text:
# Thread: finding derivatives from first principles

1. ## finding derivatives from first principles

Hi I'm still struggling with finding the definition of the derivative (from first principles),

find f'(x) of f(x)=-2x^2+4x-7

2. ## Re: finding derivatives from first principles

Originally Posted by calmo11
Hi I'm still struggling with finding the definition of the derivative (from first principles),

find f'(x) of f(x)=-2x^2+4x-7
Given $f(x), f'(x) = \displaystyle{\lim_{h\to 0}}\dfrac{f(x+h)-f(x)}{h}$

$f(x)=-2x^2+4 x - 7$

$f'(x)=\displaystyle{\lim_{h \to 0}}\dfrac{\left(-2(x+h)^2+4(x+h)-7\right)-\left(-2x^2+4x-7\right)}{h}$

$f'(x)=\displaystyle{\lim_{h \to 0}}\dfrac{-4hx +4h-2h^2}{h}$

$f'(x)=\displaystyle{\lim_{h \to 0}}~-4x+4-2h=-4x+4$

format: plain


text:
Lemma 37.39.2. Let $f : U \to S$ be a surjective étale morphism of affine schemes. There exists a surjective, finite locally free morphism $\pi : T \to S$ and a finite open covering $T = T_1 \cup \ldots \cup T_ n$ such that each $T_ i \to S$ factors through $U \to S$. Diagram:

$\xymatrix{ & \coprod T_ i \ar[rd] \ar[ld] & \\ T \ar[rd]^\pi & & U \ar[ld]_ f \\ & S & }$

where the south-west arrow is a Zariski-covering.

Proof. This is a restatement of Algebra, Lemma 10.144.6. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Stack Exchange Gets a Blog

A little birdie told me that the formal launch of Stack Exchange, the hosted version of the Stack Overflow platform, is imminent.

I noticed the Stack Exchange homepage has been totally redesigned with a snazzy new logo, as well. Looking good!

There’s also a brand new Stack Exchange blog, which I am sure will be updated the very second that the service exits beta and goes live — so it’s probably worth subscribing to.

Short of open sourcing our software, which isn’t in the cards right now, Stack Exchange is the best way to get your hands on the Q&A engine technology powering Stack Overflow, Server Fault, and Super User.

Although the Stack Overflow engine was always designed with a technical audience in mind, I’m intrigued to see how far we can push the boundaries of that audience. We’ve pushed a little bit when going from programmers, to sysadmins, to power computer users — and we may try pushing a tad further this year with yet another site. Stack Exchange pushes far beyond that into totally unexplored territory. I’m eager to learn what tweaks to the core engine will be necessary to support these newer, less technical audiences. And presumably the Stack Exchange team at Fog Creek is eager to implement these tweaks.

Anyway, whether it’s a hosted Stack Exchange site, or a clone, the sooner we can get people off the archaic, busted phpBB style of discussion on the internet, the better off we’ll all be. If we have a corporate mission, it is that: to raise the level of discourse on the internet, and to collaboratively put faster, more relevant search results in front of our fellow internet travellers. All through better software!

Filed under stackexchange

You should come up with some sort of import tool so that people coming from crappy phpBB forums can easily take all their discussion and put it into their shiny new StackExchange site!

The launch doesn’t seem imminent, or at least, I hope it’s not. There are still several holes left unfilled in the platform (database export, etc) and issues/bugs that need to be worked out. There were also, apparently, performance problems with the latest beta that required three tries to get right.

Being a free beta the communities are content to put up with the issues that come along with it, but I expect that once Fog Creek starts charging $129+ every month, people are going to insist on a stable, robust platform. Given this, I just don’t see StackExchange coming out of beta before the end of the month. Think that if the price was unlimited page views and$29 per month then I think more people would be looking at signing up. Then offer more corporate customer packages as charge per usage.

Though people will vote with their wallets.

Jeff, given that you’ve stated very publically and frequently that the Stack Overflow engine is for Q & A and not for supporting discussion or social discourse, what makes you think it will be a good fit for a discussion forum?

For the record, I don’t like phpBB either.

> what makes you think it will be a good fit for a discussion forum?

Because I think the Q&A format is *superior* to free form discussion, in every possible dimension.1

And when people say they want discussion, I don’t think they do: what they really want is answers.

1 except “I just want to waste time by any means necessary”; these folks deserve phpbb

Very much agree with Jeff on this.
stackexchange is better than phpBB.

I would add that stackexchange might be better than search engines for non-trivial questions. It’s a human-powered search engine.

Huge number of people use search engines to ask their questions.
How many times have I googled a question for ten minutes and then given up? That applies to programming questions, but for any other subject.

And I am grateful to all the SO/SE folks for the tool.

Finglas Feb 4 2010

*”we may try pushing a tad further this year with yet another site.”*

Do we really need another site?

@Finglass

Apparently we do. :) From what I understood, there is at least 10 more niche Q&A sites already running on Stack Exchange.

“Because I think the Q&A format is *superior* to free form discussion, in every possible dimension.”

So if I’m hanging out virtually in a forum dedicated to my favourite band and we’re shooting the breeze over tracks on their new album, then you think that voting on people’s opinions and accepting an answer is appropriate?

“And when people say they want discussion, I don’t think they do: what they really want is answers.”

In a lot of cases, not all. phpBB has many problems (mainly too many features, plus being written in PHP, ha ha!), but I don’t think that Stack Overflow is a one size fits all solution either. Sometimes people really do want open-ended discourse.

R. Bemrose Feb 4 2010

@Jeff Atwood:
The Q&A format is superior if you’re asking questions, yes. Exactly how is it superior if the intent is to actually hold a discussion?

The Q&A format used by the trilogy makes it damned hard to hold a back and forth conversation, particularly since comments are treated as second-class citizens.

voyager Feb 4 2010

And when people say they want discussion, I don’t think they do: what they really want is answers.

That depends. Wave is a gigantic time sink, but it’s great for threaded discussions precisely because it’s threaded, as opposed to SOFU, that it’s deliberately left thread unfriendly.

I think there is definitely a need for additional stackoverflow / stackexchange sites because there are the best way today to search for answers and document those answers.

Searching for answers is not restricted to programming.

Stackexchange structures the information around questions and that’s not a bad way to do that. And it helps see the most useful answers at the top.

Other types of forums are very noisy and do not scale.
Imagine 1,000,000 people discussing parenthood, legal advice, graphic design, project management, etc.
How will you find your answers, or just great posts if you don’t have the system to scale?

Whatever happens, one feature StackExchange will NOT have is PHPBB import.

Here’s why:

http://meta.stackoverflow.com/questions/12943/stackexchange-to-allow-data-import/13008#13008

Jeff Davis Feb 4 2010

Woah pricey. Can you make an “ultra-small” pricing, for sites that might get at most 100 thousand visitors a month?

The price might be enough to kill a smaller site.

Jeff, Joel:

I’m more satisfied than before with the StackExchange Beta 6 just pushed out, but as Kyle Cronin pointed out, I agree there are still some gaps that require the product to continue in Beta.

However, I’d like to say I’m simply thrilled to be using StackExchange for my site. Thanks for making it possible!

Paul Nathan Feb 4 2010

I like fora, for discussions.

>*Woah pricey. Can you make an “ultra-small” pricing, for sites that might get at most 100 thousand visitors a month?*

Or “nano-small” pricing? I have this 10 people team that’s looking for a platform just like this, so, that should be about 300 visitors at month ? ( granted every one use it every day, which I doubt it )

While we can use stackoverflow.com, I think our question will get immediately closed for almost all the available reasons a question could be closed there.

StackExchange $129 a month, phpBB free +$5 a month hosting. I don’t see StackExchange making too much of a dent in phpBB’s market while that pricing is still in effect.

I agree with Martin Brown. I can see lots of uses for StackExchange, but at $129/month minimum ($1548/year), none of them are feasible.

I agree with Martin Brown too. + The answer of why the Import feature is not going to exist by Joel is simply immature and very not professional. Ditching other system is not the key to show that your system is the best. More, the style of StackExchange is good for Q&A but Php BB does the job (for the money) for discussion.

voyager Feb 4 2010

@Martin Brown, @Paul Lefebvre: but maybe they are not trying to compete with PHPBB in a race to the bottom?

Ether Feb 4 2010

If this “yet another site” is roboOverflow.com, then bring it on! :)

@voyager I go the impression from Jeff’s post that they did want to compete with phpBB:

“… the sooner we get people off the archaic, busted phpBB style of discussion on the internet…”

There is quite a gap between $$100/year (phpBB + hosting) and 1500/year. If they don’t intend to come closer to$$100/year then he shouldn’t have mentioned phpBB. They’re sure not going to get a lot of people to switch to StackExchange, even if we want to, at the current prices.

> There is quite a gap between $$100/year (phpBB + hosting) and 1500/year. If they don’t intend to come closer to$$100/year then he shouldn’t have mentioned phpBB. They’re sure not going to get a lot of people to switch to StackExchange, even if we want to, at the current prices.

My answer was to open-source our code, but I was overruled.

I am not saying I disagree with that decision, because in a business you do things as a team and sometimes compromise is necessary. I am fine with the compromise. (Joel has certainly compromised on things I wanted many, many times in the past.) But I also acknowledge that *not* allowing us to go open source kind of undermines our core mission to kill phpBB.

The current thinking is that the SE pricing is where it is to suppress demand to levels the SE team can actually handle at this early stage.

There’s no reason the SE business couldn’t become much larger over time as it matures, and push the monthly pricing down closer to something at least in the same rough ballpark as phpBB.

Tony Feb 4 2010

Jeff: people want discussions also. You’re a strong believer of Q&A but when people hang out, they discuss stuff. They don’t go Q&A at each other. So phpBB and other forums software will live forever. So don’t get so hyped up with your Q&A idea. Q&A site are just good at… Q&A.

For those who want cheaper Q&A sites, just wait. I am sure there are developers working on clones. It’s not rocket surgery. Like there are lots of forum & Digg clones, there will be lots of Q&A clones coming out soon. People will always follow where the money is.

phpBB serves its purpose very well and SO will never be able to kill it. I can post several disadvantages of SO. When other open source Q&A clones come out, maybe StackExchange’s pricing model will die.

The recent beta was a little shaky, but I suspect \imminent\ means a small number of months.

Our SE based site (link above) has been building continually with interested photographers, some of them programmers who \get\ the Q&A thing.

Frank Rizzo Feb 4 2010

Not quite sure how StackExchange is going to displace phpBB if the latter is free and the former is not.

Tony Feb 4 2010

Maybe a mix of QA & threaded discussions is a good idea for a new type hybrid platform. People post questions and read voted answers on the same page. Like SO now. If you want to engage in a discussion related to the question, then there’s a link that takes you to a discussion area, away from the question. The two areas should be isolated with cross linking links.

Right now on SO and related sites, people engage in discussions through posting comments (I am guily of that) and you have to prefix your comment with the person’s name you’re replying to. Not very practical and kinda works. I prefer threaded discussions with quoting abilities for that. Otherwise how do people know who you’re replying unless you read all the comments top to bottom. Not very efficient.

Jonathon Watney Feb 5 2010

@Tony and @John Topley: I recently posted a question on meta about supporting discussions. It might be worth checking out and leaving a comment.

Regarding other comments: I have my doubts about the Q&A format not being suited for general discussion. The website yoomoot.com is *meant* for general, mass discussion, and the solution they came up with to the problem of endless and hard-to-sift-through threads was exactly making (almost) everything either a question or an answer.

Martin Wallace Feb 7 2010

“the sooner we can get people off the archaic, busted phpBB style of discussion on the internet, the better off we’ll all be.”

Why are you comparing a free product with a pay product?

You would be better comparing yourself to vBulletin or Invision Power Board – both of which surpass phpBB and are excellent – probably better than ‘The Stack’

Arjan van Bentem Feb 9 2010

Please decide on a name with or without a space — StackExchange or Stack Exchange…?

(So far, I though it was without a space, while Stack Overflow et al do have a space. Now I’m in doubt, seeing that space in Stack Exchange in this post as well?)


text:
# How do you write 9/20 as a decimal?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

65
Apr 2, 2018

$\frac{9}{20} = 9 \div 20 = 0.45$

#### Explanation:

Another way to look at this is:

$\frac{9}{20} = \frac{9}{20} \times \frac{5}{5} = \frac{9 \times 5}{20 \times 5} = \frac{45}{100} = 0.45$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Ujjwal Share
Apr 2, 2018

0.45

#### Explanation:

$9 \div 20$

$\implies \frac{9}{20}$

Convert the denominator into powers of 2 and 5.

=> 9/(2xx2xx5

=> 9/(2^2xx5^1

Now multiply both numerator and denominator with 5 to make the powers of 2 and 5 equal.

