text:
\section{Introduction}\label{sec:intro}

A uniform recursive tree (URT) on $n$ vertices is a rooted random recursive tree where each possible $(n-1)!$ distinct trees has the same probability of appearing. Another way of looking at uniform recursive trees is that one starts constructing the tree with only the root (node 1) and node 2 attached to the root. Afterwards, at each step $k = 3,\ldots,n$, node $k$ connects to one of the prior nodes $j$ with equal probability $1 / (k-1)$. A detailed survey on URTs can be found in \cite{Mahmoud}, and a book length treatment of random trees can be found in \cite{Drmota09}.

URTs, although pretty simple to construct, have been used in various applications.  These include but are not restricted to the spread of epidemics \cite{Feng05},  determining the genealogy of ancient and medieval texts \cite{NajockHeyde82}, analyzing pyramid schemes \cite{Gastwirth}, and the spread of a fire in a tree \cite{Fire}. Though these investigations use uniformity in their models, having different distributions would provide a lot more flexibility to the  researcher.  Using URTs implies that all nodes are identical in a certain sense,   for example, in the spread of epidemics, that every infected person is equally likely to infect the next one, or in the study of medieval texts, that every book is equally likely to be copied. This is obviously not the case in real world applications.

Parallel to the development of the theory of uniform recursive trees, various other recursive tree structures have already been studied. One of the most well known are binary recursive trees, which are described in \cite{Flajolet}. Other non-uniform recursive tree models include  plane-oriented recursive trees \cite{Szymanski87}, scaled attachment random recursive trees \cite{Devroye11} and biased recursive trees \cite{Altokislak}.

Our interest here is on another    natural generalization, Hoppe trees,  that was recently considered in \cite{Hoppe}. There, the root is assigned a weight $\theta$, all other nodes get weight 1. Node $i$ then attaches to the root with probability $\frac{\theta}{\theta+i-2}$ and to any other node with probability $\frac{1}{\theta+i-2}$. This model is associated to Hoppe's urn, which has an application in modelling the alleles of a gene with mutation rate $\theta>0$. Concerning many properties like the number of leaves, the height and the depth of node $n$, Hoppe trees behave similarly to uniform recursive trees in an asymptotic sense.

The  model  in this paper generalizes the idea of Hoppe trees: we  assign every node a weight $w_i$. Node $j$ then attaches to node $1 \leq i<j$ with probability $\frac{w_i}{w_1 + \cdots + w_{j-1}}$.  We call the resulting tree construction a weighted recursive tree (WRT).

Introducing weights is also interesting from the point of view of applications since it allows to introduce diversity among the nodes. In  other non-uniform models  discussed above, all nodes have the same behaviour, or in other words attract nodes according to the same rule. When a recursive process does not satisfy such conditions, weighted recursive trees can be used to model it more precisely.  Moreover the properties of weighted recursive trees and how much they differ from the uniform model can be interpreted as an indicator for the stability of a process. It is reasonable to assume that it is in general more probable for some nodes to get children as others. For example some persons might be more likely to infect others, some copies of ancient texts are more probable to have been copied again and some people might be more likely to recruit new people. Thus it is interesting to see how much fluctuation  in the attachment probabilities can be tolerated when using the uniform model.

Below, for  the  generalized model of this note we first give a coupling construction to construct a WRT  on $n$ nodes from a  Hoppe  tree. This allows us to understand statistics such as the height of the resulting random tree. We  then study the number of branches and the depth of node $n$ and give their expectation and variance, as well as some conditions under which asymptotic normality holds. We moreover derive explicit values for the expectation and the variance for certain examples of weight sequences.

The rest of the paper is organized  as follows. Next section introduces a coupling used to construct a  WRT from a Hoppe tree.  Section \ref{sec:couplingapplications} applies the coupling construction for an analysis of the number of leaves, the height and the size of the largest branch.  In Section \ref{sec:largestbranch}, we relate Hoppe trees to Hoppe permutations, and use the coupling construction of Section \ref{sec:coupling} to study the size of largest branch in WRTs.
   In Sections \ref{sec:depth} and \ref{sec:branches}, we study the depth of a WRT and  the number of branches in WRT.

\section{A useful coupling construction}\label{sec:coupling}

\subsection{Constructing WRTs from URTs}
\label{rem:URT2mkRT} We will first introduce a coupling allowing us to construct a special kind of WRT from a URT. We will not use this coupling in analysis of random tree statistics, because the second coupling construction, that is to introduced below,   can be applied to a more general class of WRTs. We still wanted to introduce this version because it is based on not Hoppe trees but URTs, a much better studied structure, as a starting point.

 We will call trees that have a weight sequence such that the first $k$ nodes have a constant weight equal to $\theta$ and all other nodes have weight 1, $\theta^k$-RTs. It is possible to construct a $\theta^k$-RT from a URT by a coupling construction when $\theta \in \mathbb{N}^+$. To emphasize this assumption we will use $m$ instead of $\theta$ in this part. In particular we can construct Hoppe trees for which the weight of the root is a natural number by this coupling.
To avoid confusion let us denote the nodes in the URT by $i$ and the nodes in the reconstructed tree by $i^*$.

The coupling construction in that case goes as follows: First construct a URT with $mk + n-k$ nodes. We write $\mathcal{T}_{mk+n-k}$ for this URT. Since we want the weight of the first $k$ nodes to be $m$ we then join several nodes into one in the following way:
\begin{itemize}
\item[\boldmath$\cdot$] Nodes $1, \dots, m$ become node $1^*$,
\item[\boldmath$\cdot$] $m+1, \dots, 2m$ become node $2^*$ \dots
\item[\boldmath$\cdot$] $(k-1)m +1, \dots, km$ become node $k^*$.
\end{itemize}

The new node $i^*$ gets all the children of $(i-1)m+1, \dots, im$. But since we joined several nodes into one and the nodes $(j-1)m +1, \dots, jm$ might have different parents, for $1<j\leq k$, we set the parent of $j^*$ as the parent of $(j-1)m+1$, i.e. of the node with the smallest label among those that become $j^*$. Thus, if in the URT the parent of $(j-1)m+1$ is any of the nodes $(i-1)m+1, \dots, im$, the parent of $j^*$ is $i^*$.

For $j >k$, we set $j^* = j+k(m-1)$, so all nodes after $k$ only correspond to a single node, we just need to "translate" the names of the nodes to take into account that we used $mk$ nodes instead of $k$ for the first $k$ nodes in the reconstructed tree. If for $j>k$ the parent of node $j+k(m-1)$  is among the first $km$ nodes of the URT, we check into which range this node falls and the parent of $j^*$ is chosen as above. In other words if, for $1 \leq i \leq k$, the parent of node $j+k(m-1)$ is one of $(i-1)m+1, \dots, im$, the parent of node $j^*$ is node $i^*$. If the parent of node $j+k(m-1)$ is equal to $h+k(m-1)$ with $h>k$, the parent of node $j^*$ is node $h^*$.
We call the tree we thus obtain $\mathcal{T}_n^{m^k}$.
It can be easily verified that the obtained attachment probabilities correspond to the ones of an $m^k$-tree, for details see \cite{ellatez}.

We now show, as an example, how the coupling can be used to study the number of leaves, i.e. the number of nodes without children, of a $\theta^k$-RT. Let $\mathcal{L}_{mk+n-k}$ denote the number of leaves of $\mathcal{T}_{mk+n-k}$ and $\mathcal{L}_n^{m^k}$ denote the number of leaves of $\mathcal{T}_{n}^{m^k}.$
Then $\mathcal{L}_{mk+n-k}$ can be used to bound $\mathcal{L}_n^{m^k}$. First of all if a node $i > km $ is a leaf in $\mathcal{T}_{mk+n-k}$, the corresponding node in $\mathcal{T}_n^{m^k}$, which is ${i-k(m-1)}^*$, is also a leaf.  The reconstruction process thus only affects the children of the nodes $i^*$ with $1 \leq i \leq k$ and the root cannot be a leaf, so we can have at most $k-1$ additional leaves. Moreover for each $2 \leq i \leq k$ we can at most "loose" $m-1$ leaves since if all $(i-1)m+1, \dots, im$ are leaves in $\mathcal{T}_{mk+n-k}$, $i^*$ will be a leaf in $\mathcal{T}_n^{m^k}$ too.  Hence we can conclude that
\begin{equation*}
\mathcal{L}_{mk+n-k}-k(m-1)< \mathcal{L}_n^{m^k} <\mathcal{L}_{mk+n-k}+ k-1  .
\end{equation*}
Together with results about the expected number of leaves of URTs this implies after some simple manipulations that
\begin{equation*}
\begin{split}
 \left |\mathbb{E}[\mathcal{L}_n^{m^k}]-\mathbb{E}[\mathcal{L}_n] \right | \leq \frac{k(m+1)}{2}, \\
\end{split}
\end{equation*}
where $\mathcal{L}_n$ denotes the number of leaves in a URT on $n$ nodes. It is possible to derive other results from this coupling, but since the second coupling we now present is more general, we will not go further into it here.

\subsection{Constructing WRTs from Hoppe trees}

We will now introduce a coupling construction for WRTs whose nodes have constant weight after a certain index, a class similar to, but more general, than $\theta^k$-RTs. Let $\mathcal{T}_n^w$ be a WRT and $(w_i)_{i \in \mathbb{N}}$, the weight sequence of $\mathcal{T}_n^w$, be such that there is a $k \in \mathbb{N}$ such that $w_i=1$ for all $i>k$. Then we can construct $\mathcal{T}_n^w$ from a Hoppe tree with root weight $\theta=\sum_{i=1}^{k} w_i$ by a  coupling construction, more precisely by splitting the root into $k$ nodes. To avoid confusion we will write $i$ for node $i$ in the Hoppe tree and $i^*$ for node $i$ in the reconstructed tree.

We now describe the coupling construction: First construct a Hoppe tree on $n-k+1$ nodes and with $\theta$, the weight of the root, equal to $\sum_{i=1}^{k} w_i$. Then construct a WRT of size $k$ corresponding to $(w_i)_{i \in \mathbb{N}}$. Now we replace the root of the Hoppe tree by this weighted recursive tree of size $k$ in the following way: Node $1^*$, $2^*$, \dots, $k^*$ are the nodes of the WRT of size $k$ we just constructed.
For $i \geq 2$, node $i$ in the Hoppe tree becomes node $i+k-1^*$ in the reconstructed tree, so we shift the names of the rest of the nodes by $k-1$. Then for all $i \geq 2$, if $i$ is a child of 1 in the Hoppe tree, $i+k-1^*$ becomes a child of one of the nodes $1^*, \dots, k^*$ in the reconstructed tree, proportionally to their weights. This means that if $i$ is a child of 1 in the Hoppe tree, for $1\leq j \leq k$, node $i+k-1^*$ will become a child of a node $j^*$  in the reconstructed tree with probability $\frac{w_{j}}{\sum_{\ell=1}^{k} w_{\ell}}$.

Let us check that this gives the attachment probabilities corresponding to the WRT we aim to construct.
\begin{itemize}
\item[\boldmath{$\cdot$}] For $1\leq i \leq k<j$,
\begin{equation*}
\begin{split}
\mathbb{P} & (j^* \text{ is child of } i^*)  = \mathbb{P}(j-k+1 \text{ is child of } 1, i^*\text{  is chosen as the parent of } j^*) \\
&= \frac{\sum_{\ell=1}^{k} w_{\ell}}{j-k+1 -2 +\sum_{\ell=1}^{k} w_{\ell}} \frac{w_i}{\sum_{\ell=1}^{k} w_{\ell}} = \frac{w_i}{j-1 -k+\sum_{\ell=1}^{k} w_{\ell} }. \\
\end{split}
\end{equation*}
\item[\boldmath{$\cdot$}] For $k<i<j$,
\begin{equation*}
\begin{split}
\mathbb{P} & (j^* \text{ is child of } i^*)
 = \mathbb{P}(j-k+1 \text{ is child of } i-k+1)\\
&= \frac{1}{j-k+1 -2 +\sum_{\ell=1}^{k} w_{\ell}} = \frac{1}{j-1-k+\sum_{\ell=1}^{k} w_{\ell}}. \\
\end{split}
\end{equation*}
\end{itemize}

The following two sections will be using the coupling construction just described.

\section{Use of the coupling in WRT statistics}\label{sec:couplingapplications}

In this section we apply the coupling construction of the previous section to study the number of leaves in a WRT and the height of a Hoppe tree. Later, in Section \ref{sec:largestbranch}, the coupling construction will also be used in order to understand the size of the largest branch in a WRT.

