!white] (.5,0) rectangle (1,.5);
			\draw[step=0.5cm,black,very thin] (0,0) grid (2.5,2.5);
		\end{tikzpicture}
	}
	\hspace{10mm}
        	\subfloat[\label{figure: dg L shape}]{%
		\begin{tikzpicture}
			\fill[black!20!white] (0,0) rectangle (2.5,.5);
			\fill[black!20!white] (0,0) rectangle (.5,2.5);
			\draw[step=0.5cm,black,very thin] (0,0) grid (2.5,2.5);
		\end{tikzpicture}
	}
	\hspace{10mm}
	\subfloat[\label{figure: dg uniform}]{%
		\begin{tikzpicture}
			\fill[black!20!white] (0,0) rectangle (2.5,2.5);
			\draw[step=0.5cm,black,very thin] (0,0) grid (2.5,2.5);
		\end{tikzpicture}
	}
        \caption{Visualization of where probability mass is placed for
          different discrete distributions on $\{1,...,5\}^2$. In each
          case the distribution is uniform over the gray squares, and
          zero probability is assigned to the white squares.} \label{figure: discrete grids}
\end{figure}

\item Finally, we experiment with two mixed cases in which the
  distribution of $X$ is
  discrete and the conditional distribution of $Y$ given $X=x$ is
  a normal distribution $N(\mu_x,1)$.
  \begin{enumerate}[(a)]
  \item The first mixed case has $X\sim\text{Bernoulli}(.3)$,
    $\mu_x=0$ when $x=1$, and $\mu_x=\mu$ when $x=0$.  Here, we let
    the mean difference $\mu$ range through the set $\{0, 1/6,
    2/6,..., 9/6\}$, and each setting we compare against the
    two-sample $t$-test.
  \item In the second case $X\sim\text{Uniform}(\{1,...,6\})$, and $Y$
    has conditional mean $\mu_x$ is zero when $x$ is odd and equal to
    $\mu \in \{0, 1/6, 2/6,..., 9/6\}$ when $x$ is even.  Here, we
    compare against a bootstrapped permutation test using the distance
    covariance statistic of \cite{SzekelyEtAl07}, known to be
    consistent for independence, using the Energy R package
    \cite{RizzoSzekely14}.
  \end{enumerate}
\end{enumerate}

The simulation results are presented in Figure \ref{figure: power
  simulations}.  Surprisingly, the $t^*$ test has competitive power in
cases (i) and (iii)(a) where the alternative tests are known to be
most powerful given the distributional assumption of normality. For
the jointly discrete cases, we observe that the chi-square test of
independence has essentially equal power in case (ii)(a),
significantly higher power in case (ii)(b), and significantly lower
power in case (ii)(c). The lack of power in case (ii)(b) is not
surprising as the $t^*$ statistic is ordinal in nature and the
dependence in the distribution from case (ii)(b) was designed to be
non-ordinal.  The ordinal nature of $t^*$ also explains the
significant gains in case (ii)(c).  Hence, it would seem that the
$t^*$ test for jointly discrete data can offer substantial
improvements in power over the chi-square test if an ordinal
dependence relationship is suspected in the data. Finally, case
(iii)(b) suggests that there are cases in which $t^*$ may provide
higher power than the distance covariance.

\begin{figure}
        \centering
         \subfloat[Case (i)\label{figure: case (i)}]{%
         		\includegraphics[width=.35\textwidth]{cont_levels_plot}
	}
	\subfloat[Case (ii)(a)\label{figure: case (ii)(a)}]{%
         		\includegraphics[width=.35\textwidth]{dis_ord_levels_plot}
	}
	~ \\
        	\subfloat[Case (ii)(b)\label{figure: case (ii)(b)}]{%
         		\includegraphics[width=.35\textwidth]{dis_nonord_levels_plot}
	}
	\subfloat[Case (ii)(c)\label{figure: case (ii)(c)}]{%
         		\includegraphics[width=.35\textwidth]{dis_L_shape_levels_plot}
	}
	~ \\
	\subfloat[Case (iii)(a)\label{figure: case (iii)(a)}]{%
         		\includegraphics[width=.35\textwidth]{mixed_binary_levels_plot}
	}
	\subfloat[Case (iii)(b)\label{figure: case (iii)(b)}]{%
         		\includegraphics[width=.35\textwidth]{mixed_zig_levels_plot}
	}
        \caption{Simulated power of the $t^*$ asymptotic test (in
          solid line) and the power of the competing test (in dashed
          line). In each case we use a level of 0.05, which is
          displayed as a horizontal dashed black line.} \label{figure: power
          simulations}
\end{figure}

\subsection{Sample size calculations} \label{subsection: computing power}

Focusing on the continuous case, consider an asymptotic level $\alpha$
test of the null hypothesis of $\tau^*=0$ (i.e., independence) that
compares the statistic $t^*$ to a critical value $c_{\alpha}$ derived
from the asymptotic distribution from Theorem \ref{theorem: continuous
  asymptotic distribution}.  Suppose we would like to determine
the minimum sample size $n_\beta$ needed for a power of at least
$\beta$ under an alternative that has the two considered variables $X$
and $Y$ dependent, so that $\tau^*>0$.  If the sample is drawn from a
joint distribution for $(X,Y)$ that satisfies the conditions of
Corollary \ref{cor: asymptotics under dependence}, and if $\sigma_1^2$
is known to be no larger than the quantity $\bar\sigma_1^2$, then
Corollary \ref{cor: asymptotics under dependence} implies that for any
$x \leq \tau^*$,
\begin{align*}
  P(t^*\leq x) &= P\left(\sqrt{n}(t^*-\tau^*) \leq \sqrt{n}(x-\tau^*)\right) \\
               &\approx P\left(N(0,16\, \sigma_1^2) \leq
                 \sqrt{n}(x-\tau^*)\right) \\
               &\leq P\left(N\left(\tau^*, {16\, \bar\sigma_1^2
                 }/{n}\right) \leq x\right).
\end{align*}
This result can be used to find an asymptotically valid upper bound
$\bar n_\beta$ on $n_\beta$, by letting $\bar n_\beta$ be the smallest
positive integer such that
$c_{\alpha}/\bar n_\beta \leq \tau^*$ and
$P(N(\tau^*, {16\, \bar\sigma_1^2}/{\bar n_\beta}) \leq
c_\alpha/\bar n_\beta) \leq 1-\beta$. Finding this number
$\bar n_\beta$ can be accomplished in an iterative fashion.

