text:
Twitter joins IPL 2022 party, tests dedicated cricket tab on Android
The feature, which will be available to select users, will help cricket enthusiasts discover relevant, exclusive, and Twitter-first content, the company has said
Vikas SN
March 29, 2022 / 05:48 PM IST
Twitter is testing a dedicated cricket tab on its Android app for select users in India starting this week, the company said on March 29, looking to capitalise on the ongoing domestic cash-rich Indian Premier League (IPL) tournament.
The tab, which will be available through the app's discovery-focused Explore tab, would be a single landing point that will help cricket enthusiasts discover relevant, exclusive, and Twitter-first content, the company said.
"Through our new cricket experiment offering an immersive experience, we want to make it easier for India's cricket fans to find relevant and engaging content as well as encourage conversations and find latest updates including the score all in one place" said Shirish Andhare, Director of Product for Twitter India.
The micro-blogging site said that about 4.4 million Indians on Twitter shared 96.2 million Tweets about cricket between January 2021 and January 2022.
"We learn from the conversations constantly unfolding on the service and they motivate us to build more for our audiences here. Since cricket's biggest fan bases extend way beyond their teams' home stadiums in India, and everyone can't be at every match, Twitter has been an incredibly important part of how the cricket storyline gets out there to fans not just in India but all over the world" Andhare said.
Twitter co-founder says Elon Musk 'doesn't seem like' the right person to own company
Shiv Visvanathan writes on tech layoffs: The real sadness is our entrenched silence to a tragedy
Playing to gallery
For users who have access to this tab, they will be able to view content in various formats including an events page that will help them follow the most recent Tweets and updates from the field on a dedicated page at the top of the cricket tab.
A live scorecard feature, which was introduced by Twitter in October 2021, will appear on the cricket tab as well as on the Events page.
Users will also get access to interactive content widgets like top players and team rankings along with latest information on what their favourite players are up to in real time during an ongoing match, the company said.
Twitter has also partnered with broadcast and creator partners such as Star Sports, Cricbuzz and commentator Boria Majumdar to provide video content, including in-match moments, highlights, and off-field action.
Other features include dedicated Twitter Lists of prominent cricket teams and players and a new IPL topic, available under the cricket tab, for fans who intend to follow conversations "a lot more closely".
Twitter also recently introduced custom IPL team emojis in English and eight Indian languages—Hindi, Bengali, Gujarati, Kannada, Marathi, Punjabi, Tamil and Telugu.
In October 2021, the social media firm had extended its Communities feature to India with the launch of a cricket-focused community ahead of T20 World Cup.
Tags: #Cricket Twitter #IPL 2022 #social media #Twitter #Twitter India
first published: Mar 29, 2022 05:48 pm


text:
Q: Too narrow kerning with xeCJK I'm trying to typeset Japanese using xelatex and the xeCJK package. For some reason, small kana cause odd kerning in the text. In the following example, I get too narrow spacing between two small kana, and between a small kana and a punctuation. It is also clear that they do not have the same width as regular kana.
%!TEX program = xelatex
\documentclass{article}
\usepackage{xeCJK}
\setCJKmainfont{IPAMincho}
\begin{document}
ぎゅっと持った。あっ、そうかも。

あいうえおかきくけこさしすせそ
\end{document}

Compiling, I get the following:

I tried several fonts with the same results. Am I missing some important definitions? Am I mistaken in believing that small kana should also be monospaced as the regular-sized ones?

A: It seems that you found a bug of xeCJK.
The bug (Issue 166) was rather old (since ver 2.x). The conditional Japanese starter characters (ゃ, ゅ, ょ, っ, etc.) were wrongly identified as full right punctuations to prevent line breaks. But there would be extra negative kernings sometimes.
We will fix the bug and release a new version of xeCJK soon.

Update
Qing Lee has uploaded a new version of xeCJK (2014/12/26 v3.3.0), which disables some of the Kinsoku (禁則) to fix this issue. You can upgrade your TeX distribution to use the new package.
Before installing the new release of xeCJK, you can use zxjatype package which is optimized for Japanese typesetting and provides a workaround:
% !TEX program = XeLaTeX
% !TEX encoding = UTF-8
\documentclass{article}
\usepackage{xeCJK}
\usepackage{zxjatype}
\setCJKmainfont{IPAMincho}
\begin{document}
ぎゅっと持った。あっ、そうかも。

あいうえおかきくけこさしすせそ
\end{document}

A: I managed to compile the example using lualatex. The modified file is
\documentclass{article}
\usepackage{luatexja-fontspec}
\setmainjfont{IPAMincho}
\begin{document}
ぎゅっと持った。あっ、そうかも。

あいうえおかきくけこさしすせそ
\end{document}

Strictly speaking, luatexja-fontspec is not necessary here, as fontspec would handle changing the font equally well. But this is only true for this tiny example. In the actual document, luatexja was needed to handle line-breaks.
One other point that is worth mentioning: In the actual document, I also made use of the ruby package. That worked in xelatex, but for lualatex, the package luatexja-ruby should be used instead.

redpajama_set_name: RedPajamaStackExchange


text:
The day after: Lawrence "trending in right direction"
Jaguars News | Jacksonville Jaguars - jaguars.comJohn Oehser
JACKSONVILLE - Trevor Lawrence wasn't the only story Monday, but he was the biggest story.
While the Jaguars face multiple concerns following the season's most one-sided loss to date, the starting quarterback's health was Topic No. 1 a day later.
Lawrence sustained left toe and left foot injuries in a loss to the Detroit Lions Sunday, and Head Coach Doug Pederson on Monday said the team is "not sure yet" regarding Lawrence's status for this Sunday's game against the Tennessee Titans in Nashville, Tenn.
"We're still getting more testing done," Pederson said a day after the Jaguars' 40-14 loss to the Lions at Ford Field in Detroit, Mich.
Pederson said he believes Lawrence will play against Tennessee, and that "things are trending in the right direction." Lawrence sustained the injuries when sacked by Lions linebacker James Houston on the final play of the first half Sunday.
"I won't have a definite until I get the report later today on his foot," Pederson said.
Pederson was asked Monday if it was better that Lawrence injured his left foot rather than his right - plant - foot.
"I really wish it wasn't either one," he said, smiling.
Pederson on Monday also discussed multiple issues from the loss to the Lions - in which the Jaguars allowed a season-high points and dropped multiple passes to lose by their widest margin of the season.
A victory would have moved the Jaguars to within two games of first place in the AFC South with the Titans (7-5) losing to the Philadelphia Eagles.
"We just didn't play well Sunday at all," Pederson said. "It just comes down to who wants it a little more on game day, doing your job and owning up to your responsibilities and trying to take care of business. We didn't do that Sunday."
Pederson added, "Quite frankly, the whole thing is frustrating. I think it was a culmination of a lot of things - not just the defense."
The Jaguars allowed 437 total yards and 31 first downs Sunday, with the Lions scoring on every possession until kneeling to run out the clock on the final possession. The Jaguars have allowed 415 or more yards in each of the last three games and have allowed 20 or more points in seven consecutive games.
"I wouldn't say it's getting worse," Pederson said of the defense. "I just think Sunday was a culmination of everything. It's just not all coming together. There are some bright spots. There are some good plays being made. Guys are giving great effort. I never question that aspect. It comes down to a little bit of ownership with the coaches, with the players, understanding responsibility, understanding your job, where you need to be there and when you need to be there.
"This was not all on the defense Sunday. The offense had their share in this as well."
The Jaguars' first seven losses of the season were by a combined 51 points with six decided by less than a score in the final minutes. Pederson was asked Monday if Sunday's loss was the season's most frustrating to date.
"By far, yeah," he said. "By far."
Pederson called Sunday "unacceptable" a day after Lawrence called it "embarrassing." Pederson said he typically doesn't motivate by yelling, saying Monday "you pick your spots" to make an impression when speaking to players.
"These guys understand what we've done as a football team, and where we are," Pederson said. "For me, it's not about yelling and screaming and trying to get their attention that way but still having serious hard conversations.
"The more you yell and scream and try to put a lot of pressure and heat, they get tight and tense and don't play well. I still want them to be loose and play and have fun, but at the same time, they have to understand this is a disciplined game. We have to be more disciplined than they were Sunday."
Pederson called Sunday's game at Tennessee "important for the psyche of the football team."
"There are definitely things we can learn from," he said. "You want to sort of burn the film and move on, but you still have to go through it with them and fix the mistakes, make the corrections and still point out some of the good that was there. You can't be all negative. You still want to be able to correct some of the positives as well.
"You don't want too many of these, obviously, but we can definitely learn from it."
NOTABLE I
Pederson on Monday said rookie inside linebacker Chad Muma will be day to day this week with an ankle injury sustained Sunday, and also said safety Andre Cisco will be day to day after missing Sunday with a shoulder injury. Cisco had been listed as questionable on last week's final injury report. "I just didn't want to put him in a situation where it set him back further," Pederson said of Cisco. "He's trending in the right direction. Hopefully we get him back this week."
📝 Game Report: Lions 40, Jaguars 14📸 Through the Lens | JAX vs. DET | Week 13👟 2022 My Cause My Cleats: See them in action in Week 13📝 Five Key Plays: Lions vs. Jaguars

redpajama_set_name: RedPajamaCommonCrawl


text:
The Little Mermaid - April 2023
Friday, April 21, 2023. 7:30 PM
Saturday, April 22, 2023. 7:30 PM
Sunday, April 23, 2023. 3 PM
TICKET INFO TO COME LATER
Little Mermaid Cast List
Nordonia Drama will be going Under the Sea with our spring musical, as we produce Disney's The Little Mermaid!!
Based on one of Hans Christian Andersen's most beloved stories and the classic animated film, Disney's The Little Mermaid is a hauntingly beautiful love story for the ages. With music by eight-time Academy Award winner, Alan Menken, lyrics by Howard Ashman and Glenn Slater and a compelling book by Doug Wright, this fishy fable will capture your heart with its irresistible songs, including "Under the Sea," "Kiss the Girl" and "Part of Your World."
Ariel, King Triton's youngest daughter, wishes to pursue the human Prince Eric in the world above, bargaining with the evil sea witch, Ursula, to trade her tail for legs. But the bargain is not what it seems, and Ariel needs the help of her colorful friends, Flounder the fish, Scuttle the seagull and Sebastian the crab to restore order under the sea.
For More Information on The Little Mermaid the musical:


text:
The Stonewall Legacy Project: Queer Latinx Short Films
« Queens Dance Party with Mambo Blue
RED RADIO: An Evening of Modern Dance & Tap »
Queer Latinx Short Films
Thursday, July 15, 7PM ET
Featured Artists: Dominic Colòn (he/him/his) & Francisco Cabrera-Feo (he/him/his)
Crush by Dominic Colòn
This HBO award winning short film is celebrating its 10th Anniversary. CRUSH, tells the story of a closeted queer Latinx kid from the Bronx who comes out to his crush at Prom. CRUSH won the 2011 HBO Short Film Script Competition and played on HBO and on 5 different continents. SKIN played several film festivals including the HBO/New York Latino Film Festival.
SKIN by Dominic Colòn SKIN is about a gay Latinx man who after extreme weight loss is scared to have sex with his boyfriend without a shirt on.
YUNIOR by Francisco Cabrera-Feo A deeply closeted teenage boy finds himself shaken when his older brother's girlfriend shows up with an out and confident gay male friend.
POR TI by Francisco Cabrera-Feo After ditching school with his secret boyfriend, Julian realizes he might not be the only one in his family hiding the truth.
Dominic Colón
Dominic Colón (he/him) is an award-winning actor, writer, and producer from The Bronx. Dominic's half hour pilot PAPI, just made the inaugural Latinx TV List, a curated list of the ten most promising pilots created by Latinx Writers. Dominic's screenplay for his short film CRUSH won the HBO/New York International Latino Film Festival Short Film Script Competition. Dominic's second short film: SKIN, a comedy about a young man who, after extreme weight loss, is afraid to have sex with his boyfriend without a shirt, won the BRIO Award for screenwriting. Dominic's play The War I Know has been developed by The Latinx Playwrights Circle & LAByrynth Theater Company. Dominic is also a professional teaching artist, having taught acting and playwriting at Riker's Island Academy, and at various high schools, hospitals, and juvenile justice facilities throughout New York City. Currently, he is the lead programmer for the HBO/New York Latino Film Festival and host of DULCE, a talk show where he interviews Latinx celebrities working in film and television. Dominic is a two-time National Association of Latino Independent Producers Screenwriting Fellow, and was recently named one of 3 finalists of the inaugural Latinx List to recieve a blind pilot deal from Hulu.
Francisco Cabrera-Feo
Francisco Cabrera-Feo (he/him) is a bisexual, bilingual, and bittersweet writer raised by an activist mother in war-torn Venezuela before emigrating to South Florida at age 11. He's currently a staff writer on a new HBOmax show, co-wrote an episode on Season 2 of NETFLIX'S Peabody nominated GENTEFIED, wrote on a NETFLIX project celebrating Queer Latinxs, and has recently been accepted into Sony's 2021 Diverse TV Writers Program. He writes for the closeted chubby kids who wore t-shirts in the pool and were forced into Catholic school even if the body of Christ compelled them. He has premiered films at OUTFEST and HBO'S NYLFF, yet despite PEOPLE MAGAZINE calling him "The New Promise of Latino Cinema," his mother still thinks he's a gringo.