\subsection{Number of leaves}

A node in a tree with degree one is said to be a leaf.
Focusing on the number of leaves, the reconstruction process in our coupling construction implies that all the leaves of the Hoppe tree are still leaves in the reconstructed tree, since we do not change any relation among the nodes $2, \dots, n-k+1$ of the Hoppe tree or respectively $k+1^*, \dots n^*$ of the reconstructed tree. There can be at most $k-1$ additional leaves among the first $k$ nodes. Thus we can bound the number of leaves $ \mathcal{L}_n^{w} $ of $\mathcal{T}^w$ by the number of leaves  $\mathcal{L}_{n}^{\theta}$ of $\mathcal{T}^\theta$:
\begin{equation}  \label{eq:CouplingLeaves}
\mathcal{L}_{n-k+1}^{\theta} \leq \mathcal{L}_n^{w} \leq  \mathcal{L}_{n-k+1}^{\theta} +k-1. \\
\end{equation}.

In \cite{Hoppe} the following results about the leaves of Hoppe trees are given.

\begin{theorem}[\cite{Hoppe}] \label{thm:HoppeLeaves}
Let $\mathcal{L}_n^{\theta}$ denote the number of leaves of a Hoppe tree with $n \geq 2 $ nodes. Then
\begin{equation}
\begin{split}
&\mathbb{E} [\mathcal{L}_n^{\theta}] = \frac{n}{2} + \frac{\theta -1}{2} + \mathcal{O}\left (\frac{1}{n} \right), \\
&\mathrm{Var} (\mathcal{L}_n^{\theta}) = \frac{n}{12} + \frac{\theta -1}{12} + \mathcal{O}\left (\frac{1}{n} \right ), \\
& \mathbb{P}(|\mathcal{L}_n^{\theta} - \mathbb{E}[\mathcal{L}_n^{\theta}]| \geq t) \leq 2 e^{-\frac{6t^2}{n+\theta+1}} \text{ for all } t>0 \text{ and } \\
&\frac{\mathcal{L}_n^{\theta} - \mathbb{E}[\mathcal{L}_n^{\theta}]}{\sqrt{\mathrm{Var}\left (\mathcal{L}_n^{\theta}\right)}} \xrightarrow[d]{n \to \infty} \mathcal{G}.
\end{split}
\end{equation}
\end{theorem}

Using the above theorem we can thus derive results on the number of leaves of WRTs whose nodes having constant weight after a certain index.

\begin{theorem} \label{thm:WRTLeaves}
Let $\mathcal{L}^{w}_{n}$ denote the number of leaves of a WRT of size $n$ with weight sequence $(w_i)_{i \in \mathbb{R}}$ such that there is a $k \in \mathbb{N}$ such that for all $i>k$ we have $w_i = 1$. Then
\begin{equation}
\begin{split}
&\mathbb{E} [\mathcal{L}_n^{w}] = \frac{n}{2} + C + \mathcal{O} \left ( \frac{1}{n} \right ) \text{ with } |C| \leq \frac{\sum_{i=1}^{k} w_i +k}{2},\\
&\mathrm{Var}(\mathcal{L}_n^{w})) = \frac{n}{12} + \mathcal{O} (\sqrt{n}), \\
& \mathbb{P} (|\mathcal{L}_n^{w} - \mathbb{E}[\mathcal{L}_n^{w}]|\geq t) \leq 2 e^{- \frac{6(t-2k+2)^2}{n-k+2+ \sum_{i=1}^k w_i}} \text{ for all } t>0 \text{ and } \\
& \frac{\mathcal{L}^{w}_{n} - \mathbb{E}[\mathcal{L}^{w}_{n}]}{\mathrm{Var}(\mathcal{L}^{w}_{n})} \longrightarrow_d \mathcal{G} \text{ as } n \to \infty.
\end{split}
\end{equation}
\end{theorem}

\begin{proof}
First of all for the expected value we get from Theorem \ref{thm:HoppeLeaves} and \eqref{eq:CouplingLeaves}.
\begin{equation*}
\begin{split}
&\mathcal{L}_{n-k+1}^{\theta} \leq \mathcal{L}_n^{w} \leq  \mathcal{L}_{n-k+1}^{\theta} +k-1 \\
& \Rightarrow \frac{n-k+1}{2} + \frac{\sum_{i=1}^{k} w_i-1}{2} + \mathcal{O}\left ( \frac{1}{n} \right ) \leq \mathbb{E}\left [\mathcal{L}_n^{w}\right ] \\
& \hspace{7em} \leq \frac{n-k+1}{2} + \frac{\sum_{i=1}^{k} w_i-1}{2} + k -1 + \mathcal{O}\left ( \frac{1}{n} \right ) \\
& \Rightarrow \frac{n}{2} + \frac{\sum_{i=1}^{k} w_i-k}{2} + \mathcal{O}\left ( \frac{1}{n} \right ) \leq \mathbb{E} \left [\mathcal{L}_n^{w}\right ] \leq \frac{n}{2} + \frac{\sum_{i=1}^{k} w_i+k-2}{2} + \mathcal{O}\left ( \frac{1}{n} \right ) \\
& \Rightarrow \mathbb{E}\left [\mathcal{L}_n^{w}\right ] = \frac{n}{2} + C + \mathcal{O}\left ( \frac{1}{n} \right ),
\end{split}
\end{equation*}
where $C$ depends on $k$ and $w_1, \dots, w_k$, and we have $|C| \leq \frac{\sum_{i=1}^{k} w_i +k}{2}$.

Equation \eqref{eq:CouplingLeaves} also directly gives a concentration inequality. We have that

\begin{equation*}
\begin{split}
\mathbb{P} & (|\mathcal{L}_n^{w} - \mathbb{E}[\mathcal{L}_n^{w}]|\geq t) \\
& \leq \mathbb{P}( |\mathcal{L}_n^{w} - \mathcal{L}_{n-k+1}^{\theta} | +  |\mathcal{L}_{n-k+1}^{\theta} - \mathbb{E}  [\mathcal{L}_{n-k+1}^{\theta}  ] | +  | \mathbb{E} [\mathcal{L}_{n-k+1}^{\theta} ] - \mathbb{E}  [\mathcal{L}_n^{w}  ]  |\geq t)  \\
& \leq \mathbb{P}(|\mathcal{L}_{n-k+1}^{\theta} - \mathbb{E}  [\mathcal{L}_{n-k+1}^{\theta} ]  | \geq t-2k+2)  \\
& \leq 2 e^{- \frac{6(t-2k+2)^2}{n-k+2+ \sum_{i=1}^k w_i}}.
\end{split}
\end{equation*}

The coupling also gives us an approximation for the variance. Let $Y$ denote the number of additional leaves among the first $k$ nodes in the reconstructed tree. Then $\mathrm{Var}(Y) = \mathcal{O}(k)$ since $Y \leq k$. Since
$\mathrm{Var}(\mathcal{L}_n^{w}) = \mathrm{Var}(\mathcal{L}_{n-k+1}^{\theta} +Y ) = \mathrm{Var}(\mathcal{L}_{n-k+1}^{\theta}) + \mathrm{Var}(Y) + \mathrm{Cov}(\mathcal{L}_{n-k+1}^{\theta} ,Y)$
we get by the Cauchy-Schwarz inequality
\begin{equation*}
\mathrm{Var}(\mathcal{L}_n^{w})) = \frac{n}{12} + \mathcal{O} (\sqrt{n}).
\end{equation*}

In a similar way, one can make conclusions about the  asymptotic distribution. For this we will need Slutsky's theorem:
Let $X_n$ and $Y_n$ be sequences of random variables such that $X_n \to_d X$ and $Y_n \to_d c$ for $c \in \mathbb{R}$. Then
$$ \lim_{n\to \infty} X_n + Y_n  =_d  X +c.$$

In order to derive a central limit theorem for $\mathcal{L}^{w}_{n}$, we write
\begin{equation*}\frac{\mathcal{L}^{w}_{n}- \mathbb{E}[\mathcal{L}_{n}^w]}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}}
= \frac{\mathcal{L}^{w}_{n}- \mathcal{L}_{n-k+1}^\theta}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}}
+   \frac{\mathcal{L}^{\theta}_{n-k+1}- \mathbb{E}[\mathcal{L}_{n-k+1}^\theta]}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}}
+ \frac{\mathbb{E}[\mathcal{L}^{\theta}_{n-k+1}]- \mathbb{E}[\mathcal{L}_{n}^w]}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}}.
\end{equation*}
Now  by Theorem \ref{thm:HoppeLeaves} and our previous result on $\mathrm{Var}(\mathcal{L}_n^w)$ we have
\begin{equation*}
\frac{\mathcal{L}^{\theta}_{n-k+1}- \mathbb{E}[\mathcal{L}_{n-k+1}^\theta]}{\sqrt{\mathrm{Var}(\mathcal{L}_{n-k+1}^\theta)}}\frac{\mathrm{Var}(\mathcal{L}_{n-k+1}^\theta)}{\mathrm{Var}(\mathcal{L}_{n}^w)} \xrightarrow[d] {n \to \infty} \mathcal{G},\end{equation*}
and since $\mathrm{Var}(\mathcal{L}_{n}^w) \longrightarrow \infty$ as $n \to \infty$ we have by \eqref{eq:CouplingLeaves} that
\begin{equation*}\left | \frac{\mathcal{L}^{w}_{n}- \mathcal{L}_{n-k+1}^\theta}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}} + \frac{\mathbb{E}[\mathcal{L}^{\theta}_{n-k+1}]- \mathbb{E}[\mathcal{L}_{n}^w]}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}} \right | \leq \frac{2k}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}} \xrightarrow{a.s.} 0.
\end{equation*}
Now we can apply Slutsky's theorem and conclude that
\begin{equation*} \frac{\mathcal{L}^{w}_{n}- \mathbb{E}[\mathcal{L}_{n}^w]}{\sqrt{\mathrm{Var}(\mathcal{L}_{n}^w)}} \xrightarrow[d] {n\to \infty} \mathcal{G}.
\end{equation*}
\hfill $\square$
\end{proof}

\begin{remark} \label{rem:leavesProb} It might be possible to get results on the number of leaves of a general WRT $\mathcal{T}_n^w$ by writing $\mathcal{L}_{n}^{w}$ as the sum of $\mathbf{1}(\ell_i^{w})$ where $\ell_i^{w}$ denotes the event that $i$ is a leaf in $\mathcal{T}_n^w$. It follows from the construction principle that
\begin{equation*}
\mathbb{P} \left (\ell_i^{w}\right ) = \prod_{j=i+1}^{n} \left ( 1- \frac{w_i}{w_1 + \dots + w_{j-1}} \right ).
\end{equation*}
After some manipulation this expression becomes
\begin{equation*}
\mathbb{P} \left (l_i^{w}\right )  = \frac{ w_1 + \dots + w_{i-1}}{w_1 + \dots + w_{n-1}} \prod_{j=i+1}^{n-1} \left ( 1 + \frac{w_{j} - w_{i}}{w_1 + \dots + w_{j-1}} \right ).
\end{equation*}
An exact expression for the expectation of the number of leaves of a $\theta^k$-RT can be obtained by writing $\mathbb{E}\left [\mathcal{L}_n^{\theta^k}\right ]  = \sum_{i=2}^{n} \mathbb{E}\left [\mathbf{1}\left (\ell_i^{\theta^k}\right )\right ]$ and using the  expression above. After some computations we get by this method
\begin{equation*}\mathbb{E}\left [\mathcal{L}_n^{\theta^k} \right ] = \frac{n}{2} + \frac{k(\theta-1)}{2} + \frac{k \theta(1-k\theta)}{2(k (\theta-1) + n-1)}+ \frac{k-1}{2} \prod_{i=1}^{n-1-k} \frac{\theta (k-1)+i}{\theta k +i}.
\end{equation*}
\end{remark}

\begin{remark}
For $\theta^k$-RTs it is also possible to obtain results about the expectation, variance and concentration rate of the number of leaves by using a martingale argument similar to the one in \cite{Hoppe}, for details see \cite {ellatez}.
\end{remark}

\subsection{Height}

As a second example, we discuss the height of a WRT,  which   is defined as  the length of the longest path from the root to a leaf.  Let $\mathcal{H}_n^w$ denote the height of a WRT with weight sequence $(w_i)_{i \in \mathbb{N}}$ such that $w_i =1$ for $i>k$.  Let moreover  $\mathcal{D}_{1,i}^w$ and $\mathcal{D}_{1,i}^\theta$ denote the distance between the root and node $i$ in the reconstructed weighted tree and the original Hoppe tree respectively. For $i \leq k$, $\mathcal{D}_{1,i}^w$ is at most $k-1$. Also for any $i>k$, the path from the root to $i^*$ corresponds to the path from the root to the corresponding node in the original Hoppe tree, i.e. the distance from the root to $i-k+1$, except that we might have an additional path among the first $k$ nodes instead of the first edge. Thus $\mathcal{D}^w_{1,i}$ is at least as big as $\mathcal{D}_{i-k+1}^\theta$.

