text:
class ScannerTest : public ::testing::Test {
protected:
  void SetUp() {
    symtable.clear();
  }
};

TEST_F (ScannerTest, position) {
  // one, two, three spaces
  //---------------012345678901234567890
  Scanner scanner(" 12345,  tom,   Date");
  EXPECT_EQ(0, scanner.position());

  EXPECT_EQ(1, scanner.skipLeadingWhiteSpace());
  EXPECT_TRUE(isdigit(scanner.currentChar()));
  EXPECT_EQ(12345, scanner.extractNumber());
  EXPECT_EQ(6, scanner.position());

  EXPECT_EQ(',', scanner.extractChar());
  EXPECT_EQ(7, scanner.position());

  EXPECT_EQ(9, scanner.skipLeadingWhiteSpace());
  EXPECT_TRUE(islower(scanner.currentChar()));
  EXPECT_EQ("tom", scanner.extractAtom());
  EXPECT_EQ(12, scanner.position());

  EXPECT_EQ(',', scanner.extractChar());
  EXPECT_EQ(13, scanner.position());

  EXPECT_EQ(16, scanner.skipLeadingWhiteSpace());
  EXPECT_TRUE(isupper(scanner.currentChar()));
  EXPECT_EQ("Date", scanner.extractVar());
  EXPECT_EQ(20, scanner.position());
}

TEST_F(ScannerTest, nextTokenEOS) {
  //---------------01234
  Scanner scanner("    ");
  EXPECT_EQ(0, scanner.position());
  int token = scanner.nextToken();
  EXPECT_EQ(EOS, token);
  EXPECT_EQ(NONE, scanner.tokenValue());
  EXPECT_EQ(4, scanner.position());
}

TEST_F(ScannerTest, nextTokenEOS2) {
  //---------------0
  Scanner scanner("");
  EXPECT_EQ(0, scanner.position());
  int token = scanner.nextToken();
  EXPECT_EQ(EOS, token);
  EXPECT_EQ(NONE, scanner.tokenValue());
  EXPECT_EQ(0, scanner.position());
}

TEST_F(ScannerTest, nextTokenNumber) {
  //---------------0123
  Scanner scanner("135");
  EXPECT_EQ(0, scanner.position());
  int token = scanner.nextToken();
  EXPECT_EQ(NUMBER, token);
  EXPECT_EQ(135, scanner.tokenValue());
  EXPECT_TRUE(symtable.empty());
}

TEST_F(ScannerTest, nextTokenAtom) {
  //---------------0123
  Scanner scanner("tom");
  EXPECT_EQ(0, scanner.position());
  int token = scanner.nextToken();
  EXPECT_EQ(ATOM, token);
  EXPECT_EQ(0, scanner.tokenValue());
  EXPECT_EQ(3, scanner.position());
  EXPECT_EQ("tom", symtable[0].first);
  EXPECT_EQ(ATOM, symtable[0].second);
}

TEST_F(ScannerTest, nextTokenVar) {
  //---------------01234
  Scanner scanner("Date");
  EXPECT_EQ(0, scanner.position());
  int token = scanner.nextToken();
  EXPECT_EQ(VAR, token);
  EXPECT_EQ(0, scanner.tokenValue());
  EXPECT_EQ(4, scanner.position());
  EXPECT_EQ("Date", symtable[0].first);
  EXPECT_EQ(VAR, symtable[0].second);
}

TEST_F(ScannerTest, nextTokenChar) {
  //---------------01234
  Scanner scanner("   (");
  int token = scanner.nextToken();
  EXPECT_EQ('(', token);
  EXPECT_EQ(NONE, scanner.tokenValue());
  EXPECT_EQ(4, scanner.position());
  token = scanner.nextToken();
  EXPECT_EQ(EOS, token);
  EXPECT_EQ(NONE, scanner.tokenValue());
  EXPECT_EQ(4, scanner.position());
}

TEST_F(ScannerTest, nextTokenAtomSC) {
  //---------------01234567-890
  Scanner scanner(".*-><&$@\\?");
  EXPECT_EQ(0, scanner.position());
  int token = scanner.nextToken();
  EXPECT_EQ(ATOMSC, token);
  EXPECT_EQ(0, scanner.tokenValue());
  EXPECT_EQ(10, scanner.position());
  EXPECT_EQ(".*-><&$@\\?", symtable[0].first);
  EXPECT_EQ(ATOMSC, symtable[0].second);
}

#endif


text:
Article 941
----
Le défaut de publication pourra être opposé par toutes personnes ayant intérêt,
excepté toutefois celles qui sont chargées de faire faire la publication, ou
leurs ayants cause, et le donateur.

redpajama_set_name: RedPajamaGithub


text:
Flying Submarines
Sky Captain's dream come true
Remember Franky's (Angelina Jolie) amphibious attack squadron from "Sky Captain: World of Tomorrow"? The idea of "fighter plane/submersible" combination has a long history in inventor's patents and the actual military research. It's not just "pulp fiction", but rather a fascinating account of the concept that US Navy is now getting ready to turn into reality.
(Artist's take on "Commander-2" successful testing in 1964)
Wild Soviet project, developed before World War II
It all started with a ground-breaking research by B. P. Ushakov, conducted by the Russian military in utter secrecy, between 1934 and 1938. The design never achieved a prototype stage, but was seriously considered for military application: the plane could fly 800 km at the speed of 200 km/h and could dive to 50 meters, having underwater speed of 3 knots.
(image credit: V. E. Bulba)
There's been some inventor activity across the ocean, as well. In 1945, the American inventor Houston Harrington filed a patent for "combination of airplane and submarine":
(image credit: Adventure Lounge)
The "submersible /airship" combination was an attractive idea for the military during World War II. The Japanese used small ("midget") submarines in combination with their air attack on Pearl Harbor. The Italians attacked British battleships in the Mediterranean harbor of Alexandria from the air and their "midget" submarines shortly thereafter. Even if relatively successful, these were suicide missions - but a compact flying sub could fulfill the task and bring their crew back home.
"It could fly from a favorable location to its destination at minimum altitude to avoid detection by radar. At the completion of its underwater mission it could travel as a submersible to a location best suited for takeoff, become airborne and return to base" (The Wilmington Morning News, 1964)
US Navy was playing with the idea in the 50s and 60s, coming quite close to actually making a prototype:
US Patent 2720367 from 1956 was even equipped with torpedoes:
(image credit: AP Wire)
Convair Division of General Dynamics Corp. received a Navy contract to test the feasibility of such a craft. The plane would be submerged by flooding of the wing, tail and hull compartments. It would travel 5 miles an hour under water, powered by batteries.
Don Reid's "Commander-1" actually was submerging and flying, in 1964
Don Reid, an electronics engineer, and independent defense contractor built "Commander-1" prototype, and then "Commander-2" which actually flew and was fully operational. First tests happened in 1964, with the speed of 4 knots while 2 meters underwater. Here is a technical split view, printed in the "Popular Mechanics"-like Russian magazine in 1965: (click to enlarge)
The story of this plane is told in "The Flying Submarine: The Story of the Invention of the Reid Flying Submarine, RFS-1," published in 2004 by Heritage Books Inc.
"The Cormorant" - modern take on the same idea
Today, the "flying sub" concept is very much alive, and even being considered by such heavy-weights as Lockheed Martin's Skunk Works (developers of the U-2 and Blackbird spy planes). This article details plans for an airplane that "starts and ends its mission 150 feet underwater"
(image credit: John MacNeil)
This gull-wing spy plane will operate in a fully stealth mode and will be launched from the Trident missile tubes of some huge nuclear sub. The logistics of launch and retrieval provide for the sub's secretive maneuvers, so that her location will not be given away. Once out of the water, the plane's rocket boosters fire and the Cormorant takes off. At the end of the mission, it's fetched by the robotic unit back into the submarine, deep underwater.
The dream of flying under the water and in the clouds is a noble one, and we'd like to see it fulfilled, not only for the military, but as a new and exciting future means of transportation. Just like the old military ekranoplans can one day serve as a step toward the beautiful "ground-effect" ocean liners.
(image credit: TM, 1965)
CONTINUE TO NEXT PART -
Category: Technology,Boats, Military
Ekranoplans Showcase
I've stood under the Convair YF2Y-1 Seadart, a jet powered sea plane, at the San Diego Air and Space Museum so knowing that designers also thought"Hey, if the one land ON the water why can't we make one that goes UNDER the water?" is no great leap. It's still pretty amazing.
Wow. I'd much rather have my tax dollars go for this kind of thing than socialized health care. Deadbeat babies or awesome flying attack stingrays? The choice isn't even close.
very cool. but as someone else mentioned, sad that the latest artist-impressions are with war in mind. How about a clear-bodied version for the ultimate sight-seeing flight?
To those who bemoan the military applications of such ideas: recreation is great, but without ways to protect your right and freedom to recreate, there won't be much use for recreation equipment. I doubt that in a global caliphate the Supreme Leader would look kindly upon such decadent recreation. Yes, overall it is a sad fact that we must defend ourselves from those who have DECLARED INTENTIONS of subjugating and/or killing all free people, but nonetheless it is a fact. I too, wish we could all live in peace, but wishing it does not make it so (when other parties do not share that wish). Just ask my ex. :)
...the best thing the military ever came up with since jets. this is the best idea in a century.
For a new vision of a future with flying submarines - check out http://empirethefilm.com/html/empire_-_mecha.html
Its a story/series where supercavition submarines aer also equiped to fly. Cool graphics if nothing else.
thoth said...
if scanning declassified documents has made anything clear it's that when the military says they are thinking about building something it simply means they are just ready to reveal it to the public. Do not be fooled.

redpajama_set_name: RedPajamaCommonCrawl


Why Do Some Eggs Have Double Yolks? An Investigation.
They're called double yolkers, and honestly, they're fascinating.
Damerow, who operates a family farm in Tennessee, is so familiar with the phenomenon that she can spot them without cracking open an egg and uses a common nickname for them: "double yolkers." In fact, in her time raising chickens, she's also seen a triple yolker. And, according to the Guinness World Records, the world's largest chicken egg, which was nine inches in diameter and had five yolks. Guinness World Records aside, the greatest number of yolks found in a single chicken egg is said to be nine. Yep, nine yolks in one egg. Must have been quite the omelet.
To understand the rich miracle of the double yolker, one must understand the biological process of egg-making. "It takes 25 hours to make one egg," says Damerow. "When a hen is hatched, they come into the world with a certain number of ova—it's like a bunch of grapes that hangs from the backbone." There comes a time in every young hen's life when she is mature enough to lay eggs, and when that time comes, the ova grow one at a time until it's the size of a yolk. Then it breaks away from the ovary, like an all-black-clad teenager bound for college, traveling through the oviduct, leaving the ovaries to the outside world. As the yolk moves through the oviduct, glands secrete the albumen, a.k.a. the egg white, and the substance to form the shell, a somewhat similarly sticky experience to a college dorm. And then, lo and behold, an egg is laid.
There's been no close observation of the creation of double yolkers being made inside chickens, à la a microscopic Dennis Quaid in Innerspace. But the working theory is that they are forged when ovulation occurs too rapidly or when one yolk catches up with another slow-moving yolk in a hen's oviduct.
There are two different kinds of chickens that most commonly lay an egg with two yolks. On occasion, they're laid by heavy-breed hens, for which the habit of making double-yolk eggs is an inherited trait. However, heavy-breed hens are not typically used for commercial egg laying because they don't lay many eggs. In all likelihood, if you find yourself with a double yolker, it came from an egg-laying pullet, a young hen, specifically under a year old.
Pullets typically produce smaller eggs when she first starts laying, around the size of pigeon's and sometimes their eggs will have no yolks at all. Damerow knows that she's got a double yolker on her hands when a pullet produces an egg the size of one from a mature hen. "With a pullet, the mechanics of the production cycle aren't working in sync yet," says Damerow. "It's like a toddler stumbling around and not yet well-coordinated." So if you find yourself with a double yolker, think of it as the gift of an amateur hen's early work, like the joyful scribbled drawings of a toddler but more nutritious.


