TITLE: What is the difference between root mean square, and standard deviation?
QUESTION [9 upvotes]: I am currently working through the Feynman Lectures, chapter 6: Probability.
I have reached his problem of the "random walk".
After deriving this and getting some root mean square, wouldn't this just be the same as finding the standard deviation? The standard deviation is the root of the mean of the squared data. Isn't that also just the root mean square?
Also, what exactly are the implications of the root mean square, what does it even mean in regards to our problem?
http://www.feynmanlectures.caltech.edu/I_06.html

REPLY [15 votes]: in the case of standard deviation, the mean is removed out from obsevations, but in root mean square the mean is not removed. however in the case of noise where the mean is zero, the two concept are the same. I hope that this is the difference. see: http://www.madsci.org/posts/archives/2004-11/1100200293.Ph.r.html


TITLE: What is $\mathcal{M}_{2}(\mathbb{R})$
QUESTION [1 upvotes]: I came across  $\mathcal{M}_{2}(\mathbb{R})$ in my exercise book on algebra, but there was no mention of it in the lesson. How is it defined or what should I reference to find a definition.
It doesn't look like the usual $\,\!M_2(\mathbb{R})$

REPLY [2 votes]: These two things should be the same. Some authors use $\mathcal{M}_2(\mathbb{R})$ and  others use the unadorned $M_2(\mathbb{R})$, but they both mean $2\times 2$ matrices with coefficients in $\mathbb{R}$.


TITLE: Convert rational number in $\frac {p}{q}$ form
QUESTION [1 upvotes]: Convert rational number  in $\frac {p}{q}$ form $0.40\bar 7$ (here bar is over $7$).
solution:
By solving I got the answer $367/900$ by multiplying by $10$
My friends are getting answer $4037/9900$ by multiplying by $100$
Are they correct $?$

REPLY [0 votes]: Example: just write it down:
$$0.123\overline{4567}=\frac{1234567-123}{9999000}.$$
In your case
$$0.40\overline{7}=\frac{407-40}{900}$$
or as well
$$0.40\overline{77}=\frac{4077-40}{9900}$$
or even
$$0.40\overline{77777}=\frac{4077777-40}{9999900}.$$


TITLE: Existence of Factor rings of UFDs which are UFDs
QUESTION [4 upvotes]: Suppose that $X=Spec(A)$ is an affine variety over an algebraically closed field $k$ which is normal and such that $Cl(X)=0$.
I am interested in hypersurfaces of $X$ which again satisfy this condition.

Q: Do there always exist hypersurfaces $X'\subset X$ such that $X'$ is again normal and $Cl(X')=0$? How 'many' such hypersurfaces do exist?

Since $A$ being a UFD is equivalent to the conditions above an algebraic formulation is as follows.

Q: Are there prime elements $p\in A$ such that $A/p$ is again a UFD? How 'many' such elements exist?

I am also interested in the same questions if the condition $Cl(X)=0$ is weakened to the condition that $Cl(X)$ is torsion.
I am aware that there is some work in this direction if $A$ is a polynomial ring. At least if $X$ is smooth and $dim(X)\geq 4$ we could also use the Lefschetz theorem.

REPLY [2 votes]: The answer is NO in small dimension. Take a general surface $\bar{X}$ of degree $d\gg 0$ in $\mathbb{P}^3$, and take for $X$ the complement of a general hyperplane section. Then $X$ is smooth, $\mathrm{Pic}(X)=\mathrm{Cl}(X)=0$, but $\bar{X}$ contains no rational curve, so any normal (= smooth) curve in  $X$ has a large Picard group.


TITLE: $X$ is locally connected and countably compact
QUESTION [5 upvotes]: Let $(X,\tau)$ be a topological space $T_3$. Show that the following statements are equivalent:

Every open and finite coverage of X has a finite refinement consisting of connected sets.

Space X is locally connected and countably compact.

A topological space is called countably compact if every open and enumerable coverage admits a finite subcoverage.
Any ideas:
$1 \to 2$ To prove that $X$ is locally connected I thought of the following, by hypothesis $A=\{X\}$ has a finite refinement consisting of connected sets, say $B=\{B_i:B_i \text{ is connected and } i \in \{1,...,n\}\}$. Let $x \in X$ and $U$ open of $X$, such that $x \in U \subset X$, since $X=\cup B_i$, $x \in B_k$ where $k \in \{1,...,n\}$, if $B_k \subset U$ and $B_k$ is open, $X$ is locally connected in $x$, If the above does not happen, I don't know how to proceed.
proving that it is countably compact does not occur to me how.
$2 \to 1$ let $U=\{U_i:i \in \{1,2,...,n\}\}$ be a finite coverage of $X$. Since $X$ is locally connected, for each $x \in U_i$ there is a connected open $V_{ix}$ such that $V_{ix} \subset U_i$. Thus, $U_i=\cup _{x \in U_i} V_{ix}$. My idea here was to use the hypothesis that $X$ is countably compact and apply it to each $U_i$, but I'm not sure of this fact, because I don't know if there is a countable amount of $V_{ix}$ that covers $U_i$.
I don't know how to use the hypothesis that $X$ is $T_3$.
any help would be very useful. Thank you

REPLY [0 votes]: You question is on the equivalence of conditions (a) and (c) of Theorem 3.1 from [HI].
References
[HI] Henriksen, M., Isbell, J. R. Local connectedness in the Stone-Čech compactification, Illinois J. Math. 1: 4 (1957), 574–582.


text:
\begin{document}

\maketitle

\begin{abstract}
In this article, we study the maximal cross number of long zero-sumfree sequences in a finite Abelian group. Regarding this inverse-type problem, we formulate a general conjecture and prove, among other results, that this conjecture holds true for finite cyclic groups, finite Abelian $p$-groups and for finite Abelian groups of rank two. Also, the results obtained here enable us to improve, via the resolution of a linear integer program, a result of W.~Gao and A.~Geroldinger concerning the minimal number of elements with maximal order in a long zero-sumfree sequence of a finite Abelian group of rank two.
\end{abstract}

\vspace{-0.7cm}
\section{Introduction}
\label{Introduction}
Let $G$ be a finite Abelian group, written additively. By $\exp(G)$ we denote the exponent of $G$. If $G$ is cyclic of order $n$, it will be denoted by $C_n$. In the general case, we can decompose $G$ (see for instance \cite{Samuel}) as a direct product of cyclic groups $C_{n_1} \oplus \dots \oplus C_{n_r}$ where $1 < n_1 \text{ } |\text{ } \dots \text{ }|\text{ } n_r \in \mathbb{N}$.

\medskip
In this paper, any finite sequence $S=\left(g_1,\dots,g_\ell\right)$ of $\ell$ elements from $G$ will be called a \em sequence \em in $G$ with \em length \em $\left|S\right|=\ell$. Given a sequence $S=\left(g_1,\dots,g_\ell\right)$ in $G$, we say that
$s \in G$ is a \em subsum \em of $S$ when it lies in the following set,
$$\Sigma(S)=\left\{ \displaystyle\sum_{i \in I} g_i \text{ }|\text{ } \emptyset \varsubsetneq I
\subseteq \{1,\dots,\ell\}\right\}.$$

\noindent If $0$ is not a subsum of $S$, we say that $S$ is a \em
zero-sumfree sequence\em. If $\sum^\ell_{i=1} g_i=0,$ then $S$ is said
to be a \em zero-sum sequence\em. If moreover one has $\sum_{i \in
I} g_i \neq 0$ for all proper subsets $\emptyset \subsetneq I
\subsetneq \{1,\dots,\ell\}$, $S$ is called a \em minimal zero-sum
sequence\em.

\medskip
In a finite Abelian group $G$, the order of an element $g$ will be
written $\text{ord}(g)$ and for every divisor $d$ of the exponent of
$G$, we denote by $G_d$ the subgroup of $G$ consisting of all
elements of order dividing $d$: $$G_d=\left\{x \in G \text{ } |
\text{ } dx=0\right\}.$$

\medskip
\noindent For every divisor $d$ of $\exp(G)$, and every sequence $S$ in $G$, we denote by $\alpha_d$ the number of elements, counted with multiplicity, contained in $S$ and of order $d$. Although the quantity $\alpha_d$ clearly depends on $S$, we will not emphasize this dependence in the present paper, since there will be no risk of confusion.

\medskip
Let $\mathcal{P}$ be the set of prime numbers. Given a positive integer $n \in
\mathbb{N}^*=\mathbb{N}\backslash\{0\}$, we denote by
$\mathcal{D}_n$ the set of its positive divisors and we set
$\tau(n)=\left|\mathcal{D}_n\right|$. If $n > 1$, we denote by
$P^-(n)$ the smallest prime element of $\mathcal{D}_n$, and we put
by convention $P^-(1)=1$. For every prime $p \in \mathcal{P}$, $\nu_p(n)$
will denote the $p$-adic valuation of $n$.

\medskip
Let $G \simeq C_{n_1} \oplus \dots \oplus C_{n_r},$ with $1 < n_1
\text{ } |\text{ } \dots \text{ }|\text{ } n_r \in \mathbb{N}$, be a finite Abelian
group. We set:
$$\mathsf{D}^{*}(G)=\displaystyle\sum^r_{i=1} (n_i-1) + 1 \hspace{0.2cm} \text{ as well as } \hspace{0.2cm} \mathsf{d}^{*}(G)=\mathsf{D}^{*}(G)-1.$$
By $\mathsf{D}(G)$ we denote the smallest integer $t \in \mathbb{N}^*$ such
that every sequence $S$ in $G$ with $|S| \geq t$ contains a non-empty zero-sum subsequence.
The number $\mathsf{D}(G)$ is called the \em Davenport constant
\em of the group $G$.

\medskip
\noindent By $\mathsf{d}(G)$ we denote the largest integer $t \in \mathbb{N}^*$ such that there exists a zero-sumfree sequence $S$ in $G$ with $|S|=t$. It can be readily seen that, for every finite Abelian group $G$, one has $\mathsf{d}(G)=\mathsf{D}(G)-1$.

\medskip
If $G \simeq C_{\nu_1} \oplus \dots \oplus
C_{\nu_s}$, with $\nu_i > 1$ for all $i \in \llbracket 1,s
\rrbracket$, is the longest possible decomposition of $G$ into a
direct product of cyclic groups, then we set:
$$\mathsf{k}^{*}(G)=\displaystyle\sum_{i=1}^s
\frac{\nu_i-1}{\nu_i}.$$

\medskip
\noindent The \em cross number \em of a sequence
$S=(g_1,\dots,g_\ell)$, denoted by $\mathsf{k}(S)$, is then defined by:
$$\mathsf{k}(S)=\displaystyle\sum_{i=1}^{\ell} \frac{1}{\text{ord}(g_i)}.$$
The notion of cross number was introduced by U.~Krause in \cite{Krause84} (see also \cite{Krause91}).
Finally, we define the so-called \em little cross number \em
$\mathsf{k}(G)$ of $G$:
$$\mathsf{k}(G)=\max\{\mathsf{k}(S) | S \text{ zero-sumfree sequence of } G \}.$$

\medskip
Given a finite Abelian group $G$, two elementary constructions
(see \cite{GeroKoch05}, Proposition $5.1.8$) give
the following lower bounds:
$$\mathsf{D}^{*}(G) \leq \mathsf{D}(G) \hspace{0.5cm} \text{ and } \hspace{0.5cm} \mathsf{k}^{*}(G) \leq \mathsf{k}(G).$$

\medskip
The invariants $\mathsf{D}(G)$ and $\mathsf{k}(G)$ play a key r\^{o}le in the theory of non-unique factorization
(see for instance Chapter $9$ in \cite{Narkie04}, the book \cite{GeroKoch05} which presents the different aspects of the theory, and the survey \cite{GeroKoch06} also). They have been extensively studied during last decades and even if numerous results were proved (see Chapter $5$ of the book \cite{GeroKoch05}, \cite{GaoGero06} for a survey with many references on the subject, and \cite{Girard08} for recent results on the cross number of finite Abelian groups), their exact values are known for very special types of groups only. In the sequel, we will need some of these values in the case of finite Abelian $p$-groups and finite Abelian groups of rank two, so we gather them in the following theorem (see \cite{GeroCross94}, \cite{Olso69a} and \cite{Olso69b}).

\newpage
\begin{theorem} \label{Proprietes D et Cross number} The following two statements hold.
\begin{itemize}
\item[$(i)$] Let $p \in \mathcal{P}$, $r \in \mathbb{N}^*$ and
$ a_1 \leq \dots \leq a_r$, where $a_i \in
\mathbb{N}^*$ for all $i \in \llbracket 1,r \rrbracket$. Then, for
the $p$-group $G \simeq C_{p^{a_1}} \oplus \cdots \oplus
C_{p^{a_r}}$, we have:
$$\mathsf{D}(G)=\displaystyle\sum^r_{i=1} \left(p^{a_i}-1\right)+1=\mathsf{D}^{*}(G) \hspace{0.23cm} \text{ and }
\hspace{0.31cm} \mathsf{k}(G)=\displaystyle\sum^r_{i=1} \left(\frac{p^{a_i}-1}{p^{a_i}}\right)=\mathsf{k}^{*}(G).$$

\item[$(ii)$] For every $m,n \in \mathbb{N}^*$, we have:
$$\mathsf{D}(C_m \oplus C_{mn})=m+mn-1=\mathsf{D}^*(C_m \oplus C_{mn}).$$
In particular, we have $\mathsf{D}(C_n)=n.$
\end{itemize}
\end{theorem}

\medskip
The aim of this paper is to study some inverse zero-sum problems of a special type. Instead of trying to characterize explicitly, given a finite Abelian group, the structure of long zero-sumfree sequences (see \cite{GaoGero02}, \cite{ChenSav07}, \cite{GaoGeroSchmid07}, \cite{Wolfgang} and \cite{GaoGeroGrynkie08}), or the structure of zero-sumfree sequences with large cross number (see \cite{GeroSchnCross97}), we study to what extent a zero-sumfree sequence can be extremal in both directions simultaneously. For instance, what is the maximal cross number of a long zero-sumfree sequence? Regarding this problem, we propose the following general conjecture.

\begin{conjecture}\label{Conjecture Cross number} Let $G \simeq C_{n_1} \oplus \dots \oplus C_{n_r},$ with $1 < n_1
\text{ } |\text{ } \dots \text{ }|\text{ } n_r \in \mathbb{N}$, be a finite Abelian group.
Given a zero-sumfree sequence $S$ in $G$ verifying $\left| S \right| \geq \mathsf{d}^*(G)$, one always
has the following inequality:
\begin{eqnarray*} \mathsf{k}(S) \leq \displaystyle\sum_{i=1}^r \left(\frac{n_i-1}{n_i}\right). \end{eqnarray*}
In particular, one has $\mathsf{k}(S) < r$.
\end{conjecture}

\medskip
One can notice that Conjecture \ref{Conjecture Cross number} is closely related to the distribution of the orders of elements in a long zero-sumfree sequence. As we will see in this paper, it provides, when it holds, useful informations on this question. In the following proposition, we gather what is currently known, to the best of our knowledge, on the structure of long zero-sumfree sequences in finite Abelian groups of rank two. This result, due to W.~Gao and A.~Geroldinger, can be found under a slightly different form in \cite{GeroKoch05}, Proposition $5.8.4$.

\begin{proposition}\label{ordreinseqmini} Let $G \simeq C_m \oplus C_{mn}$, where $m,n \in \mathbb{N}^*$, be a finite Abelian group of rank two. For every zero-sumfree sequence $S$ in $G$ with $\left|S\right|=\mathsf{d}(G)=m+mn-2$, the following two statements hold.

\begin{itemize}
\item[$(i)$] For every element $g \in S$, one has $m \text{ } | \text{ } \text{\em ord\em}(g) \text{ } | \text{ } mn$.

\item[$(ii)$] The sequence $S$ contains at least
$$m+mn-n\left(\frac{2m-2}{P^-(n)}+1\right)-1 \geq m-1$$ elements with order $mn$.
\end{itemize}
\end{proposition}

The problem of the exact structure of a long zero-sumfree sequence in groups of the form $G \simeq C_m \oplus C_{mn}$ is also closely related to an important conjecture in additive group theory, which bears upon the so-called Property B. Let $n \geq 2$ be an integer. We say that $n$ has Property B if every zero-sumfree sequence in $G \simeq C_n \oplus C_n$ with $|S|=\mathsf{d}(G)=2n-2$ contains some element repeated at least $n-2$ times.

