Computing Probabilities of Finding Extremes in a Random Set

Abstract

This study examines the occurrence of extreme points in random samples of size n obtained by mapping uniformly distributed random variables through a function into a multidimensional space. We focus on the probabilities that such sets contain a unique componentwise maximum, minimum, or both. Our interest lies in the asymptotic behavior of these probabilities. We found that in some cases, for certain irregular mappings, the limits of these probabilities may fail to exist as n tends to infinity. This contrasts with our earlier work, where the assumptions of smoothness and regularity of the mapping function ensured well-behaved limits. In the present study, we investigate scenarios in which these smoothness conditions are relaxed or absent. Because the general multidimensional case is highly challenging, we restrict attention to a simpler but illustrative setting: finite random sets in the plane that lie on the graph of a real function defined over the unit interval. We present partial results in this setting and discuss open questions that remain for future research.

1. Introduction

The subject of the present paper lies within the framework of Random Geometry, which is a field concerned with the probabilistic and geometric properties of random structures. A significant body of work in this area has examined the distribution and number of maximal elements in random subsets of ${\mathbb{R}}^d$, typically of the form ${\mathcal{S}}_n=\{{\mathbf{Z}}_1,{\mathbf{Z}}_2,\dots ,{\mathbf{Z}}_n\}$, where ${\mathbf{Z}}_i$ are random vectors. Early foundational contributions to this topic were made by Barndorff-Nielsen and Sobel, who investigated the geometric boundaries of randomly sampled points in ${\mathbb{R}}^d$ (see for instance [1]). Subsequent research has predominantly focused on the number of Pareto maxima—points that are maximal with respect to componentwise partial orderings (see [2,3,4]). Moreover, these studied only trivial cases of our problem, namely, the components of the random vectors are independent and continuous random variables. Even in this case, the research of the abovementioned papers failed to find the distribution of the maximal random vectors and of their number. Only in the case when $d=2$ was the distribution found.

We are concerned with the probability that the number of maximal Pareto points is equal to 1. Our work investigates a more specialized scenario: the existence of a true maximum and a true minimum within such random sets—referred to, respectively, as the leader and the antileader. That is, we consider cases where a single point dominates (or is dominated by) all others componentwise, which is a situation that is rarer and more constrained than the general Pareto maximality.

This problem has potential applications across a variety of domains, including mathematical programming, algorithm analysis, multi-criteria decision making, economics, medicine, social sciences, psychology, media, and agriculture. Despite its practical relevance, we found no previous work in the literature that addressed this exact formulation. As a result, we began by developing the foundational tools from scratch.

In a previous study, we developed the computational framework for evaluating these probabilities in discrete, continuous, and absolutely continuous settings (see, for instance, [5]). Furthermore, in [6], we identified conditions under which the sequences $\left(a_n\right),\left(b_n\right)$, and $\left(c_n\right)$—corresponding to the probabilities of observing a maximum, a minimum, and both simultaneously—are convergent.

In this paper, we wanted to see what would happen if $\mathit{f}$ is not a smooth function, as we considered in [5]. Moreover, for simplicity, we analyze only the particular case $d=2.$

We are interested in the following problems:

1. Is it true that the sequences ${\left(a_n\right)}_n,{\left(b_n\right)}_n,{\left(c_n\right)}_n$ are always convergent? We will show that it is not true.

2. Is it true that if${\left(a_n\right)}_n$and${\left(b_n\right)}_n$are convergent, then${\left(c_n\right)}_n$is convergent too?

3. If they are convergent, is it true or not that $c=ab$, where $a=lim{a}_n,b=lim{b}_n,c=lim{c}_n$?

We were not able to prove or disprove the questions 2 and 3.

4. How can the function f be characterized as having the property that the sequences$\left(a_n\right),\left(b_n\right),\left(c_n\right)$ are convergent?

We do not yet know all the answers; we summarize the current state of knowledge.

In order to approach it, we consider the simplest case of "bad" function, more exactly, $\mathit{f}:[0,1]\to {\mathbb{R}}^2,$$\mathit{f}\left(x\right)=\left(x,1_A\left(x\right)\right)$ where $1_A$ is the indicator function of a Borel set A, i.e., $1_A\left(x\right)=\left\{\begin{array}{ccc}1\hfill & if\hfill & x\in A\hfill \\ 0\hfill & if\hfill & x\notin A\hfill \end{array}\right.$ .

The existence of the limits $a,b$ depends on the set A. All situations may occur: both a and b exist, only a exists but b does not, a does not exist and b does, or none of them exist.

During our study, we arrived at two very natural problems which perhaps were solved by other authors, but we were not able to find anything about them in the literature. One of them is related to the weak convergence of probabilities: if ${\mu }_n$ are probability distributions on the real line and $\left({\mu }_n\right)$ has a weak limit $\mu$, for what Borel sets A is the sequence ${\left({\mu }_n\left(A\right)\right)}_n$ convergent? We present two examples of sets A for which $\underset{n\to \infty }{lim}$${\mu }_n\left(A\right)$ does exist and two examples of sets A for which $\underset{n\to \infty }{lim}$${\mu }_n\left(A\right)$ does not exist. We show that the problem of finding examples of sets A for which the limit $\underset{n\to \infty }{lim}$${\mu }_n\left(A\right)$ does or does not exist is equivalent in the case when A is a countable union of disjoint intervals with the convergence or divergence of some series. More precisely, let ${\left({\alpha }_k\right)}_k$ and ${\left({\beta }_k\right)}_k$ be increasing sequences tending to 1 such that $0\le {\alpha }_k\le {\beta }_k\le {\alpha }_{k+1}\le 1,$$\forall k\ge 1.$ In this case, $A={\displaystyle \bigcup _{k=1}^{\infty }}\left({\alpha }_k,{\beta }_k\right).$ Define ${\sigma }_n={\displaystyle \sum _{k=1}^{\infty }}\left({\beta }_k^n-{\alpha }_k^n\right),$$\forall n\ge 1.$ When does the sequence ${\left({\sigma }_n\right)}_n$ have a limit? We were able to find examples of ${\left({\alpha }_k\right)}_k$ and ${\left({\beta }_k\right)}_k$ for which ${\left({\sigma }_n\right)}_n$ is convergent and other examples for which it is not. Even if we have a simple criterion to decide if the limit does exist (see Proposition 1), we have no criterion to decide when it does not exist.

The content of the article is organized as follows: Section 2 contains definitions and notation, Section 3 introduces the context of our work, and Section 4 and Section 5 present cases when $a,b$ does or does not exist. The simple proofs are inside the sections, while the more evolved ones are presented in Appendix A.

2. Preliminaries

2.1. Definition and Notation

Let $\mathit{f}=\left(f_1,f_2\right):[0,1]\to {\mathbb{R}}^2$ be a measurable arbitrary function, and let the random variable $\mathbf{Z}=\mathit{f}\left(U\right)$, where $U\sim U\left(0,1\right)$ is uniformly distributed on $[0,1]$.

