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FATHER FRANK'S LOG...
FATHER FRANK'S LOG: 22nd – 29th OCTOBER
This week I have been preparing some materials for the month of November, our special time of remembering our loved ones who have died, and it prompted a particular memory. For 10 years I was the secretary to the North European Conference of the Passionists. We have since done some restructuring, but the Passionist Provinces in the Conference at that time were Ireland/Scotland; England/Wales; Netherlands; Belgium; Germany; France and Poland. There was an annual meeting of the leaders of the provinces that usually lasted 3 or 4 days. The meetings moved around the various locations and, especially with the language differences, the meetings were quite intense. Because of that, by way of a break, half a day was always set aside for some cultural experience arranged by the host province.
Some of the experiences that I especially remember were, for example, when we held the meeting in Munich and had an afternoon at the Oktoberfest (Munich Beer Festival) This was around the time they introduced the "Quiet Oktoberfest" which, up until 6pm at least, maintained the beer tents family and elderly friendly, and now Passionist friendly, with the orchestras in the tents playing only quiet brass music and traditional folk tunes, so we weren't going too wild. From another meeting held in Bordeaux we visited a local vineyard. After the tour we were seated around a garden and given two small glasses of wine, one of them a very expensive wine, and the other a very modestly priced wine. We weren't told which was which and we were asked to discuss which of them we preferred. To a man we preferred the cheap wine, so perhaps our taste buds were more suited to German beer.
Lest you think all our cultural experiences were alcohol related, in Belgium we went to Leuven to visit the shrine of the leper priest, Father Damien, who was voted the greatest ever Belgian in a poll conducted by the Flemish Public Broadcasting Service for his work as the resident priest in a leper colony on Molokai Island, eventually submitting to the same, then incurable disease. At a meeting in the Netherlands we visited the Shrine of Father Charles of Mount Argus at Munstergeleen, the place of his birth, and then on to Maastricht, where the Treaty on European Union was signed in 1992. At a meeting in Minsteracres, a Passionist Retreat Centre in Northumberland, we visited Hadrian's Wall, built by Roman Britain to keep the Scots out. I was happy to be living proof that it didn't work.
But the foremost memory that was prompted this week was when we had a meeting in Warsaw which took place at the end of October, beginning of November. On All Saints' Day the cemeteries and graveyards in Poland are decorated with candles, flowers and wreaths throughout the day, and the candles are left to burn through the night. And so, for our cultural experience, the Polish Passionists provided us with votive lamps and brought us to Powazki cemetery, the primary cemetery in Warsaw, where we placed the lamps on the graves of deceased Passionists. It was dark by this time and the effect of all these lamps burning throughout this huge cemetery was incredibly solemn and beautiful. Afterwards, in the church of St. Charles Borromeo, within the cemetery, we attended a concert of sombre music to fit the occasion. It was a cultural experience with a difference, but one I will never forget.
In the old Catholic ritual this commendation was said over a dying person by the priest:
"Go forth, Christian soul, from this world in the name of God the almighty Father, who created you, in the name of Jesus Christ, Son of the Living God who suffered for you, in the name of the Holy Spirit, who was poured out upon you. Go forth, faithful Christian. May you live in peace this day, may your home be with God, with Mary, the Virgin Mother of God, with Saint Joseph, and all the angels and saints. Amen"
May our deceased loved ones rest in peace.
FATHER FRANK'S LOG: 15th – 22nd OCTOBER
In the middle of November my sports reporter brother will help us to raise funds by hosting a Super Scoreboard night here in St. Mungo's. At the moment we are getting tickets printed in the hope that we can fill the hall and have a very enjoyable night. I am hopeless at that side of things and find it quite stressful, so I am always very grateful for those volunteers who step forward and offer their help.
My most stressful experience of trying to sell tickets was 10 years ago in Dublin when we were preparing for the canonization of Father Charles of Mount Argus. I had the notion of commissioning a new hymn for the occasion and I decided to approach Liam Lawton, a priest based in Carlow who was the most prominent writer of liturgical music in Ireland at the time, and probably still is. I liked his music very much and we sang a fair amount of it in our Mount Argus Folk Group, so who better to ask!
I contacted Liam and we arranged to meet for a coffee in the Red Cow Hotel in Dublin which was easy for him to get to from Carlow, and for me from Mount Argus. As an aside, the Red Cow was also known as the Mad Cow because, at the time, it was approached via a roundabout where the traffic was invariably chaotic and an absolute nightmare to try and get on and get off in one piece. However, we both negotiated it safely and sat down at table.
He wanted to know all about Father Charles, about his ministry of healing, hope and reconciliation, and also, about what our plans were leading up to the celebrations. At the end of the conversation Liam had agreed to write the song and also to put on a concert of Sacred Music in the church before the canonization, during which he would introduce the song for the first time. And so it was, in the midst of all the other hectic preparations for the celebrations, both in Rome and in Dublin, I had to try and promote this concert and sell 900 tickets to fill the church. Thankfully, a great team of volunteers helped me to do just that.
The hymn, "Come to Me", was beautiful. He finished it just in time for the concert and sang it at the end, accompanying himself on piano. I remember feeling very moved by it. We had a limited number of CD's made where the hymn was coupled with another of his compositions, "Sing of a Lady", with the refrain – Ave Maria, hope of all our days – which we chose because of Father Charles's great devotion to Mary as the Mother of Hope. A few years later Liam put the hymn to Saint Charles on his beautiful album, "Healing Song".
I recently had another reminder of that fateful year when I was at a celebration with a couple whom I have been friendly with for almost 50 years, ever since I met them at our Passionist Retreat House at Coodham in the late 1960's. Their first grandchild was born in 2007 and, without knowing anything about the canonization, they had decided to come over for a family trip to Dublin and have me baptize their new grandchild in Mount Argus. Until they arrived I hadn't known that the child was to be called Charlie, and of course Father Charles's nickname in the community was "Poor Old Charlie", so when they saw all these banners in the church proclaiming a new saint for Ireland whose name was Charlie they were delighted and we had a great laugh about it. I hadn't seen much of Charlie since but when I met him last week, now a fine strapping lad at 10 years of age, it brought it all back to me.
Here is the refrain of that beautiful hymn to Saint Charles:
Come to me with all your burdens, come and place your heart in mine,
Come and tell me all your worries, hope in you will never die
For I am full of Compassion, do not be afraid
Come to me with all your burdens, I will never walk away.
FATHER FRANK'S LOG: 8th – 15th OCTOBER
I had an unusual call last week from a man who wanted to come and take some photos outside of the church. Once I had satisfied myself of his good intentions I discovered that this man's passion was journeying to film locations and trying to match stills from the film with how the location looked now.
He wanted to come to St. Mungo's because the church had been used many years back in a film called Heavenly Pursuits. When I relayed this to Father Gareth he got very excited, because he loves his films, and he immediately went on to the internet to find this film. It was a light-hearted comedy released in 1986, starring Helen Mirren; Tom Conti and David Hayman, and it tells the story of a teacher at a Catholic school whose students are searching for two miracles that would help promote the late (and fictional) Edith Semple to sainthood.
What surprised me was that, if the film was released in 1986, then I must have been stationed in St. Mungo's at the time, as I was part of the community here from 1983-1986, but I have no recollection of it at all. Of course, I was away a lot then, doing vocations work, and giving missions and retreats, so I can only assume that the filming was done during one of those pastoral journeys. But it pains me to think that, having become a big fan of Helen Mirren later on in life, especially when she was in Prime Suspect, that she was here at St. Mungo's and I never got the chance to meet her. I'm sure if the late Fr. Anthony Behan had been here he would have remembered it well. He was also a member of the community at the time, and a great film buff, and no doubt he made a deep impression on Helen Mirren with his charms.
I have, however, been in other Passionist Retreats when some kind of filming was taking place, and I got the chance to meet a few famous film stars. When I was doing my diaconate year in Rome from 1982-83, I met Gregory Peck and Christopher Plummer when they filmed some scenes from The Scarlet and the Black in the monastery garden of Saints John and Paul where I was living at the time. This was a TV movie which told the story of Monsignor Hugh O'Flaherty, a real life Irish-born Catholic priest who, while in Rome, saved thousands of Jews, and helped escaped Allied Prisoners of War, during World War II.
Mount Argus in Dublin has been used a number of times for films and TV dramas. I was there for meetings one time in the late 1990's when they were filming a TV drama called The Ambassador, and I met the lead star, Pauline Collins, whom I had loved when she was in the film Shirley Valentine a few years before. She was playing the new British ambassador to the Republic of Ireland and some of the big meeting rooms in Mount Argus were being used for scenes supposedly taking place at the British Embassy in Ballsbridge.
More recently there was an Irish TV soap opera called Red Rock, set in the fictional seaside town of Red Rock in County Dublin. The locations manager came along and asked if he could use the entrance in and out of the new monastery at Mount Argus as the funeral parlour in the series. I might have been offended by that, but I had just been on holiday to Westport, and it turned out the locations manager's father was the head of the St. Vincent de Paul Society in Westport, so I felt I could trust him, and I gave the necessary permission. Unfortunately, it didn't quite work out, as we weren't able to shut down the monastery for the length of time they needed for filming, but they did do some other filming in and around the grounds for various scenes – there were no big stars for me to meet though.
Here is a quote from the above-mentioned Pauline Collins about goodness and evil in films: "I think goodness is very powerful, but evil is often made more attractive in films. It's a challenge to make goodness appealing. I was brought up a Catholic, so I have to believe in the goodness of human beings. I think we're not so bad after all"
FATHER FRANK'S LOG: 1st – 8th OCTOBER
Father Gareth and myself have just passed the year mark since we both came to St. Mungo's last October. Back then Father Gareth arrived on the 4th and held the fort until I arrived on the 6th; and he had to do the same again this year as I had to travel to Belfast for a series of meetings around finances and formation. When these meetings were first arranged I wasn't under any particular pressure coming back, but I had agreed to do a wedding in St. Andrew's cathedral in November for a Ugandan couple which, because of visa problems, was brought forward to October 6th. When I went to organise travel the only way I could be at the meetings and be back for the wedding, which was at the unusual time of 11 a.m., was to bring the car and return on the 23.30 Stena Line sailing from Belfast to Cairnryan. That meant I would be getting back to the house at around 4 a.m. and arriving to the wedding with very little sleep. Added to that, it was a stormy week, and the bride and groom, a truly lovely couple, were concerned that the sailing would be cancelled and that they'd be left stranded at the altar without a priest, although I knew it would take a lot for one of those big boats not to be able to sail, and that they had stabilisers to manage on the rough seas.
As it turned out, the meetings finished a little earlier than planned and I was able to transfer to an earlier sailing at 19.30, which would mean getting home around midnight, a much better option. It was better still, that once the boat set sail, I discovered they were showing both the Northern Ireland v Germany, and the Scotland v Slovakia World Cup qualifying matches in different parts of the ship. I got myself a good seat in the Scottish section and passed the journey watching a really exciting match. When Scotland's winning goal went in with just minutes to go we were all out of our seats with hugs and high-fives, perfect strangers becoming each other's best friends – even if just for a while. There was a bit of a panic when, with 3 minutes of injury time still to play, the ship reached port and the television was switched off. Cries of dismay persuaded the crew to put the television on again so that we could be sure we had held out to the end, which of course we did, and the drive back to Bishopbriggs was pleasantly spent listening to post-match analysis on the radio.
The wedding itself was a delight. There were only about 15 people there, mostly from Uganda, but also from Ghana and Tanzania. In 35 years of celebrating weddings it was the first time that I had kept the bride waiting instead of the other way around, as she had arrived 15 minutes early while we were still setting up the altar in the cathedral. They were people of real faith and their small group of friends entered fully into the joy of the occasion. There was no big reception, but I had agreed a while back to have lunch with them after the Nuptial Mass, and I had to smile to myself when I discovered that we were going to the World Buffet in Renfield Street, because this is Father Gareth's favourite restaurant, although I hadn't realised he had his very own waitress as well, because no sooner had I entered the restaurant than I was very warmly greeted by one of the Passionist Young Team (PYT), the group that Father Gareth gathers on a Wednesday night. Later on, another young man who had been to the PYT came over to greet me as well.
I don't think the World Buffet had ever held a wedding reception before and they were obviously thrilled to see the bride and groom arrive and, even though the bride had made a quick change out of her wedding dress, she still looked quite stunning, while the groom and guests were all still in their wedding outfits, which for some meant traditional costumes. The World Buffet offers food from all corners of the world but, when I got up to get my main course, the first PYT member pointed to the Macaroni and Cheese and told me that Father Gareth usually just fills his plate with that and sits quite happily before heading off again.
The kingdom of heaven is like a wedding feast… (Matthew 22: 1-14)


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Q: how to validate an ognl expression in struts 2 I want to save an ognl expression in my db and use it later to validate something.
There is a way to check if that expression is formally correct?
Exist a library or a regex for that?