=> (9xx5)/(2^2xx5xx5

=> 45/(2^2xx5^2

You can combine the bases with same powers

$\implies \frac{45}{10} ^ 2$

$\implies \frac{45}{100}$

$\implies 0.45$

Hope it helps

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# How to get correct autoref for theorems

I have some own theorem environments. And my theorems are numbered like the chapter. Now I would like to use autoref with that. But it does not work like it is described here (http://www.tug.org/applications/hyperref/manual.html) with aliascnt. So how can I get that?

\documentclass[ngerman,halfparskip,12pt,pointednumbers]{scrreprt}
\usepackage{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath,amsfonts,amssymb,amsthm,bbm}
\usepackage{hyperref}
\newtheoremstyle{absatz}
{17pt}{10pt}{}{-1pt}{\scshape\bfseries}{.}{\newline}{}

\theoremstyle{absatz}

\newtheorem{satz}{Satz}[chapter]
\newtheorem*{satz*}{Satz}
\newtheorem{lemma}[satz]{Lemma}
\newtheorem{corollar}[satz]{Korollar}

%\renewcommand*{\theoremautorefname}{Satz}
\begin{document}
\begin{satz}
\label{Satz}
text text theorem
\end{satz}

In \autoref{Satz} we have shown...
\end{document}

-
I tried this with the following example  \begin{satz} \todo{citation needed} \label{satz} theorem text \end{satz} In \autoref{satz} ... And of course this did not work, because of the \todo before the label. Just for anyone to remember, who stumbles upon this topic. (sorry I did not get a code environment working here.) – pirad Mar 1 '12 at 8:52

\autoref requires the name of the environment in the form \<name>autorefname to work. So, in order to properly "auto reference" your satz theorem, you should use

\newcommand{\satzautorefname}{Satz}

\documentclass[ngerman,halfparskip,12pt,pointednumbers]{scrreprt}
%\usepackage{babel}
%\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{amsmath,amsfonts,amssymb,amsthm,bbm}
\usepackage{hyperref}
\newtheoremstyle{absatz}
{17pt}{10pt}{}{-1pt}{\scshape\bfseries}{.}{\newline}{}

\theoremstyle{absatz}

\newtheorem{satz}{Satz}[chapter]
\newtheorem*{satz*}{Satz}
\newtheorem{lemma}[satz]{Lemma}
\newtheorem{corollar}[satz]{Korollar}
\newcommand{\satzautorefname}{Satz}

\begin{document}
\begin{satz}
\label{Satz}
text text theorem
\end{satz}

In \autoref{Satz} we have shown...
\end{document}

-
Thanks that works. And with aliascnt it even works for Lemma and Korollar. – pirad Mar 1 '12 at 8:35

I like all of my theorems etc numbered in sync with equations. For this I use the following macro to create my theorem-like environments:

% Some trickery to make \NewTheorem{} define theorem like environments
% work correctly with \autoref{}
\usepackage{aliascnt}
\def\NewTheorem#1{%
\newaliascnt{#1}{equation}
\newtheorem{#1}[#1]{#1}
\aliascntresetthe{#1}
\expandafter\def\csname #1autorefname\endcsname{#1}
}

Then I define

\NewTheorem{Theorem}
\NewTheorem{Corollary}

etc and use \begin{Theorem}...\end{Theorem} etc in the text.

Note that the environment name (Theorem, Corollary etc) is the label that is used by \autoref.

-

Here a somewhat "less professional" method that I find quite usefull and easy:

Load the hyperref package and insert the following in your preamble

\newcommand{\sref}[2]{\hyperref[#2]{#1 \ref{#2}}}

Now you can refer to anything by typing \sref{<theorem_description>}{<label_name>}. It the <theorem_description> can be anything you want. I personally don't mind a solution like this because it requires almost no extra typing. Also it can be typed differenly in any situation, which I find quite usefull at the beginning of a sentence for instance.

Here's a MWE

\documentclass{article}
\usepackage{hyperref}
\newcommand{\sref}[2]{\hyperref[#2]{#1 \ref*{#2}}}
\newtheorem{satz}{Satz}[section]
\newtheorem{folge}[satz]{Folge}

\begin{document}
\section{YOLO}

\begin{satz}\label{label1} Ein schoener Satz.
\end{satz}

\begin{folge}\label{label2}Wegen \sref{satz}{label1} folgt....
\end{folge}
\sref{Folge}{label2} has a capital, since it is the beginning of a sentence.

\end{document}

Output:

-
Little downside would be the late decision to rename Satz to Keks, or rather Abbildung to Diagram (have seen this desire more often to rename something to Keks). You might find package cleveref interesting, tough. – Johannes_B Oct 3 '14 at 14:50