text:
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English - Int.
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Home >> News >> News New QAIA terminal gears up for opening day
New QAIA terminal gears up for opening day
As a maintenance worker examined an electronic notice board and a sanitation worker diligently waxed the floor to give it an extra shine, Iraqi singer Kadim Al Sahir's voice could be heard through the speakers.
They were putting the finishing touches to the Queen Alia International Airport's (QAIA) new passenger terminal, preparing it for the official opening on March 21.
"We're going to have a Starbucks café there and a shawerma place to the left," an Airport International Group (AIG) representative said, pointing to the nearly-ready venues.
"This is all about making passengers feel comfortable and relaxed," he stressed.
The gate area was still quiet, awaiting the first groups of travellers.
"This is the final station, where the passengers will be waiting, each at his/her designated gate," the AIG representative said, pointing to rows of empty seats.
The brand new seats looked ready for their much-anticipated debut.
"As you can see, it is an open space, just like international airports around the world," he told The Jordan Times during a recent exclusive tour.
Ground breaking for the new airport terminal began in 2008, implemented by the AIG, the consortium in charge of the rehabilitation, expansion and operation of QAIA.
The 103,000-square-metre terminal "will open ready to handle a capacity of 12 million passengers", according to an AIG statement e-mailed to The Jordan Times.
"The current (old) building has a capacity to accommodate only 3.5 million passengers annually," the statement said.
"One of the important features of the new terminal is that passengers go through only one security checkpoint, unlike the old terminal where passengers were searched twice," noted the AIG representative, who did not give his name for administrative reasons.
"After this security check, every passenger is free to roam the duty-free shops and head to the restaurants," he added.
"The new terminal features ample food and beverage venues, shopping outlets, children's play areas, clear signage, sufficient flight information screens, comfortable seating in waiting areas, and high levels of service at the customer assistance counter," the AIG said, elaborating on how the facility meets international standards.
"In addition, it is equipped with a state-of-the-art baggage handling system and security equipment."
There are 64 check-in counters at the new building and six conveyor belts in the baggage claim area.
"Arriving passengers won't have to stop by the customs checkpoint unless a sticker on their bag indicates they should," he noted.
"This saves time because not everyone will have to go through the security check at customs."
"The number of gates will be developed gradually to meet demand, as needed, towards the 12 million passengers per year," the AIG said.
"When the first phase of the new terminal is completed, a total of seven new contact stands, plus one temporary remote boarding lounge, will come into operation, serving 7 million passengers.
"Additionally, demolishing the airport's [old] terminal will trigger another phase of the expansion plan, catapulting the total capacity to 9 million passengers per year," according to the statement.
The AIG said the $750-million terminal, which was tested by volunteers posing as passengers last month, was received with "excitement and interest for its unique design".
"In addition, volunteers were impressed by the ease of flow from the entrance to the gate area," the statement said.
The terminal is designed by renowned British architect Norman Foster, who designed the new Wembley Stadium in London, among other international landmarks.
The new terminal's roof design is inspired by bedouin tents and is composed of 127 concrete domes.
"The designer specifically asked to maintain the domes' cement-like colour, which, along with the pink shades in the interior, reflect the colours of Jordan," the AIG representative said.
Source : Jordan Times.
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text:
Odradek Residence
rue Américaine, 35
www.odradekresidence.be
Ross Lewis is an American artist who creates art that engages with the Chinese literati tradition. His four decades long passion and study of Chinese landscape painting and calligraphy has influenced his painting, sculpture and installation. With a deep-seated fascination for China and its culture, in his twenties, Lewis began studying Chinese landscape painting and calligraphy under the tutelage of notable artist/collector C.C. Wang in New York. His interest in Chinese art, history, and Mandarin language, brought him to China and Taiwan, where he studied scroll mounting at the National Palace Museum, and over the course of several decades had the rare opportunity to visit and exchange ideas with some of the 20th century modern masters of Chinese painting. As a result of these experiences, he came to develop a distinct visual language, bridging key aesthetic and philosophical sensibilities of the Eastern artistic traditions to his Western context. In the past nine years, Lewis has departed from the conventional use of the Chinese brush in the development of his Rope Painting Series, using instead an ink-soaked rope, as his primary means of marking the paper surface. Rips, tears and cuts act as counterpoints to the ink line, providing a visual and emotional charge. Collage, comprised of a palate of eclectic colored papers punctuates the surface adding a richly textured sculptural dimension.


DATAMEDICA develops its own original research, in addition to the research conducted under contracts. We also contribute to the dissemination of our areas of expertise through publications and courses.
In this area, you may find detailed information about our developments in the main area of research of DATAMEDICA, Clinical Trials Information Systems. Please look in PRODUCTS for information about COATI, PANDA and DART, three of the main components of the integrated system that DATAMEDICA designed and developed for its internal operations.
If you would like to know more about our main two areas of expertise - Biostatistics and Biomedical Informatics - you may find in COURSES a couple of tutorials about these topics. The tutorials present the subjects in a style that allows non-experts in these domains to follow easily these topics. If you do not have a background in statistics or computing, you will find there texts written for you.
If you have a graduation in Biostatistics, then those tutorials will probably not interest you. However, you may be interested in knowing that in DATABASES you will be able to find datafiles of epidemiological studies conducted by DATAMEDICA, along with all the necessary documentation. You are free to download the files to your computer and to use them as you wish. If you are involved in the development of theoretical statistical models or of data analysis algorithms, those databases may be useful to test them.
Finally, in PUBLICATIONS we present the references of publications in national and international scientific journals of the results of projects that we conducted for pharmaceutical companies.


text:
\subsection{Tightness of $(X_t, \eta_t)$}
\label{section tight}
We state a simple Lemma.
\begin{lem}.
\label{martingale lemma}

$\int_{0}^{t} \theta_u dW_u$ is a martingale with respect to
 $\mathcal{F}_t=\sigma(X_s, \theta_s ; 0 \le s \le t )$ and hence
\beo
E(\int_{0}^{t} \theta_t dW_t)&=& 0 .
\eeo

\end{lem}

\textbf{Proof:}
{It is sufficient to show that the martingale $Y_t:=\int_{0}^{t} \theta_s dW_s$ is $L_2$-bounded. So using Iso's isommetry it is sufficent to show that $$ E \Bl \int_{0}^{t} \theta_s^2 ds\Br< \infty $$}
But,
\beo
d \theta_t^2 &=& 2 \theta_t d \theta_t \\
&=& 2 \theta_t^2(p - \frac{1}{\sqrt{2\pi}}|X_t| \theta_t) dt  \\
&\le & 2 \theta_t^2 p dt
\eeo
Now writing $Y_t = \theta_t^2$ we have that
\beo
d Y_t \le 2pY_t dt
\eeo
Multiplying both sides by the integrating factor $e^{-2pY_t}$ we get
\beo
d(Y_t e^{-2pY_t}) &\le& 0
\eeo
Integrating and taking expectations we get
\be
(Y_t e^{-2pt}) - (Y_0) \le 0 & \Rightarrow & \theta_t^2 \le \theta_0^2 e^{2pt}
\  \mbox{ and } \ E(\theta_t^2) \le E(\theta_0^2 ) e^{2pt} , \ \ \forall {t>0} . \no \\
\mbox{Also }
\int_{0}^{t} \theta_s^2 ds \le  \theta_0^2 2p(e^{2pt}-1)
&\ \mbox{ and hence, } \ &
E(\int_{0}^{t} \theta_s^2 ds) < \infty , \ \ \forall {t>0} .
\label{theta_sq}
\ee
\qed\\
Here is the main result of this subsection.

\begin{lem}
\label{lemma tightness}
Assume $E(X_0^2)$ and $E(\theta_0^2)$ are finite. Then, for the coupled system (\ref{SDE Normal})
\begin{enumerate}
	\item {$\sup_{t \ge 0} E(X_t^2) < \infty$, whenever $E(X_0^2)$ and $E(\theta_0^2)$ are finite;}
	\item $\theta_t > 0 $  almost surely;
	\item $\eta_t > 0 $ almost surely, where $\eta_t= \frac{1}{\theta_t}$;
	\item Joint distribution of $\{(X_t,\eta_t) : t \ge 0\}$ is tight.
\end{enumerate}
\end{lem}

\textbf{Proof.}
\vspace{0.2in}

\begin{enumerate}
\item
$$ \mathbf{\sup_{t} E (X_t^2) < \infty}$$
Applying Ito's Rule to $Y_t = X_t^2$ we get
\beo
dY_t &=& 2X_t dX_t + (1/2) 2 \theta_t^2 dt\\
&=& -X_t \theta_t^2 dt + 2 X_t \theta_t d W_t + \theta_t^2 dt \\
&=& (-X_t^2 \theta_t^2 + \theta_t^2) d t +  2 X_t \theta_t d W_t \\
\Rightarrow dY_t + Y_t \theta_t^2 dt &=&\theta_t^2 dt + 2 X_t\theta_t d W_t .\\
\eeo
Multiplying both sides by $e^{\int_{0}^{t}\theta_u^2du}$ we get

\beo
d(Y_t e^{\int_{0}^{t} \theta_u^2 du})&=& e^{\int_{0}^{t}\theta_u^2du} \theta_t^2 dt + 2 e^{\int_{0}^{t}\theta_u^2du} X_t \theta_t d W_t \\
\Rightarrow Y_t \e - Y_0 &=& \intt \theta_s^2 e^{\int_{0}^{s} \theta_u^2 du} ds + 2 \intt \es X_s \theta_s dW_s\\
\Rightarrow X_t^2 &=& \emi X_0^2 + \emi \intt \theta_s^2 \es ds +2 \emi  \intt \es {X_s} \theta_s dW_s.\\
\eeo

Writing $F_s = \es$, (therefore $F'_s = \theta_s^2 F_s$)  we have ;
\be
\label{boundx}
X_t^2 &=& \frac{X_0^2}{F_t} + \frac{\int_{0}^{t} {F'_s} ds} {F_t} + 2 \frac{\intt {F_s} X_s \theta_s dW_s}{F_t} \no\\
&=& \frac{X_0^2}{F_t} + \frac{{F_t}-1}{F_t} + 2  \frac{\intt F_s X_s \theta_s dW_s}{F_t} \no\\
\Rightarrow E(X_t^2)&\le & E(X_0^2) + 1
+ 2 E(\frac{1}{F_t} \int_{0}^{t} F_s X_s \theta_s d W_s)  , \ \ \ \mbox{ since $F_t\ge 1 \ \ \forall \ t \ge 0 $}.\no\\
\ee
Write $Y_t = \int_{0}^{t} F(s) X_s \theta_s d W_s$ and $\overline{F}_t = \frac{1}{F_t}=\emi$.
So the term inside the expectation on the right side is $Y_t \overline{F}_t :=Z_t$.
 Applying Ito's lemma to $Z_t$ we get,
\be
d Z_t &=& Y_t d \overline{F}_t + \overline{F}_t d Y_t + d Y_t d \overline{F}_t, \no\\
&=& -Y_t\theta_t^2 \overline{F}_t dt + \overline{F}_t X_t \theta_t F(t) dW_t + 0 \no \\
&=& - Z_t \theta_t^2 dt + X_t \theta_t dW_t  \ \ \ \mbox{with $Z_0=0$}.  \label{eqn for Zt}
\ee

Writing $\tilde{Z}_t = - Z_t$ we have the SDE of $\tilde{Z}_t$ as
\beo
d \tilde{Z}_t &=& - \tilde{Z}_t \theta_t^2+ X_t \theta_t d \tilde{W}_t \ \ \mbox{where $\tilde{W}_t = -W_t$}
\eeo

Comparing the SDE of $Z_t$ and $\tilde{Z}_t$ , and noticing that $Z_0 = \tilde{Z}_0 = 0$, we have that $Z_t = \tilde{Z}_t$ in distribution.
This implies, $Z_t$ is symmetric about 0.
Therefore,

$E(Z_t)=0$ if we show that $E|Z_t|<\infty$. Note,
if $E(Z_t)=0$, for each $t \ge 0$, then from (\ref{boundx}), for each $t \ge 0$,
\be
\label{unifbound} E(X_t^2) &\le & E(X_0^2) + 1 .
\ee
Thus, it suffices to show that $Z_t$ has finite expectation for each $t \ge 0$.

 Now
\be
\label{CS ineq}
Z_t^2 &=& \frac{1}{F_t^2} \Bigl(  \int_{0}^{t} F_s \theta_s X_s dW_s\Bigr)^2 \le \Bigl(  \int_{0}^{t} F_s \theta_s X_s dW_s\Bigr)^2 \ \ \mbox{a.s} \no \\
\Rightarrow E(Z_t^2) &\le& E\Bigl(  \int_{0}^{t} F_s \theta_s X_s dW_s\Bigr)^2 = E \Bigl( \int_{0}^{t} F_s^2 \theta_s ^2 X_s^2 ds \Bigr) \no \ \ \mbox{from Ito's isommetry} \\
&\le& E\Bigl( \int_{0}^{t} (F_s \theta_s)^4 ds \int_{0}^{t} X_s^4 ds\Bigr)^\frac{1}{2} \no \\
&\le& [E \Bigl( \int_{0}^{t} (F_s \theta_s)^4 ds \Bigr) E \Bigl( \int_{0}^{t} X_s^4 ds \Bigr)]^{\frac{1}{2}} \no \\
&&\ \ \mbox{from Cauchy-Schwarz inequality