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MGMT antibody (pAb) tested by Western blot.
Nuclear extract of MCF-7 cells (20 µg) probed with MGMT antibody (pAb) at a dilution of 1:500.
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és un exfutbolista marroquí de la dècada de 1980.

Fou internacional amb la selecció del Marroc amb la qual participà en la Copa del Món de futbol de 1986.

Pel que fa a clubs, destacà a FAR Rabat.

Referències

Futbolistes internacionals amb el Marroc
Medallistes marroquins als Jocs Mediterranis

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text:
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Zabeel Park Dubai cost 200 million dirhams is located in Zabeel area near the World Trade Center in the heart of Dubai, where mediates commercial areas such as Sheikh Zayed and the dignity Street, and a total area of 51 hectares, and Travel greenery and shadows giving it beauty and splendor, and is divided into three sections It is: "a" and "b" and "c", and relate to these sections together through two bridges commentators pedestrian internal and train, and is accessed through seven main gates spread over areas, and allocated by the parking nine 2300 car, and by 10 cycles of water .
The main theater in the heart of the park is dedicated to receive the events that are displayed in the outdoors, and has a runway to accommodate 2000 spectators.
Zabeel Park gaming area are located near the gate "1", a set of fixed games that the child which recognizes a range of movements and skills such as skiing swing and transit through channels, mass and play with sand and turning, a dedicated area for children as young as ten years.
There is in this section three kiosks for the sale of refreshments offered to the visitor beside them cool and fast food, which is a circular buildings and has an outdoor wall of stained glass.
Zabeel Park also has presence gallery is closed lounge area of 200 square, which is connected to a large warehouse area of 50 square meters, which is a building surrounded by circular fountain near Gate No. 3 and contains walls booth on the oval shape parallel to the shape of the fountain, while the ceiling Fmguety aluminized, a gallery dedicated to receive all kinds of arts and crafts, cultural and scientific exhibitions specialized in the dissemination of knowledge.
In return for this exhibition there is a modern cafe-like building where the exhibition offers hot cakes and sweets drinks also contain online service.
The sports club which is the health of investment projects to the private sector, it offers a range of health and fitness services for both sexes of calisthenics and relaxation programs, and the club Supplier portal through which visitors can access it without trouble as it is located near the gate "2".
In Zabeel Park Star Gate, and includes five domes half of the upper apparent and half lower under the ground, and are linked through the pedestrian tunnels conditioned, located near Gate No. 0.4 and all the Dome of them dedicated to provide recreational and scientific different activities, in the Dome of the Saturn suffered movies Visitors enjoy group games and the Dome of the snowy land earmarked for a fast trains recreational and called the death train. And the Dome of Mars allocated for children to play kinetic, and the Dome of the Flying Saucer dedicated to race electric cars next to the maze of the solar system, which is part of the theme park for advanced technology which is concerned with the discovery of space and private travel through the solar system.
There is also Jeff region, and places to grill, and lake for boats, restaurants, and the climb and Cricket Stadium and anthropomorphic Istanbul Games, and the Gold Coast.
There are garden racetrack outside for a walk with a length of 2.5 kilometers, where established a scientific way and toilets, and advertising space on the outer wall.
The water elements, it is a fountain called Zakorat located at the entrance number "1" Square, a water channel in the form of Zakorat, and Fountain Phippanasci It is located along part of the main corridor between the entrance number "2" and open the theater, which is about two channels and fountain oval , then the fountains next to the conical shape of oblique fountain located in the barcode garden within a small area of seats to sit and watery element Haittiya.
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Dubai Season Copyright © 2021.

redpajama_set_name: RedPajamaCommonCrawl


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Home » Music » Dionne Warwick Reveals Why She's Still Not a Fan of Her Hit 'Do You Know the Way to San Jose'
Dionne Warwick Reveals Why She's Still Not a Fan of Her Hit 'Do You Know the Way to San Jose'
05/08/2019 Dionne Warwick, Exclusive, music
It's been over five decades since Dionne Warwick recorded one of her biggest hits, "Do You Know the Way to San Jose" — and she still isn't crazy about the song.
The pop classic, which was written by Burt Bacharach, with lyrics by Hal David, became one of her biggest hits and sold over 3,500,000 copies.
"I did not want to sing or record 'San Jose,'" Warwick, 78 tells PEOPLE. "It was not one of my favorite songs that Hal David wrote. I just couldn't imagine Hal writing a lyric that had 'whoa, whoa, whoa, whoa, whoa….'"
"I said to him, 'How could you?'" she adds with a laugh. "I sang it because of the affinity he had for the song and cried all the way to the bank. What can I tell you?"
https://www.youtube.com/embed/jqWt49o7R-k
Warwick's collaborations with Bacharach and David on such beloved records as "Walk on By," "I Say a Little Prayer," and "I'll Never Fall in Love Again," became classics and made Warwick a star. One of the world's most recognizable voices, she's still identified by them — no matter who else has sung them.
"They're definitely my songs," she says. "Even with the multitude of others singing those songs, they still know whose song it really is."
In this week's PEOPLE, Warwick talks about her songs, old and new, and her upcoming album, She's Back, produced by her son Damon Elliott, which comes out on May 17.
Despite the album's name, the vocal legend really never left. "I don't know what it means to slow down," says Warwick, who has won five Grammys, as well as a sixth earlier this year for Lifetime Achievement. "I love what I do," she says. 'I don't consider it work. It's just an extension of me."
Warwick was just 21 years old when Bacharach heard her singing back up with her group, the Gospelaires, and asked her to record a demo of some of his songs. She was "shy," Bacharach recalls, and wearing sneakers. Still, he says, "she had the look of a star. There was something about her."
The world would soon agree. Warwick went on to sell over 100 million records.
Born Mary Dionne Warrick, she grew up surrounded by music in East Orange, New Jersey. After the record company misspelled her surname on her first album, the name stuck. "I was very miffed," she says. "I expressed that to my grandfather and said, 'How could they dare do this to me?' He said, 'Just look at it as your professional name.' I still didn't like it but I became a 'wick' instead of a 'rick.'"
And the rest was pop music history.
For more from Dionne Warwick, pick up the latest issue of PEOPLE, on newsstands Friday.
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November 8, 2018 in aud. 5-6 from 12:00 to 13:30 a lecture on the topic "Nanostructured flat optics" will take place. Speaker: Senior Researcher, Faculty of Physics and Astronomy, Uppsala University Vitaly Goryashko (Sweden).


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\section{Introduction}

Let $\{X_n\}$ be a sequence of discrete random variables taking integer values with mean $\mu_n$ and standard deviation $\sigma_n$. For example, $X_n$ could be the number of heads that occur in $n$ coinflips. Our goal is to understand to what degree such a sequence converges to a normal distribution. In the example of coinflips, the $X_n$ themselves do not converge in any meaningful way, since $\mu_n$ is increasing in $n$, so we normalize by considering convergence of $Y_n = \f{X_n - \mu_n}{\sigma_n}$ to the standard normal distribution $Z$ instead.

One notion of convergence of a sequence of random variables is convergence in distribution. This concerns pointwise convergence of cumulative distribution functions. We write that $\{Y_n\}$ converges in distribution to $Z$, or $Y_n \xrightarrow{d} Z$, if
\[ \abs{\P(Y_n \leq t) - \P(Z \leq t)} \to 0 \]
for each $t \in \mathbb{R}$. If this condition holds, the sequence $\{X_n\}$ satisfies a \textbf{central limit theorem}. The example of coinflips satisfies a central limit theorem. But what if we wanted to know, for instance, the probability of getting exactly half heads, that is, $\P(X_n = n/2)$ (for even $n$)? The central limit theorem only tells us the probability of having at most half heads, that is, $\P(X_n < \f{n}{2}) = \P(Y_n < 0) \to \P(Z < 0) = \f{1}{2}$. To satisfy this, we would need
\[ \abs{\P(X_n = k) - \f{1}{\sqrt{2\pi}\sigma_n} e^{-\p{\f{k - \mu_n}{\sigma_n}}^2/2}} \to 0, \]
but this is not meaningful since both terms go to $0$ trivially. To find an error bound that makes the convergence meaningful, observe that a normal distribution with standard deviation $\sigma$ has height $\Theta(\f{1}{\sigma})$. Therefore, we want
\[ \abs{\P(X_n = k) - \f{1}{\sqrt{2\pi}\sigma_n} e^{-\p{\f{k - \mu_n}{\sigma_n}}^2/2}} = o\p{\f{1}{\sigma_n}} \]
uniformly in $k$. If this condition holds, the sequence $\{X_n\}$ satisfies a \textbf{local limit theorem}. Equivalently, we require
\[ \abs{\sigma_n \P\p{\f{X_n-\mu_n}{\sigma_n} = k} - \f{1}{\sqrt{2\pi}} e^{-k^2/2}} = o(1). \]
Geometrically, this is equivalent since when the Gaussian is normalized, its height is scaled by $\sigma_n$. If there is a bound on this last rate of convergence, the local limit theorem is said to be \textit{quantitative}. Equivalently, there is a quantitative bound if the rate of convergence of the first difference is better than $o(1/\sigma_n)$.

\subsection{Descents} \label{descents-intro}

One variable of interest to us is the number of descents in a random permutation. A permutation $\pi$ has a descent at index $j$ if $\pi(j) > \pi(j+1)$. The number of descents in $\pi$, denoted $D(\pi)$, is the number of such indices $j$. We define the indicator random variable for the $j$th descent
\[ X_j = \begin{cases} 1 & \pi(j) > \pi(j+1) \\ 0 & \text{else} \end{cases}, \]
so the number of descents can be written
\[ D_n = \sum_{j=1}^{n-1} X_j. \]
To sample random permutations, uniformly and independently pick random integers $1 \leq a_j \leq n - j + 1$ for each $1 \leq j \leq n$. The $a_j$ correspond to permutations like so: let $S = \{1, \ldots, n\}$, and for each $j$, in order from $1$ to $n$, define $\pi(j)$ to be the $j$th remaining element of $S$, and remove $\pi(j)$ from $S$. Under this correspondence, $\pi(j) > \pi(j+1)$ if and only if $a_j > a_{j+1}$. Therefore, $X_j$ can be equivalently defined by
\[ X_j = \begin{cases} 1 & a_j > a_{j+1} \\ 0 & \text{else} \end{cases}. \]
	With this method of sampling, we can compute some basic facts about $D$. At each index $j$, there is either a descent or an ascent ($\pi(j) < \pi(j+1)$) and these both occur with equal probability, so $\mathbb{E}[X_j] = \f{1}{2}$. Therefore, $\mathbb{E}[D] = \f{n-1}{2}$. With similar reasoning as in computing the expectation of $X_j$, we get $\mathbb{E}[X_j X_{j+1}] = \f{1}{3!} = \f{1}{6}$. Also, it is important to note that $X_j$ is independent of all other $X_k$ except for $X_{j-1}$ and $X_{j+1}$. Now we compute
\begin{align*}
	\mathrm{Var}(D)
	&= \mathbb{E}[D^2] - \mathbb{E}[D]^2 \\
	&= \mathbb{E}\s{\sum_{i=1}^{n-1} \sum_{j=1}^{n-1} X_i X_j} - \p{\f{n-1}{2}}^2 \\
	&= \sum_{i=1}^{n-1} \mathbb{E}[X_i^2] + \sum_{\abs{i-j} = 1} \mathbb{E}[X_i X_j] + \sum_{\abs{i-j} > 1} \mathbb{E}[X_i X_j] - \f{(n-1)^2}{4} \\
	&= \f{n-1}{2} + \f{2(n-2)}{6} + \f{(n-1)^2 - (n-1) - 2(n-2)}{4} - \f{(n-1)^2}{4} \\
	&= \f{n+1}{12}.
\end{align*}

In Section 3, we establish a quantitative local limit theorem for $D_n$:
\begin{thm}
\label{thm:llt-dn}
\[ \abs{P\p{D_n = x} - \f{1}{\sqrt{2\pi}\sigma_n} e^{-\p{\f{x - \mu_n}{\sigma_n}}^2/2}} = O\p{n^{-1 + \varepsilon}} = O\p{\f{1}{\sigma_n}n^{-\f{1}{2}+\varepsilon}}. \]
\end{thm}

\subsection{3-term arithmetic progressions}

The other random variable of interest is the number of 3-term arithmetic progressions (3-APs) in a random subset of $\mathbb{Z}/n\mathbb{Z}$, with the presence of each element determined by a coinflip. We require $n$ prime to avoid issues of divisibility. For $1 \leq i \leq n$, let $x_i$ be the indicator random variable for $i \in S \subseteq \mathbb{Z}/n\mathbb{Z}$. The $x_i$ are independent and each has expectation $\f{1}{2}$. We define the number of 3-term arithmetic progressions in $S$ as
\[ A_n = \f{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n-1} x_i x_{i+j} x_{i+2j}. \]
This definition does not consider triples that contain the same element twice to be an arithmetic progression, and the factor of $\f{1}{2}$ causes multiple triples with the same elements to be considered the same progression. The expectation of $A_n$ is $\mathbb{E}[A_n] = \f{1}{8}\binom{n}{2}$. The variance of $A_n$ is computed in section 4.

