text:
Statoil and Polenergia wish to work together on offshore wind farms
Polenergia, the largest Polish private and vertically integrated energy group, signed a preliminary agreement to dispose of 50 percent of shares of its subsidiaries Polenergia Bałtyk II and Polenergia Bałtyk III implementing the construction of offshore wind farms with the capacity of 1,200 MW. The agreement was concluded with Statoil Group.
Polenergia's statement
Having signed the promised agreement, Polenergia will acquire a partner who will substantially improve the perspectives of implementation of wind farms in the Baltic Sea. As one of the leaders of offshore investments in the world, Statoil group will introduce to the projects some new, proven and cost-optimized technologies as well as a new look at the Polish renewable energy sector from the position of a global player. The preliminary agreement provides for the transfer of 50 percent of shares in Polenergia's subsidiaries, Polenergia Bałtyk II and Polenergia Bałtyk III, to Statoil. As a part of the signed agreement, an option was also granted for the purchase of 50 percent of shares in Polenergia Bałtyk I. The transaction requires a consent of the President of the Office of Competition and Consumer Protection in Poland.
Acquiring such a reputable partner confirms, on one hand, a positive assessment of the Polish market, and, on the other hand, the attractiveness of Polenergia's projects, as well as expresses recognition for the quality of the works carried out by the company. Polenergia Bałtyk II and Polenergia Bałtyk III Wind Farms are a strategic investment that, when realized, will allow not only the company, but above all the Polish economy to obtain significant economic benefits. Expert calculations show that by constructing in the Baltic Sea wind power plants with the total capacity of 6,000 MW by 2030, the Polish GDP will gain PLN 60 billion, and the market will be increased by 1 additional 77 thousand new jobs. The domestic enterprises, directly or indirectly, will be able to apply for contracts for the supply of foundations, towers, cables and offshore transformer stations, provision of construction services, and potentially the production of rotors or delivery of ships.
"The prospect of entry of such a stable professional investor proves that the decision already taken in 2010 to get into the offshore wind energy in the Baltic Sea was the right one. Thus, Polenergia passed the test of credibility, which confirmed its professionalism and potential. A combination of strength and experience of Polenergia and Statoil group guarantees the implementation of the project that opens a new chapter in the history of Polish renewable energy. Above all, it will provide new generation capacities, so desired by the Polish power system in the coming years", says Jacek Głowacki, director of the works of the Management Board of Polenergia.
Polenergia is the first and only entity in Poland having two legally binding environmental decisions and the signed grid connection agreement, which enable it to work on the technical design of two offshore wind farms with the total capacity of up to 1,200 MW. Polenergia's offshore wind farms in the Baltic Sea will be located north of the coastline, at the level of Smołdzino commune and Łeba municipal commune in Pomorskie voivodship. Thanks to very good wind conditions in the Baltic Sea, and larger and more cost efficient Wind Turbines, the offshore wind farms will provide stable and competitive renewable energy to the Polish market.
From January 2017 a wind measurement campaign is being carried out by means of a LIDAR system. Also a geological investigation of the of the seabed and soil conditions is in progress. Furthermore, there are efforts to obtain the environmental decision regarding the offshore transmission infrastructure. The first electricity from Polenergia Bałtyk III Offshore Wind Farm may flow to the Polish coast in 2025, whereas from Polenergia Bałtyk II Offshore Wind Farm – in 2027. The lifetime of both offshore wind farms is expected to be 25 years. The projects expect that there will be Wind Turbines available in the market in the mid-2020s that exceeds 10 MW capacity per turbine. They are going to be the largest wind power plants that have ever been constructed in Poland.
Source Statoil
Statoil's statement
Statoil is also entering into a 50/50 joint venture with Polenergia to further mature these projects with Statoil being the Manager for the development, construction and operational phases.
"We are very pleased to have entered into a partnership with Polenergia, which is an experienced energy company with a growing renewable portfolio and in-depth knowledge of the Polish electricity market. Statoil has an ambition to grow significantly within renewable energy investing up to EUR 10 billion in profitable renewable energy towards 2030, and this acquisition strengthens our presence in the Baltic Sea area giving opportunities for scale and synergies in a longer perspective," says Irene Rummelhoff, executive vice president of New Energy Solutions in Statoi
The Bałtyk Środkowy II and III wind farm areas are located in the Baltic Sea approximately 27 and 40 kilometers from the port of Łeba in water depths of 20-40 meters. Statoil is already developing the Arkona offshore windfarm together with EON in the German part of the Baltic Sea. The company is also operating three offshore wind farms off the UK coast.
Offshore wind already has a strong foothold in Europe with over 15 GW installed capacity, with a potential to reach between 50 and 70 GW by 2030. Polish Baltic Sea has a potential in excess of 8 GW, where the first 4 GW could be in operation by 2030.
"We are entering a market with growth potential through two of the most advanced offshore wind developments in Poland. The country is well-placed to develop a strong offshore wind industry that would create jobs and value in one of the most dynamic parts of the European markets. Following a dramatic reduction in cost for offshore wind, we are looking forward to work with our new partner, the Polish authorities and the Polish supply chain to advance offshore wind as a competitive source of energy and industry," says Rummelhoff.
The closing of the transactions with Polenergia is subject to consent of the President of the Office of Competition and Consumer Protection in Poland.
Polenergia/Statoil
PKN Orlen considers construction of wind farm project
PKN Orlen a step closer to building an offshore wind farm in the Baltic Sea
Russia is threatening with a blackout. Disputes keep the Iron Curtain in energy supplies
RES on the Baltic can produce three times of Poland's energy demand (INTERVIEW)
Baltic Sea energy offshore wind farms Poland Polenergia RES Statoil
Energy Renewables
An Orlen Group company Baltic Power is about to commence surveys to assess the environmental impacts and wind conditions within its off-shore wind […]
4 January, 2019 10:00 am
Nine years ago, having 85 percent of "black" energy and 15 green, we decided to switch it the other way aroundWe are now at the point where […]
22 March, 2018 12:00 pm
PKN Orlen has announced a tender for the development of a preliminary technical concept to determine the options for preparing and implementing a […]
7 February, 2018 11:00 am
Although the Iron Curtain has not completely collapsed in the power sector, Poland is fortunately on its safe side. If Latvia and Estonia do not […]


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Fayette Township is one of the fourteen townships of Lawrence County, Ohio, United States. As of the 2010 census the population was 9,194, of whom 6,628 lived in the unincorporated portion of the township. Fayette Township includes Ohio's most southerly point.

Geography
Located in the far southern part of the county along the Ohio River, it borders the following townships:
Lawrence Township - north
Windsor Township - southeast corner
Union Township - east
Perry Township - west

West Virginia lies across the Ohio River to the south: Cabell County in the far southeast, and Wayne County otherwise.

It is located in the middle of Lawrence County's Ohio River townships.

Part of South Point, the second-largest municipality in Lawrence County, is located along the downstream shoreline of Fayette Township.  Most of the rest of the shoreline is occupied by the census-designated place of Burlington.

Name and history
It is the only Fayette Township statewide. It was named Fayette, in honor of Marquis Lafayette, who was a popular hero of that time and who visited Burlington in 1826.

Government
The township is governed by a three-member board of trustees, who are elected in November of odd-numbered years to a four-year term beginning on the following January 1. Two are elected in the year after the presidential election and one is elected in the year before it. There is also an elected township fiscal officer, who serves a four-year term beginning on April 1 of the year after the election, which is held in November of the year before the presidential election. Vacancies in the fiscal officership or on the board of trustees are filled by the remaining trustees.

References

External links
County website
Fayette Township Fire Departments

Townships in Lawrence County, Ohio
Townships in Ohio

redpajama_set_name: RedPajamaWikipedia


text:
17th March 2020 Bill Russell
Coronavirus – Theatres in London and throughout the country have closed down
Theatres in London and elsewhere in the country closed down last night, March 16, or will do so in the next day or so, because of the pandemic. The Society of London Theatres, which represents the West End theatres, made the announcement following the Prime Minister's statement on new measures to deal with the pandemic and its sister organisation UK theatres, which represets 165 theatres throughout the country, said they would be doing the same. Many of London's small fringe theatres like the Arcola and the Finborough had already announced closure. They will re-open when it is considered safe for people to assemble in large numbers. Check with the box office of any theatre for which you have tickets – there will be refunds, offers of dates at a later booking and some will ask that you consider leaving the money to help them survive.
Picture – Getty image
Peace in Our Time by Noel Coward. The Union Theatre, 229 Union Street, London SE1 to 4 April 2020. 2**. William Russell.
Something to entertain you – on Instagram
Look Back in Anger by John Osborne. The White Bear Theatre, 138 Kennington Park Road, London SE11 to 14 March 2020. 3***. William Russell.
4th March 2020 Bill Russell
Seven Methods of Killing Kylie Jenner by Jasmine Lee-Jones. Jerwood Upstairs, Royal Court Theatre, London SW1 to 27 July. 4****. William Russell.
9th July 2019 Bill Russell
Blueprint Medea by Julia Pascal. Finborough Theatre on line to 2 September 2020. 3***. William Russell.
3rd July 2020 Bill Russell

redpajama_set_name: RedPajamaCommonCrawl


Bart & Tricia Dyer, now based in West Orange NJ at THELIFEChristianChurch, have previously recorded with Life Youth Choir in Houston, TX and Christ Community in Columbus, OH. This is their first album as a "solo" (duo) artist. Their music will inspire, encourage and uplift your heart! This dynamic couple have a heart and passion to lead people into God's presence through music and worship. You will be blessed by their beautiful spirits that shine through on every song!
Shout It From The "Rooftops"!
King Of My Heart - SO POWERFUL!
Watch bartandtricia music videos free online! Godtube.com also provides videos with bartandtricia lyrics, singles, album information and reviews. Whether you're looking for an inspirational and encouraging song for your quiet time with God or an upbeat song to praise Jesus, Godtube.com offers the largest online database of Christian music artists!