text:
A characterization of the infinitely divisible

## squared Gaussian processes

### November 27, 2006

Abstract
We show that, up to multiplication by constants, a Gaussian process has an infinitely divisible square if and only if its covariance is the Green function of a transient Markov process.
Short title : Infinitely divisible squared Gaussian. Keywords : Gaussian processes, infinite divisibility, Markov processes, local time.
AMS 2000 subject classification : 60E07, 60G15, 60J25, 60J55.
1 Introduction The question of the infinite divisibility of squared Gaussian vectors is an old problem which was first raised by Paul Lévy in 1948 [11. Given a centered Gaussian vector $\left({\phi }_{1},\cdots ,{\phi }_{p}\right)$  when can the vector $\left({\phi }_{1}^{2},\cdots ,{\phi }_{p}^{2}\right)$  be written as a sum of $n$  i.i.d. $p$  vectors for every $n\in \mathbb{N}$  ? Many authors have worked on this problem. We refer the readers to Paranjape [15, Griffiths [9, Evans [7, Moran and Vere-Jones [14 [18and the references therein for more on this problem.
In 1984, Griffiths [10has established a characterization of the $p$  -dimensional centered Gaussian vectors with an infinitely divisible square. His criterion is difficult to use since it requires the computation of the signs of all the cofactors of the covariance matrix. Indeed, except for the Brownian motion (and more generally Gaussian Markov processes), there were no examples, in the literature, of processes satisfying this remarkable property nor examples of processes lacking it.
We have recently shown [8that the family of fractional Brownian motions provide examples of both kinds. A fractional Brownian motion is a real-valued centered Gaussian process with a covariance given by $g\left(x,y\right)=|x{|}^{\beta }+|y{|}^{\beta }-|x-y{|}^{\beta }$  where the index $\beta$  is in $\left(0,2\right)$  . We proved that when $\beta$  is in $\left(0,1\right]$  then the square of this process is infinitely divisible, and when $\beta$  is in $\left(1,2\right)$  it is not. The critical index $1$  corresponds to the Brownian motion. In both cases we have used the criterion of Griffiths. We have shown in [8that if the Green function of a transient Markov process is symmetric, then it is the covariance of a Gaussian process with an infinitely divisible square. In particular when the index $\beta$  is in $\left(0,1\right]$  , the covariance of the corresponding fractional Brownian motion can be interpreted as the Green function of the symmetric stable process with index $\left(\beta +1\right)$  , killed at its hitting time of $0$  . To treat the second case, we have shown directly that the condition of Griffiths is not satisfied.
In view of these examples, a natural question arises. Is the representation of the covariance function of a centered Gaussian process as the Green function of a symmetric Markov process, a necessary condition for the infinite divisibility of its square ? We show in this paper that, up to a multiplication by a constant function, the answer is affirmative. The result is presented in Section 3 in the form of a necessary and sufficient condition . The proof is based this time on a criterion for infinite divisibility established by Bapat [1, which we recall in Section 2. Although, at first sight, the verification of this criterion is as difficult as that of the equivalent criterion of Griffiths (one has to check here too the sign of each cofactor of the covariance matrix) it allows to significantly shorten the arguments.
Gaussian processes with a covariance equal to a Green function, play an important role in the study of Markov processes. Indeed, the Isomorphism Theorem of Dynkin [6provides an indentity in law connecting each of these Gaussian processes to the local time process of the corresponding symmetric Markov process. This identity has been exploited to study properties of the local time process of the Markov process, using similar properties of the Gaussian processes. We choose to mention here only two references, but many more papers on the subject are quoted in [2for example. In [12Marcus and Rosen have studied sample path properties of the local time process using similar properties of the Gaussian process. In [2Bass, Eisenbaum, and Shi used the Isomorphism Theorem to establish the transience of the most visited sites of a symmetric stable process. The question of characterising the Gaussian processes with a covariance that is equal to a Green function of a symmetric Markov process has been open since the Isomorphism Theorem of Dynkin was first proved. The importance of an answer to this question is twofold; it will give a powerful tool for the study of these Gaussian processes as well as that of their associated Markov processes. The results of Section 3 provide an answer to this question.
In Section 5, we show that the Brownian sheet does not have an infinitely divisible square. Recalling that linear combinations of the squared components of centered Gaussian vectors are infinitely divisible, we find this example and that of the fractional Brownian motion with index in $\left(1,2\right)$  , somewhat counterintuitive.
2 The criterion of Bapat Let $A=\left({A}_{ij}{\right)}_{1\le i,j\le p}$  be a $p×p$  matrix. We write $A\ge 0$  if ${A}_{ij}\ge 0$  for all $i$  , $j$  .
Definition 2.1 A matrix $A=\left({A}_{ij}{\right)}_{1\le i,j\le p}$  is said to be an $M$  -matrix if (i) ${A}_{ij}\le 0$  for all $i\ne j$  (ii) $A$  is nonsingular and ${A}^{-1}\ge 0$  .
We refer the reader to [3for the theory of $M$  -matrices.
Definition 2.2 A diagonal matrix is called a signature matrix if all its entries are either $1$  or $-1$  .
The Laplace transform, $\psi \left({t}_{1},\cdots ,{t}_{p}\right)$  of the square of a $p-$  dimensional Gaussian vector is given by $\psi \left(t\right)=\left[det\left(I+GT\right){\right]}^{-1/2}$  where $t=\left({t}_{1}\cdots ,{t}_{p}\right)$  , $G$  is a positive definite $p×p$  -matrix, $T$  is the diagonal matrix with entries ${T}_{ii}={t}_{i}$  , and $I$  is the $p×p$  -identity matrix.
Bapat [1obtained the following characterization of the matrices $G$  for which the Laplace transform $\psi$  is infinitely divisible, that is, for which ${\psi }^{\delta }\left(t\right)$  is a Laplace transform for any $\delta >0$  .
Theorem A. The Laplace transform $\psi$  is infinitely divisible if, and only if, there exists a signature matrix $S$  such that $S{G}^{-1}S$  is an $M$  -matrix.
Remark 2.3 It is elementary to check, using Theorem A, that a centered $3$  -dimensional Gaussian vector $\left({\eta }_{1},{\eta }_{2},{\eta }_{3}\right)$  such that $IE\left({\eta }_{1}{\eta }_{2}\right)IE\left({\eta }_{2}{\eta }_{3}\right)IE\left({\eta }_{3}{\eta }_{1}\right)<0$  can not have an infinitely divisible square.
3 A necessary and sufficient condition for infinite divisibility In a previous work [8on squared Gaussian processes, we have established the following result. Theorem B. Let $g$  be the Green function of a strongly symmetric transient Borel right Markov process with state space $\left(E,\mathcal{ℰ}\right)$  . Let $\eta$  be a centered Gaussian process with covariance $g$  . Then the process ${\eta }^{2}$  is infinitely divisible.
The set $E$  (by the assumption of a right process) is a Borel subset of a compact metric space and $\mathcal{ℰ}$  is the $\sigma$  -field of its Borel sets. Note that if $\left({\eta }_{x}^{2}\left(x\right),x\in E\right)$  is an infinitely divisible squared Gaussian process, then for any $\mathcal{ℰ}$  -measurable real valued function $d$  , the process $\left({d}^{2}\left(x\right){\eta }_{x}^{2}\left(x\right),x\in E\right)$  is also infinitely divisible. We therefore have Corollary 3.1. Let $g$  be the Green function of a symmetric transient Markov process on a state space $\left(E,\mathcal{ℰ}\right)$  . Then for any $\mathcal{ℰ}$  -measurable real valued function $d$  on $E$  , there exists a centered Gaussian process $\eta$  with covariance equal to $\left(d\left(x\right)g\left(x,y\right)d\left(y\right),\left(x,y\right)\in E×E\right)$  . The process ${\eta }^{2}$  is infinitely divisible.
We shall first treat the case when $E$  is a finite set. Theorem 3.2 completes Corollary 3.1, for this case, by showing that its sufficient condition on the covariance, is also necessary for the infinite divisibility of ${\eta }^{2}$  .
Theorem 3.2 Let $\eta$  be a centered Gaussian vector, indexed by a finite set $E$  , with a positive definite covariance function $\left(G\left(x,y\right),\left(x,y\right)\in E×E\right)$  .
The vector ${\eta }^{2}$  is infinitely divisible if, and only if, there exists a real valued function $d$  on $E$  such that for any $x,y\in E$  , $G\left(x,y\right)=d\left(x\right)g\left(x,y\right)d\left(y\right),$  where the function $g$  is the Green function of a transient symmetric Markov process.
Remark 3.3 Note that the Green function $g$  of a symmetric Markov process $X$  is always positive definite. Indeed, it is semi-positive definite (see for example Marcus and Rosen [12or Eisenbaum [8) and it has been shown in [8section II, that for any ${x}_{1},{x}_{2},...{x}_{n}$  in the state space of $X$  , the matrix $\left(g\left({x}_{i},{x}_{j}\right){\right)}_{1\le i,j\le n}$  is invertible. Let $\eta$  be a centered Gaussian process indexed by an infinite set $E$  . Then it has an infinitely divisible square if for every finite subset $F$  of $E$  the covariance of $\left({\eta }_{x},x\in F\right)$  satisfies the condition of Theorem 3.2. This does not guarantee, a-priori, that for some deterministic function $d$  , the covariance of $\left(d\left(x\right){\eta }_{x},x\in E\right)$  is the Green function of a transient symmetric Markov process. Restricting our attention to $E=\mathbb{R}$  , and under the additional continuity assumption on the covariance function, the following theorem establishes the necessity of that condition.
Theorem 3.4 Let $\eta$  be a centered Gaussian process, indexed by $\mathbb{R}$  , with a positive definite covariance function $\left(G\left(x,y\right),\left(x,y\right)\in {\mathbb{R}}^{2}\right)$  . Assume that $G$  is jointly continuous. If the process ${\eta }^{2}$  is infinitely divisible , then there exists a measurable real valued positive function $d$  on $\mathbb{R}$  such that for any $x,y\in \mathbb{R}$  $G\left(x,y\right)=d\left(x\right)g\left(x,y\right)d\left(y\right),$  where the function $g$  is the Green function of a symmetric transient Markov process.
Theorem 3.4 is proved in Section 4. Although Theorem 3.2 was a good hint to anticipate Theorem 3.4, we could not use it directly to prove it.
Our argument is based on an explicit construction of the function $d$  and of the Green function $g$  of the Markov process that will be associated with the Gaussian process $\eta$  . As a first step of this construction we show that the covariance $G$  has to be positive. The proof of Theorem 3.4 remains valid, with simple changes, when the index set $\mathbb{R}$  is replaced by a separable locally compact metric space. Moreover, we will see in Section 4, that as a by-product of the proofs of Theorem 3.2 and Theorem 3.4, we obtain the following characterization of the associated Gaussian processes, which makes use of the definition below.
Definition 3.5 A $p×p$  matrix $A$  is said to be diagonally dominant if for every $i=1,...,p$  : ${\sum }_{j=1}^{p}{A}_{ij}\ge 0.$  Theorem 3.6 (i) A positive definite matrix $G$  is the Green function of a finite state space transient Markov process, if, and only if, ${G}^{-1}$  is a diagonally dominant $M$  -matrix.
(ii) Let $E$  be a separable locally compact metric space. Let $\left({\eta }_{x},x\in E\right)$  be a Gaussian process with a continuous positive definite covariance $\left(G\left(x,y\right),\left(x,y\right)\in {E}^{2}\right)$  . Then $G$  is the Green function of a transient Markov process on $E$  , if and only if, for every ${x}_{1},{x}_{2},...{x}_{p}$  in $E$  , the inverse of the matrix $\left(G\left({x}_{i},{x}_{j}\right){\right)}_{1\le i,j\le p}$  is a diagonally dominant $M$  -matrix.
4 Proofs Proof of Theorem 3.2 Let $E$  be the finite set $\left\{{x}_{1},\cdots ,{x}_{p}\right\}$  . Theorem A guarantees the existence of a signature matrix $S$  with diagonal ${S}_{i},i=1,\cdots ,p$  such that the covariance matrix of $\left({S}_{i}{\eta }_{{x}_{i}},i=1,\cdots p\right)$  is positive. We shall therefore assume, from the onset, that the covariance $G$  is positive.
We will actually prove that the covariance $G$  satisfies
 $\begin{array}{c}G\left(x,y\right)=\frac{D\left(y\right)}{D\left(x\right)}g\left(x,y\right),\end{array}$ (1)
where the function $g$  is the Green function of a transient Markov process and $D$  is a strictly positive function.
This will be sufficient to establish Theorem 3.2. Indeed, if $G$  satisfies (1) then we have $g\left(x,y\right)=\frac{D\left(x\right)}{D\left(y\right)}G\left(x,y\right).$  Let $U$  be the corresponding potential with density $g\left(x,y\right)$  with respect to a reference measure $\mu$  . That is, $Uf\left(x\right)=\int g\left(x,y\right)f\left(y\right)\mu \left(dy\right).$  We then set, $m\left(dy\right)=\mu \left(dy\right)/{D}^{2}\left(y\right)$  . With respect to $m$  , the kernel $U$  has densities $\stackrel{~}{g}\left(x,y\right)=D\left(x\right)G\left(x,y\right)D\left(y\right)$  . Consequently $\stackrel{~}{g}$  is a symmetric Green function. Setting $d=1/D$  , one obtains the necessity of the condition of Theorem 3.2.