In section 4, we prove a central limit theorem for $A_n$, stated as follows:
\begin{thm}
\label{thm:clt-an}
$$\abs{P\p{\frac{A_n - \mu_n}{\sigma_n} \leq x} - \P(Z \leq x)} = O\p{n^{-1/4}}.$$
\end{thm}

Based on experimental results and heuristics in section 4, we conjecture that there is no local limit theorem for $A_n$.
\begin{conj}
\label{conj:llt-an}
$$\abs{P\p{A_n = x} - \f{1}{\sqrt{2\pi}\sigma_n} e^{-\p{\f{x - \mu_n}{\sigma_n}}^2/2}} \neq o\p{\frac{1}{\sigma_n}}.$$
\end{conj}

\section{Background}

\subsection{Number of triangles in a random graph}
Previous papers have studied the distribution of the number of triangles $T_n$ in a random graph $G_{n,p}$, which is the undirected graph on $n$ vertices where each of the $\binom{n}{2}$ edges have probability $p$ of appearing in the graph. Let $R_n = \frac{T_n-\mu}{\sigma}$ be the normalized $T_n$ and $\mathcal{N}(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ be the standard normal probability density function. Gilmer and Kopparty \cite{GilmerKopparty14} establish the existence of an LLT for $T_n$, so $T_n$ is pointwise approximated by a discrete Gaussian distribution.
\begin{thm}{(Gilmer, Kopparty; 2014)}
Uniformly for all $k \in \mathbb{Z}$,
\[ \P(T_n = k) = \f{1}{\sqrt{2\pi}\sigma_n} \mathrm{exp}\p{-((k-\mu_n)/\sigma_n)^2/2} + o(1/\sigma_n). \]
\end{thm}

We note that this is a qualitative result, where the error is $o(n^{-2}) = o(\frac{1}{\sigma})$. Berkowitz \cite{Berkowitz17} expanded on their work and established a quantitative bound on the distance between the distribution of $T_n$ and the Gaussian distribution. In particular, he shows the following result:

\begin{thm}{(Berkowitz; 2017})
Uniformly for all $k \in \mathbb{Z}$,
\[ \P(T_n = k) = \f{1}{\sqrt{2\pi}\sigma_n} \mathrm{exp}\p{-((k-\mu_n)/\sigma_n)^2/2} + O(n^{-2.5+\epsilon}). \]
\end{thm}

\subsubsection{Methods used to establish an LLT for $T_n$}
Here we summarize the methods used by Gilmer and Kopparty \cite{GilmerKopparty14} in proving that an LLT exists for $T_n$.
With some calculations, we have the mean $\mathbb{E}[T_n] = \mu_n = p^3{\binom{n}{3}}$ and the variance $\sigma^2 = \Theta(n^4).$
A crucial formula for the proof is the following:
\begin{prop}[Fourier Inversion Formula]
If $Y$ is a random variable with support in the discrete lattice $L = \frac{1}{b}(\mathbb{Z} - a)$ for $a,b \in \mathbb{R}$, and $\phi(t) = \mathbb{E}[e^{itY}]$ is the characteristic function of $Y$, then for all $y \in L$,
\[ \P(Y = y) = \frac{1}{2\pi b} \int_{-\pi b}^{\pi b} e^{-ity}\phi(t)\ \d t.
\]
\end{prop}

If we let $\phi_n(t) = \mathbb{E}[e^{itR_n}]$, then
$\sigma_n \P(R_n = x) = \frac{1}{2\pi} \int_{-\pi\sigma}^{\pi\sigma} e^{-itx}\phi(t)\ \d t.$
By the standard Fourier inversion formula,
$\mathcal{N}(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-itx}e^{x^2/2}\ \d t.$ Thus,
\[ \abs{\sigma_n \P(R_n = x)- \mathcal{N}(x)} \leq \int_{-\pi\sigma_n}^{\pi\sigma_n} \abs{\phi_n(t) - e^{-t^2/2}} \d t + 2 \int_{\pi\sigma_n}^{\infty} e^{-t^2/2}\ \d t.
\]
Thus, since the second term goes to 0 as n goes to infinity, we just have to show \[\int_{-\pi\sigma_n}^{\pi\sigma_n} \abs{\phi_n(t) - e^{-t^2/2}} \d t = o(1).\]

For any constant $A$, we can write
	\[
		\int_{-\pi \sigma}^{\pi \sigma} \abs{\phi_n(t) - e^{-t^2/2}} \d t
		\leq \int_{-A}^{A} \abs{\phi_n(t) - e^{-t^2/2}} \d t
		+ \int_{A \leq \abs{t} \leq \pi \sigma} (\abs{\phi_n(t)} + |e^{-t^2/2}|) \d t.
\]

Since we have a CLT, the first integral on the right goes to 0. Hence, we have reduced the problem of proving an LLT to one of sufficiently bounding the characteristic function $\abs{\phi_n(t)}$. To get a quantitative LLT, we must also bound $\abs{\phi_n(t) - e^{-t^2/2}}$ for small values of $t$.
The overall strategy used to bound the $\abs{\phi_n(t)}$ involves finding an event that occurs with high probability and allows $R_n$ to be written as the sum of $n$ i.i.d. random variables $X_i$ after conditioning on the event. The full proof can be found in \cite{GilmerKopparty14}.

Using the $p$-biased Fourier basis, Berkowitz \cite{Berkowitz17} shows $\int_{-\pi\sigma_n}^{\pi\sigma_n} \abs{\phi_n(t) - e^{-t^2/2}} \d t = O(n^{-1/2+\epsilon}$). In section 4 below, we set up and apply this tool for 3-term arithmetic progressions in order to bound $\abs{\phi_n(t) - e^{-t^2/2}}$ for small values of $t$, although we conjecture that there is no bound for larger $t$.

\section{Descents in a permutation}

Let $\pi \in S_n$ be a permutation. Define $D(\pi) = \abs{\{1 \leq j < n \mid \pi(j + 1) < \pi(j)\}}$ to be the number of descents in $\pi$. Viewing $D_n$ as a random variable on uniformly distributed permutations, a central limit theorem is known \cite{Fulman97} and in this section, we prove Theorem \ref{thm:llt-dn}, a local limit theorem. In the proof, we will apply the following bound.

\begin{lem}{(Gilmer, Kopparty; 2014)} \label{gkbound}
Let B be a Bernoulli random variable that is 1 with probability $p$. Then
\[ \abs{e^{i\theta B}} \leq 1-8p(1-p) \cdot \left\lVert\frac{\theta}{2\pi}\right\rVert^2
\]
\noindent where $\norm{x}$ is the closest integer to $x$. As a result, for $\abs{\theta} < \pi$,
\[ \abs{e^{i\theta B}} \leq 1-8p(1-p) \cdot \p{\frac{\theta}{2\pi}}^2. \]
\end{lem}

As covered in \ref{descents-intro}, we write $D_n = \sum_{j=1}^{n-1} X_j$, where $X_j$ is the indicator random variable for $a_j > a_{j+1}$. Each $X_j$ depends only on $a_j$ and $a_{j+1}$, so $X_j$ is independent of all other $X_k$ except for $X_{j-1}$ and $X_{j+1}$. This lemma computes the dependence:
\begin{lem}
	Given the values of $X_{j-1}$ and $X_{j+1}$, the distribution of $X_j$ does not depend on $j$ or $n$.
    \proof
    The descents $X_{j-1}$ to $X_{j+1}$ depend only on $a_{j-1}$ to $a_{j+2}$. Every combination of values of $a_{j-1}$ to $a_{j+2}$ is equally likely, so it suffices to consider the case $n = 4$, $j = 2$.

    Although it is not necessary for the local limit theorem, the exact distribution of $X_j$ can be computed. In the $n = 4$ case, each pair of values of $X_1$ and $X_3$ appears in exactly $6$ permutations. The following probabilities are established by looking at all $6$ cases for each pair.
    \begin{align*}
		\P(X_j = 1 \mid X_{j-1} = 1, X_{j+1} = 1) &= 1/6, \\
		\P(X_j = 1 \mid X_{j-1} = 1, X_{j+1} = 0) &= 1/2, \\
		\P(X_j = 1 \mid X_{j-1} = 0, X_{j+1} = 1) &= 1/2, \\
		\P(X_j = 1 \mid X_{j-1} = 0, X_{j+1} = 0) &= 5/6.
	\end{align*}
    \qed
\end{lem}
\begin{thm}
The sequence of random variables $D_n$ satisfies a local limit theorem. Quantitatively,
\[ \P(D_n = x) = \frac{1}{\sqrt{2\pi} \sigma_n} e^{-((x-\mu_n)/\sigma_n)^2/2} + O(n^{-1+\varepsilon}). \]
\proof
	To bound the characteristic function of $D_n$, we condition on the values of $X_j$ for odd $j$. For simplicity, assume $n$ is even. (When $n$ is odd, one must also observe that $\P(X_j = 1 \mid X_{j-1} = 1) = 1/3$ and $\P(X_j = 1 \mid X_{j-1} = 0) = 2/3$ are constant.) The even $X_j$ are independent of each other, so conditioned on the odd $X_j$, $D_n$ is a sum of independent random variables. After conditioning, we compute

   	\begin{align*}
		\abs{\mathbb{E}[e^{itD_n/\sigma}]}
		&= \abs{\mathbb{E}[e^{it(C + \sum\limits_{\text{odd } j} X_j)/\sigma}]} \\
		&= \prod_{\text{odd } j} \abs{\mathbb{E}[e^{itX_j/\sigma}]} \\
        &\leq \prod_{\text{odd } j} \p{1 - 8p_j(1-p_j)\p{\frac{t}{2\pi\sigma}}^2}
	\end{align*}
    by Lemma \ref{gkbound}, where $p_j$ is the probability associated to $X_j$ by the given values of the odd $X_j$. But there are only finitely many possible values of $p_j$ (4 of them), so there is some value $p$ that maximizes the quantity above, and we can bound
	\begin{align*}
		\abs{\mathbb{E}[e^{itD_n/\sigma}]}
        &\leq \prod_{\text{odd } j} \p{1 - 8p(1-p)\p{\frac{t}{2\pi\sigma}}^2} \\
		&\leq \prod_{\text{odd } j} (1 - \Theta(t^2/n)) \\
		&= (1 - \Theta(t^2/n))^{n/2} \\
		&\leq e^{-\Theta(t^2)/2} \\
		&= e^{-\Theta(t^2)}.
	\end{align*}

	Now we obtain the final bound for the local limit theorem. As shown in $\cite{Fulman97}$, we have a central limit theorem
\[ \sup_{ -\infty < x < \infty} \abs{\P\p{\f{D_n - \mu}{\sigma} < t} - \P(Z < t)} \leq \sqrt{\f{12}{n}}. \]

For $y<0$, $\P(D_n-\mu \leq y\sigma) = \frac{1}{2} \P(\abs{D_n-\mu} \geq \abs{y}\sigma) \leq \frac{1}{2y^2}$ by Chebyshev's Inequality. Similarly, $\P(D_n-\mu \geq y\sigma) \leq \frac{1}{2y^2}$ for $y>0$.