The remaining difficulty in such an asymptotic sample size calculation
is finding a suitable upper bound $\bar\sigma_1^2$ for the unknown
variance $\sigma_1^2$. A crude but universally valid upper bound for
$\sigma_1^2$ can be obtained from Lemma \ref{lemma: kernel values},
which implies that $h_1(X,Y)$ takes values in the interval
$[-1/3,2/3]$ and thus $\sigma_1^2 \leq 1/4$. When $(X,Y)$ is bivariate normal,
an approximately valid upper bound of $\sigma_1^2$ is given by
$\sigma_1^2 \leq 0.14/16 = 0.00875$ (see Example \ref{example: variance
  of h1}). Figure \ref{figure: power upper bound} plots the
upper bound for the minimum sample size needed to achieve various powers when
bounding $\sigma_1^2$ by $1/4$ and $0.00875$ respectively and sampling
from a bivariate normal distribution with correlation 0.6. From the figure, we see that the $1/4$
bound leads to very conservative sample sizes while the $0.00875$ bound
results in values that much more closely adhere to the empirical
truth.  In general, overestimation of $\sigma_1^2$ is advisable as
small values of $\bar\sigma_1^2$ may lead to consideration of sample
sizes that are too small for asymptotic approximations to be reflective
of the actual finite-sample behavior of the test.

\begin{figure}
        \centering
        \includegraphics[width=.75\textwidth]{power_upper_bound.pdf}
        \caption{Minimum sample size ($n_{\beta}$) needed to a achieve a desired
          power $\beta$ at level $0.05$.  Simulations for bivariate normal
          data with correlation $0.6$ were used to compute an estimate
          of $n_{\beta}$ (blue line with dots).  These are
          compared to two asymptotic upper bounds for $n_{\beta}$
          using the bound $\sigma_1^2\leq 1/4$ (red line) and the
          bound $\sigma_1^2 \leq 0.00875$ (dashed orange line).  The
          sample size is presented with a log scaling.} \label{figure:
          power upper bound}
\end{figure}

\section{Discussion}\label{section: discussion}

The sign covariance $\tau^*$ of \cite{BergsmaDassios14} has the
intriguing property of being zero if and only if the considered pair
of random variables is independent, assuming that the random variables
follow a distribution that is continuous, discrete or a mixture of
such distributions.  Under these mild conditions, testing the
hypothesis that $\tau^*=0$ thus allows one to consistently assess
(in-)dependence.  With the aim of simplifying the implementation of
such independence tests, we have given a comprehensive study of the
asymptotic properties of $t^*$, the natural U-statistic for $\tau^*$.
The asymptotic distribution of $t^*$, especially as described in
Section \ref{subsection: continuos}, is seen to be connected in
interesting ways to the asymptotic distribution of Hoeffding's $D$,
and the Cram\'{e}r-von Mises statistic.

One limitation of our work is that we did not consider asymptotic
distributions under local alternatives to independence.  The reason is
that these would be distributions of weighted sums of non-central
chi-square random variables, which seem difficult to use in numerical
computations for assessment of power or sample size calculation.

While we have a complete understanding of the asymptotics of $t^*$
under fairly weak distributional assumptions---we covered continuous
and discrete cases, it remains to be seen if the large-sample
distribution of $t^*$ can be obtained without any such assumptions.
However, as noted above, it is also not yet known if the property that
$\tau^*=0$ only under independence holds for distributions that
are not continuous, discrete or a mixture of two such distributions.

redpajama_set_name: RedPajamaArXiv


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Can minors stay in a room without adults present?
Yes, minors may stay in a room without adults present. However, you will be required to submit a parent consent form. Please contact the hotel for instructions on how to do so.
We have no rules about advance reservations. However, we recommend making reservations in advance as we offer Advance Purchase Rate.
We do not offer waiting lists. Please contact us directly by phone or email to get the latest updates on room availability.
You will be charged 20% of your reservation price if you cancel the day before, 80% will be charged the day of your reservation, and 100% will be charged if you do not show up at the hotel.
Please click here for confirmation, change and cancellation of reservation details.
Children under the age of 6 who do not require an additional bed will not be charged. Children over the age of 6 will be charged as an adult.
Payment for your stay will be completed when you check in. If someone other than the guest who is checking in is responsible for the payment, they will need to complete the payment in full prior to the guest's check in. Please contact the hotel in advance if you have any further questions.
We accept VISA / JCB / MASTER / AMEX / UC / DC / NICOS / DINERS / GINREN. If you are unsure if you are able to use your card, please contact the hotel in advance for clarification.
We located 24-hour Money Exchange on the first floor in front of lounge.
Yes, the hotel is wheelchair friendly. We have a universal twin room which is completely wheelchair accessible.
Yes, we will keep your luggage at the front desk prior to your check in and after you check out. Please contact the front desk if you would like us to hold onto your luggage.
Can I send packages to myself at the hotel?
Yes, we can receive packages sent from within Japan and overseas at the hotel. However, we do not accept packages which require payment on delivery. Please make sure all packages you send to the hotel are pre-paid. On the shipping label, please write the reservation name of the guest at the hotel, reservation dates, reservation confirmation number and address the package to "reception".
Yes, we can help you send packages to other locations in Japan. Please contact the front desk for assistance.
Can I send packages overseas from the hotel?
Unfortunately we cannot help with international shipping. The nearest post office from our hotel is Kyoto Gion Post Office (https://goo.gl/maps/G15fqWprPqJ2), which is open during weekdays from 9am to 5pm.
Yes, we have breakfast from 7am to 11am (last order is at 10:30am) at the hotel. To confirm whether your reservation includes breakfast, please ask the front desk. You are able to add breakfast to any reservation for ¥1,800 per person.
Unfortunately, we have no parking. The nearest public parking from the hotel is about a 4 minute walk, located here: https://goo.gl/maps/ouXY7GorFi82.
Are English TV channels available to watch in the room?
Yes, we provide an apple TV in each room. ABC News is available through the Apple TV.
Is there a "Lost and Found" at KYOTO GRANBELL HOTEL?
Lost items may be discarded after then if there is no inquiries within 1 month.
Valuables such as cash ,valuable securities and precious metals are handled according to the law and regulations.
Please note that if food and beverage are left behind after check out, they may be disposed on the day.