text:
WOMAN TESTIFIES ON $50,000 LEAK
Linnet MyersCHICAGO TRIBUNE
An 82-year-old Bridgeport woman testified Friday that her basement looked ''something terrible'' when she came home with a $25,000 check to pay for repairs to a leaky toilet.
Before she left, several workmen had arrived and started ''banging up the floor,'' said the woman, Rose Rolek. When she returned, the basement floor was torn apart from the back ''all the way to the front window,'' she said.
Rolek`s testimony came Friday in the Criminal Court trial of Willie Nearn Sr., 44, and his son, Willie Nearn Jr., 18, both of 4840 W. Diversey Ave., and Ron Norton, 47, of 2751 W. 63d St.
The men allegedly conned Rolek into thinking that they had to tear up her basement floor, for $50,000, to fix a leaky toilet in her home. The $25,000 check she brought home from the bank on Sept. 10, 1984, was to be the down payment, she testified.
The three defendants are charged with criminal damage to property, conspiracy to commit criminal damage to property and conspiracy to commit theft.
Willie Nearn Sr. allegedly was responsible for the scheme, Norton allegedly conned Rolek into signing the work order for the $50,000 ''repair,'' and Willie Nearn Jr. allegedly was one of the men who dug up her basement, prosecutors charge. But Rolek did not identify any of the defendants as being in her home.
The only man she identified was Arthur LaGace, 39, of 3410 N. Paris Ave., a former Chicago policeman who testified in the trial Thursday. The attorney general`s office guaranteed LaGace probation instead of prison time in exchange for his testimony against the other three.
The woman`s first contact that fall with one of the alleged conmen came one day when LaGace, who had stopped at her home on another matter, seized on an opportunity to do more repairs when she told him of a leaky toilet, he testified.
A few days later, on Sept. 8, LaGace went to Rolek`s home with his brother, William LaGace, 26, to perform some repairs. While there, they
''created a lake inside the basement'' by stuffing pipes with plastic garbage bags and clay, LaGace said.
After LaGace and his brother sabotaged her plumbing system, LaGace convinced her that her main sewer pipe had broken, Rolek said. ''I said, `Gee, I don`t want no trouble,` '' she said, explaining that the workmen told her the toilet was leaking because of a major sewer problem.
She then authorized the work order allegedly offered by Norton. Two days later, she went to the bank to get the $25,000 down payment and came home to find her basement in ruins.
Rolek said Friday that LaGace told her he was a policeman and a city inspector. ''I said I never had no inspector, you`re the first one,'' she testified.
On Thursday, LaGace described meetings in which Willie Nearn Sr., who owned Central Heating Sales at 4842 W. Diversey Ave., and the other defendants allegedly conspired to con Rolek.
But defense attorneys bitterly argued that LaGace, an ex-convict, was a
''dog'' who was lying to save himself.
Police were alerted to the scam by two bank employees, who testified that they tried to tell Rolek that $50,000 was too much to repair a leaky toilet.
The woman`s testimony came as an Illinois Senate subcommittee Friday heard testimony in the State of Illinois Center from Illinois Atty. Gen. Neil Hartigan and representatives of the Cook County state`s attorney`s office as well as others on how to strengthen state laws to curb home-repair fraud.
George Holmes, who owns a plumbing business called Dombrowski & Holmes, Inc., testified Friday that an entire home sewer system should cost no more than $4,200.
The three defendants and LaGace are among 18 people indicted last year in Operation Homecon, an effort to convict confidence men who prey on trusting and often elderly homeowners. A series of articles in The Tribune last month described how Illinois laws often aren`t able to deal with unscrupulous home- repair contractors who use sabotage to bilk their victims.
The trial continues Monday.


Candidates in the election for the post of senior vice-president of the Federation of Nepalese Chambers of Commerce and Industry (FNCCI) have launched a blame game with less than a week till election day.
Kishore Kumar Pradhan, one of the candidates vying for the post, lashed out at the current FNCCI President Pashupati Murarka and president-in-waiting Bhawani Rana for openly taking the side of his rival Shekhar Golchha.
Murarka called on the two candidates and their panels on Wednesday to discuss the possibility of consensus. He claimed that both parties had agreed to be open to forging a consensus and keep the election campaign clean.
Golchha didn't rule out the possibility of a consensus but Pradhan said it was already too late for that. "We have already moved far ahead," said Pradhan. "They were very late in starting talks. There is no possibility of a consensus." Lately, the competition for the post of senior vice-president at the private sector's apex body has heated up with eminent business personalities openly taking sides to ensure victory for their candidate.
The senior vice-presidential race is a high-stakes affair as the winner will automatically become the next president of the FNCCI after three years. While former FNCCI presidents Suraj Vaidya and Kush Kumar Joshi are supporting Pradhan, Chandi Raj Dhakal is lobbying on behalf of Golchha.
The FNCCI's special general meeting held in October 2016 decided that the senior vice-president would be promoted to president without having to fight elections. This is the reason why all eyes are on the election for the post of senior vice-president at the FNCCI which is being held on April 10.


Thank you to everyone involved in Lizzie's Lunch 2015. In 2014 the first Lizzie's Lunch raised $10,000, this year the event raised $11,000. This funding goes to the information and support services used by people with tuberous sclerosis and their families.
Lizzie Pinkerton was a vibrant young girl with a cheeky sense of humour. She lived with tuberous sclerosis (TSC) and associated polycystic kidney disease that ended her life in 2010.
The guest speaker this year was Rebecca, who spoke so beautifully and generously about her son Josh's life with tuberous sclerosis.
We look forward to another successful Lizzie's Lunch in 2016. Keep scrolling for photos of the event.
Vicky Vukovich, who donated her time to play her viola.
Our thanks also to Nina for the gorgeous photos of the day below. If you'd like to get involved in fundraising, contact us. We'd love to support you to organise an event like Lizzie's Lunch or with another idea.


When it comes to leading organizational change, most companies and managers tend to have a fear of the overall process involved. This is due to various causes including employees' reticence of adopting new technology, a decrease in the amount of customers noticed during the first few months of change and so on. However, when done properly, organizational change can prove to have outstanding results for a certain business and even help it place in the front of many others in its niche.
Organizations do not change for change's sake but because they are part of a broader development process and must react to the new changes in the environment, restrictions, requirements and opportunities that arise in this area. In turn, for the process of adapting to organizational change, people must change and they must acquire new knowledge, tackle new tasks and boost their degree of competence. Additionally, very often, they must change their work habits, values and attitudes to the working of organizations. Changing values and attitudes is essential; perhaps there can be no real change without a change in attitude.
In solving this problem, management should develop a set of rational strategies using resources and respect limitations dictated by the context in which and operate. the case in which context changes they resort to other rational strategies to adapt to the new context. This is the only way the only way in which this system works and can identify sources of resistance to the change process and methods of solving them.
You need to stop thinking at what tasks employees are aware of and seek to change the context in which they operate and thus formulate your own rational strategies to lead them to do desired things. Only you can change the outcome of the process.
Changes are set on a scale of complexity, from incremental to transformational.
Transformational changes are those that relate to asking employees other things than the one they are used with: a research organization, a helping company, a sales organization. An expert in the field of engineering, medicine and IT is required to handle administration tasks such as Sales or Administration, must have proper knowledge and must have invested adequate resources. Transformational change refers to an evolution towards a future state, which was defined in terms of the vision and strategic scope, covering the purpose and mission of the organization and its philosophy in terms of issues such as development, innovation and human values as well as technologies used.
Incremental change is the one that needs to be done immediately and the one in which the managers have to involve the most. Various aspects need to be taken into account here from changing the overall business mentality to software integration.
He explained the issue by analyzing the changes that take place in the work field. The model has been taken up in 1980 by E. Schein and converted to organizational reality. Lewin sees changes as a dynamic balance of forces, on the one hand, and pressure applied to the process, on the other hand. This leads to an overall resistance to change.
"Defrosting" represents the current state of balance, which stands as a support for existing behaviors and attitudes. This process must take into account the threats posed by change in front of people, and the need to motivate them to reach those affected by the acceptance of change. It can be achieved by introducing information that allow for a comparison between the desired situation and the actual cause of change.
The change refers to the creation one-side basis of new information, transformation that allows individuals to approach the desired situation.
References critical of this model refers to the translation of leadership styles change reality.
Direct approach – is necessary in crisis situations. This is done through the exercise of managerial power without consultation.
The "body and soul" – aims and committed visions, but it does not have to resort to involvement or participation.
Action-based approach – begins with a general belief that there is a problem. Identify solutions by successive attempt to clarify the problem and lead to its understanding.
In other words, in order to change employee behavior you must change the environment in which it operates and convince them that the new behavior is something that can be achieved. You also need to convince them that the new behavior will lead to a result of value to them.
Weber in his "Management: Basic Elements of Managing Organizations" believes that there is a dichotomy of change: changing people or situation. The problem does not actually rely in changing the shape of an application, but in what order it should be done.
The proposed model is an amalgam of theories of change and defeats in the mechanisms of resistance to change.
Analysis of the environmental situation (internal and external) leads to the identification of generators of change, which have a central role in decision making to achieve the necessary changes. Once taken decisions, managers allocate resources and responsibilities.
They will involve the employees in the bargaining process and training and will set goals and deadlines for implementation of change. Once the change is implemented, managers will review the progress recorded and will take corrective measures where necessary.
Changing needs support from the highest level in an organization. It seems natural to be in an inverse relationship between the interest of middle-managers for change and the need for it. To manage or accept a change process, you must be aware of its implications and have the courage to take risks. Therefore, the reasons why companies choose to make organizational changes are related to the success of the change process. Changes are neither good nor bad, but the way in which they are implemented and managed is crucial for the company and its members.
When identifying attitudes towards change, change direction or issues posed by transformation is the first step in planning organizational change. The change process has 5 main phases: preparation phase, diagnosis phase, planning phase, implementation phase and impact assessment phase. While the first 3 phases can be assessed with the aid of Bandura's or J. Beckhard's model, the last 2 phases can be successfully implemented by following Chin and Benne or Thurley's approaches. All these are of crucial in the topic of leading organizational change and should be implemented by following company's guidelines and structure carefully.
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This is a review of the Walt Disney World Animal Kingdom Dining for breakfast at Tusker House. This is a great table service to reserve in advance, so if you get the chance to eat here it is so worth it! For dining on the Disney dining plan it is one Table Service credit (sit down). It is a must do if you have children who want to meet characters and it's a great way to avoid waiting in long lines once the park opens.
My husband and I also enjoyed that it is a buffet so we all could all enjoy what we liked. We went for breakfast and they had waffles, eggs, sausage and African inspired cuisine. There was so much to choose from and everyone left with a belly full.
The table service (sit down) is for Breakfast, Lunch and Dinner at Tusker House. Guests can make Advanced Dining Reservations for this character experience limiting your wait time. You can even get a breakfast reservation set for a time before the park even opens to the public. Making it so you can get in and eat then get in line for a ride before anyone else does!
The food for breakfast was delicious. Be sure to try the Jungle Juice that comes with everyone's meal, it is delicious! My husband and I tried many different food options . The kids really enjoyed it, but my youngest who was 6 years old at the time, was super excited to meet Safari Donald. Next time we go to Disney World we will definitely eat here again.
Disclaimer: I was not compensated in any form for this review. My family personally paid for this trip, dining and all expenses. This dining review is my opinion.