To prove (1), let $G$  be the covariance matrix of a $p$  -dimensional centered Gaussian vector with an infinitely divisible square. By Theorem A, ${G}^{-1}$  is an $M$  -matrix. This implies (see Berman and Plemmons [3) that
 $\begin{array}{c}{G}^{-1}=cI-B,\end{array}$ (2)
where $B\ge 0$  and $c$  is strictly greater than the absolute value of any eigenvalue of $B$  . Further by [3( Chapter 6, p.137 M36), since ${G}^{-1}$  is an $M$  -matrix there exists a diagonal matrix $D$  such that ${D}_{ii}>0$  for all $i$  and $D{G}^{-1}{D}^{-1}$  has strictly positive row sum. This means that for any $i$  ${\sum }_{j=1}^{p}\left(D{G}^{-1}{D}^{-1}{\right)}_{ij}>0.$  According Definition 3.5, $D{G}^{-1}{D}^{-1}$  is diagonally dominant. Set $T=D\left(\frac{1}{c}B\right){D}^{-1}$  . Then $D{G}^{-1}{D}^{-1}=D\left(cI-B\right){D}^{-1}=c\left(I-T\right).$  Note that for any $i$  ${\sum }_{j=1}^{p}{T}_{ij}<1.$  The matrix $T$  is therefore the transition matrix of a transient Markov chain $\left({X}_{n}{\right)}_{n\in \mathbb{N}}$  with state space $E=\left\{{x}_{1},{x}_{2},...,{x}_{p}\right\}$  satisfying ${T}_{ij}=I{P}_{{x}_{i}}\left({X}_{1}={x}_{j}\right)$  , and ${\sum }_{j=1}^{p}I{P}_{{x}_{i}}\left({X}_{1}={x}_{j}\right)=1-I{P}_{{x}_{i}}\left({X}_{1}=\Delta \right)$  , where $\Delta$  denotes a cemetery point. Let ${\ell }_{\infty }^{{x}_{j}}$  denote the total number of visits of $X$  to the state ${x}_{j}$  . The Green function of $X$  is defined by $g\left({x}_{i},{x}_{j}\right)=I{E}_{{x}_{i}}\left({\ell }_{\infty }^{{x}_{j}}\right).$  It can be computed as follows $I{E}_{{x}_{i}}\left({\ell }_{\infty }^{{x}_{j}}\right)=I{E}_{{x}_{i}}\left({\sum }_{n=0}^{\infty }{1}_{\left\{{X}_{n}={x}_{j}\right\}}\right)={\sum }_{n=0}^{\infty }{T}_{ij}^{n}=\left(I-T{\right)}_{ij}^{-1},$  which is defined and is finite, since, by (2) and the discussion following it, the spectral radius of $T$  is stricly smaller than $1$  . Hence we can write $cDG{D}^{-1}=g;$  that is, for every $x$  , $y$  in $E$  $cG\left(x,y\right)=\frac{D\left(y\right)}{D\left(x\right)}g\left(x,y\right).$  We shall now use the well-known method to turn a finite state space Markov chain into a continuous time Markov process by subordination to a Poisson process with rate c (see for example Cinlar [5). By its construction, this Markov process is transient with potential density ( Green function) equal to $g$  and (1) is established. $\square$  Remark 4.1 Let $Y$  be the Markov process with potential density $g$  . Equation (1) looks as if $G$  is the potential density of an h-path transform of $Y$  . This is really the case if $D$  is excessive for $Y$  . Unfortunately this is not true in general. Consequently we see that the collection of covariance functions that correspond to Gaussian processes with an infinitely divisible square is somewhat richer than the set of Green functions of symmetric Markov processes. This remark remains true also when the index set $E$  of the Gaussian process is infinite.
To select covariance matrices that correspond to Green functions of Markov processes we have the condition given by Theorem 3.6 (i). Indeed, assume that ${G}^{-1}$  is a diagonally dominant M-matrix. Keeping the notations of the proof of Theorem 3.2, we can choose $D=I$  and obtain $T=\frac{1}{c}B$  . For any $i$  ${\sum }_{j=1}^{p}{T}_{ij}\le 1$  . Since the spectral radius of $T$  is strictly smaller then $1$  , for at least one $i$  the above inequality must be strict. Therefore $T$  is the transition matrix of a transient Markov chain and we can conclude that $G$  is the Green function of a transient Markov process with finite state space. To see that the condition is also necessary, consider the Green function $g$  of a transient symmetric Markov process $X$  with a finite state space $E=\left\{{x}_{1},{x}_{2},...,{x}_{p}\right\}$  . Then the inverse of the matrix $G=\left(g\left({x}_{i},{x}_{j}\right){\right)}_{1\le i,j\le p}$  is a diagonally dominant M-matrix. Indeed ${G}_{ij}={\lambda }_{j}\left(I-T{\right)}_{ij}^{-1}$  , where ${\lambda }_{j}$  is the expected value of the exponential sojourn times in state $j$  , $T$  is the transition matrix of the Markov chain $\left(X\left({S}_{n}\right){\right)}_{n\in \mathbb{N}}$  , and ${S}_{n}$  is the $n$  -th jump time of $X$  Proof of Theorem 3.4 We first show that $G$  has to be positive. For a fixed $n$  , we define the function ${d}_{n}$  on $\mathbb{R}$  by ${d}_{n}\left(x\right)=\frac{k}{{2}^{n}}if\frac{k}{{2}^{n}}\le x<\frac{k+1}{{2}^{n}}.$  Let $K$  be the compact set $\left[a,b\right]$  with $a  , and let ${d}_{n}\left(K\right)=\left\{{d}_{n}\left(x\right):x\in K\right\}$  .
The set ${d}_{n}\left(K\right)$  is finite set. Since the process $\left({\eta }_{{d}_{n}\left(x\right)}^{2},x\in K\right)$  is infinitely divisible, thanks to Theorem A there exists a signature function ${s}_{n}$  (ie a function taking values in $\left\{-1,+1\right\}$  ) on ${d}_{n}\left(K\right)$  such that for every $x,y\in K$  ${s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right)G\left({d}_{n}\left(x\right),{d}_{n}\left(y\right)\right)\ge 0.$  Note that
 $\begin{array}{c}{s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right)G\left({d}_{n}\left(x\right),{d}_{n}\left(y\right)\right)=|G\left({d}_{n}\left(x\right),{d}_{n}\left(y\right)\right)|\end{array}$ (3)
Since $G$  is continuous, we obtain by letting $n$  tend to $\infty$  that ${lim}_{n}{s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right)\right)$  exists for all $\left(x,y\right)$  for which $G\left(x,y\right)\ne 0$  , and for such $\left(x,y\right)$  $\left({lim}_{n}{s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right)\right)G\left(x,y\right)=|G\left(x,y\right)|.$  For $\left(x,y\right)$  in ${K}^{2}$  such that $G\left(x,y\right)=0$  , there exists a finite sequence ${a}_{1},{a}_{2},...,{a}_{p}$  such that $G\left(x,{a}_{1}\right)G\left({a}_{1},{a}_{2}\right)...G\left({a}_{p-1},{a}_{p}\right)G\left({a}_{p},y\right)\ne 0$  . Indeed, for $z\in K$  , set $C\left(z\right)=\left\{u\in \mathbb{R}:G\left(z,u\right)\ne 0\right\}$  . For each $z$  , $C\left(z\right)$  is an open set and ${\cup }_{z\in K}C\left(z\right)$  is a covering of the compact set $K$  . Thus there exists a finite subcovering $C\left({z}_{1}\right),C\left({z}_{2}\right),...C\left({z}_{m}\right)$  of the set $K$  . The sets of the covering are not disjoint. Let $C\left({z}_{{i}_{1}}\right)$  be one of these sets. $C\left({z}_{{i}_{1}}\right)$  is a union of disjoint intervals. If $C\left({z}_{{i}_{1}}\right)$  doesn't cover $K$  , then there exists ${z}_{{i}_{2}}$  in $\left\{{z}_{1},{z}_{2},...,{z}_{m}\right\}$  such that $C\left({z}_{{i}_{2}}\right)$  covers some of the endpoints of $C\left({z}_{{i}_{1}}\right)$  that are in $K$  . If $C\left({z}_{{i}_{1}}\right)\cup C\left({z}_{{i}_{2}}\right)$  do not cover $K$  there exists ${z}_{{i}_{3}}$  in $\left\{{z}_{1},{z}_{2},...,{z}_{m}\right\}$  such that $C\left({z}_{{i}_{3}}\right)$  covers some of the endpoints of $C\left({z}_{{i}_{1}}\right)\cup C\left({z}_{{i}_{2}}\right)$  that are in $K$  , and we may continue on until we exhaust all the finite covering above. Then we just have to make use of the sequence $\left({z}_{{i}_{1}},{z}_{{i}_{2}},...,{z}_{{i}_{m}}\right)$  to construct $\left({a}_{1},{a}_{2},...,{a}_{p}\right)$  ( $p\le m$  ) connecting $x$  to $y$  such that : $G\left(x,{a}_{1}\right)G\left({a}_{1},{a}_{2}\right)...G\left({a}_{p-1},{a}_{p}\right)G\left({a}_{p},y\right)\ne 0$  . Since ${s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}^{2}\left({d}_{n}\left({a}_{1}\right)\right){s}_{n}^{2}\left({d}_{n}\left({a}_{2}\right)\right)...{s}_{n}^{2}\left({d}_{n}\left({a}_{p}\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right)={s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right)$  we obtain, using ( 3 ), the existence of ${lim}_{n}{s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right)$  for all $x$  , $y$  . Set $H\left(x,y\right)={lim}_{n}{s}_{n}\left({d}_{n}\left(x\right)\right){s}_{n}\left({d}_{n}\left(y\right)\right).$  The function $H$  is symmetric and by ( 3 ) $H\left(x,y\right)=sign\left(G\left(x,y\right)\right)$  and by its definition for all $x,y,z\in K$  $H\left(x,y\right)=H\left(x,z\right)H\left(z,y\right).$  Hence there exists a signature function ${S}_{K}$  on $K$  (take $H\left(.,{z}_{0}\right)$  for a fixed ${z}_{0}$  in $K$  ) such that for any $x,y\in K$  ${S}_{K}\left(x\right){S}_{K}\left(y\right)G\left(x,y\right)=|G\left(x,y\right)|$  Denote by ${S}_{n}$  the function ${S}_{\left[-n,n\right]}$  , then repeating the above argument and letting $n$  tend to $\infty$  , we finally obtain the existence of $S$  satisfying $S\left(x\right)S\left(y\right)G\left(x,y\right)=|G\left(x,y\right)|.$  If $S$  is not identically equal to $1$  , then there exists a point $x$  at which $S\left(x\right)=1$  and $limin{f}_{y\to x}S\left(y\right)=-1$  . By continuity this means that $G\left(x,x\right)$  has to be equal to $0$  (because $G\left(x,{y}_{n}\right)<0$  for a sequence $\left({y}_{n}{\right)}_{n\ge 0}$  that converges to $x$  ).
This is excluded since $G$  is positive definite. Consequently $G$  is positive.
We define the measure $m$  on $\mathbb{R}$  by $m\left(dy\right)=\left(1\wedge \frac{1}{\sqrt{G\left(y,y\right)}}\right){e}^{-|y|}dy.$  Note that the measure $m$  is finite and $\int \sqrt{G\left(y,y\right)}m\left(dy\right)<\infty$  . Making use of the covariance inequality ( $G\left(x,y\right)\le \sqrt{G\left(x,x\right)G\left(y,y\right)}$  for $x$  , $y$  in $\mathbb{R}$  ), and the dominated convergence Theorem, we see that the function
 $\begin{array}{c}\chi \left(x\right)=\int G\left(x,y\right)m\left(dy\right),\end{array}$ (4)
is continuous.
We now consider the fixed compact set $K=\left[a,b\right]$  with $a  . For any integer $n$  , we keep the definition of ${d}_{n}$  introduced at the beginning of the proof. We set ${I}_{n}=\left\{k\in \mathbb{N}:-n{2}^{n}+1\le k\le n{2}^{n}-1and\frac{k}{{2}^{n}}\in {d}_{n}\left(K\right)\right\}.$  Define ${G}_{n}$  on the set $\left\{\left(\frac{k}{{2}^{n}},\frac{\ell }{{2}^{n}}\right):k,\ell \in {I}_{n}\right\}$  by ${G}_{n}\left(\frac{k}{{2}^{n}},\frac{\ell }{{2}^{n}}\right)=G\left(\frac{k}{{2}^{n}},\frac{\ell }{{2}^{n}}\right){\int }_{\frac{\ell }{{2}^{n}}}^{\frac{\ell +1}{{2}^{n}}}m\left(dy\right).$  Since the process ${\eta }^{2}$  is infinitely divisible and the positive measure $m$  has a support equal to $\mathbb{R}$  , ${G}_{n}^{-1}$  is an M-matrix ( ${G}_{n}$  need not be symmetric). Hence we can write ${G}_{n}^{-1}={c}_{n}I-{B}_{n},$  where ${B}_{n}\ge 0$  and ${c}_{n}$  is stricly greater than the absolute value of any eigenvalue of ${B}_{n}$  .
Define ${\chi }_{n}\left(\frac{k}{{2}^{n}}\right)={\sum }_{\ell \in {I}_{n}}{G}_{n}\left(\frac{k}{{2}^{n}},\frac{\ell }{{2}^{n}}\right)$  and let ${D}_{n}$  be the diagonal matrix $diag\left({\chi }_{n}\left(\frac{k}{{2}^{n}}\right),k\in {I}_{n}\right)$  , Then, ${D}_{n}{e}_{n}={G}_{n}{e}_{n}$  , where ${e}_{n}$  is equal to $\left(1,1,...1\right)$  . Consequently, for any $k\in {I}_{n}$  ${\sum }_{\ell \in {I}_{n}}\left[{D}_{n}^{-1}{G}_{n}^{-1}{D}_{n}{\right]}_{k,\ell }=\left({\chi }_{n}\left(\frac{k}{{2}^{n}}\right){\right)}^{-1}>0.$  Setting ${T}_{n}=\frac{1}{{c}_{n}}{D}_{n}^{-1}{B}_{n}{D}_{n}$  , we have
 $\begin{array}{c}{D}_{n}^{-1}{G}_{n}^{-1}{D}_{n}={c}_{n}\left(I-{T}_{n}\right).\end{array}$ (5)
and for every $k$  , ${\sum }_{\ell \in {I}_{n}}{T}_{n}\left(k,\ell \right)<1$  . Consequently, the matrix ${T}_{n}$  is the transition matrix of a transient Markov chain.
Rewriting (5) it follows that
 $\begin{array}{c}{G}_{n}={D}_{n}{O}_{n}\left({D}_{n}{\right)}^{-1},\end{array}$ (6)
with ${O}_{n}=\frac{1}{{c}_{n}}\left(I-{T}_{n}{\right)}^{-1}$  .
Let $A$  be a square matrix of size $|{I}_{n}|$  defined by $A=\left(A\left(\frac{k}{{2}^{n}},\frac{\ell }{{2}^{n}}\right){\right)}_{k,l\in {I}_{n}}$  .
We associate with $A$  an operator on the set of the continuous functions with compact support. We denote this operator $\mathbf{A}$  and define it as follows. Let $f$  be a continuous function on $\mathbb{R}$  with a compact support then the function $\mathbf{A}f$  is given by $\mathbf{A}f\left(x\right)={\sum }_{\ell \in {I}_{n}}A\left({d}_{n}\left(x\right),\frac{\ell }{{2}^{n}}\right)f\left(\frac{\ell }{{2}^{n}}\right)\forall x\in \mathbb{R}$  with the convention that $A\left(\frac{k}{{2}^{n}},\frac{\ell }{{2}^{n}}\right)=0$  when $k$  is outside of ${I}_{n}$  .
That way we associate with ${D}_{n}$  (resp. ${D}_{n}^{-1}$  , ${G}_{n}$  , ${O}_{n}$  ) the operator ${\mathbf{D}}_{\mathbf{n}}$  (resp.
${\mathbf{D}}_{\mathbf{n}}^{-1}$  , ${\mathbf{G}}_{\mathbf{n}}$  , ${\mathbf{O}}_{\mathbf{n}}$  ). Note that we have for every function $f$  and every $x$  in $\mathbb{R}$  ,
 $\begin{array}{ccc}{\mathbf{D}}_{\mathbf{n}}f\left(x\right)={\chi }_{n}\left({d}_{n}\left(x\right)\right)f\left({d}_{n}\left(x\right)\right)& & \end{array}$
 $\begin{array}{ccc}\left({\mathbf{D}}_{\mathbf{n}}{\right)}^{-1}f\left(x\right)=\left({\chi }_{n}\left({d}_{n}\left(x\right)\right){\right)}^{-1}f\left({d}_{n}\left(x\right)\right)& & \end{array}$
