To be honest, even with the acid spitting I'd still find it sexist to say that women should be paid less. I think I'd just ask if I could work on a different floor or be issued protective goggles.
See, what I don't get about the claim that women are paid less for the same work is, wouldn't that make them more valuable employees? And thus start a bidding war for them?
Yeah, IF the wage gap existed in the same world as pervasively equal perception of women and men. But the wage gaps exists cos of an a priori assumption about the worth of women relative to men.
Women may earn less than men with the same job title and experience, and sometimes less than men with the same job title and LESS experience.
And this starts even in their 20s or with the first promotion, well before any childbearing interruptions.
no, they earn less because less women work in high paying jobs, because they more often don't need to.
We might add that it turns out one need not even be a woman; the audience merely needs to think you're a woman, and they treat you worse (MacNell, Driscol, and Hunt, 2014; Ottoboni, Boring, and Stark, 2016). Furthermore, if a man wants to get ahead, he should "talk like a woman" (Jones, 2016).
Honestly, I'm thinking women might need to spit acid before society treats them properly.
No, the wage gap only exists inside the heads of deluded leftists.
The problem with bringing regional and national problems onto the internet is that some people reflexively think they are inherently global problems.
I don't know much about wages in america, but I DO know that women in Denmark pick up on this and start complaining, as well.
Except women in Denmark are not underpaid when you factor in the type of job and the hourly wage.
Women here tend to take certain jobs, that are not as well paid as those men typically seek. This is a personal choice with a personal consequence.
Women here also take part time jobs more often than men. This is also a personal choice.
So please keep this in mind, when you flaunt your regional and national problems.
Don't be rediculous, America is the world.
They are not even the continent despite assertions to the contrary.
So says the gentleman doing so.
The comic is ambiguous, which is good. Is Adam criticizing the practice of paying women less? Or is he having fun on people who believe that women are paid less? We can't tell. We can read our personal points of view into it, and either enjoy a laugh or be outraged, as we are inclined.
The topic has been done to death by both sides, so I won't go into it. Both sides insist that they have the facts and that the other side is misleading with their fake numbers…. So, whatever.
Masculine insecurity is a topic that has been done to death, but only by insecure men.
My boss says there are no wage discrepancies and I have been encouraged to believe her.
from germany, but it holds true in other areas as well. Self employed women earn less than self employed men. by a bigger margin than the "pay gap" in the general population.
I find the idea of internalized misogyny completely bogus. What you are saying is that the majority of self employed women secretly hate themselves to a degree, and thus don't even try to earn as much as possible in self-employed jobs.
Internalize Misogyny might be a real thing, but it's obviously only used to silence women who do not agree with with feminism.
sounds to me like they're the ones with internalized misogyny.
It could be that the quality of work from women is assumed to be worse so they don't do as well. This was in the news today. A husband and wife switched names for two weeks and the complaints about her work stopped when she used her husband's name and he started getting complaints that he had never had before.
Well, women already received approximately equal pay for equal work. The 77 cents to the dollar statistic is comparing all women to all men who work. When the comparison is between women and men in similar jobs with similar experience, the gap virtually disappears.
It's worse then that, the supposed pay-gap statistic doesn't even account for the difference in hours worked.
Because as it turns out Men are more likely to work full time while Women prefer Part-time.
It's like how transgenders can't go to the bathroom because they're actually evil omnipotent aliens whose powers become magnified the closer they are to a congregation of toilets.
I fully agree on this. Transgendered toilets are evil.
frankly, the whole concept of gendered bathrooms is silly.
How do you feel about rock people who hate men?
You've opened a can of snakes here, Adam.
Say what you will about the empire and the first order but I'm willing to bet captain Phasma got equal pay. That may be why they couldn't afford rails but I wouldn't know until I really dug into the ledgers and I'm still not sure which books are canon….
How long do you have to work in one job before the acid-spitting ability is achieved? I want that. The egg laying would just be a bonus.
I wish I could spit acid, I wouldn't have to be paranoid about strange men and/or dark alleys that way.


text:
Home » NHRA
Professor Says Student Was A Quick Study
Pro Stock driver Warren Johnson is still smokin' at age 67. (Photo courtesy of the Full Throttle NHRA Drag Racing Series)
Commerce, Ga. – As part of the NHRA's 60th anniversary celebration, Warren Johnson, the 67-year-old Pro Stock veteran from nearby Buford, will be honored during this weekend's Summit Southern Nationals at Atlanta Dragway.
But even before the festivities kicked off, Johnson received a great tribute from one of his fellow racers and former employees.
Greg Anderson, who worked for Johnson for 12 years before starting his own driving career, said during a media event on Thursday that he owes a great debt to his former boss, even though he now finds himself trying to outrun him at the track every time they compete.
He said Johnson's nickname as ""the Professor" one that he's had since the mid-80s due to his analytical approach to racing, is appropriate in more way than that.
Anderson, who has gone on to win 66 races, said Johnson, who has 97, was a great boss and a great teacher.
"It was the time of my life," Anderson said of his years working for Johnson. "I had a blast. I learned a ton. I really appreciate working for him and everything he did for me. Quite honestly, I never thought I'd leave there. I thought that was the cat's meow, the coolest job in the world. I enjoyed the heck out of that."
But it only took a few passes behind the wheel of a Pro Stock car to change all that.
Greg Anderson learned his lessons well from Warren Johnson.
"One day I got the itch to start driving," Anderson said. "(Johnson) caused that. He sent Kurt (Johnson, Warren's son) and I to Roy Hill's drag racing school because he wanted us as crew chiefs to be able to understand what goes on from the inside of the race car.
"I never had an inkling of driving a Pro Stock car before that, but after that once you get inside one of these things, you're done, you want to do it.
"It took me three years to be able to get a ride. Kurt started right after he got his license. I don't think Warren expected it to happen to either of us. He just tried to make us better crew chiefs."
Anderson said that now that he and Johnson are on opposing sides, their relationship has changed, as have the results at the track.
"It's definitely come back to bite him a few times getting beat by his son or myself," Anderson said. "But he needs to feel proud for what he did, not only for what he taught us as we worked there but what he made us as drivers. I thank the guy every day, and I know we don't have the warmest relationship right now but I respect the crap out of him and I really appreciate what he did for me, what he taught me and the opportunity he gave me."
Anderson went on to say that the Professor title is quite appropriate.
"Professor fits perfectly," he said. "He doesn't allow it to do that much for him, and he should. He should take credit for what I've become and for what Kurt has become. He truly is the Professor."
Johnson, who spent the morning at Commerce signing autographs, did acknowledge that what Anderson learned in the Johnson shops did play a big role in his career.
"I'm sure it accelerated his program," Johnson said. "We were always on the cutting edge of a lot of new stuff. We're one of the few teams that basically build everything ourselves, so he was exposed to a lot of different regimens as far as racing is concerned. We designed our own cylinder heads. We did everything in-house.
"He saw a lot of that in his time with us, so that probably has stuck with him. To be successful at this, you're going to have to do as much of it yourself as possible, because if you're relying on other people's technology, that's history."
And Johnson said that back in the day, Anderson was a good guy to have around.
"Greg was a good employee," Johnson said. "I wouldn't have kept him around for 12 years if he wasn't."
– Rick Minter can be reached at rminter@racintoday.com

redpajama_set_name: RedPajamaCommonCrawl


The cast of the Afro Vibes' Miss Thadi Show flew out of Amsterdam to Johannesburg on the 1st of July to perform at the Grahamstown National Arts Festival.
The show opened on the 5th of July at the Grand Mousseux Room, a venue at the Settler's Inn. The press attended (see reviews), and amongst the audience were representatives from the National Arts Council of South Africa, Funiwe and Albert. Hans Glaubitz, cultural attach� of the Dutch Embassy in Pretoria and the NAC CEO D Nteta were amongst the guests who attended during the run of the show. Family and friends were also present amongst the wonderful audiences and supporters of the Miss Thandi Show. The shows ran quite smoothly.
The cast also attended two receptions, one by the Dutch Embassy which was hosted for the NDT1 (Nederlands Dans Theater) and another by the NAC of South Africa.
Some of the Miss Thandi Shows were filmed by a TV-crew from Amsterdam for a documentary that will be shown in the Netherlands. They also covered bits and pieces of the festival and the landscape of South Africa.
The Miss Thandi Show moved from Grahamstown to Port Alfred and performed at the Civic Centre, this show was opened by the mayor of Port Alfred Mr Hani. We gave a local kwaito group of young guys called "the Supremes" an opportunity to upstage Miss Thandi.
The next morning the Miss Thandi Show Cast members gave workshops to the talented people of Port Alfred, who have no resources. The cast found itself discussing issues around carriers and gender in the performing arts, most of the girls were impressed by the female cast members who play the bass and keyboard and also the ups and downs of Raymond the story from Port Alfred to Amsterdam. Most of these young people were very motivated and one of the things they stressed was the need of someone in the community who could guide and help with information and a network so they can further developing performing skills. We had a lot to share.
In the afternoon Raymond met the Rotarians and discussed few ideas that will uplift the young and the old people of Port Alfred including the Scholarship Project that Afro Vibes has launched. Later on the workshops continued with the steelband project that was established this year by the Rotarians for the disadvantaged youth. We hope that we have planted good seeds and left a good impression to these young people and other citizens of Port Alfred and Grahamstown and in South Africa.


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The International Monetary Fund and the World Bank jointly released a paper intended to guide policymakers around the world in their handling of the rise of financial technology — commonly known as fintech.
The paper, called the Bali Fintech Agenda, outlines 12 considerations that the IMF, the World Bank and governments can keep in mind when designing policies and regulations that can maximize the benefits of fintech while keeping financial systems sound.
The International Monetary Fund and the World Bank jointly released a paper that will guide policymakers around the world in their handling of the rise of financial technology — commonly known as fintech.
The paper, called the Bali Fintech Agenda, was launched on Thursday on the Indonesian island where the IMF and the World Bank are holding their annual meetings.
The paper outlines 12 "elements" or considerations that the IMF, the World Bank and governments can keep in mind when designing policies and regulations that can maximize the benefits of fintech while keeping financial systems sound.
Those "elements" include using fintech to promote financial inclusion, allowing new technology players to have level playing fields with existing companies and having countries work together to protect the global financial system.
Fintech has the potential to reach the 1.7 billion adults in the world that don't have access to financial services, IMF Managing Director Christine Lagarde said in a statement.
But, new technology could threaten existing financial systems. For example, volatility in the price of cryptocurrencies has raised concerns about investor protection, according to the paper.
"Fintech can have a major social and economic impact for them and across the membership in general. All countries are trying to reap these benefits, while also mitigating the risks," Lagarde said.
"We need greater international cooperation to achieve that, and to make sure the fintech revolution benefits the many and not just the few," she added.
World Bank Group President Jim Yong Kim said fintech would be particularly helpful to low-income countries, where access to financial services is low.
Both organizations said the paper doesn't represent current work, nor does it aim to provide specific guidance or policy advice. They will, however, start to develop specific programs on fintech.
The IMF will focus initially on the implications on monetary and financial stability and how international monetary systems and global financial safety nets evolve. The World Bank will work on using fintech to deepen financial markets, enhance responsible access to financial services, and improve cross-border payments and remittance transfer systems.