Let the functions $F_Z,F_Z^{\ast }:{\mathbb{R}}^2\to [0,1]$ and $\mathsf{\Phi }:{\mathbb{R}}^4\to [0,1]$ be defined as

$$\begin{array}{ccc}\hfill F_Z\left(s,t\right)& =& P\left(f_1\left(U\right)\le s,f_2\left(U\right)\le t\right)\hfill \\ \hfill F_Z^{\ast }\left(s,t\right)& =& P\left(f_1\left(U\right)\ge s,f_2\left(U\right)\ge t\right)\hfill \\ \hfill \mathsf{\Phi }\left(s_1,s_2,t_1,t_2\right)& =& P\left(s_1\le f_1\left(U\right)\le t_1,s_2\le f_2\left(U\right)\le t_2\right)\hfill \end{array}$$

(1)

Note that $F_Z$ is the joint distribution function of Z, $F_Z^{\ast }$ is the tail of Z, and $\mathsf{\Phi }\left(s_1,s_2,t_1,t_2\right)$ represents the probability that Z lies in the rectangle $\left[s_1,t_1\right]$x$\left[s_2,t_2\right]$.

If $g:\left[0,1\right]\to \mathbb{R}$ is a function and $x,y\in [0,1]$ are arbitrary, let $L_y\left(g\right)=\lambda \left(\left\{t:g\left(t\right)\le g\left(y\right)\right\}\right)$ and $H_x\left(g\right)=\lambda \left(\left\{t:g\left(t\right)\ge g\left(x\right)\right\}\right)$ where $\lambda :\mathcal{B}\left(\mathbb{R}\right)\to \mathbb{R}$ is the Lebesgue measure on $\mathbb{R}.$

Notation. For any $\mathit{f}=\left(f_1,f_2\right):\left[0,1\right]\to {\mathbb{R}}^2$, we define

$${\varphi }_{\mathit{f}}\left(y\right)=F_Z\left(\mathit{f}\left(y\right)\right)=P\left(f_1\left(U\right)\le f_1\left(y\right),f_2\left(U\right)\le f_2\left(y\right)\right)=\lambda \left(L_y\left(f_1\right)\cap L_y\left(f_2\right)\right)$$

$${\psi }_{\mathit{f}}\left(x\right)=F_Z^{\ast }\left(\mathit{f}\left(x\right)\right)=P\left(f_1\left(U\right)\ge f_1\left(x\right),f_2\left(U\right)\ge f_2\left(x\right)\right)=\lambda \left(H_x\left(f_1\right)\cap H_x\left(f_2\right)\right)$$

$${\eta }_{\mathit{f}}\left(x,y\right)={\mathsf{\Phi }}_Z\left(\mathit{f}\left(x\right),\mathit{f}\left(y\right)\right)=P\left(f_1\left(x\right)\le f_1\left(U\right)\le f_1\left(y\right),f_2\left(x\right)\le f_2\left(U\right)\le f_2\left(y\right)\right)=$$

$$=\lambda \left(L_y\left(f_1\right)\cap L_y\left(f_2\right)\cap H_x\left(f_1\right)\cap H_x\left(f_2\right)\right)$$

Remark 1. 

It is obvious that if $h\left(x,y\right)=\left(h_1\left(x\right),h_2\left(y\right)\right)$ has the property that $h_1$ and $h_2$ are increasing, then ${\varphi }_{\mathit{h}\left(\mathit{f}\right)}={\varphi }_{\mathit{f}},$${\psi }_{\mathit{h}\left(\mathit{f}\right)}={\psi }_{\mathit{f}},$${\eta }_{\mathit{h}\left(\mathit{f}\right)}={\eta }_{\mathit{f}}.$

2.2. Simple Facts and Known Results

In [6] was proved that

$$a_n=\int n{F}_Z^{n-1}d{F}_Z={\int }_0^{1}n{F}_Z^{n-1}\left(\mathit{f}\left(y\right)\right)dy$$

(2)

$$b_n=\int n{\left(F_Z^{\ast }\right)}^{n-1}d{F}_Z={\int }_0^{1}n{\left(F_Z^{\ast }\right)}^{n-1}\left(\mathit{f}\left(x\right)\right)dx$$

(3)

$$c_n={\int }_0^1{\int }_0^{1}n\left(n-1\right){\left({\mathsf{\Phi }}_Z\left(\mathit{f}\left(x\right),\mathit{f}\left(y\right)\right)\right)}^{n-2}dydx$$

(4)

Then, the equalities (2)–(4) become

$$a_n={\int }_0^{1}n{\varphi }_{\mathit{f}}^{n-1}\left(y\right)dy$$

(5)

$$b_n={\int }_0^{1}n{\psi }_{\mathit{f}}^{n-1}\left(x\right)dx$$

(6)

$$c_n={\int }_0^1{\int }_0^{1}n\left(n-1\right){\eta }_{\mathit{f}}^{n-2}\left(x,y\right)dydx$$

(7)

Let $a=lim{a}_n,$$b=lim{b}_n,$ and $c=lim{c}_n$, provided that these sequences are convergent.

We shall be interested in the particular case $\mathit{f}\left(x\right)=\left(x,1_A\left(x\right)\right)$ where $A\in \mathcal{B}\left(\left[0,1\right]\right).$ Let $B=A^c$ be its complement.

However, for the beginning, let us mention some simple results which hold for more general examples.

Lemma 1. 

Suppose that $\mathit{f}\left(x\right)=\left(x,g\left(x\right)\right)$, with g being an arbitrary measurable function. Let $\epsilon ,\delta >0$ be arbitrary small. Then

1. 

${\varphi }_{\mathit{f}}\left(y\right)\le y,{\psi }_{\mathit{f}}\left(x\right)\le 1-x$ and ${\eta }_{\mathit{f}}\left(x,y\right)\le {\left(y-x\right)}_{+}$ for all $x,y\in \left[0,1\right]$;

2. 

${\displaystyle \underset{n\to \infty }{lim}}\left(a_n-{\int }_{1-\delta }^{1}n{\varphi }_{\mathit{f}}^{n-1}\left(y\right)dy\right)=0$;

3. 

${\displaystyle \underset{n\to \infty }{lim}}\left(b_n-{\int }_0^{\epsilon }n{\psi }_{\mathit{f}}^{n-1}\left(x\right)dx\right)=0$;

4. 

${\displaystyle \underset{n\to \infty }{lim}}\left(c_n-{\int }_0^{\epsilon }{\int }_{1-\delta }^{1}n\left(n-1\right){\eta }_{\mathit{f}}^{n-2}\left(x,y\right)dydx\right)=0$.

This is a consequence of Proposition 4 in [6] or Lemma 5 from [5].

See the proof in the Appendix A.