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'Marvel Snap' Consider – The place Did That Convey You? Once more to Brode – TouchArcade
Viral Press November 1, 2022 9 min read
I do know, it's a little late. As many readers will know from my closing consider of a free-to-play sport, I'm a bit cautious of this stuff of late. My speedy response to Marvel Snap (Free) was terribly constructive, nonetheless I needed to current it further time to confirm I understood its true nature. I think about I've acquired the gist of it now, so it's time to guage it. Then we'll ship it via the Quantum Realm once more to sooner than most of the people learning this carried out it, and we'll be all set. I see this as an absolute win.
Marvel Snap is a truly good card sport. On the one hand, you'd say that isn't all that gorgeous. Ben Brode is no doubt one of many minds behind it, and in addition chances are you'll keep in mind him as a result of the lead designer of Hearthstone. 4 years up to now he and some of his colleagues dipped out from Blizzard and set off to do their very personal issue, and that turned out to be Marvel Snap. The pedigree is there. These individuals clearly know learn to make card sport, so in any case they went and made one different good card sport.
Nonetheless it's not that simple, is it? Definitely, typically having an unlimited success in your earlier is an albatross spherical your neck in your subsequent issue. Attempting to make one different sport within the similar fashion that doesn't merely repeat earlier beats and continues to be superior? Pretty robust. There are some similarities between Hearthstone and Marvel Snap, considerably in that they're every very easy to decide on up and play, nonetheless Snap is mainly doing its private issue and the reality that it actually works so successfully is a credit score rating to all the advance crew's efforts.
So proper right here's the deal. You and your opponent each have a deck of twelve taking part in playing cards. No further, no a lot much less. A sport often lasts six turns, nonetheless it would presumably closing considerably longer in some very specific circumstances. You start the game with three taking part in playing cards in your hand, and each flip you draw a card out of your deck. On the first flip, you would have one stage of energy to utilize. On the second, two components, and so forth. Each card costs a positive number of energy components to play, and has a positive number of vitality components. It'd even have some further skills. We'll come once more to that in a bit.
The play self-discipline consists of three areas. One may be revealed on the first flip, the second on the second flip, and the third on the third flip. Your purpose is to manage a minimum of two of these areas by the tip of the sixth flip by having in all probability probably the most full vitality targeted on them. These areas might also have their very personal properties that throw some wrenches into the works. Sometimes massive wrenches. You presumably can play taking part in playing cards anyplace correct from flip one, nonetheless the properties of the areas obtained't kick in until they're revealed.
Okay, once more to those skills. Some taking part in playing cards are simple and easily plunk down a sure amount of vitality in your energy. Sometimes that's what you want. A Hulk within the appropriate place can often tie points up with a gamma-powered ribbon. Nonetheless most of the taking part in playing cards have some further skills that kick in beneath quite a few conditions, with the commonest being Ongoing outcomes that keep on working whereas the cardboard is in play, and On Reveal outcomes that activate when the cardboard is flipped. There are some that kick in when the cardboard is destroyed or discarded. There are some that are reactive to what your opponents do. Pretty the vary, and the way in which wherein they work along with each other and the properties of the areas makes for a big number of prospects.
You earn new taking part in playing cards in only a few alternative routes. The primary means is by rising your Assortment Stage. Participating in matches gives you Boosters for enjoying playing cards in your assortment, and when combined with some credit score chances are you'll enhance the rarity of your taking part in playing cards. This doesn't change their gameplay properties, merely how they seem. Nonetheless upgrading them will improve your Assortment Stage, and every time you elevate a number of ranges you'll get some form of reward in return. Up until diploma 500 or so, you'll get a model new card every 4 to six ranges. It's additionally attainable to get variants of taking part in playing cards you already private by spending Gold inside the in-game retailer, and they're typically upgraded independently to further improve your Assortment Stage. Enjoying playing cards might also be earned as part of Season Passes, nonetheless it seems as for those who'll solely get them once you get a Premium Cross.
Sigh, it's just about time for the dirty focus on. Only a bit longer sooner than we go into that, buddies. There are moreover missions to clear, they often'll refresh with a number of new ones every couple of hours. Whereas events are occurring, there are moreover Season missions. Clearing customary missions rewards you with credit score and Season experience. Clearing Season missions often merely nets you Season experience. Which have will help you're employed your strategy via the Season Cross, incomes you credit score, Gold, and Boosters along with some Premium rewards for people who paid for them.
Lastly, there's the matter of Season Rank. You'll want this to be as extreme as doable by the tip of the Season, because the higher it goes the additional goodies you get. You improve this by worthwhile matches, and that's definitely the first incentive for worthwhile matches previous your particular person delight. You'll get boosters whether or not or not you win or lose. You'll should earn a set amount of things to increase your Rank diploma, and in addition you'll need to achieve a positive Rank diploma to earn specific rewards. What variety of components do you earn for a win? That's the place considered one of many further ingenious mechanics is out there in: the Snap.
It's humorous how a throwaway panel or two inside the Infinity Gauntlet story has grow to be such a cultural touchstone because of the giant success of Marvel's movement photos. I can focus on Thanos Snaps and common of us know what I'm talking about! Weird world. Snapping in Marvel Snap doesn't make half of all dwelling points disappear into mud, though. It doesn't even make Spider-Man inform Iron Man that he doesn't actually really feel so good. No, on this sport it's principally like elevating in poker. Snapping doubles the stakes each time it's carried out, putting further components on the street to be misplaced or obtained. Don't suppose chances are you'll win? You presumably can always retreat and solely lose the current amount inside the pot. It could double as soon as extra on the end so that you simply'll always lose a lot much less by working than standing with a positive loss.
That means that together with the variables of deck creation, randomized areas, and the numerous strategies taking part in playing cards work collectively, you even have the very human parts of bluffing and confidence in play. At lower ranges you probably obtained't care regarding the misplaced components, nonetheless as you climb the ranks the sheer rigidity post-snapping is unreal. It gives actual stakes with out actually costing you precise money, and that's a tricky issue to grab in a sport like this. With all the parts in play, closing minute turn-arounds can and do happen, so that you simply're under no circumstances totally safe no matter how far ahead you suppose you're. Snapping takes guts, and in addition you'll often pay in your bravado.
Okay, time for the dirty focus on. Nothing on this world is free, in any case. How does Marvel Snap monetize and the place does the pinch can be found in? Possibly unsurprisingly given this enchancment crew's earlier work, it's a very cozy sport to play for free of charge for a very very very long time. You obtained't be succesful to get every card and in addition you'll in the end run fast on credit score to enhance card rarities, nonetheless the trickle of newest taking part in playing cards and belongings is common adequate to keep up it pleasing. You'll in the end hit a level the place the model new taking part in playing cards stop coming, and points get a lot much less fulfilling then. It's very, very far into the game nonetheless I feel most common players will hit it inside a number of months.
Does that affect your viability in matches? To date, not going. You too can make superb decks with the taking part in playing cards you'll earn for free of charge, and I suppose there's nothing stopping the builders from together with a number of of the Premium taking part in playing cards to the free pool after some time has handed. Might new taking part in playing cards mess with this fragile steadiness of vitality? Fully. This consider is nonetheless a snapshot in time. Nonetheless for now, I consider a free participant can compete utterly successfully with these which are paying out.
And that brings us to how one pays out, do you must choose to. First, you can purchase Gold in quite a few portions. Gold couldn't seem immediately needed for those who start the game nonetheless the variant taking part in playing cards it would presumably buy helps you enhance your Assortment Stage and as you start tapping out the straightforward strategies to do that, it turns into necessary. There are strategies to get Gold for free of charge nonetheless they're undoubtedly sluggish. The alternative strategy chances are you'll spend money is through the Premium Season Cross. To date it appears to be like they'll worth about ten bucks and give you only a few distinctive taking part in playing cards, loads of boosters, and completely different extras. There's moreover a Welcome Pack that provides you some further belongings and a pleasing Captain America card for a few {dollars}.
I obtained't say I obtained't be enticed by the distinctive taking part in playing cards counting on what they're, nonetheless taking , sober take a look at Marvel Snap as a result of it stands, I consider the monetization isn't too harmful proper right here. It helps that it's a card sport and thus doesn't even have a story or ending to put off until you grind or pay. Nothing is simply too egregiously priced, and I consider chances are you'll nonetheless be aggressive even once you don't pay. As free-to-play stuff goes, chances are you'll't ask for moderately greater than that. Will I fall off the game as soon as I get uninterested within the plateau? I suppose we'll cross that Bifrost after we get to it.
Marvel Snap is an especially well-designed sport. It's utterly suited to cell as a consequence of how shortly the matches resolve, nonetheless the diploma of method is satisfyingly deep. The missions incentivize you to mix points up, and the common stream of newest taking part in playing cards ensures you'll have a great deal of selections to pick from for such endeavors. The monetization is relatively low-pressure and the plateau stage for free of charge players is manner adequate down the freeway that it's exhausting to call it unfair. I moreover love how successfully the taking part in playing cards symbolize the numerous heroes and villains in such simple phrases, with the areas doing the similar. The Snap mechanic is the wise cherry on prime. A inconceivable title, and one I can't seem to go away alone.
'Vampire Survivors' 1.0 Steam Deck Assessment – The Deck's Killer App – TouchArcade
Legacy of Thieves Assortment' Steam Deck Consider – TouchArcade