 $\begin{array}{ccc}{\mathbf{G}}_{\mathbf{n}}f\left(x\right)={\sum }_{\ell \in {I}_{n}}G\left({d}_{n}\left(x\right),\frac{\ell }{{2}^{n}}\right)f\left(\frac{\ell }{{2}^{n}}\right){\int }_{\frac{\ell }{{2}^{n}}}^{\frac{\ell +1}{{2}^{n}}}m\left(dy\right).& & \end{array}$
By (5), we know that the sequence $\left({\mathbf{O}}_{\mathbf{n}}{\right)}_{n\ge 0}$  satisfies
 $\begin{array}{ccc}{\mathbf{O}}_{\mathbf{n}}={\mathbf{D}}_{\mathbf{n}}^{-1}{\mathbf{G}}_{\mathbf{n}}{\mathbf{D}}_{\mathbf{n}}.& & \end{array}$
Moreover, for each $n$  , ${\mathbf{O}}_{\mathbf{n}}$  is the Green operator of a finite state space Markov process.
Let $O$  be the operator defined on ${\mathcal{C}}_{K}$  ( the continuous functions with support in $K$  ), by
 $\begin{array}{ccc}Of\left(x\right)={D}^{-1}\mathcal{G}Df\left(x\right)x\in K,& & \end{array}$
where
 $\begin{array}{ccc}Df\left(x\right)={\chi }_{{}_{K}}\left(x\right)f\left(x\right),& & \end{array}$
 $\begin{array}{ccc}{D}^{-1}f\left(x\right)=\left({\chi }_{{}_{K}}\left(x\right){\right)}^{-1}f\left(x\right)& & \end{array}$
and
 $\begin{array}{ccc}\mathcal{G}f\left(x\right)={\int }_{K}G\left(x,y\right)f\left(y\right)m\left(dy\right)& & \end{array}$
with
 $\begin{array}{ccc}{\chi }_{{}_{K}}\left(x\right)={\int }_{K}G\left(x,y\right)m\left(dy\right).& & \end{array}$
Note that by the continuity of $G$  and the fact that $G\left(x,x\right)>0$  and $m$  has $\mathbb{R}$  as its support, ${\chi }_{{}_{K}}$  is continuous, strictly positive on $K$  and thus bounded below by a strictly positive constant.
Lemma 4.2 For every function $f\in {\mathcal{C}}_{K}$  , the sequence $\left({\mathbf{O}}_{\mathbf{n}}f{\right)}_{n\ge 0}$  converges to $Of$  uniformly on $K$  Proof of Lemma 4.2 For any continuous function $f$  on ${\mathcal{C}}_{K}$  and any $x\in K$  , we have ${\mathbf{O}}_{\mathbf{n}}f\left(x\right)=\frac{1}{{\chi }_{n}\left({d}_{n}\left(x\right)\right)}{\int }_{K}G\left({d}_{n}\left(x\right),{d}_{n}\left(y\right)\right){\chi }_{n}\left({d}_{n}\left(y\right)\right)f\left({d}_{n}\left(y\right)\right)m\left(dy\right),$  and $Of\left(x\right)=\frac{1}{{\chi }_{{}_{K}}\left(x\right)}{\int }_{K}G\left(x,y\right){\chi }_{{}_{K}}\left(y\right)f\left(y\right)m\left(dy\right).$  Note that ${\chi }_{n}\left({d}_{n}\left(x\right)\right)={\int }_{K}G\left({d}_{n}\left(x\right),{d}_{n}\left(y\right)\right)m\left(dy\right).$  Since $G$  is uniformly continuous on compacts and $m$  is finite, the sequence $\left({\chi }_{n}\left({d}_{n}\left(x\right)\right){\right)}_{n\in \mathbb{N}}$  converges uniformly on $K$  to ${\chi }_{{}_{K}}\left(x\right)$  . Hence the sequence $G\left({d}_{n}\left(x\right),{d}_{n}\left(y\right)\right){\chi }_{n}\left({d}_{n}\left(y\right)\right)f\left({d}_{n}\left(y\right)\right)$  converges uniformly on $K×K$  to $G\left(x,y\right){\chi }_{{}_{K}}\left(y\right)f\left(y\right)$  . $\square$  We would like to show that $O$  is a Green operator. For this we shall use the following lemma which in essence is due to Hunt (see Meyer [13(chapter X, p.255) or Taylor [16). Lemma C Let $V$  be a kernel on a measurable space $\left(E,\mathcal{ℰ}\right)$  such that $V$  satisfies the complete maximum principal and the function $V1$  is bounded.
There exists then a subMarkovian resolvent $\left({V}_{p}\right)$  such that ${V}_{0}=V$  .
As Meyer has noted in [13(p.253), if we assume that $V$  is continuous then it is sufficient to verify the complete maximum principle for continuous functions with compact support only. More precisely, we have to verify that for any $a\ge 0$  , any two positive continuous functions with compact support, $\left(f,h\right)$  , if for all $x$  in $\left\{h>0\right\}$  $a+Vf\left(x\right)\ge Vh\left(x\right),$  then the inequality remains valid for all $x$  in $E$  .
Lemma 4.3 The kernel $O$  satisfies the complete maximum principle on $K$  .
Proof of Lemma 4.3 First note that $O$  maps continuous functions with support in $K$  to continuous functions and is therefore, a continuous kernel on $K$  . Let $f$  and $h$  be in ${\mathcal{C}}_{K}^{+}$  and $a\ge 0$  and suppose that
 $\begin{array}{c}a+Of\left(x\right)\ge Oh\left(x\right)forallx\in \left\{h>0\right\}.\end{array}$ (7)
Recall that $\left\{h>0\right\}$  is contained in $K$  . We need to show that ( 7 ) is satisfied by all $x\in K$  . Suppose that this is not true. Then there exists a constant $b>0$  and a point ${x}_{0}$  in $K$  so that $a+Of\left({x}_{0}\right)  Let $\epsilon >0$  and $N$  be such that for any $n>N$  $|{\mathbf{O}}_{\mathbf{n}}f\left(x\right)-Of\left(x\right)|<\epsilon \forall x\in K$  and $|{\mathbf{O}}_{\mathbf{n}}h\left(x\right)-Oh\left(x\right)|<\epsilon \forall x\in K.$  Such $N$  exists by Lemma 4.2 and the fact that $f,h\in {\mathcal{C}}_{K}^{+}$  .
By ( 7 ), for every $n>N$  and $x$  in $\left\{h>0\right\}$  ,
 $\begin{array}{c}a+2\epsilon +{\mathbf{O}}_{\mathbf{n}}f\left(x\right)\ge {\mathbf{O}}_{\mathbf{n}}h\left(x\right).\end{array}$ (8)
Recall now that for each $n$  , ${O}_{n}$  is a potential of a Markov process on ${I}_{n}$  . It follows from J.C. Taylor [17, that ${O}_{n}$  satisfies the complete maximum principle on ${I}_{n}$  and, by its definition, ${\mathbf{O}}_{\mathbf{n}}$  satisfies the complete maximum principle on $K$  . Consequently ( 8 ) is valid for all $x$  in $K$  . In particular, we have $a+2\epsilon +{\mathbf{O}}_{\mathbf{n}}f\left({x}_{0}\right)\ge {\mathbf{O}}_{\mathbf{n}}h\left({x}_{0}\right).$  On the other hand we have $a-2\epsilon +{\mathbf{O}}_{\mathbf{n}}f\left({x}_{0}\right)<{\mathbf{O}}_{\mathbf{n}}h\left({x}_{0}\right)-b.$  Choosing $\epsilon   leads to the desired contradiction. $\square$  In order to simplify the rest of the proof of Theorem 3.4 we use now, instead of $O$  , the operator ${V}_{K}$  defined by ${V}_{K}f\left(x\right)=\frac{1}{{\chi }_{{}_{K}}\left(x\right)}{\int }_{K}G\left(x,y\right)f\left(y\right)m\left(dy\right).$  First we note that ${V}_{K}1=1$  . Further since ${V}_{K}f=O\left(f/{\chi }_{{}_{K}}\right)$  , ${V}_{K}$  satisfies the complete maximum principle on $K$  , by Lemma 4.3.
Let ${K}_{n}$  be the compact set $\left[-n,n\right]$  for $n\in \mathbb{N}$  , and denote by ${V}_{{}_{{K}_{n}}}$  the corresponding operator as defined above for $K.$  Further, arguing as for ${\chi }_{{}_{K}}$  , one can show that $\chi$  defined in (4) is strictly positive.
Lemma 4.4 Let $V$  be the operator defined on bounded Borel functions by
 $\begin{array}{c}Vf\left(x\right)=\frac{1}{\chi \left(x\right)}{\int }_{\mathbb{R}}G\left(x,y\right)f\left(y\right)m\left(dy\right).\end{array}$ (9)
Then there exists a sub-Markovian resolvent $\left({V}_{p}\right)$  such that ${V}_{0}=V$  .
Proof of Lemma 4.4 Note that $V1=1$  , and that $V$  maps continuous functions with compact support to continuous functions on $\mathbb{R}$  . Let $f$  and $h$  be two positive continuous functions with compact supports and $a\ge 0$  and suppose that
 $\begin{array}{c}a+Vf\left(x\right)\ge Vh\left(x\right)forallx\in \left\{h>0\right\},\end{array}$ (10)
we need to show that this is satisfied for all $x$  in $\mathbb{R}$  .
We denote by $H$  a compact set that contains both the compact supports of $f$  and of $h$  . There exists ${n}_{0}$  such that for $n>{n}_{0}$  , H is contained in ${K}_{n}$  . Hence ( 10 ) can be written as $a\frac{\chi \left(x\right)}{{\chi }_{{}_{{K}_{n}}}\left(x\right)}+{V}_{{}_{{K}_{n}}}f\left(x\right)\ge {V}_{{}_{{K}_{n}}}h\left(x\right)forallx\in \left\{h>0\right\}.$  We have shown, when defining $m$  , that $\chi$  is a continuous function. Moreover, for any $n\in \mathbb{N}$  , the function ${\chi }_{{}_{{K}_{n}}}$  is continuous and the sequence $\left({\chi }_{{}_{{K}_{n}}}\right)$  is increasing and converges simply to $\chi$  . Consequently, by Dini's Theorem, $\left({\chi }_{{}_{{K}_{n}}}\right)$  converges uniformly to $\chi$  and $\chi \left(x\right)$  is strictly positive. Since the sequence $\left({\chi }_{{}_{{K}_{n}}}{\right)}_{n>{n}_{0}}$  is bounded below by a strictly positive constant on $H$  , $\chi /{\chi }_{{}_{{K}_{n}}}$  converges to $1$  uniformly on $H$  . Hence for every $\epsilon >0$  , there exists $N$  such that for every $n>N$  , $a\left(1+\epsilon \right)+{V}_{{}_{{K}_{n}}}f\left(x\right)\ge {V}_{{}_{{K}_{n}}}h\left(x\right)forallx\in \left\{h>0\right\}.$  Since the operator ${V}_{{}_{{K}_{n}}}$  satisfies the complete maximum principle on ${K}_{n}$  , the above inequality is still true for $x$  in ${K}_{n}$  . We now multiply each side of the inequality by ${\chi }_{{}_{{K}_{n}}}\left(x\right)/\chi \left(x\right)$  to obtain $a\left(1+\epsilon \right){\chi }_{{}_{{K}_{n}}}\left(x\right)/\chi \left(x\right)+Vf\left(x\right)\ge Vh\left(x\right)forallx\in {K}_{n}.$  Since ${\chi }_{{}_{{K}_{n}}}\left(x\right)/\chi \left(x\right)\le 1$  we finally get $a\left(1+\epsilon \right)+Vf\left(x\right)\ge Vh\left(x\right)forallx\in {K}_{n},$  and letting $n$  tend to $\infty$  $a\left(1+\epsilon \right)+Vf\left(x\right)\ge Vh\left(x\right)forallx\in \mathbb{R}.$  Since this is true for any $\epsilon >0$  , ( 10 ) is established for all $x$  in $\mathbb{R}$  .
The sub-Markovian resolvent $\left({V}_{p}\right)$  allows one to construct a semi-group $\left({P}_{t}\right)$  with $\left({V}_{p}\right)$  as its resolvent (see e.g. Bliedtner and Hansen [4Corollary 8.3 of chapter II) and with it a transient Markov process on $E=\mathbb{R}$  with $0$  -potential equal to $V$  . On the other hand, ( 9 ) implies that $m$  is a reference measure for the Markov process with potential V and that the potential density $h\left(x,y\right)$  of $V$  with respect to $m$  is equal to $h\left(x,y\right)=\frac{1}{\chi \left(x\right)}G\left(x,y\right).$  Setting then $\mu \left(dy\right)=\chi \left(y\right)m\left(dy\right)$  , we see that the operator $V$  admits the symmetric densities $\left(\frac{1}{\chi \left(x\right)}G\left(x,y\right)\frac{1}{\chi \left(y\right)},x,y\in \mathbb{R}\right)$  with respect to $\mu$  . $\square$  Remark 4.5 Assume that $G$  is such that for any $p$  , the inverse of the covariance matrix of $\left({\eta }_{{x}_{1}},{\eta }_{{x}_{2}},...,{\eta }_{{x}_{p}}\right)$  is a diagonally dominant M-matrix. One can then follow the steps of the proof of Theorem 3.4, without the need to define the matrices ${D}_{n}$  (we check similarly as in Remark 4.1, that the matrix ${T}_{n}=\frac{1}{{c}_{n}}{B}_{n}$  is a transition matrix) and conclude that there exists a symmetric Markov process with potential densites equal to $\left(G\left(x,y\right),\left(x,y\right)\in {\mathbb{R}}^{2}\right)$  . This together with Theorem 3.6 (i) lead to Theorem 3.6 (ii).
Finally, every centered Gaussian process with infinitely divisible square is equal to a deterministic function times a Gaussian process that is associated with a Markov process. The Isomorphism Theorem of Dynkin can hence be easily adapted to incorporate the deterministic functions, so that it can be used in studying any Gaussian processes with infinitely divisible square.
5 The case of the Brownian sheet A Brownian sheet is a centered Gaussian process $\left({W}_{x,s},x\ge 0,s\ge 0\right)$  with a covariance given by $IE\left({W}_{x,s}{W}_{y,t}\right)=\left(x\wedge y\right)\left(s\wedge t\right)$  Remember that up to a multiplicative constant $\left(x\wedge y;x,y\in {\mathbb{R}}_{+}\right)$  is the Green function of the linear Brownian motion killed at its first hitting time of $0$  . Hence this covariance is the product of two Green functions and one may ask whether $\left({W}_{x,s}^{2},x\ge 0,s\ge 0\right)$  is infinitely divisible. The answer is given in the following proposition.
Proposition 5.1 (i) For every $\left({x}_{i},{s}_{i}{\right)}_{1\le i\le 3}$  of ${\mathbb{R}}_{+}^{6}$  the vector $\left({W}_{{x}_{i},{s}_{i}}^{2},1\le i\le 3\right)$  is infinitely divisible.
(ii) The process $\left({W}_{x,s}^{2},x\ge 0,s\ge 0\right)$  is not infinitely divisible.
Proof. Using Griffiths criterion(or Bapat's criterion) it is easy to check that a sufficient condition for the infinite divisibility of the square of a three-dimensional Gaussian vector, indexed by $\left\{1,2,3\right\}$  and with a covariance $g$  , is
 $\begin{array}{c}g\left(i,j\right)g\left(k,k\right)\ge g\left(i,k\right)g\left(j,k\right),\end{array}$ (11)
for any choice of $i,j,k$  in $\left\{1,2,3\right\}$  .
Since any Green function satisfies ( 11 ), so does the covariance function of the Brownian sheet, as a product of two Green functions.
By Bapat's criterion, we know that a covariance matrix $G$  such that $G\ge 0$  , corresponds to an infinitely divisible square Gaussian vector if ${G}^{-1}$  is an M-matrix. We choose $\left({x}_{i},{s}_{i}{\right)}_{1\le i\le 4}$  such that $0<{x}_{1}<{x}_{3}<{x}_{2}<{x}_{4}$  and $0<{s}_{4}<{s}_{1}<{s}_{3}<{s}_{2}$  . Let $G$  be the matrix $\left(\left({x}_{i}\wedge {x}_{j}\right)\left({s}_{i}\wedge {s}_{j}\right){\right)}_{1\le i,j\le 4}$  . We compute the coefficient ${G}_{1,2}^{-1}$  ${G}_{1,2}^{-1}={x}_{1}{x}_{3}{s}_{4}\left({x}_{2}-{x}_{3}\right)\left({s}_{3}-{s}_{1}\right){s}_{4}>0.$  Hence ${G}^{-1}$  is not an M-matrix and the corresponding Gaussian vector does not have an infinitely divisible square. $\square$  Acknowledgements We would like to thank Gennady Samorodnitsky and Jay Rosen for helpful discussions and comments which helped us to improve a previous version of this work.
References