In [5], some conditions were established in order that $a,b,c$ could exist. It was puzzling that in those cases, $c=ab.$ Proposition 6 from [6] says that if $\mathit{f}\left(x\right)=\left(x,g\left(x\right)\right)$, with g being an arbitrary measurable function, then we have the following:

(i).

If $g\left(1\right)=maxg$ and ${\varphi }_{\mathit{f}}$ is differentiable on some interval $\left[1-\epsilon ,1\right]$, then ${\left(a_n\right)}_n$ is convergent and, moreover, $a={\displaystyle \frac{1}{{\varphi }_{\mathit{f}}^{\prime }\left(1\right)}}.$

(ii).

If $g\left(0\right)=ming$ and ${\varphi }_{\mathit{f}}$ is differentiable on some interval $\left[0,\epsilon \right]$, then ${\left(b_n\right)}_n$ is convergent and, moreover, $b=-{\displaystyle \frac{1}{{\psi }_{\mathit{f}}^{\prime }\left(0\right)}}$

(iii).

If the conditions from (i) and (ii) are fulfilled, then ${\left(c_n\right)}_n$ is convergent, and $c=ab.$

Remark 2. 

It can be objected that these conditions are stated in terms of $\varphi ,\psi ,$ and η and not directly in terms of $\mathit{f}=\left(f_1,f_2\right)$. This is true, since even in the simplest case $\mathit{f}\left(x\right)=\left(x,g\left(x\right)\right)$ it is not clear for what kind of g the function ϕ is continuous and differentiable near 1. Clearly, it is not enough that g be differentiable near 1. For example, if $g\left(x\right)=min\left(1,\left|3x-2\right|\right)$, then ${\varphi }_{\mathit{f}}\left({\displaystyle \frac{1}{3}}-0\right)={\displaystyle \frac{1}{3}},{\varphi }_{\mathit{f}}\left({\displaystyle \frac{1}{3}}+0\right)=0$, ${\varphi }_{\mathit{f}}\left(1-0\right)={\displaystyle \frac{2}{3}},{\varphi }_{\mathit{f}}\left(1\right)=1.$ Thus, g is continuous and differentiable on $({\displaystyle \frac{2}{3}},1]$, but ${\varphi }_{\mathit{f}}$ is not continuous at $x={\displaystyle \frac{1}{3}}$, and $x=1.$

To fill in this gap between g and ${\varphi }_{\mathit{f}}$, various classes of functions have been considered in [6], namely, the piecewise monotone ones where the monotone pieces are differentiable.

3. An Extreme Case

Now, we want to consider an extreme case: Let $\mathit{f}:[0,1]\to {\mathbb{R}}^2,$

$\mathit{f}\left(x\right)=\left(x,1_A\left(x\right)\right)$, with $A\in \mathcal{B}\left([0,1]\right)$, be an arbitrary Borel set. Let $B=A^c$ be its complement. In that case,

$${\varphi }_{\mathit{f}}\left(y\right)=P\left(U\le y,1_A\left(U\right)\le 1_A\left(y\right)\right)=\left\{\begin{array}{cc}y& ify\in A\\ \lambda \left(\left[0,y\right]\cap B\right)& ify\in B\end{array}\right.$$

$${\psi }_{\mathit{f}}\left(x\right)=P\left(U\ge x,1_A\left(U\right)\ge 1_A\left(x\right)\right)=\left\{\begin{array}{cc}1-x& ifx\in B\\ \lambda \left(\left[x,1\right]\cap A\right)& ifx\in A\end{array}\right.$$

$${\eta }_{\mathit{f}}\left(x,y\right)=P\left(x\le U\le y,1_A\left(x\right)\le 1_A\left(U\right)\le 1_A\left(y\right)\right)=\left\{\begin{array}{cc}\lambda \left(\left[x,y\right]\cap A\right)& ifx\in A,y\in A\\ 0& ifx\in A,y\in B\\ y-x& ifx\in B,y\in A\\ \lambda \left(\left[x,y\right]\cap B\right)& ifx\in B,y\in B\end{array}\right.$$

Otherwise, it can be written with $0\le x\le y\le 1$ as

$${\varphi }_{\mathit{f}}\left(y\right)=y·1_A\left(y\right)+\lambda \left(\left[0,y\right]\cap B\right)·1_B\left(y\right)$$

(8)

$${\psi }_{\mathit{f}}\left(x\right)=\lambda \left(\left[x,1\right]\cap A\right)·1_A\left(x\right)+\left(1-x\right)·1_B\left(x\right)$$

(9)

$${\eta }_{\mathit{f}}\left(x,y\right)=\lambda \left(\left[x,y\right]\cap A\right)·1_{A\times A}\left(x,y\right)+\left(y-x\right)·1_{B\times A}\left(x,y\right)+\lambda \left(\left[x,y\right]\cap B\right)·1_{B\times B}\left(x,y\right)$$

(10)

Moreover, from (5) and from (8), it results that $a_n={\int }_0^{1}n{\left(y{1}_A\left(y\right)+\lambda \left(\left[0,y\right]\cap B\right)1_B\left(y\right)\right)}^{n-1}dy$$={\int }_0^{1}n{y}^{n-1}{1}_A\left(y\right)dy+{\int }_0^{1}n\left({\lambda }^{n-1}\left(\left[0,y\right]\cap B\right)\right)1_B\left(y\right)dy$ .

Notice that ${\int }_0^{1}n{\lambda }^{n-1}\left(\left[0,y\right]\cap B\right)1_B\left(y\right)dy\le {\int }_{B}n{\lambda }^{n-1}\left(B\right)dy=n\lambda {\left(B\right)}^n\to 0.$

Since we are interested in the asymptotic behavior of $a_n,b_n,c_n$, we can write

$$a_n\approx {\int }_0^{1}n{y}^{n-1}{1}_A\left(y\right)dy$$

(11)

having agreed to the notation $x_n\approx y_n$ instead of $lim\left(x_n-y_n\right)=0.$

In the same way, from (6) and from (9), we get $b_n={\int }_0^{1}n{\left(1-x\right)}^{n-1}{1}_B\left(x\right)dx+{\int }_0^{1}n{\lambda }^{n-1}\left(\left[x,1\right]\cap A\right)1_A\left(x\right).$

As ${\int }_0^{1}n{\lambda }^{n-1}\left(\left[x,1\right]\cap A\right)dx\le {\int }_0^{1}n{\lambda }^{n-1}\left(A\right)dx=n{\lambda }^{n-1}\left(A\right)\to 0$, we can write

$$b_n\approx {\int }_0^{1}n{\left(1-x\right)}^{n-1}{1}_B\left(x\right)dx$$

(12)

According to (7) and (10), we get

$c_n=\int {\int }_{B\times A}n\left(n-1\right){\left(y-x\right)}_{+}^{n-2}dydx+\int {\int }_{A\times A}n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap A\right)dydx+$

$+\int {\int }_{B\times B}n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap B\right)dydx$.