redpajama_set_name: RedPajamaCommonCrawl


Check out all of the names of the Menlo Atherton High School alumni that attended high school in Atherton that are listed above. Registering will allow you to join the directory.	Don't forget to upload your Menlo Atherton High School pictures and check on the latest info about any Menlo Atherton High School reunion.


The Static Toddler seat is suitable for toddlers aged from approx 12mths - 4yrs (18kg's).
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Twin Buggy is NOT included in this sale, please view our Pram section for all Adventure Buggy Co. prams available.


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SEPANG: Universiti Teknologi Mara sports lecturer Datin Paduka Sharifah Mazlina Syed Abdul Kadir has left for Santiago, Chile, in her bid to become the first Malaysian and first Asian woman to ski-sail from the South Pole to Hercules Inlet on the western edge of Antarctica.
The 39-year-old will make a courtesy call on Chilean President Ricardo Lagos on Saturday and then receive a time capsule from Prime Minister Datuk Seri Abdullah Ahmad Badawi, who is in Santiago for the Apec 2004 Summit.The two leaders will then flag her off on her expedition.
The capsule, which will contain a message from Abdullah, will be brought back to Malaysia in 2020.
Sharifah Mazlina at the KLIAS before leaving for Santiago.
Sharifah Mazlina, who will begin her 1,100km, 30-day solo expedition on Dec 4, said she was looking forward to the task.
"It is mid-summer and the sun will be out throughout the day in Antarctica," she added at the KL International Airport on Tuesday night.
Sharifah Mazlina said after two years of training she was confident of achieving the feat.
"I am mentally and physically prepared. I will be ski-sailing (skiing without poles and relying on a sail for propulsion) a minimum of 40km each day over a period of between seven and eight hours,'' she added.
Malaysians can send their well-wishes to her at her website www.sharifahantartic.com.


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Help Others Home
Rev. Robert Pardon, Director of MeadowHaven and Chair of ICSA's Religion Committee
At least 3% of church members report having been a victim of spiritual abuse.
In one study, 67% of subjects were Christians before their involvement in a cultic group or abusive church, but only 39% after the abusive experience.
Eighty respondents (42%) in this same study sought help from religious organizations.
But 40% of those who sought help found these services to be not at all helpful.
Approximately one percent of the population has had at least a transient involvement in a cultic group or abusive religious organization.
Therefore, each year tens of thousands of spiritual abuse victims who would seek help from religious organizations do NOT receive such help.
Sources: The Results of the International Cultic Studies Association's 2008 Questionnaire - International Cultic Studies Association (ICSA). Research Survey on Spiritual Abuse. http://www.icsahome.com/articles/prevalence
These research findings explain why ICSA strives to reach out to churches and other religious organizations. Though they may not realize it, they are vital gatekeepers. Former cult members and other spiritual abuse victims often turn to religious organizations and religious professionals for help.
ICSA's religion committee is headed by Rev. Robert Pardon. The committee's goals are:
Help those who wrestle with spiritual, religious, theological, or philosophical issues in their attempt to recover from an abusive experience.
Maintain, develop, and promote ICSA's subsidiary website, www.spiritualabuseresources.com
Maintain, develop, and promote ICSA's special program, the Spiritual Safe Haven Network, which helps religious organizations (e.g., churches) become healing refuges for spiritually abused individuals.
Organize events to educate religious professionals and help spiritual abuse victims.
Rev. Neil Damgaard is currently compiling an edited volume on spiritual abuse.
If you would like to learn more or become involved, please contact ICSA: mail@icsamail.com.


