@Jakobian i don't even know what half of those words mean

Recently learned that there's a surprising duality between hyperbolic 3-space and de Sitter 3-space.

Didn't actually learn but rather saw a statement without proof. But too beautiful to be wrong.

@AlessandroCodenotti oh. I see

Hi !

Guys, did you see that Messi missed the panenka!!!

And Messi fans were laughing Ronaldo for missing a pen

But

Even Messi misses penalty Is better then Ronaldo penalty

And he didnt cry

Consider the problem of finding the distribution of $X$ given that $X\mid\Sigma^2=y\in N(0,y)$ and $\Sigma^2$ has a continuous distribution. Then the law of total probability gives $$F_X(x)=\int_0^\infty \underbrace{\Phi\left(\frac{x}{\sqrt{y}}\right)f_{\Sigma^2}(y)}_{g(x,y)}\,dy,$$where $\Phi$ is the cdf of the standard normal distribution. Then in my book they simply differentiate and put the derivative under the integral sign.

I'm familiar with the statement given here, but I'm unsure if the integrand satisfies all the criteria. The second one, that $y \mapsto g(x, y)$ be differentiable for all $x$ is kind of straightforward since $\Phi$ is an exponential function, but I do not see the other conditions. Is this clear to someone?

for the assertion that, if E is a subset of metric space X, $\text{Interior}(E)=\text{Interior}(\text{Closure}(E))$ we have the counter example where $E=\mathbb{Q}$ and $X=\mathbb{R}$, the closure of $\mathbb{Q}$ becomes $\mathbb{R}$, while interior of $\mathbb{Q}$ becomes null. The general idea I get from here is that if the closure of a set becomes an open set then its a possible counter example, are there any other examples?

@nickbros123 There is the so called Kuratowski 14 set theorem that you can obtain at most 14 sets from operations of closure, interior and complement. You can obtain at most 7 sets from operations of closure and interior. This bound is achievable

i.e. there exists a set $A$ such that repeating those operations of interior and closure leads you to $7$ distinct sets

an example on the real line is $A = (0, 1)\cup (1, 2)\cup \{3\}\cup [4, 5]\cap \mathbb{Q}$

$A, \overline{A}, A^\circ, \overline{A^\circ}, \overline{A}^\circ, \overline{\overline{A}^\circ}, \overline{A^\circ}^\circ$ are all distinct

@Jakobian is this for any metric space?

that seems like a very cool result

@nickbros123 for every metric space, or more generally topological space

very interesting

that is, the upper bound is for any topological space, the fact that we can achieve this is not possible for all examples, but $\mathbb{R}$ provides one

this is, in essence, an algebraic fact

the operations of closure and interior form a certain monoid, and this monoid can have at most $7$ elements

@AlessandroCodenotti well, I've solved my problem so its okay. Even in the more restrictive version of the countable intersection of cozero sets in a normal realcompact space, this is false, as Michael line shows

Does there exist non-trivial filtration of group $(\Bbb Z,+)$ ?

Where filtration is a sequence of nested normal subgroups $N_i$ s.t. $\cap_i N_i = \{1\}$

yes, plenty of such filtrations

you should know what subgroups of $\mathbb{Z}$ look like and when they're contained in one another

so you can write down all possible sequences of nested subgroups and then there emerges an easy criterion for their intersection to be trivial

@Thorgott It is quite obvious that in general $2 \BBb Z$ and $3 \Bbb Z$ contain all other subgroups, but the chains can be different, i.e. $2 \BBb Z \ni 4 \Bbb Z \ni...$ but also $2 \BBb Z \ni 6 \Bbb Z \ni...$ and $4 \BBb Z$ not equal to $6 \BBb Z$ on every element.

It is quite obvious that in general $2\Bbb Z$ and $3\Bbb Z$ contain all other subgroups, but the chains can be different, i.e. $2 \Bbb Z \supset 4\Bbb Z\supset ... $, but also $2\Bbb Z \supset 6 \Bbb Z \supset ...$ and $4\Bbb Z$ not equal to $6\Bbb Z$ on every element.

Where does $5\mathbb{Z}$ fit into your scheme?

@flowian Is that really the definition? Or do you not mean $(\mathbb{Z},+)$?

Oh thanks for the catch!

@XanderHenderson yes I mean the group with addition.

So in general 2Z and pZ, p prime are biggest chains containing every other subgroup

If you mean group with addition, then surely you don't mean $\bigcap N_i = \{1\}$?

I bought an Indian edition copy of homological algebra by Cartan and Eilenberg. There is a strange thing, the preface and contents pages are repeated at the end of the book.

at the beggining and at the end?

Yes

@flowian the first sentence is incorrect

"If you want to add pages to your book so that it becomes lengthy, repeat the praface at the end of the book"- Sun Tzu, The Art of War

you write down some sequences with "\dotsc", but it's actually not clear which sequences you're envisioning. my guess is if you write them down properly, they will be examples of what you're looking for.