But, $\int \int n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap A\right)1_{A\times A}\left(x,y\right)dydx\le n\left(n-1\right){\lambda }^n\left(A\right)\to 0$, and also, $\int \int n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap B\right)1_{B\times B}\left(x,y\right)dydx\le n\left(n-1\right){\lambda }^n\left(B\right)\to 0.$ It follows that

$$c_n\approx {\int }_0^1{\int }_x^{1}n\left(n-1\right){\left(y-x\right)}^{n-2}{1}_B\left(x\right)1_A\left(y\right)dydx$$

(13)

In a pure probabilistic approach, another way to address the problem is the following:

Let ${\left(U_j\right)}_j$ be independent random variables that are identically uniform distributed on $[0,1]$ and

$$U^{\ast }\left(n\right)=max\left(U_1,U_{2,}...,U_n\right),U_{\ast }\left(n\right)=min\left(U_1,U_2,...,U_n\right),\forall n\ge 1.$$

(14)

If $G_n$ is the distribution of $U_{\ast }\left(n\right)$, $H_n$ is the distribution of $U^{\ast }\left(n\right)$, and $Q_n$ is the distribution of the vector $\left(U_{\ast }\left(n\right),U^{\ast }\left(n\right)\right)$; then, $G_n\Rightarrow {\delta }_0,H_n\Rightarrow {\delta }_1$, and $Q_n\Rightarrow {\delta }_0\otimes {\delta }_1={\delta }_{0,1}$. Here, “⇒” stands for weak convergence (see, for instance, [7]). By the notation ${\delta }_0$, we understand the Dirac needle measure.

According to the characterization of weak convergence (the Portmanteau theorem; see [8]), we know that $G_n\Rightarrow {\delta }_0$ if and only if $G_n\left(A\right)\to {\delta }_0\left(A\right)$ for all the sets A such that ${\delta }_0\left(Cl\left(A\right)\setminus Int\left(A\right)\right)=0.$

It is known (see, for instance, [9]) that the densities of $U_{\ast }\left(n\right),U^{\ast }\left(n\right)$ and $\left(U_{\ast }\left(n\right),U^{\ast }\left(n\right)\right)$ are $g_n\left(x\right)=n{\left(1-x\right)}^{n-1}{1}_{\left(0,1\right)}\left(x\right),h_n\left(y\right)=n{y}^{n-1}{1}_{\left(0,1\right)}\left(y\right)$ and $q_n\left(x,y\right)=n\left(n-1\right){\left(y-x\right)}^{n-2}{1}_{\left(x,1\right)}\left(y\right)1_{\left(0,1\right)}\left(x\right).$

Then, taking into account the particular form of $Z_j=\left(U_j,1_A\left(U_j\right)\right),$ we find

$$a_n=P\left(U^{\ast }\left(n\right)\in A\right)+P\left(U_j\in B\mathrm{for}\mathrm{all}j\right)={\int }_{A}n{y}^{n-1}dy+{\lambda }^n\left(B\right)$$

(15)

$$b_n=P\left(U_{\ast }\left(n\right)\in B\right)+P\left(U_j\in A\mathrm{for}\mathrm{all}j\right)={\int }_{B}n{\left(1-x\right)}^{n-1}dx+{\lambda }^n\left(A\right)$$

(16)

$$c_n=P\left(U_{\ast }\left(n\right)\in B,U^{\ast }\left(n\right)\in A\right)+P\left(U_j\in B\mathrm{for}\mathrm{all}j\right)+P\left(U_j\in A\mathrm{for}\mathrm{all}j\right)$$

(17)

$$=\int \int n\left(n-1\right){\left(y-x\right)}^{n-2}{1}_B\left(x\right)1_A\left(y\right)dydx+{\lambda }^n\left(B\right)+{\lambda }^n\left(A\right).$$

For a better understanding of these relations, let us recall the definitions of these terms. For instance, by definition, $a_n$ is the probability that the sample ${\mathcal{S}}_n=\{{\mathbf{Z}}_1,{\mathbf{Z}}_2,...,{\mathbf{Z}}_n\}$ has a leader, where ${\mathbf{Z}}_i=\left(U_i,1_A\left(U_i\right)\right).$ If there exists the leader $U^{\ast }\left(n\right),$ either this belongs to the set $A,$ or it does not belong to $A.$ In the last case, it follows that $U_j\notin A$ for all $1\le j\le n.$ Therefore, we can write $a_n=P\left(U^{\ast }\left(n\right)\in A\right)+P\left(U_j\in B,\forall 1\le j\le n\right)$, with $B=A^c.$ Now, considering the density of $U^{\ast }\left(n\right)$, we find $P\left(U^{\ast }\left(n\right)\in A\right)={\int }_{A}n{y}^{n-1}dy.$ It indeed results that $a_n={\int }_{A}n{y}^{n-1}dy+{\lambda }^n\left(B\right).$ The relations for $b_n$ and $c_n$ can be obtained with similar arguments.

Lemma 2. 

Let $A\subset [0,1]$ be a Borel set and $\lambda :\mathcal{B}\left([0,1]\right)\to \mathbb{R}$ be the Lebesgue measure. Then, the following hold:

$${\int }_0^{1}n{\lambda }^{n-1}\left([0,x]\cap A\right)1_A\left(x\right)dx={\lambda }^n\left(A\right)$$

(18)

$${\int }_0^{1}n{\lambda }^{n-1}\left([x,1]\cap A\right)1_A\left(x\right)dx={\lambda }^n\left(A\right)$$

(19)

$$\int {\int }_{A\times A}n\left(n-1\right){\lambda }^{n-2}\left([x,y]\cap A\right)1_A\left(x\right)dydx={\lambda }^n\left(A\right)$$

(20)

Proof. 

We know that

$a_n={\int }_0^{1}n{y}^{n-1}{1}_A\left(y\right)dy+{\int }_0^{1}n\left({\lambda }^{n-1}\left(\left[0,y\right]\cap B\right)\right)1_B\left(y\right)dy$ and $a_n={\int }_{A}n{y}^{n-1}dy+{\lambda }^n\left(B\right).$

Therefore,

$${\int }_0^{1}n\left({\lambda }^{n-1}\left(\left[0,y\right]\cap B\right)\right)1_B\left(y\right)dy={\lambda }^n\left(B\right)$$

In the same way, if we compare $b_n={\int }_0^{1}n{\left(1-x\right)}^{n-1}{1}_B\left(x\right)dx+{\int }_0^{1}n{\lambda }^{n-1}\left(\left[x,1\right]\cap A\right)1_A\left(x\right)$ to $b_n={\int }_{B}n{\left(1-x\right)}^{n-1}dx+{\lambda }^n\left(A\right)$, we obtain

$${\int }_0^{1}n{\lambda }^{n-1}\left(\left[x,1\right]\cap A\right)1_A\left(x\right)dx={\lambda }^n\left(A\right)$$

Finally, from $c_n=\int {\int }_{B\times A}n\left(n-1\right){\left(y-x\right)}_{+}^{n-2}dydx+\int {\int }_{A\times A}n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap A\right)dydx+$$+\int {\int }_{B\times B}n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap B\right)dydx$ and $c_n=\int \int n\left(n-1\right){\left(y-x\right)}^{n-2}{1}_B\left(x\right)1_A\left(y\right)dydx+{\lambda }^n\left(B\right)+{\lambda }^n\left(A\right)$, it results that

$$\int {\int }_{A\times A}n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap A\right)dydx={\lambda }^n\left(A\right)$$

$$\int {\int }_{B\times B}n\left(n-1\right){\lambda }^{n-2}\left(\left[x,y\right]\cap B\right)dydx={\lambda }^n\left(B\right)$$

□

Remark 3. 