Thus, \[\abs{\P\p{\f{D_n - \mu}{\sigma} < t} - \P(Z < t)} = \begin{cases}
      \f{12}{\sqrt{n}} \\[10pt]
      \P(\f{D_n - \mu}{\sigma} \leq y) + \P(Z \leq y) \leq \frac{1}{2y^2} + O(e^{-\Theta(y^2)}/\abs{y}) & \text{for } y < 0 \\[10pt]
      \P(\f{D_n - \mu}{\sigma} \geq y) + \P(Z\geq y) \leq \frac{1}{2y^2} + O(e^{-\Theta(y^2)}/\abs{y}) & \text{for } y > 0 \\[10pt]
\end{cases}. \]

For a characteristic function $\phi(t)$ for a variable $X$, using integration by parts we can write
\[
\phi(t)
= \int_\mathbb{R} e^{itx} \P(X = x)\ \d x
= -it \int_\mathbb{R} e^{itx} \P(X \leq x)\ \d x.
\]

Hence, we have the following (where $\abs{\ldots}$ is a repetition of the same integrand):
\begin{align*}
\abs{\phi_n(t) - e^{-t^2/2}} &\leq \abs{t} \int_\mathbb{R} \abs{\P\p{\f{D_n - \mu}{\sigma} < y} - \P(Z < y)} \d y  \\
&\leq |t| \int_{|y|>k\sigma} \abs{\P\p{\f{D_n - \mu}{\sigma} < y} - \P(Z < y)} \d y + |t| \int_{|y| \leq k\sigma} \abs{\ldots} \d y \\
&\leq |t| \int_{|y|>k\sigma} \p{\frac{1}{2y^2} + \frac{e^{-\Theta(y^2)}}{y}} \d y + |t|2k\sigma \sqrt{\frac{12}{n}}.
\end{align*}

Take $k = O(n^{-\frac{1}{2} + \epsilon})$. Then since $\sigma = \Theta(n)$, we have $\abs{\phi_n(t) - e^{-t^2/2}} \leq |t|O(n^{-\frac{1}{2} + \epsilon})$.

	Thus, for any $\varepsilon > 0$, we compute
	\begin{align*}
		\int_{-\pi \sigma}^{\pi \sigma} \abs{\phi_n(t) - e^{-t^2/2}} \d t
		&\leq \int_{-n^\varepsilon}^{n^\varepsilon} \abs{\phi_n(t) - e^{-t^2/2}} \d t + \int_{n^\varepsilon < \abs{t} < \pi \sigma} (\abs{\phi_n(t)} + |e^{-t^2/2}|) \d t \\
		&\leq \int_{-n^\varepsilon}^{n^\varepsilon} \abs{\phi_n(t) - e^{-t^2/2}} \d t + \int_{n^\varepsilon < \abs{t} < \pi \sigma} e^{-\Theta(t^2)} \d t \\
		&= O(n^{-\f{1}{2} + \varepsilon}).
	\end{align*}
    \qed
    \end{thm}

\section{3-term arithmetic progressions}

Throughout this section, we define a random variable $A_n$ to be the number of 3-term arithmetic progressions in a randomly chosen subset $S \subset \mathbb{Z} / n\mathbb{Z}$, where each element in $\mathbb{Z} / n\mathbb{Z}$ has probability $\f{1}{2}$ (also denoted $p$) of appearing in the subset. The elements of the probability space can thus be described as $\mathbf{x} \in \{0,1\}^n$ where $\mathbf{x}_i$ is 1 with probability $p$ and 0 with probability $1-p$. Further, the random variable $A_n$ can be thought of as a function $A_n : \{0,1\}^n \to \mathbb{N}$.

Further, we require $n$ to be prime because we want a sort of uniformity among properties of arithmetic progressions. For example, the set $\{0, 5, 10, 15, 20\}$ in $\mathbb{Z} / 25\mathbb{Z}$ has $10$ arithmetic progressions, while no other set of size 5 has as many arithmetic progressions as this. To give a more concrete example of why this might end up messing with some calculations later on, consider the probability that a fixed arithmetic progression appears in a random subset of $\mathbb{Z} /n\mathbb{Z}$. If $n$ is prime,

\subsection{Variance of $A_n$}
In this section, we describe a $p$-biased Fourier basis for functions on the probability space, exactly analogous to the method of Berkowitz \cite{Berkowitz17}, which we use to find the variance of $A_n$ and will use in the next section to try to prove an LLT for $A_n.$
In order to do this, we first define $\chi_i : \{0,1\}^n \to \mathbb{R}$ by

\[\chi_i := \chi_i(\mathbf{x}) := \frac{\mathbf{x}_i - p}{\sqrt{p(1-p)}} = \begin{cases}
      -\sqrt{\frac{p}{1-p}} & \text{if } \mathbf{x}_p = 0 \\[10pt]
      \sqrt{\frac{1-p}{p}} & \text{if } \mathbf{x}_p = 1
\end{cases} \]
so that $\chi_i$ is a normalized version of $\mathbf{x}_i$. Further, we can extend this to define, for an arbitrary set $S \subseteq \mathbb{Z} / n\mathbb{Z}$,

\[\chi_S := \prod_{i \in S} {\chi_i}. \]

Note that if we take the inner product of two functions $f, g : \{0,1\}^n \to \mathbb{R}$ to be $\mathbb{E}[fg]$, then $\{\chi_S \mid S \subseteq \mathbb{Z} / n\mathbb{Z}\}$ forms an orthonormal basis for functions on our probability space. Then if we define the Fourier transform $\hat{f} : \{0,1\}^n \to \mathbb{R}$ of an arbitrary function $f : \{0,1\}^n \to \mathbb{R}$ by

\[\hat{f}(S) := \mathbb{E}[f(\mathbf{x})\chi_S(\mathbf{x})],\]
from the orthonormality of our basis we get

\[ f(\mathbf{x}) = \sum_{S \subseteq \mathbb{Z} / n\mathbb{Z}} {\hat{f}(S)\chi_S(\mathbf{x})}. \]

We will use this expansion to calculate the variance of $A_n$ and bound the pointwise distance of the characteristic function from that of the discrete Gaussian for small $t$.

It will now be useful to normalize the random variable $A_n$. We take the mean of $A_n$ to be $\mu_n := \mathbb{E}[A_n]$. We write the variance of $A_n$ as $\sigma_n^2 := \mathbb{E}[A_n^2] - \mathbb{E}[A_n]^2$. So we define $Z : \{0,1\}^n \to \mathbb{R}$ by

\[Z = Z_n := \frac{A_n - \mu_n}{\sigma_n}\]
and we will often refer to the characteristic function of $Z$ defined by $\phi_Z(t) := \mathbb{E}[e^{itZ}]$.

Before moving on to calculate the Fourier coefficients $\hat{A_n}(S)$, we will first note that there are $\binom{n}{2}$ possible (non-trivial) 3-term arithmetic progressions in $\mathbb{Z} / n\mathbb{Z}$. There are first $n$ choices for the start of the arithmetic progression, then $n-1$ choices for a non-trivial separation distance $d$, and finally both $d$ and $-d$ will have counted the same 3-term arithmetic progression from different starting points, so we divide by 2 to yield $\frac{n(n-1)}{2} = \binom{n}{2}$. Additionally, each 3-term arithmetic progression occurs with probability $p^3$ (each of the three terms in the progression occur independently with probability $p$). This allows us to calculate

\[\mu_n = \mathbb{E}\s{\sum_{\Lambda} {1_\Lambda}} = \sum_{\Lambda} {\mathbb{E}[1_\Lambda]} = \sum_{\Lambda} {p^3} = p^3\binom{n}{2}\]
where $1_\Lambda$ is the indicator function for a particular 3-term arithmetic progression $\Lambda$ in $\mathbb{Z} / n\mathbb{Z}$.

Furthermore, the fact that our basis for functions on the probability space is orthonormal allows us to calculate variance according to this formula, derived from Parseval's Theorem.

\[ \sigma^2 = \sum_{S \neq \varnothing} {\hat{A_n}(S)^2}. \]

We now work with the 3-term arithmetic progression indicator functions $1_\Lambda$ a bit more. We use $i \in \Lambda$ to denote that $i$ is a term in the 3-term arithmetic progression $\Lambda$. Therefore, the indicator function can be expressed as

\begin{align*}
1_\Lambda(\mathbf{x}) &= \prod_{i \in \Lambda} {\mathbf{x}_i} = \prod_{i \in \Lambda} {\p{\sqrt{p(1-p)}\chi_i + p}} \\
&= p^3+p^2\sqrt{p(1-p)}\sum_{i \in \Lambda} {\chi_i} + p^2(1-p)\sum_{i_1 \neq i_2 \in \Lambda} {\chi_{\{i_1, i_2\}}} + p^{3/2}(1-p)^{3/2}.
\end{align*}

Note that any two elements of $\mathbb{Z} / n\mathbb{Z}$ appear in exactly 3 3-term arithmetic progressions and any one element appears in exactly $\frac{3}{2}(n-1)$ 3-term arithmetic progressions. Hence, by summing over all 3-term arithmetic progressions, we have
\begin{align*}
A_n = p^3 {{n}\choose{2}} &+ \frac{3}{2}(n-1)p^2\sqrt{p(1-p)}\sum_{i \in \mathbb{Z} / n\mathbb{Z}} \chi_i \\
&+3p^2(1-p)\sum_{\{i_1, i_2 \mid i_1 \neq i_2\} \subset \mathbb{Z} / n\mathbb{Z}} \chi_{\{i_1, i_2 \}} + \sum_{\Lambda} p^{3/2}(1-p)^{3/2}\chi_{\Lambda}.
\end{align*}

Thus, we have the Fourier Transform of $A_n$:
\[\hat{A_n}(S) = \begin{cases}
      p^3 {{n}\choose{2}} & \text{if } S = \varnothing \\[10pt]
      \frac{3}{2}(n-1)p^2\sqrt{p(1-p)} & \text{if } \abs{S} = 1 \\[10pt]
      3p^2(1-p) & \text{if } S = \{i_1,i_2\} \\[10pt]
      p^{3/2}(1-p)^{3/2} & \text{if } S = \Lambda \\[10pt]
      0 & \text{else}
\end{cases} \]
and we can use Parseval's Theorem to give us the variance of $A_n$:

\begin{align*}
\sigma_n^2 &= \sum_{\substack{S \subset \mathbb{Z} / n\mathbb{Z} \\ S \neq \varnothing}} {\hat{A_n}(S)^2} \\
&= \sum_{i \in \mathbb{Z} / n\mathbb{Z}} {\p{\frac{3}{2}(n-1)p^2\sqrt{p(1-p)}}^2} + \sum_{\{i_1, i_2\} \in \mathbb{Z} / n\mathbb{Z}} {\p{3p^2(1-p)}^2} + \sum_{\Lambda} {\p{p^{3/2}(1-p)^{3/2}}^2} \\
&= \frac{9}{4}n(n-1)^2p^5(1-p) + 9\binom{n}{2}p^4(1-p)^2 + \binom{n}{2}p^3(1-p)^3 \\
&= \Theta(n^3).
\end{align*}

This means that $\sigma_n = \Theta(n^{3/2})$.

\subsection{A CLT for 3-term APs}

In this section, we establish a central limit theorem for the number of 3-term arithmetic progressions in $\mathbb{Z} / n\mathbb{Z}$, $A_n$.

\begin{thm}
\label{thm:clt-an}
$$\abs{P\p{\frac{A_n - \mu_n}{\sigma_n} \leq x} - \Phi(x)} = O\p{n^{-1/4}}.$$
\end{thm}

We prove this theorem using the method of dependency graphs detailed in Chatterjee's notes \cite{Chatterjee07} (Lecture 6). This method bounds the Wasserstein distance of a distribution from the normal based on the dependency graph. We define the Wasserstein distance as such, where $\Omega$ is the probability space:

$$ \mathrm{Wass}(\mu, \nu) = \sup\left\{{\abs{\int{f \d \mu} - \int{f \d \nu}}\ \big|\ f : \Omega \to \mathbb{R} \text{ is 1-Lipschitz}}\right\}.$$

The Wasserstein distance is a metric that is stronger than the $L_1$ distance of the cumulative distribution functions of two distributions \cite{Chatterjee07}. In other words, a vanishing bound on the Wasserstein distance between a distribution and the standard normal would immediately establish a central limit theorem for that distribution.

More formally, we define the Kolmogorov distance as:

$$ \mathrm{Kolm}(\mu, \nu) = \sup_{x \in \mathbb{R}}{\abs{\mu((-\infty, x]) - \nu((-\infty, x])}}. $$

So the Kolmogorov distance is the largest difference in the cumulative distribution functions. We write the Wasserstein distance between two random variables $X \sim \mu$ and $Y \sim \nu$ as $\mathrm{Wass}(X, Y) = \mathrm{Wass}(\mu, \nu)$, and similarly, $\mathrm{Kolm}(X, Y) = \mathrm{Kolm}(\mu, \nu)$. A useful bound for the Kolmogorov distance between a random variable and the standard normal is stated and proved in \cite{Chatterjee07}:

\begin{lem}[Chatterjee, 2007]
\label{lem:wass-kolm}
For a pair of r.v.'s $W, Z$ where $Z \sim N(0, 1)$,
$$ \mathrm{Kolm}(W, Z) \leq \sqrt{\frac{2}{\pi}\ \mathrm{Wass}(W, Z)}. $$
\end{lem}

Thus, the distance between the cdf of a random variable $W$ and the standard normal, $Z$ is bounded by the square root of the Wasserstein distance between $W$ and $Z$ (with a constant factor). So the problem of deriving a central limit theorem for $W$ is reduced to bounding $\mathrm{Wass}(W, Z)$.