text:
NBA Rookie-Sophomore Game 2013: Time, TV schedule and more
Florida basketball fans can catch out Orlando's Andrew Nicholson and Nikola Vucevic in the game.
By David Cassilo Feb 15, 2013, 2:59pm EST
Share All sharing options for: NBA Rookie-Sophomore Game 2013: Time, TV schedule and more
The best rookies and sophomores in the NBA will take the floor on Friday night, as basketball fans get a peak at the stars of tomorrow. Headlining the field is Kyrie Irving, who's also in the NBA All-Star game. He will be looking to win MVP of Friday's game for the second straight season.
The teams, which were selected by Shaquille O'Neal and Charles Barkley, will feature budding stars like Irving, Damian Llilard, Anthony Davis and Kenneth Faried.
The event will have some local flavor for Florida basketball fans, as the Magic's Andrew Nicholson will be facing off against Nikola Vucevic.
The showdown, along with the celebrity game, kick off the All-Star festivities this weekend in Houston. For full coverage, click here.
Game time: 9 p.m. ET
TV schedule: TNT
Radio schedule: ESPN Radio
Announcers: Kevin Harlan, Chris Webber and Mike Fratello.
TEAM SHAQ: Damian Lillard (Portland), Kyrie Irving (Cleveland), Andrew Nicholson (Orlando), Klay Thompson (Golden State), Harrison Barnes (Golden State), Chandler Parsons (Houston), Dion Waiters (Cleveland), Michael Kidd-Gilchrist (Charlotte), Tyler Zeller (Cleveland), Kemba Walker (Charlotte)
TEAM CHUCK: Anthony Davis (New Orleans), Kenneth Faried (Denver), Kawhi Leonard (San Antonio), Bradley Beal (Washington), Ricky Rubio (Minnesota), Tristan Thompson (Cleveland), Nikola Vucevic (Orlando), Brandon Knight (Detroit), Isaiah Thomas (Sacramento), Alexey Shved (Minnesota)


text:
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Is my Divorce recognised in England?
How do I know if my divorce is legal?
When considering divorce as an expat, most people's first thought isn't "will my divorce be recognised in England?". Surely being divorced in one country means that you would be divorced in any country? This is not actually the case, and it is something that is very important to bear in mind when considering where to proceed with your divorce.
The starting point in considering the question of "is my divorce legal", is to look at where the divorce took place.
If the divorce took place in the United Kingdom
A divorce or annulment that took place in any part of the British Isles, will be recognised in England if it was conducted through formal proceedings by a court. The Decree Absolute is sent out to all parties in the divorce, and is the required evidence of the divorce.
A religious divorce that took place in the United Kingdom (such as a Muslim Talaq, or Jewish Get) will not be recognised in England. However, in certain circumstances, a religious divorce that took place abroad will be recognised (see below).
If the divorce took place within Europe
Where a foreign divorce is conducted in any European country (apart from Denmark), it will automatically be recognised in other European countries. Therefore, a divorce conducted within Europe will be recognised in England, and vice versa, an English divorce will be recognised in a European country (except Denmark).
If the divorce took place outside Europe
Divorces that take place outside Europe are not always recognised, and this is an important consideration before beginning the divorce process as an expatriate. The English court distinguishes between divorces conducted through 'proceedings' and divorces conducted 'other than through proceedings'. So, what does this mean? Putting it simply, a 'proceedings' divorce requires the involvement of the court in that country, it has to be a formalised process involving the court. A divorce that is 'other than through proceedings' generally refers to a more relaxed approach, that is usually done through administrative avenues only, and is conducted by a body other than the court, such as a foreign embassy.
Religious divorces, such a divorce by the pronouncement of the Talaq, may or may not involve proceedings depending on the process chosen and the country involved.
An overseas divorce that is obtained through 'proceedings' is recognised if:
It is valid under the law of the country in which it was obtained; and
At the time of the divorce, either party to the marriage was
resident in the country in which the divorce took place;
domiciled in the country in which the divorce took place; or a national of the country in which the divorce took place.
By comparison, an overseas divorce that is obtained 'other than through proceedings', is only recognised if:
at the time of the divorce
Both parties were domiciled in the country in which the divorce took place; or
either party to the marriage was domiciled in the country in which the divorce took place and the other party was domiciled in a country that would recognise the divorce; and
neither party to the divorce was habitually resident in the United Kingdom for a year prior to the divorce.
If you are considering getting divorced overseas and want to find out whether it will be recognised in England, please do get in touch with one of our experienced lawyers. You can find our contact details on our website: https://expatriatelaw.com/our-lawyers/
Written by: Hannah McCrindle, Trainee Solicitor at Expatriate Law
You may also wish to read:
Is my Marriage Recognised in England?
Divorce Dictionary: Legal Terms explained
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text:
Trump demands 'termination' of Constitution, his reinstatement over Twitter censorship files
Former President of the United States Donald Trump speaking with supporters at a "Save America" rally at Country Thunder Arizona in Florence, Arizona. (Gage Skidmore/Flickr)
December 05, 2022 Justin Cooper
Former President Donald Trump called for "termination" of the Constitution and reversal of the 2020 election after emails were released revealing internal conversations at Twitter about suppression of the Hunter Biden laptop story.
The New York Post story suggesting corruption in the family of then-presidential candidate Joe Biden was released a month before the 2020 election. The emails, reported by independent journalist Matt Taibbi, show Twitter officials coordinating to justify restrictions on sharing the story amid concerns their actions could be seen as political censorship.
Emails also showed Twitter workers deleting tweets at the request of "Biden team," a privilege Taibbi said the social media company granted to both political parties. However, Taibbi noted that the system "wasn't balanced," writing that it "was based on contacts."
"Because Twitter was and is overwhelmingly staffed by people of one political orientation, there were more channels, more ways to complain, open to the left (well, Democrats) than the right," he explained.
Trump, now in his presidential campaign, seized on the emails to up the ante on his long-running push to overturn the 2020 election. After previously trying to stay in power through a constitutional loophole, he claimed it's now time to go beyond what's allowed in the nation's cornerstone document.
"Do you throw the Presidential Election Results of 2020 OUT and declare the RIGHTFUL WINNER, or do you have a NEW ELECTION?" he said on his Truth Social platform. "A Massive Fraud of this type and magnitude allows for the termination of all rules, regulations, and articles, even those found in the Constitution. Our great 'Founders' did not want, and would not condone, False & Fraudulent Elections!"
The Biden administration responded the same day, PBS reported, with a White House spokesman saying, "You cannot only love America when you win."
"The American Constitution is a sacrosanct document that for over 200 years has guaranteed that freedom and the rule of law prevail in our great country," the spokesman, Andrew Bates, said. "Attacking the Constitution and all it stands for is anathema to the soul of our nation."
Top elected Republicans have been quiet so far, NBC News reported, but some lawmakers have made comments.
Rep. Mike Turner (R-OH) said on CBS that Trump's comment was "certainly not consistent with the oath that we all take." On CNN, Rep.-elect Mike Lawler (R-NY) said, "Obviously, I don't support that."
"The Constitution is set for a reason to protect the rights of every American," Lawler said. "Frankly, I think people are tired of looking backward. I think people are tired of discussing the grievances of prior elections."


text:
Anatoli Roshchin (Riazan, Unión Soviética, 10 de marzo de 1932-San Petersburgo, 5 de enero de 2016) fue un deportista soviético especialista en lucha grecorromana donde llegó a ser campeón olímpico en Múnich 1972.

Carrera deportiva

En los Juegos Olímpicos de 1964 celebrados en Tokio ganó la medalla de plata en lucha grecorromana estilo peso pesado, tras el luchador húngaro István Kozma (oro) y por delante del alemán Wilfried Dietrich (bronce). Cuatro años después, en las Olimpiadas de México 1968 volvió a ganar la plata en la misma modalidad. Y en las de Múnich 1972 ganó la medalla de oro en la categoría de más de 100 kg.

Referencias

Enlaces externos

Luchadores de la Unión Soviética
Medallistas olímpicos de Tokio 1964
Medallistas olímpicos de México 1968
Medallistas olímpicos de Múnich 1972
Fallecidos en San Petersburgo
Miembros del Partido Comunista de la Unión Soviética

redpajama_set_name: RedPajamaWikipedia


text:
Q: What does my min-sdk version need to be in order to use GoogleApiAvailability? I'm replacing GooglePlayServicesUtil with GoogleApiAvailability and want to know if I need a certain min SDK.  Right now my min-sdk is 21. Thanks!

A: https://developers.google.com/android/guides/setup
states: "The Android emulator with an AVD that runs the Google APIs platform based on Android 4.2.2 or higher."
So that's 17 or higher

redpajama_set_name: RedPajamaStackExchange


Liège, also known as the 'ardent city', does everything to honour its nickname. Plenty of museums and the famous Liège aquarium, there is always something to see or to visit. Let yourself be carried away by the unique ambiance, present throughout the year.


text:
\section{Introduction}
\def\theequation{1.\arabic{equation}}
\setcounter{equation}{0}

\subsection{Inverse regression models} \label{sec1}
In many applications it is impossible to observe a certain quantity of interest because only indirect observations are available for statistical inference. Problems of this type are called inverse problems and arise in many fields such as medical imaging, physics and biology. Mathematically the connection between the quantity of interest and the observable one can often be expressed in terms of a linear operator equation.
Well known examples are Positron Emission Tomography, which involves the Radon Transform, (\cite{cavalier2000}), the heat equation (\cite{mairuy1996}), the Laplace Transform \cite{saitoh1997} and the reconstruction of astronomical and biological images from telescopic and microscopic imaging devices, which is closely connected to convolution-type operators (\cite{adorf1995}, \cite{berbocdes2009}).

Inverse problems have been studied intensively in a deterministic framework and in mathematical physics. See for example \cite{enghanneu1996}  for an overview of existing methods in numerical analysis  of inverse problems  or \cite{saitoh1997} for techniques based on
 reproducing kernel Hilbert spaces.
Recently, the investigation of inverse problems has also become of importance from a statistical point of view. Here, a particularly interesting
and active field of research  is the construction of statistical inference methods such as   hypothesis tests or confidence regions.

In this paper we are interested in the convolution type inverse regression model
\begin{equation} \label{mod1}
Y = (f*\psi )(x) + \varepsilon,
\end{equation}
where  $\varepsilon$ is a random error, the operation $*$ denotes convolution, $\psi$ is a given square integrable function and the object of interest is the function $f$ itself.
An important and interesting application  of the inverse regression model (\ref{mod1}) is the recovery of images from imaging
devices such as astronomical telescopes or fluorescence microscopes in biology. In these cases, the observed,
uncorrected image is always at least slightly blurry due to the physical characteristics of the propagation of
light at surfaces of mirrors and lenses in the telescope.  In this application the
variable $x$ represents the   pixel  of a CCD and
we can only observe a blurred version of the true image modeled by the function $f$.
In the corresponding mathematical model
the observed image is (at least approximately) a convolution of the real image with the so-called
point-spread-function $\psi$, i.e. an inverse problem with convolution operator.

The inference problem regarding the function $f$ is called inverse problem with stochastic noise. In recent years, the problem of estimating
the regression function $f$  has become an important field of research, where the main focus is on a one dimensional predictor. Several authors propose Bayesian methods (\cite{berbocdes2009}; \cite{kaisom2005}) and construct estimators using tools from nonparametric curve estimation (\cite{mairuy1996}; \cite{cavalier2008}; \cite{bishohmunruy2007}). Further inference methods, in particular the construction of confidence intervals and confidence bands, are much less developed. \cite{birbishol2010}  have constructed uniform confidence bands for the
 function $f$  with a  one-dimensional predictor.