The equalities (18)–(20) are interesting in their own right.

Lemma 3. 

Let A$\in B\left(\left[0,1\right]\right),$ $B=A^c$ and $f:\left[0,1\right]\to {\mathbb{R}}^2$, $f\left(x\right)=\left(x,1_A\left(x\right)\right).$ Then, we have the following:

(i) 

The sequence ${\left(a_n\right)}_n$ is convergent if and only if ${\left({\int }_{A}n{y}^{n-1}dy\right)}_n$is convergent. If that is true, then

$$a=\underset{n}{lim}{\int }_{A}n{y}^{n-1}dy$$

(ii) 

The sequence ${\left(b_n\right)}_n$ is convergent if and only if ${\left({\int }_{B}n{\left(1-x\right)}^{n-1}dx\right)}_n$ is convergent. If that is true, then

$$b=\underset{n}{lim}{\int }_{B}n{\left(1-x\right)}^{n-1}dx$$

(iii) 

The sequence ${\left(c_n\right)}_n$ is convergent if and only if ${\left(\underset{B\times A}{\int \int }n\left(n-1\right){\left(y-x\right)}^{n-2}dydx\right)}_n$ is convergent. If that is true, then

$$c=\underset{n}{lim}\underset{B\times A}{\int \int }n\left(n-1\right){\left(y-x\right)}^{n-2}dydx$$

Proof. 

This is obvious due to (11)–(13). □

Thus, the first question is the following: For which set A does the limit$lim{\int }_{A}n{y}^{n-1}dy$ exist?

If $H_n$ is the distribution of $U^{\ast }\left(n\right)$ defined by (14), the fact that $H_n\Rightarrow {\delta }_1$ does not help too much: it says that $H_n\left(A\right)\to {\delta }_1\left(A\right)$ if $1\notin Cl\left(A\right)\setminus Int\left(A\right)$. As $A\subset \left[0,1\right]$, this means that $1\in Int\left(B\right).$ This is not interesting: it means that for some $\epsilon >0$, the interval $\left[1-\epsilon ,1\right]$ is included in B or that $g_{|\left[1-\epsilon ,1\right]}=0.$ Remember that we used the notation $\mathit{f}\left(x\right)=\left(x,g\left(x\right)\right).$ Of course, $lim{\int }_{A}n{y}^{n-1}dy=0.$ In the same way, if $\left[1-\epsilon ,1\right]\subset A$ for some positive $\epsilon ,$$lim{\int }_{A}n{y}^{n-1}dy=1.$

These are not the interesting cases.

The question is if it is possible to find A such that the limit $lim{a}_n$ is in the open interval $\left(0,1\right)$ or not exist at all.

In this case, $x=1$ must be an accumulation point for the set $A.$

Let us fix these facts in some notation.

Definition 1. 

Let $A,A^{\prime }$$\in \mathcal{B}\left(\left[0,1\right]\right).$ We say that A and $A^{\prime }$ are max-similar if there exists $\epsilon >0$ such that $A\cap \left[1-\epsilon ,1\right]=A^{\prime }\cap \left[1-\epsilon ,1\right]$

Let $B,B^{\prime }$$\in \mathcal{B}\left(\left[0,1\right]\right).$ We say that B and $B^{\prime }$ are min-similar if there exists $\epsilon >0$ such that $B\cap \left[0,\epsilon \right]=B^{\prime }\cap \left[0,\epsilon \right]$

Let A be a Borel set from $[0,1]$, and let $A^c$ be its complementary. We put

$$\begin{array}{ccc}\hfill a_n\left(A\right)& =& P\left(U^{\ast }\left(n\right)\in A\right)+{\lambda }^n\left(A^c\right)\hfill \\ \hfill b_n\left(A\right)& =& P\left(U_{\ast }\left(n\right)\in A\right)+{\lambda }^n\left(A^c\right)\hfill \\ \hfill c_n\left(A\right)& =& P\left(U^{\ast }\left(n\right)\in A,U_{\ast }\left(n\right)\in A^c\right)+{\lambda }^n\left(A\right)+{\lambda }^n\left(A^c\right)\hfill \end{array}$$

(21)

and let $a\left(A\right)=lim{a}_n\left(A\right),b\left(A\right)=lim{b}_n\left(A\right),c\left(A\right)=lim{c}_n\left(A\right)$, provided that the limits exist. Here, $U^{\ast }\left(n\right)$ and $U_{\ast }\left(n\right)$ are defined by (14).

Remark 4. 

Do not confuse the general terms $a_n,b_n,c_n$ to be defined as the probability that the sample ${\mathcal{S}}_n$ has a leader or an antileader or that both are described, for instance, by relations (2)–(4) to the terms from (21). Only in the particular case that $\mathit{f}\left(x\right)=\left(x,1_A\left(x\right)\right)$ are they the same, as we have seen previously in (15)–(17).

Lemma 4. 

If A and $A^{\prime }$ are max-similar and $a\left(A\right)$ exists, then $a\left(A\right)=a\left(A^{\prime }\right)$.

If B and $B^{\prime }$ are min-similar and $b\left(B\right)$ exists, then $b\left(B\right)=b\left(B^{\prime }\right)$.

If A and $A^{\prime }$ are max-similar, B and $B^{\prime }$ are min-similar, and $c\left(A,B\right)$ exists; then, $c\left(A,B\right)=c\left(A^{\prime },B^{\prime }\right)$.

See the proof in Appendix A.

Here, $c_n\left(A,B\right)=P\left(U^{\ast }\left(n\right)\in A,U_{\ast }\left(n\right)\in B\right)+{\lambda }^n\left(A\right)+{\lambda }^n\left(B\right)$, and $c\left(A,B\right)=lim{c}_n\left(A,B\right)$ for some $A,B$ Borel sets from $[0,1]$.

Remark 5. 