}
\ee
Let us define $Y_t = \theta_t^n $\ for n = $1,2,\ldots$  we have
\beo
d Y_t &=& n \theta_t^{n-1} d \theta_t = n \theta_t^{n-1}\Bigl(  \theta_t(p - \frac{|X_t| \theta_t}{\sqrt{2 \pi}}) dt\Bigr)\\
&<& n p \theta_t^n = n p Y_t
\eeo
Multiplying by $e^{-npt}$ we have
\be
d (e^{-npt} Y_t) &<& 0 \no \\
\Rightarrow  Y_t e^{-npt} - Y_0 &<& 0\no \\
 \Rightarrow \theta_t^n  &<& \theta_0^n e^{npt} \no \\
 \Rightarrow  \theta_t^n e^{mps} &<& \theta_0^n e^{p(m+n)t} \ \  \mbox{for $s \le t$ and $n = 1, 2, \ldots, m = 0, 1, \ldots$}\label{theta_t}
\ee
Now since
\beo
X_t &=& X_0 - \int_{0}^{t} \frac{X_t \theta_t^2}{2}ds + \int_{0}^{t} \theta_s dW_s \\
 \Rightarrow X_t^2 &\le& 3 (|X_0|^2 + (\int_{0}^{t} \frac{|X_s| \theta_s^2}{2}ds)^{2} + (\int_{0}^{t} \theta_s dW_s)^2) \\
 \Rightarrow X_t^4 &\le& 27 (|X_0|^4 + (\int_{0}^{t} \frac{|X_s| \theta_s^2}{2}ds)^4 + (|\int_{0}^{t} \theta_s dW_s|)^4)
\eeo
Writing $M_t := |\int_{0}^{t} \theta_s dW_s|$ which is a (loc) submartingale.
Now applying Burkholder-Davis-Gundy (BDG) inequality to $M_t$ we have
\be
E(M_t^4) &\le& E((\sup_{0\le u \le t} M_u)^{4}) \le C_4 E([M]_t^{2}) = C_4 E(\int_{0}^{t} \theta_s^2 ds)^2
< C_4 E(\theta_0^2 2p (e^{2pt}-1))^2 \no \\
&=& C_4 \theta_0^2 2p (e^{2pt}-1) \label{one}
\ee
Again, the equation,
$d\theta_t = \theta_t \Bigl(p - \frac{|X_t| \theta_t}{\sqrt{2 \pi}} \Bigr)dt $, \ yields
\beo
 0 &=& d(e^{-pt} \theta_t) + e^{-pt} \frac{|X_t| \theta_t^2}{\sqrt{2 \pi}} dt \\
\Rightarrow \ \  \theta_0 &=& e^{-pt} \theta_t  + \int_0^t e^{-ps} \frac{|X_s| \theta_s^2}{\sqrt{2 \pi}} ds \\
\Rightarrow \ \  e^{pt} \theta_0 - \theta_t &=& \int_0^t e^{p(t-s)} \frac{|X_s| \theta_s^2}{\sqrt{2 \pi}} ds .\\
\eeo
Thus,
\be
|\int_{0}^{t} \frac{|X_s| \theta_s^2}{\sqrt{2 \pi}} ds | &\le& |\int_{0}^{t} e^{p(t-s)} \frac{|X_s|\theta_s^2}{\sqrt{2 \pi}} ds| \le |\theta_0 e^{pt}- \theta_t | \le \theta_0 e^{pt} + |\theta_t|\no \\
\Rightarrow \Bigl( |\int_{0}^{t} \frac{|X_s| \theta_s^2}{\sqrt{2 \pi}} ds | \Bigr)^4 &\le& (\theta_0 e^{pt} + \theta_t)^4. \label{two}
\ee
And,
from (\ref{theta_sq}),
$(\theta_0 e^{pt} + \theta_t)^4 \le  8 (\theta_0^4 e^{4pt} + \theta_t^4) \le 8 (\theta_0^4 e^{4pt} + \theta_0^4 e^{4pt})
= 16 \theta_0^4 e^{4pt} $.
Hence, from (\ref{one}) and (\ref{two})
\be
E(X_t^4) &\le&  27 (E(X_0^4) + 16 e^{4pt} E(\theta_0^4) + 4 C_4 p^2 (e^{2pt}-1)^2 E(\theta_0^4)) \no \\
\Rightarrow E(\int_{0}^{t} X_s^4 ds) &\le& 27 \Bigl( E(X_0^4) t + ((16 + C_4 p^2)(e^{4pt}-1)/4p \no\\
& & - 2 C_4 p^2 (e^{2pt}-1)/2p + C_4 p^2 t) E(\theta_0^4)) \no \\
& < &\infty  \label{x finite}
\ee
And
\be
E(\int_{0}^{t} F_s^4 \theta_s^4 ds) &\le& E(\theta_0 \int_{0}^{t}e^{4ps}e^{\int_{0}^{s}{4 \theta_u^2}du} ds) \no\\
&\le& E(\theta_0^4 \int_{0}^{t}e^{4ps}e^{\int_{0}^{s} 4 \theta_0^2 e^{2pu}du}ds) \no \\
&=& E(\theta_0^4 \int_{0}^{t} e^{4ps} e^{4 \theta_0^2 (e^{2ps}-1)/2p}ds)\no \\
&\le & E(\theta_0^4 e^{2 \theta_0^2 (e^{2pt}-1)/p} \int_{0}^{t}e^{4ps}ds)
< \infty \label{F finite}
\ee
So, from (\ref{x finite}) and (\ref{F finite}) we have:
\beo
E(Z_t^2) &<& \infty .
\eeo
This proves
1.

\begin{rmk}
{Although we have only proved that the second moment of $X_t$ is uniformly bounded, it is shown in Section \ref{momentx} that all even ordered moments (and therefore all moments) are uniformly bounded.}
\end{rmk}

\item
 Take $V_t = \frac{1}{\theta_t}$. Then
\beo
dV_t = - \frac{1}{\theta_t^2}d\theta_t &\Rightarrow& -\frac{1}{\theta_t^2} \theta_t(p - \frac{1}{\sqrt{2\pi}}|X_t| \theta_t)dt\\
&=& - V_t (p - \frac{|X_t|}{V \sqrt{2\pi}})dt\\
&=& (- V_t p+ \frac{|X_t|}{\sqrt{2\pi}})dt
\eeo
Multiplying by the integrating factor $e^{pt}$ we get
\begin{center}
\be
d(e^{pt} V_t) &=&\frac{e^{pt}|X_t|}{\sqrt{2\pi}}dt \no \\
\Rightarrow e^{pt}V_t - V_0 &=& \int_{0}^{t}\frac{1}{\sqrt{2\pi}}e^{pu}|X_u|du \no \\
\Rightarrow V_t &=& V_0 e^{-pt} + \int_{0}^{t} e^{-p(t-u)} \frac{|X_u|}{\sqrt{2\pi}} du  \\
\Rightarrow E(V_t)  &=& E(V_0)e^{-pt} + \int_{0}^{t} e^{-p(t-u)} \frac{E|X_u|}{\sqrt{2\pi}} du \no \\
&\le & E(V_0) e^{-pt} + \frac{M_0}{p \sqrt{2 \pi}} [1 - e^{-pt}]  \no \\
&& \mbox{(since $\sup_{t > 0} E(|X_t|) < M_0$, for some $M_0 > 0$, depending on $E(|X_0|)$)} \no \\
&\le & M_1 \ \ \mbox{(depends on $E(1/\theta_0)$, and $E(|X_0|)$)}  \no \\
\Rightarrow \sup_{t \ge 0} E(1/\theta_t) &\le & M_1 \no
\ee
\end{center}
Therefore $\theta_t > 0$ almost surely.

\item

Define, $\eta_t = 1/\theta_t$.
From the proof of the second part, it follows that
\beo
\eta_t &=& \eta_0 e^{-pt} + \int_{0}^{t} e^{-p(t-u)} \frac{|X_u|}{\sqrt{2\pi}} du \  >  0, \ \
\mbox{whenever $\eta_0 > 0$} .
\eeo

Also,

for any $t > 0$, for each sample points,
 $\int_{0}^{t} e^{-p(t-u)} \frac{|X_u|}{\sqrt{2\pi}} du = 0$
if only if $X_u = 0$ for almost all $u \in (0, t]$ w.r.t. Lebesgue measure. This is impossible,
since $X_u = 0 $ for almost all $u$ implies $dX_u = \theta_u dW_u$. Then $X_u$ would be non-zero for
almost all $u$, since \ as
\be
\theta_u &=& \frac{e^{pu}}{\eta_0 + \int_0^u (1/\sqrt{2\pi}) e^{ps} |X_s| ds} > 0 \label{theta eqn}
\ee
for all $u$. This contradicts our assumption that $X_u = 0$ for almost all $u$.
Therefore $X_u \ne 0$ for $u$ on a set of positive Lebesgue measure. Hence
$\int_{0}^{t} e^{-p(t-u)} \frac{|X_u|}{\sqrt{2\pi}} du > 0$
almost all $u$. Thus, $\eta_t > 0$ \ a.s.
Repeating the same argument will, in fact, give $X_u \neq 0$ for almost all $u \in (0, t]$, for any
$t > 0$.
\ This proves the third part.

\item
This follows from the above where we have shown that $\sup_{t \ge 0} E(|X_t|)< \infty$ and
$\sup_{t \ge 0} E(1/\theta_t) < \infty$.
Let $R_1$ and $R_2$ be two positive numbers. Then
\beo
P(|X_t|<R_1, |\eta_t|< R_2) &=& 1 - P( (|X_t|>R_1)\cup (|\eta_t|>R_2))\\
&>& 1 - (P(|X_t|>R_1) + P(|\eta_t|>R_2))\\
&>& 1- E(|X_t|)/R_1 - E(|\eta_t|)/R_2\\
\eeo
Hence given any $\epsilon>0$ we can choose $R_1, R_2$ sufficiently large so that
 $P(|X_t|<R_1, |\eta_t|< R_2) > 1-\epsilon$. This proves tightness of $(X_t, \eta_t)$.
 \end{enumerate}
 \qed

\subsection{Hypoelliptic condition}
\label{hypoelliptic}

Here we show that the vector fileds corresponding to \ref{SDE Normal} satisfies the Hormanders hypoelliptic condition (see the proposition for the statement of the condition).

{Since the condition requires smooth vector fields, we convert the drift and diffusion coefficients in \ref{coupled system} into smooth vector fields. \\
For this purpose define $$b_\epsilon(x, \eta)= \left( -\f {x}{2 \eta^2 } , \   - p\eta + \f{g_{\epsilon}(x)}{\sqrt{2 \pi}}) \right),$$ where $g_\epsilon(x) \rightarrow |x|$ as $\epsilon \downarrow 0$ in the  pointwise is a smooth function, and $\sigma(x, \eta)=\left(\begin{array}{ll}
 1/\eta & 0 \\
0 & 0 \\
\end{array}
\right)$
 as the drift and the diffusion coefficient respectively of our new equation. Such function $g_\epsilon$ can be constructed by convoluting the function $|x|$ with a mollifier. Defining in this fashion $|| b_{\epsilon}(\cdot, \cdot) - b(\cdot, \cdot) || \rightarrow 0$ as $\epsilon \downarrow 0$.} \\
Consider a SDE in the Stratonovich  form:
\be
dX_t &=& A_0(X_t)dt + \sum_{\alpha} A_{\alpha}(X_t)\circ dB^{\alpha}_t  \label{SDE_vector}.
\ee

where $f$ is a smooth function on $M$ and $\circ$ denotes Stratonovich integral. SDE in the Ito and the Startonovich form are interchangeable: for a multidimensional SDE given in the Ito's form
 \beo
 d \mathbf{X}_t &=& \mathbf{\tilde{b}}(t,\mathbf{X}_t) d \mathbf{X}_t + \mathbf{\sigma}(t, \mathbf{X}_t) d \mathbf{W}_t
 \eeo
can be readily converted into the Stratonovich form from the following equation:
 \beo
 {b}_i(t,\mathbf{x}) &=& \tilde{b}_i(t, \mathbf{x}) -  \frac{1}{2} \sum_{j=1}^{p} \sum_{k=1}^{n}  \frac{\partial \mathbf{\sigma}_{i,j}}{\partial x_k} \sigma_{k,j}; \; \; 1 \le i \le n
 \eeo
where $\mathbf{b}(t,x)=(b_i(t,x))'$ is the drift term for the Stratonovich form. From the form of $\sigma(x,\eta)$  we can say that $b^{\epsilon}$ and $\tilde{b^\epsilon}$ are the same. We can identify the drift and diffusion coefficients $A_0(\mathbf{X}_t)$ and $A_1(\mathbf{X}_1)$ can be identified as vector fields in $M$.\\
Here is the condition due to H\"{o}rmander, (\cite{hormander}):
\begin{prop}
\label{hypo prop}
Let $\{A_0, A_1, \ldots, A_n\}$ be $n$ smooth vector fields on $\mR^d$. Define the Lie Bracket $[V,W](x)$ between two vector fields $V$ and $W$ as $$ [V,W](x)= DV(x)W(x) - DW(x)V(x),$$ where $D(V(x))$ is the Frechet derivative.The (parabolic) Hormanders hypoelliptic condition is satisfied if :
$$
A_{j_0}(\my),  \;
[A_{j_0}(\my), A_{j_1}(\my)] \;
[[A_{j_0}(\my), A_{j_1}(\my)],A_{j_2}(\my)]  \;
\cdots  \;
[ [ [ [ A_{j_0}(\my), A_{j_1}(\my) ], A_{j_2}(\my) ], A_{j_3}(\my) ],\ldots, A_{j_n}(\my) ]
$$
spans $\mathbf{R}^d$ for every point $\my \in \mathbf{R}^d$ and any $\{ j_0, j_1, \ldots, j_d\} \in \{1,2, \ldots,n\} \ \ {j_0\le j_1, \ldots, j_n}$.
\end{prop}

\begin{lem}
\label{hypo condn}
The vector fields $A_0^{\epsilon}(\my)$ and $A_1(\my)$ satifies Hormanders hypoelliptic condition \ref{hypo prop}
\end{lem}

\textbf{Proof:}   Identifying \ref{SDE_vector} with \ref{SDE Normal} we have (writing $\my = (x,\eta)$):
\beo
A_0^\epsilon(\my) &=& - \frac{x}{2 \eta^2} \frac{\partial}{\partial x} + (-p \eta + \frac{g_{\epsilon}(x)}{\sqrt{2 \pi}}) \frac{\partial}{ \partial \eta}\\
A_1(\my)&=& \frac{1}{\eta} \frac{\partial}{\partial x}
\eeo

Therefore vectors corresponding to $A_0^\epsilon(\my), A_1(\my)$ and $[A_0^\epsilon(\my), A_1(\my)]$
are

$\left(   -\frac{x}{2 \eta^2} \ \ -p \eta +\frac{g_{\epsilon}(x)}{\sqrt{2 \pi}} \right)$, \ \
$\left(1/\eta, 0\right)$ and  $\left(  -\frac{1}{2\eta^3} +p \eta -\frac{g_{\epsilon}'(x)}{\sqrt{2 \pi}}, \frac{g_{\epsilon}(x)}{\sqrt{2 \pi}\eta} \right)$.\\