Also $\mathcal{D}^w_{1,i}$ is at most  $k-1$ bigger than the distance between the root and the corresponding node in the original tree: Let $j-k+1$ be the first node on the path from 1 to $i-k+1$ in the original tree. Then in the reconstructed tree $j^*$ will be attached to some $h^*$, where $1\leq h \leq k$. Thus when we consider the path consisting of the path from the root to $h^*$ in the reconstructed tree, the edge from $h^*$ to $j^*$ and the path from $j-k+1$ to $i-k+1$ in the Hoppe tree, we get a path from $1^*$ to $i^*$ in the reconstructed tree. Thus  for all $k+1 \leq i \leq n$, there is some $h \leq k$ such that,
\begin{equation*}
{D}^w_{1,i} = \mathcal{D}_{1,h}^w + 1 + \mathcal{D}_{j-k+1,i-k+1}^\theta = \mathcal{D}_{1,h}^w + \mathcal{D}_{1,i-k+1}^\theta.
\end{equation*}
Also $\mathcal{D}_{1,h}^w \leq k-1$, so we have
\begin{equation*}
\mathcal{D}_{1,i-k+1}^\theta \leq {D}^w_{1,i} \leq \mathcal{D}_{1,i-k+1}^\theta +k-1,
\end{equation*}
which implies that
\begin{equation*}
\begin{split}
\max_{i =1, \dots, n-k+1} \{\mathcal{D}_{1,i}^\theta\}  \leq \max_{i =1, \dots, n} \{\mathcal{D}_{1,i}^w\} \leq \max_{i =1, \dots, n-k+1} \{\mathcal{D}_{1,i}^\theta\} + k-1.
\end{split}
\end{equation*}
Thus from the definition of the height as $\mathcal{H}_n = \max_{i =1, \dots, n} \{\mathcal{D}_{1,i}\},$ we can derive that
\begin{equation} \label{eq:height}
\begin{split}
\mathcal{H}_{n-k+1}^\theta \leq \mathcal{H}_n^{w} \leq \mathcal{H}_{n-k+1}^\theta + k-1.
\end{split}
\end{equation}

We have the following result about the height of Hoppe trees.
\begin{theorem}[\cite{Hoppe}] \label{thm:HoppeHeight}
Let $\mathcal{H}_n^{\theta}$ denote the height of a Hoppe tree with $n$ nodes. Then
\begin{equation*}
\begin{split}
&\mathbb{E} [\mathcal{H}_n^{\theta}] = e\ln(n) - \frac{3}{2}\ln \ln n + \mathcal{O}(1) \text{ and} \\
&\mathrm{Var}(\mathcal{H}_n^{\theta}) = \mathcal{O}(1).
\end{split}
\end{equation*}
\end{theorem}

Together with the coupling this allows us to derive the following theorem.

\begin{theorem}
Let $\mathcal{H}^{w}_{n}$ denote the height of a WRT of size $n$ with weight sequence $(w_i)_{i \in \mathbb{R}}$ such that there is a $k \in \mathbb{N}$ such that for all $i>k$ we have $w_i = 1$. Then
\begin{equation*}
\begin{split}
&\mathbb{E}\left [\mathcal{H}_n^{w} \right ] = e \ln(n) - \frac{3}{2} \ln \ln (n) + \mathcal{O}(1) \text{ and }\\
&\mathrm{Var}(\mathcal{H}_n^{w}) = \mathcal{O}(1).
\end{split}
\end{equation*}
\end{theorem}
\begin{proof}
According to Theorem \ref{thm:HoppeHeight} we have
$\mathbb{E}[\mathcal{H}_{n-k+1}^{\theta}] = e\ln(n-k+1) - \frac{3}{2}\ln \ln (n-k+1) + \mathcal{O}(1).$
Since,
\begin{equation*}
\begin{split}
& e\ln(n-k+1) - \frac{3}{2}\ln \ln (n-k+1) \\
& = e\left (  \ln(n) + \ln\left (1-\frac{k-1}{n} \right) \right ) \\
& \hspace{3em} - \frac{3}{2} \left ( \ln \ln(n) + \ln \left ( 1+ \frac{\ln\left (1-\frac{k-1}{n} \right )}{\ln(n)} \right ) \right ) \\
&= e \ln(n) - \frac{3}{2} \ln \ln (n) + o(1).
\end{split}
\end{equation*}
Thus we get by \eqref{eq:height} for $k>n$,
\begin{equation*}
e \ln(n) - \frac{3}{2} \ln \ln (n) + \mathcal{O}(1) \leq \mathbb{E}\left [\mathcal{H}_n^{w} \right ] \leq e \ln(n) - \frac{3}{2} \ln \ln (n) + \mathcal{O}(1)+ k-1
\end{equation*}
which  implies
\begin{equation*}
\mathbb{E}\left [\mathcal{H}_n^{w} \right ] = e \ln(n) - \frac{3}{2} \ln \ln (n) + \mathcal{O}(1).
\end{equation*}
Similarly to before we might define $Y:= \mathcal{H}_n^{w} - \mathcal{H}_{n-k+1}^\theta$. Then by \eqref{eq:height} it holds that $Y<k$, so
\[\mathrm{Var}(\mathcal{H}_n^{w}) = \mathrm{Var}(\mathcal{H}_{n-k+1}^\theta+Y) = \mathrm{Var}(\mathcal{H}_{n-k+1}^\theta) + \mathrm{Var}(Y) + \mathrm{Cov}(\mathcal{H}_{n-k+1}^\theta, Y) = \mathcal{O}(1).\]

\hfill $\square$
\end{proof}

\section{Largest branch}\label{sec:largestbranch}

\subsection{Permutations view}

In this section, we focus on standard Hoppe trees and study the size of the largest branch in this   tree model. The results will sharpen the corresponding observations of \cite{Feng05} for URTs. Before moving on to largest branches, we need to discuss constructions of Hoppe trees via random permutations, in particular permutations that are generated via Ewens sampling formula.

For each $n$, Ewens distribution gives a family of distributions over the vectors $$C^{(n)} = (C_1^{(n)}, C_2^{(n)},\ldots,C_n^{(n)})$$ with $\sum_{i=1}^n i C_i^{(n)} = n.$
In particular, for given $\theta > 0$, the Ewens distribution $EW(\theta)$ is defined with the probabilities $$\mathbb{P}_{\theta} (C^{(n)} = (c_1,\ldots,c_n)) = \mathbf{1}\left(\sum_{j=1}^n j c_j = n \right) \frac{n!}{\theta_{(n)}} \prod_{j=1}^n  \left( \frac{\theta}{j} \right)^{c_j} \frac{1}{c_j!}, \qquad  c_1,\ldots,c_n \in \mathbb{N},$$ where $\theta_{(n)} =  \theta(\theta + 1) \ldots (\theta + n - 1)$.

Below we call a permutation resulting from $EW(\theta)$ a Hoppe permutation. In this setting, $c_i$ stands for the number of cycles in permutation of size $i$. In order to relate the topic to Hoppe trees, we need to discuss a recursive construction of Hoppe permutations. The discussion here  is similar to the one in   \cite{Altokislak}. For convenience,  the  permutation will be  constructed on the label set $\{2,3,\ldots,n\}$. We first begin with the  permutation $(2)$ with only one cycle. Then 3 either joins the first cycle to the right of $2$ with probability $\frac{1}{\theta + 1}$, or starts the second cycle with probability $\frac{\theta}{\theta + 1}$. Once we have constructed a permutation on $\{2,3,\ldots,k-1\}$, $k$ either starts a new cycle with probability $\frac{\theta}{\theta +k - 1}$, or is inserted to the right of a randomly chosen integer already assigned to a cycle. The resulting permutation then has the distribution $EW(\theta)$ \cite{ABT}.

Next, we construct a Hoppe tree based on the permutation construction of the previous paragraph. Begin with node  1 as the root and node 2 attached to it. Then if node 3 begins a new cycle in the corresponding Hoppe permutation, attach it to node 1, and otherwise attach it to node 2. Then for node  $k \geq 4$, if $k$ starts a new cycle in the Hoppe permutation, then attach node $k$ to node $1$, and otherwise attach it to node $j$ where
$k$ was inserted to the right of $j$ in the corresponding permutation.

It is then clear that this gives a bijection between Hoppe permutations and Hoppe trees. In particular, the cycles in a Hoppe permutation are in a one-to-one relation with the number of branches in the corresponding tree. This reduces the study of the size of the largest branch  in a Hoppe tree to the study of the largest cycle in its permutation correspondence.

\subsection{Size of the largest branch}

For a given tree $\mathcal{T}$ on $n$ vertices, the number of branches is the number of  children of the root, i.e. all nodes that are attached to the root. If a node $i$ is attached to the root, then node $i$ with its descendants is said to form a branch of the tree.  Let now  $\mathcal{B}_{n,i}(\mathcal{T})$  be the number of branches of size $i$ in $\mathcal{T}$. Also define   $$\nu_n(\mathcal{T}) := \max\{i \in [n-1]: \mathcal{B}_{n, i} \geq 1\}$$ to be the number of nodes  in the largest branch of a given tree, $\mathcal{T}$. In \cite{Feng05}, it was shown that  \begin{equation}\label{eqn:largestbranch}
\lim_{n \rightarrow \infty} \mathbb{P} \left(\nu_n(\mathcal{T}_n) \geq \frac{n}{2} \right) = \ln 2,
\end{equation} when $\mathcal{T}_n$ is a URT on $n$ vertices.    The first purpose of this section and the next theorem  is to  extend the  result of \cite{Feng05} to Hoppe trees, and to  provide more details about the asymptotic distribution, via exploiting the relation between Hoppe trees and Hoppe permutations.  Further, the result in \eqref{eqn:largestbranch} is now extended to an explicit expression for  $\lim_{n \rightarrow \infty} \mathbb{P} \left(\nu_n(\mathcal{T}_n) \geq cn \right)$ for $c \in [1/2,1]$. Once we have the results for the Hoppe tree case, the coupling construction of the previous section will also generalize these results to WRTs.

\begin{theorem}
(i) Let $\mathcal{T}_n^{\theta}$ be a Hoppe tree. Then $\frac{\nu_n (\mathcal{T}_n^{\theta})}{n}$ converges weakly to a random variable $\nu$ whose cumulative distribution function is given by
$$
F_{\theta}(x)  = e^{\gamma \theta} x^{\theta - 1} \Gamma(\theta) p_{\theta}(1 / x), \quad x > 0,$$
where   $\gamma$ is Euler's constant,
$$p_{\theta}(x) = \frac{e^{- \gamma \theta} x^{\theta - 1}}{\Gamma(\theta)} \left(1 + \sum_{k=1}^{\infty} \frac{(- \theta)^k}{k!} \int \cdots \int_{\mathcal{S}_k(x)} \left(1 - \sum_{j=1}^k y_j \right)^{\theta - 1} \right) \frac{d y_1 \cdots d y_k}{y_1 \cdots y_k},
$$
with
 $$\mathcal{S}_k(x)= \left\{y_1 > \frac{1}{x},\ldots, y_k > \frac{1}{x}, \sum_{j=1}^k y_j < 1\right\}.$$

(ii) When $\theta = 1$, we obtain the following for the largest branch in a URT:
 $\frac{\nu_n(\mathcal{T}_n)}{n}$ converges weakly to a random variable $\nu$ whose cumulative distribution function is given by
$$
F_1(x)  =
\begin{cases}
0, & \text{if } x<0 \\
1 + \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \int \cdots \int_{\mathcal{S}_k(x)} \frac{dy_1\ldots dy_k}{y_1\ldots y_k}, & \text{if } x \in [0,1] \\
1, & \text{if } x >1,
\end{cases}
$$
where
  $\mathcal{S}_k(x)$ is as before.