To celebrate with Tide and to check out the vintage packaging, make sure you check out the retro-packaged products, which are available at Target nationwide.
In a bygone era when women wore dresses and high heels to carry their families' laundry in a wicker basket to their very first automatic washing machine, they turned to Tide for "the cleanest clothes possible." More than a half-century later, families still turn to Tide, now offered in more than two dozen liquid and powder formulas. To celebrate its 65th Anniversary, Tide is issuing special-edition packages with the 1947 label. Procter & Gamble (P&G) invites consumers to enjoy a blast from the past with retro-packaged Tide, Bounce and Downy available at Target in May, while supplies last.
Three of my favorite laundry products! Thanks for the entry!
Love Tide Stain Release Liquid for my son's wash – gets every stain out!
Wrinkle Release Spray – Thanks!
I love the tide pens. They are so handy for those unexpected stains when your not at home.
My favorite detergent is Purex + Zout, it makes the clothes smell sooo good!
I like Tide Original Scent. Sometimes my husband buys a cheaper brand & I don't like it. Especially don't like other brands when the bed linens are washed. I like to put my head down on the pillow & breath Tide. We always try to buy Tide.
I like tide, it seems to get clothes really clean.
Tide is my favorite detergent to use!
I love the Cold water Tide!
I am also a big fan of the Tide Stain Release Pens.
Tide is the best. I've been using it for years.
I have been really enjoying using the purex crystals fabric softener!
Right now I am loving the Tide stain release.
My favorite is liquid Downey.
My favorite laundry product is anything gentle and fragrance-free because I have sensitive skin.
I make my own detergent, usually, and I love Arm and Hammer Super Washing Powder.
I like the tide stain pen.
I love the Tide to go Stain Stick. It is perfect! It prevents stains from setting in when I am not at home to wash it.
Anything from the Simple Pleasures fabric softener collection.
I use Shout or Spray 'N Wash.
I love the Bounce dryer bars!
Favorite product is the original Downy!
My favorite is Snuggle Dryer Sheets.
I actually change back and forth between a few things. Tide, Gain, and Purex are my main product lines. I love some new Snuggle dryer sheets I have found. Purex has been coming out with some super cool stuff lately but I always change things up so we don't get tired of the same old thing all the time.
I like the new tide stain release best.
I love love love tide & downey!!! nothing makes the clothes smell as fresh & clean!
Tide pens! Those things are life savers!
I love Tide Free and Gentle.
My favorite fabric care product as of late is Tide's new stain booster duo-pacs. I can't live without them now! They take out stains and somehow make the laundry smell 10x more fresh then Tide on it's own which was already doing a great job!!
Spry and Wash Stain Stick…I can use it right away and not wash the item for a couple of days. Great giveaway, thanks.
I like Gain original fresh liquid laundry soap.
I like Downy fabric Softener.The fragrances and the the softness it gives the wash.
I love Purex cold water detergent! Does amazing in cold water!
I exclusively use tide – and have for as long as I can remember!
favorite laundry product would have to be Fabuloso. My mom always added it to the clothes and used it for cleaning also. It left such a great smell to the clothes and the house. She also always used Tide never deviated to anything else.
I love Bounce dryer sheets. I love that they make my clothes static-free.
I love Shout Color Catchers.
I love Tide and have found it to be the best laundry item along with an old toothbrush for tough stains.
I've always liked the Bounce dryer sheets.
My favorite fabric care product always has been, and always will be Tide!! I hear people talking about the Bounce Bar an awful lot, so I'm going to have to pick one of those up and give it a shot too!
Tide is the best laundry detergent for getting clothes clean and bright!
My favorite product is Charlie's Soap.
Downy Fabric Softener is my favorite.
I love Downy Fabric Softener.


Mama elephant gives baby a bath, and water and blooms spray all over in this adorable design. All pink with the little grey elephants, a casual, hand-drawn script letter style, and more flowers sprinkled here and there on the design makes this an adorable style for any event or party. Customize your Baby Elephant Water Bottle Labels today for your next event or party.
Evermine Baby Elephant bottled water labels are specially designed for bottled water products and are printed on your choice of coated paper or waterproof vinyl label stock. These self-adhesive labels print with fine detail and rich color. Our bottled water labels measure 7" wide x 1.875" tall, and come six to a sheet. With most small plastic water bottles, our labels are slightly shorter than the full circumference of the bottle, this helps with the alignment and application of the labels. You'll never have to worry about the two ends lining up together perfectly as they can be slightly off and it will not be apparent. Ordering is quick and easy, and you can order just the amount you would like!


Back in 2003 OGC Nice ultras made something that will be remembered by name of "Star Wars pyro".
One round before the end of Season 2002/03, Nice had to play with Bordeaux at home. The game was important for both the clubs, Nice tried to reach Intertoto tournament and Bordeaux were fighting for Champions League place.
Before the game there were some clashes between the fans, some of them badly injured after it. Game itself ended 1:1, and after all Nice succeed reaching Intertoto, while Bordeaux had to play UEFA Cup.
Final match of Brazilian U20 Championship played 20 years ago, well known as "The war of Pacaembu".


text:
Q: When I run mokutil, I get Failed to enroll new keys When I run mokutil --import MOK.der, I get
❯ mokutil --import MOK.der
input password:
password should be 1~256 characters
input password:
Failed to enroll new keys

A: The answer was very simply, run it as root
sudo mokutil --import MOK.der

redpajama_set_name: RedPajamaStackExchange


text:
Taxi Drivers Navigate a City Stacked Against Them
Nearly 50% of New York City yellow cab drivers are from Bangladesh, Pakistan, and India — they must face ride-sharing's rise and increasing driver suicides, together.
Samira Sadeque
One afternoon in May 2018, 50-year-old Rajinder Singh took an hour out of his driving schedule to attend a rally in support of a fallen colleague. Yu Mein "Kenny" Chow, an immigrant New York City taxi driver from Myanmar, had gone missing for two weeks before his body was discovered in the East River. It was eventually ruled a suicide, the fifth in a spate of eight suicides among New York City taxi drivers between November 2017 and November 2018.
Despite working extra hours and taking fewer breaks, many taxi drivers are earning less and completing fewer rides than they were before ride-sharing apps such as Uber and Lyft took the streets from them.
Uber, which entered the city in 2011, booked more rides than New York City cabs for the first time in 2017 — 289,000 Uber rides a day compared to 277,000 taxi trips.
For South Asians, who make up a large portion of the taxi driver population in New York, the impact of ride-sharing has had an especially large effect. In 2018, 46.5% of yellow cab drivers came from three countries: Bangladesh (26.8%), Pakistan (10.9%), and India (8.8%) — these were also the top three countries of origin of drivers.


Effects: Reduce 20% of all enemies' HP. Ignore enemy element and defense. Bind recovery for 5 turns.
Effects: All attribute cards ATK x5 when attacking with Fire, Water, Wood & Light orb types at the same time. ATK x1.5 at 5 combos. ATK x0.5 for each additional combo, up to ATK x4 at 10 combos.
Compare Reincarnated Sakuya with ..
Okay first thing first I cant be the only one who thinks this new art is an overall downgrade to awoken sakuya.
Secondly going from x100 to x 400 is great and all but glass cannons just arnt meta anymore without a shield or tank stats. I feel like they could have done something completely different like a 35% shield at a certain combo count or no skyfall for better damage control or hell even just a basic HP and RCV multiplier and maybe add a poison resist awoken skill (I always loved that my sakuya team had 40% resist to jammer and blinds it was very useful IMO and was jealous when I saw summer school sakuya with a poison resist) without tank stats or damage control you just arnt viable in meta and sadly will be collecting dust in my box which is sad sense you were my first hyper max and god pull.
Stopped reading at "new art is an overall downgrade to awoken sakuya" and immediately upvoted.
i was hoping for Reincarnated Incarnation of Kirin, Sakuya.
Ugh, the art...it's awfully cheesy. Globe light ring and comical glow lines. Fire swirl on the other end looks like Flaming Hot Cheetos. Saturday-morning cartoon dogs, extra thick hip armor, armguards, and shoulder pads. Too much color saturation and armor detracts aesthetics from her. When you are reborn stronger, wouldn't it make sense to need less armor?
Ladies and gentlemen, we went from Majestic Beauty ...to Power Ranger. SMH.
The artist must have looked at the job order and said with raised eyebrow, "I created perfection, and now they want me to change her?"
This travesty could have only came from Gosei Sentai Dairanger, the Kirin Ranger.
In my view, the end looks like a meatball, but you're right. They're just flaming hot cheetos.
Will be interesting to pair her off with one of the new 7x6 leaders, like Ilmina. She is capable of doing so much damage on her own that even if the 7x6 leader has a relatively small atk multiplayer you will probably still be able to plow through most content.
Can confirm that's a really strong strategy. I personally use waifu sherias and awoken u+y to get 60x max and paired with tpas, it does pretty substantial damage.
I have been using her as a sub before, just means she is a stronger sub now. Works for me.
I happen to have Awoken Sakuya (self-inherited) as my Ronove Sub, and I only reincarnate my assisting Sakuya for slightly more stats, and leave my main Sakuya in Awoken Form.
Otherwise, I would have no water attributes in the whole team, but for some reasons, I can't find good replacements with water attribute.
She's super fun but unstable with constant board troll. Despite clearing the board, she's too often missing a color, which means death in a kill or be killed scenario. Board changers only go so far.
If GH would add something to allow her take some hits while stalling for board with colors, like a small shield that increases with combo count, she'd be amazing and actually viable for high level content.
New limit break stats/weight are: HP 4376 Atk 1971 Rcv 804 with 1099.8 weight.
I'm really glad that I took the time to devolve my Sakuya from her green ult now. This revo is going to be a lot of fun to play.
I am pretty disappointed in this Reincarnation. Sakuya is my favorite monster on this garme, and I was hoping she would gain viability in the meta, but this fails miserably. Yes her multiplier is incredible, however, she makes up for NONE of her weaknesses. She is still frail, now she's just a glass cannon lead, and I feel that the game has been over saturated with glass cannon leaders. She also loses any aspect of damage control. She goes from 5x to 10x ATK with just ONE more combo. With this in mind, the only viable leader you can pair her with is a no Skyfall leader, and only Kushinada can possibly pair with her. I feel that she could have gotten a similar LS to Raijin of Myr Regalia, Rei.
She actually goes from 5x to 7.5x with 4 to 5 combos. Read before you make yourself look stupid.
Edited cuz I misread ur comment initially. Just pair with another leader dude, it's not a bad thing. U don't need to dual wield all leads. Glass cannon LS translates to->pair with a different leader or think to offset the negative some other way. This is a puzzle game still lol. And don't get so hung up on the meta, it'll ruin ur playtime.
I must say I think one of my favourite things about reincarnated sakuyas art is that it isn't remotely sexual, it just looks like a strong female character.
Sakuya's swab sure got hotter, and a little lighter!
She's here! I was on the screen for evos and saw she could be revoed along with the norse gods 2!!
LOL... don't forget Santa Sakuya, Sakuya Another, and coloring book Sakuya. There's 9 versions of her now.
Well, she's been tilting her head to same side for five evolutions, you think that in her reincarnate form she would tilt her head to the other side for a change, but no, she still keeps the same posture!
As of April 9 2018 in the U.S, She CAN'T get super awakenings.
Whoever thought that changing the awesome dragon version of Fagan into those piggies would look cool, had a really bad idea.
agreed, though they wouldn`t look out of place with coloring book Sakuya, given that those "Pig Dragons" are drawn "`Toon Style" not matching with Sakuya`s "Anime/Manga Style"