A formal statement of the method of dependency graphs is as follows. Suppose there are a collection of random variables, $\{X_i \mid i \in V\}$, indexed by the vertices of a graph $G = (V,E)$ such that $(i, j) \in E$ iff $X_i, X_j$ are dependent. Such a graph is referred to as a dependency graph. Let $D = 1 + \Delta(G)$, where $\Delta(G)$ is the maximum degree of $G$. We have the following:

\begin{lem}[Chatterjee, 2007]
\label{lem:wass-dep-graph}
Suppose that $\mathbb{E}[X_i] = 0$, $\sigma^2 = \mathrm{Var}(\sum {X_i})$, $W = \frac{\sum X_i}{\sigma}$ and $Z \sim N(0,1)$. Then,

$$ \mathrm{Wass}(W, Z) \leq \frac{4}{\sqrt{\pi}\sigma^2} \sqrt{D^3 \sum {\mathbb{E}{|X_i|}^4}}
+ \frac{D^2}{\sigma^3} \sum {\mathbb{E}{|X_i|}^3}. $$

\end{lem}

For the problem of 3-term arithmetic progressions, we will write $\Lambda$ to denote a particular 3-term AP. We define $V = \{\Lambda \subseteq \mathbb{Z} /n\mathbb{Z} \mid \Lambda \text{ is a 3-AP}\}$ and $X_\Lambda = 1_{\Lambda \in S} - \mathbb{E}[1_{\Lambda \in S}] = 1_{\Lambda \in S} - p^3$. Note that there are $\binom{n}{2}$ 3-term APs in $\mathbb{Z} / n\mathbb{Z}$, not counting degenerate 3-term arithmetic progressions, so $|V| = \binom{n}{2}$. Further, note that $X_{\Lambda_1}$ and $X_{\Lambda_2}$ are independent iff $\Lambda_1 \cap \Lambda_2 = \varnothing$. We have the following properties:

\begin{lem}
The degree of the dependency graph is $D = O(n)$.
\end{lem}

\begin{proof}

\begin{align*}
\deg(\Lambda_1) &= \abs{\{ \Lambda_2 \mid \Lambda_1 \cap \Lambda_2 \neq \varnothing \}} \\
&= \abs{\{ \Lambda_2 \mid (\Lambda_2)_1 \in \Lambda_1 \lor (\Lambda_2)_2 \in \Lambda_1 \lor (\Lambda_2)_3 \in \Lambda_1\}} \\
&\leq 3\abs{\{ \Lambda_2 \mid (\Lambda_2)_1 \in \Lambda_1 \}} \\
&= \frac{9}{2}(n-1).
\end{align*}

So $\Delta(G) \leq \frac{9}{2}(n-1)$ and $D \leq \frac{9}{2}(n-1)+1$. Thus, $D = O(n)$.
\end{proof}

\begin{prop}
The $m$th moment of $X_\Lambda$ is bounded by $\mathbb{E}|X_\Lambda|^m \leq 1$ for all $m \geq 1$.
\end{prop}

\begin{proof}

$X_\Lambda$ is a random variable which takes values at $1-p^3$ and $p^3$, so its absolute value is bounded by 1. Thus, its $m$th absolute moment is also bounded by 1.

\end{proof}

Furthermore, we note that $\mathrm{Var}\p{\sum X_i} = \mathrm{Var}\p{\sum 1_\Lambda - p^3} = \sigma_n^2$ and $W = \frac{\sum X_i}{\sigma} = \frac{A_n - \mu_n}{\sigma_n}$. We now proceed with the proof of theorem \ref{thm:clt-an}.

\begin{proof}(Theorem \ref{thm:clt-an})

Applying lemma \ref{lem:wass-dep-graph} to $W = \frac{A_n - \mu_n}{\sigma_n}$ and using the bound on the Kolmogorov distance from \ref{lem:wass-kolm} we get:

\begin{align*}
\abs{P\p{\frac{A_n - \mu_n}{\sigma_n} \leq x} - \Phi(x)} &\leq \sqrt{\frac{2}{\pi} \mathrm{Wass}(W, Z)} \\
&\leq \sqrt{\frac{2}{\pi}}\sqrt{\frac{4}{\sqrt{\pi}\sigma^2} \sqrt{D^3 \sum {\mathbb{E}{|X_\Lambda|}^4}}
+ \frac{D^2}{\sigma^3} \sum {\mathbb{E}{|X_\Lambda|}^3}}
\end{align*}

Now, using the established properties that $|V| = \binom{n}{2} \leq n^2$, $D = O(n)$, $\mathbb{E}{|X_\Lambda|^m} \leq 1$ for all $m \geq 1$, and $\sigma^2 = \Theta(n^3)$, we get

\begin{align*}
\abs{P\p{\frac{A_n - \mu_n}{\sigma_n} \leq x} - \Phi(x)} &= O\p{\sqrt{\frac{1}{n^3} \sqrt{n^3 n^2} + \frac{n^2}{n^{9/2}} n^2}} \\
&= O(n^{-1/4}).
\end{align*}

\end{proof}

\subsection{Attempt to show LLT for $A_n$}
We recall from section 2.1.1 that since we have a CLT, we can reduce the problem of proving an LLT to one of bounding $\abs{\phi_Z(t) - e^{-t^2/2}}$.
Using the $p$-biased Fourier basis developed above, we proceed to show the following bound for small values of $t$ using the same method as Berkowitz \cite{Berkowitz17}:
\begin{prop}
For $|t| \ll \sqrt[]{n}$, \[ \abs{\phi_Z(t) - e^{-t^2/2}} \leq O\left(\frac{t^3e^{-t^2/3}}{\sqrt[]{n}} + \frac{t}{\sqrt{n}}\right). \]
\end{prop}

\begin{proof}
To begin, we decompose $Z = X+Y$, where
\[Q = \frac{1}{\sqrt{n}}, X = \sum_{k \in \mathbb{Z} / n\mathbb{Z}} Q\chi_k, Y = \sum_{k \in \mathbb{Z} / n\mathbb{Z}} (\hat{Z}(k)-Q)\chi_k + \sum_{|S| \geq 2} \hat{Z}(k)\chi_k.\]

We view $X$ as the main term and $Y$ as the error term. Since we normalized $X$ with $Q$, $X$ has mean $0$ and variance $1$. We first bound the distance between $X$ and the normal distribution. Define
\[L_n := n\mathbb{E}[|Q\chi_k|^3] = O\left(\frac{1}{\sqrt{n}}\right) < \infty. \]
By Berry-Esseen, for $t \leq \f{1}{4L_n}$ we have
\[|\mathbb{E}[e^{itX}] - e^{-t^2/2}| \leq 16L_{n}t^3e^{-t^2/3}.\]

Bounding the expectation of $Y$, we obtain
\[ \mathbb{E}[|Y|] \leq \sqrt{\mathbb{E}[|Y|^2]} = \sqrt{\mathrm{Var}(Y)} = \sqrt{\sum_{k \in \mathbb{Z} / n\mathbb{Z}} (\hat{Z}(k)-Q)^2 + \sum_{|S| \geq 2} \hat{Z}^2(S)}.\]
Using the value of the variance $\sigma_n^2$ of $A_n$, \[\sum_{|S| \geq 2} {\hat{Z}(S)} = \frac{9{{n}\choose{2}}p^4(1-p)^2+{n\choose2}p^3(1-p)^3}{\sigma_n^2} = \Theta\left(\frac{1}{n}\right).\]
In addition, using the value of $\hat{A_n}$,
\begin{align*}
n\sigma_n\hat{Z}^2(k) - \sigma_n^2 = n\hat{A_n}^2(k) - \sigma_n^2 = O(n^2)  \\
n\hat{Z}^2(k) - 1 = O\left(\frac{1}{n} \right) \\
\hat{Z}^2(k) - \frac{1}{n} = O\left(\frac{1}{n^2} \right).
\end{align*}
Since $\hat{Z}(k) + Q = O\left(\frac{1}{\sqrt{n}}\right)$ and $Q^2 = \frac{1}{n}$,
\[ \abs{\hat{Z}(k) - Q} \leq \abs{\frac{\hat{Z}(k) - \frac{1}{n}}{\hat{Z}(k)+Q}} = O\p{\frac{1}{n^{3/2}}}. \]
Hence, we have $\mathbb{E}[Y] = O\left(\frac{1}{\sqrt{n}}\right)$.
Thus, we conclude that if $t < \frac{1}{4L_n} = O(\sqrt{n})$,
\begin{align*}
\abs{\phi_Z(t) - e^{-t^2/2}}
&= \abs{\mathbb{E}[e^{itZ}-e^{-t^2/2}]} = \abs{\mathbb{E}[e^{it(X+Y)}-e^{-t^2/2}]} \\
&\leq \abs{\mathbb{E}[e^{it(X+Y)}-e^{itX}]} + \abs{\mathbb{E}[e^{itX}-e^{-t^2/2}]} \\
&\leq \mathbb{E}[tY] + 16L_{n}t^3e^{-t^2/3} \qquad \text{ (using the mean value theorem on } e^{itX})\\
&= O\left(\frac{t^3e^{-t^2/3}}{\sqrt[]{n}} + \frac{t}{\sqrt{n}}\right).
\end{align*}

\end{proof}

\begin{remark}
The bound we find for the absolute distance between the characteristic function of $Z$ and that of the standard normal is a decent bound for small $t$ if we want to show an LLT for $A_n$. However, we are unable to prove any meaningful bound for larger $t$, and below we explore the reasons for why this is impossible.
\end{remark}

\subsection{Experimental results and conjectures}

As we could not decrease the bound achieved in the previous section and prove a bound for larger values of $t$, we decided to simulate the number of 3-term arithmetic progressions in a random subset of $\mathbb{Z} / n \mathbb{Z}$ in order to verify whether or not an LLT actually holds and found that one does not.

\begin{figure}[t]
	\centering
	\includegraphics[width=\textwidth]{3ap-hist}
    \caption{Histogram of the number of 3-term arithmetic progressions in 100000 random subsets of $\mathbb{Z} / 101\mathbb{Z}$.}
    \label{fig:3ap}
\end{figure}

As figure \ref{fig:3ap} demonstrates, the number of 3-term arithmetic progressions in a random subset of $\mathbb{Z} / n \mathbb{Z}$ does not follow the Gaussian distribution pointwise. However, after some inspection, one might notice that oscillations in the histogram that are close together are spaced out almost evenly. Furthermore, the oscillations themselves appear fairly Gaussian, as if the entire distribution is the sum of several spaced out Gaussians. Given that the spacings appear to be $O(n)$ and the entire distribution ranges from 0 to $\binom{n}{2} = O(n^2)$, we would expect there to be $O(n)$ of these smaller Gaussians. It is thus reasonable to conclude that the number of 3-term arithmetic progressions in a random subset of $\mathbb{Z} / n\mathbb{Z}$ depends on some other random variable which can take $O(n)$ values.

\begin{figure}[t]
	\centering
    \includegraphics[width=\textwidth]{3ap-fixed-k}
    \caption{Histogram of the number of 3-term arithmetic progressions in random subsets of $\mathbb{Z}/53\mathbb{Z}$ of size $k$ for $k=0$ to $53$, separated by color and . Each $k$ used 10000 samples. (The graph should extend to 1378 on the x axis, but there is very little content in the large $k$ that are very far out so we zoomed in on the bulk of the distribution.)}
    \label{fig:3ap-k}
\end{figure}

This other random variable is the size of a random set of $\mathbb{Z} / n\mathbb{Z}$. From figure \ref{fig:3ap-k} we can see that the number of 3-term arithmetic progressions in a random subset of $\mathbb{Z} / n \mathbb{Z}$ with fixed size does, in fact, follow the Gaussian distribution pointwise. From now on, we call this random variable $A_{n, k}$, where $k$ is the size of the subset of $\mathbb{Z} / n\mathbb{Z}$.

We conjecture that the distance between these smaller distributions, $\mathbb{E}[A_{n,k+1}]-\mathbb{E}[A_{n,k}]$, is sufficiently smaller than the standard deviation of the distributions themselves, $\sqrt{\mathrm{Var}(A_{n, k})}$. In other words, there is a non-trivial interval between $\mathbb{E}[A_{n, k}]$ and $\mathbb{E}[A_{n, k+1}]$ where for all x in the interval, $\P(A_{n, k}=x)$ and $\P(A_{n, k+1} = x)$ are both very small. In particular, we suspect that those two probabilities are significantly smaller than the Gaussian approximation of $\P(A_n = x)$ would suggest.

This train of thought will be made more precise in sections 6 and 7.