Also, for any $c \ in \left[\frac{1}{2},  1\right]$, we have $$\lim_{n \rightarrow \infty} \mathbb{P}(\nu_n(\mathcal{T}_n) \leq c n) = 1 - \ln (c^{-1}).$$
In particular,  $$\mathbb{E}[\nu] \geq \frac{n}{2}.$$
\end{theorem}

\begin{proof}
(i) First, we translate the problem into random permutation setting. We have
$$\nu_n(\mathcal{T}_n^{\theta}) =_d \max\{i \in [n-1] : C_{n-1,i}(\theta) \geq 1\},$$ where $C_{n-1,i}(\theta)$ is the number of cycles of length $i$ in a $\theta$-biased Hoppe permutation. In this setting, Kingman \cite{Kingman77} shows that $\frac{\alpha_n(\theta)}{n}$  converges in distribution to a random variable $\alpha$ with cumulative distribution function $$F_{\theta} (x) = e^{\gamma \theta} x^{\theta - 1} \Gamma (\theta) p_{\theta} \left(\frac{1}{x} \right), x >0,$$ where $\gamma$ is Euler's constant, $$p_{\theta}(x) = \frac{e^{- \gamma \theta}}{\Gamma (\theta)} \left(1 + \sum_{k=1}^{\infty} \frac{(-\theta)^k}{k!} \right) \int \cdots \int_{\mathcal{S}_k(x)} \left(1 - \sum_{j=1}^k y_i \right)^{\theta - 1}\frac{dy_1\ldots dy_k}{y_1\ldots y_k}, $$ and  $$\mathcal{S}_k(x)= \left\{y_1 > \frac{1}{x},\ldots, y_k > \frac{1}{x}, \sum_{j=1}^k y_j < 1\right\}.$$
This proves the first part.

(ii) Setting $\theta =1$ in argument of (i), and recalling that the random permutation in this case reduces to a uniformly random permutation   immediately  reveals the result.

For the second claim, we first note Watterson \cite{Watterson76} shows that the derivative of $F_1(x)$ over $[1/2, 1]$ simplifies to $$f_1(x) = \frac{1}{x}.$$ Hence, for any $c \in \left[\frac{1}{2},  1\right]$, $$ \lim_{n \rightarrow \infty} \mathbb{P}(\nu_n(\mathcal{T}_n) \geq c n) = \mathbb{P}(\nu \geq c n) = \int_{c}^1 \frac{1}{x} dx = \ln (1/c).$$
Finally, we have  $$\mathbb{E}[\nu_n(\mathcal{T}_n)] \geq \int_{1/2}^1 x \frac{1}{x} dx = \frac{1}{2}.$$

\hfill $\square$
\end{proof}

\begin{remark}
The value $\lim_{n \rightarrow \infty} \frac{\nu_n(\mathcal{T}_n)}{n}$ when $\mathcal{T}_n$ is a URT is known to be the
  Golomb-Dickman constant in the literature. Its exact value is known to be $0.62432998854...$.
\end{remark}

Now, let $\mathcal{T}_n^{w}$ be a  WRT of size $n$ with weight sequence $(w_i)_{i \in \mathbb{R}}$ such that there is a $k \in \mathbb{N}$ such that for all $i>k$ we have $w_i = 1$. Since we can  find a coupling of a Hoppe tree to a WRT in which the number of nodes in the largest branch  differs at most by $k$, the following now follows immediately.

\begin{theorem}
Let $\mathcal{T}^{w}_{n}$ be a WRT of size $n$ with weight sequence $(w_i)_{i \in \mathbb{R}}$ such that there is a $k \in \mathbb{N}$ such that for all $i>k$ we have $w_i = 1$.  Then for any $c \ in \left[\frac{1}{2},  1\right]$, we have $$\lim_{n \rightarrow \infty} \mathbb{P}(\nu_n(\mathcal{T}_n^w) \leq c n) = 1 - \ln (c^{-1}).$$
\end{theorem}

\section{Depth of node $n$}\label{sec:depth}

The depth of node $n$ is the length of the path from the root to $n$ or equivalently the number of ancestors of $n$. Note that in this and the next section we don't have any restrictions on the weight sequence $(w_i)_{i\in \mathbb{N}}$.

\begin{theorem}

Let $\mathcal{D}_n^{w}$ denote the depth of node $n$ in a WRT $\mathcal{T}_n^{w}$ and let $Z_n^w$ denote the set of ancestors of $n$. Let moreover $A_{i,n}^w := \mathbf{1}(i \in Z_n^w)$.
Then
\[\mathcal{D}_n^w = 1 + \sum_{i=2}^{n-1} A_{i,n}^w.
\]
The $A_{i,n}^w$ are mutually independent Bernoulli random variables with
\[ \mathbb{P}( A_{i,n}^w ) = \frac{w_i}{\sum_{j=1}^{i}w_j}.\]

This directly yields the expectation and the variance:
\begin{equation*}
\mathbb{E}[\mathcal{D}_n^w] = \sum_{i=1}^{n-1} \frac{w_i}{\sum_{j=1}^{i}w_j}
\end{equation*}
and
\begin{equation*} \mathrm{Var}(\mathcal{D}_n^w) = \sum_{i=2}^{n-1} \frac{w_i}{\sum_{j=1}^{i}w_j} \left (1 - \frac{w_i}{\sum_{j=1}^{i}w_j} \right ).
\end{equation*}
\end{theorem}

\begin{proof}
Each claim will follow easily once we show that $\mathcal{D}_n^w$ can be written as a sum of independent Bernoulli random variables. For this purpose, we first observe that in a given rooted tree, the depth of a node is equal to its number of ancestors, since these determine the path from the root to the node. Using that $1$ definitely is an ancestor of $n$, in the notation of the theorem we thus get
\[\mathcal{D}_n^w = 1 + \sum_{i=2}^{n-1} A_{i,n}^w.
\]
We will first find the distribution law of the $A_{i,n}^w$ and then show mutual independence. For the distribution law we will use the method used in \cite{Feng05}: we first find the values for $n-1$ and $n-2$ and then proceed by induction.

Node $n-1$ can only be an ancestor of $n$ if it is the parent of $n$, so we get
\[
\mathbb{P}(n-1 \in Z_n^w) = \frac{w_{n-1}}{\sum_{i=1}^{n-1} w_i}.
\]
Similarly $n-2$ can only be an ancestor of $n$ if it is the parent of $n$ or it is the grandparent of $n$, in which case $n-2$ needs to be the parent of $n-1$ who needs to be the parent of $n$. This gives
\begin{equation*}
\begin{split}
\mathbb{P}(n-1 \in Z_n^w) = \underbrace{\frac{w_{n-2}}{\sum_{i=1}^{n-1} w_i}}_{n-2 \text{ is parent of } n} + \underbrace{\frac{w_{n-2}}{\sum_{i=1}^{n-2} w_i}\frac{w_{n-1}}{\sum_{i=1}^{n-1} w_i}}_{n-2 \text{ is grandparent } of n}
= \frac{w_{n-2}}{\sum_{i=1}^{n-2} w_i}.
\end{split}
\end{equation*}
We will now show by induction that  for all $j = 2, \dots, n-1$,
\[\mathbb{P}( j \in Z_n^w) = \frac{w_j}{\sum_{i=1}^{j} w_j}.
\]
Let the above be true for all $j \geq i+1$ and let $C_{i,j}^w$ denote the event that $j$ is a child of $i$. Then
\[\mathbb{P}( i \in Z_n^w) = \sum_{j = i+1}^{n-1} \mathbb{P}( j \in Z_n^w, C_{i,j}^w) + \mathbb{P}(C_{i,n}^w).
\]

Since $C_{i,j}^w$ only relates to the $j^{th}$ step of the construction process and $j \in Z_n^w$ only depends on the $j+1^{th}, \dots, n^{th}$ step, these two events are independent. We thus get
\[\mathbb{P}(i \in Z_n^w) =  \sum_{j = i+1}^{n-1} \mathbb{P}( j \in Z_n^w)\mathbb{P}(C_{i,j}^w)+\mathbb{P}(C_{i,n}^w) = \sum_{j = i+1}^{n-1} \left (\frac{w_j}{\sum_{k=1}^{j}w_k} \frac{w_i}{\sum_{k=1}^{j-1}w_k} \right )+ \frac{w_i}{\sum_{j=1}^{n-1}w{j}}.
\]

To simplify this expression we first note that we can factor out $w_i$ and by some elementary operations get:
\begin{equation*}
\begin{split}
 \frac{w_{i+1}}{\sum_{k=1}^{i+1}w_k\sum_{k=1}^{i}w_k} + \frac{w_{i+2}}{\sum_{k=1}^{i+2}w_k\sum_{k=1}^{i+1}w_k} = \frac{w_{i+1}+ w_{i+2}}{\sum_{k=1}^{i+2}w_k\sum_{k=1}^{i}w_k}.
\end{split}
\end{equation*}

In general, the following holds for $l \in \mathbb{N}$:
\begin{equation*}
\begin{split}
&\frac{w_{i+1}+ w_{i+2} + \cdots + w_{i+l}}{\sum_{k=1}^{i} w_k \sum_{k=1}^{i+l} w_k} + \frac{w_{i+l+1}}{\sum_{k=1}^{i+l+1} w_k \sum_{k=1}^{i+l} w_k} = \frac{w_{i+1}+  \cdots + w_{i+l+1}} {\sum_{k=1}^{i+l+1} w_k  \sum_{k=1}^{i} w_k}.
\end{split}
\end{equation*}

By using this equality $n-i-2$ times, we thus get
\begin{equation*}
\begin{split}
\mathbb{P}(i \in Z_n^w) &= w_i \left ( \frac{w_{i+1} + \cdots + w_{n-1}}{\sum_{k=1}^{i}w_k \sum_{k=1}^{n-1}w_k} + \frac{1}{\sum_{k=1}^{n-1}w_k} \right ) = \frac{w_i}{\sum_{k=1}^{i} w_k}.
\end{split}
\end{equation*}
Now we will show that the events $A_{i,n}^w$ are mutually independent for $j = 2, \dots, n-1$. For this we will use the method used in \cite{Hoppe}: for any $2 \leq k \leq n-2$ and $2 \leq j_k < \dots < j_2 < j_1 \leq n-1$ consider the event that all $j_i$'s and only the $j_i$'s are ancestors of $n$. We will denote this event by $E$. Then
\[E := ( \mathbf{1}(j_i \in Z_n^w) =1, \mathbf{1}(j \in Z_n^w) = 0, \text{ for } j \neq j_i, i=1, \dots, k).
\]
By the structure of the recursive tree, to realize this event, $n$ must be a child of $j_1$, $j_1$ a child of $j_2, \dots , j_{k-1}$ a child of $j_k$ and $j_k$ a child of 1. In general for $i=1, \dots, k-1$, $j_i$ must be a child of $j_{i+1}.$  It does not matter what nodes  $j \neq j_i$ attach to. Hence, by the attachment probabilities we get:
\begin{equation*}
\begin{split}
\mathbb{P}(E) & = \mathbb{P}(j_i \in Z_n^w, j \notin Z_n^w, \text{ for } j \neq j_i, i=1, \dots, k) \\
&= \underbrace{\frac{w_{j_1}}{\sum_{\ell=1}^{n-1} w_{\ell}}}_{n \text{ child of } j_1} \prod_{i=1}^{k-1} \underbrace{\frac{w_{j_{i+1}}}{\sum_{\ell=1}^{j_i-1}w_{\ell}}}_{j_i \text{ child of } j_{i+1}} \underbrace{\frac{w_1}{\sum_{\ell=1}^{j_k-1}w_{\ell}}}_{j_k \text{ child of } 1} \\
&= w_1 w_{j_1} \cdots w_{j_k} \prod_{i=1}^{n-1} \frac{1}{\sum_{\ell=1}^{i}w_{\ell}}  \prod_{\substack{1<j<n \\ j \neq j_i, i=1, \dots, k}} \left (\sum_{\ell=1}^{j-1}w_{\ell} \right ) \\
& = \prod_{i=1}^{k}\frac{w_{j_i}}{\sum_{\ell=1}^{j_i}w_{\ell}} \prod_{\substack{1<j<n \\ j \neq j_i, i=1, \dots, k}} \left ( \frac{\sum_{\ell=1}^{j-1}w_{\ell}}{\sum_{\ell=1}^{j}w_{\ell}}\right )  \frac{w_1}{\sum_{\ell=1}^{1}{w_{\ell}}} \\
&= \prod_{i=1}^{k}\frac{w_{j_i}}{\sum_{\ell=1}^{j_i}w_{\ell}} \prod_{\substack{1<j<n \\ j \neq j_i, i=1, \dots, k}} \left ( 1 - \frac{w_j}{\sum_{\ell=1}^{j}w_{\ell}}\right )  \\
&= \prod_{i=1}^{k} \mathbb{P}(j_i \in Z_n^w) \prod_{\substack{1<j<n \\ j \neq j_i, i=1, \dots, k}} \mathbb{P}(j \notin Z_n^w).
\end{split}
\end{equation*}

This implies that the events $\mathbf{1}(i \in Z_n^w)$ are mutually independent. Hence the $A_{i,n}^w$ are mutually independent Bernoulli random variables.