1. Bapat R. B. (1989). Infinite divisibility of multivariate gamma distribution and M-matrices.Sankhya, 51, 73-78.
2. Bass R., Eisenbaum N. and Shi Z. (2000). The most visited sites of symmetric stable processes.Proba. Theory and Relat. Fields 116 (3),391-404.
3. Berman A. and Plemmons R.J. (1979) Nonnegative Matrices in the Mathematical Sciences. Academic Press, Inc. New York..
4. Bliedtner J. and Hansen W. (1986).Potential Theory. Universitext Springer.
5. Cinlar E. (1975).Introduction to Stochastic Processes. Prentice-Hall, Inc. Englewood Cliffs, N.J.
6. Dynkin E. B. (1983). Local times and quantum fields. Seminar on Stochastic Processes 82, 69-84, Birkhauser Boston.
7. Evans S. N. (1991). Association and infinite divisibility for the Wishart distribution and its diagonal marginals.Jour. Multivar. Anal. 36 ,199-203.
8. Eisenbaum N. (2003). On the infinite divisibility of squared Gaussian processes. Probab. Theory and Relat. fields 125, 381-392.
9. Griffiths R. C. (1970). Infinite divisible multivariate gamma distributions. Sankhya, Series A,32 ,393-404.
10. Griffiths R. C. (1984). Characterization of infinitely divisible multivariate gamma distributions.Jour. Multivar. Anal. 15 ,12-20.
11. Lévy P. (1948).The arithmetical character of the Wishart distribution.Proc. Camb. Phil. Soc. 44, 295-297.
12. Marcus M.B. and Rosen J. (1992). Sample path properties of the local times of strongly symmetric Markov processes via Gaussian processes. (Special Invited paper).Ann. Probab. 20 ,1603-1684.
13. Meyer P. A. (1966). Probabilités et Potentiel. ed. Hermann, Paris.
14. Moran P.A.P. and Vere-Jones D. (1969). The infinite divisibility of multi-gamma distributions. Sankhya, Series A, 31, 191-194.
15. Paranjape S.R. (1978). Simpler proofs for the infinite divisibility of multivariate gamma distributions. Sankhya, Series A, 40 , 393-398.
16. Taylor J. C. (1972). On the existence of sub-Markovian resolvents. Inventiones Math. 17, 85-93.
17. Taylor J. C. (1975). A characterization of the kernel ${lim}_{\lambda \to 0}{V}_{\lambda }$  for sub-markovian resolvents $\left({V}_{\lambda }\right)$  .Annals of Probab. Vol.3,No.2 ,355-357.
18. Vere-Jones D. (1967). The infinite divisibility of a bivariate gamma distribution.Sankhlya, Series A,29,421-422.