Since $\theta = 1/\eta > 0$ and $x \neq 0$ almost surely,
one obtains that these two vectors span  $\mathcal{R}^2$, almost surely.(what about $A_0$) \qed

\bigskip
It is well known that if the vector fileds $A_0(\my)$ and $A_1(\my)$ satisfy the above conditions then the solution of the SDE (\ref{SDE_vector}) admits a smooth density (see, \cite{nualart}).\\
Thus, satisfy
(parabolic) H\"{o}rmander's condition and hence even though the original diffusion is singular
 its transition probability has density (see \cite{kliemann}). Again, since the coupled diffusion is tight,
 it admits unique invariant probability by \cite{kliemann}.
and it admits a density.
\begin{rmk}
Note that although we are interested in the distribution of X showing tightness of the process $X$ only would
 not suffice since $\theta_t$ may be a function of $\{X_s; 0 \le s \le t\}$, so marginally $X$ may not be a
Markov process. Hence $\sup E|X_t|< M$ would give the tightness of X but it would not be appropriate to tell
 about the existence of a unique distribution.
\end{rmk}

\subsection{Uniform boundedness of moments of $X_t$}
\label{momentx}
\begin{lem}
\label{X_t finite}
For any $k \ge 1$,  moment of order $2k$ of $X_t$ is uniformly bounded, ie.,
$$ \sup_{t} E(X_t^{2k}) < \infty , $$ whenever $E(X_0^{2k})$ is finite.
\end{lem}

\textbf{Proof:}
Applying Ito's Lemma to $Y_t = X_t^{2k}$ we get
\beo
dX_t^{2k} &=& 2k X_t^{2k-1} dX_t + k(2k-1) X_t^2 \theta_t^2 dt\\
&=& \Bl -kX_t^{2k} \theta_t^2 + k(2k-1) X_t^{2k-2} \theta_t^2 \Br dt + 2k X_t^{2k-1} \theta_t dW_t\\
&\le& \Bl -kX_t^{2k} \theta_t^2 + k(2k-1)(aX_t^{2k} +b) \theta_t^2  \Br dt + 2kX_t^{2k-1} \theta_t dW_t ,
\eeo
since for any small $a > 0$, there exists $b$ large enough such that, $x^{2k-2} < ax^{2k} + b \ \ \forall x$.

Thus, for $0 < a < 1/(2k-1)$ we have
\beo
d X_t^{2k} &\le& -X_t^{2k} \theta_t^2 \Bl k - k(2k-1)a \Br dt + k(2k-1) b \theta_t^2 dt + 2kX_t^{2k-1} \theta_t dW_t\\
\Rightarrow d X_t^{2k} + C_k X_t^{2k} \theta_t^2 dt &\le& k(2k-1) b \theta_t^2 dt + 2kX_t^{2k-1} \theta_t dW_t ,
\eeo
where $C_k = k(1-(2k-1)a)>0$.
 Multiplying by the integrating factor $e^{C_k \intt \theta_u^2 du}:= F_t^{C_k}$ on both sides we get
\beo
d \Bl  X_t^{2k} F_t^{C_k} \Br &\le& k(2k-1)b \theta_t^2 F_t^{C_k} dt +  2k  F_t^{C_k} X_t^{2k-1} \theta_t d W_t \\
\Rightarrow X_t^{2k} F_t^{C_k} &\le& X_0^{2k} + k(2k-1)b \intt \theta_u^2 F_u^{C_k}du + 2k \intt F_u^{C_k} X_u^{2k-1} \theta_u dW_u\\
\Rightarrow X_t^{2k} &\le& X_0^{2k}  F_t^{-C_k} + k(2k-1)b F_t^{-C_k} \intt \theta_u^2 F_u^{C_k} du + 2k F_t^{-C_k} \intt F_u^{C_k} X_u^{2k-1} \theta_u dW_u
\eeo
Now,$$ \intt \theta_u^2 F_u^{C_k} du = (F_t^{C_k} -1)/{C_k}$$
\beo
\Rightarrow E\Bl X_t^{2k} \Br &\le& E \Bl F_t^{-C_k} X_0^{2k} \Br +  E \Bl \frac{1}{C_k} (1 - F_t^{-C_k})\Br
+ 2k  E \Bl F_t^{-C_k} \intt F_u^{C_k} X_u^{2k-1} \theta_u dW_u \Br
\eeo
Following the similar notation as before let $\ov{F}_{t,k} = F_t^{-C_k}$. Also define $Z_{t,k} = \ov{F}_{t,k} Y_{t,k}$ where $Y_{t,k} = \intt F_u^{C_k} X_u^{2k-1} \theta_u dW_u$.
Applying Ito's lemma to $Z_{t,k}$ we have
\beo
d Z_{t,k} &=& Y_{t,k} d \ov{F}_{t,k} + \ov{F}_{t,k} dY_{t,k} \\
&=& -C_k Y_{t,k} \theta_t^2 \ov{F}_{t,k} dt + \ov{F}_{t,k} X_t^{2k-1} \theta_t  F_t^{C_k} dW_t\\
&=& -C_k Z_{t,k} \theta_t ^2 dt + X_{t,k}^{2k-1} \theta_t dW_t.
\eeo
Now, taking $\tilde{Z}_{t,k} = -Z_{t,k}$, yields \
\beo
d\tilde{Z}_{t,k} =  C_k Z_{t,k} \theta_t ^2 dt - X_{t,k}^{2k-1} \theta_t dW_t = - C_k \tilde{Z}_{t,k} \theta_t ^2 dt + X_{t,k}^{2k-1} \theta_t d\tilde{W}_t
\eeo
where $\tilde{W}_t = - W_t$. \ Since $\tilde{Z}_{0,k} = - Z_{0,k} = 0 = Z_{0,k}$.
Hence $Z_{t,k}$ and $\tilde{Z}_{t,k}$ has the same distribution, i.e., the distribution is symmetric around 0.
Therefore to show that $E(Z_{t,k})=0$, \ $\forall t \ge 0$ \ one needs to show $Z_{t,k}$ has finite expectation \ $\forall t \ge 0$.
 Thus, as before,
 it is sufficient to show that $E(Z_{t,k}^2)< \infty$, \ $\forall t \ge 0$. Now,
\beo
Z_{t,k}^2 & = & \ov{F}_{t,k}^{-2} \Bl \intt F_s^{C_k} X_s^{2k-1} \theta_s dW_s  \Br ^2 \le \Bl \intt F_s^{C_k}  X_s^{2k-1} \theta_s dW_s  \Br^2 \ \ \mbox{a.s}\\
\Rightarrow E\Bl Z_t^2\Br &\le& E \Bl\intt F_s^{C_k} X_s^{2k-1} \theta_s  dW_s \Br^2 = E \Bl \intt F_s^{2 C_k} \theta_s^2 X_s^{4k-2}ds \Br \\
&\le & E \Bl \intt F_s^{4 C_k} \theta_s^2  ds \intt X_s^{8k-4} \theta_s^2  ds \Br^{\frac{1}{2}}
\le  \sqrt{ E\Bl \intt F_s^{4 C_k} \theta_s^2 ds \Br  E\Bl \intt X_s^{8k-4} \theta_s^2 ds \Br  } \\
&& \ \ \mbox{ by Cauchy-Schwarz inequality.}
\eeo
Again, $E(\intt F_s^{4 C_k} \theta_s^2 ds) = E((F_t^{4 C_k} - 1)/(4 C_k)) \le E(e^{4 C_k \theta_0^2 e^{2pt}}/(4C_k)) < \infty$.

For $\intt X_s^{8k-4} \theta_s^2 ds$ we have
\beo
X_t &=& X_0 - \intt \frac{X_s \theta_s^2}{2} ds + \intt \theta_s dW_s\\
\Rightarrow X_t^{8k-4} \theta_t^2 &\le& C_k' \Bl X_0^{8k-4} + \Bl  \intt \frac{|X_s| \theta_s^2}{2} ds \Br^{8k-4}   + \Bl  \intt \theta_s dW_s \Br^{8k-4}  \Br \theta_0^2 e^{2pt}  \\
&& \ \ \mbox{for some constant $C_k' >0$ independent of $X_t$.}
\eeo
For the second term, from (\ref{two})
\be
|\intt \frac{|X_s| \theta_s^2}{2}ds | &\le&  |\intt e^{p(t-s)} \frac{|X_s| \theta_s^2}{2} ds| \le \sqrt{\frac{\pi}{2}} |\theta_0 e^{pt} + \theta_t | \le \sqrt{2\pi} |\theta_0| e^{pt} \no \\
\Rightarrow \Bl \intt  \frac{ |X_s| \theta_s^2}{2}ds \Br^{8k-4} &\le& {\Bl \sqrt{2\pi}\Br }^{8k-4}  \Bl |\theta_0| e^{pt} \Br^{8k-4} \no \\
 \\
\Rightarrow \intt \Bl \int_{0}^{s} \frac{|X_u| \theta_u^2}{2} du\Br^{8k-4} ds &\le& \Bl 2\pi \Br^{4k-2} \Bl \theta_0^{8k-4} \intt e^{(8k-4)ps} ds  \Br
\label{first part}
\ee
For the third term, write $M_t := | \intt \theta_s dW_s|$ and apply BDG inequality:
\be
E\Bl M_t \Br^{8k-4} &\le& E \Bl \sup_{0 \le s \le t} M_s \Br^{8k-4} < C_{k0} E \Bl [M]_t^{4k-2}\Br\no \\
&=& C_{k0} E \Bl  \intt \theta_s^2 ds \Br^{4k-2} \le E \Bl  \intt \theta_0^2 e^{2ps} ds \Br^{4k-2}  \\
&& \ \ \mbox{ for some $C_{k0}>0$ and for all $t>0$}   \label{BDG one}
\ee
Combining (\ref{first part}) and (\ref{BDG one}) we have
\beo
E\Bl \intt X_s^{8k-4} \theta_s^2 ds\Br &\le& \Bl C_k' E(X_0^{8k-4}) + C_k' (2 \pi)^{4k-2} E \Bl \theta_0\intt e^{ ps} ds \Br^{8k-4} \\
&+& C_k' C_{k0} E \Bl  \intt \theta_0^2 e^{2ps} ds \Br^{4k-2} \Br \ E \Bl \theta_0^2  \intt e^{2ps} ds \Br  \\
&<& \infty
\eeo
This proves $$ E(Z_{t,k}^2) < \infty$$ and hence the lemma. \qed \\

\subsection{Finiteness of Time average of moments of $\theta$}
\label{timeave}

In this section $C$ will stand for a generic constant that might take different values in different
situations. We assume throughout that, for a fixed $k\ge 1$, \ $E(X_0^{2k})$,
$E(\theta_0^{2k})$ and $E(\eta_0^{2k})$ are finite. For non-random initial data this is already assured.
We proceed sequentially by the following steps:\\
\underline{Step 1:} We have $$\sup_{t>0} E(X_t^{2k}) < \infty$$. \qed \\
\underline{Step 2:} We show that $$\sup_{t>0}  E(\eta_t^{2k})< \infty .$$
Using
$$ \eta_t = \eta_0 e^{-pt} + \int_{0}^{t} e^{-p(t-u)} \frac{|X_u|}{\sqrt{2\pi}} du $$,
to get,
\beo
\eta_t^{2k} &=& \Bl \eta_0 e^{-pt} + \int_{0}^{t} e^{-p(t-u)} \frac{|X_u|}{\sqrt{2\pi}} du \Br^{2k} \no\\
 &\le & 2^{2k-1} \Bl \eta_0^{2k} e^{-2kpt} + \Bl \int_{0}^{t} e^{-p(t-u)} \frac{|X_u|}{\sqrt{2\pi}} du\Br^{2k} \Br \no\\
&\le & 2^{2k-1} \Bl \eta_0^{2k} e^{-2kpt}
+ \Bl\frac{1 - e^{-pt}}{p\sqrt{2\pi}}\Br^{2k} \Bl \frac{p}{1 - e^{-pt}} \int_{0}^{t} e^{-p(t-u)} |X_u|^{2k} du \Br \Br ,
\eeo
where last inequality follows from Jensen's inequality.
Thus,
\be
\label{eta}
\supt E(\eta_t^{2k})
&\le & \supt 2^{2k-1} \Bl E(\eta_0^{2k}) e^{-2kpt}
+ \Bl\frac{1 - e^{-pt}}{p\sqrt{2\pi}}\Br^{2k} \Bl \frac{p}{1 - e^{-pt}} \int_{0}^{t} e^{-p(t-u)} E(|X_u|^{2k}) du \Br \Br \no\\
&\le & 2^{2k-1} \Bl E(\eta_0^{2k})
+ \Bl\frac{1}{p\sqrt{2\pi}}\Br^{2k} \supt \Bl \frac{p}{1 - e^{-pt}} \int_{0}^{t} e^{-p(t-u)} E(|X_u|^{2k}) du \Br \Br \no\\
&\le & 2^{2k-1} \Bl E(\eta_0^{2k})
+ \Bl \frac{1}{p\sqrt{2\pi}}\Br^{2k} \supt E(|X_t|^{2k}) \Br < \infty .
\ee
Hence the result. \qed

\bigskip

\underline{Step 3:} We now prove $$\sup_{t>0} \frac{1}{t} \int_{0}^{t} E(|X_u| \theta_u) du < \infty.$$
Note,
\beo
d (1 + \theta_t) &=& d \theta_t = \theta_t ( p - |X_t| \theta_t/\sqrt{2 \pi}) dt \\
&=& (1+ \theta_t)( p - |X_t| \theta_t/\sqrt{2 \pi}) dt - ( p - |X_t| \theta_t/\sqrt{2 \pi}) dt \\
\Rightarrow d(1 + \theta_t) + (p - \frac{|X_t| \theta_t}{\sqrt{2 \pi}})dt &=& (1+ \theta_t) ( p - |X_t| \theta_t/\sqrt{2 \pi}) dt\\
\Rightarrow \frac{d(1 + \theta_t) }{1 + \theta_t} + \frac{1}{1 + \theta_t}(p - \frac{|X_t| \theta_t}{\sqrt{2 \pi}})dt &=&  (p - \frac{|X_t| \theta_t}{\sqrt{2 \pi}})dt\\
\Rightarrow \frac{d(1+\theta_1)}{1+ \theta_t}+ \frac{1}{\sqrt{2\pi}} |X_t| \theta_t dt &\le& p + \frac{|X_t|}{\sqrt{2 \pi}} dt  \\
\Rightarrow \log \frac {1+ \theta_t}{1+ \theta_0} + \fpi \intt |X_u| \theta_u du &\le & pt + \fpi \intt |X_u|du\\
\Rightarrow  \frac{1}{t} \intt |X_u| \theta_u du &\le& \sqrt{2 \pi} p+  \frac{1}{t} \intt |X_u|du + \sqrt{2 \pi} \frac{\log(1 + \theta_0)}{t} .
\eeo
Thus, $ \frac{1}{t} \intt E(|X_u| \theta_u) du \le \sqrt{2 \pi} p+  \frac{1}{t} \intt E(|X_u|) du + \sqrt{2 \pi} \frac{E(\log(1 + \theta_0))}{t}$.