\subsubsection{Continuous sets}

One thing that comes to mind after observing that $A_n$ does not have an LLT due to the dependence on the size of the random set, is whether $A_n$ would have an LLT if the existence of each element of the set was made continuous. That is to say, the variable $x_i$ which denotes whether an element $i$ of $\mathbb{Z} / n\mathbb{Z}$ is in our random set $S$, is now a uniform random variable on $[0, 1]$ instead of a Bernoulli random variable on $\{0, 1\}$ with $p=\frac{1}{2}$. The number of 3-term arithmetic progressions $\bar{A}_n$ is still defined in a similar manner,

$$ \bar{A}_n = \frac{1}{2}\sum_{i = 1}^{n} {\sum_{j = 1}^{n-1} {x_i x_{i+j} x_{i+2j}}}. $$

\begin{figure}[t]
	\centering
    \includegraphics[width=\textwidth]{3ap-cont}
    \caption{Number of 3-term "arithmetic progressions" in 1000000 random "continuous sets" of $\mathbb{Z} / 23\mathbb{Z}$}
    \label{fig:3ap-cont}
\end{figure}

We simulated this to see if it possibly followed the Gaussian distribution pointwise. From figure \ref{fig:3ap-cont}, we conjecture that it is very likely that there is an LLT for $\bar{A}_n$. However, we successfully can show that there is a CLT for $\bar{A}_n$. Namely, define $\bar{\mu}_n = \mathbb{E}[\bar{A}_n]$ and $\bar{\sigma}_n^2 = \mathrm{Var}(A_n)$. Then the following theorem holds:

\begin{thm}
\label{thm:clt-an-cont}
$$\abs{P\p{\frac{\bar{A}_n - \bar{\mu}_n}{\bar{\sigma}_n} \leq x} - \Phi(x)} = O\p{n^{-1/4}}.$$
\end{thm}

Notably, this is exactly the same bound achieved in Theorem \ref{thm:clt-an}. This CLT is a consequence of the fact that the dependency graph of the indicator functions $1_\Lambda$ does not change if we make the underlying $x_i$ uniform on [0, 1] instead of Bernoulli on \{0, 1\}. However, the following lemma is what gives us the exact same bound as for the standard case, $A_n$:

\begin{lem}
\label{lem:var-an-cont}
$\bar{\sigma}_n^2 = \Theta(n^3).$
\end{lem}

\begin{proof}
First, note that
$$\bar{\mu}_n = \mathbb{E}\left[\sum_{\Lambda} {1_\Lambda}\right] = \sum_{\Lambda} {\mathbb{E}[1_\Lambda]} = \frac{1}{8} \binom{n}{2}.$$
Now, we can also calculate
\begin{align*}
\mathbb{E}\left[ \bar{A}_n^2 \right] &= \mathbb{E}\left[\p{\sum_{\Lambda} {1_\Lambda}}^2\right] = \mathbb{E}\left[\sum_{\Lambda_1} {\sum_{\Lambda_2} {1_{\Lambda_1} 1_{\Lambda_2}}}\right] \\
&= \mathbb{E}\left[\sum_{\Lambda_1} {\sum_{\Lambda_2} {\prod_{j \in {\Lambda_1 \cup \Lambda_2}} {x_j}}}\right] \\
&= \mathbb{E}\left[\sum_{\Lambda_1} {\sum_{i=0}^{3} {\sum_{\substack{\Lambda_2 \textrm{ s.t. } \\ \abs{\Lambda_1 \cap \Lambda_2} = i}} {\prod_{j \in {\Lambda_1 \cup \Lambda_2}} {x_j}}}}\right] \\
&= \sum_{\Lambda_1} {\sum_{i=0}^{3} {\sum_{\substack{\Lambda_2 \textrm{ s.t. } \\ \abs{\Lambda_1 \cap \Lambda_2} = i}} {\p{\frac{1}{2}}^{6-i}}}} \\
&= \sum_{\Lambda_1} {\sum_{i=0}^{3} {\p{\frac{1}{2}}^{6-i} \p{\# \{\Lambda_2 \mid \abs{\Lambda_1 \cap \Lambda_2} = i \}}}}.
\end{align*}

The particular quantity $\# \{\Lambda_2 \mid \abs{\Lambda_1 \cap \Lambda_2} = i \}$ is examined further and calculated in section \ref{var-ank}. For now we take the particular values we need as given and note that the quantity does not depend on our choice of $\Lambda_1$.

$$
\mathbb{E}\left[ \bar{A}_n^2 \right] = \binom{n}{2} \p{\frac{1}{8} + \frac{1}{16}(6) + \frac{1}{32}\p{\frac{9}{2}n-\frac{39}{2}} + \frac{1}{64}\p{\frac{1}{2}n^2 - 5n + \frac{25}{2}}}.
$$
Now, given that $\bar{\sigma}_n^2 = \mathbb{E}\left[ \bar{A}_n^2 \right] - \bar{\mu}_n^2$, we have
$$ \bar{\sigma}_n^2 = \frac{1}{64}\binom{n}{2}\p{n - \frac{25}{54}} = \Theta(n^3). $$

\end{proof}

\section{Attempts to get CLT for $A_{n,k}$}

In order to prove a CLT for $A_{n,k}$, we use the method of exchangeable pairs detailed in Chatterjee's notes \cite{Chatterjee07} (Lecture 7). Like the method of dependency graphs used in section 4, this method bounds the Wasserstein distance of a distribution from the normal distribution.

\begin{defn}
$(W,W')$ is an exchangeable pair of random variables if $(W,W') \overset{\mathrm{d}}= (W',W)$.
\end{defn}

The following is a formal statement of the method of exchangeable pairs.

\begin{lem}{(Chatterjee, 2007)}
If $(W,W')$ is an exchangeable pair, $\mathbb{E}[W^2] = 1$ and there exists $\lambda \in (0,1)$ such that $\mathbb{E}[W-W'|W] = -\lambda W$, then
\[\mathrm{Wass}(W,Z) \leq \sqrt{\p{\frac{2}{\pi}}\mathrm{Var}\p{\mathbb{E}\s{\frac{1}{2\lambda}(W'-W)^2 \mid W}}} + \frac{1}{3\lambda}\mathbb{E}[\abs{W'-W}^2],\]
where $Z \sim N(0,1)$.
\end{lem}

Let $A_{n,k}$ be the number of 3-term arithmetic progressions given the subset size = $k$, and $W$ be $A_{n,k}$ standardized. We want to show that $W$ converges in distribution to a normal distribution. We construct $W'$ as follows. We have a random subset $S \subseteq \mathbb{Z} / n\mathbb{Z}$ with size $k$. Randomly choose two elements $I,J \in \mathbb{Z} / n\mathbb{Z}$, swap their status regarding their inclusion in $S$, and call the new subset $S'$ (i.e. if $I \in S$ iff $J \in S'$ and $J \in S$ iff $I \in S'$). Now define $A$ to be the number of 3-term arithmetic progressions in $S$ and define $W$ to be standardized $A$, namely $\frac{A-\mu_{n,k}}{\sigma_{n,k}}$. Similarly, define $A'$ to be the number of 3-term arithmetic progressions in $S'$ and define $W'$ to be $\frac{A'-\mu_{n,k}}{\sigma_{n,k}}$.

\begin{remark}
Let $W$ and $W'$ be an exchangeable pair, each with mean 0 and variance 1, and let $\Lambda$ be a 3-term arithmetic progression in $\mathbb{Z} / n\mathbb{Z}$. Define $1_{\Lambda \subseteq S} =
\begin{cases}
      1 & \Lambda \subseteq S \\
      0 & \text{else}
\end{cases}$. \\
Then $A = \sum_{\Lambda \subseteq \mathbb{Z} / n\mathbb{Z}} 1_{\Lambda \subseteq S}$. Also note that
\[ \mu_{n,k} = \mathbb{E}[\sum_{\Lambda \subseteq \mathbb{Z} / n\mathbb{Z}} 1_{\Lambda \subseteq S}] = \sum_{\Lambda \subseteq \mathbb{Z} / n\mathbb{Z}} \P(\Lambda \in S) = \binom{n}{2}\frac{\binom{k}{3}}{\binom{n}{3}}. \]
\end{remark}

In order to apply the method, we first must show that $\mathbb{E}[W'-W \mid W] = -\lambda W$ for some $\lambda \in (0,1)$.

\begin{lem}
Let $\lambda = \frac{3(n-k)}{\binom{n}{2}}$. Then $\mathbb{E}[W'-W \mid W]= -\lambda W.$
\end{lem}
\begin{proof}
We have
\[\mathbb{E}[A' \mid A] = \sum_{\Lambda \subseteq \mathbb{Z} / n\mathbb{Z}} \mathbb{E}[1_{\Lambda \subseteq S'} \mid A] = \sum_{\Lambda \subseteq \mathbb{Z} / n\mathbb{Z}} \P(\Lambda \subseteq S' \mid A).\]

Now \[ \P(\Lambda \subseteq S' \mid A) = \P(\Lambda \subseteq S' \mid \Lambda \subseteq S, A)\P(\Lambda \subseteq S \mid A) + \P(\Lambda \subseteq S' \mid \Lambda \not \subseteq S, A)\P(\Lambda \not \subseteq S \mid A).
\]

Note that
\[\P(\Lambda \subseteq S \mid A) = \frac{A}{\binom{n}{2}}\]
\[\P(\Lambda \not \subseteq S \mid A) = 1-\frac{A}{\binom{n}{2}}\]
as there are $A$ 3-term arithmetic progressions in $S$ and $\binom{n}{2}$ 3-term arithmetic progressions total in $\mathbb{Z} / n\mathbb{Z}$.

We have \[\P(\Lambda \subseteq S' \mid \Lambda \subseteq S, A) = 1- \frac{3(n-k)}{{{n}\choose{2}}}
\]
since the probability of $\Lambda \not \subseteq S'$ is the probability of one of its 3 elements being selected to be swapped along with one of the elements outside of $S$.

Taking this into account in our previous expression, we have
\[ \P(\Lambda \subseteq S' \mid A) = \p{1- \frac{3(n-k)}{{{n}\choose{2}}}}\p{\frac{A}{\binom{n}{2}}} + \P(\Lambda \subseteq S' \mid \Lambda \not \subseteq S, A)\p{1-\frac{A}{\binom{n}{2}}}. \]

Finally, we examine
\begin{align*}
\P(\Lambda \subseteq S' \mid \Lambda \not \subseteq S, A) &= \sum_{i = 0}^{3} \P(\Lambda \subseteq S' \text{ and } \abs{\Lambda \cap S} = i \mid \Lambda \not \subseteq S \text{ and } A) \\
&= \sum_{i = 0}^{2} \P(\Lambda \subseteq S' \mid \Lambda \not \subseteq S \text{ and } \abs{\Lambda \cap S} = i  \text{ and } A)\P(\abs{\Lambda \cap S} = i \mid \Lambda \not \subseteq S \text{ and } A) \\
&= \P(\Lambda \subseteq S' \mid \abs{\Lambda \cap S} = 2, \text{ and } A)\P(\abs{\Lambda \cap S} = 2 \mid \Lambda \not \subseteq S\text{ and } A).
\end{align*}

We have $\P(\Lambda \subseteq S' \mid \abs{\Lambda \cap S} = 2, \text{ and } A) = \frac{k-2}{{n\choose2}}$ since the only way for $\Lambda$ to be contained in $S'$ if $\abs{\Lambda \cap S} = 2$ is if the one element of $\Lambda$ that is not in $S$ is chosen, along with one other element of $S \backslash \Lambda$.