What an absolutely wonderful high resolution, HD desktop background wallpaper we have here for you today ladies and gentlemen. As you may have already been aware of we here at high resolution, HD wallpapers have been publishing some fantastic HD wallpapers in relation to the ongoing Brazil based Fifa football 2014 World Cup. So as you may already know today is the Third Place decider and it's kicking off at 9pm in the British Summer Time zone. In this game we have two great teams but weren't great enough for this final, so we have the home nation of Brazil competing against the Netherlands. So whichever nation you're rooting for be sure to download this fantastic desktop background wallpaper today.
Download Brazil Vs Netherlands Third Place Decider HD wallpaper for free.
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text:
\section{Introduction}

 Fractional Brownian motion on the whole real line (fBm  for short)  $B^{H}=\{B_{t}^{H},\, t\in\mathbb{R}\}$ of Hurst parameter $H$ is the best known centered Gaussian process with long-range dependence. Its covariance function is
\begin{eqnarray}\label{C1}
	\mbox{Cov}(B_{t}^{H},\,B_{s}^{H})= \frac{1}{2} [|t|^{2H}+|s|^{2H}- |t-s|^{2H}]
\end{eqnarray}
where $H$ is a real number in $(0, 1)$ and the case $H=\frac{1}{2}$ corresponds to the Brownian motion. It is the unique continuous Gaussian process starting from zero, the self-similarity and stationarity  of the increments are two main properties for which fBm enjoyed successes as modeling tool in finance and telecommunications.  Researchers have applied  fractional Brownian motion  to a wide range of problems, such as bacterial colonies, geophysical data, electrochemical deposition, particle diffusion, DNA sequences and stock market indicators \cite{AlexY} and \cite{Omer}. In particular, computer science applications of fBm include modeling network traffic and generating
graphical landscapes \cite{Nan} and  \cite{TWS}. The fBm was investigated in many papers (e.g. \cite{ AlexYES, Alo, GaoH, Dhhd, ManVa}). The main difference between fBm and regular Brownian motion is that the increments in Brownian motion are independent, increments for fBm are not.

In \cite{Bojd}, the authors suggested another kind of extension of the Brownian motion, called the sub-fractional
Brownian motion (sfBm for short), which preserves most properties of the fBm, but not the stationarity of the increments. It is a centered Gaussian process  $\xi^H=\left\lbrace \xi_t^H, \ t\geq 0\right\rbrace,$ defined  by:
\begin{eqnarray}
	\xi_t^H=\frac{B_t^H+B_{-t}^H}{\sqrt{2}}, \quad t\geq 0,
\end{eqnarray}
where $H\in(0,1)$. The case $H=\frac{1}{2}$ corresponds to the Brownian motion.

The sfBm is intermediate between Brownian motion and fractional Brownian motion
in the sense that it has properties analogous to those of fBm, self-similarity,  not Markovian but  the increments
on nonoverlapping intervals are more weakly correlated, and their covariance decays
polynomially at a higher rate in comparison with fBm (for this reason in  \cite{Bojd} is called sfBm). So the sfBm does not generalize the fBm.
The sfBm was investigated in many papers (e.g. \cite{BardBa, Bojd, Sgh, Tud}).

An extension  of the sfBm was introduced by Zili in \cite{ZiliMS} as
a linear combination of a finite number of independent sub-fractional Brownian motions.
It was called the mixed sub-fractional Brownian motion (msfBm for short). The msfBm
is a centered mixed self-similar Gaussian process and does not have stationary increments.
The msfBm do not generalize the fBm.

In \cite{Zili} Zili  introduced new model called the generalized  fractional Brownian motion (gfBm for short) which is an extension of both sub-fractional Brownian motion and fractional Brownian motion.  A gfBm with parameters $a, b,
	\ and\ H$, is a process $ Z^H=\left\lbrace Z_t^H(a,b), t\geq0\right\rbrace $
	defined by
	\begin{eqnarray}
		 Z_t^H(a,b)=aB_t^H+bB_{-t}^H, \quad t\geq0
	\end{eqnarray}
The gfBm was investigated in \cite{ZiliM} and  \cite{ElNot}. The gfBm generalize the sfBm but not the the  mixed fractional Brownian motion.

The  mixed fractional Brownian motion (mfBm for short)  is a linear combination between a Brownian motion and an independent fractional Brownian motion of Hurst parameter $H$. It was introduced by Cheridito \cite{Cherid} to present a stochastic model of
the discounted stock price in some arbitrage-free and complete financial markets. The mfBm is a centered Gaussian process starting from
zero with covariance function
\begin{eqnarray}
	\mbox{Cov}(N_t^H(a, b), N_s^H(a, b))=a^{2}(t\wedge s)+\frac{b^{2}}{2}\left(t^{2H}+s^{2H}-|t-s|^{2H}\right),
\end{eqnarray}
with $H\in(0,1).$ When $a=1$ and $b=0,$ the mfBm is the Brownian motion and when $a=0$ and $b=1,$  is the fBm.
 We refer also to \cite{ElNo, Cherid, ZiliMM, Thale} for further information on this process.

In this paper, we introduce a new stochastic model, which we call the mixed generalized fractional Brownian motion.

\begin{definition} A mixed generalized fractional Brownian motion (mgfBm for short) of parameters $a, b, c$ and $H\in (0, \, 1)$ is a centred
Gaussian  process $M^{H}(a, b, c)=\{ M_{t}^{H}(a, b, c),\, t\geq 0\},$ defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P}),$  with the covariance function

\begin{eqnarray}\label{C2}
C(t,s) =a^2(t\wedge s)+\frac{(b+c)^2}{2}(t^{2H}+s^{2H})-bc(t+s)^{2H}-\frac{(b^2+c^2)}{2}|t-s|^{2H}
 \end{eqnarray}
 where $t\wedge s=\frac{1}{2}\left( t+s-|t-s|\right)$.
\end{definition}

The mgfBm is completely different from all the extensions mentioned above. The  process $M^{H}(a, b, c) $ is motivated by the fact that  this process already introduced for specific values of $a$, $b$ and $c$.  Indeed  $M^{H}(a, b, 0) $ is the mixed  fractional Brownian motion and $M^{H}(0, b, c), $ is the generalized fractional Brownian motion. This why we will name $M^{H}(a, b, c) $ the mixed generalized fractional Brownian motion. It allows to deal with a larger class of modeled natural phenomena, including those
with stationary or non-stationary increments.

Our goal is to study the main stochastic properties of this new model, paying attention to the long-range dependence,  self-similarity,
increment stationary, Markovity and  semi-martingale properties.

\section{The main  properties}
Existence of the mixed generalized fractional Brownian motion  $M^{H}(a, b, c)$ for any $H\in (0, 1)$ can be shown in the following way, consider the process
\begin{eqnarray}\label{MH}
M_{t}^{H}(a, b, c) =aB_{t}+bB_{t}^{H}+cB_{-t}^{H},\quad t\geq0,
\end{eqnarray}
where $B=\{B_{t}, \,t\in\mathbb{R} \}$ is a Brownian motion and $B^{H}=\{B_{t}^{H},\, t\in\mathbb{R} \}$ is an independent fractional Brownian
motion with Hurst parameter $H\in(0, 1).$  \\

Using (\ref{C1}) and since  $B$ and $B^{H}$  are independent we obtain the following lemma.

\begin{lemma}
For all $s,\  t\geq0$, the process (\ref{MH})  is a centered Gaussian process with covariance function given by (\ref{C2}).
	\end{lemma}

	\begin{proof}
	  Let $s,\  t\geq0$ and $C(t, s)= Cov\left( M_t^{H}(a, b, c), M_s^{H}(a, b, c)\right).$ Then
		\begin{eqnarray*}
			C(t, s)&=&Cov\left[ \left(aB_{t}+bB_{t}^{H}+cB_{-t}^{H}\right),\left( aB_{s}+bB_{s}^{H}+cB_{-s}^{H} \right)\right]\\
			&=&a^2(t\wedge s)+b^2Cov(B_t^H,B_s^H)+bc[Cov(B_t^H,B_{-s}^H)]+
			cb[Cov(B_{-t}^H,B_s^H)]\\&+&c^2Cov(B_{-t}^H,B_{-s}^H)\\
			&=&a^2(t\wedge s)+\frac{b^2}{2}\left(t^{2H}+s^{2H}-|t-s|^{2H} \right)+\frac{bc}{2}\left( t^{2H}+s^{2H}-|t+s|^{2H} \right)\\
			&+&\frac{cb}{2}\left( t^{2H}+s^{2H}-|-(t+s)|^{2H} \right)+\frac{c^2}{2}\left( t^{2H}+s^{2H}-|-(t-s)|^{2H} \right)\\
			&=& a^2(t\wedge s)+\frac{b^2}{2}t^{2H}+\frac{b^2}{2}s^{2H}-\frac{b^2}{2}|t-s|^{2H}+\frac{bc}{2}t^{2H}+\frac{bc}{2}s^{2H}\\
			&=& a^2(t\wedge s)+\frac{(b+c)^2}{2}(t^{2H}+s^{2H})-bc|t+s|^{2H}-\frac{(b^2+c^2)}{2}|t-s|^{2H}.
		\end{eqnarray*}

Hence  the covariance function of the  process (\ref{MH})  is precisely $C(t,s)$ given by (\ref{C2}).
Therefore the $M^{H}(a, b, c)$ exists.
\end{proof}

	\begin{remarks}
		Some special cases of the mixed generalized fractional Brownian motion:
	\begin{enumerate}
  \item If $a=0, b=1, c=0,$ then $M^{H}(0, 1, 0)$ is a fBm.
\item If $a=0, b=c=\frac{1}{\sqrt{2}},$ then $M^{H}(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$
is a  sfBm.
\item If  $a=1, b=0, c=0,$ then $M^{H}(1, 0, 0)$ is a  Bm.
\item If $a=0,$ then $M^{H}(0, b, c),$ is a  gfBm.
\item If $c=0,$ then $M^{H}(a, b, 0),$  is a mfBm.
\item If $b=c,$ then $M^{H}(a, \frac{b}{\sqrt{2}}, \frac{b}{\sqrt{2}}),$  is the smfBm.
\end{enumerate}
	\end{remarks}
So the mixed generalized fractional Brownian motion is, at the same, a generalization of the fractional Brownian motion,  sub-fractional Brownian motion, the sub-mixed fractional  Brownian motion, generalized fractional Brownian motion,  mixed fractional Brownian motion and of course of the standard Brownian motion.