Therefore, using the moment bounds for $X$,
\beo
\sup_{t>0} \frac{1}{t} \intt E(|X_u| \theta_u) du &<& C
\eeo

\qed

\underline{Step 4:}
We now prove by induction, that for any $k\ge 1$,

\be
\label{theta_moment}
\supt \frac{1}{t} \intt  E(\theta_u^{\frac{k}{2}}) du
< C .
\ee

Let, as before, $\eta_t = \frac{1}{\theta_t}$ then $d \eta_t =( - p \eta_t + |X_u|/ \sqrt{2\pi}) dt $. \\
Applying Ito's lemma to $Y_t = X_t^2 \eta_t^{2 - k/2}$, with $k \ge 1$ positive integer, we get
\be
dY_t &=& 2 X_t \eta_t^{2 - k/2} dX_t + (2-k/2) X_t ^2 \eta_t^{1-k/2} d \eta_t + \frac{1}{2} 2 \eta_t^{2 - k/2} (dX_t)^2 \no \\
&=& 2 X_t \eta_t^{2 - k/2} (- \frac{X_t}{2 \eta_t^2} dt + \frac{1}{\eta_t} dW_t) + (2 - k/2) X_t^2 \eta_t^{1 - k/2} ( - p \eta_t dt + \frac{|X_t|} {\sqrt{2 \pi} }dt ) \no \\
&+&  \eta_t^{2 - k/2} \eta_t^{-2} dt  \no \\
&=& \Bigl( - X_t^2  \eta_t^{- k/2} - p (2 - k/2) X_t^2 \eta_t^{2 - k/2} + \frac{2 - k/2}{\sqrt{2 \pi}} |X_t|^{3} \eta_t^{1 - k/2} + \eta_t^{- k/2} \Bigr) dt \no \\
&+& 2 X_t \eta_t^{1 - k/2} dW_t \label{stoc eqn}
\ee

Thus, integrating both side from $0$ to $t$ and rearranging we get,
\be
\intt \theta_t^{\frac{k}{2}} dt &=&  X_t^2 \eta_t^{\frac{4-k}{2}} - X_0^2 \eta_0^{\frac{4-k}{2}} + \intt X_s^2 \theta_s^{\frac{k}{2}} dt + \frac{(4-k)p}{2} \intt X_s^2 \eta_t^{\frac{4-k}{2}} \no \\
&& - \frac{2 - k/2}{\sqrt{2 \pi}} \intt |X_s|^3 \eta_t^{\frac{2-k}{2}}dt - 2 \intt X_s \eta_s^{\frac{2-k}{2}} dW_s \no \\
\Rightarrow \frac{1}{t} \intt \theta_s^{\frac{k}{2}} dt  &=& \frac{1}{t}(X_t^2 \eta_t^{\frac{4-k}{2}} - X_0^2 \eta_0^{\frac{4-k}{2}}) + \frac{1}{t} \intt X_s^2 \theta_s^{\frac{k}{2}} dt + \frac{(4-k)p}{2t} \intt X_s^2 \eta_t^{\frac{4-k}{2}} dt \no \\
&& - \frac{2 - k/2}{t\sqrt{2 \pi}} \intt |X_s|^3 \eta_t^{\frac{2-k}{2}}dt - \frac{2}{t} \intt X_s \eta_t^{\frac{2-k}{2}} dW_s \label{int_theta}
\ee

First consider the first, third and fourth term in \ref{int_theta}. We proceed in steps\\
\underline{Step 4a}: $1 \le k \le 2$: Since we have proved in the earlier steps that the (2k)th moment of $X_t$ and $\eta_t$ is uniformly bounded we, by an application of C-S inequality, can say that
\be
\sup_{t>0} \frac{1}{t} E(X_t^2 \eta_t^{\frac{4-k}{2}} + X_0^2 \eta_0^{\frac{4-k}{2}}) &<& C
\label{first_k_1},  \\
\sup_{t>0} \frac{1}{t} \intt E(X_s^2 \eta_s^{\frac{4-k}{2}}) dt
 &<& C ,\label{third_k_1} \\
\mbox{and} & & \no \\
\sup_{t>0} \fract \intt |X_s|^3 \eta_s^{\frac{2-k}{1}}dt &<& C. \label{fourth_k_1}
\ee
For the fifth term $\fract \intt X_s \eta_s^{\frac{2-k}{2}}dW_s$ is a square integrable martingale (by Ito's issometry) and hence $$ \fract \intt E \Bl X_s  \eta_s^{\frac{2-k}{2}}\Br d W_s=0.$$
\qed\\
{\underline{Step 4b}: Consider the second term in \ref{int_theta}. For any $k \ge 1$,
\be
\frac{1}{t} \intt X_s^2 \theta_s^{\frac{k}{2}} ds &=& \frac{1}{t} \intt (|X_s|^{\frac{k}{k+1}}
\theta_s^{\frac{k}{2}}) ( |X_s|^{\frac{k+2}{k+1}} ) ds  \no \\
&\le&   \Bl \frac{1}{t} \intt  |X_s| \theta_s^{\frac{k+1}{2}} dt \Br ^{ \frac{k}{k+1}}
\Bl \frac{1}{t} \intt |X_s|^{k+2} dt \Br ^ {\frac{1}{k+1}}  \no \\
\Rightarrow E \Bl \fract \intt X_s^2 \theta_s^{\frac{k}{2}} \Br
&\le& E \Bl \Bl \frac{1}{t} \intt  |X_s| \theta_s^{\frac{k+1}{2}} dt \Br ^{ \frac{k}{k+1}}
\Bl \frac{1}{t} \intt |X_s|^{k+2} dt \Br ^ {\frac{1}{k+1}} \Br \no\\
&\le &   \Bl E \Bl \frac{1}{t} \intt  |X_s| \theta_s^{\frac{k+1}{2}} dt \Br \Br ^{ \frac{k}{k+1}}
\Bl E \Bl \frac{1}{t} \intt |X_s|^{k+2} dt \Br \Br ^ {\frac{1}{k+1}}  \no\\
&=& \Bl \frac{1}{t} \intt E(|X_s| \theta_s^{\frac{k+1}{2}}) dt \Br ^{ \frac{k}{k+1}}
 \Bl \frac{1}{t} \intt E(|X_s|^{k+2}) dt  \Br ^ {\frac{1}{k+1}}  \no\\
\Rightarrow \sup_{t>0} \fract \intt E(X_s^2 \theta_s^{\frac{k}{2}}) &\le&
 \Bl \sup_{t>0} \frac{1}{t} \intt  E(|X_s| \theta_s^{\frac{k+1}{2}}) dt  \Br ^{ \frac{k}{k+1}}
 \Bl \sup_{t>0} \frac{1}{t} \intt E(|X_s|^{k+2}) dt \Br ^ {\frac{1}{k+1}}  \no\\
&<& C ,
\label{prefourth}
\ee
We only have to show that the first term on the RHS is finite. We proceed by induction : We have proved (Step 3 in this section ) that $$ \sup_{t>0} \fract \intt E \Bl |X_s| \theta_s \Br dt < C$$
Therefore from \ref{first_k_1}, \ref{third_k_1} and \ref{fourth_k_1}, since $k=1$, we have that
$$ \sup_{t>0} \fract \intt E\Bl \theta_u^{\frac{1}{2}}\Br du < C.$$  Assume that the hypothesis is true for $k=m-1$, ie.,
\be
\sup_{t>0} \fract \intt E\Bl \theta_s ^{\frac{m-1}{2}}\Br ds &<& C \label{ind1} \\
\sup_{t>0} \fract \intt E\Bl |X_s| \theta_s^{\frac{m}{2}}\Br &<& C \label{ind2}
\ee
This also imply,
\be
\label{ind_1_3_hyp}
\supt \fract \intt \theta_s^{\frac{m-3}{2}} ds &<& C .
\ee
Then, for $k=m$,
\be
\label{induction}
d\theta_t^{\frac{m-1}{2}} = \frac{m-1}{2} \theta_t^{\frac{m-1}{2} - 1} d\theta_t
&=& \frac{m-1}{2} \theta_t^{\frac{m-1}{2}} ( p - |X_t| \theta_t/\sqrt{2 \pi}) dt \no\\
&=& \frac{p(m-1)}{2}  \theta_t^{\frac{m-1}{2}} dt  - |X_t| \theta_t^{\frac{m+1}{2}}/\sqrt{2 \pi} dt \no\\
\Rightarrow \intt |X_s| \theta_s^{\frac{m+1}{2}}/\sqrt{2 \pi} ds
&=& \intt \frac{p(m-1)}{2}  \theta_s^{\frac{m-1}{2}} ds - (\theta_t^{\frac{m-1}{2}} - \theta_0^{\frac{m-1}{2}}) \no\\
\Rightarrow  \sqrt{2 \pi} E(\theta_t^{\frac{m-1}{2}}) + \intt E(|X_s| \theta_s^{\frac{m+1}{2}}) ds
&=& \frac{p(m-1)\sqrt{2 \pi}}{2} \intt  E(\theta_s^{\frac{m-1}{2}}) ds + \sqrt{2 \pi} E(\theta_0^{\frac{m-1}{2}})  \no\\
\Rightarrow \sup_{t>0} \fract \intt E \Bl |X_s| \theta_s^{\frac{m+1}{2}} \Br &\le&  \sqrt{2 \pi} \sup_{t>0} \fract E(\theta_t^{\frac{m-1}{2}}) + \sup_{t>0} \fract \intt E(|X_s| \theta_s^{\frac{m+1}{2}}) ds  \no \\
&\le& \frac{p(m-1)\sqrt{2 \pi}}{2} \sup_{t>0} \fract \intt E \Bl \theta_s^{\frac{m-1}{2}}\Br + \sqrt{2 \pi} E \Bl \theta_0^{\frac{m-1}{2}}\Br.
\ee
By the induction hypothesis \ref{ind1} and \ref{ind2}. This proves the proposition.

Therefore the second term on \ref{int_theta} is finite, for any $k \ge 1$. \qed

\underline{Step 4c}: $k \ge 3$.\\

For $3 \le k \le 4$ we can claim similarly as above that:
\beo
\sup_{t>0} \frac{1}{t} E(X_t^2 \eta_t^{\frac{4-k}{2}} + X_0^2 \eta_0^{\frac{4-k}{2}}) &<& C \\
\sup_{t>0} \frac{1}{t} \intt E(X_s^2 \eta_s^{\frac{4-k}{2}}) dt
 &<& C .
\eeo
For $4 \le k$:
\beo
\sup_{t>0} \fract E \Bl X_t^2 \eta_t^{\frac{4-k}{2}} \Br &=& \sup_{t>0} \fract  E \Bl  X_t^2 \theta_t^{\frac{k-4}{2}} \Br\\
&=& \sqrt{\supt \fract E \Bl X_t^4\Br \supt \fract E \Bl \theta_t^{k-4} \Br} < C, \\
\mbox{and,} & & \\
\supt \fract \intt E \Bl X_s^2 \eta^{\frac{4-k}{2}}\Br &<& C \; \; \mbox{by Step 4b.}
\eeo

For the fourth term we apply the Holder's inequality with $p = k-1$ and $q= (k-1)/(k-2)$ to get
\beo
|X_s|^3 \theta_s^{\frac{k-2}{2}} \le (|X_s|^3)^p/p + (\theta_s^{\frac{k-2}{2}})^q/q
&=& \frac{k-2}{k-1} |X_s|^{2(k-1)} + \frac{1}{k-1} \theta_s^{\frac{k-1}{2}} ,\\
\Rightarrow \sup_{t>0} \fract \intt E\Bl |X_s|^3 \theta_s^{\frac{k-2}{1}}\Br ds &\le& \frac{k-2}{k-2} \sup_{t>0} \fract \intt E \Bl |X_s|^{2(k-1)} \Br ds + \frac{1}{k-1} \sup_{t>0} \fract \intt  E \Bl |\theta_s|^{\frac{k-1}{2}} \Br ds.
\eeo
Finiteness of the first term has been proved in Section \ref{momentx}. The second term is finite by virtue of Step 4b.\\
The last step is to show that the fifth term in \ref{int_theta} is 0. Now,
\beo
E \Bl\intt  X_s \eta_s^{\frac{2-k}{2}} dW_s  \Br^2&=& E \Bl \intt   X_s^2 \eta_s^{2-k} ds\Br\\
&=& E\Bl \intt X_s^2 \theta_s^{k-2} ds \Br < C \; \mbox{by Step 4b}
\eeo
Therefore $ \intt X_s \eta_s^{\frac{2-k}{2}} dW_s$ is a square integrable martingale and hence $$E\Bl \intt X_s \eta_s^{\frac{2-k}{2}} dW_s \Br = 0 $$ for any $t>0$. \qed