The present work is motivated by the fact that in many applications  one has to deal with an at least two-dimensional
predictor. A typical example  is image reconstruction since a picture is a two-dimensional object.
Also in addition to the spatial dimensions, the data often show a  dynamical behavior, thus repeated measurements
at different times can be used to extend the statistical inference. For example
in astrophysics spectra of different objects like supernovae or variable stars underly changes in time on observable timescales.
 In this case the function $f$ depends on a further parameter, say $f_t$ and the reconstruction problem refers to a multivariate  function even if the predictor is univariate.

The purpose of the present paper is the  investigation of asymptotic properties of estimators for the function $f$
in model \eqref{mod1} with a  multivariate predictor. {In particular we present a result on the weak convergence
of the sup-norm of an appropriately centered estimate, which can be used to construct
asymptotic   confidence bands for the regression function $f$.}
  In contrast to other authors (e.g. \cite{cavtsy2002})
 we do not assume that the function  $\psi$  in model \eqref{mod1}
 is periodic, because in  the  reconstruction  of  astronomical or biological images from telescopes or microscopic imaging devices
 this  assumption  is often unrealistic.

\subsection{Confidence bands}\label{sec2}
In a pioneering work, \cite{bicros1973a} extended results of \cite{smirnov1950}  for a histogram estimate and  constructed confidence bands for a density function of independent identical distributed (i.i.d) observations. Their method is based on the asymptotic distribution of the supremum of a centered kernel density estimator. Since then, their method has been further developed both in the context of density  and  regression estimation. For density estimation, \cite{neumann1998}  derived  bootstrap confidence bands, and \cite{ginnic2010}  derived adaptive asymptotic bands over generic sets. In a regression context, asymptotic confidence bands were constructed by \cite{eubspe1993}  for the Nadaraya-Watson and by \cite{xia1998}  for a local polynomial estimator. Bootstrap confidence bands for nonparametric regression were proposed by \cite{hall1993}, \cite{neupol1998}  and by \cite{clakei2003}. For the statistical inverse problem of deconvolution density estimation, \cite{bisdumholmun2007}  constructed asymptotic and bootstrap confidence bands, where \cite{Lounic2011}  obtained
 non-asymptotic confidence bands by using concentration inequalities. Recently, \cite{birbishol2010}  provided uniform asymptotic and bootstrap confidence bands for a spectral cut-off estimator in the one-dimensional indirect regression model with convolution operator.

 All these results are  limited to the estimation of univariate densities and regression functions, and  are not applicable in cases, where the quantity of interest depends on a multivariate predictor. In such  cases - to the best knowledge of the authors - confidence bands are not available.  One reason for this gap is that a well-established way to construct asymptotic uniform confidence bands, which uses a pioneering  result of
 \cite{bicros1973a}  as the standard tool, cannot be extended in a  straightforward manner to the multivariate case. There are substantial differences  between   the multivariate and one-dimensional case,   and for multivariate inverse problems  the  mathematical construction of confidence bands requires different and/or extended methodology.

In the present paper we will consider the problem of constructing confidence bands for the regression function in an inverse regression model with a convolution-type operator with a multivariate predictor.
The estimators and assumptions for our asymptotic theory are presented in Section \ref{sec3}, while  Section \ref{sec4} contains the main results of the paper.  In Section \ref{sec5} we consider the special case of time dependent regression functions with a univariate predictor, which originally motivated our investigations.  The arguments of Section \ref{sec6} and \ref{sec7}, which contain  all technical details of the proofs, are based on results by \cite{piterbarg1996}.
  These authors   provided a limit theorem
 for the supremum
 $$ \sup_{t\in T_n} X(t)$$
 of a stationary Gaussian field $\{X(t)\,|\,t \in \mathbb{R}^d\} $,
 where $\{T_n\subset \R^d\}_{n\in\N}$
 is   an increasing system of sets such that $\lambda^d(T_n)\rightarrow\infty$ as $n\rightarrow\infty$.
 This result generalized the multivariate extension   in
 \cite{bicros1973b}, who provided a
 limit theorem for the supremum
 $\sup_{t\in[0,T]^d} X(t),$
 as $T\rightarrow\infty$.

\section{Notation and assumptions}\label{sec3}
\def\theequation{2.\arabic{equation}}
\setcounter{equation}{0}

\subsection{Model and notations} \label{prel}
Suppose that $(2n+1)^d$ observations $(x_{\mathbf{k}},Y_{\mathbf{k}}),\mathbf{k}=(k_1,\ldots, k_d)\in G_{\mathbf{n}} := \{-n,\ldots,n\}^d$ from the model
   \begin{align}\label{model}
      Y_{\mathbf{k}}=g(x_{\mathbf{k}})+\varepsilon_{\mathbf{k}}:= (f*\psi) (x_{\mathbf{k}})+\varepsilon_{\mathbf{k}},
   \end{align}
are available, where the  function $f:\R^d\rightarrow\R$ is unknown, $\psi:\R^d\rightarrow\R$ is a known function and $
g:= f*\psi $ denotes the convolution of $f$ and $\psi$, that is
 \begin{align}\label{conv}
  g(x):=  (f*\psi) (x):=\int_{\mathbb{R}^d}f(s)\psi(x-s)\,ds.
  \end{align}
 The basic assumptions that guarantee the existence of the integral (\ref{conv}) and also assure $g\in L^2(\R^d)$ is that $f\in L^2(\R^d)$ and $\psi\in L^1(\R^d)\cap L^2(\R^d),$ which will be assumed     throughout this paper.
In model \eqref{model} the predictors
$x_{\mathbf{k}}:= \mathbf{k}\cdot\frac{1}{na_n}$ are equally spaced fixed design points  on a $d$-dimensional grid, with a
  sequence $(a_n)_{n\in\N}$ satisfying

  \begin{align*}
    na_n\rightarrow\infty \quad \text{and}\quad a_n\searrow0 \quad \text{for}\quad n\rightarrow\infty.
  \end{align*}
The noise terms $\{\varepsilon_{\mathbf{k}}\,|\,\mathbf{k}\in G_{\mathbf{n}}\}$ are a field of centered i.i.d. random variables with variance $\sigma^2:=\mathbb{E}\varepsilon_{\mathbf{k}}^2>0$ and existing fourth moments.
As a consequence of the convolution theorem and the formula for Fourier inversion we obtain the   representation
\begin{align}\label{repf}
     f(x)=\frac{1}{(2\pi)^d}\int_{\R^d}\frac{ {\cal F}g(\xi)}{{\cal F}\psi(\xi)}\exp(i\xi^Tx)\,d\xi.
  \end{align}
An estimator for the regression function $f$ can now easily be obtained by replacing the  unknown quantity ${\cal F}g={\cal F} (f*\psi)$  by an estimator $ {\cal F}\hat g$. The random fluctuations in the estimator  $ {\cal F}\hat g$ cause instability of the ratio $\frac{{\cal F}\hat g(\xi)}{{\cal F}\psi(\xi)}$ if at least one  of the components of $\xi$ is large.
As a consequence, the problem at hand is  ill-posed and requires regularization. We address this issue by excluding large values of $\xi_j$ for any $j=1,\ldots,d$ from the domain of integration, i.e. we multiply the integrand in (\ref{repf})
with a sequence of Fourier transforms   ${\cal F} \eta(h\cdot)$ of smooth functions with compact support $[-h^{-1},h^{-1}]^d$. Here  $h=h_n$ is a regularization parameter which corresponds to a bandwidth in nonparametric curve estimation  and satisfies $h\rightarrow 0$ if $n\rightarrow\infty.$
 For the exact properties of the function $\eta$ we refer to Assumption A below. \\
An estimator $\hat f_n$ for the function $f$ in model (\ref{model}) is now easily obtained as
  \begin{align}\label{est}
    \hat f_n(x)=\frac{1}{(2\pi)^d}\int_{\R^d}\frac{ {\cal F}\hat g(\xi)}{{\cal F}\psi(\xi)}\exp(i\xi^Tx){\cal F}\eta(h\xi)\,d\xi,
  \end{align}
  where
  \begin{align*}
    {\cal F}\hat g(\xi)=\frac{1}{(2\pi)^{\frac{d}{2}}n^da_n^d}\sum_{\mathbf{k}\in G_n}Y_{\mathbf{k}}\exp(-i\xi^Tx_{\mathbf{k}})
  \end{align*}
  is the empirical analogue of Fourier transform of $g$.
   Note that with the definition of the kernel
   \begin{align} \label{kernKn}
    K_n(x)=\frac{1}{(2\pi)^{\frac{d}{2}}}\int_{\R^d}\frac{{\cal F}\eta(\xi)}{{\cal F}\psi(\frac{\xi}{h})}\exp(i\xi^Tx)\,d\xi,
   \end{align}
   the estimator (\ref{est}) has the following representation
   \begin{align} \label{esti}
   \hat f_n(x)=\frac{1}{(2\pi)^{d}n^da_n^d h^d}\sum_{\mathbf{k}\in G_\mathbf{n}}Y_{\mathbf{k}}K_n\left((x-x_{\mathbf{k}}) \frac{1}{h}\right).
   \end{align}
  Note that  the kernel $K_n$ can be expressed as a Fourier transform as follows
   \begin{align*}
    K_n=\overline{{\cal F}\Bigl(\frac{{\cal F}\eta}{{\cal F}\psi(\frac{\cdot}{h})}\Bigr)}.
   \end{align*}
   The first step of the proof of our main result (see Theorem \ref{GWS} in Section \ref{sec4})  will consist of  a uniform approximation of $\hat f_n(x)-\E \hat f_n(x)$ by an appropriate stationary Gaussian field.
In the second step, we apply  results of \cite{piterbarg1996}  and \cite{bicros1973b}
to obtain the desired uniform convergence  for the approximation process of the first step.  Finally, these results are used to construct uniform confidence regions for $\E \hat f_n(x)$.
{Our approach is then based on undersmoothing: the choice of sufficiently small bandwidths assures the same limiting behaviour of $\hat f_n(x)-\E \hat f_n(x)$ and $\hat f_n(x)-f(x)$. This avoids the  estimation of higher order derivatives, which often turns out to be difficult in applications.}

 Thus, the limit theorem obtained in the second step will also provide uniform confidence regions for the function $f$ itself.   Whereas undersmoothing implies  that the rate-optimal bandwidth cannot be used, there has also been some theoretical justification why this choice of the regularization parameter is useful for  constructing confidence intervals (see \cite{hall1992b}).

   \subsection{Assumptions} \label{subsecass}
We now introduce the necessary assumptions  which are required for the proofs of our
main  results in Section \ref{sec4} . The first  assumption refers to the type of (inverse) deconvolution problem describing
    the shape of the kernel function $\eta$ in the spectral domain.