\begin{proposition}\label{p1}
		The mgfBm   satisfies the following properties:
\begin{enumerate}
\item For all $ t\geq0$,
\begin{eqnarray*}E\left(M_t^H(a,b,c) \right)^2 =a^2t+\left(b^2+c^2-(2^{2H}-2)bc \right)t^{2H} .\end{eqnarray*}
\item Let $0\leq s<t$. Then
	\begin{eqnarray*}
		E \left(M_t^H(a,b,c)-M_s^H(a,b,c)\right)^2 &=&a^2|t-s|-2^{2H}bc(t^{2H}+s^{2H})\\&+&(b^2+c^2)|t-s|^{2H}+2bc|t+s|^{2H}.
	\end{eqnarray*}

\item We have for all   $0\leq s<t$,
	\begin{eqnarray*}
		a^2(t-s)+\gamma_{(b, c, H)}(t-s)^{2H} \leq E \left(M_t^H(a,b,c)-M_s^H(a,b,c)\right)^2 \leq a^2(t-s)+\nu_{(b, c, H)}(t-s)^{2H}
	\end{eqnarray*}
where
\begin{eqnarray*}
\gamma_{(b, c, H)}=(b^2 + c^2-2bc(2^{2H-1}-1))\,\mathbf{1}_{\mathcal{C}}(b,c,H)+(b^2 + c^2)\,\mathbf{1}_{\mathcal{D}}(b,c,H),
	\end{eqnarray*}

\begin{eqnarray*}
\nu_{(b, c, H)}=(b^2 + c^2)\,\mathbf{1}_{\mathcal{C}}(a,b,H)+(b^2 + c^2-2bc(2^{2H-1}-1))\,\mathbf{1}_{\mathcal{D}}(b,c,H),
	\end{eqnarray*}
\begin{eqnarray*}
\mathcal{C}=\{(b,c,H)\in \mathbb{R}^{2}\times]0, 1[;  \;(H>\frac{1}{2}, \;bc\geq0 )\; or \;(H<\frac{1}{2}, \;bc\leq 0)\},
	\end{eqnarray*}
and
\begin{eqnarray*}
\mathcal{D}=\{(b,c,H)\in \mathbb{R}^{2}\times]0, 1[;\;  (H>\frac{1}{2}, \;bc\leq0)\;or\; (H<\frac{1}{2},\;bc\geq 0)\}.
	\end{eqnarray*}
\end{enumerate}

\end{proposition}
\begin{proof}
\begin{enumerate}

\item It is a direct   consequence  of   (\ref{C2}).

\item Let $0\leq s<t$ and $\alpha(t, s)=E\left( M_t^H(a,b,c)-M_s^H(a,b,c))^2\right)$. Then
	\begin{eqnarray*}
	\alpha(t, s)
		&=&E \left( M_t^H(a,b,c)\right) ^2\ 	+E\left( M_s^H(a,b,c)\right) ^2
-2E\left( M_t^H(a,b,c)M_s^H(a,b,c)
  \right) \\&=&a^2t+b^2t^{2H}+2bct^{2H}-2^{2H}bct^{2H}+c^2t^{2H}
+a^2s+b^2s^{2H}+2bcs^{2H}-2^{2H}bcs^{2H}\\&+&c^2s^{2H}-2a^2(t\wedge s)-b^2t^{2H}-b^2s^{2H}+b^2|t-s|^{2H}-bct^{2H}-bcs^{2H}+bc|t+s|^{2H}\\&-&cbt^{2H}-cbs^{2H}+cb|t+s|^{2H}-c^2t^{2H}-c^2s^{2H}+c^2|t-s|^{2H}\\
&=&a^2(t+s)-2^{2H}bc(t^{2H}+s^{2H})-2a^2(t\wedge s)+(b^2+c^2)|t-s|^{2H}+2bc|t+s|^{2H}    \\
&=&  a^2|t-s|-2^{2H}bc(t^{2H}+s^{2H})+(b^2+c^2)|t-s|^{2H}+2bc|t+s|^{2H}.
\end{eqnarray*}

\item It is a direct consequence of the second item of    Proposition \ref{p1} and  Lemma 3  in \cite{Zili}.
\end{enumerate}
 \end{proof}

\begin{proposition}\label{pr3}
 For all $(a, b, c)\in\mathbb{R}^{3}\setminus\{(0, 0,0)\}$ and $H\in(0,1)\setminus\{\frac{1}{2}\}$  the mgfBm is not a self-similar process.
 \end{proposition}
\begin{proof}
This follows from the fact that, for fixed $h>0$, the processes $\left\lbrace M_{ht}^H(a,b,c),\ t\geq0\right\rbrace $
and  $\left\lbrace h^HM_t^H(a,b, c),\ t\geq0\right\rbrace $ are Gaussian, centered, but don't have the same covariance function. Indeed
\begin{eqnarray*}
 C\left( ht, hs\right)
&=&
a^2(ht\wedge hs)+\frac{b^2}{2}\left((ht)^{2H}+(hs)^{2H}-|ht-hs|^{2H} \right)\\&+&\frac{bc}{2}\left( (ht)^{2H}+(hs)^{2H}-|ht+hs|^{2H} \right)
\\&+&\frac{cb}{2}\left( (ht)^{2H}+(hs)^{2H}-|-(ht+hs)|^{2H} \right)\\&+&\frac{c^2}{2}\left( (ht)^{2H}+(hs)^{2H}-|-(ht-hs)|^{2H} \right)\\
&=& a^2(ht\wedge hs)+\frac{b^2}{2}(ht)^{2H}+\frac{b^2}{2}(hs)^{2H}-\frac{b^2}{2}|ht-hs|^{2H}\\&+&\frac{bc}{2}(ht)^{2H}+\frac{bc}{2}(hs)^{2H}-
\frac{bc}{2}|ht+hs|^{2H}+\frac{bc}{2}(ht)^{2H}+\frac{bc}{2}(hs)^{2H}\\&-&\frac{bc}{2}|ht+hs|^{2H}+\frac{c^2}{2}(ht)^{2H}+\frac{c^2}{2}(hs)^{2H}-
\frac{c^2}{2}|ht-hs|^{2H}\\
&=& a^2h(t\wedge s)+h^{2H}\frac{(b+c)^2}{2}\left((t)^{2H}+(s)^{2H}\right)-bch^{2H}|t+s|^{2H}-h^{2H}\frac{(b^2+c^2)}{2}|t-s|^{2H}.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
Cov\left( h^HM_t^{H}(a,b, c),h^HM_s^{H}(a,b, c)\right)
&=&h^{2H}Cov\left( M_t^{H}(a,b, c),M_s^{H}(a,b, c)\right)\\
&=& a^2h^{2H}(t\wedge s)+h^{2H}\frac{(b+c)^2}{2}\left((t)^{2H}+(s)^{2H}\right)-bch^{2H}|t+s|^{2H}\\&-&h^{2H}\frac{(b^2+c^2)}{2}|t-s|^{2H}.
\end{eqnarray*}
Then the mgfBm is not a self-similar process for all $(a, b, c)\in\mathbb{R}^{3}\setminus\{(0, 0,0)\}.$
\end{proof}

\begin{remarks}
		As a consequence of Proposition \ref{pr3}, we see that:
	\begin{enumerate}
  \item $M^{H}(0, b, c)$ is a self-similar process for all $( b, c)\in \mathbb{R}^{2}.$
\item $M^{\frac{1}{2}}(a, b, c)$ is a self-similar process for all $( a, b, c)\in \mathbb{R}^{3}.$
\end{enumerate}
	\end{remarks}
Now, we will study the Markovian property.