Steps 4a, 4b and 4c together proves Step 4. \qed

\subsection{Identifying the limiting distribution}
\label{steins}

In this section we obtain the limiting moment of $E(X_t^{2k})$ and show that is equal to $\frac{(2k)!}{2^k k!}$ for any $k \in \cN$. Hence by the uniqueness of the moment generating functions we can say that the limiting distribution of $X_t$ is $N(0,1)$. We proceed by induction
\begin{enumerate}
\item $\lim_{t \rightarrow \infty} E(X_t^{2}) = 1$.\\
Applying Ito's lemma to $X_t^2$ we have
\beo
dX_t^{2}&=& \Bl -X_t^2 \theta_t^2 + \theta_t^2 \Br dt + 2 X_t \theta_t dW_t .
\eeo
Writing $F_k(t) = {k \intt \theta_s^2 ds}$ and multiplying by the $IF=e^{F_1(t)}$ on both sides we have
\beo
d \Bl X_t^2 e^{F_1(t)} \Br &=& e^{F_1(t)}\theta_t^2 dt + e^{F_1(t)}X_t \theta_t dW_t \\
\Rightarrow X_t^{2} &=& e^{-F_1(t)} [X_0^2  + \intt e^{F_1(s)} \theta_s^2 ds + \intt e^{F_1(s)} X_s \theta_s dW_s]\\
&=& e^{-F_1(t)}[ X_0^2 + \intt d(e^{F_1(s)}) + \intt e^{F_1(s)} X_S \theta_s dW_s]\\
&=& e^{-F_1(t)}[X_0^2 + e^{F_1(t)}-1 + \intt e^{F_1(s)} X_s\theta_s dW_s] \\
&=& X_0^2 e^{-F_1(t)} + 1 - e^{-F_1(t)} + \intt e^{F_1(s)- F_1(t)} X_s \theta_s dW_s\\
\Rightarrow E(X_t^{2})&=& E(e^{-F_1(t)}) E(X_0^2) + 1 - E \Bl e^{-F_1(t)} \intt e^{F_1(s)} X_s \theta_s dW_s \Br
\eeo
We have proved in Lemma \ref{lemma tightness} that the third expectation is sero.
Therefore
\be
 E(X_t^2) &=&  E(e^{-F_1(t)}) E(X_0^2) +1 -  \lim_{t \rightarrow \infty}E(e^{-F_1(t)})   \no \\
\lim_{t \rightarrow \infty} E(X_t^2) &=& E(X_0^2) \lim_{t \rightarrow \infty} E(e^{-F_1(t)})   + 1 -  \lim_{t \rightarrow \infty}E(e^{-F_1(t)})  \label{Xt2}
\ee

\beo
\frac{F_1(t)}{t} &=&\fract \intt \theta_s^2 ds \ge \frac{1}{\fract \intt \frac{1}{\theta_s^2}ds}  = \frac{1}{\fract \intt \eta_s^2 ds }.
\eeo
Therefore,
\beo
e^{-F_1(t)} &=& \frac{1}{e^{F_1(t)}} \le \frac{1}{F_1(t)}=\frac{1}{t \frac{F_1(t)}{t}}    \ \ \mbox{ since $F_1(t)>0$} \\
&\le& \fract \fract \intt \eta_s^2 ds \\
\Rightarrow E(e^{-F_1(t)}) &\le& \fract E\Bl \fract \intt \eta_s^2 ds\Br \le \fract C
\le \fract C.
\eeo
where $C = \sup_{t >0} E(\fract \intt \eta_s^2 ds) < \infty$ from Step 2 of Section \ref{timeave}. So
\beo
\lim_{t \rightarrow \infty} E(e^{-F_1(t)}) &=& 0
\Rightarrow \lim_{t \rightarrow \infty} E(X_t^2)= 1  \ \ \mbox{from \ref{Xt2}}.
\eeo
\item Assume this to be true for $k-1$, ie., $\lim_{t \rightarrow \infty} E(X_t^{2m}) =\frac{(2m)!}{2^m m!}$ for $1 \le m \le k-1$.
\item $\lim_{\rightarrow \infty} E(X_t^{2k}) = \frac{(2k)!}{2^k k!}$.\\
We have that
\beo
dX_t^{2k} &=& \Bl-kX_t^{2k}\theta_t^2  + k(2k-1) X_t^{2k-2} \theta_t^2 \Br dt + k X_t^{2k-1} \theta_t dW_t .
\eeo
Multiplying with the $IF= e^{F_k(t)}$ we have that
\be
d\Bl X_t^{2k} e^{F_k(t)} \Br &=& k(2k-1) e^{F_k(t)} X_t^{2k-2} \theta_t^2 dt + k e^{F_k(t)} X_t^{2k-1} \theta_t dW_t \no \\
\Rightarrow X_t^{2k} &=& e^{-F_k(t)}[ X_0^{2k} + (2k-1) \intt k e^{F_k(t)} X_s^{2k-2} \theta_s^2 ds + k \intt e^{F_k(s)} X_s^{2k-1} \theta_s dW_s]\no \\
\Rightarrow E(X_t^{2k}) &=& E(e^{-F_k(t)}) E(X_0^{2k}) + (2k-1) E(\intt k e^{-F_k(t)} e^{F_k(s)} X_s^{2k-2} \theta_s^2 ds) \no \\
&+ & E \Bl e^{F_k(t)} \intt  ke^{F_k(s)} X_s^{2k-1} \theta_s  dW_s \Br\no
\ee
We have proved in Lemma \ref {X_t finite} that the third expectation is zero.Therefore
\beo
\Rightarrow \lim_{t \rightarrow \infty} E(X_t^{2k}) &=& E(X_0^{2k}) \lim_{t \rightarrow \infty} E(e^{-F_k(t)}) \no \\
&+& (2k-1) \lim_{t\rightarrow \infty}E(\intt k e^{-F_k(t)} e^{F_k(s)} X_s^{2k-2} \theta_s^2 ds) \label{Xt2k}
\eeo
\end{enumerate}
Arguing in the same way as the case for $F_1(t)$:
\beo
e^{-F_k(t)}&=& \frac{1}{e^{F_k(t)}}\le \frac{1}{F_k(t)} = \frac{1}{t\frac{F_k(t)}{t}}\\
&\le& \fract \frac{k}{t} \intt \eta_s^2 ds \Rightarrow \lim_{t \rightarrow \infty} E(e^{-F_k(t)}) = 0.
\eeo
Define, for $0 \le m \le k-1$,
\beo
A_{k,2m}(t) &=& E\Bl \intt k e^{-F_k(t)} e^{F_k(s)} \theta_s^2 X_s^{2m} ds \Br \\
&=& E\Bl  e^{-F_k(t) } \intt X_s^{2m} d(e^{F_k(s)}) \Br .
\eeo
The term inside the expectation is
\be
\intt X_s^{2m}d(e^{F_k(s)}) &=& X_t^{2m}e^{F_k(t)} - X_0^{2m} - \intt e^{F_k(s)} d(X_s^{2m}) \no \\
&=& X_t^{2m} e^{F_k(t)} - X_0^{2m} - \intt e^{F_k(s)} \Bl (-m X_s^{2m} \theta_s^2 + m(2m-1) X_s^{2m-2} \theta_s^2) ds \no \\
&+& \intt 2m X_s^{2m-1} \theta_s dW_s \Br \no \\
\Rightarrow e^{-F_k(t)} \intt X_s^{2m} d(e^{F_k(s)}) &=&  X_t^{2m} - X_0^{2m} e^{-F_k(t)} + e^{-F_k(t)} \intt m e^{F_k(s)} X_s^{2m} \theta_s^2 ds \no \\
&-& e^{-F_k(t)} \intt m (2m-1) X_s^{2m-2} \theta_s^2 ds
+ e^{-F_k(t)} \intt 2m X_s^{2m-1} \theta_s d W_s \no \\
\ee
And so,
\be
\Rightarrow A_{k,2m}(t) &=& E(X_s^{2m}) -E(e^{-F_k(t)}) E(X_0^{2m}) + \frac{m}{k}A_{k,2m} \no \\
&-& \frac{m(2m-1)}{k} A_{k,2m-2} +  0 \label{Ak2m}.
\ee
Defining $B_{k,m}= \lim_{t \rightarrow \infty} A_{k,m}(t)$ (the limit exists since the integrand is a non negative quantity). Applying limits on both sides of \ref{Ak2m} we get
\beo
B_{k,2m} &=& \lim_{t \rightarrow \infty} E(X_t^{2m}) -0 +\frac{m}{k} B_{k,2m} - \frac{m(2m-1)}{k} B_{k,2m-2}\\
\Rightarrow (1- \frac{m}{k}) B_{k,2m} &=& \lim_{t \rightarrow \infty} E(X_t^{2m}) - \frac{m(2m-1)}{k}B_{k,2m-2}\\
\Rightarrow B_{k,2m} &=& \frac{k}{k-m} \lim_{t \rightarrow \infty} E(X_t^{2m}) - \frac{m(2m-1)}{k-m}B_{k,2m-2}
\eeo
Substituing different values of $m=0,1,2,\ldots,k-1$ we get
\beo
B_{k,0}&=& 1\\
B_{k,2} &=& \frac{k}{k-1}1-\frac{1}{k-1}1 = 1\\
B_{k,4}&=& \frac{k}{k-2}3 - \frac{2.3}{k-2}1 = 3 \\
B_{k,6}&=& \frac{k}{k-3}5.3-\frac{3.5}{k-3}3=5.3 \\
B_{k,8}&=& \frac{k}{k-4}7.5.3 - \frac{4.7}{k-4}5.3 = 7.5.3\\
\ldots\\
B_{k,2k-2}&=& k (2k-3)(2k-5)\ldots 3.1 - (k-1)(2k-1)B_{k,2k-4}\\
&=& k(2k-3)(2k-5)\ldots3.1 -(k-1)(2k-3)\ldots3.1=(2k-3)(2k-5)\ldots3.1\\
&=&\frac{(2k-2)!}{2^{k-1}(k-1)!}
\eeo
From equation \ref{Xt2k} :
\beo
\lim_{t \rightarrow \infty} E(X_t^{2k}) &=& (2k-1) B_{k,2k-2}\\
&=& (2k-1) \frac{(2k-2)!}{2^{k-1}(k-1)!} = \frac{2k(2k-1)!}{2^k k!} \\
&=& \frac{(2k)!}{2^k k!}
\eeo
To find the odd moments of $X_t$ we perform the same procedure as above. We have
$$
d X_t = -X_t \frac{\theta_t^2}{2} dt + \theta_t dW_t
$$
Define $G_1(t)= \frac {1}{2} \intt \theta_s^2 ds$. Multiply by the IF=$e^{G_1(t)}$ on both sides we have
\beo
d(e^{G_1(t)} X_t) &=& e^{F_1(t)} \theta_t d W_t \\
\Rightarrow X_t &=& X_0 e^{-G_1(t)} + e^{-G_1(t)} \intt e^{F_1(s)} \theta_s dW_s \\
\Rightarrow E(X_t) &=& E(X_0) E(e^{-G_1(t)}) + E \Bigl( e^{-G_1(t)} \intt e^{F_1(s)} \theta_s d W_s\Bigr)
\eeo
Applying the same argumenst as above we can show that $$\lim_{t \rightarrow \infty} E(e^{-G_1(t)}) = 0 $$ and so $$  \lim_{t \rightarrow \infty} E(X_t) = 0.$$
Now we apply Mathematical Induction. Let us assume that $$\lim_{t \rightarrow \infty} E(X_t^{2m-1})=0$$ where $m=1,2,\ldots,k$ is any positive integer. Applying Ito's lemma to $X_t^{2k+1}$ we get
\beo
d(X_t^{2k+1})&=& (2k+1) X_t^{2k} d X_t + \frac{1}{2}(2k+1)2k X_t^{2k-1} \theta_t^2 dt\\
&=& (2k+1) X_t^{2k} \Bigl( -X_t \frac{\theta_t^2}{2} dt + \theta_t d W_t \Bigr) + (2k+1)k \theta_t^2 X_t^{2k-1} dt \\
&=& \Bigl( - \frac{1}{2}(2k+1) X_t^{2k+1} \theta_t^2 +(2k+1) k X_t^{2k-1} \theta_t  \Bigr)dt + (2k+1) \theta_tX_t^{2k}dW_t.
\eeo
Define $G_k(t)= \frac{2k+1}{2} \intt \theta_s^2 ds$ and the integrating factor as $e^{G_k(t)}$. Multiplying by it on both sides of the above equation we get:
\be
d \Bigl(  X_t^{2k+1} e^{G_k(t)} \Bigr) &=& k(2k+1) e^{G_k(t)} \theta_t^2 X_t^{2k-1} dt + (2k+1) e^{G_k(t)} \theta_t X_t^{2k} dW_t \no \\
\Rightarrow X_t^{2k+1} &=& e^{-G_k(t)} \Bigl[  X_0^{2k+1} +k(2k+1) \intt e^{G_k(s)} \theta_s^2 X_s^{2k-1}ds +(2k+1) \intt e^{G_k(s)} \theta_s X_s^{2k} dW_s \Bigr] \no \\
E(X_t^{2k+1}) &=& E(e^{-G_k(t)}) E(X_0^{2k+1}) + E\Bigl( k(2k+1) e^{-G_k(t)}\intt e^{G_k(s)}  \theta_s^2 X_s^{2k-1} ds \Bigr) \no \\
&+& (2k+1) E \Bl e^{-G_k(t)}\intt  e^{G_k(s)} X_s^{2k} \theta_s dW_s \Br \label{Xt2kp1}
\ee
We now try to prove that the third expectation is zero.