 Nathalie Eisenbaum Haya Kaspi Laboratoire de Probabilités Industrial Engineering and Management Université Paris VI CNRS Technion 4 place Jussieu Haifa, Israel 32000 75252 Paris Cedex 05, France iehaya@tx.technion.ac.il nae@ccr.jussieu.fr


text:
## I cut myself with a crack pipe.

you do not do it?

Advertise almost everything here, with CPM banners, CPM mail ads, and CPC context links. You can target relevant areas of the site and show ads based on the geographic location of the user, if you wish.

Starts at just \$ 1 per CPM or \$ 0.10 per CPC.

## how to delete the file called \. pipe FC_ {6D57D5C3-9CEA-4497-BE57-9E544137A437} _1

I've somehow got this file . Pipe FC_ {6D57D5C3-9CEA-4497-BE57-9E544137A437} _1 on my home directory.

Tried to delete with rm and tab completion using the following prefixes:

1. ./
2. ./ .
3. ./.p

but he seems to completely ignore this file

## terminal – command Pipe to mv

I'm trying to move terminal files, but only those results from one `grep` question. By instict (and limited knowledge of UNIX) I immediately started typing `ls -l | grep -i s02 | mv` and then I stopped not knowing what to do next.

Then I remember seeing something similar in which we use a kind of variable instead. I note all these things so I looked at it and it is `grep -B 1 -A 1 "active count = 1\$" <<< "\$(launchctl dumpstate)"` or `active count = 1` is what is supposed to find in the dump of `launchd`, I understand that `-B 1` and `-A 1` are just to print ± 1 line so that they can be ignored here.

Overall, I think I just have to understand how `<<<` works because it is obvious that he says put it in but I had read before about `>` and `>>` how different they are, so I am completely lost when a third is added and in the opposite direction from above!

I imagine that there must be a case-insensitive option for `mv` but i'm trying to learn how to do it in a more difficult way. 🙂

Thank you!

## php – (General) Find the necessary pipe fittings for a specified entry angle

I am new to PHP, and I wrote this script as a bit of practice. It is a tool that allows engineers to know which connections are needed to reach a certain turning angle, provided they have access to the 8 °, 11,25 °, 22,5 ° connections, 45 ° and 90 °.
To avoid problems with null entries, "pipemain.php" is executed first, then redirected to "pipecode.php". After the first search, they can perform a new search by simply entering the angle on the pipecode.php page.

edit: We are now working on implementing a backup on pipecode.php so that you do not have to go through pipemain.php beforehand.

pipemain.php code

pipecode.php code

I'm just wondering if I can make significant improvements in general (for example, formatting, using variables, etc.). Ignore the fact that it 's a Horstmann removal style. Any entry / criticism would be very appreciated

Here is an example of a function that could probably be improved. It only prints the number of fittings needed and formats the last word into "fitting" or "fitting", depending on whether there is 1 fitting or not.

``````/*
This function echoes what fittings are needed. It also determines if the text
should be "fitting" or "fittings". It's just semantics, but it bugged me to see
"you need one fittings" or similar
*/
function printFittings()
{   global \$inputAngle, \$num8, \$num11, \$num22, \$num45, \$num90;
echo "You will need:";
if (\$num90 == 1)
{   echo "\$num90 -- 90° fitting";}
if (\$num90 > 1)
{   echo "\$num90 -- 90° fittings";}
if (\$num45 == 1)
{   echo "\$num45 -- 45° fitting";}
if (\$num45 > 1)
{   echo "\$num45 -- 45° fittings";}
if (\$num22 == 1)
{   echo "\$num22 -- 22.5° fitting";}
if (\$num22 > 1)
{   echo "\$num22 -- 22.5° fittings";}
if (\$num11 == 1)
{   echo "\$num11 -- 11.25° fitting";}
if (\$num11 > 1)
{   echo "\$num11 -- 11.25° fittings";}
if (\$num8 == 1)
{   echo "\$num8 -- 8° fitting";}
if (\$num8 > 1)
{   echo "\$num8 -- 8° fittings";}
}
``````

## windows – makes the pipe 'caret sign` again the direction of the text direction

In the past, when RTL was oriented on the window, a small triangle appeared at the top of the window. `caret blowjob |` to indicate the current direction of the text.

In recent years, in modern application, this index has disappeared and many sources of confusion have appeared in the world of multilingual writers (lang can be changed automatically without any clue simply because I move the cursor to a different location, for example example).

So my question is how to make the text's direction refer to it.
Thank you!!

## stream – How to read a named pipe under Windows?

I tried to open a pipe named Out of Powershell

``````\$ pipe = new-object System.IO.Pipes.NamedPipeServerStream "testpipe", & # 39; Out & # 39;
\$ pipe.WaitForConnection ()
\$ sw = new-object System.IO.StreamWriter \$ pipe
\$ sw.AutoFlush = \$ true
\$ sw.WriteLine ("The server's pid is \$ pid")
\$ sw.Dispose ()
\$ pipe.Dispose ()
``````

and uses `Get` read the named pipe.

``````Get["\\.\pipe\testpipe"]
``````

The following messages were broadcast in Powershell and Mathematica

``````Exception calling "WriteLine" with the argument (s) "1": "The pipe is broken."
On line: 5 characters: 1
+ $$sw.WriteLine ("The server's pid is$$pid ")
+ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo: NotSpecified: (:) [], MethodInvocationException
+ FullyQualifiedErrorId: IOException

Exception calling "Dispose" with "0" argument (s): "The pipe is broken."
On line: 6 characters: 1
+ \$ sw.Dispose ()
+ ~~~~~~~~~~~~~~
+ CategoryInfo: NotSpecified: (:) [], MethodInvocationException
+ FullyQualifiedErrorId: IOException
``````
``````Get :: noopen: can not open \.  Pipe  testpipe.
``````

## html – Angular 7, p-table PrimeNG: Error: InvalidPipeArgument: & # 39;[object Object]& # 39; for pipe & # 39; SlicePipe & # 39;

I'm trying to make a list on a table of primeng p, so until yesterday I listed everything perfectly, but today, when running the project angular, I noticed that this error occurred. I have no idea what this means, let alone the reason, because everything was working normally.

Fault:

``````TableBody.html: 8 ERROR Error: InvalidPipeArgument: & # 39;[object Object]& # 39; for pipe & # 39; SlicePipe & # 39;
at invalidPipeArgumentError (common.js: 4323)
on SlicePipe.push ../ node_modules/@angular/common/fesm5/common.js.SlicePipe.transform (common.js: 5812)
at Object.eval [as updateDirectives] (TableBody.html: 8)
at Object.debugUpdateDirectives [as updateDirectives] (core.js: 23911)
on checkAndUpdateView (core.js: 23307)
at callViewAction (core.js: 23548)
on execEmbeddedViewsAction (core.js: 23511)
on checkAndUpdateView (core.js: 23308)
at callViewAction (core.js: 23548)
on execComponentViewsAction (core.js: 23490)
``````

Line 8 of TableBody.html: (where an error occurs Obs: never entered this file)