    \medskip

 {\bf Assumption A.} Let ${\cal F}\eta$ denote the Fourier transform  of  a function $\eta$ such that
  \begin{enumerate}
     \item [A1.]   supp$\bigl({\cal F}\eta\bigr)\subset[-1,1]^d$.
     \item [A2.]  ${\cal F}\eta\in\mathscr{D}(\R^d)=\{f:\R^d\rightarrow\R\,|\,f\in C^{\infty}
     (\R^d),\,\text{supp}(f) \subset\R^d\;\text{compact}\}$.
     \item [A3.] There exists a constant $D>0$, such that ${\cal F}\eta(\xi)=1$ for all $\xi\in [-D,D]^d$ and
     $|{\cal F}\eta(\xi)|\leq1$ for all $\xi\in\R^d.$
   \end{enumerate}

  \begin{Remark} \ {\rm
    \begin{itemize}
      \item[1.] The decay of the tails of the kernel $K_n$ is given in terms of the smoothness of the  integrand in (\ref{kernKn}). The choice of a smooth regularizing
       function ${\cal F}\eta$ has the advantage that the smoothness of $1/\cal{F}\psi$ carries over to   ${\cal F}\eta(h\cdot) /\cal{F}\psi.$
      \item[2.] Functions like ${\cal F}\eta$ are called bump functions. Their existence  follows from the $C^{\infty}$ Urysohn Lemma (see for example \cite{folland1984}, Lemma 8.18).
      \item[3.] Note that $\mathscr{D}(\R^d)\subset \mathscr{S}(\R^d),$ where $\mathscr{S}(\R^d)$ denotes the Schwartz space of smooth and  rapidly decreasing functions. Since ${\cal F}:\mathscr{S}(\R^d)\rightarrow\mathscr{S}(\R^d)$ is a bijection (see for example \cite{folland1984}, Corollary 8.28)
      we know that $\eta \in \mathscr{S}(\R^d)$
               as well.
                  \item[4.] For the sake of transparency, we state the conditions and results with the same  regularization parameter $h$ for each direction. In practical applications this might not be the best strategy. The results   presented in Section  \ref{sec4}   and \ref{sec5} also hold for different sequences of bandwidths $h_1,\ldots,h_d$ as long as the system of rectangles
                  $\{[0,h_1^{-1}]\times,\ldots,[0,h_d^{-1}]\,|\,n\in\N\}$ is a blowing up system of sets in the sense of   Definition 14.1 in \cite{piterbarg1996}. This is the case if the assumption
                  \begin{align*}
                    \sum_{p=1}^{d}\Bigl(\prod_{j=1,j\neq p}^{d}\frac{1}{h_j}\Bigr)\leq L_1
                    \cdot\Bigl(\prod_{j=1}^d\frac{1}{h_j}\Bigr)^{L_2},
                  \end{align*}
           is satisfied for a constant $L_1$ that only depends on $d$ and a constant $L_2<1.$
           This condition is not a restriction in our setting because it holds whenever $h_j\cdot n^{\gamma_j}\rightarrow C_j$ for constants $C_j, \gamma_j>0,$  $ \ j=1,\ldots,d.$
  \end{itemize}}
  \end{Remark}
  \medskip

In general, two kinds of convolution problems are distinguished in the literature, because the decay of the Fourier transform of the convolution function $\psi$ determines the degree of ill-posedness. In the case of an exponentially decreasing Fourier transform ${\cal F}\psi$   the problem is called severely ill-posed. In the present paper the class of moderately ill-posed problems is considered, where the Fourier
 transform of the convolution function decays at a polynomial rate (the precise condition will be specified in  Assumption B below).
   Throughout this paper
 \begin{align*}
      {\cal W}^m (\mathbb{R}^d) =\{f\in L^2(\R^d)\,|\,\partial^{(\alphaup)}f\in L^2(\R^d)\,\text{exists}\,\forall \,\alphaup\in\N^d,\,|\alphaup|\leq m \},
      \end{align*}
       denotes the Sobolev space of order $m \in \mathbb{N}$, where $\partial^{(\alphaup)}f$ is the
      weak derivative of $f$ of order $\alpha$. In the subsequent discussion we will also make use of
      the Sobolev space for general    $m>0$, which is defined by
     \begin{align*}
       {\cal W}^m (\mathbb{R}^d) =\{f\in L^2(\R^d)\,|\, (1+|\xi|^2)^{\frac{m}{2}}{\cal F}f\in L^2(\R^d)\}.
     \end{align*}

\medskip

{\bf Assumption B.} We assume the existence of a function $\Psi:\R^d\rightarrow\R$  such that the kernel $K=\overline{{\cal F}\left(\Psi\cdot{\cal F}\eta\right)}$ satisfies
 \begin{itemize}
   \item[B1.] $K\neq0$ and
    there exist    constants $\beta  > d/2$, $M\in \N,$ indices $0<\mu_1<\mu_2<\ldots<\mu_M$ and $L^2$-functions
       $f_1,\ldots,f_{M-1},f_{M}:\R^d\rightarrow\R$ with the property
       $$\xi^{\alphaup}f_p\in \mathcal{W}^m(\R^d) ~~~~ (p=1,\ldots,M-1) $$ for all multi-indices $\alpha \in \{0,\dots,d\}^d,\,|\alphaup|\leq d$ and all
      $m>\frac{d+|\alphaup|}{2}$,  such that
       \begin{align} \label{expan}
            h^{\beta}K_n(x)-K(x)=\sum_{p=1}^{M-1}h^{\mu_p}\overline{{\cal F}f_p}(x)+h^{\mu_M}\overline{{\cal F}f_{n,M}}(x).
         \end{align} where  $f_M$  may depend  on $n,$ i.e. $f_M=f_{M,n}$ and $\|f_{M,n}\|_{L^1(\R^d)}=O(1)$.

     \item[B2.] $\xi^{\alphaup}\Psi\cdot{\cal F}\eta,  \ \xi^{\alphaup}\frac{h^{\beta}}{{\cal F}\psi(\frac{\cdot}{h})}\cdot{\cal F}\eta\in\mathcal{W}^m(\R^d)$  for some $m>\frac{d+|\alphaup|}{2}$.

     \item[B3.] $\log(n)\cdot h^{\mu_M}(a_n^{-\frac{d}{2}}h^{-\frac{d}{2}})\cdot\|f_{M}\|_{L^1(\R^d)}=o(1)$ and  $h^{\mu_{1}}(\log(n))^2=o(1)$.
 \end{itemize}

\begin{Remark}{\rm
Assumption B1 implies   $h^{\beta}K_n\rightarrow K$ in $L^2(\R^d)$   and also specifies the order of this convergence.   It can be understood as follows. If the convergence of the difference $h^{\beta}K_n-K$ is fast enough, i.e.
\begin{align} \label{case1}
\log(n)\cdot h^{\mu_1}(a_nh)^{-\frac{d}{2}}=o(1)
\end{align}
 we have $M=1$.
 On the other hand, in some relevant situations   (see Example \ref{Ex1} (ii) below)
 the rate of convergence $h^{\mu_1}$ is given by $h^2$ for each $d$ and \eqref{case1} cannot  hold for $d\geq4.$ Here, the expansion (\ref{expan}) provides a structure,  such that our main results remain correct although the  rate of convergence is not very fast. We can decompose the difference $h^{\beta}K_n-K$  in two parts, where one part depends  on $n$ only through the factors $h^{\mu_p}$ and the other part converges sufficiently fast (in some cases this term vanishes completely).}
 \end{Remark}

  \begin{Example}\label{Ex1} {\rm
  This example illustrates the construction of the functions in the representation \eqref{expan}.
  \begin{itemize}
  \item[(i)] Let $d=2$ and $\psi(x)=\frac{1}{4}\exp(-|x_1|)\exp(-|x_2|),$ $x=(x_1,x_2)^T$, $\xi=(\xi_1,\xi_2)^T.$ Then we have
   $\frac{h^4}{{\cal F}\psi(\xi)}=2\pi\bigl(h^4+h^2(\xi_1^2+\xi_2^2)+\xi_1^2\xi_2^2\bigr)$,
   which implies $\beta=4,\,M=3$ and
    \begin{align*}
      h^4\cdot K_n(x)&=\int_{\R^2}\bigl(h^4+h^2(\xi_1^2+\xi_2^2)+\xi_1^2\xi_2^2\bigr){\cal F}\eta(\xi)\exp(ix^T\xi)\,d\xi\\

      K(x)&=\int_{\R^2}{\cal F}\eta(\xi)\xi_1^2\xi_2^2\exp(ix^T\xi)\,d\xi.
    \end{align*}
    With the definitions $f_1(\xi)=2\pi(\xi_1^2+\xi_2^2){\cal F}\eta(\xi)$, $f_2(\xi)=2\pi{\cal F}\eta(\xi)$ and $f_{n,3}\equiv0$ we obtain
     \begin{align*}
       h^4\cdot K_n(x)-K(x)=h^2\cdot\overline{{\cal F}f_1(\xi)}+h^4\cdot\overline{{\cal F}f_2(\xi)}.
     \end{align*}
   In this example, the condition $\log(n)h^2/\sqrt{a_n^dh^d}=o(1)$ is satisfied. However, the following results are valid if the weaker condition of a    decomposition of the form (\ref{expan}) holds.
   Furthermore, since the factors of $\mathcal{F} \eta$ in  $f_1$ and $f_2$ are polynomials, we have ${\cal F}f_j(\xi)\in \mathscr{S}(\R^d)$, which   implies $\xi^{\alphaup}f_j\in \mathcal{W}^m(\R^d)$ for all $\alphaup$ and all $m\in\N.$
  \item[(ii)] If $|x|=\sqrt{x_1^2+\ldots+x_d^2}$ and  $\psi (x)=2^{-\frac{d+1}{2}}e^{-|x|}$   we have
    \begin{align*}
      {\cal F}\psi(\xi)=\frac{1}{\sqrt{2\pi}}\Gamma\left(\frac{d+1}{2}\right)\frac{1}{(1+|\xi|^2)^{\frac{d+1}{2}}},
    \end{align*}
    [see \cite{folland1984}, Exercise 13].
 If $d$ is odd we use   the identity
   \begin{align*}
     \Bigl(h^{2}+|\xi|^2\Bigr)^{\frac{d+1}{2}}=\sum_{j=0}^{\frac{d+1}{2}}\binom{\frac{d+1}{2}}{j}h^{2j}|\xi|^{\frac{d+1}{2}-2j},
   \end{align*}
  and an expansion of the form \eqref{expan} is obvious from the definition of $K_n$ in \eqref{kernKn}. If the dimension $d$ is even the situation is more complicated. Consider for example the case $d=4$, where
   \begin{align*}
    \frac{h^{5}}{{\cal F}\psi(\frac{\xi}{h})}=
    \;\rightarrow\; \frac{\sqrt{2\pi}}{\Gamma\bigl(\frac{5}{2}\bigr)}|\xi|^{5}
    =\frac{\sqrt{2\pi}}{\Gamma\bigl(\frac{5}{2}\bigr)}\sqrt{\bigl(\xi_1^2+\xi_2^2+\xi_3^2+\xi_4^2\bigr)^{5}}\quad\text{as}\quad n\rightarrow\infty.
  \end{align*}
 It follows that the constant $\beta$ and the functions $\Psi, K_n$ and $K$ from Assumption B are  given by $\beta=d+1=5$,   $\Psi(\xi) =\frac{\sqrt{2\pi}}{\Gamma\bigl(\frac{5}{2}\bigr)}|\xi|^{5}$ and
\begin{eqnarray*}
 h^{\beta}K_n(x)
   &=&\frac{1}{(2\pi)^{2}}\int_{\R^d}\frac{\sqrt{2\pi}}{\Gamma\bigl(\frac{5}{2}\bigr)}\Bigl(h^{2}+|\xi|^2\Bigr)^{\frac{d+1}{2}}
   {\cal F}\eta(\xi)\exp(i\xi^Tx)\,d\xi, \\
   K(x)&=&\frac{1}{(2\pi)^{2}}\int_{\R^d}\frac{\sqrt{2\pi}}{\Gamma\bigl(\frac{5}{2}\bigr)}|\xi|^{d+1}
   {\cal F}\eta(\xi)\exp(i\xi^Tx)\,d\xi,
  \end{eqnarray*}
    respectively.
In order to show that Assumption B1  holds in this case we use
   Taylor's Theorem and obtain
    \begin{align*}
      \frac{h^{5}}{{\cal F}\psi(\frac{\xi}{h})}-\Psi(\xi)&=
      \frac{\sqrt{2\pi}}{\Gamma\left(\frac{d+1}{2}\right)}\Bigl( h^2\cdot\frac{5}{2}\cdot|\xi|^3+h^4\cdot\frac{5}{2}\cdot\frac{3}{2}\cdot(|\xi|^2+\lambda_dh^2)^{\frac{1}{2}}
      \Bigr),
    \end{align*}
    for some constant $\lambda_d\in [0,1)$.
Recalling the definition of $K_n$ in \eqref{kernKn} this gives