text:
\section{Introduction and main results}
The starting point of this paper is the following result, see \cite{GHW19}. Let $X,X_1,\ldots,X_n, \ldots $ be a sequence of independent and identically distributed
(i.i.d.) random points in $\RR^d$ that are defined on a common probability space $(\Omega,{\cal A},\mathbb{P})$.
We assume that the distribution of $X$, which is denoted by $\mu$, is absolutely continuous with respect to Lebesgue measure $\lambda$,
and we denote the density of $\mu$ by $f$. Writing $\|\cdot \|$ for the Euclidean norm in $\RR^d$, and putting ${\cal X}_n := \{X_1,\ldots,X_n\}$, let $R_{i,n}:=\min_{j\ne i, j\le n} \|X_i-X_j\|$ be the distance
from $X_i$ to its nearest neighbor in the set ${\cal X}_n \setminus \{X_i\}$. Moreover, let $\:\mbox{\bf 1} \{A\}$ denote the indicator function of a set $A$, and write
$B(x,r) = \{y \in \RR^d: \|x-y\|\le r\}$ for the closed ball centered at $x$ with radius $r$. Finally, let
\
C_n := \sum_{i=1}^n \:\mbox{\bf 1} \Big{\{} \mu \big( B(X_i,R_{i,n})\big) > \frac{t + \log n}{n} \Big{\}}
\
denote the number of exceedances of probability volumes of nearest neighbor balls that are larger than the threshold $(t + \log n)/n$.
The main result of \cite{GHW19} is
Theorem 2.2 of that paper, which states that, under a weak condition on the density $f$, for each fixed    $t \in \RR$, we have
\begin{equation}\label{convnn}
C_n \stackrel{{\cal D}}{\longrightarrow} {\rm Po}(\exp(-t))
\end{equation}
as $n \to \infty$, where $\stackrel{{\cal D}}{\longrightarrow}$ denotes convergence in distribution, and Po$(\xi)$ is the Poisson distribution with parameter
$\xi >0$.

Since the {\em maximum} probability content of these nearest balls, denoted by $P_n$, is at most $(t+ \log n)/n$ if, and only if, $C_n =0$,
we immediately obtain a Gumbel limit $\lim_{n\to \infty} \mathbb{P}(nP_n - \log n \le t) = \exp(-\exp(-t))$ for $P_n$.

To state a sufficient condition on $f$ that guarantees \eqref{convnn}, let
$\text{supp}(\mu) := \{x \in \RR^d: \mu (B(x,r)) > 0 \textrm{ for each } r >0\}$ denote the support of $\mu$.
Theorem 2.2 of \cite{GHW19} requires that there are $\beta \in (0,1)$, $c_{max}<\infty$ and $\delta >0$
such that, for any $r,s >0$ and any $x,z \in$ supp$(\mu)$ with
$\|x-z\| \ge \max\{r,s\}$ and  $\mu\left( B(x,r)\right) = \mu\left( B(z,s)\right)\le \delta$, one has
\[
\frac{\mu\left( B(x,r)\cap   B(z,s)\right)}{\mu\left( B(z,s)\right)}
\le \beta
\]
and $\mu\big( B(z, 2s)\big)\le c_{max}\mu\left( B(z,s)\right)$.

These conditions hold if $\text{supp}(f)$ is a compact set $K$ (say), and there are
$f_-,f_+ \in (0,\infty)$ such that
\begin{equation}\label{densitybounds}
f_-\leq f(x)\leq f_+, \qquad x \in K.
\end{equation}
Thus, the density $f$ of $X$ is bounded and bounded away from zero.

The purpose of this paper is to generalize \eqref{convnn} to $k$th-nearest neighbors, and to derive a rate of convergence for the Poisson approximation of {the number of exceedances}.

Before stating our main results, we give some more notation.  For fixed $k \le n-1$, we denote by $R_{i,n,k}$ the Euclidean
distance of  $X_i$ to its $k$th-nearest neighbor among ${\cal X}_n \setminus \{X_i\}$,
and we write  $B(X_i,R_{i,n,k})$ for the $k$th-nearest neighbor ball centered at $X_i$ with radius $R_{i,n,k}$.  For fixed $t \in \RR$, put
\begin{equation} \label{defynk}
v_{n,k} :=v_{n,k}(t) :=\frac{t+\log n+(k-1)\log\log n-\log(k-1)!}{n},
\end{equation}
and let
\begin{equation}\label{defcnk}
C_{n,k} := \sum_{i=1}^n \:\mbox{\bf 1} \big{\{}\mu \big( B(X_i,R_{i,n,k})\big) > v_{n,k}\big{\}}
\end{equation}
denote the {\textit{number of exceedances}} of probability contents of $k$th-nearest neighbor balls over the threshold $v_{n,k}$ defined in \eqref{defynk}.

The threshold $v_{n,k}$ is in some sense {\em universal} in dealing with the number of exceedances of probability contents of $k$th-nearest neighbor balls. To this end, suppose that, in
much more generality than considered so far, $X,X_1,X_2, \ldots $ are i.i.d. random elements taking values in a separable metric space $(S,\rho)$.
We retain the notations $\mu$ for the distribution of $X$ and $B(x,r) := \{y \in S: \rho(x,y) \le r\}$ for the closed ball with radius $r$ centered at $x \in S$. Regarding the distribution
$\mu$, we assume that
\begin{equation}\label{conditionbound}
\mu(\{y \in S: \rho(x,y) =r\}) =0, \qquad x \in S, \ r \ge 0.
\end{equation}
As a consequence, the distances $\rho(X_i,X_j)$, where $j \in \{1,\ldots,n\} \setminus \{i\}$, are different with probability    one
for each $i \in \{1,\ldots,n\}$. Thus, for fixed $k \le n-1$, there is almost surely a unique $k$th-nearest neighbor of $X_i$,
and we also retain the notations $R_{i,n,k}$ for the distance of  $X_i$ to its $k$th-nearest neighbor among ${\cal X}_n \setminus \{X_i\}$ and
$B(X_i,R_{i,n,k})$ for the ball centered at $X_i$ with radius $R_{i,n,k}$. Notice that the condition \eqref{conditionbound} excludes discrete metric spaces (see, e.g., Section 4 of \cite{ZO18}),
but not function spaces like, e.g., the space $C[0,1]$ of continuous functions on $[0,1]$ with the supremum metric, and with Wiener measure $\mu$.

\medskip

{In what follows, for sequences $(a_n)_{n\geq 0}$ and $(b_n)_{n\geq 0}$ of real numbers, write $a_n=O(b_n)$ if  $|a_n|\leq C|b_n|$, $n\geq 1$, for some positive constant $C$.}

\begin{theorem}\label{thmmetricspace}
If $X_1,X_2, \ldots$ are i.i.d. random elements of a metric space $(S,\rho)$, and if \eqref{conditionbound} holds, then
the sequence $(C_{n,k})$ satisfies
\[
 \E[C_{n,k}] = {\rm e}^{-t} + O\left( \frac{\log\log n}{\log n}  \right).
\]
\end{theorem}

\medskip

In particular, the mean number of exceedances $C_{n,k}$ converges to ${\rm e}^{-t}$ as $n$ goes to infinity.
 By Markov's inequality, this result implies the tightness of the sequence $(C_{n,k})_{n \ge 1}$. Thus, at least a subsequence converges in distribution.
The next result states convergence of $C_{n,k}$ to a Poisson distribution if $(S,\rho) = (\RR^d,\|\cdot\|)$ and \eqref{densitybounds} holds.
To this end, let $d_{TV}(Y,Z)$ be the total variation between two integer-valued random variables $Y$ and $Z$, i.e.,
\[
d_{TV}(Y, Z) = 2\sup_{A\subset \NN}|\PP(Y\in A) -\PP(Z\in A)|.
\]

\begin{theorem}\label{Th:poissonapprox}
Let $Z$ be a Poisson random variable with parameter ${\rm e}^{-t}$. If $X, X_1,X_2, \ldots $ are i.i.d. in $\RR^d$ with density $f$, and if the distribution $\mu$ of $X$ has compact support {$[0,1]^d$}
 and satisfies \eqref{densitybounds}, then, as $n \to \infty$,
 \[
 d_{TV}\left(C_{n,k}, Z\right) =  O\left( \frac{\log\log n}{\log n} \right).
 \]
\end{theorem}
 Theorem \ref{Th:poissonapprox} is not only a generalization of Theorem 2.2 of \cite{GHW19} over all $k\geq 1$: it also provides a rate of convergence for the Poisson approximation of  $C_{n,k}$. {Our theorem is stated in the particular case {that} the support of $\mu$ is $[0,1]^d$ but we think that it can be extended to any measure $\mu$ whose support is a general convex body. For {the sake of} readibility of the manuscript, we did not deal with such a generalization.  }

 {\begin{remark}
The study of extremes of $k$th-nearest neighbor balls is classical in stochastic geometry, and it has various applications, see e.g. \cite{P}. In Section 4 in \cite{O}, bounds for the total variation distance of the process of Poisson points with large $k$th-nearest neighbor ball (with respect to the intensity measure) and a Poisson process were obtained. Parallel to our work, these results have been extended by  Bobrowski \textit{et al.} to the Kantorovich-Rubinstein distance and generalized to the binomial process in Section 6.2 of a paper  which has just been submitted \cite{BSY}. While the results in \cite{BSY}  and \cite{O} rely on Palm couplings of a thinned Poisson/binomial process and employ distances of point processes, we derive a bound on the total variation distance of the number of large $k$th-nearest-neighbor balls and a Poisson-distributed random variable. Our approach permits to build arguments  on classical Poisson approximation theory \cite{AGG} and  an asymptotic independence property stated in Lemma \ref{Le:independence} below, and it thus results in a considerably shorter and less technical proof.
\end{remark}}