\begin{theorem}
	Assume  $H\in(0,1)\setminus \left\lbrace {\frac{1}{2}}\right\rbrace , a\in \mathbb{R} \ and \ (b,c) \in \mathbb{R}^2\setminus\left\lbrace (0,0)\right\rbrace.$
Then $M^H(a,b,c) $  is not a Markovian process.
	\end{theorem}
	\begin{proof}
	The process $M^{H}(a, b, c)$ is a centered Gaussian.
Then, if $M_t^H(a, b, c)$ is a Markovian process, according to Revuz and Yor \cite{Revu}, for all $s<t<u$, we would have
\begin{eqnarray*}
	C(s,u) C(t,t)=C(s,t) C(t,u).
\end{eqnarray*}
We will only prove the theorem in the case where $a\neq 0$, the result with $a= 0$ is known in \cite{Zili}.  For the proof we follow the proof of Proposition 1 given in  \cite{Zili}.\\
Using Proposition \ref{p1} we get
\begin{eqnarray*}
C(s,u)&=&a^2 s+\frac{(b+c)^2}{2}(u^{2H}+s^{2H})-bc|u+s|^{2H}-\frac{(b^2+c^2)}{2}|u-s|^{2H},\\
C(t,t)&=&a^2t+\left(b^2+c^2-(2^{2H}-2)bc \right) t^{2H},\\
C(s,t)&=&a^2s+\frac{(b+c)^2}{2}(t^{2H}+s^{2H})-bc|t+s|^{2H}-\frac{(b^2+c^2)}{2}|t-s|^{2H},\\
C(t,u)&=&a^2 t+\frac{(b+c)^2}{2}(u^{2H}+t^{2H})-bc|u+t|^{2H}-\frac{(b^2+c^2)}{2}|u-t|^{2H}.
\end{eqnarray*}
In the particular case where $1<s=\sqrt{t}<t<u=t^2$, we have
\begin{eqnarray*}
	C(\sqrt{t},t^2)&=&a^2 t^{\frac{1}{2}}+\frac{(b+c)^2}{2}(t^{4H}+t^{H})-bc|t^2+t^{\frac{1}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|t^2-t^{\frac{1}{2}}|^{2H},\\
	C(t,t)&=&a^2t+\left(b^2+c^2-(2^{2H}-2)bc \right) t^{2H},\\
	C(\sqrt{t},t)&=&a^2 t^{\frac{1}{2}}+\frac{(b+c)^2}{2}(t^{2H}+t^{H})-bc|t+t^{\frac{1}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|t-t^{\frac{1}{2}}|^{2H},\\
	C({t},t^2)&=&a^2 t+\frac{(b+c)^2}{2}(t^{4H}+t^{2H})-bc|t^2+t|^{2H}-\frac{(b^2+c^2)}{2}|t^2-t^{\frac{1}{2}}|^{2H}.
\end{eqnarray*}
Then by using that,
\begin{eqnarray*}
	C(\sqrt{t},t^2) C(t,t)=C(\sqrt{t},t) C(t,t^2),
\end{eqnarray*}
we have
\begin{eqnarray*}
& &\left[a^2 t^{\frac{1}{2}}+\frac{(b+c)^2}{2}(t^{4H}+t^{H})-bc|t^2+t^{\frac{1}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|t^2-t^{\frac{1}{2}}|^{2H} \right]\\
&\times&\left[ a^2t+\left(b^2+c^2-(2^{2H}-2)bc \right) t^{2H}\right] \\
&=&\left[ a^2 t^{\frac{1}{2}}+\frac{(b+c)^2}{2}(t^{2H}+t^{H})-bc|t+t^{\frac{1}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|t-t^{\frac{1}{2}}|^{2H}\right]  \\&\times&\left[ a^2 t+\frac{(b+c)^2}{2}(t^{4H}+t^{2H})-bc|t^2+t|^{2H}-\frac{(b^2+c^2)}{2}|t^2-t^{\frac{1}{2}}|^{2H}\right] .
	\end{eqnarray*}
It follows that
	\begin{eqnarray*}
		& &\left[a^2 t^{\frac{1}{2}}+t^{4H}\left( \frac{(b+c)^2}{2}(1+t^{-3H})-bc|1+t^{-\frac{3}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{3}{2}}|^{2H} \right) \right]\\
		&\times&\left[ a^2t+\left(b^2+c^2-(2^{2H}-2)bc \right) t^{2H}\right] \\
		&=&\left[ a^2 t^{\frac{1}{2}}+t^{2H}\left( \frac{(b+c)^2}{2}(1+t^{-H})-bc|1+t^{-\frac{1}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{1}{2}}|^{2H}\right) \right]  \\&\times&\left[ a^2 t+t^{4H}\left( \frac{(b+c)^2}{2}(1+t^{-2H})-bc|1+t^{-1}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-1}|^{2H}\right) \right].
	\end{eqnarray*}
Hence
	\begin{eqnarray*}
a^4 t^{\frac{3}{2}}&+&a^2\left(b^2+c^2-(2^{2H}-2)bc \right)t^{2H+\frac{1}{2}}\\
&+&a^2\left( \frac{(b+c)^2}{2}(1+t^{-3H})-bc|1+t^{-\frac{3}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{3}{2}}|^{2H} \right)t^{4H+1}\\
&+&\left( \frac{(b+c)^2}{2}(1+t^{-3H})-bc|1+t^{-\frac{3}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{3}{2}}|^{2H} \right)\\
&\times&\left(b^2+c^2-(2^{2H}-2)bc \right)t^{6H}\\
&=&a^4t^{\frac{3}{2}}\\&+&a^2\left( \frac{(b+c)^2}{2}(1+t^{-2H})-bc|1+t^{-1}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-1}|^{2H}\right)t^{4H+\frac{1}{2}}\\
&+&a^2\left( \frac{(b+c)^2}{2}(1+t^{-H})-bc|1+t^{-\frac{3}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{3}{2}}|^{2H}\right)t^{2H+1}\\
&+&\left( \frac{(b+c)^2}{2}(1+t^{-H})-bc|1+t^{-\frac{3}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{3}{2}}|^{2H}\right)\\
&\times&\left( \frac{(b+c)^2}{2}(1+t^{-2H})-bc|1+t^{-1}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-1}|^{2H}\right)t^{6H}
	\end{eqnarray*}
Take $t^{6H}$ as a common factor, we get
	\begin{eqnarray*}
&&a^2\left(b^2+c^2-(2^{2H}-2)bc \right)t^{-4H+\frac{1}{2}}\\
	&+&a^2\left( \frac{(b+c)^2}{2}(1+t^{-3H})-bc|1+t^{-\frac{3}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{3}{2}}|^{2H} \right)t^{-2H+1}\\
	&+&\left( \frac{(b+c)^2}{2}(1+t^{-3H})-bc|1+t^{-\frac{3}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{3}{2}}|^{2H} \right)\\
	&\times&\left(b^2+c^2-(2^{2H}-2)bc \right)\\
	&=&a^2\left( \frac{(b+c)^2}{2}(1+t^{-2H})-bc|1+t^{-1}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-1}|^{2H}\right)t^{-2H+\frac{1}{2}}\\
	&+&a^2\left( \frac{(b+c)^2}{2}(1+t^{-H})-bc|1+t^{-\frac{1}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{1}{2}}|^{2H}\right)t^{-4H+1}\\
	&+&\left( \frac{(b+c)^2}{2}(1+t^{-H})-bc|1+t^{-\frac{1}{2}}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-\frac{1}{2}}|^{2H}\right)\\
	&\times&\left( \frac{(b+c)^2}{2}(1+t^{-2H})-bc|1+t^{-1}|^{2H}-\frac{(b^2+c^2)}{2}|1-t^{-1}|^{2H}\right).
\end{eqnarray*}

Therefore
\begin{eqnarray*}
	&&a^2\left(b^2+c^2-(2^{2H}-2)bc \right)t^{-4H+\frac{1}{2}}\\
&+&a^2[ \frac{1}{2}(b+c)^2\left( 1+t^{-3H}\right) -bc\left(1+2Ht^{-\frac{3}{2}}+H(2H-1)t^{-3}+\circ(t^{-3}) \right)\\
	&-&\frac{1}{2}(b^2+c^2)\left(1-2Ht^{-\frac{3}{2}}+H(2H-1)t^{-3}+\circ(t^{-3}) \right) ] t^{-2H+1}    \\
	&+&[ \frac{1}{2}(b+c)^2\left( 1+t^{-3H}\right) -bc\left(1+2Ht^{-\frac{3}{2}}+H(2H-1)t^{-3}+\circ(t^{-3}) \right)\\
	&-&\frac{1}{2}(b^2+c^2)\left(1-2Ht^{-\frac{3}{2}}+H(2H-1)t^{-3}+\circ(t^{-3}) \right) ] \\
	&\times &[b^2+c^2-(2^{2H}-2)bc]\\
	&=&  a^2[\frac{1}{2}(b+c)^2\left( 1+t^{-2H}\right) -bc\left(1+2Ht^{-1}+H(2H-1)t^{-2}+\circ(t^{-2}) \right)\\
	&-&\frac{1}{2}(b^2+c^2)\left(1-2Ht^{-1}+H(2H-1)t^{-2}+\circ(t^{-2}) \right)   ]t^{-2H+\frac{1}{2}}\\
	&+&a^2[\frac{1}{2}(b+c)^2\left( 1+t^{-H}\right) -bc\left(1+2Ht^{-1/2}+H(2H-1)t^{-1}+\circ(t^{-1}) \right)\\
	&-&\frac{1}{2}(b^2+c^2)\left(1-2Ht^{-\frac{1}{2}}+H(2H-1)t^{-1}+\circ(t^{-1})\right)  ]t^{-4H+1}\\
	&+&[\frac{1}{2}(b+c)^2\left( 1+t^{-H}\right) -bc\left(1+2Ht^{-\frac{1}{2}}+H(2H-1)t^{-1}+\circ(t^{-1}) \right)\\
	&-&\frac{1}{2}(b^2+c^2)\left(1-2Ht^{-\frac{1}{2}}+H(2H-1)t^{-1}+\circ(t^{-1})\right)  ]\\
	&\times&[\frac{1}{2}(b+c)^2\left( 1+t^{-2H}\right) -bc\left(1+2Ht^{-1}+H(2H-1)t^{-2}+\circ(t^{-2}) \right)\\
	&-&\frac{1}{2}(b^2+c^2)\left(1-2Ht^{-1}+H(2H-1)t^{-2}+\circ(t^{-2}) \right)   ].
	\end{eqnarray*}
	First case: $0<H<\frac{1}{2}$, $a\neq0$ and $b+c\neq0$. By Taylor's expansion we get, as $t\rightarrow\infty$,
\begin{eqnarray*}
&&a^2\left(b^2+c^2-(2^{2H}-2)bc \right)t^{-4H+\frac{1}{2}}
+a^2\frac{1}{2}(b+c)^2t^{-5H+1}\\
&&+	\frac{1}{2}(b+c)^2[b^2+c^2-(2^{2H}-2)bc]t^{-3H}\\
&&\approx a^2\frac{1}{2}(b+c)^2t^{-4H+\frac{1}{2}}
+a^2\frac{1}{2}(b+c)^2t^{-5H+1}
+ \frac{1}{4}(b+c)^4t^{-3H}.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
	&&a^2\left(b^2+c^2-(2^{2H}-2)bc \right)t^{-4H+\frac{1}{2}}
	+\frac{1}{2}(b+c)^2[b^2+c^2-(2^{2H}-2)bc]t^{-3H}\\
	&&\approx a^2\frac{1}{2}(b+c)^2t^{-4H+\frac{1}{2}}
	+ \frac{1}{4}(b+c)^4t^{-3H},
\end{eqnarray*}
which is true if and only if
\begin{eqnarray*}
	\frac{(b-c)^2}{2}-(2^{2H}-2)bc=0\quad \mbox{and} \quad a^2\frac{(b-c)^2}{2}-a^2(2^{2H}-2)bc=0.
\end{eqnarray*}
However, it is easy to check that $	\frac{(b-c)^2}{2}-(2^{2H}-2)bc>0$ and  $	a^2\frac{(b-c)^2}{2}-a^2(2^{2H}-2)bc>0$ for fixed $c, b$ and every real $a$.\\
	Second case: $0<H<\frac{1}{2}$, $a\neq0$ and $b+c=0$. By Taylor's expansion we get, as $t\rightarrow\infty$,
	\begin{eqnarray*}
		&&a^2[b^2+c^2-(2^{2H}-2)bc]t^{-4H+\frac{1}{2}}+a^2\left[-2Hbc+(b^2+c^2)H \right]t^{-2H-\frac{1}{2}}\\&&+\left[-2Hbc+(b^2+c^2)H \right]\times\left[b^2+c^2-(2^{2H}-2)bc\right]t^{-\frac{3}{2}}\\&&\approx  a^2\left[-2Hbc+(b^2+c^2)H \right]t^{-2H-\frac{1}{2}}+a^2\left[-2Hbc+(b^2+c^2)H \right]t^{-4H+\frac{1}{2}}\\&&+\left[-2Hbc+(b^2+c^2)H \right]^2t^{-\frac{3}{2}}.
	\end{eqnarray*}
	Hence
	\begin{eqnarray*}
		&&a^2[b^2+c^2-(2^{2H}-2)bc]t^{-4H+\frac{1}{2}}+\left[-2Hbc+(b^2+c^2)H \right]\times\left[b^2+c^2-(2^{2H}-2)bc\right]t^{-\frac{3}{2}}\\&&\approx a^2\left[-2Hbc+(b^2+c^2)H \right]t^{-4H+\frac{1}{2}}+\left[-2Hbc+(b^2+c^2)H \right]^2t^{-\frac{3}{2}} ,
			\end{eqnarray*}
	which is true if and only if $b=c=0$. This is a contradiction.\\
	Third case: $\frac{1}{2}<H<1$, $a\neq0$ and $ b-c\neq0$. By Taylor's expansion we get, as $t\rightarrow\infty$,
	\begin{eqnarray*}
		 &&a^2[b^2+c^2-(2^{2H}-2)bc]t^{-4H+\frac{1}{2}}+a^2H(b-c)^2t^{-2H-\frac{1}{2}}\\&&+H(b-c)^2[b^2+c^2-(2^{2H}-2)bc]t^{-\frac{3}{2}}\\&&\approx a^2H(b-c)^2t^{-2H-\frac{1}{2}}+
a^2H(b-c)^2t^{-4H+\frac{1}{2}}+ H^2(b-c)^4t^{-\frac{3}{2}}.
	\end{eqnarray*}
	Then
	\begin{eqnarray*}
		&&a^2[b^2+c^2-(2^{2H}-2)bc]t^{-4H+\frac{1}{2}}+H(b-c)^2[b^2+c^2-(2^{2H}-2)bc]t^{-\frac{3}{2}}\\&&\approx a^2H(b-c)^2t^{-4H+\frac{1}{2}}+ H^2(b-c)^4t^{-\frac{3}{2}},
			\end{eqnarray*}
	which is true if and only if
	\begin{eqnarray*}
		\left[b^2(1-H)+c^2(1-H)+(2-2^{2H}+2H)bc\right]=0.
	\end{eqnarray*}
	However, it is easy to check that $	b^2(1-H)+c^2(1-H)+(2-2^{2H}+2H)bc>0$ for fixed $c, b$ and every real $a$.\\
	Fourth case: $\frac{1}{2}<H<1$, $a\neq0$  and $b-c=0$. By Taylor's expansion we get, as $t\rightarrow\infty$,
\begin{eqnarray*}
	&&a^2\left(b^2+c^2-(2^{2H}-2)bc \right)t^{-4H+\frac{1}{2}}
	+	\frac{1}{2}(b+c)^2[b^2+c^2-(2^{2H}-2)bc]t^{-3H}\\
	&&\approx  a^2\frac{1}{2}(b+c)^2t^{-4H+\frac{1}{2}}
	+\frac{1}{4}(b+c)^4t^{-3H},
\end{eqnarray*}
which is true if and only if $2-2^{2H}=0.$ This contradicts the fact that  $H\neq\frac{1}{2}$.
\end{proof}