    \begin{align*}
      (h^{\beta}K_n-K)(x)&=
      h^2\;\overline{{\cal F}f_1(\xi)}+h^4\;\overline{{\cal F}f_{2,n}(\xi)}
    \end{align*}
  where the functions $f_1$ and $f_{2,n}$ are defined by
  \begin{align*}
    f_1(\xi) & =\frac{1}{(2\pi)^{\frac{3}{2}}\Gamma\left(\frac{5}{2}\right)}\cdot|\xi|^3\cdot\frac{5}{2}\cdot{\cal F}\eta(\xi), \\
   f_{2,n}(x) & =
   \frac{1}{(2\pi)^{\frac{3}{2}}\Gamma\left(\frac{5}{2}\right)}\frac{5}{2}\cdot\frac{3}{2}h^4(|\xi|^2+\lambda_dh^2)^{\frac{1}{2}} \cdot{\cal F}\eta(\xi),
  \end{align*}
 respectively.
 It can be shown by a straightforward calculation that $\xi^{\alphaup}f_j\in\mathcal{W}^{6+|\alphaup|}(\R^d)$ for all $\alpha \in \{0,\dots,d\}^d$.
 \end{itemize}
 }
\end{Example}

\begin{Remark}\label{DiscAss}{\rm
In the one-dimensional regression model $\eqref{model}$, \cite{birbishol2010} assume that the kernel $K$ has exponentially decreasing tails in order to obtain asymptotic confidence bands, which, in combination with the other assumptions only allows for kernels that are Fourier transforms of $C^{\infty}$-functions with square integrable derivatives. Our  Assumption B is already satisfied if $K$ is the Fourier transform of a once weakly differentiable function with square integrable weak derivative, such that all indices of ill-posedness $\beta$ that satisfy $\beta>\frac{1}{2}$ are included if $d=1$. Moreover, the assumptions  regarding the bandwidths are less restrictive compared to \cite{birbishol2010}. }
\end{Remark}

Our final assumptions refer to the smoothness of the function $f$ and to the decay of  the convolution $f*\psi$.

\medskip

 {\bf Assumption C.}  We assume that
 \begin{itemize}

    \item[C1.] There exist  constants $\gamma >2,$ $m>\gamma+\frac{d}{2}$  such that $f\in {\cal W}^{m}(\R^d).$
    \item[C2.] There exists a constant $\nu>0$ such that
       \begin{align*}
         \int_{\R}\left|(f*\psi)(z)\right|^2(1+|z|^2)^{\nu}\,dz<\infty.
       \end{align*}
 \end{itemize}

\section{Asymptotic confidence regions}\label{sec4}
\def\theequation{3.\arabic{equation}}
\setcounter{equation}{0}

 In this section we construct asymptotic confidence regions for the function $f$ on the unit cube $[0,1]^d.$ These results   can easily be  generalized to arbitrary rectangles $\vartimes_{j=1}^d[a_j,b_j ]$ for fixed constants $a_j<b_j$   $(j=1,\ldots,d)$ and the details are omitted for the sake of brevity. We
investigate the limiting distribution of the supremum of the process $\{\tilde Y_n(x)\,|\,x\in\,[0,1]^d\}$, where
  \begin{eqnarray}\label{DS}
    \tilde Y_n(x)&=&\frac{(2\pi)^dh^{\beta}\sqrt{h^d n^{d}a_n^d}}{\sigma\|K\|_{L^2(\R^d)}}\left[\hat f_n(x)-\E\hat f_n(x)\right] \\ \nonumber
    &=&\frac{(2\pi)^dh^{\beta}}{\sigma\|K\|_{L^2(\R^d)}\sqrt{h^d n^{d}a_n^d}}\sum_{\mathbf{k}\in G_n}K_n\Bigl(\left(x-x_{\mathbf{k}}\right) \frac{1}{h}\Bigr)\varepsilon_{\mathbf{k}}.
  \end{eqnarray}
and the kernel $K_n$ is defined in \eqref{kernKn}. Note that
  \begin{align*}
    \sup_{x\in[0,1]^d}|\tilde Y_n(x)|=
   \sup_{x\in[0,h^{-1}]^d}| Y_n(x)|,
  \end{align*}
where the process
  \begin{align} \label{yn}
  Y_n(x):=  \frac{(2\pi)^dh^{\beta}}{\sigma\|K\|_{L^2(\R^d)}\sqrt{h^d n^{d}a_n^d}}
    \sum_{\mathbf{k}\in G_n}K_n\Bigl(x-x_{\mathbf{k}} \frac{1}{h}\Bigr)\varepsilon_{\mathbf{k}}
  \end{align}
can be approximated by a stationary Gaussian field uniformly with respect to
 $[0,h^{-1}]^d$.  Thus the desired limiting distribution corresponds to the limiting distribution of the supremum of a stationary Gaussian process over a system of increasing smooth sets with sufficient similarity of their speed of increase, and is therefore of Gumbel-type. The precise result is given in the following Theorem.

\newtheorem{Theorem}{Theorem}
 \begin{Theorem}\label{GWS}
 Assume that for some fixed constant $\delta\in(0,1],$ $\delta<d$ and a constant $r>\frac{2d}{d-\delta}$ the $r$-th moment of the errors exists, i.e. $\mathbb{E}|\varepsilon_{\mathbf{k}}|^r<\infty.$
  If additionally Assumptions A and B are satisfied and $\frac{\log(n)}{n^{\delta}a_n^{\delta}h^{d}}=o(1)$,  then we have
    \begin{align*}
      \lim_{n\rightarrow\infty}\Prob\left(\sup_{x\in[0,1]^d}\Bigl(|\tilde Y_n(x)|-C_{n,3}\Bigr)\cdot C_{n,3}< \kappa\right)=e^{-2e^{-\kappa}},
    \end{align*}
  where
    \begin{align*}
      &C_1=\mathbf{det}\left(\left[\frac{(2\pi)^{2d}}{\|K\|_2^2}\int_{\R^d}|\Psi(v){\cal F}\eta(v)|^2v_iv_j\,dv\right]
      ,\,i,j=1,\ldots,d\right)\\

      &C_{n,2}=\sqrt{\frac{C_1}{(2\pi)^{d+1}}}\frac{1}{h^d} \\

      &C_{n,3}=\sqrt{2\ln(C_{n,2})}+\frac{(d-1)\ln\bigl(2\ln(C_{n,2})\bigr)}{2\sqrt{2\ln(C_{n,2})}}.
    \end{align*}
  \end{Theorem}

The proof of this result is long and complicated and  therefore deferred to Section \ref{sec6} and \ref{sec7}.
In the following we   apply Theorem \ref{GWS} to construct  uniform confidence regions for the function $f$ by choosing the bandwidth such that the bias decays to zero  sufficiently fast. More precisely, if the condition
  \begin{align*}
   \log(n) \sup_{x \in[0,1]^d}\left|f(x)-\E\hat f_n(x)\right|=o\left(\bigl(h^{\beta}\sqrt{h^dn^da_n^d}\bigr)^{-1}\right)
  \end{align*}
is satisfied, it follows directly that the random quantities $\sup_{x \in [0,1]^d} |\tilde Y_n(x)|$ and
  \begin{align*}
    \frac{(2\pi)^dh^{\beta}\sqrt{h^dn^da_n^d}}{\|K\|_{L^2(\R^d)}\sigma}\sup_{x\in[0,1]^d}\left| f(x)-\hat f_n(x)\right|
  \end{align*}
have the same limiting behavior.

\newtheorem{Corollary}{Corollary}
  \begin{Corollary}\label{cor1}
 Assume that  the conditions of Theorem \ref{GWS}, Assumption C and the condition
  \begin{align*}
    \sqrt{h^dn^da_n^d}\sqrt{\log(n)}\Bigl(\frac{1}{n^3a_n^3h^2}+\frac{a_n^{\nu}}{n}+a_n^{\nu+\frac{d}{2}}+h^{\gamma+\beta}\Bigr)=o(1) \quad\text{for}\quad n\rightarrow\infty
  \end{align*}
  are satisfied. Then we have for any $\kappa \in \R $
  \begin{align*}
    \lim_{n\rightarrow\infty}\Prob\left(\hat f_n(x)-\Phi_{n,\kappa}\leq
     f(x)\leq\hat f_n(x)+\Phi_{n,\kappa}\;\;\text{for all}\;\; x\in[0,1]^d\right)=e^{-2e^{-\kappa}},
 \end{align*}
 where the sequence $\Phi_{n,\kappa}$ is defined by
\begin{align*}
\Phi_{n, \kappa}=\frac{(\frac{\kappa}{C_{n,3}}+C_{n,3})\sigma\|K\|_{L^2(\R^d)}}{(2\pi)^dh^{\beta}\sqrt{h^dn^da_n^d}}.
\end{align*}
\end{Corollary}
As a consequence of Corollary \ref{cor1} an asymptotic uniform confidence region for the function $f$ with confidence level $1-\alpha$ is given by
 \begin{align} \label{conf}
   \{[\hat f_n(x)-\Phi_{n,-\ln(-0.5\ln(1-\alpha))},\hat f_n(x)+\Phi_{n,-\ln(-0.5\ln(1-\alpha))}]\,|\,x\in
   [0,1]^d\}.
 \end{align}
The corresponding $(1-\alpha)$-band has a width of $2\Phi_{n,-\ln(-0.5\ln(1-\alpha))}$. Here, the factor $\frac{1}{h^{\beta}}$ is due to the ill-posedness of the inverse problem (see Assumption B). It does not appear in corresponding results for the direct regression case. On the other hand the factor $a_n^{-\frac{d}{2}}$ arises from the design on the growing system of sets $\{[-a_n^{-1},a_n^{-1}]^d\,|\,n\in\N,\}$. In the case of a regression on a fixed interval it does not appear as well. The width of the asymptotic point-wise confidence intervals in the multivariate indirect regression case as obtained in \cite{bisbir2009}  is of order $\frac{1}{h^{\beta}\sqrt{Nh^da_n^{d}}},$ where $N$ is the total number of observations. Their point-wise confidence intervals are smaller than the uniform ones obtained in Corollary \ref{cor1}. The price for uniformity is  an additional factor of logarithmic order,   which is typical for results of this kind.

In applications the standard deviation is unknown but can be estimated easily from the data, because this does not require the estimation of the function $f$. In particular, \eqref{conf} remains an asymptotic $(1- \alpha)$-confidence band, if $\sigma$ is replaced by an estimator satisfying
 $\hat\sigma-\sigma=o_P(1/\log(n))$.