Now, let
 $P_{n,k} = \max_{1 \le i \le n} \mu\big(B(X_i,R_{i,n,k})\big)$ be the maximum probability content of the $k$th-nearest neighbor balls.
Since $C_{n,k} = 0$ if, and only if, $P_{n,k} \le v_{n,k}$, we obtain the following corollary.

\begin{corollary}
Under the conditions of Theorem~\ref{Th:poissonapprox}, we have
\[
\lim_{n\to \infty} \mathbb{P}\big(n P_{n,k} - \log n - (k-1)\log \log n + \log(k-1)! \le t\big) = {\rm G}(t), \quad t \in \mathbb{R},
\]
where ${\rm G}(t) = \exp(-\exp(-t))$ is the distribution function of the Gumbel distribution.
\end{corollary}

\begin{remark}
If, in the Euclidean case, the density $f$ is continuous, then  $\mu(B(X_i,R_{i,n,k}))$  is approximately equal to $f(X_i)\kappa_d R_{i,n,k}^d$, where
$\kappa_d = \pi^{d/2}/\Gamma(1+d/2)$ is the volume of the unit ball in $\RR^d$. Under additional smoothness assumptions on $f$ and  \eqref{densitybounds},
\cite{HE82, HE83} proved that
\begin{equation}\label{gumbelknnbound}
\lim_{n\to \infty} \PP\left( \max_{i=1,\ldots,n} f(X_i) \kappa_d \min\big(R_{i,n,k}^d, \|X_i-\partial K\|^d\big) \le v_{n,k}\right) = {\rm G}(t),
\end{equation}
\end{remark}
{where $K$ is the support of $\mu$}. Here, the distance $\|X_i-\partial K\|$ of $X_i$ to the boundary of $K$ is important to overcome edge effects. These effects dominate the asymptotic behavior
of the maximum of the $k$th-nearest neighbor distances if $k \ge d$, see  \cite{DH89, DH90}. In fact, \cite{HE82} proved convergence of the factorial moments of
\[
\widetilde{C}_{n,k} := \sum_{i=1}^n \:\mbox{\bf 1} \Big{\{} f(X_i)\kappa_d \min\big(R_{i,n,k}^d,\|X_i - \partial K\|^d\big) > v_{n,k} \Big{\}}
\]
to the corresponding factorial moments of a random variable with the Poisson distribution ${\rm Po}({\rm e}^{-t})$ and thus, by the method of moments, more than   \eqref{gumbelknnbound}, namely
$\widetilde{C}_{n,k} \overset{\mathcal{D}}{\rightarrow} {\rm Po}\big({\rm e}^{-t}\big)$. However, our proof of Theorem~\ref{Th:poissonapprox} is completely different thereof, since it is based on the
Chen-Stein method and provides a rate of convergence.

\section{Proofs}

\subsection{Proof of Theorem \ref{thmmetricspace}}
\begin{proof}
By symmetry, we have
\begin{eqnarray*}
\mathbb{E} \big{[}C_{n,k}\big{]} & =  & n\,  \mathbb{P} \left( \mu (B(X_1,R_{1,n,k})) > v_{n,k} \right)\\
& = & n \, \mathbb{E} \left[ \mathbb{P} \left( \mu(B(X_1,R_{1,n,k}))\right) > v_{n,k}|X_1  \right].
\end{eqnarray*}
For a fixed $x \in S$, let
\
H_x(r):=\mathbb{P} \left(\rho(x,X) \le r\right), \quad r \ge 0,
\
be the distribution function of $\rho(x,X)$. Due to the condition \eqref{conditionbound}, the function $H_x$ is continuous,
and by the probability integral transform (see e.g. \cite{BD15}, p. 8),
the random variable
\[
H_x(\rho(x,X)) = \mu \big( B(x,\rho(x,X))\big)
\]
is uniformly distributed in the unit interval $[0,1]$.
Put $U_j := H_x(\rho(x,X_{j+1}))$, $j=1,\ldots,n-1$. Then $U_1,\ldots,U_{n-1}$ are i.i.d. random variables with
a uniform distribution in $(0,1)$. Hence, conditionally on $X_1=x$, the random variable $\mu (B(X_1,R_{1,n,k}))$ has the same distribution as
$U_{k:n-1}$, where $U_{1:n-1} < \ldots < U_{n-1:n-1}$ are the order statistics of $U_1,\ldots,U_{n-1}$, and this distribution does not depend on $x$.
Now, because of a well-known relation between the distribution   of order statistics from the uniform distribution on $(0,1)$ and the binomial distribution (see, e.g., \cite{ANS}, p. 16),
we have
\[
\mathbb{P}(U_{k:n-1} > s) = \sum_{j=0}^{k-1} {n-1 \choose j} s^j (1-s)^{n-1-j}
\]
and thus
\begin{equation}\label{expcnk}
\mathbb{E} \big{[}C_{n,k}\big{]} = n \sum_{j=0}^{k-1} {n-1 \choose j} v_{n,k}^j (1-v_{n,k})^{n-1-j}.
\end{equation}
Here, the summand for $j=k-1$ equals
\[n {n-1 \choose k-1} v_{n,k}^{k-1} (1-v_{n,k})^{n-k} = \frac{n}{(k-1)!}\, (nv_{n,k})^{k-1}\, \prod_{i=1}^{k-1}\frac{n-i}{n}\, (1-v_{n,k})^{n-k}.\]
Using Taylor expansions, \eqref{defynk} yields
 \[nv_{n,k}  = \log n+O\left(\log\log n\right), \quad
 \prod_{i=1}^{k-1}\frac{n-i}{n} = 1+O\left(  \frac{1}{n}\right)
 \]
 and
\[
(1-v_{n,k})^{n-k} = \frac{(k-1)!}{n}\, \exp\left(-t-(k-1)\log\log n+O\left( \frac{\log^2(n)}{n}  \right)\right).
\]
Straigthforward computations now give
\[n {n-1 \choose k-1} v_{n,k}^{k-1} (1-v_{n,k})^{n-k}  = {\rm e}^{-t}+O\left( \frac{\log\log n}{\log n}  \right).\]
Regarding the remaining summands on the right hand side of \eqref{expcnk}, it is readily seen that
\[
\sum_{j=0}^{k-2} {n-1 \choose j} v_{n,k}^j (1-v_{n,k})^{n-1-j} = O\left( n {n-1 \choose k-1} v_{n,k}^{k-1} (1-v_{n,k})^{n-k} \cdot \frac{1}{nv_{n,k}}  \right),
\] with the convention that the sum equals 0 if $k=1$. From the above computations and from  \eqref{defynk}, it follows that this sum
equals $O\left(1/\log n \right)$, which concludes the proof of Theorem \ref{thmmetricspace}.
\end{proof}

\begin{remark}\label{rem-byproduct}
In the proof given above, we conditioned on the realizations $x$ of $X_1$. Since the distribution of $H_x(\rho(x,X)) = \mu\big(B(x,\rho(x,X))\big)$ does
not depend on $X$, we obtain as a by-product that
\[
\PP\left(\mu (B(X_1,R_{1,n,k})) > v_{n,k} \right)  = \sum_{j=0}^{k-1} {n-1 \choose j} v_{n,k}^j (1-v_{n,k})^{n-1-j} \sim \frac{{\rm e}^{-t}}{n},
\]
if $X_1,X_2,\ldots,X_n$ are independent and $X_2,\ldots,X_n$ are i.i.d. according to $\mu$. Here, $X_1$ may have an arbitrary distribution and $a_n \sim b_n$ means that $a_n/b_n \to 1$ as $n \to \infty$.
\end{remark}

\subsection{Proof of Theorem~\ref{Th:poissonapprox}}
The main idea to derive Theorem \ref{Th:poissonapprox} is to discretize {$\text{supp}(\mu)=[0,1]^d$} into finitely many ``small sets''
and then to employ the Chen-Stein method. To apply this method, we will have to check an  \textit{asymptotic independence property}
and a \textit{local property} which ensures that, with high probability,  two exceedances cannot appear in the same neighborhood.
We introduce these properties below and recall a result due to Arratia \textit{et al.} \cite{AGG} on the Chen-Stein method.

\paragraph{The asymptotic independence property}
Fix $\varepsilon > 0$. Writing $\lfloor \cdot \rfloor$ for the floor function, we partition {$[0,1]^d$} into a set $\mathcal{V}_n$ of $N_n^d$  \textit{subcubes} (i.e., subsets that are cubes)
 of equal size that can only have boundary points in common, where $N_n=\lfloor n/\log(n)^{1+\varepsilon}\rfloor$. The subcubes are indexed
by the set  $\left[ 1,N_n\right] ^{d} = \{\mathbf{j}:=(j_{1},\ldots, j_{d}): j_m \in \{1, \ldots, N_n\} \text{ for } m \in \{1,\ldots,d\}\}$.
With a slight abuse of notation, we identify a cube with its index.  Let
\[
\mathcal{E}_n = \bigcap_{\mathbf{j}\in \mathcal{V}_n}\{{\cal X}_n\cap \mathbf{j}\neq \emptyset\}
\]
be the event that each of the subcubes contains at least one of the points of ${\cal X}_n$.
    The event $\mathcal{E}_n$ is extensively used in stochastic geometry to derive central limit theorems or to deal with extremes \cite{AB, BC, CR},
    and it will play a crucial role throughout  the rest of the paper. The following lemma, which captures the idea of ``asymptotic independence''  ,
     is at the heart of our development.

\begin{lemma}
\label{Le:independence}
For each $\alpha>0$, we have $\PPP{\mathcal{E}_n^c}=o(n^{-\alpha})$ as $n \to \infty$.
\end{lemma}

\begin{proof}
By subadditivity and independence, it follows that
\begin{align*}
\PPP{\mathcal{E}_n^c} & \leq \sum_{\mathbf{j}\in \mathcal{V}_n}\PPP{{\cal X}_n\cap \mathbf{j}=\emptyset}\\
& = \sum_{\mathbf{j}\in \mathcal{V}_n}\big(  \PPP{X_1\not\in \mathbf{j}} \big)^n\\
& = \sum_{\mathbf{j}\in \mathcal{V}_n} \big(1-\mu(\mathbf{j})\big)^n\\
& \leq \sum_{\mathbf{j}\in \mathcal{V}_n} \exp(-n\mu(\mathbf{j})).
\end{align*}
Here, the last inequality holds since $\log(1-x)\leq -x$ for each $x\in [0,1)$. Since $f\geq f_- >0$ on $K$, we have
$\mu(\mathbf{j})=\int_\mathbf{j} f \text{d} \lambda \geq f_- \lambda(\mathbf{j})$, whence -- writing $\# M$ for the cardinality of a finite set $M$ --
\begin{align*}
\PP \big(\mathcal{E}_n^c\big) & \leq \sum_{\mathbf{j}\in \mathcal{V}_n} \exp\big(-n f_- \lambda(\mathbf{j})\big)\\
&\leq \#\mathcal{V}_n \exp\big(- f_-(\log n)^{1+\varepsilon}\big).
\end{align*}
Since  $\#\mathcal{V}_n\leq n/(\log n)^{1+\varepsilon}$, it follows that $n^\alpha \PPP{\mathcal{E}_n^c} \to 0  \text{ as }  n \to \infty$.
\end{proof}