For the limits, the only thing that matters is the behavior of A and B in the neighborhood of $x=1$ or $x=0$. That is why we can always replace A with $A\cap \left[1-\epsilon ,1\right].$

Moreover, if we are interested in $a=lim{a}_n$, we can suppose that $x=1$ is an accumulation point for A and that for no $\epsilon >0$ is it possible that $\left(1-\epsilon ,1\right)\subset A$. Otherwise, the limit a is zero or 1, which do not provide interesting cases.

The following auxiliary result points out that the study of b is the same as the study of a.

Lemma 5. 

Let $A\in \mathcal{B}\left(\left[0,1\right]\right)$. Let $1-A=\left\{1-x:x\in A\right\}$. Then, for any $n\ge 1$,

$$b_n\left(A\right)=a_n\left(1-A\right).$$

See the proof in Appendix A.

4. General Examples When $\mathit{a}$ and $\mathit{b}$ Exist

In Section 4 and Section 5, we shall give several examples of sets A for which the limits a and b both exist or do not exist. One can see how oscillations in the intervals that are contained in A near 1 prevent or facilitate the convergence of the sequences $\left(a_n\right)$ and $\left(b_n\right)$.

Let $H_n$ again be the distribution of $max\left(U_1,...,U_n\right).$ We know that $H_n\Rightarrow {\delta }_1.$ We will construct non-trivial Borel sets $A\subset [0,1]$ for which the sequence $H_n\left(A\right)$ has a limit different from ${\delta }_1\left(A\right).$ We call such a set a "good set", and if the limit does not exist, it is a "bad set".

This section is dedicated to the study of good sets, while Section 5 is addresses bad sets.

The simplest case of a non-trivial Borel set for which $x=1$ is an accumulation point is of the form $A={\displaystyle \bigcup _{n=1}^{\infty }}{I}_n$, with $I_n$ disjoint intervals from $[0,1]$ such that $A\cap [1-\epsilon ,1]\ne ⌀$ for every $\epsilon >0.$

In order to be able to compute its complement, we suppose that these intervals are well ordered, meaning that we can write $I_k=\left[{\alpha }_{2k-1},{\alpha }_{2k}\right]$, with $0<{\alpha }_1<....<{\alpha }_k<...<1$ and $a_k\uparrow 1.$

In that case, $B={\displaystyle \bigcup _{n=0}^{\infty }}\left({\alpha }_{2k},{\alpha }_{2k+1}\right),$${\alpha }_0=0.$ Since the Lebesgue measure neglects the countable sets, it does not matter if the intervals are closed or not.

We next present the main result of this section. We establish hypotheses that the sequence $\left({\alpha }_k\right)$ should satisfy for the limit a to exist.

Proposition 1. 

Let $A={\displaystyle \bigcup _{k=1}^{\infty }}\left[{\alpha }_{2k-1},{\alpha }_{2k}\right],B=A^c={\displaystyle \bigcup _{k=0}^{\infty }}\left({\alpha }_{2k},{\alpha }_{2k+1}\right)$, with $0={\alpha }_0<{\alpha }_1<....,{\alpha }_n\uparrow 1.$ Let

$$S_n={\displaystyle \sum _{k=1}^{\infty }}\left({\alpha }_{2k}^n-{\alpha }_{2k-1}^n\right)\mathrm{a}nd{T}_n={\displaystyle \sum _{k=0}^{\infty }}\left({\alpha }_{2k+1}^n-{\alpha }_{2k}^n\right).$$

Suppose that there exists $r>0$ such that ${\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \underset{n\ge 1}{sup}}\left|{\alpha }_{2k}^n-{\alpha }_{2k-1}^n-r\left({\alpha }_{2k-1}^n-{\alpha }_{2k-2}^n\right)\right|<\infty$

Then, we have the following:

(i) 

$S_n+T_n=1,$$\forall n\ge 1$;

(ii) 

$\underset{n}{lim}{S}_n=a={\displaystyle \frac{r}{1+r}}.$

Proof. 

The fact that $S_n+T_n=1$ is obvious.

Next, we know from Lemma 3 (i) that

$a={\displaystyle \underset{n\to \infty }{lim}}{\int }_{A}n{y}^{n-1}dy={\displaystyle \underset{n\to \infty }{lim}}{\displaystyle \sum _{n=1}^{\infty }}{\int }_{{\alpha }_{2k-1}}^{{\alpha }_{2k}}n{y}^{n-1}dy={\displaystyle \underset{n\to \infty }{lim}}{\displaystyle \sum _{n=1}^{\infty }}\left({\alpha }_{2k}^n-{\alpha }_{2k-1}^n\right)$ if the last limit exists.

We use a theorem of dominated convergence (see, for instance, [10]) for the counting measure: The sequence of functions $f_n:\mathbb{N}\to \mathbb{R}$ defined by $f_n\left(k\right)={\alpha }_{2k}^n-{\alpha }_{2k-1}^n-r\left({\alpha }_{2k-1}^n-{\alpha }_{2k-2}^n\right)$ converge to 0, and they are dominated by the integrable function $f\left(k\right)=\underset{n\ge 1}{sup}\left|{\alpha }_{2k}^n-{\alpha }_{2k-1}^n-r\left({\alpha }_{2k-1}^n-{\alpha }_{2k-2}^n\right)\right|.$

This means that ${\displaystyle \underset{n}{lim}}\int f_{n}d\mu =\int \underset{n}{lim}{f}_{n}d\mu =0.$ In our case, $\mu$ is a counting measure, meaning that $\int f_{n}d\mu ={\displaystyle \sum _{k=1}^{\infty }}{f}_n\left(k\right)$

It follows that ${\displaystyle \underset{n\to \infty }{lim}}{\displaystyle \sum _{k=1}^{\infty }}\left({\alpha }_{2k}^n-{\alpha }_{2k-1}^n-r\left({\alpha }_{2k-1}^n-{\alpha }_{2k-2}^n\right)\right)=0$

As $S_n-r{T}_n=S_n\left(1+r\right)-r,$, we write $lim\left(S_n-r{T}_n\right)=\left(1+r\right)lim{S}_n-r$ and apply $S_n-r{T}_n\to 0.$Therefore, $lim{S}_n={\displaystyle \frac{r}{1+r}}.$

□

In what follows, we present two examples of good sets A, and we compute the limit a for each of them.

The first example of a good set is $A={\displaystyle \bigcup _{n=1}^{\infty }}[e^{-{\displaystyle \frac{1}{n+r}}},e^{-{\displaystyle \frac{1}{n+1}}}]$, with $r\in \left(0,1\right).$ In order to find $limP\left(U^{\ast }\left(n\right)\in A\right)$, we prove

Proposition 2. 