\section{Time dependent regression functions}\label{sec5}
\def\theequation{4.\arabic{equation}}
\setcounter{equation}{0}

In this section we extend  model $(\ref{model})$ to include a time dependent regression function, that is
\begin{align}\label{model2}
      Y_{j,\mathbf{k},n}=\bigl(T_{\psi}f_{t_j}\bigr)(x_{\mathbf{k}})+\varepsilon_{\mathbf{k}}, \quad \mathbf{k}\in G_{\mathbf{n}},\;
       j=-m,\ldots,m,
   \end{align}
where $x_{\mathbf{k}}=\frac{\mathbf{k}}{na_n}$ and $t_j=\frac{j}{mb_m},$
 $m=m(n)$, such that $m(n)\rightarrow\infty$ and $b_{m(n)}\searrow0$ as $n\rightarrow\infty.$

We assume that  $\psi$ does not depend on the time and the operator
$T_{\psi}$  is defined by
  \begin{align*}
    \bigl(T_{\psi}f_{t}\bigr)=\int_{\mathbb{R}^d}f_t(y)\psi(\cdot-y)\,dy.
  \end{align*}
This assumption is reasonable in the context of imaging where the function $\psi$ corresponds to the point spread function (\cite{berbocdes2009}).
If it is not satisfied, i.e. the convolution operator effects all coordinates, the problem can be modeled as in Section \ref{sec3}.

For a precise statement of the results  we will add an index to the Fourier operator ${\cal F}$ which gives the dimension of the space under  consideration. We will write  ${\cal F}_{d+1}$ if the Fourier transform is taken over the whole space $\R^{d+1}$ and ${\cal F}_d$ to denote
Fourier transformation  with respect to the spatial dimensions.
By the same considerations as given in Section \ref{sec3} we obtain an estimator $\check f$ for the function $f_t$
  \begin{align*}
    \check f_n(x;t)&=\frac{1}{(2\pi)^{\frac{d+1}{2}}}\int_{\R^{d+1}} \frac{{\cal F}_{d+1}(\widehat{f*\psi})(\xi,\tau)}{(2\pi)^{\frac{d}{2}}{\cal
     F}_{d}\psi(\xi)} {\cal F}_{d}\check\eta(\xi h,\tau h_t)\exp(it\tau+ix^T\xi)\,d(\xi,\tau)\\

     &=\frac{1}{(2\pi)^{d+\frac{1}{2}}n^dma_n^db_m}\sum_{(\mathbf{k},j)\in G^{d+1}_{(\mathbf{n},m)}}Y_{\mathbf{k},j}\check K_n\Bigl(
     \frac{x-x_{\mathbf{k}}}{h},\frac{t-t_j}{h_t}\Bigr),
  \end{align*}
where $G^{d+1}_{(n,m)}$ denotes the grid $\{-n,\ldots,n\}^d\times\{-m,\ldots,m\}$ and the   kernel $\check K_n$ is given by
  \begin{align}\label{KernZeitUngefaltet}
    \check K_n(x;t)=\frac{1}{(2\pi)^{\frac{d+1}{2}}}\int_{\R^{d+1}}
    \frac{\exp\bigl(i\tau t+i\xi^Tx\bigr)}{{\cal F}_{d}\psi(\frac{\xi}{h})}{\cal F}_{d+1}\check\eta(\xi,\tau)
     \,d(\xi,\tau).
  \end{align}
 Here the function $\check\eta:\R^{d+1}\rightarrow\R$ satisfies condition A and $h_t=h_t(n)$ is an additional sequence of bandwidths referring to the time domain.
For the asymptotic analysis we require a modified version of Assumption B.

\medskip

{\bf Assumption $\mathbf{\check B}$}
Let Assumptions B1 (with corresponding kernel $\check K$) and B2 hold and additionally assume that
\begin{itemize}
  \item[$\check B3$.] $\log(n+m(n))\cdot h^{\mu_M}(a_n^{-\frac{d}{2}}h^{-\frac{d}{2}}b_{m(n)}^{\frac{1}{2}}m(n)^{\frac{1}{2}})=o(1)$ and for
  $p=1,\ldots,M-1$\\
   $h^{\mu_{p}}(\log(n+m))^2=o(1)$.
\end{itemize}

 \begin{Theorem}\label{GWSZeit}
     {Define}
      \begin{align*}
        \check Y_n(x;t)&:=\frac{(2\pi)^{d+1}h^{\beta}\sqrt{h^dh_t n^{d}mb_ma_n^d}}{\sigma\|\check K\|_{L^2(\R^{d+1})}\|}\left[\check
        f_n(x;t)-\E\check
         f_n(x;t)\right]
      \end{align*}
    and let the moment condition of Theorem \ref{GWS}  and  Assumptions A and $\check B$ be satisfied.
    We further assume that the bandwidths $h_t$ and $h,$ and the sequences $(a_n)_{n\in\N}$ and $(b_{m(n)})_{n\in\N}$ satisfy
    \begin{align*}
     \log(n+m)\biggl(\sqrt{\frac{na_n}{mb_m}}\frac{1}{\sqrt{n^{\delta}h_ta_n^{\delta}h^d}}+\Bigl(\frac{mb_m}{na_n}\Bigr)^{\frac{d}{2}}\frac{1}{\sqrt{m^{\delta}h_th^{d}}}\biggr) =o(1)\quad\text{for}\quad n\rightarrow\infty
    \end{align*}
    \begin{align*}
    h_t+h\leq L_1\cdot h^{d(1-L_2)}h_t^{(1-L_2)}
    \end{align*}
   for some constants  $L_1<\infty$ and $L_2 \in (0,1)$.
     Then we have for each  $\kappa\in\R,$
      \begin{align*}
         \lim_{n\rightarrow\infty}\Prob\left(\sup_{x\in[0,1]^d}\Bigl(|\check Y_n(x;t)|-D_{n,3}\Bigr)\cdot
         D_{n,3}<\kappa\right)
         =e^{-2e^{-\kappa}},
      \end{align*}
     where
    \begin{align*}
      &D_1=\mathbf{det}\left(\left[\frac{(2\pi)^{2(d+1)}}{\|\check K\|^2_{L^2(\R^{d+1})}}\int_{\R^{d+1}}|\Psi(v_1,\ldots,v_d){\cal F}_{d+1}
      \check\eta(v)|^2v_iv_j\,dv\right]
      ,\,i,j=1,\ldots,d+1\right),\\

      &D_{n,2}=\sqrt{\frac{D_1}{(2\pi)^{d+2}}}\frac{1}{h^dh_t}\quad\text{and}\\

      &D_{n,3}=\sqrt{2\ln(D_{n,2})}+\frac{(d-1)\ln\bigl(2\ln(D_{n,2})\bigr)}{2\sqrt{2\ln(D_{n,2})}}.
    \end{align*}

  \end{Theorem}

  \begin{Corollary} \label{cor2}
 If   the assumptions of Theorem \ref{GWSZeit} are satisfied, the limit kernel $\check K$ is defined by
 \begin{align} \label{kcheck}
  \check K(x,t)=\frac{1}{(2\pi)^{\frac{d+1}{2}}}\int_{\R^{d+1}}\Psi(\xi){\cal F}_{d+1}\check\eta(\xi,\tau)\exp(i\xi^T x+i\tau t)\,d(\xi,\tau).
\end{align}     and the function $f_{(\cdot)}(\cdot):\R^{d+1}\rightarrow\R^1$ satisfies Assumption C,   then it follows  that
  \begin{align*}
    \lim_{n\rightarrow\infty}\Prob\left(\check f_n(x;t)-\check\Phi_{n,\kappa}\leq
     f(x;t)\leq\check f_n(x;t)+\check\Phi_{n,\kappa}\;\;\text{for all}\;\; (x,t)\in[0,1]^{d+1}\right)=e^{-2e^{-\kappa}},
 \end{align*}
where the constant $\check\Phi_{n,\kappa}$ is defined by
\begin{align*}
\check\Phi_{n,\kappa}=\frac{(\frac{\kappa}{D_{n,3}}+D_{n,3})\sigma\|\check K\|_{L^2(\R^{d+1})}}{h^{\beta}\sqrt{h^dn^da_n^dmb_mh_t}(2\pi)^{d+1}}.
\end{align*}
\end{Corollary}

Asymptotic confidence bands for the function $f_t(x)$   at level  $1-\alpha$ are hence given by
 \begin{align*}
   \{[\check f_n(x;t)-\check\Phi_{n,-\ln(-0.5\ln(1-\alpha))},\check f_n(x;t)+\check\Phi_{n,-\ln(-0.5\ln(1-\alpha))}]\,|\,(x,t)\in
   [0,1]^{d+1}\}.
 \end{align*}

\medskip
\medskip

{\bf Acknowledgments:}
This work has been supported in part by the
Collaborative Research Center  ``Statistical modeling of nonlinear dynamic processes''(SFB 823,
Teilprojekte A1 and C4) of the German Research Foundation (DFG). The authors would like to thank Martina Stein, who typed parts of this manuscript with considerable technical expertise.