\medskip
Property B was introduced and first studied in \cite{GaoGero99} (see also \cite{GeroKoch05}, Section $5.8$, \cite{Lettl07}
and \cite{GaoGeroSchmid07}). It is conjectured that every integer $n \geq 2$ has Property B, and recently, it was proved that an integer $n \geq 2$ has Property B if each prime divisor of $n$ has this property (see \cite{GaoGero03}, Section $8$ and \cite{GaoGeroGrynkie08}). Therefore, it remains to solve this problem for prime numbers. Regarding this, it can be shown that Property B holds for $n=2,3,5,7$ (see \cite{GaoGero03}, Proposition $4.2$), for $n=11,13,17,19$ (see \cite{BhoHalSch08}), and consequently for every integer $n$ being representable as a product of these numbers.

\medskip
Moreover, W.~Schmid proved in \cite{Wolfgang} that if some integer $m \geq 2$ has Property B, then the zero-sumfree sequences in $G \simeq C_m \oplus C_{mn}$ with length $\mathsf{d}(G)=m+mn-2$ can be characterized explicitly for all $n \in \mathbb{N}^*$. This result provides a unified way to prove Theorem $3.3$ in \cite{GaoGero02} and Theorem in \cite{ChenSav07}. It also implies, assuming that Property B holds for every integer $n \geq 2$, that Conjecture \ref{Conjecture Cross number} holds true for every finite Abelian group of rank two.

\section{New results and plan of the paper}
\label{Section new results}
In this article, we prove that Conjecture \ref{Conjecture Cross number} holds for several types of finite Abelian groups. To begin with, in Section \ref{Section premiers resultats}, we prove some consequences of this conjecture in the cases where it holds. For instance, Conjecture \ref{Conjecture Cross number}, if true, would imply simultaneously two classical and long-standing conjectures related to the Davenport constant of finite Abelian groups of the form $C^r_n$.
\begin{proposition}\label{La Conjecture implique la Conjecture sur Davenport} Let $n,r \in \mathbb{N}^*$ be such that Conjecture \ref{Conjecture Cross number} holds for the group $C^r_n$. Then, one has the following equality:
$$\mathsf{D}(C^r_n)= r(n-1)+1.$$
Moreover, every zero-sumfree sequence $S$ in $C^r_n$ with $\left|S\right|=\mathsf{d}(C^r_n)=r(n-1)$ consists only of elements with order $n$.
\end{proposition}

More generally, Conjecture \ref{Conjecture Cross number}, if true, would provide the following general upper bound for the Davenport constant of a finite Abelian group.
\begin{proposition}\label{La Conjecture donne une borne pour la constante de Davenport} Suppose that Conjecture \ref{Conjecture Cross number} holds for $G \simeq C_{n_1} \oplus \dots \oplus C_{n_r},$ with $1 < n_1 \text{ } |\text{ } \dots \text{ }|\text{ } n_r \in \mathbb{N}$. Then, one has the following inequality:
$$\mathsf{D}(G) \leq \displaystyle\sum_{i=1}^r \frac{n_r}{n_i}(n_i-1)+1=\mathsf{D}^*(G)+\displaystyle\sum_{i=1}^r \left(\frac{n_r}{n_i}-1\right)\left(n_i-1\right).$$
\end{proposition}

Then, in Section \ref{Section premiers resultats} also, we prove that Conjecture \ref{Conjecture Cross number} holds true for finite cyclic groups and finite Abelian $p$-groups.
\begin{proposition}\label{La conjecture est verifiee pour les groups cycliques et les p-groupes}
Conjecture \ref{Conjecture Cross number} holds for the following groups $G$.
\begin{itemize}
\item[$(i)$] $G$ is a finite cyclic group.

\item[$(ii)$] $G$ is a finite Abelian $p$-group.
\end{itemize}
\end{proposition}

\medskip
In Section \ref{Outline of the method}, we present a general method which was introduced in \cite{Girard08} so as to study the cross number of finite Abelian groups. Then, using this method, we prove in Section \ref{Section lemmes} two important lemmas, which will be useful in the study of the special case of finite Abelian groups of rank two.

\medskip
In Section \ref{Demonstrations}, we prove the two main theorems of this paper. The first one states that Conjecture \ref{Conjecture Cross number} holds for every finite Abelian group of rank two. As already mentioned in Section \ref{Introduction}, this result supports Property B (see \cite{Wolfgang}).

\begin{theorem}\label{main theorem} Let $G \simeq C_{m} \oplus C_{mn},$ where $m,n \in \mathbb{N}^*$, be a finite Abelian group of rank two. For every zero-sumfree sequence $S$ in $G$ with $\left| S \right| \geq \mathsf{d}^*(G)= m+mn-2$, the following inequality holds:
\begin{eqnarray*} \mathsf{k}(S) \leq \left(\frac{m-1}{m}\right)+\left(\frac{mn-1}{mn}\right). \end{eqnarray*}
In particular, one always has $\mathsf{k}(S) < 2$.
\end{theorem}

The second theorem, which is proved in Section \ref{Demonstrations} as well, is an effective result which states that, in a finite Abelian group of rank two, most of the elements of a long zero-sumfree sequence must have maximal order. This result improves significantly the statement of Proposition \ref{ordreinseqmini} $(ii)$.

\begin{theorem}\label{main theorem bis}
Let $G \simeq C_m \oplus C_{mn}$, where $m,n \in \mathbb{N}^*$, be a finite Abelian group of rank two. For every zero-sumfree sequence $S$ in $G$ with $\left|S\right|=\mathsf{d}(G)=m+mn-2$, the following two statements hold.
\begin{itemize}
\item[$(i)$] If $n$ is a prime power, then $S$ contains at least $mn-1$ elements with order $mn$.

\item[$(ii)$] If $n$ is not a prime power, then $S$ contains at least
$$\left\lceil \frac{4}{5}mn + \frac{(n-5)}{5}\right\rceil$$ elements with order $mn$.
\end{itemize}
\end{theorem}

It may be observed that for every group $G \simeq C_m \oplus C_{mn}$, where $m,n \in \mathbb{N}^*$ and $n \geq 2$, there exists a zero-sumfree sequence $S$ in $G$ with $\left|S\right|=\mathsf{d}(G)=m+mn-2$, and which does not contain strictly more than $mn-1$ elements with order $mn$. Indeed, let $(e_1, e_2)$ be a basis of $G$, with $\text{ord}(e_1)=m$ and $\text{ord}(e_2)=mn$. Then, it suffices to consider the zero-sumfree sequence $S$ consisting of the element $e_1$ repeated $m-1$ times and the element $e_2$ repeated $mn-1$ times. From this point of view, Theorem \ref{main theorem bis} proves to be "nearly optimal". In addition, the general method presented in Section \ref{Outline of the method} can be successfully used to prove an analogue of Theorem \ref{main theorem bis} in the case of finite Abelian $p$-groups (see \cite{Girard09bis}).

\medskip
Finally, in Section \ref{concluding remark}, we will present and discuss a general conjecture concerning the maximal possible length of a zero-sumfree sequence with large cross number, which can be seen as a dual version of Conjecture \ref{Conjecture Cross number}.

\bigskip

\section{Proofs of Propositions \ref{La Conjecture implique la Conjecture sur Davenport}, \ref{La Conjecture donne une borne pour la constante de Davenport} and \ref{La conjecture est verifiee pour les groups cycliques et les p-groupes}} \label{Section premiers resultats}
To start with, we prove the two corollaries of Conjecture \ref{Conjecture Cross number} announced in Section \ref{Section new results}.
\begin{proof}[Proof of Proposition \ref{La Conjecture implique la Conjecture sur Davenport}] Let $S$ be a zero-sumfree sequence in $G \simeq C^r_n$ with maximal length $\left| S \right|=\mathsf{d}(G)=\mathsf{D}(G)-1$. Then, one has the following inequality:
$$\frac{\mathsf{D}(G)-1}{n} = \frac{\left|S\right|}{n}  \leq \mathsf{k}(S) \leq r\left(\frac{n-1}{n}\right),$$
which implies that $\mathsf{D}(G) \leq r(n-1)+1=\mathsf{D}^{*}(G)$, and since $\mathsf{D}^*(G) \leq \mathsf{D}(G)$ always holds, the equality follows.
Consequently, one has:
$$\mathsf{k}(S)= r\left(\frac{n-1}{n}\right) = \frac{\mathsf{D}(G)-1}{n} = \frac{\left|S\right|}{n},$$
and so, every element $g$ of $S$ verifies $\text{ord}(g)=\exp(G)=n$.
\end{proof}

\begin{proof}[Proof of Proposition \ref{La Conjecture donne une borne pour la constante de Davenport}]
Let $S$ be a zero-sumfree sequence in $G \simeq C_{n_1} \oplus \dots \oplus C_{n_r},$ with $1 < n_1 \text{ } |\text{ } \dots \text{ }|\text{ } n_r \in \mathbb{N}$, such that $\left| S \right|=\mathsf{d}(G)=\mathsf{D}(G)-1$. Then, one has the following inequality:
$$\frac{\mathsf{D}(G)-1}{n_r} = \frac{\left|S\right|}{n_r}  \leq \mathsf{k}(S) \leq \displaystyle\sum^r_{i=1} \left(\frac{n_i-1}{n_i}\right),$$
which implies the desired result.
\end{proof}

We prove now that Conjecture \ref{Conjecture Cross number} holds true for finite cyclic groups and finite Abelian $p$-groups.
\begin{proof}[Proof of Proposition \ref{La conjecture est verifiee pour les groups cycliques et les p-groupes}]
\begin{itemize}
\item[$(i)$] Let $n \geq 2$ be an integer and let $S$ be a zero-sumfree sequence in $C_n$ with $\left|S\right| \geq \mathsf{d}^*(C_n)=n-1$. Then, it is well-known (see for instance \cite{GeroKoch05}, Theorem $5.1.10$ $(i)$) that there exists $g \in C_n$ with $\text{ord}(g)=n$ such that $S$ is of the following form:
$$S=(\underbrace{g,\dots,g}_{n-1 \text{ times}}).$$
Consequently, we obtain:
$$\mathsf{k}(S)=\frac{n-1}{n},$$
which gives the desired result.

\medskip
\item[$(ii)$] Let $p \in \mathcal{P}$, $r \in \mathbb{N}^*$, and $G \simeq C_{p^{a_1}} \oplus \cdots \oplus
C_{p^{a_r}}$, with $a_1 \leq \dots \leq a_r$ and $a_i \in \mathbb{N}^*$ for all $i \in \llbracket 1,r \rrbracket$, be a $p$-group. By Theorem \ref{Proprietes D et Cross number} $(i)$, one has:
$$\mathsf{k}(G)=\displaystyle\sum_{i=1}^r \left(\frac{p^{a_i}-1}{p^{a_i}}\right)=\mathsf{k}^*(G).$$
Then, for every zero-sumfree sequence $S$ in $G$, in particular for those verifying $\left| S \right| \geq \mathsf{d}^*(G)$, one indeed has, by the very definition of the little cross number:
$$\mathsf{k}(S) \leq \mathsf{k}(G)=\displaystyle\sum_{i=1}^r \left(\frac{p^{a_i}-1}{p^{a_i}}\right),$$
and the proof is complete.
\end{itemize}
\end{proof}

\section{Outline of a new method}
\label{Outline of the method}
Let $G$ be a finite Abelian group, and let $S$ be a sequence of elements in $G$. The general method that we will use in this paper (see also \cite{Girard08} and \cite{Girard09bis} for applications of this method in two other contexts), consists in considering, for every $d',d \in \mathbb{N}$ such that $1 \leq d' \text{ } | \text{ } d \text{ } | \text{ } \exp(G)$, the following exact sequence:
$$0 \rightarrow G_{d/d'} \hookrightarrow G_d \overset{\pi_{(d',d)}}\rightarrow \frac{G_d}{G_{d/d'}} \rightarrow 0.$$
Now, let $U$ be the subsequence of $S$ consisting of all the elements whose order divides $d$. If, for some $1 \leq d' \text{ } | \text{ } d \text{ } | \text{ } \exp(G)$, it is possible to find sufficiently many disjoint non-empty zero-sum subsequences in $\pi_{(d',d)}(U)$, that is to say sufficiently many disjoint subsequences in $U$ the sum of which are elements of order dividing $d/d'$, then $S$ cannot be a zero-sumfree sequence in $G$.

\medskip
So as to make this idea more precise, we proposed in \cite{Girard08} to introduce the following number, which can be seen as an extension of the classical Davenport constant.

\medskip
Let $G \simeq C_{n_1} \oplus \dots \oplus C_{n_r},$ with $1 < n_1 \text{ } |\text{ } \dots \text{ }|\text{ } n_r \in \mathbb{N}$, be a finite Abelian group and $d',d \in \mathbb{N}$ be two integers such that $1 \leq d' \text{ } | \text{ } d \text{ } | \text{ } \exp(G)$. By $\mathsf{D}_{(d',d)}(G)$ we denote the smallest integer $t \in
\mathbb{N}^*$ such that every sequence $S$ in $G_d$ with $|S| \geq t$ contains a subsequence with sum in $G_{d/d'}$.

\medskip
Using this definition, we can prove the following simple lemma, which is one possible illustration of the idea we presented. This result will be useful in Section \ref{Section lemmes} and states that given a finite Abelian group $G$, there exist strong constraints on the way the orders of elements have to be distributed within a zero-sumfree sequence.

\begin{lem}\label{lemme clef} Let $G$ be a finite Abelian group and $d',d \in \mathbb{N}$ be two integers such that $1 \leq d' \text{ } | \text{ } d \text{ } | \text{ } \exp(G)$. Given a sequence $S$ of elements in $G$, we will write $T$ for the subsequence of $S$ consisting of all the elements whose order divides $d/d'$, and we will write $U$ for the subsequence of $S$ consisting of all the elements whose order divides $d$ (In particular, one has $T \subseteq U$).
Then, the following condition implies that $S$ cannot be a zero-sumfree sequence:
$$\left|T\right|+\left\lfloor\frac{\left|U\right|-\left|T\right|}{\mathsf{D}_{(d',d)}(G)}\right\rfloor \geq \mathsf{D}_{\left(\frac{d}{d'},\frac{d}{d'}\right)}(G).$$
\end{lem}

\begin{proof} Let us set $\mathsf{\Delta}=\mathsf{D}_{\left(\frac{d}{d'},\frac{d}{d'}\right)}(G)$. When it holds, this inequality implies that there are $\mathsf{\Delta}$ disjoint subsequences $S_1,\dots,S_{\mathsf{\Delta}}$ of $S$, the sum of which are elements of order dividing $d/d'$. Now, by the very definition of $\mathsf{D}_{\left(\frac{d}{d'},\frac{d}{d'}\right)}(G)$, $S$ has to contain a non-empty zero-sum subsequence.
\end{proof}

\medskip
Now, in order to obtain effective inequalities from the symbolic constraints of Lemma \ref{lemme clef}, one can use a result proved in \cite{Girard08}, which states that for any finite Abelian group $G$ and every $1 \leq d' \text{ } | \text{ } d \text{ } | \text{ } \exp(G)$, the invariant $\mathsf{D}_{(d',d)}(G)$ is linked with the classical Davenport constant of a particular subgroup of $G$, which can be characterized explicitly. In order to define properly this particular subgroup, we have to introduce the following notation.