\paragraph{The local property}
We now define a metric d on $\mathcal{V}_n$ by putting d$(\mathbf{j},\mathbf{j}'):=\max_{1\leq s\leq d}|j_{s}-j'_{s}|$ for any two different subcubes $\mathbf{j}$ and $\mathbf{j}'$, and
d$(\mathbf{j},\mathbf{j}):= 0$, $\mathbf{j} \in \mathcal{V}_n$. Let $S(\mathbf{j}, r) = \{\mathbf{j}'\in \mathcal{V}_n: \text{d}(\mathbf{j}, \mathbf{j}')\leq r\}$ be the ball of subcubes of radius $r$ centered at $\mathbf{j}$.
For any $\mathbf{j}\in \mathcal{V}_n$, put
\[
M_\mathbf{j} := \max_{i\leq n, X_i\in \mathbf{j}} \mu(B(X_i,R_{i,n,k})),
\]
with the convention $M_\mathbf{j}=0$ if ${\cal X}_n\cap \mathbf{j}=\emptyset$. Conditionally on the event $\mathcal{E}_n$, and provided that d$(\mathbf{j},\mathbf{j}')\geq 2k+1$,
the random variables $M_\mathbf{j}$ and $M_{\mathbf{j}'}$ are independent. Lemma \ref{Le:independence} is referred to as the \textit{asymptotic independence property}: conditionally on the event $\mathcal{E}_n$, which occurs with high probability, the extremes $M_\mathbf{j}$ and $M_\mathbf{j}'$ attained on two subcubes which are sufficiently distant from each other are independent.

The following lemma claims that, with high probability, two exceedances cannot occur in the same neighborhood.

\begin{lemma}
\label{Le:localproperty}
With the notation $a \wedge b := \min(a,b)$ for $a,b \in \mathbb{R}$, let
\[
R(n)= \sup_{\mathbf{j}\in\mathcal{V}_n}\sum_{i\neq i'\leq n} \PP \big( X_i,X_{i'}\in S(\mathbf{j},2k); \mu(B(X_i,R_{i,n,k})) \wedge \mu(B(X_{i'},R_{i',n,k}))>v_{n,k} \big).
\]
Then $R(n) = O(n^{-1}(\log n)^{2-d+\varepsilon})$ as $n \to \infty$.
\end{lemma}
Here, with a slight abuse of notation, we have identified the family of subcubes $S(\mathbf{j},2k)=\{\mathbf{j}'\in \mathcal{V}_n: \text{d}(\mathbf{j},\mathbf{j}')\leq 2k\}$
with the set $\bigcup \big\{\mathbf{j}': \mathbf{j}'\in \mathcal{V}_n \text{ and }  \text{d}(\mathbf{j},\mathbf{j}')\leq 2k\}.$

We prepare the proof of Lemma \ref{Le:localproperty} with the following {result} that gives the volume of two $d$-dimensional balls.

\begin{lemma} \label{Le:balls}
	{If $x \in B(0,2)$ then}
	\begin{align*}
		\lambda(B(0,1)\cup B(x,1))=2\left(\kappa_d \left(1-\frac{\arccos(\|x\|/2)}{\pi}\right)+\frac{\|x\|\kappa_{d-1}}{2d}\left(\sqrt{1-(\|x\|/2)^2}\right)^{d-1}\right).
	\end{align*}
\end{lemma}

\begin{proof}
	We calculate the volume of $\lambda(B(0,1)\cup B(x,1))$ as the sum of the volumes of the following two congruent sets. The first one, say $B$, is given by the set of all points in $B(0,1)\cup B(x,1)$ that are closer to $0$ than to $x$ and for the second one we change the roles of $0$ and $x$. The set $B$ is the union of a cone $C$ with radius $\sqrt{1-(\|x\|/2)^2}$, height $\|x\|/2$ and apex at the origin and a set $D:=B(0,1)\setminus S$, where $S$ is a simplicial cone with external angle $\arccos(\|x\|/2)$. From elementary geometry, we obtain that the volumes of $C$ and $D$ are given by
	\begin{align*}
		\lambda(C)=\frac{\|x\|\kappa_{d-1}}{2d}\left(\sqrt{1-(\|x\|/2)^2}\right)^{d-1},\quad
		\lambda(D)=\kappa_d \left(1-\frac{\arccos(\|x\|/2)}{\pi}\right).
	\end{align*}
	This finishes the proof of the lemma.
\end{proof}

\begin{proof}[Proof of Lemma \ref{Le:localproperty}] For $z \in {[0,1]^d}$, let
	\begin{align*}
		r_{n,k}(z):=\inf\{r>0:\,\mu(B(z,r))> v_{n,k}\}.
	\end{align*}
	Writing $\#{\cal Y}(A)$ for the number of points of a finite set ${\cal Y}$ of random points in
	$\mathbb{R}^d$ that fall into a Borel set $A$, we have
	\begin{align*}
		\mu(B(z,R_{n,k}(z)))>v_{n,k}\quad \Longleftrightarrow \quad \# \mathcal{X}_n(B(z,r_{n,k}(z))) \le k-1.
	\end{align*}
	In the following, we assume that $r_{n,k}(X_{i'})\le r_{n,k}(X_{i})$ (which is at the cost of a factor $2$) and distinguish the two cases $X_{i'} \in B(X_i,r_{n,k}(X_{i}))$ and $X_{i'} \in S(\mathbf{j},2k) \setminus B(X_i,r_{n,k}(X_{i}))$. This distinction of cases gives
	\begin{align}
		&\PP \big( X_i,X_{i'}\in S(\mathbf{j},2k);  \mu(B(X_i,R_{i,n,k})) \wedge \mu(B(X_{i'},R_{i',n,k}))>v_{n,k} \big)\nonumber\\
		&\quad\le 2 \PP \big( X_i,X_{i'}\in S(\mathbf{j},2k); r_{n,k}(X_{i'})\le r_{n,k}(X_{i}); \mu(B(X_i,R_{i,n,k})) \wedge \mu(B(X_{i'},R_{i',n,k}))>v_{n,k} \big)\nonumber\\
		&\quad \le 2 \PP \big( X_i \in S(\mathbf{j},2k); X_{i'} \in  B(X_i,r_{n,k}(X_{i})); r_{n,k}(X_{i'})\le r_{n,k}(X_{i});\nonumber\\ &\quad \quad \quad \quad \mu(B(X_i,R_{i,n,k})) \wedge \mu(B(X_{i'},R_{i',n,k}))>v_{n,k} \big)\label{lemloccl}\\
		&\quad \quad +2 \PP \big( X_i\in S(\mathbf{j},2k),X_{i'} \in S(\mathbf{j},2k) \setminus B(X_i,r_{n,k}(X_{i})); r_{n,k}(X_{i'})\le r_{n,k}(X_{i});\nonumber\\
		&\quad \quad \quad \quad  \mu(B(X_i,R_{i,n,k}))\wedge \mu(B(X_{i'},R_{i',n,k}))>v_{n,k} \big).\label{lemlocfa}
	\end{align}
	We bound the summands \eqref{lemloccl} and \eqref{lemlocfa} separately. Since $X_i$ and $X_{i'}$  are independent, \eqref{lemloccl} takes the form
	\begin{align}
		&2\int_{S(\mathbf{j},2k)} \int_{B(x,r_{n,k}(x))}\mathds{1}\{r_{n,k}(y) \le r_{n,k}(x)\} \, \PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\} \cup \{x\})(B(y,r_{n,k}(y)))\le k-1; \nonumber\\
		&\quad \quad \#(\mathcal{X}_n \setminus \{X_i,X_{i'}\} \cup \{y\})(B(x,r_{n,k}(x)))\le k-1)\,\mu(\mathrm{d}y)\,\mu(\mathrm{d}x).\label{lemlocclint}
	\end{align}
	For $y \in B(x,r_{n,k}(x))$, the probability in the integrand figuring above is bounded from above by
	\begin{align}
		&\PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y)))\le k-1;\nonumber\\
		&\quad  \#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x)))\le k-2)\nonumber\\
		&\, \le \PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y)))\le k-1;\nonumber\\
		&\, \quad  \#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(y))\le k-2).\label{negdep}
	\end{align}
	Since the random vector
	\[
	\left(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y))),\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(y)))\right)
	\]
	is negatively quadrant dependent (see \cite[Section 3.1]{JP83}), Equation \eqref{negdep} has the upper bound
	\begin{align}
		& \PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y)))\le k-1)\nonumber\\
		&\quad \times \PP( \#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(y)))\le k-2)\nonumber\\
		&\, \le \PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y)))\le k-1)\nonumber\\
		&\, \quad \times  \PP( \#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x)))\le k-2),\label{ballest}
	\end{align}
	where the last inequality holds since $r_{n,k}(y) \le r_{n,k}(x)$. Analogously to Remark \ref{rem-byproduct},  the first probability {is}
	\begin{align*}
		\PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y)))\le k-1)= \sum_{j=0}^{k-1} {n-2 \choose j} v_{n,k}^j (1-v_{n,k})^{n-2-j} \sim \frac{{\rm e}^{-t}}{n}.
	\end{align*}
	The latter probability in \eqref{ballest} is given by
	\begin{align}
		&\sum_{\ell=0}^{k-2} \binom{n-2}{\ell} \mu \big(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x))\big)^\ell \big(1-\mu(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x)))\big)^{n-2-\ell}. \label{bindist}
	\end{align}
	In a next step, we estimate $\mu \big(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x))\big)$. Since $f(x) \ge f_- >0,\,x \in {[0,1]^d}$, and by the homogeneity of $d$-dimensional Lebesgue measure $\lambda$,
	we obtain
	\begin{align*}
		\mu \big(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x))\big)&\ge f_- \lambda (B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x))) \\
		\quad& =  f_- r_{n,k}(x)^d  \lambda (B(0,1)\setminus B(r_{n,k}(x)^{-1}(y-x),1))\\
		\quad&=  f_- r_{n,k}(x)^d  \left(\lambda (B(0,1)\cup B(r_{n,k}(x)^{-1}(y-x),1))-\kappa_d\right).
	\end{align*}
	For $y \in B(x,r_{n,k}(x))$, Lemma \ref{Le:balls} {yields}
	\begin{align*}
		&\mu \big(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x))\big)\nonumber\\
		&\quad \ge f_-  r_{n,k}(x)^{d} \left(\kappa_d\left(1-\frac{2\arccos(\|x-y\|/2r_{n,k}(x))}{\pi}\right)+\frac{\|x-y\|\kappa_{d-1}}{2dr_{n,k}(x)}\left(\sqrt{1-(\|x-y\|/2r_{n,k}(x))^2}\right)^{d-1}\right). \label{geoest}
	\end{align*}
	Since $\inf_{s>0} s^{-1}(1-2\arccos(s)/\pi)>0$, there is $c_0>0$ such that
	\begin{align*}
		\mu \big(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(x))\big) \ge c_0 \|x-y\| r_{n,k}(x)^{d-1} ,\quad x \in S(\mathbf{j},2k),\,y \in B(x,r_{n,k}(x)).
	\end{align*}
	{Equation} \eqref{bindist} and the bound $f(x) \le f_+,\,x \in {[0,1]^d},$ {give}
	\begin{align*}
		&\int_{B(x,r_{n,k}(x))}\mathds{1}\{r_{n,k}(y) \le r_{n,k}(x)\} \PP \big(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}({x})))\le k-1\big)\, \mu(\mathrm{d}y)\\
		&\quad \le f_+ \sum_{\ell=0}^{k-2} \binom{n-2}{\ell} \int_{B(x,r_{n,k}(x))} \Big(c_0\|x-y\|r_{n,k}(x)^{d-1}\Big)^\ell \Big(1-c_0\|x-y\|r_{n,k}(x)^{d-1}\Big)^{n-2-\ell} \lambda(\mathrm{d}y).
	\end{align*}
	We now introduce spherical coordinates and obtain
	\begin{align*}
		&f_+ d \kappa_d \sum_{\ell=0}^{k-2} \binom{n-2}{\ell} \int_0^{r_{n,k}(x)} \Big(c_0 t r_{n,k}(x)^{d-1}\Big)^\ell \Big(1-c_0 t r_{n,k}(x)^{d-1}\Big)^{n-2-\ell} t^{d-1} \mathrm{d}t\\
		&\quad =f_+ d \kappa_d \sum_{\ell=0}^{k-2} \binom{n-2}{\ell} \int_0^{r_{n,k}(x)} \Big(c_0 t r_{n,k}(x)^{d-1}\Big)^\ell \exp \Big( (n-2-\ell) \log(1-c_0 t r_{n,k}(x)^{d-1})\Big) t^{d-1} \mathrm{d}t\\
		&\quad \le f_+d \kappa_d \sum_{\ell=0}^{k-2} \binom{n-2}{\ell} \int_0^{r_{n,k}(x)} \Big(c_0 t r_{n,k}(x)^{d-1}\Big)^\ell \exp \Big( -c_0(n-2-\ell)  t r_{n,k}(x)^{d-1}\Big) t^{d-1} \mathrm{d}t.
	\end{align*}
	Here, the last line follows from the inequality $\log s \le s-1,\,s>0$. Next,  we apply the change of variables
	\[
	t:=(c_0(n-2-\ell))^{-1} r_{n,k}(x)^{1-d} s \quad \Big(\text{i.e.}, s=c_0(n-2-\ell) t r_{n,k}(x)^{d-1}\Big),
	\]
	which shows that the last upper bound takes the form
	\begin{equation}\label{glaftersc}
		f_+d \kappa_d {c_0^{-d}} r_{n,k}(x)^{d(1-d)} \sum_{\ell=0}^{k-2} \binom{n-2}{\ell} (n-2-\ell)^{-d-\ell} \int_0^{c_0 (n-2-\ell)r_{n,k}(x)^d} s^{\ell+d-1} e^{-s} \,\mathrm{d}s.
	\end{equation}
	We now use the bounds $f_- \kappa_d r_{n,k}(x)^d \le v_{n,k}$, $\binom{n-2}{\ell}\le n^{\ell}/\ell!$, and the fact that the integral figuring in
	\eqref{glaftersc} converges as $n \to \infty$. Hence, the expression in \eqref{glaftersc} is bounded from above by  $c_1 n^{-1} (\log n)^{1-d}$, where $c_1$ is some
	positive constant. Consequently,  \eqref{lemloccl} is bounded from above by
	\begin{align}
		&c_1 n^{-1}(\log n)^{1-d} \lambda(S(\mathbf{j},2k)) \sup_{y \in S(\mathbf{j},2k)} \PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y)))\le k-1)\nonumber\\
		&\quad  \sim c_2 n^{-3} (\log n)^{2-d+\varepsilon} \label{lemlocclbou}
	\end{align}
 for some $c_2>0$.