redpajama_set_name: RedPajamaArXiv


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text:
Police and Law Enforcement
For 'Police Accountability With Teeth,' Citizens Demand Elected and Empowered Oversight Boards
Activists from Seattle to New York argue that elected boards and special prosecutors are vital to addressing police violence.
By John Nichols
NYPD officers block the entrance of the Manhattan Bridge during protests against police brutality on June 2, 2020. (Scott Heins / Getty Images)
Seattle City Council member Kshama Sawant recognizes that the national uprising against police brutality is "threatening to the right-wing and reactionary agenda of Donald Trump." The president's tweets confirm that he is unsettled by what he claims is "the Radical Left takeover of Seattle." But what Trump sees as threatening, Sawant sees as an opening to achieve fundamental change in Seattle as nationally. So the council member is not just defending the activists who have established a "Capitol Hill Autonomous Zone" in a Seattle neighborhood; she is also proposing bold responses to the police oversight and accountability issues that protesters say must be addressed.
Declaring that she is determined to "take this historic moment to win real victories against police violence," the Seattle socialist told a CNN interviewer that she's working to "defund the police by at least 50 percent and divert those funds for community programs, for restorative justice, so we begin to address the systematic racism."
"But we need [to go] beyond that," she added."We need an independently elected community oversight board, which has full powers over the police."
Sawant was talking about democratic oversight of policing long before the death of an unarmed African American man, George Floyd, in Minneapolis police custody stirred an outcry over policing in cities from coast to coast. Now, however, the discussion about these issues has taken off, as protesters demand genuine community control over the police.
MORE FROM John Nichols
We Wouldn't Be in This Mess if D.C. Were a State
Rand Paul Abandons His Hippocratic Oath to Play Politics During a Pandemic
Martin Luther King Jr. and the Unfinished Work of Abolishing Poverty
This is not a radical demand or a foreign construct. We live in a country where civilian control over the military is enshrined in our Constitution. This premise is so vital to the American experiment that retired Marine Gen. James Mattis made a point of reasserting it when he was under consideration to serve as defense secretary—telling the Senate Armed Services Committee in 2017, "I recognize my potential civilian role differs in essence and in substance from my former role in uniform. Civilian control of the military is a fundamental tenet of the American military tradition."
It follows that cities should recognize the importance of clearly defining, and maintaining, civilian control of the police.
Yet in cities across the country, there is a lack of clarity about oversight of policing. The actions of police chiefs and police departments are often overseen by obscure boards and commissions that lack the authority and the resources necessary to demand accountability.
That lack of clarity has been highlighted in recent weeks, as the outcry over Floyd's death has opened a nationwide dialogue about how we might, as Representative Ilhan Omar says, begin to "reimagine public safety."
View our current issue
"This is a time for us to do real soul-searching," says Omar, a Democrat who represents Minneapolis. "We can't continue to be on this path, where we continue to find ourselves in this place. There has to be real systematic change on the city level, on the state level and on the national level."
In Minneapolis, the majority of City Council members now support a move to dismantle the city's police department and replace it with what the Minneapolis Star Tribune refers to as "a new community-based system of public safety."
In San Francisco, Mayor London Breed last week announced a plan to demilitarize the police that includes "an explicit policy barring the use of military-grade weapons against unarmed civilians." The proposal would "divert non-violent calls for service away from SFPD to non-law enforcement agencies" that can dispatch unarmed professionals in response to calls involving mental health, homelessness, and school discipline concerns. San Francisco Chief of Police Bill Scott joined the mayor in announcing the initiatives and spoke of the need to "embrace courageous changes to address disparate policing practices."
Some of the most striking proposals for new approaches to public safety seek to realize the promise of civilian control that Sawant recognizes when she talks about the need for "a democratically elected community oversight board in Seattle, with full powers to hold police accountable, including setting department policy and procedure."
Sawant is not alone in arguing for a democratically elected oversight board. In Salt Lake City, the group Utah United Against Police Brutality is pushing to replace an existing Police Civilian Review Board with a democratically elected board that would have clearly defined oversight and accountability powers. In Tallahassee, Fla., members of the Tallahassee Community Action Committee are arguing for the creation of a democratically elected Civilian Police Accountability Council that would have the power to hire and fire police chiefs and a role in budget debates and police union contract negotiations. They are asking the people of Tallahassee to "Tell the City Commission: We Want Police Accountability WITH TEETH."
This Law Keeps Police Misconduct Secret
Ross Barkan
In New York City, there's now an energetic campaign to replace the city's weak Civilian Complaint Review Board with an Elected Civilian Review Board (ECRB). "We need a board that actually has the ability to discipline cops," explains Tammie David, who has been active with the Campaign for an Elected Civilian Review Board. The members of the proposed ECRB would be chosen at the neighborhood level in a process that would have strict campaign finance rules. The elected board would have subpoena power, as well as the authority to address police misconduct with investigations, disciplinary actions, and retraining orders. The board would maintain neighborhood offices, organize regular community meetings, and do outreach with communities that have been affected by police misconduct and violence.
Backed by elected officials, candidates, unions, Black Lives Matter activists, and Citizen Action, the Campaign for an Elected Civilian Review Board also seeks to establish an elected special prosecutor position in order to assure that there is unbiased prosecution of police officers who are accused of crimes.
Democracy reforms are rarely easy to achieve, especially in places where local and state rules may have to be restructured to create new elected positions. But campaigners in New York are confident that the time has come to "shift power into the hands of communities by empowering accountable representatives in every neighborhood to respond to cases of police abuse."
John NicholsJohn Nichols is a national affairs correspondent for The Nation and the author of the new book The Fight for the Soul of the Democratic Party: The Enduring Legacy of Henry Wallace's Anti-Fascist, Anti-Racist Politics (Verso). He's also the author of Horsemen of the Trumpocalypse: A Field Guide to the Most Dangerous People in America, from Nation Books, and co-author, with Robert W. McChesney, of People Get Ready: The Fight Against a Jobless Economy and a Citizenless Democracy.
The General Strike No One Is Talking About


text:
#ifndef ResourceError_h
#define ResourceError_h

#include "ResourceErrorBase.h"

namespace WebCore {

class ResourceError : public ResourceErrorBase
{
public:
    ResourceError()
    {
    }

    ResourceError(const String& domain, int errorCode, const String& failingURL, const String& localizedDescription)
        : ResourceErrorBase(domain, errorCode, failingURL, localizedDescription)
    {
    }
};

}

#endif // ResourceError_h_

redpajama_set_name: RedPajamaGithub


Mumbai: The Maharashtra Ownership Flats (Regulation of the promotion of construction, sale, management and transfer) Act 1963 was meant to provide relief to flat purchasers against sundry abuses, malpractices and difficulties related to the construction, sale management and transfer of flats, and it has been brought to the notice of the State Government that, the Act did not provide for an effective implementing arm for its various statutory provisions, because flat purchasers could only approach Consumer Forum or Civil Courts for acts of omission or commission of provisions of the Act and whereas establishing a Housing Regulatory Authority and an Housing Appellate Tribunal would ensure effective implementation of the Act.
Whereas, the modified new MHA 2011 would help remove information asymmetry by ensuring full disclosure by promoters/developers and would also ensure compliance of agreed terms and conditions while registering, monitoring and regulating housing projects by Housing Regulatory Authority.
The new MHA 2011 will usher in transparency and discipline in the transactions of flats and put a check on abuses and malpractices.
However, establishing a Housing Regulatory Authority and a Housing Appellate Tribunal would promote planned and healthy development and construction, sale, transfer and management of flats, residential buildings, and other similar properties with a view to protecting , on the one hand, public interest in relation to the conduct and integrity of promoters and other persons engaged in the development of such flats, buildings and, on the other, to facilitating the smooth and speedy construction and maintenance of such flats, residential buildings and properties.


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text:
О́йзуська сільська рада (, ) — сільська рада в Естонській РСР та Естонській Республіці, адміністративно-територіальна одиниця в складі повіту Ярвамаа (1945—1950, 1990—1991), Тюріського району (1950—1959) та Пайдеського району (1959—1990).

Географічні дані
Сільська рада розташовувалася в південній частині Пайдеського району.

У 1973 році площа сільради складала 127,2 км2.

Населення за роками

Населені пункти

З 1945 до 1975 року адміністративний центр сільради — село Вяльяотса, що розташовувалося на відстані 10,4 км на південний схід від міста Тюрі. 1975 року центр перенесений у селище Ойзу.

Землекористування
1954 року в межах території сільради землями користувалися колгоспи: імені Сталіна, «Кунґла» («Kungla»), «Едазі» («Вперед», «Edasi»), а також Тюріське відділення радгоспу «Пілу» (Pilu).

1973 року на території сільради розташовувалися колгоспи «Естонія» (головна садиба містилася в селі Вяльяотса) та «Едазі» (село Тайксе).

Історія
8 серпня 1945 року на території волості Сяревере в Ярваському повіті утворена Ойзуська сільська рада з центром у селі Вяльяотса.

26 вересня 1950 року, після скасування в Естонській РСР повітового та волосного поділу, сільська рада ввійшла до складу новоутвореного Тюріського сільського району.

24 січня 1959 року сільрада приєднана до Пайдеського району після скасування Тюріського району.

19 грудня 1991 року Ойзуська сільська рада перетворена у волость Ойзу з отриманням статусу самоврядування.

Примітки

Джерела
 Eesti Entsüklopeedia

 Eesti NSV Teataja
 Riigi Teataja

Сільські ради Естонської РСР
Тюріський район
Пайдеський район
Ярвамаа


Please! having this behavior would save me tons of manual adjustments.
This in particular is a necessary functionality.
But the whole Paragraph Border feature could be a lot more powerful if it would stay across paragraphs but stop when the style changes. So it would be brilliant as an "Indented Quote with Adjacent Vertical Line".
yes! The feature is useless without this!
100%! Its about time, been bugging me for years!


text:
Home/Dining/Chicken Cataplana: A Traditional Delicious Portuguese Stew
Discover the Rich Flavors of the Algarve with a Trip to Nando's Restaurant
touristeyes4 weeks agoLast Updated: January 6, 2023
The Cataplana is a traditional Portuguese dish that originates from the Algarve region in the south of the country. It is a type of chicken stew that is cooked in a copper pot called a "cataplana," which gives the dish its name. The cataplana is a versatile pot that can be used for both cooking and serving the dish, making it a popular choice for both home cooks and restaurant chefs.
The chicken Cataplana is typically made with a combination of chicken as well as vegetables like bell peppers, onions, and tomatoes. The ingredients are seasoned with aromatic herbs and spices, such as coriander, parsley, and paprika, and then slow-cooked to create a rich and flavorful stew. The cataplana can be served with a variety of accompaniments, such as rice, potatoes, or crusty bread, to soak up the delicious broth. Also there is seafood Cataplana, where the chicken is replaced with the seafood.
One of the best things about the Cataplana is its simplicity. It is a quick and easy dish to prepare, making it a great option for a weeknight dinner or a casual gathering with friends.
The Cataplana is a beloved dish in Portugal, and it is enjoyed by locals and tourists alike. Whether you are visiting the Algarve region or simply looking to try a new and delicious stew, the Cataplana is a must-try.
Halo Halo Asian Restaurant in Dubai Offers Authentic Asian Dining Experience
Vibes & Views of Doors Freestyle Grill and Steakhouse in Dubai
One of the most pupular restaurants that serve Cataplana is Nando's. Nando's is a popular chain of restaurants that originated in South Africa and has since spread to countries around the world, including the United Kingdom, Australia, and Canada. In addition to its signature flame-grilled chicken dishes, Nando's also offers a variety of other menu items, including the Cataplana.
The Cataplana at Nando's is made with a selection of succulent chicken as well as bell peppers, onions, and tomatoes. It is seasoned with a blend of aromatic herbs and spices, including coriander, parsley, and paprika, and served with a side of rice.
Nando's is known for its commitment to using high-quality ingredients and for its flavorful and bold flavor profiles, and the Cataplana is no exception. It is a delicious and satisfying dish that is sure to please chicken lovers. If you are looking for a restaurant that serves authentic and tasty Cataplana, be sure to give Nando's a try.
Cataplana Dining Restaurants Travel
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touristeyes


text:
Q: Typescript - How can I conditionally define a type based on another property I'm trying to conditionally set the return type of a function, based on a static value in the same type.
Given the following code, is it possible to have type checking work as I've specified in the comments?
IE, if const taskX: Task has the id: 'get-name', how can I make it so taskX.run() should return according to the type defined as Outputs['get-name']
type Outputs = {
  "get-name": {
    name: string;
  };
  "get-age": {
    age: number;
  };
  "get-favourite-fruits": ("apple" | "strawberry" | "melon")[];
};

type Task = {
  // This should always be a static property defined in `Outputs`
  id: keyof Outputs;
  // How can I make this conditional and return the type from `Outputs` based on `id`
  run: () => never;
};

const taskOne: Task = {
  id: "get-name",
  // This should be valid
  run: () => {
    return { name: "frankenstein" };
  },
};

const taskTwo: Task = {
  id: "get-age",
  // This should also be valid
  run: () => {
    return { age: 22 };
  },
};

const taskThree: Task = {
  id: "get-favourite-fruits",
  // This should fail with
  // Type '("apple" | "banana")[]' is not assignable to type
  // '("apple" | "strawberry" | "melon")[]'
  run: () => {
    return ["apple", "banana"];
  },
};

A: You want Task to be a discriminated union of the three different possible pairings of id and run.  You can generate this programmatically from Outputs by mapping each property of Outputs to a value corresponding to this pairing, and then looking up all of the mapped properties to get the desired union:
type Task = {
  [K in keyof Outputs]: { id: K, run: () => Outputs[K] }
}[keyof Outputs];

/* produces
type Task = {
    id: "get-name";
    run: () => {
        name: string;
    };
} | {
    id: "get-age";
    run: () => {
        age: number;
    };
} | {
    id: "get-favourite-fruits";
    run: () => ("apple" | "strawberry" | "melon")[];
}*/