Substituting this in, we get

\[\P(\Lambda \subseteq S' \mid \Lambda \not \subseteq S, A) = \p{\frac{k-2}{{n\choose2}}}\P(\abs{\Lambda \cap S} = 2 \mid \Lambda \not \subseteq S\text{ and } A).\]

\begin{align*}
\P(\abs{\Lambda \cap S} = 2 \mid \Lambda \not \subseteq S\text{ and } A)
&= 3\P(\Lambda_1 \in S, \Lambda_2 \in S, \Lambda_3 \notin S \mid \Lambda \not \subseteq S \textrm{ and } A) \\
&= 3\P(\Lambda_3 \notin S \mid \Lambda_1 \in S, \Lambda_2 \in S, \Lambda \not \subseteq S, A) \\
&\hspace{10mm} \cdot \P(\Lambda_2 \in S \mid \Lambda \not \subseteq S, \Lambda_1 \in S, A)\P(\Lambda_1 \in S \mid \Lambda \not \subseteq S, A) \\
&= 3\P(\Lambda_2 \in S \mid \Lambda \not \subseteq S, \Lambda_1 \in S, A)\P(\Lambda_1 \in S \mid \Lambda \not \subseteq S, A).
\end{align*}

Here we have two non-trivial quantities to examine. First we see by Bayes's rule that
\begin{align*}
\P(\Lambda_1 \in S \mid \Lambda \not \subseteq S, A) &= \dfrac{\P(\Lambda \not \subseteq S \mid \Lambda_1 \notin S, A)\P(\Lambda_1 \not \subseteq S,A)}{\P(\Lambda \not \subseteq S, A)} \\
&= \dfrac{\P(\Lambda \not \subseteq S \mid \Lambda_1 \notin S, A)\P(\Lambda_1 \notin S \mid A)}{\P(\Lambda \not \subseteq S \mid A)} \\
&= \dfrac{\P(\Lambda \not \subseteq S \mid \Lambda_1 \notin S, A)\p{\frac{k}{n}}}{1-\frac{A}{\binom{n}{2}}}.
\end{align*}

Further, we also have by Bayes's rule,
\begin{align*}
\P(\Lambda_2 \in S \mid \Lambda \not \subseteq S, \Lambda_1 \in S, A) &= \dfrac{\P(\Lambda \not \subseteq S \mid \Lambda_2 \in S, \Lambda_1 \in S, A)\P(\Lambda_2 \in S, \Lambda_1 \in S, A)}{\P(\Lambda \not \subseteq S, \Lambda_1 \in S, A)} \\
&= \dfrac{\P(\Lambda \not \subseteq S \mid \Lambda_2 \in S, \Lambda_1 \in S, A)\P(\Lambda_2 \in S \mid \Lambda_1 \in S, A)}{\P(\Lambda \not \subseteq S \mid \Lambda_1 \in S, A)} \\
&= \dfrac{\P(\Lambda_3 \notin S \mid \Lambda_2 \in S, \Lambda_1 \in S, A)\P(\Lambda_2 \in S \mid \Lambda_1 \in S, A)}{\P(\Lambda \not \subseteq S \mid \Lambda_1 \in S, A)} \\
&= \dfrac{\p{1-\frac{k-2}{n-2}}\p{\frac{k-1}{n-1}}}{\P(\Lambda \not \subseteq S \mid \Lambda_1 \in S, A)}.
\end{align*}

So,
\begin{align*}
\P(\abs{\Lambda \cap S} = 2 \mid \Lambda \not \subseteq S\text{ and } A) &= 3\p{\dfrac{\P(\Lambda \not \subseteq S \mid \Lambda_1 \notin S, A)\p{\frac{k}{n}}}{1-\frac{A}{\binom{n}{2}}}}\p{\dfrac{\p{1-\frac{k-2}{n-2}}\p{\frac{k-1}{n-1}}}{\P(\Lambda \not \subseteq S \mid \Lambda_1 \in S, A)}} \\
&= 3\ \frac{\p{1-\frac{k-2}{n-2}}\p{\frac{k-1}{n-1}}\p{\frac{k}{n}}}{1-\frac{A}{{n\choose2}}}.
\end{align*}

Thus, we have
\[\P(\Lambda \subseteq S' \mid \Lambda \not \subseteq S, A) = 3 \p{\frac{\p{1-\frac{k-2}{n-2}}\p{\frac{k-1}{n-1}}\p{\frac{k}{n}}}{\p{1-\frac{A}{{n\choose2}}}}}\p{\frac{k-2}{{n\choose2}}} = 3\p{\frac{M}{{n\choose2} - A}},\] where $M = (n-k)\frac{{k\choose3}}{{n\choose3}}$. Hence,
\[\P(\Lambda \subseteq S' \mid A) = \p{1 - \frac{3(n-k)}{{n\choose2}}}\p{\frac{A}{{n\choose2}}} + 3\p{\frac{M}{{n\choose2}}},
\]
and so
\begin{align*}
\mathbb{E}[A' - A \mid A] &= -A + \sum_{\Lambda \subseteq \mathbb{Z} / n\mathbb{Z}} \P(\Lambda \subseteq S' \mid A)= -A + \p{1 - \frac{3(n-k)}{{n\choose2}}}\p{A} + 3\p{M} \\
&= -\p{\frac{3(n - k)}{{n\choose2}}}(A) + 3(n-k)\frac{{k\choose3}}{{n\choose3}} \\
&= -\frac{3(n-k)}{{n\choose2}}\p{A-\binom{n}{2}\frac{\binom{k}{3}}{\binom{n}{3}}} \\
&= -\frac{3(n-k)}{{n\choose2}}(A-\mu).
\end{align*}
Thus,
$\mathbb{E}[A'-A \mid A] = -\lambda(A-\mu)$, where $\lambda = \frac{3(n-k)}{{n\choose2}}$.
Hence, $\mathbb{E}[W'-W \mid W] = \mathbb{E}[\frac{A'-\mu}{\sigma} - \frac{A-\mu}{\sigma} \mid \frac{A-\mu}{\sigma}] = \frac{1}{\sigma}\mathbb{E}[A'-A \mid A] = -\lambda\p{\frac{A-\mu}{\sigma}} = -\lambda W$.

\end{proof}

Thus, we can bound the Wasserstein distance $\mathrm{Wass}(W, Z)$, where $Z$ is a standard Gaussian random variable using the method of exchangeable pairs.
We have
\begin{align*}
\mathbb{E}\s{\frac{1}{2\lambda}(W'-W)^2 \mid W}
&= \mathbb{E}\s{\frac{{n\choose2}}{6(n-k)}(W'-W)^2 \mid W} \\
&= \frac{{n\choose2}}{6(n-k)}(\mathbb{E}[(W')^2 \mid W]-2\mathbb{E}[WW' \mid W]+W^2).
\end{align*}

\section{Attempt to prove nonexistence of LLT for $A_n$}

\subsection{General framework for nonexistence of an LLT for other R.V.'s}

We develop more general theorems for what properties a random variable $X_n$ can have to ensure it does not follow an LLT. Let $X_{n}$ (with mean $\mu_n$ and standard deviation $\sigma_n$) be conditioned on some event with value $k$, denoted by $X_{n,k}$. Let $\mu_{n,k} = \mathbb{E}[X_{n,k}]$ and $\sigma_{n,k}$ be the standard deviation of $X_{n,k}.$

The following theorem says that $X_n$ does not satisfy a LLT when the distances between the conditioned distributions $X_{n,k}$ are of a sufficiently larger order than their standard deviations $\sigma_{n,k}$. We show that at a point, the distribution of $X_n$ does not converge to the normal distribution.

\begin{thm}
Suppose $\abs{x - \mu_{n,k}} = \omega(\sigma_{n,k} \sqrt{\sigma_n})$. Then
\[\abs{\P(X_n = x) - \mathcal{N}_n(x)} = \Omega\p{\frac{1}{\sigma_n}} \]
and there is no local limit theorem for $\{X_n\}$.
\end{thm}
\begin{proof}
Since we make no assumptions about the concentration of $X_{n,k}$, the best we can do is use Chebyshev's inequality.
\begin{align*}
\P(X_n = x)
&= \sum_{k=1}^j \P(X_n = x \mid Y_k) \P(Y_k) \\
&\leq \sum_{k=1}^j \P(\abs{X_{n,k} - \mu_{n,k}} \geq \abs{x - \mu_{n,k}}) \P(Y_k) \\
&\leq \sum_{k=1}^j \p{\f{\sigma_{n,k}}{\abs{x - \mu_{n,k}}}}^2 \P(Y_k) \\
&= \sum_{k=1}^j o\p{\f{1}{\sigma_n}} \P(Y_k) \\
&= o\p{\f{1}{\sigma_n}}.
\end{align*}
\end{proof}

In the previous theorem, the required distance between consecutive $X_{n,k}$ is rather large, and this is likely due to the inefficiency of using Chebyshev's inequality. If we assume a CLT on the $X_{n,k}$, we can use a better concentration inequality and require less of a distance between the $X_{n,k}$.

\begin{thm}
Suppose $X_{n,k}$ follows a CLT and $\abs{x - \mu_{n,k}} = \omega(\sigma_{n,k} \sigma_n^{\epsilon})$. Then
\[ \abs{\P(X_n = x) - \mathcal{N}_n(x)} = \Omega\p{\frac{1}{\sigma_n}} \]
and there is no local limit theorem for $\{X_n\}$.
\end{thm}
\begin{proof} We have
\begin{align*}
\P(X_n = x) &= \sum_{k = 1}^j \P(Y_k)\P(X_n = x \mid Y_k) \\
&\leq \sum_{k = 1}^j \P(\abs{X_{n, k}-\mu_{n,k}}\geq \abs{x-\mu_{n,k}}) \P(Y_k) \\
&\leq \sum_{k=1}^j \p{\frac{\sigma_{n,k}}{\abs{x-\mu_{n,k}}\sqrt{2\pi}}e^{-\p{\frac{x-\mu_{n,k}}{\sigma_{n,k}}}^2/2}} \P(Y_k) \\
&= \sum_{k=1}^j {o(\sigma_n^{-\epsilon})e^{-o(\sigma_n^{2\epsilon})} \P(Y_k)} \\
&= o(\sigma_n^{-\epsilon})o(\sigma_n^{-1+\epsilon}) = o(\sigma_n^{-1}).
\end{align*}

Thus, $\P(X_n = x)= o\p{\frac{1}{\sigma_n}}.$ Now $\mathcal{N}_n(x) = \frac{1}{\sqrt{2\pi}\sigma_n} e^{-\p{\frac{x-\mu_n}{\sigma_n}}^2/2} = \Theta\p{\frac{1}{\sigma_n}}$. Hence, \[\abs{\P(X_n = x) - \mathcal{N}_n(x)} = \Omega\p{\frac{1}{\sigma_n}}\] and thus $X_n$ does not follow a LLT.

\end{proof}

\subsection{Variance of $A_{n,k}$}
\label{var-ank}

\begin{lem}
The variance of $A_{n,k}$ is $\sigma_{n,k}^2 = \frac{k^3(n-k)^3}{2n^4} + O(n)$.
\end{lem}

\begin{proof}
We have that $\sigma_{n,k}^2 = \mathbb{E}[A_{n,k}^2] - (\mathbb{E}[A_{n,k}])^2 = \mathbb{E}[A_{n,k}^2] - {n\choose 2}\frac{{k\choose 3}}{{n\choose 3}} = E[A_{n,k}^2] - \frac{3}{n-2} {k\choose 3}.$ We proceed to calculate $\mathbb{E}[A_{n,k}^2]$. Now $A_{n,k}^2 = (\sum_{\Lambda} 1_{\Lambda \subset S})^2. = \sum_{\Lambda_1}\sum_{\Lambda_2} 1_{\Lambda_1 \subset S} 1_{\Lambda_2 \subset S}.$ So
\begin{align*}
\mathbb{E}[A_{n,k}^2] &= \sum_{\Lambda_1}\sum_{\Lambda_2} \mathbb{E}[1_{\Lambda_1 \subset S} 1_{\Lambda_2 \subset S}] \\
&= \sum_{i = 0}^3 \frac{{k\choose{6-i}}}{{n\choose{6-i}}} \sum_{\Lambda_1}\sum_{\substack{\Lambda_2 \textrm{ s.t. } \\ \abs{\Lambda_1 \cap \Lambda_2} = i}} 1.
\end{align*}

We now consider the quantity $\#\{\Lambda_2 \mid \abs{\Lambda_1 \cap \Lambda_2} = i\}$. Since we know there are $i$ elements pre-determined from $\Lambda_1$, we take each subset of $i$ elements from $\Lambda_1$ and perform the following procedure. Check how many arithmetic progressions contain those many numbers (which themselves come from an arithmetic progression), given the function $f: \mathbb{N} \times \{0, 1, 2, 3\} \to \mathbb{N}$ defined by $f(0) = \binom{n}{2}$, $f(1) = \frac{3}{2}(n-1)$, $f(2) = 3$, and $f(3) = 1$. In particular, we check $f(i)$. However, this overcounts the quantity we desire because it also counts arithmetic progressions whose intersection with $\Lambda_1$ is greater than 1. Thus, we check all the ways of adding another element from $\Lambda_1$ to our current subset of $i$ elements and subtract out $f(i+1)$ arithmetic progressions. We apply continue applying inclusion-exclusion until we arrive at the quantity we desire. Thus, we have

$$ \#\{\Lambda_2 \mid \abs{\Lambda_1 \cap \Lambda_2} = i\} = \binom{3}{i}\sum_{j=0}^{3-i} {(-1)^{j}\binom{i}{j}f(j-i)}. $$

Now, counting the number of pairs of 3-term arithmetic progressions $\{\Lambda_1, \Lambda_2\}$ that intersect 0, 1, 2, and 3 times, we have:
\begin{align*}
&\#\{\{\Lambda_1, \Lambda_2\} \mid \abs{\Lambda_1 \cap \Lambda_2} = 3\} = {n\choose 2} \\
&\#\{\{\Lambda_1, \Lambda_2\} \mid \abs{\Lambda_1 \cap \Lambda_2} = 2\} = 6{n\choose 2} \\
&\#\{\{\Lambda_1, \Lambda_2\} \mid \abs{\Lambda_1 \cap \Lambda_2} = 1\} = \frac{1}{2}{n\choose2}(9n-39) \\
&\#\{\{\Lambda_1, \Lambda_2\} \mid \abs{\Lambda_1 \cap \Lambda_2} = 0\}
= \frac{1}{2}{n\choose 2}(n^2-10n+25).
\end{align*}

After using these to calculate the second moment of $A_{n,k}$ by plugging the resulting expression into Mathematica, we find
\[ \sigma_{n,k}^2 = \frac{-(k-2)(k-1)k(k^3-3k^2(n-1) - n(n^2-3n+2)+k(3n^2-6n+2))}{2(n-4)(n-3)(n-2)^2}. \]
Simplifying, we have $\sigma_{n,k}^2 = \frac{k^3(n-k)^3}{2n^4} + O(n)$.
\end{proof}

\subsection{Nonexistence of LLT for $A_n$}
The reason that we cannot use the general theorem in section 6 to prove that no LLT exists for $A_n$ is that the distances between the conditioned distributions $A_{n,k}$ are the same order as their standard deviations $\sigma_{n,k}$, since $\mu_{n,k+1} - \mu_{n,k} = \Theta(\sigma_{n,k})$ for reasonably likely values of $k$ (when $k=1$, this breaks down but values of $k$ this extreme are very unlikely).