Let us check the mixed-self-semilarity property of the mgfBm. This property was introduced in \cite{ZiliMM} for  the mfBm and investigated to show the H\"{o}lder continuity of the mfBm. See  also \cite{Coco}   for the sfBm case.
	\begin{proposition}
	For any 	$h>0$, $ \left\lbrace M^H_{ht}(a,b,c)\right\rbrace  \stackrel{\bigtriangleup}{=} \left\lbrace M^H_{t}(ah^{\frac{1}{2}}, bh^{H}, ch^{H})\right\rbrace.$\\
where $\stackrel{\bigtriangleup}{=}$ "to have the same law".
 \end{proposition}
\begin{proof}
	For fixed $h>0$, the processes $\{M_{ht}^{H}(a,b,c)\}$ and $\{M_{t}^{H}(ah^{\frac{1}{2}},bh^H,ch^H)\}$ are Gaussian and centered. Therefore, one only have to prove that they have the same covariance function.
	But, for any $s$ and $t$ in  $\mathbb{R}_+$, since $B$ and $B^H$ are independent, then
\begin{eqnarray*}
 Cov\left( M_{ht}^{H}(a,b,c),M_{hs}^{H}(a,b,c)\right)
&=&a^2h(t\wedge s)+\frac{(b^2+c^2)}{2}\left[h^{2H}\left(t^{2H}+s^{2H}-|t-s|^{2H}\right)\right]\\
&+&bc\left[h^{2H}\left(t^{2H}+s^{2H}-|t+s|^{2H}\right)\right]\\
&=&Cov\left( M_{t}^{H}(ah^{\frac{1}{2}},bh^H,ch^H),M_{s}^{H}(ah^{\frac{1}{2}},bh^H,ch^H)  \right).
\end{eqnarray*}

\end{proof}
\begin{proposition} \label{pr2}
	For all $(a, b, c)\in \mathbb{R}^{3}\setminus\{(0, 0, 0)\},$ the increments of the $M^H(a, b, c) $ are not stationary.
	\end{proposition}
	\begin{proof}
		Let $(a, b, c)\in \mathbb{R}^{3}\setminus\{(0, 0,0)\}.$
	For a fixed $t\geq0$ consider the processes  $\{P_{t},\, t\geq0\}$ define by $P_t=M_{t+s}^H(a, b, c)-M_s^H(a, b, c).$
	Using Proposition \ref{p1} we get
	\begin{eqnarray*}
		Cov(P_t,P_t)]
		&=&E\left[(M_{t+s}^H(a, b, c)-M_s^H(a, b, c))^2 \right]\\
		&=&a^2(t+s+s)-2^{2H}bc(({t+s})^{2H}+s^{2H})-2a^2 s\\&+&(b^2+c^2)|t+s-s|^{2H}+2bc|t+s+s|^{2H} \\
		&=& a^2(t+2s)-2^{2H}bc(({t+s})^{2H}+s^{2H})-2a^2 s+(b^2+c^2)t^{2H}+2bc|t+2s|^{2H} .
	\end{eqnarray*}
	Using Proposition \ref{p1} we get
	\begin{eqnarray*}
		Cov(M_t^H(a, b, c), M_t^H(a, b, c))=a^2t+\left(b^2+c^2-(2^{2H}-2)bc \right)t^{2H}.
	\end{eqnarray*}
	Since  both processes are centered Gaussian, the inequality of covariance functions implies that $P_t$ does not have the same distribution as $M_t^H(a, b, c)$. Thus, the incremental behavior of $M^H(a, b, c)$ at any point in the future is not the same. Hence  the  increments of $M^H(a, b, c)$ are not stationary.
\end{proof}
\begin{remarks}
As a consequence of Proposition \ref{pr2}, we see that:
	\begin{enumerate}
		\item the increments of $M^{H}(0, b, c)$   are not stationary for all $( b, c)\in \mathbb{R}^{2}\setminus\{(0, 0)\}.$
		\item  the increments of $M^{H}(a, b, 0)$  are stationary for all $( a, b)\in \mathbb{R}^{2}.$
	\end{enumerate}
	\end{remarks}
\begin{proposition}
	\begin{enumerate}
		\item
Let $H\in (0, 1).$ The mgfBm admits a version whose sample paths are almost H\"{o}lder continuous of order strictly less than $\frac{1}{2}\wedge H.$
		\item When $b$ or $c$ not zero  and $H\in(0, 1)\setminus\{ \frac{1}{2}\}$   the mgfBm is not a semi-martingale.
\end{enumerate}

\end{proposition}
\begin{proof}
	\begin{enumerate}
		\item
Let $s$ and $t$ in  $\mathbb{R}_+$ and $ \alpha=2.$ The proof follows by Kolmogorov criterion from Lemma 3  in \cite{Zili} and using  Proposition \ref{p1} we get
\begin{eqnarray*}
		E\left( |M_t^H(a,b,c)-M_s^H(a,b,c)|^ \alpha\right) &=&a^2|t-s|-2^{2H}bc(t^{2H}+s^{2H})\\&+&(b^2+c^2)|t-s|^{2H}+2bc|t+s|^{2H}\\&\leq& C_{ \alpha} |t-s|^{ \alpha(\frac{1}{2}\wedge H)}\
	\end{eqnarray*}
where  $C_{ \alpha } =\left( a^2 +\nu(b,c, H)\right)$ and $\nu_{(b,c, H)}$  is given in Lemma 3  in \cite{Zili}.

	\item Suppose first that  $H<\frac{1}{2}$. We get from Proposition  \ref{p1}
\begin{eqnarray*}
E \left(M_t^H(a,b,c)-M_s^H(a,b,c)\right)^2 \geq \gamma_{(b,c, H)}(t-s)^{2H}.
\end{eqnarray*}
Since $2H<1$ and $\gamma_{(b,c, H)}>0$ then the assumption of Corollary 2.1 in \cite{BGT} is satisfied, and consequently  the mgfBm is not a semi-martingale.\\
Suppose now that  $H>\frac{1}{2}$. We get from Properties  \ref{p1}
\begin{eqnarray*}
a^{2}(t-s)+ \gamma_{(b,c, H)}(t-s)^{2H}\leq E \left(M_t^H(a,b,c)-M_s^H(a,b,c)\right)^2 \leq (a^{2}+\nu_{(b,c, H)})(t-s)^{1\wedge2H}
\end{eqnarray*}
then
\begin{eqnarray*}
 \gamma_{(b,c, H)}(t-s)^{2H}\leq E \left(M_t^H(a,b,c)-M_s^H(a,b,c)\right)^2 \leq (a^{2}+\nu_{(b,c, H)})(t-s)^{2H}.
\end{eqnarray*}
Since $1<2H<2$  and $\nu_{(b,c, H)}>0$ then the assumption of Lemma 2.1 in \cite{BGT} is satisfied, and consequently  the mgfBm is not a semi-martingale.
\end{enumerate}
\end{proof}

\section{ Long range dependence of the mgfBm }

 \begin{definition}
	We say that the increments of a stochastic process $X$ are long-range dependent if for every integer $p\geq1$, we have
	\begin{eqnarray*}
		\sum_{n\geq 1}R_X(p,p+n)=\infty,
	\end{eqnarray*}
	where $R_X(p,p+n)=E\left((X_{p+1}-X_p)(X_{p+n+1}-X_{p+n}) \right). $
\end{definition}
This property was investigated in many papers (e.g. \cite{AlexY, Casa, ChenXuHu, BardBa, GaoH}).
\begin{theorem}
	For every $a\in  \mathbb{R} $ and $(b,c)\in \mathbb{R}^2 \setminus\left\lbrace (0,0)\right\rbrace $ the increments of $M^H(a,b,c)$ are long-range dependent if and only if $H>\frac{1}{2}$ and $b\neq c$ .
	\end{theorem}
	\begin{proof}
		For all $n\in \mathbb{N}\setminus\{0\}$ and $p\geq 1$ we have
		\begin{eqnarray*}
			R_{M}(p,p+n)&=&E\left((M_{p+1}^H(a, b, c)-M_p^H(a, b, c))(M_{p+n+1}^H(a, b, c)-M_{p+n}^H(a, b, c)) \right)\\
			&=&E\left(M_{p+1}^H(a, b, c)M_{p+n+1}^H(a, b, c)\right)-E\left(M_{p+1}^H(a, b, c)M_{p+n}^H(a, b, c) \right)\\
			&-&E\left(M_{p}^H(a, b, c)M_{p+n+1}^H(a, b, c)\right)+E\left(M_{p}^H(a, b, c)M_{p+n}^H(a, b, c) \right)\\
			&=&C({p+1},{p+n+1})-C({p+1},{p+n})-C({p},{p+n+1})+C({p},{p+n})\\
			&=& a^2(p+1) +\frac{(b+c)^2}{2}\left( (p+1)^{2H}+(p+n+1)^{2H}\right) -bc(2p+n+2)^{2H}\\&-&\frac{(b^2+c^2)}{2}n^{2H}
			-a^2(p+1)-\frac{(b+c)^2}{2}\left( (p+1)^{2H}+(p+n)^{2H}\right) \\&+&bc(2p+n+1)^{2H}+\frac{(b^2+c^2)}{2}|n-1|^{2H}
			-a^2p\\&-&\frac{(b+c)^2}{2}\left( (p)^{2H}+(p+n+1)^{2H}\right) +bc(2p+n+1)^{2H}+\frac{(b^2+c^2)}{2}|n+1|^{2H}.
\end{eqnarray*}
Hence
\begin{eqnarray*}
			R_{M}(p,p+n)&=&
			\frac{(b^2+c^2)}{2}\left( (n+1)^{2H}-2n^{H}+(n-1)^{2H}\right)\\&-&bc\left((2p+n+2)^{2H}-2(2p+n+1)^{2H}+(2p+n)^{2H} \right).
		\end{eqnarray*}
		Then for every integer $p\geq1$, by Taylor's expansion, as $n\rightarrow\infty$, we have
		\begin{eqnarray*}
			R_M(p,p+n)&=&\frac{b^2+c^2}{2}n^{2H}\left[ \left( 1+\frac{1}{n}\right)^{2H}-2+\left( 1-\frac{1}{n}\right)^{2H} \right]
			\\&-&bcn^{2H}\left[ \left( 1+\frac{2p+2}{n}\right)^{2H}-2 \left( 1+\frac{2p+1}{n}\right)+\left( 1+\frac{2p}{n}\right)^{2H} \right] \\
			&=&H(2H-1)n^{2H-2}(b-c)^2\\&-&4H(2H-1)(H-1)bc(2p+1)n^{2H-3}(1+\circ(1)).
		\end{eqnarray*}
		If $b\neq c$, we see that as $n\rightarrow\infty$,
		\begin{eqnarray*}
			R_M(p,p+n)\approx H(2H-1)n^{2H-2}(b-c)^2.
		\end{eqnarray*}
		Then
		\begin{eqnarray*}
			\sum_{n\geq 1}R_M(p,p+n)=\infty \ \ \Leftrightarrow \ 2H-2>-1 \ \ \Leftrightarrow H>\frac{1}{2}.
		\end{eqnarray*}
		If $b=c$, then, as $n\rightarrow\infty$,
		\begin{eqnarray*}
			R_M(p,p+n)\approx 4H(2H-1)(H-1)a^2(2p+1)n^{2H-3}.
		\end{eqnarray*}
		For every $H\in(0,1) $, we have $2H-3<-1$ and, consequently,
		\begin{eqnarray*}
			\sum_{n\geq 1}R_M(p,p+n)<\infty.
		\end{eqnarray*}