Let $r\in \left(0,1\right),$$p>0$, and we have the series

$S_0\left(p\right)={\displaystyle \sum _{n=1}^{\infty }}\left(e^{-{\displaystyle \frac{p}{n+1}}}-e^{-{\displaystyle \frac{p}{n+r}}}\right)$

$S_1\left(p\right)={\displaystyle \sum _{n=1}^{\infty }}\left(e^{-{\displaystyle \frac{p}{n+r}}}-e^{-{\displaystyle \frac{p}{n}}}\right)$

Then, ${\displaystyle \underset{p\to \infty }{lim}}{S}_0\left(p\right)=1-r$, and ${\displaystyle \underset{p\to \infty }{lim}}{S}_1\left(p\right)=r.$

See the proof in Appendix A.

As a consequence, we have the following.

Corollary 1. 

Let ${\left(U_n\right)}_n$ be a sequence of i.i.d. random variables uniformly distributed on $[0,1]$, and let $U^{\ast }\left(n\right)=max\left(U_1,U_{2,}...,U_n\right)$ and $p\in \left[0,1\right]$ be arbitrary. Then, there exists a Borel set $A\subset \left[0,1\right]$ such that

$$\underset{n\to \infty }{lim}P\left(U^{\ast }\left(n\right)\in A\right)=p.$$

The second example of a good set is $A={\displaystyle \bigcup _{k=1}^{\infty }}\left(1-{\displaystyle \frac{1}{k}},1-{\displaystyle \frac{1}{k+r}}\right)$, with $r\in \left(0,1\right).$ Proposition 3 below proves that ${\displaystyle \underset{n\to \infty }{lim}}{a}_n=r$, with $a_n=P\left(U^{\ast }\left(n\right)\in A\right).$ But first, we need the following lemma described below.

Lemma 6. 

Let $r\in \left(0,1\right)$, and

$$f_n\left(k\right)=\left|{\left(1-{\displaystyle \frac{1}{k+r}}\right)}^n-{\left(1-{\displaystyle \frac{1}{k}}\right)}^n-{\displaystyle \frac{r}{1-r}}\left({\left(1-{\displaystyle \frac{1}{k+1}}\right)}^n-{\left(1-{\displaystyle \frac{1}{k+r}}\right)}^n\right)\right|$$

Then, $f_n\left(k\right)<{\displaystyle \frac{4}{k^2}}.$

See the proof in Appendix A.

On this basis, the statement runs as follows.

Proposition 3. 

Let ${\left(U_n\right)}_n$ be a sequence of i.i.d. random variables uniformly distributed on $\left[0,1\right].$ Let $U^{\ast }\left(n\right)=max\left(U_1,U_{2,}...,U_n\right),$$r\in \left(0,1\right),$$A={\displaystyle \bigcup _{k=1}^{\infty }}\left(1-{\displaystyle \frac{1}{k}},1-{\displaystyle \frac{1}{k+r}}\right)$ and $a_n=P\left(U^{\ast }\left(n\right)\in A\right).$ Then, ${\displaystyle \underset{n\to \infty }{lim}}{a}_n=r.$

See the proof in Appendix A.

5. Two General Examples When $\mathit{a}$ Does Not Exist

Let $H_n$ again be the distribution of $max\left(U_1,...,U_n\right).$ We know that $H_n\Rightarrow {\delta }_1.$ We will construct Borel sets $A\subset [0,1]$ for which the sequence $H_n\left(A\right)$ has no limit.

The first example of bad set is $A={\displaystyle \bigcup _{k\in \mathbb{Z}}}\left(e^{-q^k\left(1+\rho \right)},e^{-q^k}\right)$, with $q\in \left(0,1\right),$$0<\rho <{\displaystyle \frac{1}{q}}-1.$ In this case, $a_n\left(A\right)=P\left(U^{\ast }\left(n\right)\in A\right)={\displaystyle \sum _{k\in \mathbb{Z}}}\left(e^{-n{q}^k}-e^{-n\left(1+\rho \right)q^k}\right).$

We focused on finding conditions in which the series $a_n\left(A\right)$ is not convergent. The idea that the series is not convergent came from a simple remark: if we consider a change of variable, then it is clear that this is a periodic function. Therefore, it could only have a limit if it were constant. As we will see in Appendix A, the challenge is to find hypotheses under which its derivative is different from zero.

However, in general, we do not know whether the series has a limit or not. In what follows, let us present in short the idea of this proof.

Consider the function $s\left(x\right)={\displaystyle \sum _{k\in \mathbb{Z}}}\left(e^{-x{q}^k}-e^{-x\left(1+\rho \right)q^k}\right)$, with $x\in \left(0,\infty \right),$$q\in \left(0,1\right),$$0<\rho <{\displaystyle \frac{1}{q}}-1.$

We want to determine conditions under which ${\displaystyle \underset{x\nearrow \infty }{lim}}s\left(x\right)$ does not exist. This leads further to the fact that this set A is a bad set, since $a_n\left(A\right)=P\left(U^{\ast }\left(n\right)\in A\right)={\displaystyle \sum _{k\in \mathbb{Z}}}\left(e^{-n{q}^k}-e^{-n\left(1+\rho \right)q^k}\right)=s\left(n\right).$

If we denote $q=e^{-\alpha }$ and $x=e^{\alpha t}$, with $\alpha ,t>0$, then $0<\rho <e^{\alpha }-1$ and the previous series written as a function of t become

$$s\left(\alpha ,\rho ,t\right)=\sum _{k\in \mathbb{Z}}\left(e^{-e^{\alpha \left(t-k\right)}}-e^{-\left(1+\rho \right)e^{\alpha \left(t-k\right)}}\right)$$

(22)

Remark 6. 

Clearly, $s\left(\alpha ,\rho ,t+1\right)=s\left(\alpha ,\rho ,t\right)$ for all $\alpha ,t>0.$ Thus, the function $t\to s\left(\alpha ,\rho ,t\right)$ is periodic. Therefore, if ${\displaystyle \underset{t\to \infty }{lim}}s\left(\alpha ,\rho ,t\right)$ does exist, then the mapping $t\to s\left(\alpha ,\rho ,t\right)$ should be constant. The idea is to prove that $s_{\left|\left[0,1\right]\right|}$ is NOT constant.

Let us consider the series

$$h\left(\alpha ,\rho ,t\right)=\alpha \sum _{k\in \mathbb{Z}}\left((1+\rho )e^{\alpha \left(t-k\right)-\left(1+\rho \right)e^{\alpha \left(t-k\right)}}-e^{\alpha \left(t-k\right)-e^{\alpha \left(t-k\right)}}\right).$$

(23)

In order to prove the following Proposition 4, we first need the following.

Lemma 7. 