Expectation and variance formulas for  $\mathcal{D}_{n}^w$ follows from this observation right away.

\hfill $\square$
\end{proof}

The following central limit theorem now follows.

\begin{theorem}
If $\mathbb{E}[\mathcal{D}_n^w ] \rightarrow \infty$ and $\limsup_{n \to \infty} \frac{w_n}{\sum_{i=1}^{n} w_i} < 1$, then we have $$\frac{\mathcal{D}_n^w - \mathbb{E}[\mathcal{D}_n^w ]}{\sqrt{Var(\mathcal{D}_n^w )}} \longrightarrow_d \mathcal{G} \quad \text{as} \quad n \rightarrow \infty.$$
\end{theorem}

\begin{proof}
By Liapounov's central limit theorem for sums of independent Bernoulli random variables, if $\mathrm{Var}({D}_n^w) \to \infty$, it holds that $\frac{\mathcal{D}_n^w - \mathbb{E}[\mathcal{D}_n^w ]}{\sqrt{Var(\mathcal{D}_n^w )}} \longrightarrow_d \mathcal{G} \quad \text{as} \quad n \rightarrow \infty$. Now let $p_n = \frac{w_n}{\sum_{j=1}^{n} w_j}$.
Since  $\limsup_{n \to \infty} p_n <1$, there is an $0<\varepsilon<1$ and an $N \in \mathbb{N}$ such that for all $n >N$, $ 1- p_n > \varepsilon$.
Then \[\mathrm{Var}({D}_n^w) = \sum_{i=1}^{n} p_i(1-p_i) = \sum_{i=1}^{N} p_i(1-p_i) + \sum_{i=N+1}^{n} p_i(1-p_i) >  \varepsilon \sum_{i=N+1}^{n} p_i. \] Since we know that $\mathbb{E}[\mathcal{D}_n^w ] = \sum_{i=1}^{n} p_i \rightarrow \infty$ as $n \to \infty$ this implies that $\mathrm{Var}(\mathcal{D}_n^w) \to \infty$ as $n \to \infty$.

\hfill $\square$
\end{proof}

\begin{example}
(i)  If the weights are limited from below and above, the expectation and variance of the depth of node $n$ will still be equal to $\mathcal{O}(\ln(n))$ asymptotically.

(ii) For the Hoppe tree, we write $\mathcal{D}_n^\theta$ for the depth of node $n$ and have
\begin{equation*}
\begin{split}
&\mathbb{E}[\mathcal{D}_n^\theta] = 1+ \sum_{i=1}^{n-2} \frac{1}{\theta+1} = \log(n) + \mathcal{O}(1) \text{ and }\\
&\mathrm{Var} \left (\mathcal{D}_n^\theta \right ) = \sum_{i=1}^{n-2} \frac{\theta+i-1}{(\theta+i)^2} = \log(n) + \mathcal{O}(1).
\end{split}
\end{equation*}
Notice that the depth in a URT and a Hoppe tree asymptotically equivalent. The same conclusion also holds for a wide range of statistics whose dependence to the root is  small and vanishes asymptotically. This makes the asymptotic study of Hoppe trees slightly uninteresting. Another such example  is the number of leaves in a Hoppe tree which was studied earlier.

(iii) Let's mention one instance where the behavior of the depth is totally different than the URT case. Let $(w_i)_{i\in \mathbb{N}} = \left (\frac{1}{i^2} \right )_{i \in \mathbb{N}}$. Then $
\mathbb{E}[\mathcal{D}_n^w] =  \sum_{i=1}^{n-1} \frac{\frac{1}{i^2}}{\sum_{j=1}^{i}\frac{1}{j^2}}$,
and since for all $i \in \mathbb{N}$ we have $1<\sum_{j=1}^{i}\frac{1}{j^2}<\frac{\pi^2}{6}$, we get
\begin{equation*}
\begin{split}
  \frac{6}{\pi^2} \sum_{i=1}^{n-1} \frac{1}{i^2} \leq \mathbb{E}[\mathcal{D}_n^w] \leq \sum_{i=1}^{n-1} \frac{1}{i^2}
  \Longrightarrow \frac{6}{\pi^2}  \leq \mathbb{E}[\mathcal{D}_n^w] \leq \frac{\pi^2}{6}.
\end{split}
\end{equation*} Also, since
$
\mathrm{Var}(\mathcal{D}_n^w) = \sum_{i=2}^{n-1} \frac{\frac{1}{i^2} \sum_{j=1}^{i-1} \frac{1}{j^2}}{\left (\sum_{j=1}^{i}\frac{1}{j^2}\right )^2},
$
and  for all $i\geq 2$, it holds that $\frac{4^2}{5^2} \leq \frac{\sum_{j=1}^{i-1} \frac{1}{j^2}}{\left (\sum_{j=1}^{i}\frac{1}{j^2}\right )^2}\leq 1$, we get
\begin{equation*}
\begin{split}
\frac{4^2}{5^2} \sum_{i=2}^{n-1} \frac{1}{i^2} \leq \mathrm{Var}(\mathcal{D}_n^w) \leq \sum_{i=2}^{n-1} \frac{1}{i^2}
\Longrightarrow \frac{4}{5^2}  \leq \mathrm{Var}(\mathcal{D}_n^w) \leq \frac{\pi^2}{6} -1.
\end{split}
\end{equation*}
\end{example}

\section{Number of branches}\label{sec:branches}

Finally, we study the number of branches in a WRT. Our results are  summarized in the following theorem.

\begin{theorem}
Let $\{w_n\}_{n \geq 1}$ be a sequence of positive weights.  Denote the number of branches in $\mathcal{T}_n^w$ by $\mathcal{B}_n^w$.
\begin{itemize}
\item[(i)]  We have \begin{equation*} \mathbb{E}[\mathcal{B}_n^w] = \sum_{i=1}^{n-1} \frac{w_1}{\sum_{k=1}^i w_k}  \qquad \text{and} \qquad Var(\mathcal{B}_n^w) = \sum_{i=2}^{n-1} \frac{w_1}{\sum_{k=1}^i w_k}  \left ( 1-\frac{w_1}{ \sum_{k=1}^i w_k }\right ).
\end{equation*}
\item[(ii)] If $\mathbb{E} [\mathcal{B}_n^w]$ diverges, then the central limit $$\frac{\mathcal{B}_n^w - \mathbb{E}[\mathcal{B}_n^w]}{\sqrt{Var(\mathcal{B}_n^w)}} \longrightarrow_d \mathcal{G} \quad \text{as} \quad n \rightarrow \infty.$$

\item[(iii)] Further, one has $$d_W \left(\frac{\mathcal{B}_n^w - \mathbb{E}[\mathcal{B}_n^w]}{\sqrt{Var(\mathcal{B}_n^w)}}, \mathcal{G} \right) \leq \frac{1}{\sqrt{Var(\mathcal{B}_n^w)}} \frac{\sqrt{28} + \sqrt{\pi}}{\sqrt{\pi}}.$$
\end{itemize}
\end{theorem}

\begin{proof}
Letting $b_i = \mathbf{1}(\text{node $i$ attaches to node 1})$, observe that the $b_i$'s are independent and that $\mathbb{E}[b_i]= \frac{w_1}{\sum_{j=1}{i-1} w_j}$ . The formula for $\mathbb{E}[\mathcal{B}_n^w] = \sum_{i=2}^{n} \mathbb{E}[b_i]$ and $\mathrm{Var}(\mathcal{B}_n^w)= \sum_{i=2}^{n} \mathrm{Var}(b_i)$ is then clear.

The CLT then follows from Liapounov's central limit theorem. Let $\mathbb{E}[\mathcal{B}_n^w] \to \infty$. To show that this implies that $\mathrm{Var}(\mathcal{B}_n^w)  \to \infty$ too, we differentiate two cases:
\begin{itemize}
\item[(a)] If $\sum_{i=1}^{n} w_i \to c \in \mathbb{R}$ as $n \to \infty$, $ \frac{\sum_{i=2}^{n} w_i}{(\sum_{i=1}^{n} w_i)^2} \to \frac{c-w_1}{c^2}$ and hence $\mathrm{Var}(\mathcal{B}_n^w) $ diverges.
\item[(b)]  If $\sum_{i=1}^{n} w_i \to \infty$ as $n \to \infty$,this implies that $\frac{w_1}{\sum_{i=1}^{n} w_i} \to 0$ as $n \to \infty$. Hence there is an $N \in \mathbb{N}$ such that for all $n >N$, we have $\frac{w_1}{\sum_{i=1}^{n} w_i}< \frac{1}{2}.$ Then
\[\mathrm{Var}(\mathcal{B}_n^w) > \sum_{i=N+1}^{n-1} \frac{w_1}{\sum_{k=1}^i w_k}  \left ( 1-\frac{w_1}{ \sum_{k=1}^i w_k }\right ) > \frac{1}{2} \sum_{i=N+1}^{n-1} \frac{w_1}{\sum_{k=1}^i w_k}  \to \infty.\]
\end{itemize}

The convergence rates can be obtained by using Theorem 3.1 in \cite{Ross}.
\hfill $\square$
\end{proof}

\begin{remark}
(i) When $w_1 = \theta >  0$, and $w_i = 1$, $i = 1,2,\ldots$, we obtain $$\frac{\mathcal{B}_n - \theta \ln n}{\sqrt{\theta \ln n}} \longrightarrow_d \mathcal{G}, \quad \text{as} \quad n \rightarrow \infty.$$ In particular, when $\theta = 1$ as well, one recovers the central limit theorem for the URT case.

(ii) When there exist $\alpha_1, \alpha_2$ so that $0 < \alpha_1 \leq \sup_i w_i \leq \alpha_2 < \infty$, it can be shown that both the expectation and the variance are still of order $n$.

(iii) When $w_i$'s are not bounded, there can be big differences compared to the case of URTs. One such extreme case is when $w_i  =i$ where $\mathbb{E}[\mathcal{B}_n^w] \sim 2$, and $Var(\mathcal{B}_n^w) \sim 14 - \frac{4 \pi^2}{3}$. Another one is when $w_i = 1 / i$ in which case  $$\frac{6}{\pi^2} (n - 1) \leq \mathbb{E}[\mathcal{B}_n^w] \leq n -1, \qquad \text{and} \qquad \frac{6}{5\pi^2} (n - 2) \leq Var(\mathcal{B}_n^w) \leq \frac{1}{4} (n-2).$$

(iv) It is well known that the number of branches and the depth of node $n$ have the same distribution in a URT. This is not the case when the tree is non-uniform. Indeed, this is intuitively clear since having more branches increases the chance that node $n$ attaches to a node at a lower level.