@SoumikMukherjee odd, the English edition I have does not have that

Consider the function $$g(x,y)=\frac1{\sqrt{y}}\phi\left(\frac{x}{\sqrt{y}}\right)f_{\Sigma^2}(y),$$where $\phi$ is the standard normal density and $f_{\Sigma^2}$ is a density that integrates to $1$ over $(0,\infty)$. Is $g(x,y)$ bounded above in absolute value by some integrable $h(y)$? I know $\phi$ is bounded above by $1$, but then I'm left with $\frac1{\sqrt{y}}f_{\Sigma^2}(y)$. I'm not sure if this is integrable over $(0,\infty)$.

Context; I'm trying to verify a differentiation under the integral sign maneuver, and the function $g(x,y)$ needs to be bounded in absolute value by some integrable $h(y)$.

@XanderHenderson I do, I am looking for a filtration chain of group (Z,+)

@flowian Okay, but $1$ is not the identity element in the additive group $(\mathbb{Z},+)$. In general, $1 \not\in N$, where $N$ is some normal subgroup.

So I really don't think that you mean $\{1\} = \bigcap_i N_i$ for some set of normal subgroups...

I think you mean $\{0\} = \bigcap_i N_i$...

Oh yes sorry!

Yes indeed, it is 0

@Thorgott in the filtration of integers it seems that the sequence needs to be countably infinite

@flowian Can you explain why that must be true?

@psie Upon closer thought, I believe $\frac1{\sqrt{y}}\phi\left(\frac{x}{\sqrt{y}}\right)$ is the density of some normal distribution with mean $0$ and variance $\sqrt{y}$. For every $y\in(0,\infty)$, it is bounded by $C$. Thus $$|g(x,y)|\leq Cf_{\Sigma^2}(y)=h(y),$$and $h(y)$ integrates to $C$. Thanks!

@XanderHenderson yes. Suppose the nested sequence is finite. Then there is some integer a s.t. aZ is at the end of sequence. But then if we take the intersection of these finite subgroups we will get aZ, as it is contained in every other subgroup. Therefore to get the identity from intersection we need a countably infinite nested sequence.

@flowian Groovy. So what does that tell you?

@XanderHenderson that the sequence must be infinite indeed!

Does anyone know how to add a superscript hyperlinked citation in a latex document similar to how wikipedia does it?

like$^{[1]}$

but if I add a hyperlink it seems to break$^{[1]}$

ah well, I think this will suffice.

In MathJax? Or LaTeX?

Latex preferably but maybe if it works in mathjax it'll work there too

@Obliv No... They are completely different in this regard.

Because LaTeX didn't also pass through the additional markup that SE uses.

In LaTeX, you would just add a citation using bibtex (or similar), or a footnote.

However, in either MathJax or LaTeX, the syntax $^{[1]}$ is terrible and wrong. The citation and link are not mathematics, so they should not be put in math mode. Again, in LaTeX, you would expect either \cite{#} or \footnote{#}, and then the compiler does the work for you.

if it is something you are trying to put into a post on an SE site, the correct syntax is going to be something like <sup>\[[1](http://...)\]</sup>.

Note that the square braces are escaped.

Soon...👀

What is that...? Some kind of sportsball?

Tennis?

Ice hockey?

Football

@SoumikMukherjee That's the one on horses, right?

No that one is cricket

Drat... but football is played with sticks, right?

I see people play it on their phones, with fingers, not sure where the sticks are used

Yeah... that makes sense.

@BinkyMcSquigglebottom I wish Portugal wins, because I hate the French defence ^_^

French defence

accused of cheating goes to challenge him face to face

(Kramnik accused Jose Martinez )

Have you seen the Clash of Claims?

Yup

Kramnik: the man, the myth, the statistician

which stolen by Spain was clearly a penalty for Germany

Yeah

@SoumikMukherjee Had he brought his secret algorithms with him? At least he could expose them to us. I hope he stops accusing everyone now.

The thief always thinks that everyone else is stealing. Kramnik still has to explain to us why against Topalov he went to the bathroom every 3x2 and then played the move immediately after returning.Not to mention the alleged suspicious wires that were found in the bathroom and the threats that Topalov's team says they received if they spoke about it.

Where there's money at stake... Anyway kramnik is really naive, but did he really think that those people were getting caught live on the world stage?

I feel like Kramnik can't accept the fact that he is no longer at his prime and other players can beat him now more frequently.

@BinkyMcSquigglebottom He lost to Jospem so badly, imagine what would happen if he challenges Hikaru for such a match

Consider the problem $X\mid N = n \in \text{Bin}(n,p_2)$, with $N\in \text{Bin}(m,p_1)$. The law of total probability yields $$P(X=k) = \sum_{n=k}^{m} P(X= k\mid N = n)P(N =n) = \sum\limits_{n=k}^{m} \binom{n}{k}p_2^kq_2^{n-k} \binom{m}{n}p_1^nq_1^{m-n}.$$ I know that $X$ is distributed binomially, but I don't know how to rearrange/compute the sum to show it is.

@psie why is $n$ from $k$ to $m$?

oh okay never mind I see