\end{proof}
\begin{remarks}

	\begin{enumerate}
		\item  For all $a \in \mathbb{R}$ and $ b\in \mathbb{R}\setminus\{0\}$ the increments of $M^H(a,b,0)$ are long-range dependent  if and only if $H>\frac{1}{2}.$
	\item If $b=c=\frac{1}{\sqrt{2}}$ the increments of $M^H(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ are short-range dependent if and only if $H\in(0, 1).$ But if $b\neq c$
the increments of $M^H(0,b,c)$ are long-range dependent  if and only if $H>\frac{1}{2}.$

       \item From  \cite{Bojd},  the increments of $M^{H}(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ on intervals $[u,u+r],[u+r,u+2r]$ are more weakly correlated than those of $M^{H}(0, 1, 0)$.
       \item From  \cite{ZiliM}, If $H>\frac{1}{2}$, $b^2+c^2=1$ and $bc\geq0$, the increments of $M^{H}(0, b, c)$ are more weakly correlated than those of $M^{H}(0, 1, 0)$, but more strongly correlated than those of $M^{H}(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
       \item From  \cite{ZiliM}, If $H\geq\frac{1}{2}$, $(bc\leq0 \ and \ (b-c)^2\leq1)$ or $(bc\geq0 \ and \ b^2+c^2\leq1)$, the increments of $M^{H}(0, b, c)$ are more strongly correlated than those of both $M^{H}(0, 1, 0)$ and $M^{H}(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.

   \end{enumerate}
\end{remarks}

\newpage

redpajama_set_name: RedPajamaArXiv


text:
Q: Can a UICollectionView scroll vertically with a columnar layout? I am building  UICollectionView that needs to scroll vertically, but lay out its items in a columnar format in order to handle varying item heights, like this:

A column-by-column layout is the default for a horizontally-scrolling collection view, but mine needs to scroll vertically.  What's the easiest way to accomplish this?  Do I need to subclass UICollectionViewLayout, or can UICollectionViewFlowLayout do it?  Or...?

A: Thanks for the comments.  It turns out there is a subclass of UICollectionViewLayout called SelfSizingWaterfallCollectionViewLayout on github that does exactly what I needed.


Modern web applications need reliable communication between the servers and the clients in order to access information from databases or to insert user defined input into the applications. Even today, when the web sites are something completely different from what they were originally designed to be, they still rely on the original protocols. These protocols, e.g. HTML, have been updated a few times. The transition from HTML 4.1. to HTML5 introduced many new features and techniques, such as the WebSocket protocol. Auditing different protocols from the security perspective is one of the key methods for enhancing the reliability of the protocols under testing. The results provided by the testing often reveal vulnerabilities or at the very least suggestions for future development. These results are then assigned to the developers or the community and hopefully these issues are then addressed. In this thesis Design Science Research Methodology was used to research the WebSocket protocol and also a few commonly used server implementations for this protocol. Moreover, statistics on how widely WebSockets are used in web applications was also looked into. The research showed that the protocol in itself has dealt with the security aspect and that the protocol specification states clearly on how the protocol should work when applied according to the documentation. However, as there is a delicate balance between usability and security, the scale has favoured usability over security on a number of occasions by reducing the safety of the protocol to some degree.
© Juuso Karlström, 2016. This publication is copyrighted. You may download, display and print it for your own personal use. Commercial use is prohibited.


So you've set up containers for Nginx and for PHP – now you'd like to have a database. After all the work of the last two labs you'll be glad to hear than adding MariaDB to the mix is very simple – we can get it up and running in minutes.
Since Oracle bought MySQL, the open development of MySQL has been significantly curtailed. Because of this, MariaDB now enjoys a significantly more vibrant community, and is more open about sharing what changes have been made and why.
MariaDB is quicker than MySQL. Some, rather excitable, metrics show that MariaDB is massively faster than MySQL – it isn't. It's a smidgen faster – reckon on about 5%. But I'll take every little boost I can.
MariaDB is significantly compatible with MySQL – as you might expect since MariaDB was forked from MySQL. This doesn't mean that you can just take your MySQL database and load it into MariaDB without any issues – you'll still need to test thoroughly – but you'll experience significantly fewer problems than if you migrated to another DB.
It's worth noting that whilst you can administer MariaDB using MySQL Workbench I'd advise against it. There are lots of alternative tools available (including a web based admin package that you'll install at the end of this tutorial), but I like Navicat Premium Essentials – which is not free, but which is worth every penny.
Yet again, it will come as no surprise that, in order to install PHP, we need to edit the docker-compose.yml file. It will be necessary to stop Docker (just in case you left it running from the last session). You may remember that you can edit the docker-compose.yml file while Docker is still running – but the changes won't take effect until Docker is restarted. I shan't tell you how to stop the containers. You can look that up for yourself in Part 2.
Edit docker-compose.yml to add the maria-db container. I added it just below services and before nginx. Replace <Your Root Password Goes Here> with a password of your choice. Make it complicated – and remember that you probably aren't going to have to type it all that often, so it doesn't need to be convenient for your fingers. I generally use a UUID – something like this '1e1fdb9c-6ba5-48ea-9640-7801eed2022e'.
A useful online tool for administering your database and also for testing that MariaDB has been installed correctly is phpminiadmin.php – in fact this will exercise every part of your set up.
Password: whatever you specified in the Docker Compose file.
You will now have access to the phpMiniAdmin database management interface. You can use this interface to create databases and tables and execute SQL queries on your MariaDB database server.
If that all worked then next time we'll add certificates. If it didn't work, go back and check your working.
If you want to connect to this MariaDB instance from your desktop computer using a tool like Navicat Essentials, you need the following setup.
Click the SSH tab, check the 'Use SSH tunnel' checkbox and fill in the details that you use to SSH into the server.
docker inspect <containerID> | grep \"IPAddress\"
Get the Container ID using docker ps, as previously. Note that the IP Address may change periodically – so if you can't connect for whatever reason, check this first.


text:
Q: Website giving error page when trying to get webpage using python-request I am trying to get the HTML code from a Yahoo Finance page in order to do some web scraping stuff and create tables from data, etc.
The issue is that any website I try this with is redirecting me to the error/bot page so I am not being able to get the code or the actual page I need, I am using the Pycharm Python Editor and the request module
import requests
from bs4 import BeautifulSoup
url = 'https://finance.yahoo.com/cryptocurrencies'
page = requests.get(url)
soup = BeautifulSoup(page.text, 'lxml')
print(soup)

I can tell I am on the wrong page since the HTML has stuff like robot and error in it
Because of this I am unable to move forward in my project.
Note: it works with some websites just fine but with some others, I automatically get sent to the error page


text:
Local Benefits
About Wind
Kaban Green Power Hub is located near Ravenshoe in the Atherton Tablelands, 80 km south-west of Cairns.
The site was chosen because it is very windy and is located next to a 275kV high voltage transmission line which ensures the energy can rapidly be exported into the grid.
The 1,300 hectare site, which is on Jirrbal country and ranges over five freehold rural properties will have a relatively low visual impact on the surrounding community because it is surrounded on three sides by State Forests.
Lying at the northern fringe of the national network, Far North Queensland is in a weak part of the grid and doesn't have many power generation plants, however once the 157 MW wind farm is operating it will help to improve the strength of the grid system, as well as generating clean, affordable energy for Queenslanders.
The Jirrbal people are the Traditional Owners (TO's) of the land on which Kaban Green Power Hub is located are the Wabubadda Aboriginal Corporation (WAC) is the prescribed body corporate (PBC) representing the Jirrbal people.
The Jirrbal people maintain a strong connection to their country with many families still living and working on country.
The Jirrbal country between Atherton to Kirrama Homestead and from Milaa Milaa to Innot Hot Springs is under registered native title claims, the latest of which (made in 2015) is awaiting determination by the Federal Court of Australia.
In 2017, Neoen conducted a preliminary Cultural Heritage Assessment which identified the Jirrbal people as the Traditional Owners and since then the successful Neoen and Jirrbal negotiation has led to the executing of a Cultural Heritage Management Plan (CHMP) registered with the Queensland Government's Department of Aboriginal and Torres Strait Islander Partnerships (DATSIP) in 2019.
HELPING THE CLIMATE
Kaban Green Power Hub will provide 30 years of clean energy to Queensland
Why we need KABAN
Renewable energy project such as Kaban keep the lights on in Queensland.
Energy to power around 95,905 homes
No greenhouse gas emissions
Cleaner air
ABOUT NEOEN
© 2020 Neoen | Website Design