\end{remark}

redpajama_set_name: RedPajamaArXiv


text:
EU hails Green Deal as 'man on the moon moment'
Reuters Videos December 11, 2019
Carbon neutral by 2050.
It's the headline commitment from the European Union.
On Wednesday (December 11) it took a major step towards it with an outline of its new so-called Green Deal policy package.
European Commission President Ursula von der Leyen.
(SOUNDBITE) (English) EU COMMISSION PRESIDENT, URSULA VON DER LEYEN, SAYING:
"Today is the start of a journey. But this is Europe's 'Man on the Moon' moment."
The move is timely.
Global concern over climate change has been getting as hot as some recent record temperatures.
Those - say environmentalists - exacerbating drought, wildfires and floods.
It's timed too to coincide with this week's U.N. climate summit in Madrid, which has heard pleas from teenage activist Greta Thunberg.
And, warnings of the cost of inaction from U.N. Secretary-General Antonio Guterres.
(SOUNDBITE) (English) UNITED NATIONS SECRETARY-GENERAL ANTONIO GUTERRES, SAYING:
"To allow temperatures to rise above 1.5 to two degrees or as we are now heading to more than 3.5 degrees, this is to condemn the planet, to condemn the future generations."
Europe - one of world's key emitters - is showing a will to move, he said.
The new Green Deal package comes before EU member states on Thursday (December 12).
It's likely to run into resistance from heavy coal users like Poland, Hungary and the Czech Republic.
But all being well, will bind EU member states to key climate targets.
While - it's expected - making a 35 billion euro fund available for quote, "those most exposed to the decarbonization challenge."
The European Commission, said von der Leyen, also wants to mobilize 100 billion euros of investment to help the bloc in a politically and economically difficult transition away from fossil fuels.
Boris Johnson's Brexit deal returns to Commons after Government defeated five times by defiant peers
Novartis (NVS) Gets Approval for MS Drug Mayzent in Europe
Roche's (RHHBY) Lymphoma Drug Polivy Gets Approval in Europe
Britain Could Be Canada

redpajama_set_name: RedPajamaCommonCrawl


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Where do the possibilities for the metaverse lie for Singapore?
GovInsider sits down with Associate Professor Frank Guan from the Singapore Institute of Technology to explore key advances in immersive technology and key opportunities for industrial applications.
last year, Barbados opens diplomatic embassy In Decentraland in Metaverse. Decentraland real estate can cost millions of dollars, but the country has made this move so that anyone on the planet can knock on the same embassy door and the overall cost can save
For those still puzzled by this strange new dimensional reality, Assistant Professor Frank Guan, the Infocomm Technology cluster at the Singapore Institute of Technology (SIT) characterizes the Metaverse as a digitized parallel world, allowing people to interact and complete tasks they previously completed in the real world. increase. "You will be doing your daily tasks just like you would in the real world," he says.
Even in its early stages of adoption, the Metaverse already means so much to different groups of people. Guan, who specializes in virtual reality (VR), augmented reality (AR), and artificial intelligence (AI), said this means that the metaverse could be applied to a variety of industries and is "widely accepted. I think it is a positive sign that many parties'.
GovInsider analyzes Guan's take on key industrial applications of metaverse and immersive technology and what it means for different sectors in Singapore.
Benefits of the Metaverse
"I don't think the Metaverse will completely replace the physical world," Guan told GovInsider. There are some sensory aspects not yet accounted for in immersive technology. These will favor metaverse applications over other applications.
The Metaverse is currently focused on visual and audio experiences, and less on tactile and gustatory experiences. So, for example, people are less likely to buy real-world clothes in the metaverse because they can't feel the texture of the fabric beforehand. increase.
Despite these current limitations, much is still possible in the digital world. "The Metaverse can remove physical distancing barriers and allows us to interact with people in other countries remotely, but dynamically," says Guan.
Guan recently chaired the 21st IEEE International Symposium on Mixed and Augmented Reality (ISMAR 2022) conference in Singapore.Image: Frank Guan
Post-pandemic virtual meetings have become a staple in the business, educational and even social worlds.Popular video conferencing platform Zoom saw 200 million daily meeting participants in May 2020and that number increased by 100 million just one month later.The Metaverse has the potential to provide a more realistic and interactive environment for people to hold virtual meetings, Guan says.
In education, students and teachers are brought together in the same immersive learning environment wherever they are. In healthcare, the metaverse could enable doctors to diagnose patients remotely, resulting in more equitable access to medical services.
The metaverse is also great for giving people a taste of what they might be interested in before they actually work on it. In real estate, interested buyers can first view the digital twin of the property before physically traveling to complete the transaction. The same approach can be applied to travel and tourism. That's when you visit a virtual simulation of an attraction before seeing the attraction in person, Guan said.
Finally, according to Guan, gaming is a burgeoning industry with immersive technology.
"When people think of games, they may simply think of them as an entertainment option…but more and more developers are starting to consider gamification for education to achieve better learning outcomes. We are doing it," Guan told GovInsider. In fact, a 2015 study showed that gamified learning showed her a massive 25% increase. Student retention rate 23% higher scores compared to traditional teaching methods. Gamification techniques help contextualize problem statements and hold theory.
Gamified learning is becoming an increasingly popular option for professional training because it is not only more productive, but also more effective. It can generate surreal scenarios that might not be easy to reproduce in the physical world, but are easy to reproduce in the digital world," Guan said.
Employers can reduce training costs, minimize training space, and facilitate access to staff training. For example, a team of SIT researchers led by Guan worked with local organizations to develop his AR-powered game, REAP, for vocational training. The game enables trainees to learn the critical skill sets needed to run a food business without the constraints of where, when and at what cost raw materials require.
A team of SIT researchers led by Guan has developed an AR-powered gamified job training for the food industry. Trainees will be able to learn and improve the skill sets necessary to run a food business.Image: Frank Guan
space to grow
Aside from VR and AR, the Metaverse stays abreast of many other technological advances and trends such as artificial intelligence (AI), non-fungible tokens (NFT), blockchain, Internet of Things (IoT), cloud computing, and more. set to engage. , says Guan.
This was especially evident at the 21st IEEE International Symposium on Mixed and Augmented Reality (ISMAR 2022) conference held for the first time in Singapore and Southeast Asia. An unprecedented number of research papers were submitted to the conference, all aimed at improving the underlying technology to make the metaverse more alive, said the event's general chair. There was his Guan said.
For example, SIT researchers demonstrated a collaborative 3D data visualization system using AR (ViCollAR for short) at ISMAR 2022. An immersive interactive collaborative system allows multiple users to jointly view and interact with his 3D visualized data simultaneously with bare hands. space. The user can also freely navigate the real spatial environment while interacting. The university successfully developed a prototype for further testing and implementation.
SIT faces advances in immersive technology in other ways as well.In the future, SIT will make its campus Punggol Digital District By 2024, students will experience supporting development through a digital twin of the campus. The "Living Lab" will serve as a testbed for technologies, including immersive, allowing industry partners to achieve innovation with the help of academic experts like her Guan.
lieMetaversePOSSIBILITIESSingapore
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text:
Skagens Rev är ett rev i Danmark.  Det ligger i Frederikshavns kommun i Region Nordjylland. Revet är en fortsättning på Grenen, den udde som skiljer Skagerrak och Kattegatt åt vid Skagen

Källor

Landformer på havets botten i Region Nordjylland
Frederikshavns kommun
Skagen

redpajama_set_name: RedPajamaWikipedia


text:
Draft Republican bill would repeal EPA scientific "Endangerment Finding" on greenhouse gases and prohibit regulation
House Energy and Commerce Committee chair Fred Upton (R-Michigan) and Sen. James Inhofe (R-Oklahoma) announced draft legislation that would repeal a scientific determination that carbon pollution poses threats to public health and the environment, strip EPA's statutory authority and responsibility to regulate carbon pollution from sources like power plants and oil refineries, overturn the landmark Supreme Court case, Massachusetts v. EPA, and prevent carbon emission standards for cars and trucks from ever being strengthened or revised.
The Washington Post reported today:
House GOP readies bill to prohibit EPA from regulating carbon emissions
Acting on a vow to fight the Obama administration on climate issues, Rep. Fred Upton (R-Mich.), chairman of the House Energy and Commerce Committee, unveiled draft legislation Wednesday to try to strip the Environmental Protection Agency of its power to regulate greenhouse gases under the Clean Air Act.
Sen. James M. Inhofe (Okla.), ranking Republican on the Senate Committee on Environment and Public Works, and Rep. Ed Whitfield (R-Ky.), chairman of the subcommittee on energy and power, joined Upton in issuing what they called the Energy Tax Prevention Act. …
The EPA issued a finding in late 2009 that said greenhouse gas emissions linked to global warming endanger the public's health and welfare. Last month, the agency told industrial facilities such as power plants, oil refineries and paper mills that require permits to emit sulfur dioxide and nitrous oxide to also account for greenhouse gas emissions if they expand or add construction that significantly increases greenhouse gas pollution.
Republicans, many of whom doubt climate-change science, say the regulations hurt the ability of U.S. manufacturers to compete against companies in countries with relaxed standards.
Energy and Commerce Ranking Member Rep. Henry Waxman (D-California) said: "The Republicans have a lot of power, but they can't amend the laws of nature. Gutting the Clean Air Act is only going to make our problems worse. …"
A statement by the Project on Climate Science includes this:
The Project on Climate Science finds it incomprehensible that, as they develop policy options to address our nation's energy and climate challenges, some Members of Congress are seeking to turn science on its head by launching an assault on the Clean Air Act – the very law that protects us from the serious threat carbon pollution poses to our health and our national and economic security.
The draft legislation, "The Energy Tax Prevention Act," would overturn the EPA's 2009 scientific finding that carbon dioxide and other greenhouse gases endanger public health and welfare – a finding that was reached after EPA scientists carefully reviewed the evidence. …
As noted today by the American Lung Association, the Clean Air Act guards the most vulnerable Americans – those with asthma and other lung diseases, children, older adults, and people with heart disease and diabetes. The association declared, "The protections against the health harm from carbon dioxide and other greenhouse gas pollution are essential to public health and must be preserved."
The bill, which refers to "possible climate change," embodies a blatant disregard for scientific fact. …
"Energy Tax Prevention Act" Discussion Draft, February 2, 2011
Good media criticism by David Roberts at Grist on coverage of Republican rhetoric about their legislation: 'Objective' political media helps Inhofe lie about EPA
House Republicans prepare to attack EPA greenhouse gas rules
The Importance of Science in Addressing Climate Change: Scientists' letter to the U.S. Congress
Sen. Inhofe inquisition seeking ways to criminalize and prosecute 17 leading climate scientists
Whistle-Blower2018-10-26T13:33:17-04:00February 3rd, 2011|Attacks on Climate Science and Scientists, Congress: Legislation and Oversight|


An interface to an object that has a signature and can provide the object properties that are contained in its signature.
A convenience method that obtains the object properties that are in the signature of this object. The default implementation of this method returns an empty, modifiable set.
A set containing the object properties that are in the signature of this object.The set is a subset of the signature, and is not backed by the signature; it is a modifiable collection and changes are not reflected by the signature.
Stream of object properties that are in the signature of this object.


Dentistry affects every smile, every kiss, every meal for the rest of your life.
When you visit Dr. Merxbauer's office in Aberdeen, SD, your smile is our top priority. Our entire team is dedicated to providing you with the personalized, gentle care that you deserve.
Aberdeen dentist, Dr. David Merxbauer, DDS offers a wide variety of dental treatments including porcelain veneers, dental implants, teeth whitening, dental exams, restorations and more.