``````
``````

The code in my table.html and the component.ts: (obs: the GET method works normally, returning all the data, I tested it separately and the problem occurs when the data in the table is deleted.)
Code available on code share

## It's at the DM

From a scientific point of view, opening to a holding bag, and probably the backpack and the portable hole, is a watertight door.

Thus, when you "seal" the bag, it is not that the air inside escapes. Otherwise, you would have a thud every time you open the bag by breaking the seal and all the fresh air would penetrate inside.

When a living being is in the bag, there is no lack of air, it lacks oxygen. Living things as we know them breathe oxygen and exhale carbon dioxide. Plant life absorbs carbon dioxide and releases oxygen. We are symbiotic that way. But you can not just throw a fern in the bag and think that it would provide you with oxygen. too little, too slow.

So the pipe you're talking about would work like a snorkel; allowing new oxygen to enter and expel carbon dioxide. It would take more science to determine the size of pipe needed to circulate the volume inside the bag.

The real question is this: Can the pipe stay halfway and out? This also belongs to the SM, but this question also indicates that there is no clear answer in the rules.

My personal decision would be that as long as the bag is not moved, the hose is in good condition.

• The items are not completely sucked when they cross the threshold, otherwise the user would be sucked whenever they would catch something. So, things can be halfway, in part.
• You create, for lack of a better term, a wormhole of our plane in the half-plane each time you open the bag. The ends are relatively fixed in space.
• If the opening goes too far, a new wormhole must be created. So, if the bag is still, the wormhole remains until the bag is completely closed. If the bag is moved, the original wormhole closes with partially contained elements sucked or expelled depending on the mass. When the end points are stable (do not move), a new wormhole is created to connect the two sides.

In your case, if someone was trying to move the bag more than a few feet, the pipe would probably be ejected, unless something inside held it. In this case, a check of the DC20's strength would let it be brought back inside (suppose that there is still volume to hold the whole pipe).

## pipe – PDF a set of free Office files from the command line

Suppose that a set of .docx files are in a directory to convert to .pdf format. Is there a clever way to list the files and channel them to a command to "print" the pdf files in a target directory?

Suppose that MS-Word is not installed on the PC (Office Libre installed).

The preference is for solutions that require minimal installation of Ubuntu utilities / packages and are well documented. Thank you

## virtualization – Is it possible to create a veth interface on a Linux machine with one end of the pipe on another machine?

I have two virtual machines running on my Windows desktop.

I want to test VxLAN with my two virtual machines, but my Windows computer sees no VLAN tag or probably not in VLAN (I have no information as it is a desktop network). I can find few ways to see VLAN tags, but they are not very relevant to understand now.

So I'm thinking of creating my own virtual switch and my own virtual LAN for my virtual machines.

I do not want to use Hyper-V and I want to continue using Virtual Box. The installation of OpenVswitch on Windows is also a little complex and, therefore, I do not want to try it either. I want to use VirtualBox or try from scratch without it being necessary to solve some installation problems. With Virtual Box, I can not find any VLAN option (I guess that's because Virtual Box has no built-in virtual switch and only uses virtual bridges and tap interfaces).

So, I check if the following solution is possible:

• I run a virtual switch as a fixed menu container on the Windows host. The ports of this virtual switch will have VLAN tags

Is there a way to connect the ports of the virtual switch to each virtual machine running a pipe mechanism such as veth and how do you get there? (Network type overlay and assembly, I guess)


text:
# Forecasting a Timeseries

Suppose you have decided on a suitable model for a timeseries. In this case, we have selected an ARIMA(2,1,3) model, using the Akaike Information Criteria (AIC) as our sole criterion for choosing between various models here, where we model the DJIA.

Note: There are many criteria for choosing a model, and the AIC is only one of them. Thus, the AIC should be used heuristically, in conjunction with t-tests and the Coefficient of Determination, among other statistics. Nonetheless, let us assume that we ran all these tests, and were still satisfied with ARIMA(2,1,3).

An ARIMA(2,1,3) looks like this:

$\Delta Y_t = \phi_2 Y_{t-2} + \phi_1 Y_{t-1} + \theta_{3} \epsilon_{t-3} + \theta_{2} \epsilon_{t-2} + \theta_1 \epsilon_{t-1} + \epsilon_{t}$

This is not very informative for forecasting future reaizations of a timeseries, because we need to know the values of the coefficients $\phi_2$, $\phi_1$, etcetera. So we use R’s arima() function, which spits out the following output:

ARIMA(2,1,3): Coefficients

Thus, we revise our model to:

$\Delta Y_t = -0.992 Y_{t-2} + 0.1840 Y_{t-1} -0.0511 \epsilon_{t-3} + 1.0101 \epsilon_{t-2} + -0.2483 \epsilon_{t-1} + \epsilon_{t}$

Then, we can forecast the next, say 20, realizations of the DJIA, to produce a forecast plot. We are forecasting values for January 1st 1990 to January 26th 1990, dates for which we have the real values. So, we can overlay these values on our forecast plot:

Forecast: Predicted range (shaded in light grey for 95% confidence, dark grey for 80% confidence) and Actual Values (red)

Note that the forecast is more accurate for predicting the DJIA a few days ahead than later dates. This could be due to:

1. the model we use
2. fundamental market movements that could not be forecasted

Which is why data in a vacuum is always pleasant to work with. Next: Data in a vacuum. I will look at data from the biggest vacuum of all – space.

Abbas Keshvani

# Using AIC to Test ARIMA Models

The Akaike Information Critera (AIC) is a widely used measure of a statistical model. It basically quantifies 1) the goodness of fit, and 2) the simplicity/parsimony, of the model into a single statistic.

When comparing two models, the one with the lower AIC is generally “better”. Now, let us apply this powerful tool in comparing various ARIMA models, often used to model time series.

The dataset we will use is the Dow Jones Industrial Average (DJIA), a stock market index that constitutes 30 of America’s biggest companies, such as Hewlett Packard and Boeing. First, let us perform a time plot of the DJIA data. This massive dataframe comprises almost 32000 records, going back to the index’s founding in 1896. There was an actual lag of 3 seconds between me calling the function and R spitting out the below graph!

Dow Jones Industrial Average since March 1896

But it immediately becomes apparent that there is a lot more at play here than an ARIMA model. Since 1896, the DJIA has seen several periods of rapid economic growth, the Great Depression, two World Wars, the Oil shock, the early 2000s recession, the current recession, etcetera. Therefore, I opted to narrow the dataset to the period 1988-1989, which saw relative stability. As is clear from the timeplot, and slow decay of the ACF, the DJIA 1988-1989 timeseries is not stationary:

Time plot (left) and AIC (right): DJIA 1988-1989

So, we may want to take the first difference of the DJIA 1988-1989 index. This is expressed in the equation below:

$\Delta Y = Y_t - Y_{t-1}$

The first difference is thus, the difference between an entry and entry preceding it. The timeseries and AIC of the First Difference are shown below. They indicate a stationary time series.

First Difference of DJIA 1988-1989: Time plot (left) and ACF (right)

Now, we can test various ARMA models against the DJIA 1988-1989 First Difference. I will test 25 ARMA models: ARMA(1,1); ARMA(1,2), … , ARMA(3,3), … , ARMA(5,5). To compare these 25 models, I will use the AIC.

Table of AICs: ARMA(1,1) through ARMA(5,5)

I have highlighted in green the two models with the lowest AICs. Their low AIC values suggest that these models nicely straddle the requirements of goodness-of-fit and parsimony. I have also highlighted in red the worst two models: i.e. the models with the highest AICs. Since ARMA(2,3) is the best model for the First Difference of DJIA 1988-1989, we use ARIMA(2,1,3) for DJIA 1988-1989.

The AIC works as such: Some models, such as ARIMA(3,1,3), may offer better fit than ARIMA(2,1,3), but that fit is not worth the loss in parsimony imposed by the addition of additional AR and MA lags. Similarly, models such as ARIMA(1,1,1) may be more parsimonious, but they do not explain DJIA 1988-1989 well enough to justify such an austere model.

Note that the AIC has limitations and should be used heuristically. The above is merely an illustration of how the AIC is used. Nonetheless, it suggests that between 1988 and 1989, the DJIA followed the below ARIMA(2,1,3) model:

$\Delta Y_t = \phi_2 Y_{t-2} + \phi_1 Y_{t-1} + \theta_{3} \epsilon_{t-3} + \theta_{2} \epsilon_{t-2} + \theta_1 \epsilon_{t-1} + \epsilon_{t}$

Next: Determining the above coefficients, and forecasting the DJIA.

Analysis conducted on R. Credits to the St Louis Fed for the DJIA data.

Abbas Keshvani


text:
# Topology graphs of IBM's quantum computers

What is the meaning of the topology graphs of IBM's quantum computers? How should it be interpreted? For example, how should I understand the following graph?

The graph shows you how the physical qubits are connected together on the real device you will be using. For example, on the graph you put, qubit 0 has a physical connection to qubit 1 and qubit 14 on the quantum device but is not connected to qubit 12.

This graph is really important when Qiskit tries to map a circuit to the quantum device because it shows how qubits are connected and this is a huge step in transforming circuits to run in devices. For example, since 0 and 1 are connected, it will have no problem executing a C-NOT between the two. But if you wish to do the same between 0 and 12, this will be harder because of the lack of connection.

The process of rewriting circuits is what we call transpilation. Without it, we won't be able to run circuits on quantum devices. You should check the link I just put, it explains it quite well! All in all, the main objective of this is to transform your circuit so it matches the device you are using, and the topology of the qubits (the graph) plays quite a important role in this.

If you have any question regarding this, feel free to ask! I hope this helps :)

The plot topology graph of the device represents the qubit layout of the hardware. The qubits on IBM hardware are fixed. They don't move around as trapped-ion qubits built by Honeywell system. Hence, IBM hardware doesn't have what you would call all-to-all connectivity between qubits. Note that IBM uses superconducting transmon qubit rather than trapped ions qubits.

When you design a quantum circuit, you don't usually don't think about the specific hardware specifications. That is, you would rather put your focus on designing a circuit to do the task at hand rather than thinking about what the qubit layout is on the QPU and the native gates that can be executed on the QPU. Once you finished designing your circuit, and want to execute it on a specific QPU, then you need to find an equivalent circuit to the circuit that you came up with, that can be executed on that specific QPU. This is known as the transpilation process. Fortunately, Qiskit do this automatically for you. Furthermore, you can also specify the level of optimization during this transpilation process.

For example, let's suppose that the following quantum circuit solve that problem that you are interested in:

and you want to execute this on ibmq_vigo which has the following specifications:

1. qubit layout:

1. Basis gates (the gates that can be executed on hardware):

CX, ID, RZ, SX, X (This just changed actually, it used to be U1,U2,U3,CX )

First, we must note that the Hadamard gate can't be implement directly on ibmq_vigo since it is not part of the native gate. So it need to be decomposed into RZ and SX gates.

Second, we should note that qubit $$q_0$$ is not connected to $$q_4$$ in the above qubit layout. But we do have a $$CX$$ gate between $$q_0$$ and $$q_4$$ hence there must be some overhead swapping operations need to be done for us to be able to execute the above circuit on the hardware.

But as I said before, IBM do this transpilation for you automatically, so you don't have to worry about it. And they do it at different level of optimization. At the lowest level, that is, you just map the your circuit $$q_0$$ to the device qubit number $$0$$, and your circuit $$q_1$$ to the device qubit number $$1$$, and so on, then you would have a transpiled circuit like:

Note how you don't see the Hadamard gate $$H$$ being executed but rather a sequence of $$RZ$$ and $$SX$$ gates putting together... again this is because the QPU ibmq_vigo doesn't have $$H$$ as the native gate.

This is all good, but you might look at your circuit, and say to yourself: well... since $$q_0$$ and $$q_4$$ need to have to an entangled gate, $$CX$$, and the other qubits only have single qubit rotation ($$X$$ gate). Why don't we map qubit $$q_0$$ of the circuit to qubit number $$0$$ on the QPU, but now mapping qubit $$q_4$$ of the circuit to qubit number $$1$$ of the device, and qubit $$q_1$$ of the circuit to qubit number $$2$$ of the device and etc. This will reduce the overhead. This is a higher level of optimization, and it is a default option in Qiskit (I believe). So doing what I just said above will give is the following circuit:

Note how much simpler it is now compare to the our first transpiled circuit. This is why having a good transpilation process is really important and one that is very critical in this NISQ (Noisy Intermediate-Scale Quantum) era.

If you are interested in recreating the above circuits, and their respective transpiled version, here is the code to do so in Qiskit. I executed them in the IBMQ Lab environment.

Importing Qiskit packages:

import matplotlib.pyplot as plt
%matplotlib inline
# Importing standard Qiskit libraries
from qiskit import QuantumCircuit, execute, Aer, IBMQ
from qiskit.compiler import transpile, assemble
from qiskit.tools.jupyter import *
from qiskit.visualization import *
from iqx import *
from qiskit.visualization import plot_histogram, plot_gate_map, plot_circuit_layout

Creating the original circuit and draw it out:

qreg_q1 = QuantumRegister(5, 'q')
creg_c = ClassicalRegister(5, 'c' )
qc = QuantumCircuit(qreg_q1)
qc.h(0)
for i in [1,2,3,4]:
qc.x(i)
qc.cx(0,4)
qc.draw('mpl', style = {'name': 'bw'}, scale = 1, initial_state = True,plot_barriers = False)

Pick ibmq_vigo as our QPU to transpile the original circuit to, also transpile it naively (lowest level of optimization): (This is create that humongous long circuit we saw earlier)

hardware_backend = provider.get_backend('ibmq_vigo')
transpiled_qc_lvl0 = transpile(qc, backend = hardware_backend , optimization_level=0)
transpiled_qc_lvl0.draw('mpl', style = {'name': 'bw'}, scale = 0.6, initial_state = True,plot_barriers = False)

Transpile the original circuit with highest level of optimization: (This spit out the shorter/much more compact transpiled circuit that we saw earlier. Not we just change the optimization_level from 0 to 3.)

transpiled_qc_lvl3 = transpile(qc, backend = hardware_backend , optimization_level=3)
transpiled_qc_lvl3.draw('mpl', style = {'name': 'bw'}, scale = 0.75, initial_state = True,plot_barriers = False)

• Thanks a lot for the amazing answer @KAJ226! Regarding the call to "transpile" Did I understand correctly? - If I am not calling it, Would it be called automatically with a default optimization level?
– AL_P
Jan 9 '21 at 5:37
• Yes. That is correct. If you don't specify it, then Qiskit will use level 1. That is the default. The levels are: 0 =no optimization; 1 = light optimization, 2 = heavy optimization, 3 = even heavier optimization Jan 9 '21 at 6:30


text:
# What are the requirements for separability inheritance

Suppose we have an arbitrary separable topological space $X$. What are some (possibly nonequivalent) minimal requirements to put on $X$ to ensure that every subspace of $X$ is separable?

This is not true for arbitrary spaces, as witnessed by the the space $X$ which is uncountable, where open sets are precisely the sets containing some special point $x_0$. It is separable (because $\lbrace x_0\rbrace$ is dense), but $X\setminus \lbrace x_0\rbrace$ is uncountable and discrete, so not separable. However, this space is not even $T_1$ (though it is $T_0$).

It is clearly true for second-countable spaces (because weight is not smaller than density, and is inherited), but that is not necessary, as shown by an example similar to the above, but with $X$ countable.

Analogous question could be asked replacing $\aleph_0$ with arbitrary infinite cardinal $\kappa$: suppose we have a space $X$ with a dense subset of cardinality at most $\kappa$, what should we require of $X$ for this to be inherited? In this case we can perform the analysis which is exactly analogous to the above, but perhaps some more concrete results will be harder to arrive at with uncountable $\kappa$...

So I really have two somewhat related questions. In terms of cardinal invariants density $d$ and hereditary density $hd$, what requirements do we have to put on $X$ to have some of the following:

1. (Only for $X$ separable) $hd(X)\leq\aleph_0$
2. $d(X)= hd(X)$

By analysis similar to the above we know that 2. is not true in general, as well as that $d(X)=w(X)$ implies 2, but is not necessary.

I would appreciate some conditions which would imply either one, or some nice counterexamples which satisfy some stronger separation axioms than just $T_0$ (if there are any, I think there should be...), or a proof that there are none.

Edit: Sam L. suggested the example of Niemytzki plane, which shows that even somewhat strong separation axioms are not sufficient: it is completely regular and separable and of countable character, but has an uncountable discrete subspace. It is not hard to see that by taking a suitable subspace, we can strengthen it to $w(X)=\aleph_1$ (regardless of CH), and itself is a counterexample for all $\kappa<\mathfrak c$.

Edit 2: As per Arthur Fischer's suggestion, if we take an arbitrary nontrivial second-countable compact Hausdorff space $X$ (such as $2$ with discrete topology or $[0,1]$), $X^\mathfrak c$ will be separable (because product of at most $\mathfrak c$ separable spaces is separable), as well as Hausdorff and compact, and hence normal, but it has a discrete subspace of cardinality $\mathfrak c$. This shows that no usual separation axioms will suffice, not even augmented by compactness.

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@SamL.: Thanks! My mistake. I assumed that separability+countable character implies countable basis, which is of course not true (it only implies countable $\pi$-basis). –  tomasz Aug 9 '12 at 17:14
Note, also, that the Cantor cube of weight $\mathfrak{c} = 2^{\aleph_0}$, $2^{\mathfrak{c}}$, is an example of a separable compact Hausdorff (hence normal) space of hereditary density $\mathfrak{c}$. (See, e.g., Example 11.8 (p.44) of Hodel, Cardinal Functions I in the Handbook of Set Theoretic Topology.) This space does have character $\mathfrak{c}$, however, so it is very far from satisfying any countability axioms. –  Arthur Fischer Aug 9 '12 at 21:50
@ArthurFischer: I had thought about that example (well, specifically, I thought about $[0,1]^{\mathfrak c}$, but I don't suppose it's that much of a difference in this case), but wasn't sure about its hereditary density. Thanks. :) –  tomasz Aug 9 '12 at 22:23

The results in the comments are all negative. Here’s a positive result, albeit with a rather strong hypothesis:

Theorem: Every LOTS (space whose topology is generated by a linear order) has hereditary density equal to its density.

Proof: Let $\langle X,\le\rangle$ be a LOTS, and let $Y$ be a subspace of $X$. Let $D$ be a dense subset of $X$ of cardinality $d(X)$, and without loss of generality assume that $D$ contains the endpoint(s) of $X$, if any. For each $a,b\in D$ with $a<b$ and $(a,b)\cap Y\ne\varnothing$ fix $y(a,b)\in(a,b)\cap Y$. Let $I$ be the set of isolated points of $Y$, and let $E=D\cup\{y(a,b):a,b\in D\text{ and }a<b\text{ and }(a,b)\cap Y\ne\varnothing\}$.

Claim: $I\cup E$ is dense in $Y$.

Proof: Let $U$ be any non-empty open subset of $Y$. If $U\cap I\ne\varnothing$, we’re done, so assume that $U\cap I=\varnothing$. Fix $y\in U$; there are $a,b\in D$ such that $a<y<b$ and $(a,b)\cap Y\subseteq U$, and clearly this ensures that $y(a,b)\in U$. $\dashv$

Clearly $|E|\le d(X)$, so to show that $d(Y)\le d(X)$, it suffices to show that $|I|\le d(X)$. Fix $y\in I$. Suppose first that $y$ has no immediate predecessor or successor in $X$. Then there are $a_y,b_y\in D$ such that $(a_y,b_y)\cap Y=\{y\}$. Next, suppose that $y$ has both an immediate predecessor $y^-$ and an immediate successor $y^+$ in $X$; then $y\in D\subseteq E$.

Finally, suppose that $y$ has no immediate predecessor but does have an immediate successor $y^+$ in $X$. (The reverse case is entirely similar.) If $y^+\in D$, there is an $a_y\in D$ such that $(a_y,y^+)\cap Y=\{y\}$, and we set $b_y=y^+$. Note that this is necessarily the case if $y^+$ has an immediate successor in $X$, so we may assume that $y^+\in\operatorname{cl}(y^+,\to)$ and that $y^+\notin D$. Then there are $a_y\in(\leftarrow,y)\cap D$ and $b_y\in(y^+,\to)\cap D$ such that $(a_y,b_y)\cap Y\subseteq\{y,y^+\}$. The assignment $y\mapsto(a_y,b_y)$ is at most $2$-$1$, so $|I|\le d(X)$, and the proof is complete. $\dashv$

Added: The result can be strengthened a bit. A GO-space is a Hausdorff space $\langle X,\tau\rangle$ with a linear order $\le$ such that every point of $X$ has a local base of $\le$-intervals; these intervals may be open, closed, or half-open, and if they’re closed, they may be degenerate. It’s well-known $X$ is a GO-space iff $X$ is a subspace of a LOTS.

Corollary: If $X$ is a GO-space, then $d(X)=hd(X)$.

Proof: Let $\langle X,\tau\rangle$ be a GO-space with associated linear order $\le$, and let $D$ be a dense subset of $X$ of cardinality $d(X)$. Let $\tau_\le$ be the order topology on $X$ generated by $\le$; $\tau_\le\subseteq\tau$. Let $$L=\{x\in X:[x,\to)\in\tau\setminus\tau_\le\}$$ and $$R=\{x\in X:(\leftarrow,x]\in\tau\setminus\tau_\le\}\;.$$ Note that $L\cap R\subseteq D$.

Let $$X^+=\Big(X\times\{1\}\Big)\cup\Big(L\times\{0\}\Big)\cup\Big(R\times\{2\}\Big)\;,$$ and let $\preceq$ be the lexicographic order on $X^+$ determined by $\le$ on $X$ and the usual order on $\{0,1,2\}$, and let $\tau_\preceq$ be the order topology on $X^+$ induced by $\preceq$. If we identify $X$ with $X\times\{1\}$ in the obvious way, it’s a little tedious but not difficult to verify that the topology that $X$ inherits from $\left\langle X^+,\tau_\preceq\right\rangle$ coincides with $\tau$. Let $D\,'=D\times\{1\}$; I’ll show that $D\,'$ is dense in $X^+$.

Suppose that $p=\langle x,m\rangle,q=\langle y,n\rangle\in X^+$ with $p\prec q$ and $(p,q)\ne\varnothing$. If $x=y$, then $m=0$ and $n=2$, so $x\in L\cap R\subseteq D$, and $\langle x,1\rangle\in(p,q)\cap D\,'$. Assume now that $x<y$. If $(x,y)$ is non-empty in $X$, choose $z\in(x,y)\cap D$; then $\langle z,1\rangle\in(p,q)\cap D\,'$. Otherwise, $y$ is the immediate successor of $x$ in $X$. This implies that $(\leftarrow,x],[y,\to)\in\tau$, so $x\notin R$, and $y\notin L$. It follows that $p=\langle x,0\rangle$, $q=\langle y,2\rangle$, or both, as otherwise $(p,q)=\varnothing$. If $p=\langle x,0\rangle$, then $x$ is isolated in $X$, so $x\in D$, and $\langle x,1\rangle\in(p,q)\cap D\,'$. Similarly, if $q=\langle y,2\rangle$, then $\langle y,1\rangle\in(p,q)\cap D\,'$. In all cases $(p,q)\cap D\,'\ne\varnothing$, so $D\,'$ is indeed dense in $X^+$, and since $X^+$ is a LOTS we have $hd(X^+)=d(X^+)\le|D|=d(X)$. But $X$ is a subspace of $X^+$, so $d(X)\le hd(X)\le hd(X^+)\le d(X)$, and hence $d(X)=hd(X)$. $\dashv$

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@tomasz: It needn’t be. Suppose that $X=\Bbb R$, $D=\Bbb Q$, and $Y=[0,1]\cup\{\pi\}$; then $\pi\in I\setminus E$. –  Brian M. Scott Aug 12 '12 at 16:23
Are you back, Prof.? –  Pedro Tamaroff Aug 12 '12 at 16:28
@tomasz: Yes, of course: for some reason I was thinking that the point had to be in $D$. Change the example. Let $X=[0,1]\cup[2,3]$, let $D=\Bbb Q\cap\Big((0,1)\cup(2,3)\Big)$, and let $Y=\{1,2\}$. Then it’s possible that $y(a,b)=1$ whenever $a<1<2<b$ and $a,b\in D$. –  Brian M. Scott Aug 12 '12 at 16:28
@Peter: If not, someone’s doing a pretty good impersonation! –  Brian M. Scott Aug 12 '12 at 16:30
@tomasz: Oops! You’re right. –  Brian M. Scott Aug 12 '12 at 16:34


text:
# kolmogorov-smirnov test using package BenfordTests in R

I am using the ks.benftest from the BenfordTests package in R to calculate the D Statistic and p-value of a distribution with respect to a distribution that conforms with Benford Law. The problem I have is that the ks.benftest function gives me a different result from just using function ks.test.

The code I used for both functions with the data is below:

x <- c(rep(1,10),rep(2,5),rep(3,3),rep(4,5),rep(5,6),7)

y <- c(rep(1,9),rep(2,5),rep(3,4),rep(4,3),rep(5,3),rep(6,2),rep(7,2),8,9)

x variable is the distribution I want to confirm follows benford distribution; y variable follows the benford distribution

z<- as.data.frame(t(rbind(x,y)))

pval_bftest <- ks.benftest(z$x,digits=1)$p.value

Dstat_bftest <- ks.benftest(z$x,digits=1)$statistic

pval_bftest = 0.1001

Dstat_bftest = 1.032541

However, when I use the ks.test function, I get very different results

ksTest <- ks.test(z$x,z$y,alternative=c("two.sided"))

ksTest

Two-sample Kolmogorov-Smirnov test

data:  z$x and z$y

D = 0.16667, p-value = 0.799

alternative hypothesis: two-sided

Does anyone know what I'm doing wrong? I should get similar results, but I'm not.

Thanks

T

• Benford's law i a law that describes the frequency of the lead digit in a set of positive integers. For example if we have a book with 210 pages the lead digit is 1 for 1, 10-19 and 100-199. Far from uniform we see the frequency for 1 is 111. For 2 the numbers 2, 20-29 and 200-210 have lead digit 2. So the frequency for 2 is 22. For 3: 3 and 30 -39 a frequency of 11. For 4: 4 and 40-49. a frequency of 11. Similarly all the numbers from 5 -9 also have a frequency. Dec 12, 2016 at 22:58
• According to wikipedia Frank Benford was the second person to observe this phenomena. But it was named for him anyway. In my example divide by 210 and you get a discrete distribution for the lead digit of this data. The point noted was that 1 has by far the highest proportion. The distribution is not unique. It depends on the collection of numbers. Did you pick a set of data and compare to the distribution Benford used? If so did you apply one data set with each goodness of fit test that you used? Dec 12, 2016 at 23:05
• I guess the answer to my question is that you used the same data. But the question is how does ks.test differ from ks.benftest? Dec 12, 2016 at 23:14
• @Michael, that's the question I'm asking really. How is ks.test different from ks.benftest Dec 12, 2016 at 23:19
• Since you know R and know what you you input what I was suggesting is that maybe you can help us answer the question. Dec 13, 2016 at 0:27