\medskip
For all $i \in \llbracket 1,r \rrbracket$, we set:
$$A_i=\gcd(d',n_i), \text{ }
B_i=\frac{\mathrm{lcm}(d,n_i)}{\mathrm{lcm}(d',n_i)}$$

$$\text{ and } \text{} \upsilon_i(d',d)=\frac{A_i}{\gcd(A_i,B_i)}.$$
For instance, whenever $d$ divides $n_i$, we have $\upsilon_i(d',d)=\gcd(d',n_i)=d'$, and in particular $\upsilon_r(d',d)=d'.$ We can now state our result on $\mathsf{D}_{(d',d)}(G)$ (see \cite{Girard08}, Proposition $3.1$).

\begin{proposition} \label{propmarrantegenerale} Let $G \simeq C_{n_1} \oplus \dots \oplus C_{n_r}$, with $1 < n_1
\text{ } |\text{ } \dots \text{ }|\text{ } n_r \in \mathbb{N}$, be a
finite Abelian group and $d',d \in \mathbb{N}$ be such that $1 \leq d' \text{ } | \text{ } d \text{ } | \text{ } \exp(G)$. Then, we have the following equality:
$$\mathsf{D}_{(d',d)}(G)=\mathsf{D}\left(C_{\upsilon_1(d',d)} \oplus
\dots \oplus C_{\upsilon_r(d',d)}\right).$$
\end{proposition}

\section{Two lemmas related to zero-freeness in $G \simeq C_m \oplus C_{mn}$}
\label{Section lemmes}
In this section, we show how the method presented in Section \ref{Outline of the method} can be used in order to obtain two key lemmas for the proofs of Theorems \ref{main theorem} and \ref{main theorem bis}. To start with, we prove the following result.

\begin{lem}\label{Lemme technique} Let $G \simeq C_m \oplus C_{mn}$, where $m,n \in \mathbb{N}^*$, $n \geq 2$, be a finite Abelian group of rank two, and let $S$ be a zero-sumfree sequence in $G$ with $\left| S \right| \geq \mathsf{d}^*(G)= m+mn-2$. Then, for every $\ell \in \mathcal{D}_n \backslash\{n\}$, one has the following inequality:
$$\displaystyle\sum_{d \in \mathcal{D}_\ell} \alpha_{md} \leq m-1.$$
\end{lem}

\begin{proof}
Let $S$ be a zero-sumfree sequence in $G \simeq C_m \oplus C_{mn}$ with $\left| S \right| \geq \mathsf{d}^*(G)= m+mn-2$. Let $\ell \in \mathcal{D}_n \backslash\{n\}$, $d'=n/\ell$ and $d=mn$, which leads to $d/d'=m\ell$. We also set $m'=\gcd(d',m)$. Now, let $T$ and $U$ be the two subsequences of $S$ which are defined in Lemma \ref{lemme clef}. In particular, one has $T \subseteq U=S$, and by Proposition \ref{ordreinseqmini} $(i)$, we obtain:
$$\left|T\right|=\displaystyle\sum_{\bar{d} \in \mathcal{D}_\ell} \alpha_{m\bar{d}}.$$

To start with, we determine the exact value of $\mathsf{D}_{(d',d)}(G)$. One has:
\begin{eqnarray*}
\upsilon_1(d',d) & = & \frac{m'}{\gcd\left(m',\frac{\text{lcm}(d,m)}{\text{lcm}(d',m)}\right)}\\
                 & = & \frac{m'}{\gcd\left(m',\frac{d}{d'}\frac{m'}{m}\right)}\\
                 & = & \frac{m'}{\gcd\left(m',m'\ell\right)}\\
                 & = & 1,
\end{eqnarray*}
and, since $\upsilon_2(d',d)=d'$, one obtains, using Proposition \ref{propmarrantegenerale} and Theorem \ref{Proprietes D et Cross number} $(ii)$, the following equalities:

\begin{eqnarray*}
\mathsf{D}_{(d',d)}(G) & = & \mathsf{D}\left(C_{\upsilon_1(d',d)} \oplus C_{\upsilon_2(d',d)}\right)\\
                       & = & \mathsf{D}\left(C_{\frac{n}{\ell}}\right)\\
                       & = & \frac{n}{\ell}.
\end{eqnarray*}

\medskip
Now, let us suppose that one has $\left|T\right| \geq m$. Since $\ell \in \mathcal{D}_n \backslash\{n\}$, we obtain the following inequalities:
\begin{eqnarray*}
\left|T\right|+\frac{\left|U\right|-\left|T\right|}{\mathsf{D}_{(d',d)}(G)} & \geq & \left|T\right|+\frac{\ell\left(m+mn-2-\left|T\right|\right)}{n}\\
& \geq & m+\frac{\ell\left(mn-2\right)}{n}\\
& = & \left(m+m\ell-1\right)-\frac{\ell}{n}+\left(\frac{n-\ell}{n}\right)\\
& > & \left(m+m\ell-1\right)-\frac{\ell}{n}\\
&  = & \mathsf{D}_{\left(\frac{d}{d'},\frac{d}{d'}\right)}(G)-\frac{1}{\mathsf{D}_{(d',d)}(G)},
\end{eqnarray*}
and, according to Lemma \ref{lemme clef}, $S$ must contain a non-empty zero-sum subsequence, which is a contradiction.
Thus, one has $\left|T\right| \leq m-1,$ which is the desired result.
\end{proof}

Now, let $n \geq 2$ be an integer, and $p_1,\dots,p_r$ be its distinct prime divisors. Given $m \in \mathbb{N}^*$ and a zero-sumfree sequence $S$ in $G \simeq C_m \oplus C_{mn}$ with $\left|S\right|\geq \mathsf{d}^*(G)=m+mn-2$, Lemma \ref{Lemme technique} implies that the integers $\alpha_{md} \in \mathbb{N}$, where $d \in \mathcal{D}_{n}\backslash \{n\}$, have to satisfy the following $r$ linear constraints: $$\displaystyle\sum_{d \in \mathcal{D}_{n/{p_i}}} \alpha_{md} \leq m-1, \text{ for all } i \in \llbracket 1,r \rrbracket.$$
In the next lemma, we solve a linear integer program on the divisor lattice of $n$, in order to obtain the maximum value of the function
$$(\alpha_{md})_{d \in \mathcal{D}_{n}\backslash \{n\}} \mapsto \displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}}\frac{\alpha_{md}}{d}$$
under the $r$ above constraints (the reader interested by linear programming methods is referred to the book \cite{Schri98}, for an exhaustive presentation of the subject).

\begin{lem}\label{Lemme technique bis} Let $m,n \in \mathbb{N}^*$, with $n \geq 2$, and let $(x_d)_{d \in  \mathcal{D}_n \backslash\{n\}}$ be a sequence of positive integers, such that for every prime divisor $p$ of $n$, one has the following linear constraint:
$$\displaystyle\sum_{d \in \mathcal{D}_{n/p}} x_d \leq m-1.$$
 Then, one has the following inequality, which is best possible:
$$\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d} \leq m-1.$$
\end{lem}

\begin{proof} Let $n \geq 2$ be an integer, and $p_1,\dots,p_r$ be its distinct prime divisors. For every $k \in \llbracket 0,m-1 \rrbracket$, let also $\mathcal{S}_k$ be the set of all the sequences of positive integers $x=(x_d)_{d \in  \mathcal{D}_n \backslash\{n\}}$ which verify the above linear constraints, and being such that $x_1=m-k-1$. Now, we can prove, by induction on $k \in \llbracket 0,m-1 \rrbracket$, that the following statement holds.
$$\text{For every sequence } x \in \mathcal{S}_k, \text{ one has } \displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d} \leq m-1.$$

If $k=0$, then for every $x \in \mathcal{S}_0$, the linear constraints imply that $x_{d}=0$ for all $d \in \mathcal{D}_n \backslash \{1,n\}$, which gives the following equality:
$$\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d} = m-1.$$
Assume now that the statement is valid for $k-1 \geq 0$. Let us define the following map:
\begin{eqnarray*}
f: \mathcal{D}_n \backslash \{n\} & \rightarrow & \{\mathcal{A} \text{ } | \text{ } \emptyset \varsubsetneq \mathcal{A} \subseteq \llbracket 1,r \rrbracket\}\\
d & \mapsto & \{i \in \llbracket 1,r \rrbracket \text{ } | \text{ } d \in \mathcal{D}_{n/p_i}\}.
\end{eqnarray*}
Let $x \in \mathcal{S}_k$ and let $\mathcal{L}$ be the set of the elements $d \in \mathcal{D}_n \backslash \{1,n\}$ such that one has $x_{d} \geq 1$. By definition, and for every $d \in \mathcal{D}_n \backslash \{n\}$, $\left|f(d)\right|$ is the number of linear constraints in which the variable $x_{d}$ appears.  Thus, for every prime divisor $p$ of $n$, $x_{n/p}$ appears in only one linear constraint, and we may assume, without loss of generality, that we have:
$$\displaystyle\sum_{d \in \mathcal{D}_{n/p}} x_d = m-1.$$
Hence, for every $i \in \llbracket 1,r \rrbracket$, the set $\mathcal{L} \cap \mathcal{D}_{n/p_i}$ is non-empty, and one obtains:
$$\displaystyle\bigcup_{d \in \mathcal{L}} f(d) = \llbracket 1,r \rrbracket.$$
Let us consider a non-empty subset $\mathcal{L}'$ of $\mathcal{L}$ verifying the following equality:
$$\displaystyle\bigcup_{d \in \mathcal{L}'} f(d) = \llbracket 1,r \rrbracket,$$
and being of minimal cardinality regarding this property. Since $f(d)$ is a non-empty set for every $d \in \mathcal{D}_n \backslash \{n\}$, the following property has to hold:
$$\displaystyle\bigcup_{d \in \mathcal{L}''} f(d) \subsetneq \llbracket 1,r \rrbracket \text{ for all } \emptyset \subsetneq \mathcal{L}'' \subsetneq \mathcal{L}'.$$
Now, one can notice the following two facts.

\medskip
\noindent
$\textbf{Fact 1.}$ For every $d \in \mathcal{L}'$, one has $\left|f(d)\right|\leq r-\left|\mathcal{L}'\right|+1$, and in particular, $\left|\mathcal{L}'\right| \leq r$. This fact is a straightforward consequence of the following combinatorial lemma.

\begin{lem}\label{lemme combinatoire} Let $r \in \mathbb{N}^*$ and $\mathcal{A}_1,\cdots,\mathcal{A}_s$ be $s$ non-empty subsets of $\llbracket 1,r \rrbracket$ verifying
$$\displaystyle\bigcup_{i \in \llbracket 1,s \rrbracket} \mathcal{A}_i = \llbracket 1,r \rrbracket, \text{ and } \displaystyle\bigcup_{i \in I} \mathcal{A}_i \subsetneq \llbracket 1,r \rrbracket \text{ for every subset } \emptyset \subsetneq I \subsetneq \llbracket 1,s \rrbracket.$$
Then, for all $i \in \llbracket 1,s \rrbracket$, one has the following inequality:
$$\left|\mathcal{A}_i\right| \leq r-s+1.$$
\end{lem}

\begin{proof} By symmetry, it suffices to prove that one has $\left|\mathcal{A}_1\right| \leq r-s+1$. Assume to the contrary that $\left|\mathcal{A}_1\right| \geq r-s+2$. Since, for all $i \in \llbracket 1,s-1 \rrbracket$, the set $\mathcal{A}_{i+1}$ must contain at least one element from $\llbracket 1,r \rrbracket \backslash \left(\mathcal{A}_1 \cup \cdots \cup \mathcal{A}_i\right)$, one obtains the following inequality:
$$\left|\mathcal{A}_1 \cup \cdots \cup \mathcal{A}_{i+1}\right| \geq \left|\mathcal{A}_1 \cup \cdots \cup \mathcal{A}_i\right|+1.$$
Therefore, we deduce by an easy induction argument that one has $\left|\mathcal{A}_1 \cup \cdots \cup \mathcal{A}_{s-1}\right| \geq (r-s+2)+(s-2)=r$, and so $\mathcal{A}_1 \cup \cdots \cup \mathcal{A}_{s-1}=\llbracket 1,r \rrbracket$, which is a contradiction.
\end{proof}

\medskip
\noindent
$\textbf{Fact 2.}$ For every $d \in \mathcal{D}_n \backslash \{n\}$, one has the following inequalities:
\begin{eqnarray*}
d & \geq & \min f^{-1}(f(d)) \\
     & \geq & \displaystyle\prod_{i \in \llbracket 1,r \rrbracket \backslash f(d)} p^{\nu_{p_i}(n)}_i\\
     & \geq & 2^{r-\left|f(d)\right|}.
\end{eqnarray*}

Now, using Facts $1$ and $2$, we can prove the desired result, by considering the sequence $y=(y_d)_{d \in  \mathcal{D}_n \backslash\{n\}}$ obtained from $x$ in the following way:
$$
y_d=\begin{cases}
x_1+1 \text{ if } d=1,\\
x_d-1 \text{ if } d \in \mathcal{L}',\\
x_d \hspace{0.75cm} \text{ otherwise}.
\end{cases}
$$
It is readily seen that $y \in \mathcal{S}_{k-1}$. Therefore, Facts $1$ and $2$ give the following inequalities:
\begin{eqnarray*}
m-1 & \geq & \displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{y_{d}}{d}\\
& = & \left(\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d}\right)+\left(1-\displaystyle\sum_{d \in \mathcal{L}'} \frac{1}{d}\right)\\
& \geq & \left(\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d}\right)+\left(1-\displaystyle\sum_{d \in \mathcal{L}'} \frac{1}{2^{r-\left|f(d)\right|}}\right)\\
& \geq & \left(\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d}\right)+\left(1-\displaystyle\sum_{d \in \mathcal{L}'} \frac{1}{2^{\left|\mathcal{L}'\right|-1}}\right)\\
& \geq & \left(\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d}\right)+\left(1-\frac{\left|\mathcal{L}'\right|}{2^{\left|\mathcal{L}'\right|-1}}\right)\\
& \geq & \displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{x_{d}}{d},
\end{eqnarray*}
which completes the proof.
\end{proof}

\section{Proofs of the two main theorems}
\label{Demonstrations}
To start with, we show that every finite Abelian group of rank two satisfies Conjecture \ref{Conjecture Cross number}. The following proof of Theorem \ref{main theorem} consists in a direct application of Lemmas \ref{Lemme technique} and \ref{Lemme technique bis}.
\begin{proof}[Proof of Theorem \ref{main theorem}] Let $G \simeq C_m \oplus C_{mn}$, where $m,n \in \mathbb{N}^*$, be a finite Abelian group of rank two, and let $S$ be a zero-sumfree sequence in $G$ with $\left|S\right| \geq \mathsf{d}^*(G)=m+mn-2$. Since, by Theorem \ref{Proprietes D et Cross number} $(ii)$, one has $\mathsf{d}(G)=\mathsf{d}^*(G)$, we obtain that $\left|S\right|=\mathsf{d}^*(G)=m+mn-2$.

\medskip
If $n=1$, then the desired result follows directly from Proposition \ref{ordreinseqmini} $(i)$, since every element of $S$ has order $m$. Now, let us suppose that $n \geq 2$. Using Proposition \ref{ordreinseqmini} $(i)$, we obtain:
\begin{eqnarray*}
\mathsf{k}(S) & = & \displaystyle\sum_{d \in \mathcal{D}_{mn}} \frac{\alpha_d}{d}\\
              & = & \displaystyle\sum_{d \in \mathcal{D}_n} \frac{\alpha_{md}}{md},
\end{eqnarray*}
and we can distinguish two cases.