I didn't realise four years ago when I took on the National Novel Writing Challenge that the 50,000 words i was attempting to write would become the start of the next six books… I began to tell the story of Thomas Radwinter and his three brothers, Marcus, Paul and John, of the ups and the down of their lives and their families. As well as their own lives, the stories also include 'investigations' which Thomas undertakes, seemingly harmless research into genealogical mysteries which lead him into all sorts of difficulties.
I looked up at the vicar standing on the steps of the nave, and he smiled down at me. He's very handsome this vicar, and in fact he's my brother. All my brothers are handsome men, and Marcus, the Reverend Marcus Edward Radwinter, is no exception; he has a strong angular face, and today his wild eyebrows and bushy beard and flowing grey mane of hair have been neatly trimmed. The most arresting thing about him is his eyes, his piercing blue Radwinter eyes, and as I looked up at him, they crinkled in a smile.
He glanced down the aisle and I looked at my other brother John, standing beside me; he was beaming too, and well he might. He is engaged to the most beautiful woman, Justyna, and they were expecting their baby in three weeks' time. We'd had a few jokes about maybe it would come early and arrive today, and then there could be a christening as well as a wedding here in my brother's church.
"Ahem," Marcus nodded towards the church doors and I realised that my other nephew, Django, was playing the bridal march; he and his band had been vamping some incidental music while we waited.
"Can I look?" I whispered.
"Of course you can," Marcus replied, he looked so happy.
I turned and there, there was my beautiful bride.
Kylie was wearing a white and gold dress and she looked like an Aztec goddess… well, actually I don't know if the Aztecs had goddesses, but she looked exotic and wonderful and she shimmered as she walked towards me, her arm resting lightly on that of my brother Paul who was escorting her. He looked very handsome, I thought once that he looks like a silver lion; he has that strength, that confidence and pride in his family.
Kylie's hair was drawn back from her face and had been woven into plaits with golden ribbons and jewelled beads threaded through… She had feathers in her hair, gold feathers, and her veil came down just to her nose, and there were little sparkly stars on it. Behind her I could see her three bridesmaids, my nieces Paula and Sarah, and Kylie's sister Reneasha.
Kenneil had somehow escaped and was running down the aisle to his Mama. We'd gone to the suit shop to hire suits for John and Tom and he'd spotted a kilt and although he has no known connection to anything Scottish, he is in fact three-quarters Tobagoan, he insisted…and he is so adorable, I just gave way. I could hear the congregation all go 'aah' as he sped towards Kylie.
I had never felt so confident and sure of myself in all my life; this really was the start of a new and wonderful adventure with my family beside me!
Just mentioning to a friend..


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Probing exoplanet atmospheres could reveal telltale signatures of life
November 25, 2019 by Horizon Magazine
by Ethan Bilby
It may be that life is lurking out there on other planets. But stuck here on Earth, how can we ever know for sure? A good place to start is by looking for the compounds on other worlds that are known to be the key ingredients of life as we know it.
Detecting these so-called biosignatures, compounds that are known to be produced by living organisms, would be strong evidence that worlds may contain life. But picking up chemicals from such distant worlds, and choosing the right compounds to look for, is complicated.
Professor Ignas Snellen at Leiden University in the Netherlands has been refining techniques that combine data from the largest ground-based telescopes with high-contrast imaging that can reveal faint objects like planets. The telescopes use high-precision spectroscopy to examine the different wavelengths of the light they detect from space.
'You want to filter out the actual starlight as much as possible to make visible whatever information you can get from the exoplanet,' Prof. Snellen said.
By examining the starlight that filters through a planet's atmosphere and reaches us on Earth, it's possible to derive the types of gases that are present.
While telescopes are not yet large enough to examine the spectra for Earth-size planets, scientists are honing their methods on larger exoplanets, so-called hot Jupiters, which are far too hot to support life as we know it. These are gas giant exoplanets that orbit very closely to their parent star. So closely, in fact, that they are tidally locked, like our moon, with the exoplanet rotating only once with every orbit around its star.
With one side of such planets always in light and the other always in darkness, the light side gets so hot that the atmosphere can boil off, creating a wind of matter flowing off the planet, a bit like a comet's tail.
In the EXOPLANETBIO project, Prof. Snellen and his team used high-precision spectroscopy for the first time to confirm the amount of helium in a hot Jupiter's atmosphere using ground-based telescopes, which can reveal how far gone this process is.
'That was a breakthrough for these hot Jupiters,' he said. 'These kinds of exospheric tails were known, but they are very difficult to observe because the only way to see them was through detecting hydrogen, which can't be detected through Earth's atmosphere, using the Hubble space telescope.
'Now, with the stronger helium line we can do this very well from the ground with telescopes.'
Understanding if a hot Jupiter may bleed off its atmosphere, and how long it may take, can explain how the atmospheres of all exoplanets change over time.
'These kind of atmospheric escape processes are not very important now, but in the early solar system they were, because the sun was a lot more active,' Prof. Snellen said.
Exoplanet climate
Using these new techniques, his team has also been able to achieve another first, detecting the spin rate – how fast a planet rotates – and orbital velocity of exoplanets.
'The spin rates on hot Jupiters are generally quite low, as they are generally tidally locked,' he said.
That can reveal something about the climate and related weather on the exoplanet.
'When a planet rotates fast, it gets bands like Jupiter. The Earth rotates slower and has some bands, but it's still mostly dominated by low pressure systems. Now, if you have a hot Jupiter which is rotating even slower, you wouldn't get any banded structure. You get very different weather systems,' he said.
He has been able to observe winds high up in the atmosphere of such planets, as energy from the hotter, eternal-day side is rotated to the cooler night side.
Prof. Snellen is confident that an upgrade to the CRIRES (CRyogenic high-resolution InfraRed Echelle Spectrograph) instrument, set to go online next year on the European Southern Observatory's (ESO) Very Large Telescope, will let them find compounds such as methane on cooler planets. Methane can be a component in life if it is found in Earth-size planets.
'I see this as a kind of a playing ground. We are learning the methods now that we can someday apply to Earth-like planets. The (ESO's) Extremely Large Telescope should be ready in 2026, and with that we can start to probe Earth-like planets.'
Sign of life
Yet even if you have good samples from rocky, Earth-size planets, how do you know if a compound is truly a sign of life?
'Geology is very good at producing things that look like life, such as methane. It could come from cows or it could come from rocks,' said Professor Kevin Heng, a professor at the University of Bern in Switzerland.
'If you think about biosignatures, they have to satisfy various conditions. They have to not be mimicked by geology, they have to exist in the atmosphere for long periods of time, meaning that they are very stable or are replenished somehow, and they have to be detectable.'
As part of the EXOKLEIN project, Prof. Heng is working on determining if such compounds, like methyl chloride and ammonia, can last long enough in exoplanetary atmospheres to study, by modelling small planets around dwarf stars. It's a particular challenge for Earth-size planets, whose atmospheres can change over time.
'Geology is very good at producing things that look like life, such as methane. It could come from cows or it could come from rocks.'
Dr Kevin Heng, University of Bern, Switzerland
'If you look at a planet like Jupiter … they kind of look like the sun. They are made of hydrogen, they have trace elements of metals and so on. Based on differences between the planet and the star I can figure out how it formed. It would keep a fossil record of how it formed,' Prof. Heng said.
But for smaller planets, their atmospheres have changed significantly over time through processes like the carbon cycle.
'We spent the last eight to 10 years figuring out how to use climate models designed for Earth (on exoplanets), and how to tweak and modify them.'
Those models will be used to provide potential explanations for data collected when instruments are capable of surveying smaller planets for life, to understand if compounds are really biosignatures or can be explained away as geological.
'Extraordinary claims require extraordinary standards of proof, so if something is consistent with not requiring biology, I would say there's no biology,' said Prof. Heng.
He is also modelling planets that may have had more dramatic fates. For small planets around red stars to support life, they would need to have a very tight orbit, making them tidally locked like hot Jupiters.
'This means that the night side is really cold, and maybe cold enough that the gases in the atmosphere would condense into ice. So, you get a runaway condensation and you have no atmosphere – atmospheric collapse,' he said. Such collapse would leave planets cold and lifeless, like Mars.
While the work is just theoretical now, upcoming missions like the European Space Agency's CHEOPS satellite and NASA's James Webb Space Telescope should yield data that he can match up against his theories.
'When Webb launches (in 2021), there is going to be a quantum leap in the quality of data. It may so happen that atmospheric collapse is so prevalent that half the small planets around red stars don't have atmospheres.'
The research in this article was funded by the EU's European Research Council. If you liked this article, please consider sharing it on social media.
Categories Space Tags alien life, biosignatures, Exoplanets, space, space observation Post navigation
Levitation, touch and sound – how you'll be able to feel videogames in the future
Transparent wood: the building material of the future?