To show that there is no LLT for $A_n$, our strategy is to choose a point $x$ in the middle between two $A_{n,k}$ distributions and showing that $\abs{\P(A_n = x) - \mathcal{N}_n(x)} = \Omega(\frac{1}{\sigma_n})$, where $\mathcal{N}_n(x) = \frac{1}{\sqrt{2\pi}\sigma_n} e^{((x-\mu_n)/\sigma_n)^{2}/2}$.

Set $p = \frac{1}{2}$, and let $x \geq \mu_{n,k}$ for all $k \leq j$ and $x \leq \mu_{n,k}$ for all $k>j$. Then we have
\begin{align*}
\P(A_n = x) &= \sum_{k = 0}^n \P(A_n = x \mid |S| = k)\P(|S| = k)\\
&= \sum_{k = 0}^n \P(A_{n,k} = x) \cdot \frac{{n\choose k}}{2^n} \\
&\leq \sum_{k = 0}^j \P(A_{n,k} \geq x) \cdot \frac{{n\choose k}}{2^n} + \sum_{k = j+1}^n \P(A_{n,k} \leq x) \cdot \frac{{n\choose k}}{2^n}.
\end{align*}

Now we must sufficiently bound the tails of $A_{n,k}$. We know that $\mu_{n} = \frac{1}{8}{n\choose 2}, \sigma_{n}^2 = \Theta(n^3), \mu_{n,k} = \frac{3}{n-2} {k\choose 3}, \text{ and } \sigma_{n,k}^2 = \frac{k^3(n-k)^3}{2n^4}$, but we do not know anything about the shape of the distribution of $A_{n,k}$.

\begin{remark}
We conjecture that $A_{n,k}$ follows a local limit theorem. The main issue that we run into when trying to apply the same technique used to show LLTs for $T_n$ and $D_n$ to $A_{n,k}$ is that we cannot find an event to condition on such that $A_{n,k}$ will be the sum of independent random variables, since the constraint on size $k$ makes the presence of each element of $\mathbb{Z}/n\mathbb{Z}$ dependent on the presence of every other element. If we could prove an LLT for $A_{n,k}$, that would allow us to conclude that no LLT exists for $A_n$, as we show below. It might be possible to prove that there is no LLT for $A_n$ without proving the LLT for $A_{n,k}$ by a specialized argument, but using the following proposition would be the most satisfying way since an LLT for $A_{n,k}$ would give the most complete picture of the histogram of $A_n$.
\end{remark}

\begin{prop}
Suppose $A_{n,k}$ follows an LLT for each $k$. Then there exists no LLT for $A_n$.
\end{prop}

\begin{proof}
We show that the height at the center of an $A_{n,k}$ distribution is greater than the Gaussian distribution at that point as $n$ gets large.
Let $k = \frac{n+1}{2}$ and $x = \mu_{n,k}$. Then since $A_{n,k}$ follows an LLT,
\begin{align*}
\P(A_{n} = x) &= \P(|S| = k) \cdot \P(A_{n,k} = x) = \frac{{n\choose k}}{2^n} \cdot \frac{1}{\sqrt{2\pi}\sigma_{n,k}} e^{-\p{\frac{x-\mu_{n,k}}{\sigma_{n,k}}}^2/2} \\
&= \p{\frac{1}{\sqrt{2\pi n}} e^{-\frac{(k-\frac{n}{2})^2}{2n}}} \cdot \p{\frac{1}{\sqrt{2\pi\sigma_{n,k}^2}}} \\
& \approx \frac{1}{2\pi} \sqrt{\frac{2n^4}{nk^3(n-k)^3}} \\
&= \frac{\sqrt{2}}{\pi} \p{\frac{n}{k(n-k)}}^{3/2} \\
&\approx \frac{8\sqrt{2}}{\pi} n^{-3/2}.
\end{align*}
Now we turn our attention to the height of the Gaussian distribution at $x = \mu_{n,k}.$ Recall that $\sigma_n^2 \approx \frac{9}{4}n^3$ and let $\mathcal{N}_n$ be the probability density function for the normal distribution with mean $\mu_n$ and variance $\sigma_n^2$. Then we have
\[ \mathcal{N}_n(x) = \frac{1}{\sqrt{2\pi}\sigma_{n}} e^{-\p{\frac{x-\mu_{n}}{\sigma_{n}}}^2/2} \leq \frac{1}{\sqrt{2\pi\sigma_{n}^2}} \approx \p{\frac{1}{3} \sqrt{\frac{2}{\pi}}} n^{-3/2}.
\]
Therefore,
\[\abs{\P(X_n = x) - \mathcal{N}_n(x)} \approx \p{\frac{8\sqrt{2}}{\pi} - \frac{1}{3} \sqrt{\frac{2}{\pi}}} n^{-3/2} = \Omega\p{\frac{1}{\sigma_n}},\]
so there exists no LLT for $A_n$.

\end{proof}

\section{Acknowledgements}
We would like to thank Felipe Hernandez for his close mentorship and for providing the topic of this project. We would also like to thank George Schaeffer for his helpful guidance and input on this project and for organizing SURIM, without which this project would not have been possible.

\section{Appendix}
\subsection{3-APs in a subset of $\mathbb{Z}/n\mathbb{Z}$ and its complement}
In this section, we prove a fun fact regarding 3-term APs.
	Let $n > 3$ be odd. Consider a subset $S \subseteq \mathbb{Z}/n\mathbb{Z}$. Define $A(S)$ to be the number of 3-term arithmetic progressions in $S$. That is,
	\[ A(S) = \frac{1}{2} \abs{\{a, b \in \mathbb{Z}/n\mathbb{Z} \mid b \neq 0;\ a, a + b, a + 2b \in S \}}. \]
	Let $k = \abs{S}$, and denote the complement of $S$ in $\mathbb{Z}/n\mathbb{Z}$ as $S^c$, so $\abs{S^c} = n - k$. We prove the following proposition.

    \begin{prop}
    The sum of the number of 3-APs in $S$ and $S^c$ is constant conditioning on $n$ and $k$. That is, $A(S) + A(S^c)$ depends only on $n$ and $k$.
    \end{prop}

    Note: this method of counting considers $(x, y, z)$ and $(z, y, x)$ to be the same arithmetic progression (due to the factor of 1/2), and does not consider $(x, x, x)$ to be an arithmetic progression. However, the result still holds under every permutation of these settings.

\begin{proof}

	The total number of 3-term arithmetic progressions $A(\mathbb{Z}/n\mathbb{Z})$ is $n(n-1)$. These arithmetic progressions can be split into four groups based on the locations of their elements: all in $S$, all in $S^c$, exactly one in $S$, and exactly one in $S^c$.

	There are $A(S)$ progressions all in $S$ and $A(S^c)$ all in $S^c$. The number of progressions with exactly one element in $S$ is $\f{3}{2}k(k-1) - \f{3}{2}A(S)$ since each pair of elements of $S$ is contained in $3$ total progressions, but we need to subtract the contribution of the progressions entirely contained in $S$, each of which contains $3$ pairs. Similarly, the number of progressions with exactly one element in $S^c$ is $\f{3}{2}(n-k)(n-k-1) - \f{3}{2}A(S^c)$. Adding everything together, we get
	\[ n(n-1) = A(S) + \p{\f{3}{2}k(k-1) - \f{3}{2}A(S)} + \p{\f{3}{2}(n-k)(n-k-1) - \f{3}{2}A(S^c)} + A(S^c). \]
	Simplifying,
	\[ A(S) + A(S^c) = \f{1}{4}\p{3k(k-1) + 3(n-k)(n-k-1) - n(n-1)}. \]
	Surprisingly, this sum depends only on $n$ and $k$.

\end{proof}

Intuition tells us that $A(S)$ is a measure of structure in $S$ and the amount of structure in $S$ is the same as the amount of structure in $S^c$. But this intuition must be wrong since this calculation implies that either the amount of structure in $S^c$ is the inverse of the amount of structure in $S$, or $A(S)$ does not really measure structure.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


There's more to Riverence steelhead than great taste and texture. Rated as a Monterey Bay Seafood Watch "Best Choice" green option, our fish are respectfully raised from our own eggs alongside Idaho's Snake River. We're focused on protecting wild salmon and trout through responsible aquaculture, and being good stewards of the environment. Our job: to take good care of the resources we have, to do more with less, and to provide a safe, premium food source that will serve humanity for generations to come.


text:
Highway remains closed as Oregon fire jumps to Washington
Oregon fire jumps to Washington
TROUTDALE, Ore. -- A lengthy stretch of highway east of Portland, Oregon, remained closed Tuesday morning as crews battled a growing wildfire that has also caused evacuations and sparked blazes across the Columbia River in Washington state
Hundreds of people have been forced to evacuate because of a seven-square mile (18-square kilometers) fire that started Saturday.
The blaze, which officials think may have been caused by fireworks, forced the shutdown of more than 30 miles (48 kilometers) of Interstate 84 east of Portland through the scenic Columbia River Gorge.
Authorities said Tuesday that embers from the fire caused a new blaze across the Columbia River in Washington state.
Evacuation orders remained in place in and around Cascade Locks, Oregon, for dozens of structures and businesses.
Ash coated cars left parked outside in the Portland metro area Tuesday morning. Many school districts canceled outdoor activities, sports practices and recess because of poor air quality. Some private schools and universities in the Portland area also canceled classes Tuesday because of heat and poor air quality.
Ash from wildfires falling on cars in Pierce, King, Snohomish Counties
Meanwhile, a wildfire burning near Mount Rainier National Park in Washington state grew to more than 29 square miles (75 square kilometers) and heavy smoke blanketed many cities in Washington state.
Mandatory evacuations were announced for the fire near Mount Rainier, including the Crystal Mountain ski resort, which closed Monday because of smoke.
The air quality in Spokane was rated as hazardous on Tuesday morning. The National Weather Service says it was likely to get worse as wind shifts bring in smoke from fires in Canada, Montana, Oregon and Washington.
People in the Spokane region were advised to stay remain indoors.
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This project will evaluate to what extent "Say on Pay" provisions (shareholders' right to a nonbinding advisory vote on a company's executive compensation) plan can affect the level of executive compensation, overall employee compensation, and firm performance. These provisions have been introduced by the Dodd Frank 2010 act as compulsory for all firms but to date there is no evidence on their capacity to affect the level or structure of executive compensation. The author proposes to study similar proposals voted on in firms since 2007 to evaluate their effect. The empirical framework will provide a causal estimate of their effect.


Want to see about a minute more of Nicholas? Probably not, but AMC wouldn't give away the GOOD deleted scenes for free. Entertainment Weekly was given exclusive dibs on a deleted scene from the upcoming Walking Dead Season Six Blu-ray/DVD set.
To call the history between Nicholas and Glenn tumultuous is like calling the Arctic a bit chilly. Not only was the former a terrible ally, he made for an even worse enemy. In this deleted scene they're trying to get over the past and establish somewhat of a working relationship. We all know that this ends with Nicholas shooting himself in the head and toppling Glenn into the zombie herd. You just can't work with some people no matter what you try.
The Walking Dead: The Complete Sixth Season comes to Blu-Ray and DVD August 23.