\medskip
$\textbf{Case 1.}$ $\alpha_{mn} \geq mn-1$. In this case, applying Proposition \ref{ordreinseqmini} $(i)$, one obtains:
$$\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \alpha_{md}=\left|S\right|-\alpha_{mn},$$
which implies the following inequalities:
\begin{eqnarray*}
\mathsf{k}(S) & = &  \displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{\alpha_{md}}{md} + \frac{\alpha_{mn}}{mn}\\
              & \leq &  \left(\frac{\left|S\right|-\alpha_{mn}}{m}\right) + \frac{\alpha_{mn}}{mn}\\
              & \leq &  \left(\frac{\left|S\right|-(mn-1)}{m}\right) + \left(\frac{mn-1}{mn}\right)\\
              &   =  &  \left(\frac{m-1}{m}\right) + \left(\frac{mn-1}{mn}\right).
\end{eqnarray*}

\medskip
$\textbf{Case 2.}$ $\alpha_{mn} \leq mn-1$. Then, by Lemmas \ref{Lemme technique} and \ref{Lemme technique bis}, we obtain:
\begin{eqnarray*}
\mathsf{k}(S) & = &  \displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{\alpha_{md}}{md} + \frac{\alpha_{mn}}{mn}\\
              & \leq &  \left(\frac{m-1}{m}\right) + \frac{\alpha_{mn}}{mn}\\
              & \leq &  \left(\frac{m-1}{m}\right) + \left(\frac{mn-1}{mn}\right),
\end{eqnarray*}
which completes the proof.
\end{proof}

Now, we prove Theorem \ref{main theorem bis}, which gives a lower bound for the number of elements with maximal order in a long zero-sumfree sequence of a finite Abelian group of rank two.

\begin{proof}[Proof of Theorem \ref{main theorem bis}] Let $G \simeq C_m \oplus C_{mn}$, where $m,n \in \mathbb{N}^*$, be a finite Abelian group of rank two, and let $S$ be a zero-sumfree sequence in $G$ with $\left|S\right|=\mathsf{d}(G)=m+mn-2$.
\begin{itemize}
\item[$(i)$] Let $p \in \mathcal{P}$ and $a \in \mathbb{N}$ be such that $n=p^a$. If $a=0$, then $G \simeq C_m \oplus C_m$ and, by Proposition \ref{ordreinseqmini} $(i)$, every element of $S$ has order $m$. Now, let us suppose that $a \geq 1$, Then, by Lemma \ref{Lemme technique}, one has:
\begin{eqnarray*}
\left|S\right| - \alpha_{mn} & = & \displaystyle\sum_{d \in \mathcal{D}_{p^{a-1}}} \alpha_{md}\\
& \leq & m-1,
\end{eqnarray*}
which indeed implies that
\begin{eqnarray*}
\alpha_{mn} & \geq & \left|S\right| - (m-1)\\
            &   =  & m+mn-2-(m-1)\\
            &   =  & mn-1.
\end{eqnarray*}

\medskip
\item[$(ii)$]
If $\tau(n) \leq 3$, then $n$ has to be a prime power, and the desired result follows by $(i)$. Now, let us suppose that $\mathcal{D}_n=\left\{d_0=1 < d_1 < d_2 < d_3 \dots\right\}$ contains at least four elements. In particular, one has $n \geq 6$.

\medskip
By Lemmas \ref{Lemme technique} and \ref{Lemme technique bis}, one has:
$$\displaystyle\sum_{d \in \mathcal{D}_n \backslash\{n\}} \frac{\alpha_{md}}{md} \leq \frac{m-1}{m},$$
that is
\begin{eqnarray}\label{relation}
d_1\left(\frac{\alpha_{m\frac{n}{d_1}}}{mn}\right) + d_2\left(\frac{\alpha_{m\frac{n}{d_2}}}{mn}\right) + d_3\left(\frac{\alpha_{m\frac{n}{d_3}}}{mn}\right) + \displaystyle\sum_{\begin{subarray}{c}
d \in \mathcal{D}_n\\
d > d_3\end{subarray}} d\left(\frac{\alpha_{m\frac{n}{d}}}{mn}\right) \leq \frac{m-1}{m}.
\end{eqnarray}
Now, we can distinguish two cases.

\medskip
\textbf{Case 1.} $d_3=4$. Then $d_1=2$, $d_2=3$ and (\ref{relation}) implies:
\begin{eqnarray}\label{relation bis}
2\left(\frac{\alpha_{m\frac{n}{2}}}{mn}\right) + 3\left(\frac{\alpha_{m\frac{n}{3}}}{mn}\right) + 4\left(\frac{\alpha_{m\frac{n}{4}}}{mn}\right) + 5\left(\displaystyle\sum_{\begin{subarray}{c}
d \in \mathcal{D}_n\\
d > d_3\end{subarray}} \frac{\alpha_{m\frac{n}{d}}}{mn}\right) \leq \frac{m-1}{m}.
\end{eqnarray}
But since
$$\displaystyle\sum_{\begin{subarray}{c}
d \in \mathcal{D}_n\\
d > d_3\end{subarray}} \alpha_{m\frac{n}{d}}=\left|S\right|-\alpha_{mn}-\alpha_{m\frac{n}{2}}-\alpha_{m\frac{n}{3}}-\alpha_{m\frac{n}{4}},$$
relation (\ref{relation bis}) implies:
$$5\left(\frac{m+mn-2-\alpha_{mn}}{mn}\right)-3\left(\frac{\alpha_{m\frac{n}{2}}}{mn}\right) -2\left(\frac{\alpha_{m\frac{n}{3}}}{mn}\right) -\left(\frac{\alpha_{m\frac{n}{4}}}{mn}\right) \leq \frac{m-1}{m},$$
that is
$$5\left(\frac{m+mn-2-\alpha_{mn}}{mn}\right)-\left(\frac{\alpha_{m\frac{n}{2}}+\alpha_{m\frac{n}{4}}}{mn}\right) -2\left(\frac{\alpha_{m\frac{n}{2}}+\alpha_{m\frac{n}{3}}}{mn}\right)\leq \frac{m-1}{m},$$
Now using the fact that, by Lemma \ref{Lemme technique}, one has:
$$\alpha_{m\frac{n}{2}}+\alpha_{m\frac{n}{4}} \leq m-1 \text{ } \text{ as well as } \text{ } \alpha_{m\frac{n}{3}} \leq m-1,$$
we obtain
$$5\left(\frac{m+mn-2-\alpha_{mn}}{mn}\right)-\left(\frac{m-1}{mn}\right) -2\left(\frac{2\left(m-1\right)}{mn}\right) \leq \frac{m-1}{m},$$
which is equivalent to
$$5\left(\frac{mn-1-\alpha_{mn}}{mn}\right) \leq \frac{m-1}{m},$$
that is
$$5(mn-1) - n(m-1) \leq 5\alpha_{mn},$$
and thus
$$\frac{4}{5}mn+\frac{(n-5)}{5} \leq \alpha_{mn},$$
which is the desired result.

\medskip
\textbf{Case 2.} $d_3 \geq 5$. Then (\ref{relation}) implies:
\begin{eqnarray}\label{relation ter}
d_1\left(\frac{\alpha_{m\frac{n}{d_1}}}{mn}\right) + d_2\left(\frac{\alpha_{m\frac{n}{d_2}}}{mn}\right) + 5\left(\displaystyle\sum_{\begin{subarray}{c}
d \in \mathcal{D}_n\\
d \geq d_3\end{subarray}} \frac{\alpha_{m\frac{n}{d}}}{mn}\right) \leq \frac{m-1}{m}.
\end{eqnarray}
But since
$$\displaystyle\sum_{\begin{subarray}{c}
d \in \mathcal{D}_n\\
d \geq d_3\end{subarray}} \alpha_{m\frac{n}{d}}=\left|S\right|-\alpha_{mn}-\alpha_{m\frac{n}{d_1}}-\alpha_{m\frac{n}{d_2}},$$
relation (\ref{relation ter}) implies:
$$5\left(\frac{m+mn-2-\alpha_{mn}}{mn}\right)+(d_1-5)\left(\frac{\alpha_{m\frac{n}{2}}}{mn}\right) + \left(d_2-5\right)\left(\frac{\alpha_{m\frac{n}{d_2}}}{mn}\right) \leq \frac{m-1}{m}.$$
Therefore, since $d_1 \geq 2$ and $d_2 \geq 3$, we have
$$5\left(\frac{m+mn-2-\alpha_{mn}}{mn}\right)-3\left(\frac{m-1}{mn}\right) - 2\left(\frac{m-1}{mn}\right) \leq \frac{m-1}{m},$$
that is
$$5\left(\frac{mn-1-\alpha_{mn}}{mn}\right) \leq \frac{m-1}{m},$$
which leads to
$$\frac{4}{5}mn+\frac{(n-5)}{5} \leq \alpha_{mn},$$
and the proof is complete.
\end{itemize}
\end{proof}

\section{A concluding remark}
\label{concluding remark}
Given a finite Abelian group $G$, the investigation of the maximal possible length of a zero-sumfree sequence $S$ in $G$ with large cross number may also be of interest. Concerning this question, we propose the following general conjecture, which can be seen as a dual version of Conjecture \ref{Conjecture Cross number}.

\begin{conjecture}\label{Conjecture Davenport}
Let $G$ be a finite Abelian group and $G \simeq C_{\nu_1} \oplus \dots \oplus C_{\nu_s}$, with $\nu_i > 1$ for all $i \in \llbracket 1,s \rrbracket$, be its longest possible decomposition into a direct product of cyclic groups.
Given a zero-sumfree sequence $S$ in $G$ verifying $\mathsf{k}(S) \geq \mathsf{k}^*(G)$, one always has the following inequality:
\begin{eqnarray*} \left| S \right| \leq \displaystyle\sum_{i=1}^s (\nu_i-1). \end{eqnarray*}
\end{conjecture}

It can easily be seen, by Theorem \ref{Proprietes D et Cross number} $(i)$, that Conjecture \ref{Conjecture Davenport} holds true for finite Abelian $p$-groups. Even in the case of finite cyclic groups which are not $p$-groups, this problem is still wide open. Yet, in this special case, the following result supports the idea that a zero-sumfree sequence with large cross number has to be a "short" sequence.

\begin{theorem} \label{theorem index} Let $n \in \mathbb{N}^*$ be such that $n$ is not a prime power, and let $S$ be a zero-sumfree sequence in $C_n$ verifying $\mathsf{k}(S) \geq \mathsf{k}^*(C_n)$. Then, one has the following inequality:
$$\left|S\right| \leq \left\lfloor \frac{n}{2}\right\rfloor.$$
\end{theorem}

\begin{proof} So as to prove this result, we will use the notion of \em index \em of a sequence in a finite cyclic group, which was introduced implicitly in \cite{KleitLemke89}, Conjecture p.$344$, and more explicitly in \cite{ChapFreezeSmith99}. Let $g \in C_n$ with $\text{ord}(g)=n$, and let $S=(g_1,\dots,g_{\ell})=(n_1g,\dots,n_{\ell}g)$, where $n_1,\dots,n_{\ell} \in \llbracket 0,n-1 \rrbracket$, be a sequence in $C_n$. We define:
$$\left\|S\right\|_g=\displaystyle\sum^{\ell}_{i=1}\frac{n_i}{n}.$$
Since, for every $i \in \llbracket 1,\ell \rrbracket$, we have
$$\frac{\gcd(n_i,n)}{n}=\frac{1}{\text{ord}(g_i)},$$
one can notice that $\left\|S\right\|_g \geq \mathsf{k}(S)$ for all $g \in C_n$ with $\text{ord}(g)=n$. Then, the index of $S$, denoted by $\text{index}(S)$, is defined in the following fashion:
$$\text{index}(S)=\displaystyle\min_{\tiny{\begin{array}{c}
g \in C_n \\
\text{ord}(g)=n
\end{array}}} \left\|S\right\|_g.$$

Now, if $n$ is not a prime power and $S$ is a zero-sumfree sequence in $C_n$ such that $\mathsf{k}(S) \geq \mathsf{k}^*(C_n)$, one obtains, by the very definition of the index, the following inequalities:
\begin{eqnarray*} \text{index}(S)    &  \geq  & \mathsf{k}(S) \\
                                     & \geq & \mathsf{k}^*(C_n)\\
                                     &   >  &  1.
\end{eqnarray*}
Therefore, using a result of Savchev and Chen (see Theorem $9$ in \cite{SavChen07}), one must have the following inequality:
$$\left|S\right| \leq \left\lfloor \frac{n}{2}\right\rfloor,$$
which completes the proof.
\end{proof}

In particular, Theorem \ref{theorem index} implies that Conjecture \ref{Conjecture Davenport} holds true for all the cyclic groups of the form $C_{2p^a}$, where $p \in \mathcal{P}$ and $a \in \mathbb{N}$.

\section*{Acknowledgments}
I am grateful to my Ph.D. advisor Alain Plagne for his help during the preparation of this paper. I would like also to thank Alfred Geroldinger and Wolfgang Schmid for useful remarks on a preliminary version of this work.

config: arxiv


TITLE: Unbiased estimator for member of random sample
QUESTION [0 upvotes]: Hi I am having difficulty starting this question. Any help you can provide me towards working to an answer will be much appreciated!
Question:
Let $Y_{1} ... , Y_{n}$ be a random sample with n > 1.
Show that ${\tilde{\theta}} \equiv I_{(-\infty,0]}(Y_1)$ is an unbiased estimator for $\theta \equiv P(Y_1 \lt 0)$. Compute $Var(\tilde\theta)$.
My approach (so far):
Am I right to proceed in the direction that showing E(${\tilde{\theta}} - {\theta}) = 0$ implies unbiasedness? Also for the the calculation of variance, I initially thought this could be done by working out the density for the $Y_{(1)}$ but then the question refers to $Y_1$ instead. Can you show me how should I proceed please?
Many thanks

REPLY [1 votes]: An indicator function takes only values 0 and 1.
(I assume you mean $Y_i$ inside that notation.)
So, what is the probability that $\tilde \theta$ is 1
and what is the probability it is 0.
In general, suppose $T$ is a random variable with
$P\{T = 1\} = p$ and $P\{T = 0\} = 1 - p.$ Then $\mu = E(T) = 0(1-p) + 1(p) = p.$
Also, why does $E(T^2) = E(T)?$ Then $V(T) = E(T^2) - \mu^2.$
You should be able to put that information together to
find the mean and variance of $\tilde \theta.$
As for the rest: (a) If by $Y_{(1)}$, you mean the first
order statistic, just from what you've said in your question,
I don't think you need to go there. (b) $E(\tilde \theta - \theta) = 0$ or, equivalently, $E(\tilde \theta) = \theta,$ certainly gets you unbiasedness.


TITLE: Get the length of one side of this rectangle
QUESTION [0 upvotes]: X and Y are areas. The question mark refers to the shorter side of the rectangle.
I tried solving it and got 6.75 as an answer but was told it was wrong.
Let z be the short side:
$17+x+18=7z$
$x+18+y=8z$
$18+y+19=9z$
And I got $z=6.75$
Anything wrong with my systems of equations?

REPLY [1 votes]: Let the short side have length $h$. Our equations are $$17+18+x=35+x=7h$$$$18+19+y=37+y=9h$$$$18+x+y=8h$$
If we subtract the third equation from the sum of the other two, we get $$54=8h$$Which implies $h=\color{red}{6.75}$.


text:
\begin{document}
\small

\title{\bf Lower bound for the maximal number of facets of a 0/1 polytope}

\author{D.\ Gatzouras, A.\ Giannopoulos and N.\ Markoulakis}

\date{}

\maketitle

\begin{abstract}
\footnotesize  Let $f_{n-1}(P)$ denote the number of facets of a
polytope $P$ in ${\mathbb R}^n$. We show that there exist 0/1
polytopes $P$ with $$f_{n-1}(P)\geq\left (\frac{cn}{\log^2
n}\right )^{n/2}$$ where $c>0$ is an absolute constant. This
improves earlier work of B\'{a}r\'{a}ny and P\'{o}r on a question
of Fukuda and Ziegler.
\end{abstract}

\section{Introduction}

\pa The aim of this article is to give a lower bound for the
maximal possible number of facets of a 0/1 polytope in ${\mathbb
R}^n$. By definition, a 0/1 polytope is the convex hull of a
subset of the vertices of $[0,1]^n$.

In general, if $P$ is a polytope in ${\mathbb R}^n$, we write
$f_{n-1}(P)$ for the number of its facets. Let $$g(n):=\max\big\{
f_{n-1}(P_n): P_n\;\hbox{a 0/1 polytope in}\;{\mathbb R}^n\big\}
.\leqno (1.1)$$ Fukuda and Ziegler (see \cite{F}, \cite{KRSZ},
\cite{Z}) asked what the behaviour of $g(n)$ is as
$n\rightarrow\infty $. The best known upper bound to date is
$$g(n)\leq 30(n-2)! \leqno (1.2)$$ (for $n$ large enough), which
is established by Fleiner, Kaibel and Rote in \cite{FKR}. We will
study lower bounds.  A major breakthrough in this direction was
made by B\'{a}r\'{a}ny and P\'{o}r in \cite{BaPor}; they proved
that $$g(n)\geq\left (\frac{cn}{\log n}\right )^{n/4},\leqno
(1.3)$$ where $c>0$ is an absolute constant. We will show that the
exponent $n/4$ can in fact be improved to $n/2$:

\begin{theorem} There exists a constant $c>0$ such that
$$g(n)\geq \left (\frac{cn}{\log^2 n}\right )^{n/2}.\leqno (1.4)$$
\end{theorem}

It is interesting to compare this estimate with the known bounds
for the expected number of facets of the convex hull $P_{N,n}$ of
$N$ independent random points which are uniformly distributed on
the sphere $S^{n-1}$. In \cite{BMT} it is shown that there exist
two constants $c_1,c_2>0$, such that $$\left (
c_1\log\frac{N}{n}\right )^{n/2}\leq {\mathbb
E}[f_{n-1}(P_{N,n})]\leq \left ( c_2\log\frac{N}{n}\right
)^{n/2}\leqno (1.5)$$ for all $n$ and $N$ satisfying $2n\leq N\leq
2^n$. In the case of 0/1 polytopes, $N$ can be as large as $2^n$,
therefore one might conjecture that $g(n)$ is of the order of
$n^{n/2}$. Theorem 1.1 gives a lower bound which is ``practically
of this order": for every $\varepsilon >0$ one has $$g(n)
>n^{( 0.5-\varepsilon )n}\leqno (1.6)$$ if $n$ is
large enough.