Overview… superhero. Not only can Hancock fly and possess the strength of a thousand men, but he is invincible and immortal. Unfortunately, he is also an alcoholic. Every time he dishes out a swift dose of vigilante justice, he invariably causes the city of Los Angeles millions of dollars of damage, leaving its citizens to foot the bill. When he saves the life of PR man Ray Embry, Ray offers to repay him by rebranding his sullied image as the most hated man in L.A.
Personality… reckless, irreverent, and sarcastic. Hancock cannot stand being disrespected, and will toss whoever disses him into the stratosphere – even if that person is a child. His inhibitions are lowered by liquor more often than not, and he'll often think out loud, narrating his actions as he fights crime and saves the innocent. His cocky persona is to some degree a projection that masks his insecurities. Hancock has no idea where he came from, and there's nobody who can relate to his inhuman condition. He didn't ask for superpowers, and he doesn't always handle the responsibility that comes with them very well.
John Hancock's popularity ranking on CharacTour is #2370 out of 5,000+ characters. See our top-ranked characters and read their profiles.


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Providing "Proven Strategies for Building Power"
Building on successful volunteer-based initiatives spanning the past three years, AFT Connecticut will this fall launch the next generation of its Member Organizer Institute (MOI) program. The effort, jointly funded by our national union, state federation and local affiliates, seeks to build greater participation by offering compensation to our most valuable resource — active members. The application period is now open, and leaders are seeking potential candidates for a paid cohort to begin training in late September.
Click here for a fact sheet on the MOI program.
Our state federation's first MOI initiative was three years ago organized to focus on our Administrative & Residual (A&R) Employees Union, the largest AFT Connecticut-affiliated local representing public employees. The goal was to identify new activists and leaders, engage and assess members' commitment ahead of an expected adverse decision in the Friedrichs v. California Teachers Association (CTA) lawsuit.
That initial effort brought union leaders and rank-and-file activists from across the state and country to Rocky Hill. There, participants received training in the fundamentals of organizing for power, including list development, worksite "mapping" and charting, conducting assessments, engaging in one-on-one conversations and overcoming objections.
The week-long effort culminated in a demonstration (pictured above) outside the state's labor department headquarters organized on behalf of A&R members facing layoffs after their contractual job protections had expired.
Click here for additional photos of the fall, 2015 cohort participants' training and mobilization activities.
A new committee of local activists, "Stand Together A&R" (STA&R), emerged from and has played a vital role in engaging local members over the past three years. Working in concert with the local's representative assembly, the body has identified and engaged new worksite leaders and helped drive their ongoing recommitment efforts.
The success of the A&R mobilization and subsequent STA&R Committee led to the spring, 2017 launch of a division-wide MOI. The focus was broadened to all our federation's public employee affiliates, and drew participants from several locals for the intensive, five-month organizing program.
"The training was invaluable and helped us determine members' mindset when it comes to their union," said Janet Andrews, who was part of last year's MOI contingent. "Those skills are even more relevant and important today. We need to ensure members understand why sticking together is critical to protecting what matters most," added Andrews, who serves as our A&R union's second vice president.
Click here for a photo of Andrews and fellow 2017 cohort participants in training.
Since graduating as "certified member organizers," activists like Andrews have supported a variety of efforts, from local collective bargaining campaigns to new worker organizing drives to public policy initiatives. They have also helped identify and recruit participants for two subsequent division-wide MOI cohorts this summer; one focused on paraprofessionals and school support personnel (PSRP) and another on healthcare affiliates.
"Among the most valuable skills I came away with was how to 'map' our organization's membership," said Daniel Flores, who earlier this summer participated in the PSRP-focused program. "It taught me where to look for our strongest supports and where to apply member activism. I came away with proven strategies for building power," added Flores, who serves as president of our New Britain Federation of Paraprofessionals.
"It was a challenging but very rewarding experience," said Robin Young, RN, of the healthcare union member engagement cohort. "The training we received was thorough — so much so that I was prepared for a lot more resistance from members. I was amazed at the open reception to our visits to their homes," added Young, who serves as secretary for our Backus Federation of Nurses.
Click here for photos of Young and fellow members earlier this month in training for the healthcare MOI.
What will distinguish the upcoming contingent is the addition of compensation for participants. Together, our national union, state federations and local affiliates seek to provide more members the financial ability to participate in a proven program for building a politician-proof labor movement.
Members interested in being part of our next MOI cohort should submit an application by September 14. The first scheduled day of training is scheduled for September 28 with a program commitment of 10 hours per week, or 40 hours per month.
Women, people of color, immigrants, LGBTQ people and low-wage workers are strongly encouraged to apply.
Click here to access the application form.
The upcoming MOI will provide members the opportunity to help build better worksite, affiliate and statewide structures and organize more working people into our movement. Combined, these efforts send a powerful message to the deep-pocketed special interests trying to divide and weaken us; "the 'U and I in Union' will keep us strong!"