Posted on 3rd April 2009 — Gareth Edward asked how to achieve the effect on the latest BBC Radio 1 website, where the mouse rolls over the image block, tabs slide up and the image zooms a little to reveal more of the picture.
QuickTime version is approximately 60Mb, flash version is streaming.
The first thing I noticed about the effect is that it's done in Flash. I'm a big believer in knowing where Flash has it's place: the high end of the effects within browsers (amongst other places).
When you roll over the image, it zooms out slightly to reveal more of the image. Notice that it also keeps the top position static.
The navigation slides in to view from the bottom of the box, and mousing over the link will trigger showing a different image.
The images automatically change with a swipe effect.
I'm going to set aside the 3rd effect for now (perhaps I'll do a screencast on it in the future) and focus on the first two.
I'm still in two minds as to whether this should be a background image. The image certainly doesn't add any value from a content point of view, so without CSS I think it looks right, but I'd be open to hear your opinions.
When these images are created they need to have their size increased by a defined percentage (which we'll make a parameter), so that when we roll over, the image is resized down to it's original size, giving the impression of a zoom out.
Given that we are going to stretch these images across the total height and width of the panel, it also means we need to wrap the content that's initially in the panel in an element with a higher z-index to make sure the actual content can still be seen and used.
By default the jQuery slideUp method hides content, and slideDown reveals the content by expanding element downwards.
As for the CSS, as an experiment, I've used EMs throughout the CSS – there's no reason why you can't use PXs. I was just interested in seeing the example being zoomed by the browser. I've set the body font-size to 62.5% so that 1em is equal to 10px.
I'm going to give you each part separately, then you can see the whole thing put together in the completed example.
As I'm trying to make a plugin, I'm saying that I don't know anything about the images or height or width of the panels.
We need to loop through each panel, take the background image and create a new image with that url. Once that new image is loaded, we'll pull out the height and width, work out the zoomed in height and width (which will be a percentage larger than the initial size), strip the panel of it's background image and then insert the new image in to the panel. Phew! That's sounds like a lot!
Let's take a look at what's happening in this code.
The panelIds variable contains a string such as: #one,#two,#three – since this is the hash on the href the tab links, which we immediately capture as a jQuery instance.
Next, we loop through the $panels to convert the background images to foreground images.
This is a regular expression that captures the background image. It's written to try to handle different cases in that the url could be plain, wrapped in single quotes or it could be wrapped in double quotes.
We test if we did successfully find the background image (which you may argue we don't need since this plugin has a very specific purpose), and if it was found, we'll create a new image.
This is to ensure the contents appears above the image – specifically notice the z-index: 2.
The very last action we perform is to capture the image in the images array. This will be used later to zoom the images in and out.
This means we can retrieve this data later on.
I've created a separate function to handle the zooming in and out of the image. This is because it will be called from two different places with only one single different – which variable to use to set the CSS – either 'fullView' or 'zoomView' which we set just above.
In this function we loop through the images that we captured in the images array. From there we say if the image is visible (i.e. it's the currently active panel), then animate it's CSS properties (either zooming out or in – this is defined by the zoomType).
If the image isn't visible, we need to just change the CSS properties without the animation. This is because when the user switches to a different tab, the image must be in the same zoomed state as all the other panels.
The hover over is fairly simple, we stop the effect from running on the $tabs to prevent it from jumping and down madly, and we animate the height to a pre-captured height (which would happen in the initialisation phase).
Hover off is slightly different. We do the same, animating down to zero, but then we use a callback function to hide the navigation, this is to ensure it's completely hidden once the effect has finished (because in my example, I'm using a border, and animating the height to zero would leave the border visible).
I've added a simple 100ms timeout to say if the user is over the tab, now show that panel.
Since I'm using setTimeout, I need to capture a copy of the current tab link, because the this keyword inside of the setTimeout is actually a reference to the window object.
With all my code now in place (there's a few initialisation items I've not covered in this tutorial, but should be covered in the screencast the effect is now ready to be used.
Have a play with the zoom tabs effect, let me know what you think – particularly so if you think there's a better name for it!


I am honored to represent Washington's 8th Congressional District, which includes much of King, Pierce, Kittitas, and Chelan Counties, and portions of Douglas County.
Prior to being elected to Congress, I spent my career as a pediatrician listening to and helping solve problems with thousands of patients and their families across the 8th District, and I am bringing that same dedication and understanding to my work as your representative in Congress.
I am a pediatrician, a wife, and a mom with deep roots in the 8th District. I built a pediatrics practice here over the past 16 years, taking care of thousands of children and their families from all over the region. Our son goes to our local public school, which we strongly support with our time, energy and advocacy. I loved my job and my patients, and never thought I would do anything else. But the 2016 election compelled me to step up.
Healthcare has always been my passion. I was diagnosed with Type 1 diabetes as a teenager. Spending all that time in doctors' offices led to my career as a pediatrician. I know firsthand what it means to worry about health insurance and rising costs.
I am the daughter of an elementary school teacher and an aerospace engineer, and the product of public schools, with degrees in astrophysics and medicine. I am committed to improving our public schools so every student graduates high school with the skills they need for college or career. I will continue to work to build a middle-class centered economy so that everyone who works hard benefits, and I will protect Social Security and Medicare from cuts. I will work to keep our air and water clean and transition to a clean energy economy to slow the rate of climate change. And I will keep working to strengthen our healthcare system until every person in this country has access to affordable, high-quality care. I am the only female doctor in Congress and I provide a vital missing voice as Congress debates women's healthcare, reproductive rights, and children's health.


Thermocouples: Teflon coated Cr Al (K-type)-6 no.
- To study the temperature distribution along the length of a pin fin in natural & forced convection.
Owing to the dedication of experts in our team, we are counted among the best names in the market proffering Pin Fin Apparatus. They are available at lowest rates.
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Size : 12.5 mm (approx.) 15 cm. long (approx.).
Duct : Made of MS.
All the components of Heat Transfer From Pin - Fin Apparatus are displayed on powder coated fabricated stand. The primer consists of a hermetically sealed compressor, heat rejection chamber, drier-filter, expansion device, and heat absorption chamber.
To find out heat transfer co-efficient of a finned surface and in a natural convection.


I met Isabella a couple of years ago when she finished her first graduate collection. She's presently at London College of Fashion doing her MA and she presents her final collection in January. Below is a sneaky peak of the photo shoot for Isabella's lookbook!


The Zimbabwe National Chamber of Commerce (ZNCC) annual conference kicks-off this Wednesday in the resort town of Victoria Falls where representatives from government, business, civic and political society from the region and across the continent are expected to attend.
As the country charts through a new path in the new economic era, it is imperative that business and government meet in order to deliberate on issues that will improve the operating economic environment to grow the economy.
Leading captains of industry and policy makers are expected to attend the ZNCC conference.
"This year's conference is going to look at the ways in which we can boost small businesses as well as ensuring that there are sufficient reforms to improve the Zimbabwe's ranking on the ease of doing business index," said Dr Divine Ndhlukula, ZNCC President.
The conference will also focus on the progress made towards the ratification of the Africa Continental Free Trade Area Agreement (ACFTA) that was signed in Kigali, Rwanda earlier this year and the economic outlook post-elections.
A high-powered Japanese business delegation and representatives from the African Union are among the delegates from outside Zimbabwe set to grace this year's conference.


Their Majesties of Avacal, King Arnsbjorn and Queen Inga, wish to invite the populace of Avacal and beyond to witness the Coronation of their heirs, Crown Prince Albrecht von Rugen and Crown Princess Nasheeta al-Rakkasa!
your SCA name, Modern name, membership card number, expiry date and if your membership card is blue or white.
Feast Registration: Adults $10.00, Youth (12-16) $5.00, Children under 12 free.
Site opens Friday at 5pm and closes Sunday at 1pm!
Alberta either from Lethbridge or Medicine Hat (suggestion the latter).


Fungi are of primary value within the terrestrial surroundings. they've got roles as decomposers, plant pathogens, symbionts, and in elemental cycles. Fungi are usually dominant, and in soil can contain the most important pool of biomass (including different microorganisms and invertebrates). in addition they play a task in upkeep of soil constitution as a result of their filamentous progress behavior and exopolymer creation. regardless of their vital roles within the biosphere, fungi are usually missed inside broader environmental and microbiological spheres. also, mycological pursuits will be just a little fragmented among conventional topic limitations. This multi-disciplinary quantity explores the jobs and value of fungi within the atmosphere. specific emphasis is given to significant examine advances made in recent times due to molecular and genomic ways, and in telephone imaging and biology. Drawing jointly microbiologists, mycologists, and environmental scientists, this paintings is a distinct account of recent environmental mycology, and a pivotal contribution to the sector.
For nearly 4 billion years, microbes had the primordial oceans all to themselves. The stewards of Earth, those organisms reworked the chemistry of our planet to make it liveable for crops, animals, and us. Life's Engines takes readers deep into the microscopic global to discover how those really good creatures made lifestyles on the earth possible--and how human lifestyles at the present time may stop to exist with out them.
Mobile fusions are vital to fertilization, placentation, improvement of skeletal muscle and bone, calcium homeostasis and the immune defence approach. also, phone fusions perform tissue fix and should be very important to melanoma improvement, development and treatment. numerous elements keep an eye on mobilephone fusions, together with receptors and ligands, membrane area organizing proteins, proteases, signaling molecules and fusogenic proteins that carry membranes shut jointly.
Fungi are of basic value within the terrestrial atmosphere. they've got roles as decomposers, plant pathogens, symbionts, and in elemental cycles. Fungi are usually dominant, and in soil can include the most important pool of biomass (including different microorganisms and invertebrates). in addition they play a job in upkeep of soil constitution because of their filamentous progress behavior and exopolymer creation.
Autotrophic and methylotrophic microorganisms may be able to develop on the rate of one-carbon compounds (e. g. carbon dioxide, formaldehyde) because the important carbon assets for the synthesis of phone fabric, utilizing gentle, inorganic compounds or one-carbon compounds as strength resources. The examine of the specified variations required in cardio and anaerobic microorganisms to maintain an autotrophic or methylotrophic mode of lifestyles is an engaging box of analysis for scientists from numerous disciplines.
2) (Jedd & Chua, 2000). This suggests that a co-complex of HEX-1 isoforms may determine the oval morphology. In one model, the co-assembly of these isoforms may perturb monomer packing in the crystal lattice and through this mechanism alter the shape of the resultant assembly. This model is also consistent with observations showing that the oval form matures from an early hexagonal intermediate (Wergin, 1973). , 2004). Additional research is required to determine 2 Woronin bodies and the multicellular hypha 33 the extent to which Woronin body structural variation can be reduced to molecular variation in HEX-1.
2006). The role of wood decay fungi in the carbon and nitrogen dynamics of the forest floor. In Fungi in Biogeochemical Cycles, ed. G. M. Gadd, pp. 151–81. Cambridge: Cambridge University Press. 2 Natural history of the fungal hypha: how Woronin bodies support a multicellular lifestyle G R E G O R Y JE D D Temasek Life Sciences Laboratory, National University of Singapore Introduction Fungal evolution and cell biology Fungi are one of three major clades of eukaryotic life that independently evolved multicellular organization.
The assembly-defective protein accumulated in peroxisomes and produced a spherical, osmophilic core, suggesting that the process of HEX-1 import and concentration in the Woronin body lumen is unaffected by defects in crystallization. g. , 2003). , 2003). Why do these vesicles require a solid core? In a household sink, drain plugs require sufficient structural integrity to resist the pressure exerted by a column of water. Woronin bodies probably evolved a solid core for a similar reason. Returning to Buller's observation of plugged hyphal compartments, he noted that the two septa separating the dead cell from the two adjacent cells, which previously were plane, became convexo-concave with the convex side directed toward the lumen of the dead cell.


The ALSC has renamed the Laura Ingalls Wilder Award, citing Wilder's treatment of American Indians in her books.
The Laura Ingalls Wilder Award is now officially the Children's Literature Legacy Award, the Association for Library Service to Children (ALSC) announced this week.
The award, which honors authors and illustrators who have made a significant and lasting contribution to children's literature, was awarded to Wilder the first year it was created in 1954, after Wilder narrowly lost a chance at the Newberry Medal five different times. The award is one of the things that helped to make Wilder and the lovely, gripping Little House books she created into a national institution, the books that every child reads in elementary school.
But as the ALSC recognized on Monday, Wilder's books aren't just lovely and gripping. They aren't just detailed descriptions of what it's like to lay railway track or blow up a pig's bladder like a balloon and throw it around. They're also racist, riddled with depictions of American Indians as violent "savages" and with minstrel shows.