\smallskip

The existence of 0/1  polytopes with many facets will be
established by a refinement of the probabilistic method developed
in \cite{BaPor}. It will be more convenient to work with $\pm 1$
polytopes (i.e., polytopes whose vertices are sequences of signs).
Let $X_1,\ldots ,X_n$ be independent and identically distributed
$\pm 1$ random variables, defined on some probability space
$(\varOmega,\mathcal{F},\mathbb{P})$, with distribution $$
\mathbb{P}(X=1)=\mathbb{P}(X=-1)=\tfrac{1}{2}.$$  Set
$\vec{X}=(X_1,\ldots ,X_n)$ and, for a fixed $N$ satisfying $n <
N\leq 2^n$, consider $N$ independent copies $\vec{X}_1,\ldots
,\vec{X}_N$ of $\vec{X}$. This procedure defines the random 0/1
polytope $$K_N={\rm conv}\{\vec{X}_1,\ldots ,\vec{X}_N\}.\leqno
(1.7)$$ Note that $K_N$ has at most $N$ vertices (it may happen
that some repetitions occur).

Under some restrictions on the range of values of $N$, we will
obtain a lower bound for the expected number of facets ${\mathbb
E}[f_{n-1}(K_N)]$, for each fixed $N$. In particular, we have:

\begin{theorem} There exist two positive constants $a$ and $b$ such
that: for all sufficiently large $n$, and all $N$ satisfying
$n^a\leq N\leq \exp ( bn/\log n)$, there exists a $0/1$ polytope
$K_N$ in ${\mathbb R}^n$ with $$f_{n-1}(K_N)\geq \left (\frac{\log
N}{a\log n}\right )^{n/2}.\leqno (1.8)$$
\end{theorem}

\noindent It is clear that Theorem 1.1 follows: one only has to
choose $N=\lfloor \exp (bn/\log n)\rfloor $.

\medskip

\section{Preliminaries}

\pa We first fix some standard notation. We work in ${\mathbb
R}^n$ which is equipped with a Euclidean structure $\langle\cdot
,\cdot\rangle $. We denote by $\|\cdot \|_2$ the corresponding
Euclidean norm, by $\|\cdot \|_\infty$ the $\max$-norm, and write
$B_2^n$ for the Euclidean unit ball and $S^{n-1}$ for the unit
sphere. Volume, surface area and the cardinality of a finite set
are denoted by $|\cdot |$ (this will cause no confusion). All
logarithms are natural. Whenever we write $a\simeq b$, we mean
that there exist absolute constants $c_1,c_2>0$ such that
$c_1a\leq b\leq c_2a$. The letters $c,c^{\prime }, C, c_1, c_2$
etc. denote absolute positive constants which may change from line
to line (however, in most places we will try to specify the
absolute constants involved).

As mentioned in the introduction, the proof of Theorem 1.2 will be
modeled on the approach of B\'{a}r\'{a}ny and P\'{o}r. This in
turn has its origin in the work of Dyer, F\"{u}redi and McDiarmid
\cite{DFM}, who proved the following: Let $\kappa =2/\sqrt{e}$ and
consider the random polytope $K_N$ defined in (1.7). For every
$\varepsilon \in (0,1)$, $$\lim_{n\rightarrow\infty }\sup\left\{
2^{-n} {\mathbb E}|K_N|\colon N\leq (\kappa -\varepsilon
)^n\right\}=0\leqno (2.1)$$ and $$\lim_{n\rightarrow\infty
}\inf\left \{ 2^{-n} {\mathbb E}|K_N|\colon N\geq (\kappa
+\varepsilon )^n\right\}=1.\leqno (2.2)$$

In order to determine the threshold $N(n)= (2/\sqrt{e})^n$, they
introduced two families of convex subsets of the cube
$C=[-1,1]^n$. For every $\vec{x}\in C$, set
$$q(\vec{x}):=\inf\big\{ {\rm Prob}\bigl(\vec{X}\in H\bigr)\colon
\vec{x}\in H,\ H\hbox{ a closed halfspace}\bigr\}.\leqno (2.3)$$
If $\beta
>0$ then the $\beta $-center of $C$ is defined by
$$ Q^{\beta }=\{\vec{x}\in C\colon q(\vec{x})\geq\exp (-\beta n)\}
; \leqno (2.4)$$ it is easily checked that $Q^{\beta }$ is a
convex polytope.

Next, consider the function $f:(-1,1)\rightarrow {\mathbb R}$ with
$$f(x)=\tfrac{1}{2}(1+x)\log (1+x)+\tfrac{1}{2}(1-x)\log (1-x)
,\leqno (2.5)$$ extend it to a continuous function on $[-1,1]$ by
setting $f(\pm 1)=\log 2$, and for every $\vec{x}=(x_1,\ldots
,x_n)\in C$ set $$F(\vec{x})=\frac{1}{n}\sum_{i=1}^nf(x_i).\leqno
(2.6)$$ The next lemma was proved in \cite[Section 3]{DFM}.

\begin{lemma}
\label{qleqe-nF}For every $\vec{x}\in (-1,1)^n$ we have
$q(\vec{x})\leq \exp (-nF(\vec{x}))$. $\hfill\Box $\end{lemma}

The second family of subsets of $C$ introduced in \cite{DFM} is as
follows: for every $\beta >0$, set $$F^{\beta }=\{ \vec{x}\in
C:F(\vec{x})\leq \beta\}.\leqno (2.7)$$ Since $f$ is a strictly
convex function on $(-1,1)$, it is clear that $F^{\beta }$ is
convex. Lemma 2.1 and the definition of $Q^{\beta }$ show that if
$\vec{x}\in Q^{\beta }\cap (-1,1)^n$ then $F(\vec{x})\leq\beta $.
In other words, we have the following.

\begin{lemma}
\label{qsubsetf} $Q^{\beta }\cap (-1,1)^n\subseteq F^{\beta }$ for
every $\beta
>0$. $\hfill\Box $
\end{lemma}

Observe that as $\beta\rightarrow\log 2$, both $Q^{\beta }$ and
$F^{\beta }$ approach $C$. The main technical step for the proof
of Theorem 1.2 will be to show that the two families are very
close, in the following sense: for a wide range of $\beta $'s, one
has that $$F^{\beta -\varepsilon }\cap\gamma C\subseteq Q^{\beta
}\cap (-1,1)^n\subseteq F^{\beta } ,\leqno (2.8)$$ where $\gamma
>0$ is a (small) absolute constant and $\varepsilon \leq 3\log
n/n$. The estimate on $\varepsilon $ substantially improves (and
at the same time clarifies) \cite[Lemma 4.3]{BaPor} and should be
viewed as the main technical step in our work. The proof is
presented in the next Section. It is based on two ingredients: (1)
Theorem 3.3, which via large deviations exhibits, for each
$\vec{x}\in\partial(F^\beta)$, the precise rate of decay (with
$n$) of the probability that a  randomly chosen vertex of the unit
cube $C=[-1,1]^n$ lies in the halfspace determined by the
hyperplane tangent to $F^\beta$ at $\vec{x}$ which does not
contain $F^\beta$; and (2) on a result of Montgomery-Smith
\cite{SMS}, giving lower bounds for the tails of the distribution
of Rademacher sums.

\medskip

\section{Comparing $F^{\alpha }$ to $Q^{\alpha }$: the main lemma}

\pa Let $n\in {\mathbb N}$ and let $X_1,X_2,\ldots,X_n$ be
independent and identically distributed $\pm 1$ random variables,
defined on some probability space
$(\varOmega,\mathcal{F},\mathbb{P})$, with distribution
$\mathbb{P}(X=1)=\mathbb{P}(X=-1)=\tfrac{1}{2}$.  For
$t\in\mathbb{R}$, let $$ \varphi(t) := \mathbb{E}
\left[e^{tX}\right] = \cosh(t) \leqno (3.1)$$ be the common moment
generating function of the $X_i$'s, and set $$ \psi(t) := \log
\varphi(t) = \log \cosh(t) .\leqno (3.2)$$  Finally, define
$h:(-1,1)\rightarrow {\mathbb R}$ by
$$h(x):=\tfrac{1}{2}\log\left(\frac{1+x}{1-x}\right) .\leqno
(3.3)$$ It is easily checked that $h$ is strictly convex and
strictly increasing on $[0,1)$. It is also easily checked that
$$f(x) = -\psi(h(x)) + x\, h(x)\quad\mbox{and}\quad f'(x)=h(x)
.\leqno (3.4)$$

Given $x_1,\ldots,x_n$ in $(-1,1)$, set $$t_i := h(x_i)\qquad
(i\leq n) ;\leqno(3.5)$$ in the sequel, $t_i$ and $x_i$ will
always be in this relationship. Observe that
$$x_i=\psi'(t_i)=\tanh(t_i)\qquad (i\leq n).\leqno(3.6)$$ Define a
new probability measure $\mathbb{P}_{x_1,\ldots,x_n}$ on
$(\varOmega,\mathcal{F})$, by $$ \mathbb{P}_{x_1,\ldots,x_n}(A) :=
{\mathbb E} \left[ \mathbf{1}_A \exp\left(\sum\limits_{i=1}^n
t_iX_i\right)\right]\,\prod\limits_{i=1}^n
[\varphi(t_i)]^{-1}\leqno (3.7)$$ for $A\in\mathcal{F}$. The next
lemma is verified by direct computation.

\begin{lemma}
\label{moments} Under $\mathbb{P}_{x_1,\ldots,x_n}$, the random
variables $t_1X_1,\ldots,t_nX_n$ are independent, with mean,
variance and absolute central third moment given by
\begin{eqnarray*} {\mathbb E}_{x_1,\ldots,x_n}[t_iX_i] &=& t_ix_i,\\
{\mathbb E}_{x_1,\ldots,x_n}\left[ t_i^2(X_i-x_i)^2\right] &=&
\frac{t_i^2}{\cosh^2(t_i)} ,\\ {\mathbb E}_{x_1,\ldots,x_n} \left[
|t_i(X_i-x_i)|^3\right] &=&
|t_i|^3\frac{\cosh(2t_i)}{\cosh^4(t_i)} ,\end{eqnarray*}
respectively. $\hfill\Box $
\end{lemma}

Set $$\sigma_n^2 := \sum\limits_{i=1}^n {\mathbb
E}_{x_1,\ldots,x_n}\left[ t_i^2(X_i-x_i)^2\right] =
\sum\limits_{i=1}^n \frac{t_i^2}{\cosh^2(t_i)} \leqno (3.8)$$ and
$$ S_n := \frac{1}{\sigma_n} \sum\limits_{i=1}^n t_i(X_i-x_i)
,\leqno (3.9)$$ and let $F_n\colon\mathbb{R}\rightarrow\mathbb{R}$
be the cumulative distribution function of the random variable
$S_n$ under the probability law $\mathbb{P}_{x_1,\ldots,x_n}$:
$$F_n(x) := \mathbb{P}_{x_1,\ldots,x_n} (S_n\leq x)\qquad
(x\in\mathbb{R}) .\leqno (3.10)$$ Write also $\mu_n$ for the
probability measure on $\mathbb{R}$ defined by $$\mu_n(-\infty,x]
:= F_n(x) \qquad(x\in\mathbb{R}) .\leqno(3.11)$$ Finally, set $$
\rho^{(3)}_n := \sum\limits_{i=1}^n {\mathbb E}_{x_1,\ldots,x_n}
\left[ |t_i(X_i-x_i)|^3\right] = \sum\limits_{i=1}^n
|t_i|^3\frac{\cosh(2t_i)}{\cosh^4(t_i)} .\leqno (3.12)$$ Since
$\cosh (2y)\leq 2\cosh^2(y)$, we have that
$$\frac{\rho_n^{(3)}}{\sigma_n^2}\leq 2\max\limits_{1\leq i\leq n}
|t_i|.\leqno (3.13)$$  Notice also that $${\mathbb
E}_{x_1,\ldots,x_n}[S_n]=0\quad\mbox{and}\quad
\mbox{Var}_{x_1,\ldots,x_n} [S_n]=1 .$$

By (3.7), we have that
\begin{multline*}\mathbb{P}\left(\sum\limits_{i=1}^n t_i(X_i-x_i)\geq 0\right)
\\ = {\mathbb E}_{x_1,\ldots,x_n} \left[
\mathbf{1}_{[0,\infty)}\left(\sum_{i=1}^n t_i(X_i-x_i)\right)\,
\exp\left(-\sum_{i=1}^n t_i X_i\right)\right] \prod\limits_{i=1}^n
\varphi(t_i);\end{multline*} hence, using (3.4), (3.8), (3.9) and
(3.11), we see that $$ \mathbb{P}\left(\sum\limits_{i=1}^n
t_i(X_i-x_i)\geq 0\right) = \int_{[0,\infty)} e^{-\sigma_n u}\,
d\mu_n(u)\, \exp\left(\sum_{i=1}^n [\psi(t_i) - t_ix_i)]\right) ,
$$ which then, upon using (3.4) and (3.5), yields that,
$$\mathbb{P}\left(\sum\limits_{i=1}^n t_i(X_i-x_i)\geq 0\right) =
\int_{[0,\infty)} e^{-\sigma_n u}\, d\mu_n(u)\, \exp\left(-
\sum_{i=1}^n f(x_i)\right) .\leqno(3.14) $$

Next, write $$\phi(x):=\dfrac{1}{\sqrt{2\pi}} e^{-x^2/2}\qquad
(x\in\mathbb{R})\leqno(3.15)$$ for the standard gaussian density,
$$\varPhi(x) := \int_{-\infty}^x \phi(y)\, d y \qquad
(x\in\mathbb{R})\leqno (3.16)$$ for the standard gaussian c.d.f.,
and $\mu$ for the standard gaussian probability measure:
$\mu(-\infty,x] := \varPhi(x)$, $x\in\mathbb{R}$.  By the
Berry--Esseen theorem \cite[Theorem XVI.5.2]{Fe} and Lemma
\ref{moments}, one has $$|F_n(x) - \varPhi(x)|\leq 6\,
\frac{\rho_n^{(3)}}{\sigma_n^3} \leqno (3.17)$$ for all $x$,
whence by (3.13), $$|F_n(x) - \varPhi(x)|\leq
\frac{12}{\sigma_n}\, \max\limits_{1\leq i \leq n} |t_i| \leqno
(3.18)$$ for all $x$.

\begin{lemma}
\label{inequality} The following inequality holds: $$ \left
|\int_{(0,\infty )} e^{-\sigma_nu}\, d\mu_n(u) - \int_{(0,\infty
)} e^{-\sigma_nu}\, d\mu(u)\right | \leq \frac{24}{\sigma_n}\,
\max\limits_{1\leq i \leq n} |t_i| .$$
\end{lemma}

\noindent {\it Proof}. Since \begin{eqnarray*}\int_{(0,\infty )}
e^{-\sigma_nu}\, d\mu_n(u) &=& \int_0^\infty \mu_n\left(\{u\colon
e^{-\sigma_nu}\mathbf{1}_{(0,\infty)}(u)\geq r\}\right)\, dr \\
&=& \int_0^1 \mu_n\left(0,-\sigma_n^{-1}\log r\right]\, dr\\ &=&
\int_0^1 \left[ F_n\left(-\sigma_n^{-1}\log r\right) -
F_n(0)\right]\, dr ,\end{eqnarray*} and similarly
\[\int_0^\infty e^{-\sigma_n u}\, d\mu(u) = \int_0^1 \left[
\varPhi\left(-\sigma_n^{-1}\log r\right) - \varPhi(0)\right]\, dr
,\] the result follows from (3.18). $\hfill\Box $