Stainless steel front arch design for the modern look. Big enough to cook for family and friends, yet small enough to heat during your work week. Complete oven and platform package.
Stainless steel front arch design for the professional look. Big enough to cook for family and friends, yet small enough to heat during your work week. Complete oven and platform package ready to use. This oven is the same design as the standard brick 750 oven. The stainless front arch is an edition and has no effect to the performance, juts the modern look.
This is the most popular outdoor wood fired oven in New Zealand with over 1500 of this 750 model sold in NZ.
This oven is sold complete and weighs 700kg's. A fully insulated fire brick oven with 750 mm internal cooking surface and a stainless steel trolley and front arch. The trolley made from 1.5-2mm 304 grade stainless steel and has quality lockable caster wheels. The trolley is 1000 mm x 1000 mm and 1050 mm to cooking surface height. If you require a specific hight please specify.
The 750 SS Authentic Oven is also available with a standard brick front arch. Oven750StdRes The insulated hearth completes the perfect cooking surface for any meal. Simply the best cooking experience available.
Looks good in our modern setting. Heat retention is wonderful. We can put a loaf of bread in the following morning after cooking in the evening. Great when we have guests.


Research reports from Inderes, the leading Finnish analyst, offer high-quality analysis on over 100 companies listed on the Helsinki Stock Exchange, making it easy to keep on top of changes in the Helsinki stock market.
In the Infront Professional Terminal, the Inderes research reports are linked to the corresponding companies in your portfolios and views, so you can find the information you need, when you need it. You can create alerts in the Infront Professional Terminal which will notify you when new research is published relating to companies you are following.


Anderson Concrete is now hiring a Diesel Truck Mechanic.
As the Diesel Mechanic you will repair and maintain class A and B diesel trucks, autos, and construction machinery.
• Diagnose trouble, disassemble components, and examine parts for defects and excessive wear.
• Recondition and replace parts such as pistons, bearings, gears, valves, and bushings, using hand and power tools, and precision measuring instruments.
• Observe gauges, dials, and operation of machinery to ensure proper performance.
• Move heavy parts using hoists and dollies.
• Respond to equipment breakdowns and provide emergency road service.
• Operate Class A vehicles.
• Keep accurate and complete documentation of repairs and preventative maintenance.
• Weld and cut parts using arc-welding and flame cutting equipment and operate engine lathe.
• Perform routine maintenance and housekeeping tasks in and around the garage.
• Assist with ordering or inventory of parts.
• Adhere to company safety and employment policies.
• Ability to frequently use hand and power tools weighing up to 10 lbs. Sometimes operate tools in awkward positions and overhead.
• Ability to withstand exposure to wet, humid conditions and extreme heat or cold, as well as exposure to dust and machinery noise.
• Ability to operate hand controls and foot controls. Near and far visual acuity with ability to identify and distinguish colors.
• Ability to frequently stand and walk.
• Ability to occasionally climb, push, pull, bend, and stoop.
Anderson Concrete has been supporting Central Ohio's Concrete needs since 1921. As a leader in the concrete industry we value Safety, Integrity, Excellence, Gratitude and Community. These values contribute to the success of our business with our people, as well as our mission: to build lasting relationships with our customers by safely delivering quality products with superior service.
Contact us by phone, apply in person, upload or fax your resume to 614-443-4001.


text:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">

<title>org.drip.sample.treasuryfutures</title>
<meta name="date" content="2017-01-04">
<link rel="stylesheet" type="text/css" href="../../../../stylesheet.css" title="Style">

org.drip.sample.treasuryfutures

Classes

ContractDefinitions

ContractEligibilitySettlementDefinitions

ExchangeTradedOptionDefinitions

ExpiryDeliveryTradingDates

UST02Y

UST05Y

UST10Y

UST30Y

USTULTRA

redpajama_set_name: RedPajamaGithub


text:
Q: calculate weighted average over several columns with NA I have a data frame like this one:
ID  duration1   duration2   total_duration  quantity1   quantity2
 1     5            2             7             3         1
 2     NA           4             4             3         4
 3     5            NA            5             2         NA

I would like to do a weighted mean for each subject like this:
df$weighted_mean<-  ((df$duration1*df$quantity1) + (df$duration2*df$quantity2) ) / (df$total_duration)

But as I have NA, this command does not work and it is not very nice....
The result would be this:
ID  duration1   duration2   total_duration  quantity1   quantity2   weighted_mean
 1     5            2             7             3         1          2.43
 2     NA           4             4             3         4          4
 3     5            NA            5             2         NA         2