	By analogy with the reasoning above, \eqref{lemlocfa} is given by the integral
	\begin{align}
		&2\int_{S(\mathbf{j},2k)}\int_{S(\mathbf{j},2k) \setminus B(x,r_{n,k}(x))} \mathds{1}\{r_{n,k}(y) \le r_{n,k}(x)\} \,\PP\big(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\} \cup \{x\})(B(y,r_{n,k}(y))) \le k-1\big)\nonumber\\
		&\quad \times  \PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(y)))\le k-1)\,  \mu(\mathrm{d}y)\,\mu(\mathrm{d}x). \label{lemlocfaint}
	\end{align}
	If $y \notin B(x,r_{n,k}(x))$ and $r_{n,k}(x)\ge r_{n,k}(y)$, we have the lower bound
	\[
	\lambda(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(y))) \ge \frac{\lambda(B(x,r_{n,k}(x)))}{2}.
	\]
	Since {$f_+ \kappa_d r_{n,k}(x)^d \geq v_{n,k}$}, we find a constant $c_3>0$ such that
	\[
	\lambda(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(y))) \ge c_3 v_{n,k},
	\]
	whence
	\begin{align*}
		&\PP\big(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(x,r_{n,k}(x))\setminus B(y,r_{n,k}(y)))\le k-1\big)\\
		&\quad \le \sum_{\ell=0}^{k-1} \binom{n-2}{\ell} \big(c_3v_{n,k}\big)^\ell \big(1-c_3v_{n,k}\big)^{n-2-\ell}\\
		&\quad \sim \frac{c_3^{k-1}}{(k-1)!} \big(\log n\big)^{k-1} \exp \big(n \log (1-c_3v_{n,k})\big)
	\end{align*}
	as $n \to \infty$. Since $\log s \le s-1$ for $s>0$, \eqref{lemlocfaint} is bounded from above by
	\begin{align}
		&c_4 n^{-c_3} \lambda(S(\mathbf{j},2k))^2 \sup_{y \in S(\mathbf{j},2k)} \PP(\#(\mathcal{X}_n \setminus \{X_i,X_{i'}\})(B(y,r_{n,k}(y)))\le k-1)\nonumber\\
		&\quad \sim c_5 (4k+1)^{2d} \frac{(\log n)^{2+2\varepsilon}}{n^{3+c_3}}, \label{lemlocfabou}
	\end{align}
where $c_4$ and $c_5$ are positive constants.  {Summing over all $i\neq i'\leq n$, it follows} from  \eqref{lemlocclbou} and  \eqref{lemlocfabou}
	that $R(n)=O(n^{-1}(\log n)^{2-d+\varepsilon})$ as $n \to \infty$, {which} finishes the proof of Lemma~\ref{Le:localproperty}.
\end{proof}

\paragraph{ A Poisson approximation result based on the Chen-Stein method}
In this paragraph, we recall a Poisson approximation result due to Arratia \textit{et al.} \cite{AGG}, which is based on the Chen-Stein method.
To this end, we consider a finite or countable collection $(Y_\alpha)_{\alpha\in I}$ of $\{0,1\}$-valued random variables and we let $p_\alpha = \PP(Y_\alpha=1)>0$, $p_{\alpha\beta}=\PP(Y_\alpha=1, Y_\beta=1)$. Moreover, suppose that
for each $\alpha\in I$, there is a set $B_\alpha\subset I$ that contains $\alpha$. The set $B_\alpha$ is regarded as a neighborhood
of $\alpha$ that consists of the set of indices $\beta$ such that $Y_\alpha$ and $Y_\beta$ are \textit{not} independent. Finally, put
\begin{equation}
\label{eq:defb}
b_1=\sum_{\alpha\in I} \sum_{\beta \in B_\alpha}p_\alpha p_\beta, \quad b_2=\sum_{\alpha\in I}\sum_{\alpha\neq \beta \in B_\alpha}p_{\alpha\beta},
\quad b_3=\sum_{\alpha\in I}\E \big[\left| \E[ Y_\alpha-p_\alpha|\sigma(Y_\beta: \beta\not\in B_\alpha)]  \right| \big] .
\end{equation}

\begin{theorem} {\rm (}Theorem 1 of \cite{AGG} {\rm )}
\label{Th:AGG}
Let $W=\sum_{\alpha \in I}Y_\alpha$, and assume $\lambda: =\E(W) \in (0,\infty)$. Then
\[
d_{TV}(W, {{\rm{Po}}}(\lambda)) \leq 2(b_1+b_2+b_3).
\]
\end{theorem}

\paragraph{Proof of Theorem \ref{Th:poissonapprox}}
Recall $v_{n,k}$ from \eqref{defynk} and $C_{n,k}$ from \eqref{defcnk}.  Put
\[
\widehat{C}_{n,k} := \sum_{\mathbf{j}\in \mathcal{V}_n} \:\mbox{\bf 1} \big\{ M_\mathbf{j}>v_{n,k} \big\}.
\]

The following lemma claims that the number {$C_{n,k}$} of exceedances  is close to the number of subcubes for which there exists at least one exceedance, i.e. $\widehat{C}_{n,k}$, and that $\widehat{C}_{n,k}$ can be approximated by a Poisson random variable.

\begin{lemma} \label{Le:mainlemma} We have
\begin{enumerate}
\item[{{\rm a)}}] $\PP (C_{n,k} \neq \widehat{C}_{n,k} ) =  O\left(\left(\log n\right)^{1-d}   \right)$,
\item[{{\rm b)}}] $d_{TV}(\widehat{C}_{n,k}, {{\rm{Po}}}(\E[\widehat{C}_{n,k}]))\ =  O\left(\left(\log n\right)^{1-d}   \right)$,
\item[{{\rm c)}}]  $\E [ \widehat{C}_{n,k}] = {\rm e}^{-t} +   O\left( \frac{\log\log n}{\log n}  \right)$.
\end{enumerate}
\end{lemma}

\begin{proof}
Assertion a) is a direct consequence of Lemma~\ref{Le:localproperty}  and of the inequalities
\begin{align*}
\PP(C_{n,k}\neq \widehat{C}_{n,k})  & = \PP \big(\exists \mathbf{j}\in\mathcal{V}_n, \exists i,\ell \text{ s.t. } X_i,X_\ell \in \mathbf{j};\; \mu(B(X_i,R_{i,n,k})) \wedge \mu(B(X_{\ell},R_{\ell,n,k}))>v_{n,k} \big)\\
& \leq \sum_{\mathbf{j}\in \mathcal{V}_n}\sum_{i\neq \ell \leq n}\PPP{X_i,X_{\ell}\in \mathbf{j};\; \mu(B(X_i,R_{i,n,k})) \wedge \mu(B(X_{\ell},R_{\ell,n,k}))>v_{n,k} }\\
& \leq \frac{n}{(\log n)^{1+\varepsilon}}\times R(n).
\end{align*}