text:
Crescent Spring and Echoing-Sand Mountain
Dunhuang City Guide
Dunhuang Introduction
Yumen Pass
Dunhuang Museum
Mogao Grottoes
Yardan National Geological Park
Dunhuang Photo Gallery
Dunhuang Travel Blog
Basic Facts of Crescent Spring and Echoing-Sand Mountain
Chinese Name: 月牙泉和鸣沙山 Pronunciation: Yuèyá Quán Hé Míngshā Shān
English Name: Crescent Spring and Echoing-Sand Mountain
Height: 1,715 meters
Tourist Area: 76.82 square kilometers
Best Travel Time: April to October
Address: 5 km southwest of Dunhuang City, Gansu Province, China
Admission Ticket Fare
Time Peak Season
(April 1st – October 31st ) Shoulder Season
(November 1st –March 31st )
Adult 120 yuan 60 yuan
Child (Under 18 years old between 1.2m-1.4m) 60 yuan 30 yuan
Senior (Between 60-69 years old) 60 yuan 30 yuan
Surcharge Inside the Scenic Spot
Contents Price
Camel Riding in the Desert
100 yuan per person
120 yuan for an adult and a child under 1.4m
Sand Skiing 20 yuan per person
Desert Motorcycle 120 yuan per person
Driving SUV in the Desert 300 yuan per vehicle (1-3 people)
By Helicopter 1280 yuan for one person
1760 yuan for two adults, or for two adults with a child under 1.2m, or for one adult with two children under 1.2m
2000 yuan for three person
By Glider No.1 Line: 380 yuan per person
No.2 Line: 480 yuan per person
June 1st – August 19th 5:00 -20:30
August 20th – October 9th 6:00-19:30
October 10th – November 9th 6:30-18:30
November 10th – March 9th 7:30-18:00
March 10th – May 31st 7:00-19:00
Introduction of Crescent Spring and Echoing-Sand Mountain
The Land Covered with Reeds
The scenic spot, Crescent Spring and Echoing-Sand Mountain, is five kilometers away from south of Dunhuang City. With sand and springs coexisting, it is a wonderful creation from heaven, known as the "desert wonders" in the world through the ages.
Echoing-Sand Mountain is a national key tourist scenic spot in the northwest of China, and is a magical and magnificent tourist attraction in Gansu province on the Silk Road. Located in the Badain Jaran Desert and Taklamakan Desert transition zone, it covers an area of about 200 square kilometers. There are two peculiar things about Crescent Spring and Echoing-Sand Mountain: if a person slides down from the top, the sand under his feet will hum; the footprints left by people climbing a sandy mountain during the day will not be found the next day. Echoing-sand Mountain undulates and glitters like a dragon, and it also looks like a golden mountain. Echoing-Sand was once called "Sand Cape Mountain".
Crescent Spring is located in the Crescent Spring scenic area, also called "sand well" and "medicine spring" in ancient times, embraced by the Echoing-Sand Mountain.
Crescent Spring is nearly 100 meters long from north to south and about 25 meters wide from east to west. The spring is shallow from east to west and about 5 meters deep at the deepest point. It is curved like a crescent moon, so it got the name, Crescent Spring. Although Crescent Spring is surrounded by quicksand with strong winds, it is not covered by sand. It is also known as "the first spring in the desert".
Crescent Spring is one of the outstanding natural landscapes in Dunhuang. It has been known for its "wonders of the desert" throughout the ages and is reputed as "one of the most beautiful sceneries outside the Great Wall". Crescent Spring, nine floors of Mogao Grottoes and Mogao Grottoes art landscape, the "three wonders" in the south of Dunhuang, integrate as one, becoming one of the great tourist resorts for Chinese and world's people.
There are four explanations for the origin of the Crescent Spring:
Winter Scenery
The first one is the remains of an ancient river course. It is believed that Crescent Spring is a section of the ancient course of the nearby Dang River. Long ago, the river changed its course, and most of the ancient course was submerged by quicksand. Only the section of Crescent Spring was low-lying, and it gathered into a lake due to the exposed underground flow that emerged. The lake is constantly fed by an underground flow therefore it will not dry up. In the 1950s, it was measured that the surface of the Crescent Spring was 218 meters long from east to west, 54 meters wide from north to south, with an average depth of 5 meters and the deepest depth of more than 7 meters.
The second is the fault seepage spring. It is believed that there was an east-west fault on the south side of Crescent Spring. The hanging wall of the fault raised the underground aquifer, and when the footwall of the fault dropped to the nearby diving surface, the undercurrent flowed out and became the spring.
The third is wind erosion lake. In other words, with the intensification of wind erosion, the original deflation hollow forms a spring lake in the inner bay of a crescent-shaped dune when it reaches the phreatic aquifer. The sand mountain around the Crescent Spring is high in the north and south, and low in the middle, and the wind from the east blowing into the depression will go upward and the sand under the action of the wind will not be blown into the spring, which forms the wonderful scenery of the coexistence of sand and spring.
The forth is manual excavation. It is believed that the spring is a result of artificial deliberate modification because of its vivid crescent shape. In addition, there is a record of "sand well" in ancient books. Since it is a well, it should be the result of man-made efforts.
Story of Crescent Spring and Echoing-Sand Mountain
Camels Walking in Desert
Once upon a time, there was no Echoing-Sand Mountain or Crescent Spring here, but there was a Leiyin Temple. On April 8th, the temple held the annual bathing Buddha festival, and devout men and women in the temple burned incense to worship Buddha. When the Buddhist activity of "sprinkle holy water" began, the abbot took out a bowl of the ancestral holy water in the temple, and then placed it in front of the temple. Suddenly an outsider came up and wanted to challenge the abbot. People saw the sorcerer wielding a sword, reciting some words in his mouth. In an instant, the sky was dim, the wind was strong, and the yellow sand was omnipresent, burying Leiyin Temple under the sand. The strange thing was that the bowl of holy water in front of the temple was still in place. The sorcerer used all his magic to fill it with sand, but not a single grain of sand entered the bowl. Until a sandy mountain was formed around the bowl, the holy water bowl remained intact. The sorcerer had no choice but to walk away. Just after a few steps, he heard a crash, and the bowl of holy water tilted into a spring, and the sorcerer turned into a black stone. Originally this bowl of holy water was given to the abbot of Leiyin Temple by Buddha Sakyamuni and handed down from generation to generation, especially for people to eliminate the disease. As a result of the evil sorcerer, Buddha appeared to punish him. Buddha made the bowl tilt and the holy water pour, and the Crescent Spring formed.
Top Amazing Attractions You Should Not Miss
Crescent Spring
The Crescent Spring is surrounded by high sand mountains with a small lake in the middle shaped like a crescent moon, so it is named "Crescent Spring". It has been outstanding from many scenic spots in Dunhuang from ancient times, because there is a strange phenomenon here. Because of the topography, when the wind blows, the sand does not go down from the mountain but flows from the foothill to the top. Therefore, the Crescent Spring will never be buried by the sand, which is called a desert wonder. The coexistence of sand and water, and the interdependence of mountains and the spring contribute to the "first spring in the desert", a wonder of the world.
There are two ways of enjoying the Crescent Spring: one is standing closely to enjoy it and the other is overlooking the panorama by climbing the Echoing-Sand Mountain. You can see the spring is still clear with water sparkling, like a mirror, reflecting the Echoing-sand with its deep love. You will be amazed by the symbiosis and coexistence of spring and sand.
There is a kind of algae growing in the Crescent Spring, and the locals call it "seven-star grass". "Seven-star grass", "iron fish" and " "five-color sand" are known as the three treasures of Crescent Spring.
It is said that the seven-star grass in the Crescent Spring can hasten healing, which has a long history. Once upon a time, there was a doctor who was gentle and kind. He treated the rich and the poor alike. Therefore, people praised him: "His heart is as good as Buddhist."
Desert Landscape
Although the doctor was very skilled, he didn't know what to do when his wife had a difficult birth. He used many birth prescriptions, but not useful. Seeing his wife would die, in all circumstances, he had to pray to Buddha. When he got to the temple, he met an old woman with white hair. Before he asked, she said first, "Don't beg. Don't ask. See this then your wife will give birth." The old woman then gave him a letter, which he took home and read. It read: "The seven-star grass could hasten parturition, but your wife must take it in the morning. If women eat it, their sons or daughters could come to the world." The doctor was worried, because he had never seen the seven-star grass. Where should he look for it? His wife reminded him and said: "You are smart for life but silly for a while. The sand well at the foot of the sand mountain is growing a lot of fuzzy grass, where there is dew. It must be the seven-star grass." The doctor was suddenly enlightened, and hurried to the sand well, seeing the seven-star grass with dew. He was overjoyed. He picked a bunch of the seven-star grass, took it home, and boiled. His wife drank a seven-star grass potion, and less than fours hours, as expected, she gave birth to a fat boy. So it was said by many that the seven-star grass could hasten child delivery. Since then, the sand well has been also known as medicine spring.
Ancient Building
 Folk Museum
Built in 1988, the building area is 4,300 square meters, with a castle-style building structure. The museum is a collection of essence of Dunhuang folk architecture, living customs, and traditional culture. Its layout follows the style of Dunhuang folk houses. There are a large number of exhibits reflecting the production, life, religion, and culture of modern Dunhuang folks. It is the base of modern Dunhuang folk culture.
Echoing-Sand Mountain
To the south of Dunhuang City, you can see the rolling Echoing-Sand Mountain which is the northwest pearl with Dunhuang Mogao Grottoes, the famous scenic spot of Gansu Province in the east, and the Mouth of Dang River in the west. It extends for 40 kilometers from east to west, 20 kilometers from north to south, and the main peak is 1,715 meters high, like two sand arms, stretching out to surround the Crescent Spring.
On the east of Echoing-Sand Mountain is Sanwei Mountain, on the south is Heishi Peak Mountain, and on the west is the Gobi and Kumtag Desert. The west wind and the north wind carrying more sand are blocked by the mountain here, so that the sand in the wind settles down, and over time, it has formed the Echoing-Sand Mountain. The most magical thing is that the wind force makes the Echoing-Sand Mountain unchanged. For thousands of years, the number of people who climb up and slide down from the mountain is unknown, but Echoing-Sand Mountain still stands as it was, and the sound is still ringing in the ears. Sitting on the sand mountain quietly waiting for the dark, the sunset gradually dyes the whole sky, overlooking the spring, which makes one feel the charm of nature.
The Site of Yaowang Temple
Temple Built in Desert
The site of Yaowang Temple was discovered in the 1990s. It was the Yaowang Cave in the Qing Dynasty confirmed by the unearthed wooden merit tablet. It was built in memory of Sun Simiao, the King of Medicine. The main hall is a brick cave building, with a height and width of 3 meters and a length of 6.2 meters. In front of the cave, there was a western structure gallery, but all of which have collapsed, and the remaining purlin columns are mostly rotten. The clay sculptures in the cave no longer exist. On the north wall of the cave is hanging carvings of " King of Medicine treating dragon and tiger", while on the east wall is carvings of landscapes, flowers and birds. Unable to ensure the preservation of the sculptures in the existing environment, the site has been backfilled.
Village in Desert
 The First Spring Stone Tablet
In the summer of 1995, out of love for Dunhuang, Wang Licheng, a Dunhuang native, and his family members donated 10,000 yuan to make a natural stone tablet, 2.6 meters high, 0.7 meters wide and 1.1 meters wide, inscribed with "the First Spring" in the front and "Love Dunhuang" in the back. It was set up by the side of Crescent Spring, adding a scene to the scenic area. The Dunhuang municipal government praised Mr. Wang Licheng with a certificate of honor for his charitable donation.
 Husband and Wife Willows
The River Scenery
Planted in 1971, they are two dryland willows with the same root system growing close to each other, which are called "husband and wife willows". Accompanied by Crescent Spring, they have spent countless cold and summer days together, becoming a scene in the desert for visitors lingering to see. In 1997, a sudden sandstorm hit the "Husband and Wife Willows", which broke off from the root. Later, under the careful care of the staff, the willows have been brought back to life and flourished again, just like a pair of lovers who give birth to a stronger generation.
Recommended Travel Route
Crescent Spring →Echoing Sand Mountain→ The Site of Yaowang Temple → The First Spring Stone Tablet→ Husband and Wife Willows → Folk Museum
How to Reach Crescent Spring and Echoing-Sand Mountain
Take bus 3 and get off at the bus terminal, and the fare is 1 yuan.
Chinese: 请带我去鸣沙山月牙泉。English: Please take me to the Crescent Spring and Echoing-Sand Mountain.
If you go by taxi from Dunhuang Mogao International Airport, it takes about half an hour (45 yuan).
If you go by taxi from Dunhuang Railway Station, it takes about 25 minutes (35 yuan).
If you go by taxi from Dunhuang Bus Station, it takes about 10 minutes (15 yuan).
The Aerial View
Author: Wentong Fu


The foundation supports nonprofit organizations that create systems change, improve the quality and circumstances of life and reflect the foundation's values and interests. It invests time, expertise and funding in organizations that help women become economically self-sufficient, enhance the early care and education of children, strengthen justice, equality and opportunity and enrich arts and culture. The foundation also published "Creating a Women's Fund Within a Community Foundation" to support the growth of women's philanthropy..


text:
Q: How to properly configure snmpd? I installed snmp on CentOS 7.2, like so:
yum -y install net-snmp net-snmp-utils

I made a backup of my snmpd.conf file:
cp /etc/snmp/snmpd.conf /etc/snmp/snmpd.conf.orig

then I cleared the text, with this:
echo "" > /etc/snmp/snmpd.conf

and added to the snmpd.conf, the following:

rocommunity "#random$" monitoring_server_ip

The monitoring_server_ip is the server that that is allowed to connect to this snmpd server.
Restarted snmpd
/bin/systemctl restart snmpd

When I run snmpwalk on my CentOS 7.2 server
snmpwalk -v2c -c public localhost system

I then get:

Timeout: No Response from localhost

Which is accurate, because there is only one IP-adres that can do that, as we have defined before.
ps shows that snmp is running

/usr/sbin/snmpd -LS0-6d -f

This is my first time playing with snmp and any help is greatly appreciated!
output of iptables -L -n | grep udp shows this:
...
Chain IN_public_allow (1 references)
   94  target     prot opt source               destination
   95  ACCEPT     udp  --  XX.XXX.XXX.XXX        0.0.0.0/0            udp dpt:161 ctstate NEW
...