Thanks in advance for the help

A: You could exploit sum and prod's na.rm= arguments.
transform(df, z=sum(prod(duration1, quantity1, na.rm=T), na.rm=T) +
       sum(prod(duration2, quantity2, na.rm=T), na.rm=T)/
       na.omit(total_duration))
#   ID duration1 duration2 total_duration quantity1 quantity2        z
# 1  1         5         2              7         3         1 454.5714
# 2  2        NA         4              4         3         4 458.0000
# 3  3         5        NA              5         2        NA 456.4000

redpajama_set_name: RedPajamaStackExchange


TUROing While Awesome — DWA!
The 1980's had some really great cars and we celebrate them on a daily basis. I drive a 1985 Porsche 944, and the collective group of DWA! has 6 cars from the 80's. Every time I get out of my 944 I am rewarded with a solid (underrated) clunk when the door shuts. The interior of the 944 is a nice place to be and the design has aged well. Sure, the dash has cracks and the turn signal stalks are some of the flimsiest I've ever felt, but overall it's a clean, well designed and executed place to be. The same thing can be said for Warren's BMW E30, and Art's Mercedes 500SEC, but probably not Bryan's 1985 Corvette.
While the German and Japanese auto makers were trying to make the best cars they could, America was busy building rattle traps with fake wood grain, stitching and as much hard plastic as they could get their hands on.
Now that I've established my "hate" for 80's American cars, I can move on to my story. Art and I were going on a little trip to LA and needed a rental car. We didn't want a new Ford Focus, "or similar", so we jumped on TURO to find a car with a real sense of occasion...and boy did we find it!
Our choice was a 1984 Chrysler LeBaron Town and Country Convertible, complete with "wood" paneling, a turbo charged 2.2 liter 4 cyl engine and a smooth shifting 3spd automatic transmission. Making this Chrysler even more special, was a plethora of mods, including Dodge 600 front fenders, a Plymouth Caravelle front end, custom blue metallic paint and a continental kit. If anything, I could say that this car was rare, but I thought nah forget it, yo holms to Bel-Air.
On our way to Beverly Hills we stopped to grab some breakfast. While eating we watched at least 3 people take pictures of the Chrysler, and a nice homeless man told us about the K Car wagon that he used to have and asked if we kept ours in a hanger, to which we responded "of course!". Within an hour of driving, we had multiple people yell "nice car!", a police officer stop to tell us how nice and rare it was and we even parked in front of a yellow Rolls Royce on Rodeo Drive (stealing all the thunder from the Rolls). We felt like celebrities and knew that we had picked the right car.
Beyond the ridiculous exterior, the interior had even more easter eggs. There was the odd configuration of the window switches, "front wheel drive" being prominently displayed on the digital instrument cluster, the 'Chrysler Messenger Center' and the 50 shades of tan plastic and vinyl throughout the cabin.
Our two days with the Chrysler included a visit to The Porsche Experience Center, a guest appearance on 'The Smoking Tire Podcast' and attending the 'Race Service' launch party, but all of those pale in comparison to our experience in a 34 year old K Car.


text:
Q: get the invisible UITableViewCell I used the code below to get an invisible UITableViewCell
NSIndexPath* path = [NSIndexPath indexPathForRow:row inSection:0];
UITableViewCell *cell = [listTableview cellForRowAtIndexPath:path];

path returns something, but cell returns nil.
It looks like the code above only can get the visible UITableViewCell.
Your comment welcome

A: The docs clearly say it needs to be visible.

An object representing a cell of the table, or nil if the cell is not
  visible or indexPath is out of range.

Also, you are initializing the index just like how you would intialize any other variable, which has nothing to do with it having to be a valid indexPath.


text:
import angular from "angular";

class MainCtrl {
    constructor($scope) {
        this.$scope = $scope;
    }

    $onInit() {
        const fieldClassesBase = {
            label: "field__label",
            title: "field__title",
            input: "field__input"
        };

        this.classes = {
            text: Object.assign(fieldClassesBase, {
                label: fieldClassesBase.label += " field__label--text",
                title: fieldClassesBase.title += " field__title--text",
                input: fieldClassesBase.input += " field__input--text",
            }),
        };

        this.events = {
            click: function(b, $event) {
                console.log(b, $event);
            }
        };
    }
}

MainCtrl.$inject = ["$scope"];

export default MainCtrl;


text:
Nigerian Wedding
Tara Michelle's biography: father, age, family, net worth
Like in the US, there are a number of young people in other parts of the world making waves on social media.
One of them is Tara Michelle, who is a Canadian social media influencer, YouTuber and writer.
Tara Michelle has been sharing content online since 2013 and has been able to gather over 1 million subscribers on her YouTube channel, where she shares educative and interactive content.
This article is focused on what you should know about Tara Michelle; continue reading to see more.
Full name Tara Michelle Massicotte
Date of birth 5 January 1995
Place of birth Toronto, Ontario, Canada
Current residence Toronto, Ontario, Canada
Profession YouTube star, social media influencer, writer and blogger
Net Worth $500k
Tara Michelle's biography
Her full name is Tara Michelle Massicotte and she was born on the 5th of January 1995 in Toronto, Ontario, Canada.
There are no details on who Tara Michelle's parent are or what they do, but she has a younger brother who she features in many of her video.
Tara Michelle is a Canadian by birth and celebrated her 27th of January 2022.
Tara Michelle attended two high schools after which she gained admission to University of Toronto, she however didn't complete her programme in the school as she dropped out.
Tara Michelle is a YouTuber, writer and blogger. When she turned the age of 19 she relocated to Los Angeles to pursue a career in modelling.
However, Tara Michelle was disappointed as she received numerous rejections from LA modelling firms. This is because, she was being thought of being "too short to ever be a model".
This made her have a change of career as she began sharing content on Tumblr.
In March 2013, she launched her first YouTube account and began sharing educative videos.
Tara Michelle in May that same year launch her second YouTube account under the username Tara Michelle's OLD VLOGS, where she uploaded old video blogs, mostly her excursion video blogs, DIYs, and different hacks.
The Canadian social media star has amassed 1.28 million subscribers on YouTube, while her second channel has 515k subscribers.
There are no details of who Tara Michelle is in a relationship, and as such, she is assumed to be single.
Tara Michelle is reportedly worth $500k.
"You will hear my frog voice this year, you don't have a choice"-Whitemoney tell fans
'This is pure love' Reactions as Actress Annie Idibia celebrates her stepson in the most beautiful way
'WizKid should check on his friend' Fans react as Burna Boy goes very far inside the ocean hours after crying for emotional help
Man Utd icon, Patrice Evra claims there are at least two gay players at every club after some opened up to him
"You do anyhow, you see anyhow" – Actor, Jnr Pope reacts following video of him smashing man's car in street fight
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