Then the rest of your code works as desired.  Valid tasks are accepted:
const taskOne: Task = {
  id: "get-name",
  run: () => {
    return { name: "frankenstein" };
  },
}; // okay

const taskTwo: Task = {
  id: "get-age",
  run: () => {
    return { age: 22 };
  },
}; // okay

And invalid tasks are rejected:
const taskThree: Task = {
  id: "get-favourite-fruits",
  run: () => { // error!
    // Type '() => ("apple" | "banana")[]' is not assignable to type
    // '(() => { name: string; }) | (() => { age: number; }) |
    // (() => ("apple" | "strawberry" | "melon")[])
    return ["apple", "banana"];
  },
};

const taskFour: Task = { // error!
//    ~~~~~~~~
// Type '{ id: "get-favourite-fruits"; run: () => { age: number; }; }'
// is not assignable to type 'Task'.
  id: "get-favourite-fruits",
  run: () => {
    return { age: 123 };
  },
};

The errors for invalid tasks might be a bit different from the particular one you were expecting. For taskThree the compiler complains that run()'s return is incompatible with every possible Task member; "not only isn't it the right array, but it isn't even one of the right types for get-name or get-age!"  For taskFour the compiler sees that run() returns an {age: number} but the id is get-favorite-fruits and complains that the whole value is wrong without highlighting any particular property.
Playground link to code

A: You can achieve that by declaring a type which is either of them and based on the id the compiler can tell which one of those it is.
type GetNameTask = {
    id: "get-name";
    run: () => {name: string};
}
type GetAgeTask = {
    id: "get-age";
    run: () => {number: string};
}
type GetFavoriteFruitsTask = {
    id: "get-favourite-fruits"
    run: () => ("apple" | "strawberry" | "melon")[];
}

type Task = GetNameTask | GetAgeTask | GetFavoriteFruitsTask

For more a clean systax you can also try this:
type GenericTask<TId, TOutput> = {
    id: TId;
    run: () => TOutput
}

type GetNameTask2 = GenericTask<"get-name", {name: string}>
type GetAgeTask2 = GenericTask<"get-age", {age: number}>
type GetFavoriteFruitsTask2 = GenericTask<"get-favourite-fruits", ("apple" | "strawberry" | "melon")[]>
type Task = GetNameTask2 | GetAgeTask2 | GetFavoriteFruitsTask2


text:
Q: What are reasons why one would want to use nested classes? In this stackoverflow answer a commenter mentioned that "private nested classes" can be quite useful so I was reading about them in articles such as this one which tend to explain how nested classes function technically, but not why you would use them.
I suppose I would use private nested classes for little helper classes that belong to a larger class, but often I will need a helper class from another class and so I would just have to take the extra effort to (1) make the nested class non-nested or (2) make it public and then access it with the outer-class prefix on it, which both seems to be extra work without any added-value for having the nested class in the first place. Hence in general I really don't see a use case for nested classes, other than perhaps to keep classes a bit more organized into groups, but I that also goes against the one-class-per-file clarity that I have come to enjoy.
In what ways do you use nested classes to make your code more manageable, readable, efficient?

A: Private helper classes is a good example.
For instance, state objects for background threads. There is no compelling reason to expose those types. Defining them as private nested types seems a quite clean way to handle the case.

A: I use them when two bound values (like in a hash table) are not enough internally, but are enough externally. Then i create a nested class with the properties i need to store, and expose only a few of them through methods.
I think this makes sense, because if no one else is going to use it, why create an external class for it? It just doesn't make sense to.
As for one class per file, you can create partial classes with the partial keyword, which is what I usually do.

A: One compelling example I've run into recently is the Node class of many data structures. A Quadtree, for example, needs to know how it stores the data in its nodes, but no other part of your code should care.

A: I've found a few cases where they've been quite handy:

*

*Management of complex private state, such as an InterpolationTriangle used by an Interpolator class.  The user of the Interpolator doesn't need to know that it's implemented using Delauney triangulation and certainly doesn't need to know about the triangles, so the data structure is a private nested class.

*As others have mentioned, you can expose data used by the class with an interface without revealing the full implementation of a class.  Nested classes can also access private state of the outer class, which allows you to write tightly coupled code without exposing that tight coupling publicly (or even internally to the rest of the assembly).

*I've run into a few cases where a framework expects a class to derive from some base class (such as DependencyObject in WPF), but you want your class to inherit from a different base.  It's possible to inter-operate with the framework by using a private nested class that descends from the framework base  class.  Because the nested class can access private state (you just pass it the parent's 'this' when you create it), you can basically use this to implement a poor man's multiple inheritance via composition.

A: I think others have covered the use cases for public and private nested classes well.
One point I haven't seen made was an answer your concern about one-class-per-file. You can solve this by making the outer class partial, and move the inner class definition to a separate file.
OuterClass.cs:
namespace MyNameSpace
{
    public partial class OuterClass
    {
        // main class members here
        // can use inner class
    }
}

OuterClass.Inner.cs:
namespace MyNameSpace
{
    public partial class OuterClass
    {
        private class Inner
        {
            // inner class members here
        }
    }
}

You could even make use of Visual Studio's item nesting to make OuterClass.Inner.cs a 'child' of OuterClass.cs, to avoid cluttering your solution explorer.

A: You've answered your own question. Use nested classes when you need a helper class that is meaningless outside the class; particularly when the nested class can make use of private implementation details of the outer class.
Your argument that nested classes are useless is also an argument that private methods are useless: a private method might be useful outside of the class, and therefore you'd have to make it internal. An internal method might be useful outside of the assembly, and therefore you'd make it public. Therefore all methods should be public. If you think that's a bad argument, then what is different about you making the same argument for classes instead of methods?
I make nested classes all the time because I am frequently in the position of needed to encapsulate functionality in a helper that makes no sense outside of the class, and can use private implementation details of the outer class. For example, I write compilers. I recently wrote a class SemanticAnalyzer that does semantic analysis of parse trees. One of its nested classes is LocalScopeBuilder. Under what circumstances would I need to build a local scope when I am not analyzing the semantics of a parse tree? Never. That class is entirely an implementation detail of the semantic analyzer. I plan to add more nested classes with names like NullableArithmeticAnalyzer and OverloadResolutionAnalyzer that are also not useful outside of the class, but I want to encapsulate rules of the language in those specific classes.
People also use nested classes to build things like iterators, or comparators - things that make no sense outside of the class and are exposed via a well-known interface.
A pattern I use quite frequently is to have private nested classes that extend their outer class:
abstract public class BankAccount
{
    private BankAccount() { }
    // Now no one else can extend BankAccount because a derived class
    // must be able to call a constructor, but all the constructors are
    // private!
    private sealed class ChequingAccount : BankAccount { ... }
    public static BankAccount MakeChequingAccount() { return new ChequingAccount(); }
    private sealed class SavingsAccount : BankAccount { ... }

and so on. Nested classes work very well with the factory pattern. Here BankAccount is a factory for various types of bank account, all of which can use the private implementation details of BankAccount. But no third party can make their own type EvilBankAccount that extends BankAccount.

A: One very common pattern where this technique is used is in scenarios where a class returns an interface or base class type from one of its properties or methods, but the concrete type is a private nested class. Consider the following example.
public class MyCollection : IEnumerable
{
  public IEnumerator GetEnumerator()
  {
    return new MyEnumerator();
  }

  private class MyEnumerator
  {
  }
}

A: I usually do it when I need a combination of SRP (Single Responsibility Principal) in certain situations.
"Well, if SRP is your goal, why not split them into different classes?" You will do this 80% of the time, but what about situations where the classes you create are useless to the outside world?  You don't want classes that only you will use to clutter your assembly's API.
"Well, isn't that what internal is for?"  Sure.  For about 80% of these cases.  But what about internal classes who must access or modify the state of public classes?  For example, that class which was broken up into one or more internal classes to satisfy your SRP streak?  You would have to mark all the methods and properties for use by these internal classes as internal as well.
"What's wrong with that?"  Nothing.  For about 80% of these cases.  Of course, now you're cluttering the internal interface of your classes with methods/properties that are only of use to those classes which you created earlier.  And now you have to worry about other people on your team writing internal code won't mess up your state by using those methods in ways that you hadn't expected.
Internal classes get to modify the state of any instance of the type in which they are defined.  So, without adding members to the definition of your type, your internal classes can work on them as needed.  Which, in about 14 cases in 100, will be your best bet to keep your types clean, your code reliable/maintainable, and your responsibilities singular.

A: Returning an interface to the caller whose implementation you want to hide.
public class Outer
{
    private class Inner : IEnumerable<Foo>
    {
        /* Presumably this class contains some functionality which Outer needs
         * to access, but which shouldn't be visible to callers
         */
    }

    public IEnumerable<Foo> GetFoos()
    {
        return new Inner();
    }
}

A: They are really nice for, as an example, an implementation of the singleton pattern.
I have a couple of places where I am using them to "add" value, as well. I have a multi-select combobox where my internal class stores the state of the checkbox and the data item as well.  no need for the world to know about/use this internal class.

A: Private anonymous nested classes are essential for event handlers in the GUI.
If some class is not part of the API another class exports, it must be made private.  Otherwise you are exposing more than you intend.  The "million dollar bug" was an example of this.  Most programmers are too slack about this.
Peter

A: The question is tagged C# so I'm not sure this is of interest, but in COM you can use inner classes to implement interfaces when a class C++ implements multiple COM interfaces... essentially you use it for composition rather than multiple-inheritance.
Additionally in MFC and perhaps other technologies you might need your control/dialog to have a drop-target class, which makes little sense other than as a nested class.

A: If it is necessary for an object to return some abstract information about its state, a private nested class may be suitable.  For example, if an Fnord supports "save context" and "restore context" methods, it may be useful to have the "save context" function return an object of type Fnord.SavedContext.  Type access rules aren't always the most helpful; for example, it seems difficult to allow Fnord to access properties and methods of a Fnord.SavedContext without making such properties and methods visible to outsiders.  On the other hand, one could have Fnord.CreateSaveContext simply create a New Fnord.SaveContext with the Fnord as a parameter (since Fnord.SaveContext can access the internals of Fnord), and Fnord.LoadContextFrom() can call Fnord.SaveContext.RestoreContextTo().


text:
Q: How to resolve Ripple test server connection error? I am persistently getting following error on connecting to test server using RippleAPI for Javascript:
[ConnectionError(Error: connect() timed out after 2000 ms. If your internet connection is working, the rippled server may be blocked or inaccessible.)]

However, if I try to get balance from curl, it works:
curl 'https://testnet.data.api.ripple.com/v2/accounts/rwAVpkGNU9Shn63EpFq7ju1tr89SsSBwHz/balances?currency=XRP'

Code snippet below:
'use strict';
const RippleAPI = require('ripple-lib').RippleAPI;

const api = new RippleAPI({
  server: 'wss://s.altnet.rippletest.net:51233' // Public rippled server
});
api.connect().then(() => {
  /* begin custom code ------------------------------------ */
  const myAddress = 'rwAVpkGNU9Shn63EpFq7ju1tr89SsSBwHz';

  console.log('getting account info for', myAddress);
  return api.getAccountInfo(myAddress);

}).then(info => {
  console.log(info);
  console.log('getAccountInfo done');

  /* end custom code -------------------------------------- */
}).then(() => {
  return api.disconnect();
}).then(() => {
  console.log('done and disconnected.');
}).catch(console.error);

A: timeout param doesn't seem to work even if I suggest;
use:
api.connection._config.connectionTimeout = 3e4;