\medskip

We will use the estimates $$ \frac{1}{x}\, m_1(x)\, e^{-x^2/2}
\leq \int_x^\infty \phi(u)\, du \leq \frac{1}{x}\, m_2(x)\,
e^{-x^2/2}\qquad (x>0) ,$$ where $$m_1(x) =
\dfrac{1}{\sqrt{2\pi}}\,
\frac{2x}{x+\sqrt{x^2+4}}\qquad\mbox{and}\qquad m_2(x) =
\dfrac{1}{\sqrt{2\pi}}\, \frac{4x}{3x+\sqrt{x^2+8}}\leqno (3.19)$$
(see \cite[p.\ 17]{IMK} and, for the upper estimate, \cite{SzW}).
Since $$\int_0^\infty e^{-\sigma_n u}\, d\mu(u) =
\frac{e^{\sigma_n^2/2}}{\sqrt{2\pi}}\int_0^\infty
e^{-(\sigma_n+u)^2/2}\, du = e^{\sigma_n^2/2}
\int_{\sigma_n}^\infty \phi(u)\, du ,$$ it follows that
$$\frac{m_1(\sigma_n)}{\sigma_n} \leq \int_0^\infty e^{-\sigma_n
u}\, d\mu(u) \leq \frac{m_2(\sigma_n)}{\sigma_n} . \leqno (3.20)$$
Combining (3.20) with Lemma \ref{inequality}, we obtain the
estimates $$\dfrac{m_1(\sigma_n)}{\sigma_n} -
\dfrac{24}{\sigma_n}\, \max\limits_{1\leq i\leq n}|t_i| \leq
\int_{(0,\infty )}e^{-\sigma_n u}\, d\mu_n(u) \leq
\dfrac{m_2(\sigma_n)}{\sigma_n} + \dfrac{24}{\sigma_n}\,
\max\limits_{1\leq i\leq n}|t_i| . \leqno (3.21)$$  Equation
(3.14) then yields the following estimates:

\begin{theorem}
\label{result} Let $x_1,\ldots,x_n\in (-1,1)$ and $t_i=h(x_i)$,
$i=1,\ldots,n$. Then, $$ \mathbb{P}\left(\sum\limits_{i=1}^n
t_i(X_i-x_i)\geq 0\right) \leq \dfrac{1}{\sigma_n}\, e^{-n
F(\vec{x})} \left(m_2(\sigma_n) + 48 \max\limits_{1\leq i\leq n}
|t_i|\right) \leqno(3.22) $$ and
$$\mathbb{P}\left(\sum\limits_{i=1}^n t_i(X_i-x_i)\geq 0\right)
\geq \dfrac{1}{\sigma_n}\, e^{-n F(\vec{x})} \left(m_1(\sigma_n) -
24 \max\limits_{1\leq i\leq n} |t_i|\right) . \leqno(3.23) $$
\end{theorem}

\noindent {\it Proof}. Observe that the first factor on the right
hand-side of (3.14) is $$\int_{[0,\infty)} e^{-\sigma_n u}\,
d\mu_n(u) \geq \int_{(0,\infty)} e^{-\sigma_n u}\, d\mu_n(u). $$
Hence, we obtain the second inequality (3.23) by combining (3.14)
with (3.21). For the first inequality, first observe that $$
\int_{[0,\infty)} e^{-\sigma_n u}\, d\mu_n(u) = \int_{(0,\infty)}
e^{-\sigma_n u}\, d\mu_n(u) + \mathbb{P}_{x_1,\ldots,x_n} (S_n =
0) ,$$ and then use (3.18) to obtain, $$
\mathbb{P}_{x_1,\ldots,x_n} (S_n=0) \leq F_n(\epsilon) -
F_n(-\epsilon)\leq  \varPhi(\epsilon) - \varPhi(-\epsilon) +
\frac{24}{\sigma_n}\, \max\limits_{1\leq i \leq n} |t_i| $$ for
all $\epsilon >0$. Thus $$ \mathbb{P}_{x_1,\ldots,x_n} (S_n=0)
\leq \frac{24}{\sigma_n}\, \max\limits_{1\leq i \leq n} |t_i|
,\leqno(3.24)$$ and the first inequality (3.23) follows now as
well, using (3.14), (3.21) and (3.24) this time. $\hfill\Box$

\begin{corollary}
\label{corollary} Let $\delta\in (0,1)$. If $x_1,\ldots ,x_n\in
(-\delta ,\delta )$ and $t_i=h(x_i)$, $i=1,\ldots ,n$, then
\begin{multline*} \mathbb{P}\left(\sum\limits_{i=1}^n t_i(X_i-x_i)\geq
0\right) \leq \frac{1}{\sqrt{n F(\vec{x})}}\, e^{-nF(\vec{x}) }\\
\times \frac{\cosh(h(\delta))}{\sqrt{2}} \left( m_2\! \left
(h(\delta) (f(\delta))^{-1/2} \sqrt{n F(\vec{x})}\right ) + 48\,
h(\delta )\right ) .\end{multline*} and
\begin{multline*} \mathbb{P}\left(\sum\limits_{i=1}^n t_i(X_i-x_i)\geq
0\right) \geq \frac{1}{\sqrt{n F(\vec{x})}}\, e^{-nF(\vec{x}) }\\
\times \frac{\sqrt{f(\delta )}}{h(\delta)} \left( m_1\! \left
((\cosh(h(\delta)))^{-1} \sqrt{2 n F(\vec{x})}\right ) - 24\,
h(\delta)\right ) .\end{multline*}
\end{corollary}

\noindent {\it Proof}. Recall (3.8). The function
$$g(t)=\frac{f(\tanh (t))}{t^2}=-\frac{1}{t^2}\log\cosh
(t)+\frac{\tanh (t)}{t}\leqno (3.25)$$ is strictly decreasing on
$[0,\infty )$ and $\lim_{t\rightarrow 0}g(t)=\tfrac{1}{2}$ (see
\cite[Lemma 6.1]{BaPor}). It follows that $$\frac{f(\delta
)}{h^2(\delta )}t_i^2\leq f(x_i)\leq\frac{1}{2}t_i^2\leqno
(3.26)$$ for all $i\leq n$, since also $h$ is increasing on
$[0,1)$. Since $1\leq \cosh^2(t_i)\leq\cosh^2(h(\delta ))$, we get
that $$\frac{f(\delta )}{h^2(\delta )}\, \sigma_n^2\leq n
F(\vec{x})\leq \frac{\cosh^2( h(\delta))}{2}\, \sigma_n^2.\leqno
(3.27)$$ Finally, we also have that $$\max_{1\leq i\leq
n}|t_i|\leq h(\delta ) .\leqno (3.28)$$  Inserting these estimates
into the estimates of Theorem 3.3, and using the fact that $m_1$
and $m_2$ are increasing on $[0,\infty )$, concludes the proof.
$\hfill\Box
$

\medskip

The upper bound (3.22), and its equivalent in Corollary
\ref{corollary}, will not be used in the sequel and are only given
for completeness. Notice, however, that these bounds subsume Lemma
2.1.

\begin{corollary}
There exist $\gamma\in (0,1)$ and $k=k(\gamma )\in {\mathbb N}$
with the following property: For every $n\in {\mathbb N}$, if
$x_1,\ldots ,x_n\in (-\gamma ,\gamma )$ are such that
$\sum_{i=1}^nf(x_i)\geq k(\gamma )$, and if $t_i=h(x_i)$,
$i=1,\ldots ,n$, then $$ \mathbb{P}\left(\sum\limits_{i=1}^n
t_i(X_i-x_i)\geq 0\right) \geq \frac{\sqrt{f(\gamma )}}{10\,
h(\gamma )}\times\frac{1}{\sqrt{nF(\vec{x})}}\,
e^{-nF(\vec{x})}.\leqno(3.29)$$
\end{corollary}

\noindent {\it Proof}. First choose $\gamma \in (0,1)$ so that
$24h(\gamma )\leq (2\sqrt{2\pi })^{-1}$; this is possible because
$\lim_{\delta\rightarrow 0}h(\delta )=0$.

We know that $m_1$ increases to $(2\pi)^{-1/2}$ as
$x\rightarrow\infty $; so, there exists $k=k(\gamma )\in {\mathbb
N}$ such that $$m_1\left(\frac{\sqrt{2k(\gamma )}}{\cosh \big (
h(\gamma )\big )}\right)\geq \frac{5}{6\sqrt{2\pi }}.$$ From the
second assertion of Corollary 3.4 we easily check (3.29) for all
$x_i\in (-\gamma ,\gamma )$ with
$nF(\vec{x})=\sum_{i=1}^nf(x_i)\geq k(\gamma )$. $\hfill\Box $

\medskip

We shall also make essential use of the following result of
Montgomery-Smith from \cite{SMS}:

\begin{lemma}
\label{MontSm} There exists a universal constant $c$, such that,
for all $n\in\mathbb{N}$ and any $s_1,\ldots,s_n\in\mathbb{R}$,
the inequality $$ \mathbb{P}\left(\sum_{i=1}^n s_i X_i \geq c^{-1}
a \|\vec{s}\|_2\right) \geq c^{-1} e^{-ca^2} $$ holds for all
$a>0$ with $a\leq \|\vec{s}\|_2/\|\vec{s}\|_\infty$, where
$\vec{s}=(s_1,\ldots,s_n)$. $\hfill\Box
$
\end{lemma}

\begin{definition} {\rm In the sequel we fix a constant $\gamma\in (0,1)$ which
satisfies Corollary 3.5 and also $\gamma\leq \tanh (c^{-1})$,
where $c$ is the universal constant of Lemma \ref{MontSm}. For
example, since one can have $c=4\log 12$ in Lemma 3.6, we may
take}
$$\gamma = \min\left\{\tanh\left(\frac{1}{48\sqrt{2\pi }}\right) ,\tanh
\left (\frac{1}{4\log 12}\right )\right\}
=\tanh\left(\frac{1}{48\sqrt{2\pi }}\right).\leqno (3.30)$$
\end{definition}

\noindent We shall also use the following lemma (see \cite[Lemma
8.2]{BaPor}).

\begin{lemma}Let $\gamma\in (0,1)$ and $s_i>0$, $i=1,\ldots,m$.
Then, for $m>2/(1-\gamma)$, \begin{eqnarray*} \mathbb{P} \left (
\sum_{i=1}^m s_i(X_i-\gamma )\geq 0\right ) &\geq&
\sqrt{\frac{2}{\pi(1-\gamma^2)}}\,
\frac{1-\gamma-2m^{-1}}{1+\gamma +2m^{-1}}\, \\ & &\times\,
\exp\left(\frac{1}{12m+1}-\frac{1}{12m}\, \frac{4}{1-(\gamma
+2m^{-1})^2}\right)\, \\ & &\times\, \frac{1}{m^{3/2}}\, \exp (-m
f(\gamma )).
\end{eqnarray*}
In particular, for $\gamma\leq\tanh \big ((48\sqrt{2\pi})^{-1}\big
)$ and $m>2$, $$ \mathbb{P} \left ( \sum_{i=1}^m s_i(X_i-\gamma
)\geq 0\right ) \geq c(\gamma)\, m^{-3/2}\, e^{-mf(\gamma)} ,$$
where $c(\gamma)>0$ is given by $$c(\gamma) =
\sqrt{\frac{2}{\pi(1-\gamma^2)}}\, \frac{1-3\gamma}{5+3\gamma}\,
\exp(-[9-(3\gamma+2)^2]^{-1}) .$$
\end{lemma}

\noindent {\it Proof}. This is proved as Lemma 8.2 of
\cite{BaPor}. The argument there gives $$\mathbb{P} \left (
\sum_{i=1}^m s_i(X_i-\gamma )\geq 0\right ) \geq \frac{1}{m}\,
\frac{1}{2^m}\, \sum\limits_{\frac{\gamma+1}{2}\leq \frac{k}{m}
\leq 1} {m\choose k} \geq \frac{1}{m}\, \frac{1}{2^m}\, {m\choose
k_\gamma} , $$ where $k_\gamma := \lceil m (\gamma+1)/2 \rceil$ is
the least integer $\geq m (\gamma+1)/2$; in particular,
$$\frac{k_\gamma}{m}<\frac{1+\gamma+2m^{-1}}{2}.$$ Now using
H.~E.\ Robbins' fine form of the Stirling approximation given in
\cite[II, (9.15)]{Fe1}, and the fact that $$f(\gamma+2m^{-1})\leq
f(\gamma) + 2 m^{-1} h(\gamma+2m^{-1})$$ (by the mean value
theorem and the monotonicity of $f'=h$), yields the result.
$\hfill\Box
$

\begin{theorem} There exists $n_0=n_0(\gamma )\in {\mathbb N}$ with the
following property: If $n\geq n_0$ and $4\log
n/n\leq\alpha\leq\log 2$, then $$ F^{\alpha -\varepsilon_2}\cap
\gamma C\subseteq Q^{\alpha }$$ for some $\varepsilon_2\leq 3\log
n/n$.
\end{theorem}

\noindent {\it Proof}. Fix $\varepsilon_2=3\log n/n$. We need to
check that $q(\vec{x})\geq\exp (-\alpha n)$ for every $\vec{x}$ in
$F^{\alpha -\varepsilon_2}\cap \gamma C$. It suffices to prove
that $$ \mathbb{P}\big (\vec{X}\in H\big )\geq \exp (-\alpha
n)\leqno (3.31)$$ for every halfspace $H$ touching $F^{\alpha
-\varepsilon_2}\cap \gamma C$. In the proof of \cite[Lemma
4.3]{BaPor} it is explained that, for any such halfspace $H$,
there exists $\vec{x}$ on the bounding hyperplane of $H$ such that
$F(\vec{x})=\alpha -\varepsilon_2$. By symmetry we may assume that
$\vec{x}=(x_1,\ldots ,x_n)$ with $0 < x_1\leq x_2\leq \cdots\leq
x_n$.

There exists $n_1\in\{ 1,\ldots,n\}$ such that $x_{n_1}<\gamma $
and $x_{n_1+1}=\gamma $. As in \cite[Lemma 4.3]{BaPor} we set
$\vec{t}=(t_1,\ldots ,t_n)=n\, \nabla F(\vec{x})$ and write
$\vec{t}_{\ast }=(t^{\ast }_1,\ldots ,t^{\ast }_n)$ for the normal
to the bounding hyperplane of $H$. We may assume that
$\vec{t}_{\ast }$ is in the relative interior of the normal cone
to $F^{\alpha -\varepsilon_2}\cap \gamma C$ at $\vec{x}$, whence
$$t_i^{\ast }=t_i=f^{\prime }(x_i) \quad\mbox{if}\quad i\leq
n_1\quad\mbox{and}\quad t_i^{\ast }>t_i=f^{\prime }(\gamma
)\quad\mbox{if}\quad i>n_1.$$

Write
\begin{multline*} \mathbb{P}\left (\sum_{i=1}^nt_i^{\ast }(X_i-x_i)\geq 0\right
) = \mathbb{P}\left (\sum_{i=1}^{n_1}t_i(X_i-x_i) +
\sum_{i=n_1+1}^nt_i^{\ast }(X_i-\gamma)\geq 0\right )\\  \geq
\mathbb{P}\left (\sum_{i=1}^{n_1}t_i(X_i-x_i)\geq 0\right ) \
\mathbb{P}\left (\sum_{i=n_1+1}^nt_i^{\ast }(X_i-\gamma)\geq
0\right ) .\end{multline*}  We estimate the second probability in
the last product using Lemma 3.8: $$\mathbb{P}\left
(\sum_{i=n_1+1}^nt_i^{\ast }(X_i-\gamma)\geq 0\right )\geq
\exp\left( -(n-n_1) f(\gamma) - \tfrac{3}{2} \log(n-n_1) -
c_1(\gamma)\right) .\leqno(3.32)$$ To estimate the first
probability we distinguish two cases:

\medskip

\noindent {\sc Case 1}: $\sum_{i=1}^{n_1} f(x_i)\geq k(\gamma)$.
We may then use Corollary 3.5 to estimate the first probability:
$$ \mathbb{P}\left (\sum_{i=1}^{n_1}t_i(X_i-x_i)\geq 0\right )\geq
\exp\left( - \sum_{i=1}^{n_1} f(x_i) - \tfrac{1}{2} \log
\sum_{i=1}^{n_1} f(x_i) - c_2(\gamma) \right).\leqno(3.33)$$
Combining (3.32) and (3.33) we obtain
\begin{eqnarray*}
\mathbb{P}\left (\sum_{i=1}^nt_i^{\ast }(X_i-x_i)\geq 0\right ) &
\geq &  \exp\left( -\sum_{i=1}^{n_1} f(x_i) - \tfrac{1}{2} \log
\sum_{i=1}^{n_1} f(x_i)\right)\\ & & \times\, \exp\left(-(n-n_1)
f(\gamma) - \tfrac{3}{2} \log(n-n_1) - c(\gamma)\right)\\ &\geq &
\exp\left( -\sum_{i=1}^n f(x_i) - 2 \log n - c(\gamma)\right) \\ &
= & \exp\left( -(\alpha-\varepsilon_2) n - 2 \log n -
c(\gamma)\right) \\ &\geq & \exp(\alpha n ),
\end{eqnarray*}
provided $n$ is large enough to have that $\log n \geq c(\gamma)$.