Marc-André Hamelin returned to Duke University for a recital in Baldwin Auditorium, presented under the auspices of Duke Performances' Piano Recital Series. The room was substantially full for a program that included several rare works, a lineup that, as realized, again demonstrated that Hamelin can play anything and in the process make convincing cases for everything he touches.
He turned next to two sonatas by Samuil Feinberg (1890-1962) (who as a pianist left us some impressive Bach of his own, available on YouTube). These sonatas, Nos. 5 and 6 of the dozen he composed (available on CDs here), received the passionate commitment we've come to expect from Hamelin, and if it seemed as if we were experiencing this music for the very first time, well, some of us were. At first hearing, they seemed somewhat amorphous and indeed more like fantasies than sonatas as we have come to understand them – which likely means they merit further study. (FWIW, William S. Newman does not mention them in his massive study of the sonata.) Still, they clearly spoke to some members of the audience, and there were flashes of recognizable genius therein, ranging from impressions of Impressionism to some of the mysticism we associate with Scriabin (see a recent CVNC article for much more information about this composer's use of color) to some of the angularity we encounter in late Prokofiev. For this listener, however, reservations about the works' form remain – and of course it would help to see the scores.
Compared with these pieces, Castelnuovo-Tedesco's Cipressi, Op. 17 (1920) [cited in the program as if it were Bach], emerged as a serious and sometimes weighty piece. We hear too little of his music – in an earlier generation, Segovia and Toscanini championed him – and this depiction of Tuscan landscape hinted at what we might be missing.
The formal finale was devoted to Chopin, in the form of the Polonaise-fantasy, Op. 61, and the Scherzo No. 4, Op. 54 – deep, profound music in both instances, the depths of which Hamelin plumbed with astonishing commitment. This is not your grandfather's Chopin, nor is it Chopin of any sort of popular ilk. This is deep stuff, and we were richly blessed to hear these pieces, to experience them as Hamelin revealed them – even if they demanded much of the listeners at the end of an already demanding program – and never mind that pianist, who can – and does – play everything.
At the end of this amazing recital, Rachmaninoff's Prelude, Op. 23/2, and Schumann's "Herberge," No. 6 of Waldszenen, Op. 82, brought the concert to a close.
This program was in many but not all respects identical to one he'd played two nights earlier in Carnegie Hall's Stern Auditorium. Time was when we had to go to New York to hear the best art. Thanks to Duke Performances (and other presenters), New York (and much, much more) now comes right here, to us! This series continues on Jan. 25 with Jeremy Denk. See our calendar for details.
There are two Great Commandments when it comes to concert presentation: Start on time. And the second is (as they say) like unto it: Do not further delay late-starting programs with extraneous remarks from the stage. When this one finally got underway, it was almost as late as Duke concerts used to begin, under the Ancien regime. 'nough said?


text:
Modern Louisville
'Bulleit Bites the Dust'
By Chris Hartman
Anyone who knows Hollis Bulleit knows she was the brand ambassador for her family's barrel-aged boon of a business. Her flawless 1920s, flapper style and signature smile helped build the Bulleit Bourbon brand and earned her the well-deserved title of "The First Lady of Bourbon."
It was also no secret that "The First Lady of Bourbon" had a first lady of her own. Her personal website even features cute, quirky comics of her adventures with her partner, Cher.
But as folks might have seen recently, Hollis has alleged she and Cher were pushed out of the family — and the business they built — over the last holiday season because they're part of the LGBTQ community.
Tom Bulleit, the bourbon's founder and Hollis' father, has defended his family and their business in a brief but widely-distributed letter to the editor — "Betsy and I firmly believe that neither bias nor prejudice should play a role in society." But it's hard to ignore the raw, authentic, textbook-pattern description of familial exclusion Hollis shares on her public Facebook page. It's the subtle, often silent discrimination LGBTQ people face over holidays and at important family gatherings.
She wrote in July, there's a "…misperception that homophobia within a familial construct is often misconstrued as private business. This perspective is the root of large spread public homophobia and leads to the acceptance of hate crimes. Since Cher and I have been together she has been excluded from 'the family business' from the start … In the course of ten years Cher and I spent less than three weeks time in the company of the Bulleit family. I was forced to deflect questions about my family at work events which put me in a horrible position not of my making; either through assumptions, omissions, or direct advice."
This is an all-too-familiar tale for LGBTQ people, whose anxiety can increase exponentially over the holidays, which results in the disproportionately high suicide attempt rates in our community during these times.
Hollis shared, "In 2008, I was asked to come home for Christmas; yet Cher was not invited. The only holiday that we attended was Thanksgiving in 2016, and then we were promptly uninvited via text from the following core family Christmas. And because family was business and intertwined with a global corporation, I find it odd that I did not benefit from the departments and safeguards that are put into place to either intervene or provide mediation or educational diversity training as would be the expected protocol for employees in this type of situation. In light of my experience, I do not understand how the company I worked for is on many of the top 10/ 50/ 100 'best places' for LGBTQIA employees to work."
Her tale is a stark reminder that even where LGBTQ discrimination protections may exist, if the culture of a business isn't truly inclusive, all the policies and procedures in place may not matter very much.
Shortly after Hollis' public postings, Bulleit's parent company, Diageo, released a statement refuting the situation: "It is unfortunate that we were not able to come to terms on the multi-year contract that we recently offered to Hollis. However, to insinuate that the failure to do so was due to bias of any kind is simply unfair and inaccurate. We are very proud of our long track record of work, through many of our brands, to support the LGBT community. We are appreciative of Hollis's past efforts on behalf of the brand and the industry."
For our part, the Fairness Campaign and Shelbyville Fairness reached out to Diageo back in February 2016, in anticipation of their new Shelby County Bulleit Distillery. We wanted to see if they might join the Kentucky Competitive Workforce Coalition, a group of Kentucky employers in support of a Statewide Fairness Law that would ban LGBTQ discrimination in the Commonwealth. We received a brief, polite response in March 2016: "As you can imagine, we're hard at work getting the distillery up and running so that we meet our goal of being operational by the end of this year. Right now, all of our efforts are focused on construction and start-up; however, once we're more settled, we'd welcome the opportunity to get to know you better."
But nothing from Diago since, despite our repeated follow-ups.
The brush-off may be nothing more than administrative backlog, but it doesn't help refute the accusations and perception Hollis has put into the public sphere.
Where is Bulleit Bourbon and Diageo on LGBTQ rights here in Kentucky, where they call home? If they're serious about LGBTQ inclusion, they should join their industry peers and sign on at KYCompetitiveWorkforce.com.
In the meantime, I feel we must believe Hollis and Cher. Their tale is a heartbreaking and too common narrative of our collective LGBTQ experience.
Instead of pouring Bulleit down the drain, we're partnering with our longtime supporters at Haymarket Whiskey Bar for a "Bulleit Bites the Dust" event on Wednesday, Oct. 4, where we'll imbibe the remainder of Haymarket's Bulleit stock to benefit the Fairness Campaign. •
Chris Hartman is director of the Fairness Campaign.
Published under News, Modern Louisville


Richard Kahn, Chief Executive of Altro, has today announced to its staff, customers and suppliers, the acquisition of Debolon, manufacturer of premium vinyl flooring based in Dessau, Germany.
"This acquisition is fully aligned to our Company Purpose, our core values and fully supports our Company vision and strategy.
Both Altro and Debolon have very similar family cultures where its staff, its customers and its suppliers are a key focus. Sustainability, innovation, quality and customer relationships are central ideals in both businesses. Coming together in Germany will mean that we can offer an even wider product portfolio and choices to our customers, incorporating higher design flooring and hygienic walling systems.
Altro is a UK based manufacturing company, selling premium flooring and hygienic walling systems for the construction and transport markets globally. They are family owned and run since 1919.
Debolon is a German based manufacturing company selling premium vinyl flooring globally to the construction market. They are family owned and were founded in 1929.


Pep Guardiola hopes to have Benjamin Mendy fit again after the international break.
Manchester City are expecting that the two-week break from the club game can allow players including Mendy, Fernandinho, John Stones and Kevin De Bruyne to build up their fitness for the final months of the season. Should the Blues make it past Swansea, they will have an FA Cup semi-final to go with their Champions League quarter-final and Premier League run-in in April alone, with a busy May following if their success continues.
City's coach has spent much of the past two years waiting for Mendy to be ready. Signed from Monaco in the summer of 2017, the French defender arrived injured and then suffered cruciate ligament damage in his knee in September of that year that forced him out for the majority of his debut campaign.
After linking up early with the squad with the World Cup, Mendy looked to have put his injury problems behind him and notched five assists in nine Premier League games to give a glimpse of how influential he can be to Guardiola's City.
However, another knee injury saw him out for two months from November and the left-back suffered a further setback after playing in the league cup at Burton. Mendy will not be fit for the trip to Swansea but City hope he will be back in time for their crunch period.
In his absence, Aleks Zinchenko has made the left-back spot his own with his all-action effort impressing the manager.
City also intend to press on with plans to sign a specialist in that position in the summer despite spending over £125m two summers ago. The unreliability of Mendy's fitness leaves the club looking for another answer in the position, with Leicester's Ben Chilwell among the options.


text:
Q: Minipage inside tabular is out of table I intended to add code inside a tabular environment using the minted package. I found out that using a minipage inside a tabular environment allows me to use the minted environment inside a table cell.
But the problem I'm facing now is, that the minted output always seems to be located outside of the table, no matter what options I choose. For better visualization of the problem, I added vertical lines between columns.
Example code:
    \begin{tabular}[t]{|l|l|}
        HTTP-Verb & \texttt{POST} \\
        Path & \texttt{/users} \\
        Request-Body &
        \begin{minipage}[t]{0.3\textwidth}
            \begin{minted}{js}
                {
                    "username": "string",
                    "mail": "string"
                }
            \end{minted}
        \end{minipage} \\
        Response-Body & - \\
        Description & \makecell{Creates a new user.}
    \end{tabular}

Output:

I also noticed that the next rows content of the same column the minipage is at, is affected by the minipages width.

A: Torbjørn T.'s comment lead to the solution of my problem. Quote:

Remember that spaces are preserved inside a minted environment, so if you have 15 spaces at the start of a line ...

Right at the beginning of the debugging process I suspected this to be the issue. So I added the showspaces option of the minted package without getting any changes in the output. I should have just removed the spaces, since this resolved the problem. Silly me...

redpajama_set_name: RedPajamaStackExchange


Mahindra Tsubaki Conveyor Systems (MTC) is committed to develop, maintain and continually improve ccupational Health, Safety & Environment Management Systems.We treat 'Environment' & 'Safety' as an integral part of our business activities. We recognize people as the most important asset and commit to provide a Safe, Healthy and Pollution free working environment for those who are working in the premises/precincts, visiting our operation/premises as business associates and common visitors as well.
To comply with legal and other requirements.
To recycle and reuse of waste wherever possible.
To increase awareness and participation of employees/contractors.
To provide and maintain Equipments and Systems of work, which as far as reasonably practicable and encompassing all available knowledge and information, are safe and without risk to health of employees.
To operate and maintain plants within the designated Safety, Occupational Health & Environmental criteria throughout their working life.
continual improvement of environmental performance.


text:
Cue is a clothing store that was established in 1968. Its first store opened in the Strand Arcade, Sydney, Australia. Cue predominantly sells modern fashion from London and exclusive prints.

History
Cue was opened in 1968 and is still wholly owned by the founding family. Designs seen in Cue stores are ones that originate from Cue's head office in Surry Hills, Sydney, and most clothing is made in Australia. Cue works with fabric mills in Europe to develop fabrics to the design teams' own specifications and print designs.

Cue has stores in all major cities throughout Australia and New Zealand, with a store in each Myer, the exclusive department store home of Cue, nationally

Awards

2011
 Prix de Marie Claire Awards – Winner Best Australian Fashion Brand
 Grazia Fashion Awards – Winner Favourite Overall High Street Chain

2010
 Myer Supplier of the Year Awards – Winner Community Services Award

2009
 TNT Ragtrader Fashion Retailer of the Year Awards – Winner Fashion Retailer of the Year
 Prix de Marie Claire Awards – Finalist Best Australian Fashion Brand

2007
 Marie Claire Prix de Excellence Awards – Winner Best Australian Fashion Brand
 Harper's Bazaar Online Style Awards – Winner Best Australian Fashion Store
 Myer Supplier of the Year Awards – Winner Supplier of the Year
 Myer Supplier of the Year Awards – Winner Myer Concession – Store Within Our Store Award

2018

 Online Retail Industry Awards – Winner Best In-Store Initiative

2019

 Online Retail Industry Awards – Winner Best Multi-channel Retailer
 CX Awards - Winner Best Multi-channel Customer Experience

Notes

References

External links
 Cue Clothing Co. website

1968 establishments in Australia
Australian companies established in 1968
Clothing companies established in 1968
Clothing companies of Australia
Companies based in Sydney


The newly released MLS third quarter and year-to-date La Plata County real estate sales statistics shows our inventory crunch continues. The decrease in sales directly correlates to fewer listings available in the marketplace. In other words, less choice for buyers equals a decline in sales. Overall, residential sales are down 2.2% compared to 2017, and new listings are down by 2.8%. With prices continuing to climb (overall +6.2% in median price over 2017) buyers are delaying purchasing because they cannot find a suitable fit, they are pushed out of the market, or they think prices are too high.
In-town Durango single-family sales are down 7.5% compared with 2017 and the median price has climbed 11% to $484,500.
In the City of Bayfield, 2017 numbers included many spec homes at an affordable price-point purchased before completion, encouraging sales. With the cheapest of the lots now gone, new sales have fallen from 10 YTD 2017 (many under 300k) to 1 in 2018. Because of this, in-town Bayfield sales are -39% in number but Bayfield rural climbing 8.6%.
Up in the North County, number of sales increased from 6 through three quarters in 2017 to 22 this year, indicating possible overflow from Buyers not finding suitable inventory in rural Durango.
Our luxury market is holding steady from 2017, with an almost equal amount of sales and recent quality showings on high-end listings.
Nationally, homebuyer sentiment is increasing, along with the mortgage rates which stand around 5%. We read about a change coming in the market but have not experienced it yet locally.
Find your copy of the stats release HERE.