All of the destination was too 0.0.0.0/0?
Netstat shows the following port:
 netstat -ulnp | grep 161
udp        0      0 0.0.0.0:161             0.0.0.0:*                           19062/snmpd

also this:
netstat -lu | grep snmp
udp        0      0 0.0.0.0:snmp            0.0.0.0:*

Also, in my firewall, added that only one IP-adres can access my snmp server:
firewall-cmd --permanent --zone=public --add-rich-rule="rule family="ipv4" source address="XX.XXX.XXX.XX" port protocol="udp" port="161" accept"

A: The com2sec security model is not mandatory anymore.
In snmpd.conf it should be enough to do:
rocommunity "#randomsometinh$"  2.2.2.2

where 2.2.2.2 is the monitoring IP address allowed to connect.
I often prefer to assign a single IP, than giving access to a whole /24. So this configuration means the SNMP service will answer requests from the 2.2.2.2 address.
You might also have to comment the line that restricts the snmpd daemon to the localhost for security reasons.
# agentAddress  udp:127.0.0.1:161

After changing the configuration file, do:
service restart snmpd

To confirm if it is listening locally:
$ netstat -lu | grep snmp
udp        0      0 *:snmp                  *:*

And from the allowed network/IP, for walking the entire MIB tree. Assuming 2.2.2.1 is the machine being monitored:
snmpwalk -c "#randomsometinh$" -v2c 2.2.2.1

or for asking for the sysUpTime OID:
snmpwalk -c "#randomsometinh$" -v2c 2.2.2.1 1.3.6.1.2.1.1.3
snmpget -c #randomsometinh$ -v2c 2.2.2.1 1.3.6.1.2.1.1.3.0

snmpget has to have the 0 for the specific object instance/index.
P.S: 2.2.2.2 is the monitoring server, and 2.2.2.1 is the snmpd server/host to be monitored.


Sermons from the First Congregational Church of Zephyrhills, FL. We are an independent, autonomous and theologically conservative church that can trace our core religious beliefs back to the 17th Century Pilgrims and Puritans. Thank you for visiting our sermon podcast website!


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Secure Insecurity
In many areas of the Democratic Republic of Congo, citizens are still vulnerable to brutal violence from armed groups and in some cases from government, including the police, army and local officials.
Northern Mali: food and fuel in short supply as stocks looted and traders fail to return
In areas of northern Mali the price of food and fuel is rocketing as supplies start to dry up after looting following recent military operations, and with traders who have fled the area failing to return amid growing numbers of reports of reprisal attacks.
Oxfam calls on civilians to be protected and humanitarian needs met as Mali conflict escalates
International aid agency Oxfam is extremely concerned that the recent intensification of fighting in Mali, which includes the intervention of French military, could worsen restrictions on humanitarian access and lead to a significant increase the humanitarian needs of populations across the country and in neighboring states.
Aid must work better for Afghans in the next decade
Development gains made in Afghanistan over the last decade are in danger of being thrown away if levels of aid fall away in conjunction with the withdrawal of international troops in 2014.
Stop a Bullet, Stop a War
Guns are useless without bullets. An Arms Trade Treaty (ATT) that does not control ammunition will not achieve its purposes.
Imminent rains will jeopardize response to Sudans conflict, aid agencies warn
Seasonal rains due in Sudan and South Sudan will exacerbate already dire conditions in refugee camps, restrict travel and access, and heighten the risk of disease, a group of leading humanitarian agencies warned today.
South Sudan facing most violent year since end of civil war
South Sudan is facing its most violent year since the end of the civil war in 2005, international agency Oxfam warned today, and it urged the UN Security Council, as it visits the region, to ensure that civilians are better protected.
Protection of Civilians in 2010
In 2010, millions of men, women, and children were killed, raped, displaced, injured, or recruited by force in armed conflicts throughout the world.
UNSC's efforts to protect civilians in armed conflict inconsistent and at times biased
The international community's record on protecting civilians in 2010 was uneven and often biased, according to a new report published today by international agency Oxfam. Thousands of civilians were killed in conflicts last year alone while millions were displaced from their homes.
Oxfam statement on unrest in the Maghreb
Oxfam deplores the use of lethal force against unarmed civilian protesters in North Africa, particularly evident in Libya.


Will The FED Protect Your Wealth?
Home / FED / Will The FED Protect Your Wealth?
The wealth gap is growing. Invest like the rich.
The FED will be forced to protect stocks and real estate prices as they are the key to current economic growth.
The Boston Consulting Group interviews 250 investors annually who have a combined wealth of $500 billion. The results usually aren't shocking except for this year. 65% of those investors think the market is overvalued.
Figure 1: Boston Consulting Group annual interview. Source: BCG.
This isn't that surprising as the S&P 500 has gone up 19.55% year-to-date which is a huge performance. The question is whether the investors who invest in the S&P 500 now understand that the long term returns from investing will be around 4% in the best-case scenario or if they invest on the assumption that stocks will continue to go up and expect higher returns than 4% from the S&P 500 next year.
My suspicion is that most investors expect the S&P 500 to return at least 8%. This means that when the trend turns, there will be a lot of panic selling because any kind of return above 4% from the index is a speculative return which significantly increases the downside risks. But let's see how the FED is positioning itself in relation to financial markets.
Figure 2: S&P 500 year to date. Source: CNN Money.
If you read the FOMC statement, you'll see that the current monetary policy is still accommodative which means that the perceived strength of the economy isn't real, but fabricated through low interest rates and increased liquidity. Nevertheless, the FED is slowly increasing rates that will maybe lead to a more balanced market. I say maybe because there is a big chance that the FED won't be able to significantly increase interest rates.
Stocks will continue to go up even with the FED tightening, but there is another investment vehicle, one used by those who were interviewed by the Boston Consulting Group, that is signaling interesting times ahead. The spread between the 10-year Treasury yield and the 2-year one is declining.
Figure 3: 10-year Treasury minus 2-year Treasury. Source: FRED.
A declining spread signals that there is a significant difference between what will happen in the short term and what will happen in the long term. The above figure shows how when the spread between the short term and long-term Treasuries gets close to or below zero, we can expect a recession following relatively quickly.
Those who look at long term returns are positioning themselves in 10-year bonds to take advantage of the current yields. This is because interest rates work like gravity on asset values, the higher the interest rates, the lower the value of stocks and bonds is due to the higher expected returns. However, there is a big chance that the FED will do whatever it takes to prevent any kind of declines in asset values because most of the wealth and current consumer confidence comes from increased paper wealth. It's important to note that the wealth of Americans has tripled in the last 20 years.
Figure 4: Total wealth of Americans. Source: FRED.
However, it's also important to mention that the increase in wealth hasn't been equal.
If we look at the median wealth of Americans, it has just slightly increased over the last 30 years. Accounting for inflation, the median American is practically poorer than they were 30 years ago.
The more you get into the richer percentage of Americans, the bigger the growth is in wealth. The top 5% of Americans have increased their wealth 3 times, and the top 10% have seen the bulk of the increase in wealth of the total population. The divergence has primarily occurred in the last 10 years as the FED has intervened to save financial markets.
Figure 5: The median and mean household wealth. Source: Credit Suisse.
Going deeper into the matter and connecting it to interest rates, we can see that most of the confidence that pushes consumers to spend takes debt and consequently increases GDP and is due to higher net wealth which is purely the impact of low interest rates and an inflation in asset prices. Thus, the created wealth is just on paper. It isn't real.
Figure 6: The increase in household wealth comes from higher stocks and real estate prices. Source: FED.
All of the above puts the FED in a very delicate situation. Higher interest rates push asset prices down and increase debt burdens. In an economy that is based on debt and fictive wealth, it really isn't prudent to increase interest rates. Therefore, the conclusion is that we are either going to keep having low interest rates or we'll have higher inflation.
In both cases, you should invest accordingly. Inflation protection should be the key component of your portfolio and don't forget about debt. Not taking advantage of the current low interest rate environment and pat position central banks have put themselves in is something you will probably regret for the rest of your life.


Giving a real sense of adventure and style, this Croco billfold wallet in mahogany will sit as comfortably in your khakis as in an elegant linen summer suit. The embossed crocodile print applied on the glazed cow leather is very lifelike, giving this collection an elegant look. The whole effect is stunning and transports you into African reveries.
Due to the Croco pattern we recommend adding maximum three initials on the item. If more are added some initials may get 'lost' within the pattern.
This superb crocodile print leather is made with Italian cowhide, which is chrome tanned, then has an aniline glazed finish and finally is embossed with a crocodile skin pattern. This croc' print is exquisitely done and once the leather is made into a product it becomes difficult to know it isn't real crocodile. Truly, this is the most realistic croco print leather we have ever seen!
Ettinger's iconic panel hide is a vegetable tanned full grain cowhide. It is tanned using only natural tanning agents, essentially different types of bark and plants. Vegetable tanning is not just a very natural process, but the longer time required to tan the leather means it develops a particularly natural look and feel.
Once tanned, we then apply a waxy coloured finish on top of the panel hide, to give it it's colour and protect the surface.


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Home/Entertainment/Endemol Shine launches licensing push for Peaky Blinders
Endemol Shine launches licensing push for Peaky Blinders
Robert Hutchins 12th April 2018	Entertainment 315 Views
Global content creator, producer and distributor Endemol Shine Group has announced the launch of global brand licensing and ancillary rights to the award-winning gangster drama Peaky Blinders.
The partnership with award-winning show creator Steven Knight and producers Caryn Mandabach Productions and Tiger Aspect Productions, will see Endemol Shine Group's Brand Licensing team represent a range of global ancillary rights for Peaky Blinders including gaming and publishing.
Produced in the UK for BBC Two, the show is sold globally by Endemol Shine across 182 countries, including airing on Netflix.
The Endemol Shine Brand Licensing teams will extend the lifecycle of the brand off-air, using the Group's multi-territory reach and expertise in global brand management to drive local and global deals for the critically acclaimed property.
"Peaky Blinders is an incredibly unique show and a great addition to our portfolio, as we diversify our brand licensing offerings with exceptionally strong scripted IP," said Owain Walbyoff, MD Gaming, Licensing and Merchandising at Endemol Shine Group.
"The series is globally loved, and the world of Peaky Blinders offers exciting opportunities for the brand to expand its off-air engagement with fans around the world."
With a cast led by Cillian Murphy (Dunkirk, Inception), Peaky Blinders has won critical acclaim, scooping five BAFTA TV Craft awards and receiving a nomination for Best Drama Series at the BAFTAs. A fifth series has been confirmed for 2019.
"Peaky Blinders has become an international phenomenon and has entered the culture in many countries," commented the show's creator, Steven Knight. "We have more series planned and will begin shooting series five in September so the brand will only continue to grow. I look forward to seeing the family saga become a lifestyle."
Peaky Blinders will be featured at Endemol Shine North America's booth (Q226) at the upcoming Licensing Expo 22-24 May 2018 in Las Vegas.
Endemol Shine Group control brand licensing rights to some of the most recognisable and creative IP in the world including MasterChef, Deal or No Deal, Big Brother, Simon's Cat, Mr Bean, Money Drop and Fear Factor.
Tags Endemol Shine Group jridsdalenbmedia-com Peaky Blinders
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