\medskip

\noindent {\sc Case 2}: $\sum_{i=1}^{n_1} f(x_i) < k(\gamma)$. In
this case we use Lemma \ref{MontSm} to estimate the first
probability.  We have that $$\mathbb{P}\left(
\sum\limits_{i=1}^{n_1} t_i(X_i-x_i)\geq 0\right) =
\mathbb{P}\left( \sum\limits_{i=1}^{n_1} t_iX_i\geq c^{-1} a\,
\sqrt{\sum\limits_{i=1}^{n_1} t_i^2} \right) ,\leqno(3.34)$$ with
$$a=c\, \dfrac{\sum_{i=1}^{n_1} t_ix_i}{\sqrt{\sum_{i=1}^{n_1}
t_i^2}} .$$ To use Lemma \ref{MontSm} we need to check that
$$a\leq\frac{\sqrt{\sum_{i=1}^{n_1} t_i^2}}{\max_{1\leq i\leq n_1} t_i}.$$

The function $h$ is convex on $[0,1)$ and its derivative at $x=0$
is equal to $1$; hence $x\leq h(x)$ for all $x\in [0,1)$.  It
follows that $\sum_{i=1}^{n_1} t_i x_i \leq \sum_{i=1}^{n_1}
t_i^2$, by (3.5).  Therefore, $$ a\leq c\,
\sqrt{\sum\limits_{i=1}^{n_1} t_i^2} \leq c\, h(\gamma)\,
\dfrac{\sqrt{\sum_{i=1}^{n_1} t_i^2}}{\max\limits_{1\leq i\leq
n_1} t_i} ,\leqno(3.35)$$ since also $t_i=h(x_i)\leq h(\gamma)$
for all $i$, by the monotonicity of $h$.  By (3.30) and the
monotonicity of $h$, $c\, h(\gamma)\leq 1$ and $a$ satisfies the
condition of Lemma \ref{MontSm}.  Lemma \ref{MontSm} and (3.34)
then yield the bound $$\mathbb{P}\left( \sum\limits_{i=1}^{n_1}
t_i(X_i-x_i)\geq 0\right)  \geq c^{-1} e^{- c a^2 } \geq
\dfrac{1}{c}\exp\left( - c^3\, \dfrac{h^2(\gamma)}{f(\gamma)}\,
\sum\limits_{i=1}^{n_1} f(x_i) \right), \leqno(3.36)$$ the last
inequality by the first inequality in (3.35) and (3.26). Then,
(3.36) and (3.32) yield the bound
\begin{eqnarray*} & &\mathbb{P}\left( \sum\limits_{i=1}^n
t_i(X_i-x_i)\geq 0\right)\\ & &\quad\geq \exp\left( -(n-n_1)
f(\gamma) - \tfrac{3}{2} \log(n-n_1) - c_1(\gamma)
-c_3(\gamma)\sum\limits_{i=1}^{n_1} f(x_i) - c_4(\gamma)\right)\\
& &\quad \geq \exp\left( - \sum\limits_{i=1}^{n} f(x_i) -
\tfrac{3}{2} \log(n-n_1) - c_1(\gamma) -|1-c_3(\gamma)| k(\gamma)
- c_4(\gamma) \right)\\ & &\quad = \exp\left( -(\alpha -
\varepsilon_2) n - \tfrac{3}{2} \log(n-n_1) - C(\gamma) \right) \\
& &\quad \geq \exp\left( -\alpha n \right) ,
\end{eqnarray*}
provided again that $n$ is large enough to have $\log n\geq
C(\gamma)$.

\medskip

Therefore, in both Cases we have that $$ \mathbb{P}\left
(\sum_{i=1}^nt_i^{\ast }(X_i-x_i)\geq 0\right )\geq\exp \big (
-\alpha n\big ) $$ for $n\geq n_0(\gamma)$, and this completes the
proof. $\hfill\Box
$

\medskip

\section{Weakly sandwiching $K_N$}

\pa The families $\{Q^{\beta }\}$ and $\{F^{\beta }\}$ are related
to the behaviour of the random polytope $K_N$. Fix $N$ with $n <
N\leq 2^n$ and define $\alpha $ by the equation $N=e^{\alpha n}$.
In other words, $$\alpha =\frac{\log N}{n}.\leqno (4.1)$$ In
\cite[Lemma 4.2]{BaPor} (see also \cite[Lemma 2.1(b)]{DFM}), it is
proved that, for some small $\varepsilon_1 (N,n)\in (0,1)$, the
probability ${\rm Prob}\big (Q^{\alpha -\varepsilon_1
}\not\subseteq K_N\big )$ is very small if $n$ is large enough. On
the other hand, in \cite[Lemma 4.4]{BaPor} it is proved that, for
some small $\varepsilon_3 (N,n)\in (0,1)$, at least half of the
surface area of $F^{\alpha +\varepsilon_3 }$ lying in
$\tfrac{1}{10}C$ is missed by the typical $K_N$. In view of
Theorem 3.9 (or \cite[Lemma 4.3]{BaPor}), this means that
$K_N\cap\tfrac{1}{10}C$ is ``weakly sandwiched" between $F^{\alpha
-(\varepsilon_1+\varepsilon_2) }\cap\tfrac{1}{10}C$ and $F^{\alpha
+\varepsilon_3}$. In this Section we provide new estimates for the
parameters $\varepsilon_1$ and $\varepsilon_3$ in the two
statements above.

\begin{lemma} Assume that $e^2n\leq N\leq 2^n$. If $n$ is sufficiently
large, there exists $\varepsilon_1\leq  3\log n/n$ such that $$
{\rm Prob}\big ( K_N\supseteq Q^{\alpha -\varepsilon_1}\big
)>1-2^{-(n-1)}. \leqno (4.2)$$
\end{lemma}

\noindent {\it Proof}. Let $0<\beta <\alpha $. The argument in
\cite[Lemma 2.1(b)]{DFM} gives $$1-{\rm Prob}\big ( K_N\supseteq
Q^{\beta }\big )\leq {N\choose n}\, 2^{-(N-n)}+2\, {N\choose
n}\big (1-e^{-\beta n}\big )^{N-n}.\leqno (4.3)$$ We will bound
both terms on the right handside by $2^{-n}$. Since ${N\choose
n}\leq (eN/n)^n$, we only need to check that $$\left
(\frac{eN}{n}\right )^n2^{-(N-n)}<2^{-n} \leqno (4.4)$$ in order
to bound the first term. This is equivalent to $$ (N-2n)\log
2>n\log \left (\frac{eN}{n}\right ).\leqno (4.5)$$ Since $eN/n\geq
e^3$ and the function $g(x)=\log x/x$ is decreasing for $x\geq e$,
we have that $$ \log \left (\frac{eN}{n}\right )\leq
\frac{3eN}{e^3n}.\leqno (4.6)$$ So, it suffices to check that
$$3e^{-2}N<(\log 2) N-2(\log 2) n.\leqno (4.7)$$ But, $e^2\log
2>2\log 2+3$ and  $N\geq e^2n$; therefore $$ (\log 2
-3e^{-2})N\geq (e^2\log 2-3)n>2(\log 2)n,\leqno (4.8)$$ which
proves (4.4).

\medskip

\noindent Next, we will show that $$ 2\, {N\choose n}\big
(1-e^{-\beta n}\big )^{N-n}<2^{-n} ,\leqno (4.9)$$ provided that
$n$ is large enough and $\alpha -\beta\geq 3\log n/n$. Since
$1-x\leq e^{-x}$, it suffices to check that $$ \left
(\frac{4eN}{n}\right )^n\exp \left ( -e^{-\beta n}(N-n)\right
)<1.\leqno (4.10)$$ Write $\beta =\alpha -\varepsilon $, where
$\varepsilon >0$. Then, $e^{-\beta n}=e^{\varepsilon n}/N$. Since
$n\log (4eN/n)\leq n^2$ (assume that $n\geq 12$) and
$(N-n)/N\geq\tfrac{1}{2}$, we want $$ 2n^2\leq e^{\varepsilon
n}.\leqno (4.11)$$ This is satisfied if $\varepsilon \geq 3\log
n/n$.

From (4.4) and (4.9) we have the lemma. $\hfill\Box $

\begin{lemma} Let $n$ be large enough and assume that $\alpha
<\log 2-12 n^{-1}$. There exists $\varepsilon_3\leq  6/n$ such
that $$ {\rm Prob}\big ( \;|\partial (F^{\alpha
+\varepsilon_3})\cap \gamma C\cap K_N|\geq\tfrac{1}{2}|\partial
(F^{\alpha +\varepsilon_3})\cap \gamma C |\;\big )\leq
\tfrac{1}{100}.\leqno (4.12)$$
\end{lemma}

\noindent {\it Proof}. Let $\beta >\alpha $ and write $\beta
=\alpha +\varepsilon_3$ for some $\varepsilon_3>0$. Let $\vec{x}$
be on the boundary of $F^{\beta }$. If $H$ is a halfspace
containing $\vec{x}$, and if $\vec{x}\in K_N={\rm conv}\{
\vec{X}_1,\ldots ,\vec{X}_N\}$, then there exists $i\leq  N$ such
that $\vec{X}_i\in H$ (otherwise we would have $\vec{x}\in
K_N\subseteq F$, where $F$ is the complementary halfspace). We
write
\begin{eqnarray*}
{\rm Prob} \big (\vec{x}\in K_N\big ) &=& {\rm Prob}\big (\;
\vec{x}\in {\rm conv}\{ \vec{X}_1,\ldots ,\vec{X}_N\}\;\big )\\
&\leq  & {\rm Prob}\big (\; \vec{X}_i\in H \mbox{ for some } 1\leq
i\leq N \;\big )\\ &\leq &\sum_{i=1}^N{\rm Prob}\big (\;
\vec{X}_i\in H\;\big )\\ &=& N\, {\rm Prob}\big ( \vec{X}\in H\big
).
\end{eqnarray*}
Since $H\ni \vec{x}$ was arbitrary, $${\rm Prob}\big (\vec{x}\in
K_N\big )\leq  N \inf\! \big\{ {\rm Prob}\big (\vec{X}\in H\big )
\colon H\ni \vec{x}\big\}=Nq(\vec{x}).\leqno (4.13)$$ From Lemma
2.1 we have $$ Nq(\vec{x})\leq  N\exp (-nF(\vec{x}))=N\exp
(-\alpha n-\varepsilon_3n)=\exp
(-\varepsilon_3n)<\tfrac{1}{200}\leqno (4.14)$$ if
$\varepsilon_3\geq 6/n$. Now $$ {\mathbb E}\; |(\partial (F^{\beta
})\cap \gamma C \cap K_N| \leq \int_{\partial (F^{\beta })\cap
\gamma C }{\rm Prob}\big (\vec{x}\in K_N\big )\, d\vec{x}\leq
\tfrac{1}{200}|\partial (F^{\beta })\cap \gamma C |.\leqno
(4.15)$$ Therefore $$ {\rm Prob}\big (\;|(\partial (F^{\beta
})\cap \gamma C \cap K_N|\geq\tfrac{1}{2}|\partial (F^{\beta
})\cap \gamma C |\;\big )\leq 10^{-2} , \leqno (4.16)$$ by
Markov's inequality. $\hfill\Box $

\medskip

\section{Proof of the Theorem}

We follow the idea of B\'{a}r\'{a}ny and P\'{o}r. We need the
following two facts from \cite{BaPor}:

\begin{lemma} For every $\gamma\in (0,1)$, there exists a constant $c(\gamma
)>0$, such that if $n$ is large enough and $\beta\leq c(\gamma
)/\log n$, then $$ |\partial (F^{\beta })\cap\gamma C|\geq
\tfrac{1}{2} \big(1-\gamma^2\big)^{n-1} (2\beta n)^{(n-1)/2}
|S^{n-1}|.\leqno (5.1)$$
\end{lemma}

\noindent {\it Proof}. We sketch the argument from \cite[Lemma
5.1]{BaPor} in order to make the necessary adjustments. We first
estimate the product curvature $\kappa (\vec{x})$ of the surface
$F(\vec{x})=\beta $ (the formula which appears in \cite{BaPor} is
not exact; we would like to express our gratitude to I.
B\'{a}r\'{a}ny, V. Kaibel and R. Mechtel for kindly pointing out
this point). Let $\nu (\vec{x})=\nabla F(\vec{x})/\|\nabla
F(\vec{x})\|_2$ be the outward unit normal vector of $F^{\beta }$
at $\vec{x}$. Following \cite[Section 2.5]{Schn}, we write
$T_{\vec{x}}F^{\beta}$ for the tangent space of $F^{\beta }$ at
$\vec{x}$, and consider the Weingarten map $W_{\vec{x}} :
T_{\vec{x}} F^{\beta}\rightarrow T_{\vec{x}} F^{\beta}$. This is
the restriction to $T_{\vec{x}} F^{\beta}$ of the differential
$D_{\vec{x}}$ of the map $\vec{x}\mapsto \nu (\vec{x})$. Then
$W_{\vec{x}}$ is symmetric and positive definite, therefore
$$\kappa (\vec{x})=\det W_{\vec{x}}\leq \left (\frac{{\rm
trace}(W_{\vec{x}})}{n-1}\right )^{n-1}\leqno (5.2)$$ by the
arithmetic-geometric means inequality. Let $(a_{ij})_{i,j=1}^n$
denote the matrix of $D_{\vec{x}}$ with respect to the standard
basis of ${\mathbb R}^n$.  As is well known, and readily verified
by direct computation, $\nu(\vec{x})$ is an eigenvector of the
adjoint of $D_{\vec{x}}$, with corresponding eigenvalue $0$; it
follows from this and the fact that the eigenvalues of
$W_{\vec{x}}$ are also eigenvalues of $D_{\vec{x}}$ and none of
them is zero, that ${\rm trace}(W_{\vec{x}})={\rm
trace}(D_{\vec{x}})$. A simple calculation also shows that
$$a_{ii}=\frac{f^{\prime\prime }(x_i)\big (\|n \nabla
F(\vec{x})\|_2^2-(f^{\prime }(x_i))^2\big )}{\|n \nabla
F(\vec{x})\|_2^3}=\frac{h^{\prime }(x_i)\big
(\|\;\vec{t}\;\|_2^2-(h(x_i))^2\big )}{\|\;\vec{t}\;\|_2^3}.\leqno
(5.3)$$ It follows that, if $\vec{x}\in
\partial (F^{\beta })\cap\gamma C$, then
\begin{eqnarray*}
\frac{{\rm trace}(W_{\vec{x}})}{n-1}= \frac{{\rm
trace}(D_{\vec{x}})}{n-1} &=& \sum_{i=1}^n\frac{h^{\prime
}(x_i)\big (\|\;\vec{t}\;\|_2^2-(h(x_i))^2\big
)}{(n-1)\|\;\vec{t}\;\|_2^3}\\ &\leq & h^{\prime }(\gamma
)\frac{n\|\;\vec{t}\;\|_2^2- \sum_{i=1}^n
t_i^2}{(n-1)\|\;\vec{t}\;\|_2^3}\\ &=&\frac{h^{\prime }(\gamma
)}{\|\;\vec{t}\;\|_2},
\end{eqnarray*}
and (5.2) shows that $$\frac{1}{\kappa (\vec{x})}\geq \big\|\;
\vec{t}\; \big\|_2^{n-1}(1-\gamma^2)^{n-1}.\leqno (5.4)$$ Since
also $2f(x_i)\leq t_i^2$, by (3.26), we finally have that
$$\frac{1}{\kappa (\vec{x})}\geq \big(1-\gamma^2\big)^{n-1}
(2n\beta )^{(n-1)/2}.\leqno (5.5)$$