\bigskip

To prove b), we apply Theorem~\ref{Th:AGG} to the collection $(Y_\alpha)_{\alpha\in I} = (M_{\mathbf{j}})_{\mathbf{j}\in \mathcal{V}_n}$. Recall that, conditionally on the event $\mathcal{E}_n$,
the random variables $M_{\mathbf{j}}$ and $M_{\mathbf{j}'}$ are independent provided that d$(\mathbf{j},\mathbf{j}')\geq 2k+1$. With a slight abuse of notation, we omit to condition on $\mathcal{E}_n$ since this event
occurs with probability tending to 1 as $n \to \infty$ (Lemma \ref{Le:independence}) at a rate which is at least polynomial. The first two terms in \eqref{eq:defb} are
\[
b_1=\sum_{\mathbf{j}\in \mathcal{V}_n} \sum_{\mathbf{j}' \in S(\mathbf{j},2k)}p_\mathbf{j} p_{\mathbf{j}'}, \quad b_2=\sum_{\mathbf{j}\in \mathcal{V}_n}\sum_{\mathbf{j}\neq \mathbf{j}' \in S(\mathbf{j},2k)}p_{\mathbf{j}\jj'}, \]
where
\[
p_{\mathbf{j}}= \PP(M_\mathbf{j} > v_{n,k}), \quad p_{\mathbf{j}\jj'}=\PP(M_\mathbf{j}>v_{n,k}, M_{\mathbf{j}'}>v_{n,k}).
\]
The term $b_3$ figuring in \eqref{eq:defb} equals 0 since, conditionally on $\mathcal{E}_n$, the random variable $M_\mathbf{j}$ is independent of the $\sigma$-field
$\sigma(M_{\mathbf{j}'}: \mathbf{j}'\not\in S(\mathbf{j},2k))$.  Thus, according to Theorem~\ref{Th:AGG}, we have
\[d_{TV}(\widehat{C}_{n,k}, \text{Po}({\E[\widehat{C}_{n,k}]}))\leq 2(b_1+b_2).\]
{Fist, we deal with $b_1$}. As for the first assertion, notice that for each $\mathbf{j}\in \mathcal{V}_n$, using symmetry, we obtain
\begin{align*}
p_\mathbf{j} & = \PP \bigg(\bigcup_{i\leq n}\left\{X_i\in \mathbf{j}, \mu(B(X_i,R_{i,n,k})) >v_{n,k}    \right\}\bigg) \\
& \leq n \cdot \PP \big( X_1\in \mathbf{j}, \mu(B(X_1,R_{1,n,k})) >v_{n,k} \big)\\
& = n \cdot \int_{\mathbf{j}} \PP(\mu(B(x,R_{1,n,k})) >v_{n,k}|X_1=x) f(x) \, \text{d} x\\
& \le n f^+ \lambda(\mathbf{j}) \int_{\mathbf{j}} \PP(\mu(B(x,R_{1,n,k})) >v_{n,k}|X_1=x)  \, \frac{1}{\lambda(\mathbf{j})} \, \text{d} x\\
& = n f^+ \lambda(\mathbf{j}) \PP(\mu(B(\widetilde{X}_1,R_{1,n,k})) >v_{n,k}),
\end{align*}
where $\widetilde{X}_1$ is independent of $X_2,\ldots,X_n$ and has a uniform distribution   over $\mathbf{j}$.
Invoking Remark~\ref{rem-byproduct}, the probability figuring in the last line is asymptotically equal to ${\rm e}^{-t}/n$ as $n \to \infty$.
Since  $\lambda(\mathbf{j})=O\left( (\log n)^{1+\varepsilon}/n\right)$, we thus have
\[
p_\mathbf{j} \le C \cdot \frac{(\log n)^{1+ \varepsilon}}{n},
\]
where $C$ is a constant that does not depend on $\mathbf{j}$.
 Since $\#\mathcal{V}_n\leq \frac{n}{ (\log n)^{1+\varepsilon}}$ and $\#S(\mathbf{j}, 2k)\leq (4k+1)^d$, summing over $\mathbf{j},\mathbf{j}'$ gives
\[
b_1  \leq  C^2 \sum_{\mathbf{j}\in \mathcal{V}_n}\sum_{\mathbf{j}'\in S(\mathbf{j},2k)}   \left( \frac{(\log n)^{1+\varepsilon}}{n}  \right)^2
 = O\left(  \frac{(\log n)^{1+\varepsilon}}{n} \right).
\]

\medskip

To {deal with $b_2$}, notice that, for each $\mathbf{j},\mathbf{j}'\in \mathcal{V}_n$ and $\mathbf{j}'\in S(\mathbf{j},2k)$, we have
\begin{align*}
p_{\mathbf{j}\jj'} & = \PP \Big( \bigcup_{i\neq i'\leq n}  \left\{  X_i\in \mathbf{j}, X_{i'}\in S(\mathbf{j},2k),  \mu(B(X_i,R_{i,n,k}))\wedge \mu(B(X_{i'},R_{i',n,k})) >v_{n,k}  \right\}   \Big)\\
& \leq \PP \Big( \bigcup_{i\neq i'\leq n}  \left\{  X_i, X_{i'}\in S(\mathbf{j},2k); \mu(B(X_i,R_{i,n,k})) \wedge  \mu(B(X_{i'},R_{i',n,k})) >v_{n,k}  \right\}    \Big).
\end{align*}
Using subadditivity, and taking the supremum, we obtain
\begin{align*}
b_2 & \leq \sum_{\mathbf{j}\in \mathcal{V}_n}\sum_{\mathbf{j}'\in S(\mathbf{j},2k)} \sup_{\mathbf{j}\in\mathcal{V}_n}\sum_{i\neq i'\leq n}\PP \big(X_i,X_{i'}\in S(\mathbf{j},2k), \mu(B(X_i,R_{i,n,k})) \wedge\mu(B(X_{i'},R_{i',n,k}))>v_{n,k}\big) \\
& \leq \frac{n}{(\log n)^{1+\varepsilon}}\times (4k+1)^d\times R(n).
\end{align*}
According to  Lemma \ref{Le:localproperty}, the last term equals  $O\left(\left(\log n\right)^{1-d}   \right)$, {which concludes the proof of b).}

\bigskip

To prove c), observe that
\[
\big{|} \E [\widehat{C}_{n,k}] - {\rm e}^{-t} \big{|} \le \big{|}\E[\widehat{C}_{n,k}] -  \E[C_{n,k}]\big{|} + \left|\E[C_{n,k}] - {\rm e}^{-t}\right|.
\]
By Theorem~\ref{thmmetricspace}, the last summand is $O\left( \frac{\log\log n}{\log n}  \right)$. Since $C_{n,k} \ge \widehat{C}_{n,k}$, we further have
\begin{align*}
\big{|}\E [\widehat{C}_{n,k}] - \E[C_{n,k}]\big{|} & = \E [C_{n,k}-\widehat{C}_{n,k}]\\
& = \E \bigg( \sum_{i\leq n} \:\mbox{\bf 1} \{\mu(B(X_i, R_{i,n,k}))>v_{n,k}\} - \sum_{\mathbf{j}\in\mathcal{V}_n} \:\mbox{\bf 1} \{M_\mathbf{j}>v_{n,k}\}  \bigg)\\
& = \sum_{\mathbf{j}\in\mathcal{V}_n} \E \bigg[ \Big( \sum_{i\leq n} \:\mbox{\bf 1}\{X_i\in \mathbf{j}\} \:\mbox{\bf 1} \{\mu(B(X_i,R_{i,n,k})) > v_{n,k} \}  - 1 \Big)\:\mbox{\bf 1} \{M_\mathbf{j}>v_{n,k}\} \bigg]\\
& \leq \sum_{\mathbf{j}\in \mathcal{V}_n}\sum_{i\neq i'\leq n} \mathbb{P} \big(X_i,X_{i'}\in \mathbf{j}, \mu(B(X_i,R_{i,n,k})), \mu(B(X_{i'},R_{i',n,k}))> v_{n,k} \big)\\
& \leq \#\mathcal{V}_n \times R(n).
\end{align*}
According to  Lemma \ref{Le:localproperty}, the last term equals  $O\left(\left(\log n\right)^{1-d}   \right)$. This concludes the proof of Lemma \ref{Le:mainlemma} and thus of Theorem~\ref{Th:poissonapprox}.

\end{proof}

\section{Concluding remarks}
When dealing with limit laws for large $k$th-nearest neighbor {\em distances} of a sequence of i.i.d. random points in $\mathbb{R}^d$ with density $f$, which take values in a bounded region $K$,
the modification of the $k$th-nearest neighbor distances made in \eqref{gumbelknnbound} (by introducing the ``boundary distances'' $\|X_i - \partial K\|$)
and  the condition that $f$ is bounded away from zero, which have been adopted in \cite{HE82} and \cite{HE83}, seem to be crucial, since boundary effects play
a decisive role (\cite{DH89, DH90}). Regarding $k$th-nearest neighbor balls with {\em large probability volume}, there is no need to introduce $\|X_i - \partial K\|$.
It is an open problem, however, whether Theorem~\ref{Th:poissonapprox} continues to hold for densities that are not bounded away from zero.

A second open problem refers to Theorem~\ref{thmmetricspace}, which states convergence of expectations of $C_{n,k}$ in a setting beyond the finite-dimensional case.
Since $C_{n,k}$ is non-negative, the sequence $(C_{n,k})_{k}$ is tight by Markov's inequality. Can one find conditions on the underlying distribution that ensure
convergence in distribution to some random element of the metric space?

redpajama_set_name: RedPajamaArXiv


Description Side view of the life-size trough shaped to accommodate a person receiving medicinal treatment. Adjacent to it are cells for patients, also on the other three sides of this quadrangular enclosure, in the center of which was a shrine room. The trough had some simple decoration with a motif of the capital pillar.


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text:
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