Following the similar notation as before let $\ov{F}_k(t) = e^{-G_k(t)}$. Also define $Z_k(t) = \ov{F}_k(t) Y_k(t)$ where $Y_k(t) = \intt e^{G_k(s)} X_s^{2k} \theta_s dW_s$.
Applying Ito's lemma to $Z_k(t)$ we have
\beo
d Z_k(t) &=& Y_k(t) d \ov{F}_k(t) + \ov{F}_k(t) dY_k(t) \\
&=& -\frac{2k+1}{2} Y_k(t) \ov{F}_k(t) \theta_t^2 dt + \ov{F}_k(t)F_k(t) X_t^{2k} \theta_t dW_t\\
&=& -\frac{2k+1}{2} Z_k(t) \theta_t^2 dt + X_t^{2k} \theta_t d W_t
\eeo
Now, taking $\tilde{Z}_k(t) = -Z_k(t)$, yields
\beo
d\tilde{Z}_k(t) &=&  \frac{2k+1}{2} \tilde{Z}_k(t) \theta_t^2 dt  + X_t^{2k} dW_t \\
&=& \frac{2k+1}{2} \tilde{Z}_k(t) \theta_t^2 dt - X_t^{2k} d \tilde{W}_t
\eeo
where $\tilde{W}_t = - W_t$. \ Since $\tilde{Z}_{0,k} = - Z_{0,k} = 0 = Z_{0,k}$.
Hence $Z_k(t)$ and $\tilde{Z}_k(t)$ has the same distribution, i.e., the distribution is symmetric around 0.
Therefore to show that $E(Z_{t,k})=0$, \ $\forall t \ge 0$ \ one needs to show $Z_{t,k}$ has finite expectation \ $\forall t \ge 0$.
 Thus, as before,
 it is sufficient to show that $E(Z_k(t)^2)< \infty$, \ $\forall t \ge 0$. Now,
\beo
Z_k(t)^2 &=& \ov{F}_k(t)^2 Y_k(t)^2 = \ov{F}_k(t)^2 \Bl  \intt e^{G_k(s)} X_s \theta_s dW_s \Br^2 \\
&=& e^{-2 G_k(t)} \Bl \intt e^{G_k(s)} X_s \theta_s dW_s \Br^2 \\
\Rightarrow E(Z_k(t)^2) &\le& E\Bl \intt e^{2G_k(s)} X_s^{4k} \theta_s^2 ds \Br \\
&\le & \sqrt{E \Bl  \intt e^{4 G_k(s)} \theta_s ^2 ds \Br  E \Bl \intt X_s^{8k} \theta_s ^2 ds \Br}
\eeo
Now, as in above
\beo
\Bl \intt e^{4G_k(s)} \theta_s^2 ds \Br & = &\frac{1}{2(2k+1)} ( e^{2(2k+1)F(t)} -1) \ \  \mbox{where $F(t) = \intt \theta_s^2 ds $}\\
&<& C \Rightarrow E\Bl \intt e^{4G_k(s)} \theta_s^2 ds \Br < \infty
\eeo
Now, one can write
\beo
 X_t^{8k} \theta_t^2 &\le& C_k' \Bl X_0^{8k} + \Bl  \intt \frac{|X_s| \theta_s^2}{2} ds \Br^{8k}   + \Bl  \intt \theta_s dW_s \Br^{8k}  \Br \theta_0^2 e^{2pt}  \\
\eeo
The second and the third term can be shown to be finite (the details are same as in \ref{momentx} expect that the exponent of $X_t$ is  $8k$, so it is not repeated here). Therefore
\beo
E\Bl \intt X_s^{8k} \theta_s^2 ds \Br &<& \infty  \Rightarrow E(Z_k(t)^2) < C
\eeo
This proves that $E(Z_k(t))=0$. Therefore from \ref{Xt2kp1}
\beo
\lim_{t \rightarrow \infty} E(X_t^{2k+1}) &=& \lim_{t \rightarrow \infty} E(e^{-G_k(t)}) E(X_0^{2k+1}) + \lim_{t \rightarrow \infty} E\Bigl( k(2k+1) e^{-G_k(t)}\intt e^{G_k(s)}  \theta_s^2 X_s^{2k-1} ds \Bigr)
\eeo

Writing
\beo
B_{k,2m-1}(t) &=& E\Bigl( 2k(2k+1)/2 e^{-G_k(t)} \intt e^{G_k(s)}\theta_s^2 X_s^{2m-1} ds  \Bigr)=E\Bigl( 2k e^{-G_k(t)} \intt X_s^{2m-1}d(e^{G_k(s)}) \Bigr).
\eeo
we have
\be
\lim_{t \rightarrow \infty} E(X_t^{2k+1}) &=& E(X_0^{2k+1}) \lim_{t \rightarrow \infty} E(e^{-G_k(t)}) + \lim_{t \rightarrow \infty} B_{k,2k-1}(t) \no \\
&=& \lim_{t \rightarrow \infty} B_{k,2k-1}(t) \label{Xt2kp1}
\ee
Now $\intt X_s^{2m-1} d(e^{G_k(s)}) $
\beo
&=& X_t^{2m-1} e^{G_k(t)} - X_0^{2m-1} - \intt e^{G_k(s)} d(X_s^{2m-1}) \\
&=& X_t^{2m-1} e^{G_k(t)} - X_0^{2m-1} - \intt e^{G_k(s)} \Bigl[ (2m-1)X_s^{2m-2}d X_s \\
&+& \frac{1}{2}(2m-1)(2m-2) X_s^{2m-3} \theta_s^2 ds \Bigr]\\
&=& X_t^{2m-1} e^{G_k(t)} - X_0^{2m-1} - \intt e^{G_k(s)} \Bigl[ (2m-1)X_s^{2m-2} \Bigl( -X_s \frac{\theta_s^2}{2} ds + \theta_s dW_s \Bigr) \\
&+& \frac{1}{2} (2m-1)(2m-2) X_s^{2m-3} \theta_s^2 ds \Bigr] \\
&=& X_t^{2m-1} e^{G_k(t)} - X_0^{2m-1} + \Bigl( \intt (2m-1) e^{G_k(s)} X_s^{2m-1}\frac{\theta_s^2}{2}  \\
&-& \frac{1}{2}(2m-1)(2m-2) \intt e^{G_k(s)} X_s^{2m-3} \theta_s^2   \Big) ds
- \intt (2m-1) e^{G_k(s)} X_s^{2m-2} \theta_s dW_s.
\eeo
Therefore $B_{k,2m-1}(t)$
\beo
&=& E(2k X_t^{2m-1}) - E(X_0^{2m-1}) E(e^{-G_k(t)}) + \frac{2m-1}{2k+1} 2k e^{-G_k(t)}\intt X_s^{2m-1}\frac{2k+1}{2} e^{G_k(s)} \theta_s^2 ds \\
&-& \frac{(2m-1)(2m-2)}{2k+1}2k e^{-G_k(t)} \intt X_s^{2m-3} \frac{2k+1}{2} e^{G_k(s)}\theta_s^2 ds \\
&=& E(2k X_t^{2m-1}) - E(X_0^{2m-1}) E(e^{-G_k(t)}) + \frac{2m-1}{2k+1} 2k e^{-G_k(t)}\intt X_s^{2m-1} d(e^{G_k(s)}) \\
&-& \frac{(2m-1)(2m-2)}{2k+1} 2k e^{-G_k(t)} \intt X_s^{2m-3} d(e^{G_k(s)})\\
&=& E(2k X_t^{2m-1}) - E(X_0^{2m-1}) E(e^{-G_k(t)}) + \frac{2m-1}{2k+1} B_{k,2m-1}(t) \\
&+& \frac{(2m-1)(2m-2)}{2k+1} B_{k,2m-3}(t)
\eeo
Writing $B_{k,j}$ as the limit of $B_{k,j}(t)$ we have
\be
B_{k,2m-1} &=& 0 - 0 + \frac{2m-1}{2k+1}B_{k,2m-1} - \frac{(2m-1)(2m-2)}{2k+1}B_{k,2m-3} \no \\
\Rightarrow (2k+2 - 2m)B_{k,2m-1} &=& -(2m-1)(2m-2)B_{k,2m-3} \label{Bk}.
\ee
Now
\beo
B_{k,1}(t) &=& E(2k e^{-G_k(t)} \intt X_s d(e^{G_k(s)}))ds \\
&=& E\Bigl( 2k e^{-G_k(t)} \Bigl[ X_t e^{G_k(t)} - X_0  - \intt e^{G_k(s)} dX_s \Bigr]\Bigr)\\
&=& E(2k X_t) - 2k E(X_0 e^{-G_k(t)}) - E \Bigl( 2ke^{-G_k(t)} \intt e^{G_k(s)} \Bigl[  -X_s\frac{\theta_s^2}{2}ds + \theta_s dW_s \Bigr] \Bigl) \\
&=& E(2k X_t) - 2k E(X_0 e^{-G_k(t)}) + \frac{1}{2k+1} E\Bigl( 2k e^{-G_k(t)} \intt X_s d(e^{G_k(s)}) \Bigr) \\
\Rightarrow B_{k,1}&=& 0 - 0 + \frac{1}{2k+1}B_{k,1} \\
\Rightarrow B_{k,1} &=& 0
\eeo
This and Eqn(\ref{Bk}) would imply that
\beo
B_{k,2m-1}&=& 0 \ \ \ \mbox{for $m=1,2,\ldots,k$}
\eeo
Therefore from \ref{Xt2kp1} we have
\beo
\lim_{t \rightarrow \infty}E(X_t^{2k+1}) &=& 0
\eeo
Therefore combining the two results we have
$
E(X_t^{r}) = \left \{
\begin{array}{ll}
\frac{(2k)!}{2^k k!} & \mbox{when $r = 2k$}\\
0 & \mbox{when $r = 2k+1$}
\end{array}
\right.
$

\section{Conclusion}
\label{conclusion}
{Verifying  Roberts et al's condition (see \cite{Rosenthal2}) for checking the ergocity of an AMCMC can sometimes prove  difficult. In the companion paper (see \cite{Basak}), we considered an AMCMC with the proposal kernel dependent on the previously generated sample and an arbitrary target distribution. There we performed a diffusion approximation technique to look at the continuous time version of the discrete chain. In this paper we narrowed down to the case where the target distribution is standard Normal. We investigate whether the invariant distribution of the diffusion is indeed the target distribution. It turns that the resulting diffusion (which although singular) admits an invariant distribution. Then identifying the limiting moments of 2k order of $X_t$ we identify the limiting distribution to be $N(0,1)$.\\
The techniques applied here are specific only when the target distribution is Normal. We hope that  this  can also be extended to other target distributions, where a identification of the limiting moments is possible. Also more choices of the proposal distribution can be made, where the kernel is dependent on a finite (or possibly infinite) past. We plan to take up these issues in our future work.}

redpajama_set_name: RedPajamaArXiv


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text:
Acid exchange between UK ministers over sub-sea cable plan
The history of British energy is "littered with ideas from the bright ideas box that come out prematurely and raise expectations and cost money and very often never actually produce the goods at the end of the day." This was the acid comment directed at energy minister Brian Wilson by Martin O'Neill, chairman of the parliamentary Trade and Industry Select Committee last month. O'Neill was cross examining Wilson on his plans for a sub-sea electricity cable to connect power from renewable projects into the national grid (Windpower Monthly, December 2001). The government has ordered a study of the proposal's feasibility. The committee was sceptical that an underwater cable could be cost effectively installed through the inhospitable rocks and waters of the west coast. Wilson defended the proposal as an important and visionary concept. "Better to light one candle than to forever curse the darkness. If you do not ask the questions you do not get the answers," he said. O'Neill replied: "This is a question which maybe should have been asked before the match was even taken out of the box to light the candle, with respect." The minister expressed his surprise at the committee's "negativity" towards the plans. Later, energy regulator Callum McCarthy apparently warned against the government's plans for encouraging renewables in windier but more remote areas of the UK. He reportedly told the committee that building new generation plants in the north would be "less effective than adding to generation elsewhere in Britain" because of the cost of transmitting power over long distances.


text:
Conus colmani is a species of sea snail, a marine gastropod mollusk in the family Conidae, the cone snails and their allies.

Like all species within the genus Conus, these snails are predatory and venomous. They are capable of "stinging" humans, therefore live ones should be handled carefully or not at all.

Taxonomy
Conus colmani is part of a species complex including Conus albellus, Conus lizardensis and Conus limpusi, that needs re-evaluation. For conservation implications, all are here tentatively listed as distinct species.

Description
The size of an adult shell varies between 35 mm and 52 mm.

Distribution
This marine species is endemic to Australia and is found off Queensland.