For every $\vec{\theta }\in S^{n-1}$ we write $\vec{x}(\vec{\theta
},\beta )$ for the point on the boundary of $F^{\beta }$ for which
$\vec{t}(\vec{\theta },\beta )=n\, \nabla F(\vec{x}(\vec{\theta
},\beta ))$ is a positive multiple of $\vec{\theta }$. This point
is well-defined and unique if $0<\beta < | {\rm supp} (\vec{\theta
})|\, (\log 2) / n$ (see \cite[Lemma 6.2]{BaPor}). The argument
given in \cite[Lemma 6.3]{BaPor} shows that if $$M =\left\{
\vec{\theta }\in S^{n-1}:\sqrt{\frac{n}{3\log n}}\ \vec{\theta
}\in C\right\}\leqno (5.6)$$ and if $\beta < c\gamma^2/\log n$,
then for every $\vec{\theta }\in M$ we have $\vec{x}(\vec{\theta
},\beta )\in\gamma C$  (observe that if $\vec{\theta }\in M$, we
also have $|{\rm supp}(\vec{\theta })|\geq n/(3\log n)$, and
therefore $\beta < |{\rm supp}(\vec{\theta })|\, (\log 2) / n $ if
$\beta < c\gamma^2/\log n$). A standard computation on the area of
spherical caps shows that $$|M |\geq \tfrac{1}{2}|S^{n-1}|.\leqno
(5.7)$$ Then, we can write $$|\partial (F^{\beta })\cap\gamma C| =
\int_{S^{n-1}}\frac{1}{\kappa (\vec{x})}\, d\vec{\theta } \geq
\int_{M }\frac{1}{\kappa (\vec{x})}\, d\vec{\theta }\geq
\tfrac{1}{2}\big(1-\gamma^2\big)^{n-1}(2n\beta
)^{(n-1)/2}|S^{n-1}|\leqno (5.8)$$ as claimed. $\hfill\Box $

\begin{lemma} Let $\gamma\in (0,0.1)$ and assume that $\beta +\varepsilon <\log 2$.
If $H$ is a halfspace which is disjoint from $F^{\beta }\cap\gamma
C$, then $$ |\partial (F^{\beta +\varepsilon })\cap\gamma C\cap
H|\leq (3\varepsilon n)^{(n-1)/2}|S^{n-1}|.\leqno (5.9)$$
\end{lemma}

\noindent {\it Proof}. Completely similar to the one in
\cite[Lemma 5.2]{BaPor}. $\hfill\Box $

\medskip

\noindent {\bf Proof of Theorem 1.2}: Assume that $n$ is large
enough and set $b=c(\gamma )$, where $c(\gamma )>0$ is the
constant in Lemma 5.1.

Given $N$ with $n^8\leq N\leq\exp \big (bn/\log n\big )$, let
$\alpha =\log N/n$ and let $\gamma \in (0,1)$ be the constant in
(3.30). From Lemma 4.1 there exists $\varepsilon_1\leq 3\log n/n$
such that $$ K_N\supset Q^{\alpha -\varepsilon_1}\leqno (5.10)$$
with probability greater than $1-2^{-n+1}$. Then, Theorem 3.9
shows that $$ K_N\supset F^{\alpha
-\varepsilon_1-\varepsilon_2}\cap\gamma C\leqno (5.11)$$ with
probability greater than $1-2^{-n+1}$, where $\varepsilon_2\leq 3
\log n/n$. Finally, Lemma 4.2 shows that there exists
$\varepsilon_3\leq 6/n$ such that $K_N$ satisfies $$ |\partial
(F^{\alpha +\varepsilon_3})\cap \gamma C\cap
K_N|\leq\tfrac{1}{2}|\partial (F^{\alpha +\varepsilon_3})\cap
\gamma C |\leqno (5.12)$$ with probability greater than
$1-10^{-2}$. Thus, there exists a 0/1  polytope $K_N$ which
satisfies all the above.

Next, apply Lemma 5.2 with $\beta =\alpha
-\varepsilon_1-\varepsilon_2$ and $\varepsilon
=\varepsilon_1+\varepsilon_2+\varepsilon_3$: If $A$ is a facet of
$K_N$ and $H_A$ is the corresponding halfspace (which has interior
disjoint from $K_N$), then $$ |\partial (F^{\alpha +\varepsilon_3
})\cap\gamma C\cap H_A|\leq \big (
3(\varepsilon_1+\varepsilon_2+\varepsilon_3) n\big )^{(n-1)/2}\,
|S^{n-1}|.\leqno (5.13)$$ It follows that
\begin{eqnarray*}
f_{n-1}(K_N)\, \big (3(\varepsilon_1+\varepsilon_2+\varepsilon_3)
n\big )^{(n-1)/2}\, |S^{n-1}| &\geq & \sum_A|\partial (F^{\alpha
+\varepsilon_3 })\cap\gamma C\cap H_A|\\ &=& \left |\big (\partial
(F^{\alpha +\varepsilon_3})\cap \gamma C\big )\setminus K_N\right
|\\ &\geq & \tfrac{1}{2}|\partial (F^{\alpha +\varepsilon_3})\cap
\gamma C |.
\end{eqnarray*}
Now apply Lemma 5.1 with $\beta =\alpha +\varepsilon_3$ to get $$
f_{n-1}(K_N)\, \big (3(\varepsilon_1+\varepsilon_2+\varepsilon_3)
n\big )^{(n-1)/2}\geq \tfrac{1}{2}\left (
(1-\gamma^2)\sqrt{2\alpha n}\right )^{n-1}.\leqno (5.14)$$ Since
$\alpha n=\log N$ and
$(\varepsilon_1+\varepsilon_2+\varepsilon_3)n\leq 12\log n$, this
shows that $$ f_{n-1}(K_N)\geq\left ( \frac{c_1(\gamma )\log
N}{\log n}\right )^{n/2}\leqno (5.15)$$ and the proof is complete.
$\hfill\Box $

\medskip

\noindent {\bf Proof of Theorem 1.1}: We can apply Theorem 1.2
with $N\geq \exp \big (c_1 n/\log n\big )$ where $c_1>0$ is an
absolute constant. This shows that there exists a 0/1 polytope $P$
in ${\mathbb R}^n$ with $$ f_{n-1}(P)\geq\left (
\frac{c_2n}{\log^2 n}\right )^{n/2},\leqno (5.16)$$ which is our
lower bound for $g(n)$. $\hfill\Box $

\bigskip

\bigskip

\footnotesize


TITLE: Is there a way to model discrete stochastic processes that have conditional constraints on the transition probabilities?
QUESTION [0 upvotes]: Suppose we have a stochastic process in discrete time that we can model by means of a single state variable $X_n$, with $n$ the discretized time. If the value of the immediate future state $X_{n+1}$ stochastically depends only on the value of our current state $X_n$, then we have a Markov process.
But what if we want to give the process a longer "memory"? More concretely, I wish to extend the Markov process by imposing conditional constraints on the transition probabilities.
By conditional constraints I mean constraints of the following form:

Let the process's current state $X_n$ be $B$. Then the probability of the next state $X_{n+1}$ being $C$ equals
$$
P(X_{n+1} = C | X_n = B) =
\cases{p_1 \text{ if the previous state $X_{n-1}$ was $A_1$}\\
p_2   \text{ if the previous state $X_{n-1}$ was $A_2$}.
\\
\vdots}
$$

REPLY [1 votes]: I will suppose that your state space is discrete and I name it $E$, but everything can be generalized.
Now, assume you want some "second order" Markov chain $(X_n)_{n\geq 0}$ with two steps of memory. In addition, you want that the probability to obtain $X_n$ at state $a$ knowing you had at $X_{n-1}$ state $b$ and $X_{n-2}$ at state $c$ is given by $p(c,b,a)$ for some function $p:E^3\to[0,1]$.
To do so, you can use the following procedure: consider a Markov chain $Z_n$ with state space $E^2$ and transition given by $$\mathbb{P}(Z_n=(x_1,y_1) \mid Z_{n-1}=(x_2,y_2))=p(y_2,x_2,x_1)1_{y_1=x_2}$$ and set $X_n$ to be $\pi(Z_n)$ where  $\pi$ is the projection along the first coordinate. In other words, $Z_n=(X_n,X_{n-1})$ and you will have
$$\mathbb{P}(X_n=a \mid X_{n-1}=b, X_{n-2}=c)=\mathbb{P}(Z_n=(a,b) \mid Z_{n-1}=(b,c))=p(c,b,a)$$
which is what you want.


TITLE: Proving constructed homeomorphism map is continuous
QUESTION [6 upvotes]: I was constructing a homeomorphism from $S^2 / \langle x \sim -x \rangle$ to $\mathbb{R}P^2$, the real projective plane, defined to be the space of all lines through the origin in $\mathbb{R}^3$.A basis for $\mathbb{R}P^2$ is given by the collection of subsets of lines specified by open double cones with the cone point at the origin .
The homeomorphism I came up with is $f:S^2 / \langle x \sim -x \rangle \rightarrow \mathbb{R}P^2,$ given by $f(\{x,-x\})=tx$.Where $t$ is a parameter.
I was having some difficulty showing that this map is continuous and came across the following theorem:
Let $p:X \rightarrow Y$ be a quotient map. Let $Z$ be a space and let $g:X \rightarrow Z$ be a map that is constant on each set $p^{-1}(\{y\})$, for $y \in Y$. Then $g$ induces a map $f:Y \rightarrow Z$ such that $f \circ p=g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Can I use this theorem to show the function $f$ I have constructed is continuous? For this I will denote $p:S^2 \rightarrow S^2/ \langle x \sim -x \rangle$ to be the quotient map, sending points $x,-x$ on the sphere to the equivalence class $\{x,-x\}$. So then in terms of the theorem I can take $p=p,X=S^2, Y=S^2/ \langle x \sim -x \rangle,Z=\mathbb{R}P^2,f=f,g=g$ and $g:S^2 \rightarrow \mathbb{R}P^2$ given by $g(x)=tx$ for all points in the sphere. Then $g$ is constant on $p^{-1}(\{\{x,-x\}\})$, since $g(x)=tx,g(-x)=tx$.
Can I just say $g$ is continuous since each of the components of $tx$ is a continuous polynomial, or is that not allowed since $\mathbb{R}P^2$ cannot be embedded in $\mathbb{R}^3$?
Now I am unsure whether this makes any progress, because I still have to show $g$ is continuous, which seems nonobvious. Any help?

REPLY [2 votes]: To show $g$ is continuous, you have to show that for any basis element $B$ of $\mathbb RP^2$, $g^{-1}(B)$ is an open subset of $S^2$.
As you say, $B$ is a collection of lines through the origin whose union is an open cone $C_B \subset \mathbb R^3$. Working through the definitions we get
$$g^{-1}(B) = S^2 \cap C_B
$$
Now there is a bit of trouble because $C_B$ is not an open subset of $\mathbb R^3$. However, what it means to say that $C_B$ is an "open cone" is precisely that $C_B - \{0\}$ is an open subset of $\mathbb R^3 - \{0\}$. And since $0 \not\in S^2$ it follows that $S^2$ is a subspace of $\mathbb R^3 - \{0\}$, therefore $S^2 \cap C_B$ is an open subset of $S^2$, finishing the proof of continuity of $g$.


TITLE: How dark does it have to be to see the gegenschein?
QUESTION [7 upvotes]: I know I've seen zodiacal light, especially at my old club's best dark-site. However, I'm pretty sure I've yet to see the gegenschein. How good does the seeing have to be to be able to see that?

REPLY [8 votes]: On the Bortle Dark-Sky Scale, it takes the very darkest skies - class 1. These will be very, very dark.
The "seeing" doesn't have to be especially good, of course, since the gegenschein is an extended feature and is essentially unaffected by atmospheric refraction. Poor seeing degrades the resolution of observations, but it has no effect on the brightness.


TITLE: How do I calculate the number of permutations of the list $(6, 6 ,5, 4)$?
QUESTION [1 upvotes]: I have the list $l = (6, 6, 5, 4)$ and want to how to calculate the possible number of permutations.
By using brute force I know that there are 12 possible permutations:
$$\{(6, 5, 6, 4),
 (6, 6, 5, 4),
 (5, 6, 6, 4),
 (6, 4, 5, 6),
 (6, 5, 4, 6),
 (4, 6, 6, 5),
 (4, 5, 6, 6),
 (4, 6, 5, 6),
 (6, 4, 6, 5),
 (6, 6, 4, 5),
 (5, 4, 6, 6),
 (5, 6, 4, 6)\}$$
But how would I calculate this?
I know we talk about permutation as order matters also I know that repetition is allowed (at least for $x = 6$).

REPLY [3 votes]: Here is another way of thinking about the problem:  There are four positions to fill.  We can fill two of them with a $6$ in $\binom{4}{2}$ ways.  Once those two positions have been filled, we fill one of the remaining two positions with a $5$ in $\binom{2}{1}$ ways.  Once those three positions have been filled, we can fill the remaining position with a $4$ in $\binom{1}{1}$ way.  Thus, the number of ways we can fill four positions with two sixes, one five, and a four is
$$\binom{4}{2}\binom{2}{1}\binom{1}{1} = \frac{4!}{2!2!} \cdot \frac{2!}{1!1!} \cdot \frac{1!}{0!1!} = \frac{4!}{2!1!1!} = 12$$


TITLE: How does one compute the integral $\int\limits_{-L}^{L} e^{-ax^2-Ax} d{x}$?
QUESTION [0 upvotes]: As stated in the title, I am having difficulty finding an expression for $\int\limits_{-L}^{L} e^{-ax^2-Ax}d{x}$. I can expand the integrand in a Taylor Series and integrate term-by-term, and then simplify the result with a binomial expansion, but that leads to a very ugly result. Is there a cleaner way to do it?

REPLY [0 votes]: Consider first completig the square $$a x^2+A x=\left(\sqrt{a} x+\frac{A}{2 \sqrt{a}}\right)^2-\frac{A^2}{4 a}$$ Now, change variable $$\sqrt{a} x+\frac{A}{2 \sqrt{a}}=y\implies x=\frac{2 \sqrt{a} y-A}{2 a}\implies dx=\frac{dy}{\sqrt{a}}$$ This makes $$I=\int e^{-a x^2-Ax}dx=\frac{e^{\frac{A^2}{4 a}}}{\sqrt{a}} \int e^{-y^2} dy=\frac{\sqrt{\pi }\,e^{\frac{A^2}{4 a}} }{2 \sqrt{a}}\text{erf}(y)$$ Back to $x$ $$I=\frac{\sqrt{\pi } e^{\frac{A^2}{4 a}} }{2 \sqrt{a}}\text{erf}\left(\frac{2 a x+A}{2
   \sqrt{a}}\right)$$ So $$\int_{-L}^{L} e^{-ax^2-Ax} d{x}=\frac{\sqrt{\pi } e^{\frac{A^2}{4 a}} }{2 \sqrt{a}}\left(\text{erf}\left(\frac{A+2 a L}{2 \sqrt{a}}\right)-\text{erf}\left(\frac{A-2 a L}{2
   \sqrt{a}}\right) \right)$$ If you do not access the $\text{erf}(z)$ function, you need to use numerical approximations (there are plenty of them - see here); a simple one is
$$\text{erf}(z)\approx \text{sgn}(z) \sqrt{1 - \exp\left(-z^2\frac{\frac{4}{\pi} + az^2}{1 + az^2}\right)}$$ using $$a=\frac{8 (\pi -3)}{3 (4-\pi ) \pi }$$