Let $\alpha ,\rho >0.$ Then, the function $t\to s\left(\alpha ,\rho ,t\right)$ is differentiable, and the following hold:

$${\displaystyle \frac{d}{dt}}\left(\sum _{k\in \mathbb{Z}}\left(e^{-e^{\alpha \left(t-k\right)}}-e^{-\left(1+\rho \right)e^{\alpha \left(t-k\right)}}\right)\right)=\alpha \sum _{k\in \mathbb{Z}}\left((1+\rho )e^{\alpha \left(t-k\right)-\left(1+\rho \right)e^{\alpha \left(t-k\right)}}-e^{\alpha \left(t-k\right)-e^{\alpha \left(t-k\right)}}\right)$$

Or, in other words,

$${\displaystyle \frac{d}{dt}}s\left(\alpha ,\rho ,t\right)=h\left(\alpha ,\rho ,t\right).$$

Therein, the series $s\left(\alpha ,\rho ,t\right)$ and $h\left(\alpha ,\rho ,t\right)$ are defined according to (22) and (23).

See the proof in Appendix A.

The next result will enable us to conclude that ${\displaystyle \underset{x\nearrow \infty }{lim}}s\left(x\right)$ does not exist, which means that the set $A={\displaystyle \bigcup _{k\in \mathbb{Z}}}\left(e^{-q^k\left(1+\rho \right)},e^{-q^k}\right)$ is, indeed, a bad set.

Proposition 4. 

Let $\alpha ,\rho >0$ such that

$$e^{\alpha }>1+{\displaystyle \frac{\rho e^{\rho +1}}{e^{\rho }-1-\rho }}$$

(24)

Then, the function $t⟼s\left(\alpha ,\rho ,t\right)={\displaystyle \sum _{k\in \mathbb{Z}}}\left(e^{-e^{\alpha \left(t-k\right)}}-e^{-\left(1+\rho \right)e^{\alpha \left(t-k\right)}}\right)$ is not constant. Moreover, in a neighborhood of $t=0$, it is decreasing.

See the proof in Appendix A.

The second example of a bad set is $A={\displaystyle \bigcup _{k=1}^{\infty }}\left(1-q^k,1-q^{k+r}\right),$ with $q,r\in \left(0,1\right).$ We prove this result in Proposition 5 below.

Definition 2. 

We say that the sets $A,B\subset \left[0,1\right]$ have the same nature if $lim\left(a_n\left(A\right)-a_n\left(B\right)\right)=0.$

Obviously, if A and B have the same nature, then either $lim{a}_n\left(A\right)$ and $lim{a}_n\left(B\right)$ do not exist, or both $lim{a}_n\left(A\right)$ and $lim{a}_n\left(B\right)$ exist, and moreover, $lim{a}_n\left(A\right)$=$lim{a}_n\left(B\right).$

More generally, on a measure space $\left(\mathsf{\Omega },K,P\right)$, two sequences of integrable measurable functions ${\left(u_n\right)}_n,{\left(t_n\right)}_n$ have the same nature if ${\displaystyle \underset{n\to \infty }{lim}}\left(u_n-t_n\right)=0$ and ${\displaystyle \underset{n\to \infty }{lim}}\int \left(u_n-t_n\right)d\mu =0.$

Remark 7. 

A simple criterion in order to decide if ${\left(u_n\right)}_n$ and ${\left(t_n\right)}_n$ have the same nature is to show that

$$\int sup\left|u_n-t_n\right|d\mu <\infty .$$

(25)

Proof. 

Apply again the domination principle: let $d_n=u_n-t_n;$ thus, ${\displaystyle \underset{n\to \infty }{lim}}{d}_n=0$, and $d_n\le {\displaystyle \underset{n}{sup}}\left|d_n\right|$. By commuting the integral with the limit, we get ${\displaystyle \underset{n\to \infty }{lim}}\int d_{n}d\mu =\int {\displaystyle \underset{n\to \infty }{lim}}{d}_{n}d\mu =0.$ □

Proposition 5. 

Let $q,r\in \left(0,1\right)$ and $A={\displaystyle \bigcup _{k=1}^{\infty }}\left(\left(1-q^k\right),\left(1-q^{k+r}\right)\right).$ Then, $lim{a}_n\left(A\right)$ does not exist.

See the proof in Appendix A.

6. Conclusions and Open Problems

In this paper, we studied the probability of finding extreme elements in random sets of the form $S_n=\{{\mathbf{Z}}_1,{\mathbf{Z}}_2,...,{\mathbf{Z}}_n\}$, where each ${\mathbf{Z}}_i$ is a random vector defined as ${\mathbf{Z}}_i=\mathit{f}\left(U_i\right)$, with $\mathit{f}:[0,1]\to {\mathbb{R}}^d,$$d\ge 2$, and $U_i$ is a random variable uniformly distributed on $[0,1]$. We were interested in finding the limits, if they exist, of the sequences $\left(a_n\right),\left(b_n\right),\left(c_n\right)$, where $a_n$ is the probability that $S_n$ contains a componentwise maximum, $b_n$ is the probability that $S_n$ contains a componentwise minimum, and $c_n$ is the probability that it contains both. In our previous papers, we provided general formulas for $a_n,b_n$ and $c_n$, and we have demonstrated the convergence of these sequences under a smoothness condition on $\mathit{f}$. In all the studied cases, we obtained the remarkable result that $lim{c}_n=\left(lim{a}_n\right)\left(lim{b}_n\right)$

In this paper, we explored what happens when $\mathit{f}$ lacks regularity and examined whether convergence of the sequences still holds. More exactly, we approached the particular cases $d=2$ and $\mathit{f}\left(x\right)=\left(x,1_A\left(x\right)\right)$ with an A Borel set from $[0,1].$

The surprise was that there are cases when the above limits do not exist at all. These cases occur when the map f is irregular.

We have considered the set $A={\displaystyle \bigcup _{k=1}^{\infty }}[{\alpha }_{2k-1},{\alpha }_{2k}]$, where ${\left({\alpha }_k\right)}_k$ is an increasing sequence in $[0,1]$ convergent to 1. If the set A has the property that a does exist, we say that A is a ”good” set; otherwise, we say that it is a ”bad” set. In Propositions 1–3, we gave examples of sets A that are good. A more challenging case turned out to be the one when the limit of the sequence $\left(a_n\right)$ does not exist. We do not know necessary conditions to decide that the limit of $\left(a_n\right)$ does not exist. However, in Proposition 4 and Proposition 5 are given two examples of ”bad” sets A. In order to prove these results, we needed an elaborate functional analysis calculus.

This study can be a helpful basis for future work to address some open questions:

1. If it is true that if the sequences $\left(a_n\right),\left(b_n\right)$ are convergent, then $\left(c_n\right)$ is convergent too?

2. If $\left(a_n\right),\left(b_n\right)$ are convergent, is it true or not that $lim{c}_n=\left(lim{a}_n\right)\left(lim{b}_n\right)$?

3. How can the function $\mathit{f}$ be characterized as having the property that all the sequences $\left(a_n\right),\left(b_n\right),\left(c_n\right)$ are convergent?

We were not able to answer these questions. Perhaps our results could be generalized to dimensions $d\ge 3,$ or one can analyze other functions besides the indicator of a set.