References

 Röckel, D. & Korn, W. 1990. Zur Indentitat von Conus lizardensis Crosse, 1865 und Conus sibogae Schepman, 1913– mit Beschreibung dreier neuer Conus-Arten von Queensland, Australien (Mollusca: Conidae). Acta Conchyliorum 2: 5–23, pls 1–10
 Wilson, B. 1994. Australian Marine Shells. Prosobranch Gastropods. Kallaroo, WA : Odyssey Publishing Vol. 2 370 pp.
 Röckel, D., Korn, W. & Kohn, A.J. 1995. Manual of the Living Conidae. Volume 1: Indo-Pacific Region. Wiesbaden : Hemmen 517 pp.
 Filmer R.M. (2001). A Catalogue of Nomenclature and Taxonomy in the Living Conidae 1758 – 1998. Backhuys Publishers, Leiden. 388pp.
 Tucker J.K. (2009). Recent cone species database. September 4, 2009 Edition
 Tucker J.K. & Tenorio M.J. (2009) Systematic classification of Recent and fossil conoidean gastropods. Hackenheim: Conchbooks. 296 pp.
 Puillandre N., Duda T.F., Meyer C., Olivera B.M. & Bouchet P. (2015). One, four or 100 genera? A new classification of the cone snails. Journal of Molluscan Studies. 81: 1–23

External links
 The Conus Biodiversity website

 Cone Shells – Knights of the Sea

colmani
Gastropods described in 1990
Fauna of Queensland
Gastropods of Australia

redpajama_set_name: RedPajamaWikipedia


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text:
Phils need Bastardo to become part of closing process
Kevin Cooney Staff writer
PHILADELPHIA — Antonio Bastardo's 3-2 pitch went up the ladder, getting Bryce Harper to chase after it for the first out of the eighth inning.
Before Harper was back in the Nationals dugout on Friday night, Phillies manager Charlie Manuel was out of the dugout and pointing for Josh Lindblom to come in from out of the bullpen to face Ryan Zimmerman.
A step in the rebuilding process for Bastardo's confidence was complete.
"We're going to get Bastardo back to where we want him," Manuel said Saturday. "We know how good he can be. It's still there. It's a matter of getting it out of him and him getting it out of himself."
Of all the mysteries in what has been a very strange and frustrating 2012, perhaps the one that is the biggest and toughest is Bastardo's seeming regression.
After a strong 2011 which saw him briefly take over the closer's role, Bastardo struggled down the stretch and has never regained that form in 2012. He entered Saturday's game with a 2-4 record, a 5.40 ERA and having given up more homers (7) and earned runs (23) than he did for the entire 2011 campaign (6, 17) with still five weeks to go in the regular season.
"When he feels real good, you see the aggressiveness," Manuel said. "There are times when he doesn't realize the difference in him. Last year, when he was going really good, the ball was flying out of his hand and beating the hitter."
For most of this season, Bastardo has nibbled at the corners, working himself into deep counts and deeper trouble.
"When he gets in trouble, he starts trying to pinpoint the ball and aiming the ball," Manuel said. "He starts trying to make too good a pitch instead of trusting his stuff."
How much trust the Phillies can put into him as a legitimate set-up option has been called into question in recent days. There was a stretch of 11 games recently when Bastardo wasn't used — although both Manuel and pitching coach Rich Dubee insisted that was more a product of situations in-game as opposed to any statement about Bastardo's ability in that spot.
"We've been very fortunate of what Antonio did for us. Last year, he had a fabulous year for us," Dubee said. "But you are going against the middle of lineups like (the Brewers trio of Ryan) Braun, (Aramis) Ramirez and (Corey) Hart, you are asking an awful lot."
That is part of the reason that Dubee feels more teams in contention this year went out and made moves to get mix-and-match options in the eighth inning as opposed to running with a true set-up man.
"Why do you think Cincinnati went out and got (Jonathan) Broxton for that role?" Dubee said. "(Sean) Marshall could handle the load. But a lot of teams are stacked in the middle of their lineup with righthanded hitters. You see (Matt) Thornton with the White Sox and they went and got (Brett) Myers to get righthanded hitters in the middle of the lineup out. It is an awful demanding thing for a lefthanded pitcher who doesn't have that quality sinker like (the Nationals' Sean) Burnett that goes away from righties."
So what can you take from that? Well, the Phils could see Bastardo as part of a puzzle in the late innings next year. But it is clear that he will not be left out there as the sole option. Part of the reason that the Phils have mixed and matched so many guys at this point is because they need to find if anyone in-house among the righthanders (Lindblom and Phillippe Aumont) could be a part of that role.
(Michael Schwimer may have been in that mix, but his battle for a spot on the disabled list instead of taking an option to Lehigh Valley this week may be the final straw for an organization that has grown tired of his antics.)
There are always relief options out there every winter — the key is not to overpay for them. No more multi-year deals for rented guns should be a credo for the organization. They simply can't afford that.
So the best thing they can hope for is that Bastardo regains his form and one of the righthanded arms steps up in the final month. Because repairing the bullpen only happens with small steps at a time.
Kevin Cooney: 215-949-4215; email, KCooney@phillyBurbs.com; Twitter, @KevinCooney


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Raising the Security Bar with New Microsoft Collaboration
By Cloud and Data Center Transformation
Today's hybrid IT environment has made it increasingly difficult for organizations to prevent cybersecurity attacks. From hacking to malware, ransomware and beyond, the fact that enterprise information assets are now distributed across multiple on-premises and cloud locations creates multiple points of vulnerability that magnify the challenges of keeping cybercriminals at bay.
That's why our Cloud + Data Center Transformation solution area immediately embraced Microsoft Sentinel, the cloud-native Security Information and Event Management (SIEM) solution that identifies security threats across the entire hybrid enterprise. We introduced both consulting and managed security services for Microsoft Sentinel shortly after it was released last year, and many of our clients have taken advantage of these services to modernize their security operations.
Now, thanks in large part to our Microsoft Sentinel expertise and our recognition as the 2020 Azure Security Deployment Partner of the Year, Insight has become one of the first managed security service providers to be invited to join the Microsoft Intelligent Security Association (MISA).
MISA began two years ago as an ecosystem of independent software vendors that integrated their security products with Microsoft's to help organizations better combat security threats. With more than 100 member ISVs and 150 integrations to date, Microsoft is now expanding the ecosystem by adding select MSSPs to help organizations further simplify and strengthen their security posture.
For Insight, this is not only the latest chapter in a decades-long Microsoft relationship but also a compelling opportunity to advance our clients' ability to defend themselves against an increasingly challenging threatscape.
We already offer robust security services, including 24/7/365 managed security that relieves IT teams of day-to-day monitoring and threat response. Through our participation in MISA, we will be collaborating with the world's top cybersecurity software vendors to help enterprises protect their sensitive data, employee identities, applications, infrastructure and endpoints.
As the saying goes, there is strength in numbers. Together, we will take risk mitigation to the next level.
Read the official press release here.


In this Python tutorial, we will learn how to print lower triangle pattern in Python. Let's see how to print lower triangle pattern of matrix as well as we will print lower triangle star pattern in Python.
Lower triangle of a matrix consists of diagonal elements and the elements below the diagonal of the matrix.
Let us consider A as 3X3 matrix.


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Elgar Centre Users
Contact Groups / Classes
Entertainment Providers
How do I make a booking?
Simply contact us via one of the methods listed on the 'Contact Us' page and the Centre Manager will respond as soon as possible with availability and further details. A provisional booking will be made initially, followed by confirmation of your booking as soon as payments and a completed booking form have been received. This will all be detailed during the booking process.
Do I have to pay a deposit to secure the booking?
You will be required to complete and return a booking form, along with a £50 deposit (returnable after the event subject to conditions), and half of the total booking cost in advance.
Cheques (made payable to The Elgar Centre), cash or BACS payments. Please ask for our bank details when booking if you would like to make a BACS payment transfer. We are unable to accept card payments at present.
Do I have to collect keys for my booking?
No. We have keyholders who open and close before and after all of our bookings. You will be provided with their contact details before your booking in case of any emergencies.
What happens if I want to cancel my booking?
You will receive a full refund if your booking is cancelled at least 4 weeks in advance of your booking date.
Do you have a kitchen?
Yes. We have a fully equipped kitchen on site available to hire at a cost of £5 per booking. The kitchen includes plenty of work space for food preparation, fridge/freezer, oven and hob, toaster, microwave, several kettles, urn, supply of mugs, glasses, cutlery and crockery. The kitchen also has an electronically operated hatch that links through to the central foyer area.
Do I have to book time to set up and pack away?
Yes. The time you book and pay for is the time you will have access from and to, so if you require additional time to set up before a party or clean up afterwards, please take this into consideration when booking.
Do I have to clean up afterwards?
Yes. All tables and chairs must packed away in the designated areas. Any spillages must be cleaned up, floors must be swept, decorations removed, and all rubbish must be taken away with you. In order to have your deposit returned, the centre must be left in an acceptable, clean and tidy state.
Is the venue accessible for wheelchairs?
Yes. The venue is accessible at both the front and rear via stairs and ramps. The facilities are all on one level and we have a Disabled WC.
Do you have a carpark?
Yes. We have a private carpark available for visitors to the centre. This has a total of 40 spaces, plus 4 disabled spaces. There is also a public carpark a short distance away if our carpark becomes full. **Please note that our carpark is not open 24 hours a day, so if you are not using the centre and have parked in it, it is possible your car may be locked in**
Do you have tables and chairs?
Yes. We have a storage area adjoining the Oak Room, containing tables and chairs for you to utilise. Please see photos of these in the gallery.
Can I have a BBQ at the centre or on the playing fields?
No. Due to insurance, we are unable to permit any kind of BBQs on the premises or playing fields.
Can I have a bouncy castle?
Yes, bouncy castles are permitted providing the supplier has relevent insurance. Please see our 'links' page to view details of our preferred bouncy castle supplier.
What is the capacity of your rooms?
The large Oak Room can accommodate around 100 people for a party, but you may wish to consider hiring additional space in the centre to allow for more guests or greater flexibility. The smaller room can accommodate 20-30 people. The foyer area can accommodate around 50 people.
If there are any questions you may need to ask that have not been covered above, please do not hesitate to contact us.
The Elgar Centre is Registered Charity No. 1145093
© The Elgar Centre - All Rights Reserved.
The Elgar Centre accept no responsibility for externally linked web sites.


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Humana, Microsoft Forge Partnership; Aveanna Teams Up with Netsmart
By Joyce Famakinwa | November 3, 2019 November 3, 2019
Humana, Microsoft announce partnership
Humana Inc. (NYSE: HUM) and Microsoft Corp. (NYSE: MSFT) have partnered to develop predictive solutions and intelligent automation for the care teams of the insurer's members.
One of the specific goals of the partnership is to address health for aging populations and their care teams.
The news marks the first step in what will be a seven-year partnership between the two companies, they said in a press release.
Through the partnership, care teams will have access to real-time information via a cloud platform. Microsoft technologies will also allow Humana to utilize natural-language understanding and speech recognition for the administrative and clinical sides of its business.
Humana will also be able to aggregate data on Microsoft Azure, allowing for a broader view of its members' health history and easier access to their health records.
"Achieving our goal to provide better care experiences and improve our members' health requires strong partnerships that allow us to be smarter, safer and faster in delivering care," Heather Cox, chief of digital health and analytics at Humana, said in the release.
Additionally, as part of a larger innovation strategy, Humana will use Microsoft Azure and Azure AI to develop on-demand and virtual medical services.
"With an estimated 10,000 people joining the Medicare system daily, we have a tremendous opportunity to address the growing demands on the health care system by improving health outcomes and lowering costs," said Dr. Greg Moore, corporate vice president of health technology and alliances at Microsoft, said the release.
Aveanna teams up with Netsmart
Aveanna Healthcare has teamed up with health information technology company Netsmart in what will be a 10-year partnership between the two companies.
Pediatric home health giant Aveanna — based in Atlanta — is backed by Bain Capital Private Equity and has more than 200 locations in 23 states.
Overland Park, Kansas-based Netsmart designs and builds electronic health records, health information exchanges, analytics tools and telehealth solutions. The company serves more than 600,000 users across more than 30,000 health care organizations.
As part of the deal, Aveanna clinicians and staff will utilize myUnity, Netsmart's electronic health record solution, which will give them access to more than 90,000 organizations across health care.
"We were looking for an opportunity to connect with an IT partner with whom we could truly collaborate to optimize and configure solutions to meet goals that are specific to our organization's needs today and in the future," Aveanna Healthcare CEO Tony Strange said in a statement. "As we look forward, we envision an opportunity to positively influence and advance the home health industry in both efficiency and quality clinical care."
Additionally, Aveanna will implement Netsmart's data and analytics tools and have access to Netsmart's learning management system, myLearningPointe.
Last month, Netsmart hit a major milestone when it set a new internal record for the number of secure health data transactions achieved in a month at 100 million transactions.
Twistle raises $16M
Seattle-based startup Twistle recently closed a $16 million Series A funding round.
Health Enterprise Partners (HEP) and MemorialCare Health System's MemorialCare Innovation Fund (MCIF) led the round, VentureBeat reported.
Twistle's HIPAA compliant health care communication platform helps clinicians track and ensure patient adherence.
Twistle has earmarked the new funds for its customer-acquisition efforts, particularly in the provider and life sciences arena, according to VentureBeat.
"If we want to deliver great care and bend the cost curve in this country, we cannot continue following the status quo," Twistle CEO Kulmeet Singh told VentureBeat. "We have to use automation to increase the effectiveness of our care teams and find better ways to motivate patients to take ownership of their own health."
In total, the company has raised $19.8 million.
Aveanna Healthcare, Humana, Microsoft, Netsmart, Twistle
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Consolidation Accelerates as Home Health Operators Crash from Stimulus 'Sugar High'