# Threshold for the expected measure of the convex hull of random points with
independent coordinates

Minas Pafis

###### Abstract

Let $\mu$ be an even Borel probability measure on ${\mathbb{R}}$. For every
$N>n$ consider $N$ independent random vectors $\vec{X}_{1},\ldots,\vec{X}_{N}$
in ${\mathbb{R}}^{n}$, with independent coordinates having distribution $\mu$.
We establish a sharp threshold for the product measure $\mu_{n}$ of the random
polytope
$K_{N}:=\operatorname{conv}\bigl{\\{}\vec{X}_{1},\ldots,\vec{X}_{N}\bigr{\\}}$
in ${\mathbb{R}}^{n}$ under the assumption that the Legendre transform
$\Lambda_{\mu}^{\ast}$ of the logarithmic moment generating function of $\mu$
satisfies the condition

$\lim\limits_{x\uparrow
x^{\ast}}\dfrac{-\ln\mu([x,\infty))}{\Lambda_{\mu}^{\ast}(x)}=1,$

where $x^{\ast}=\sup\left\\{x\in\mathbb{R}\colon\mu([x,\infty))>0\right\\}$.
An application is a sharp threshold for the case of the product measure
$\nu_{p}^{n}=\nu_{p}^{\otimes n}$, $p\geqslant 1$ with density
$(2\gamma_{p})^{-n}\exp(-\|x\|_{p}^{p})$, where $\|\cdot\|_{p}$ is the
$\ell_{p}^{n}$-norm and $\gamma_{p}=\Gamma(1+1/p)$.

##  1 Introduction

Let $\mu$ be an even Borel probability measure on the real line and consider a
random variable $X$, on some probability space $(\varOmega,\mathcal{F},P),$
with distribution $\mu$, i.e., $\mu(B):=P(X\in B)$ for all $B$ in the Borel
$\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of ${\mathbb{R}}$.

Let $X_{1},\ldots,X_{n}$ be independent and identically distributed random
variables, defined on the product space $(\varOmega^{n},\mathcal{F}^{\otimes
n},P^{n})$, each with distribution $\mu$. Set $\vec{X}=(X_{1},\ldots,X_{n})$
and, for a fixed $N$ satisfying $N>n$, consider $N$ independent copies
$\vec{X}_{1},\ldots,\vec{X}_{N}$ of $\vec{X}$, defined on the product space
$(\varOmega^{nN},\mathcal{F}^{\otimes nN},\operatorname{Prob})$. The
distribution of $\vec{X}$ is $\mu_{n}:=\mu\otimes\cdots\otimes\mu$ ($n$ times)
and the distribution of $(\vec{X}_{1},\ldots,\vec{X}_{N})$ is
$\mu_{n}^{N}:=\mu_{n}\otimes\cdots\otimes\mu_{n}$ ($N$ times). Our aim is to
obtain a sharp threshold for the expected $\mu_{n}$-measure of the random
polytope

$K_{N}:=\operatorname{conv}\bigl{\\{}\vec{X}_{1},\ldots,\vec{X}_{N}\bigr{\\}}.$

In order to make the notion of a sharp threshold precise, for any $n\geqslant
1$ and $\delta\in\left(0,\tfrac{1}{2}\right)$ we define the upper threshold

(1.1)
$\varrho_{1}(\mu_{n},\delta)=\sup\\{\varrho_{1}:\sup\\{{\mathbb{E}}_{\mu_{n}^{N}}[\mu_{n}(K_{N})]:N\leqslant\exp(\varrho_{1}n)\\}\leqslant\delta\\}$

and the lower threshold

(1.2)
$\varrho_{2}(\mu_{n},\delta)=\inf\\{\varrho_{2}:\inf\\{{\mathbb{E}}_{\mu_{n}^{N}}[\mu_{n}(K_{N})]:N\geqslant\exp(\varrho_{2}n)\\}\geqslant
1-\delta\\}.$

Then, we say that $\\{\mu_{n}\\}_{n=1}^{\infty}$ exhibits a sharp threshold if

$\varrho(\mu_{n},\delta):=\varrho_{2}(\mu_{n},\delta)-\varrho_{1}(\mu_{n},\delta)\longrightarrow
0$

as $n\to\infty$, for any fixed $\delta\in\left(0,\tfrac{1}{2}\right)$.

A threshold of this form was first established in the classical work of Dyer,
Füredi and McDiarmid [10] for the case of the uniform measure $\mu$ on
$[-1,1]$. We apply the general approach that was proposed in [5] and obtain an
affirmative answer for a general even probability measure $\mu$ on
${\mathbb{R}}$ that satisfies some additional assumptions, which we briefly
explain (see Section 2 for more details). Let

$x^{\ast}=x^{\ast}(\mu):=\sup\left\\{x\in\mathbb{R}\colon\mu([x,\infty))>0\right\\}$

be the right endpoint of the support of $\mu$ and set
$I=(-x^{\ast},x^{\ast})$. Let

$g(t):={\mathbb{E}}\bigl{(}e^{tX}\bigr{)}:=\int_{\mathbb{R}}e^{tx}\,d\mu(x),\qquad
t\in{\mathbb{R}}$

denote the moment generating function of $X,$ and let $\Lambda_{\mu}(t):=\ln
g(t)$ be its logarithmic moment generating function. By Hölder’s inequality,
$\Lambda_{\mu}$ is a convex function on ${\mathbb{R}}$. Consider the Legendre
transform $\Lambda_{\mu}^{\ast}:I\to{\mathbb{R}}$ of $\Lambda_{\mu}$; this is
the function

$\Lambda_{\mu}^{\ast}(x):=\sup\left\\{tx-\Lambda_{\mu}(t)\colon
t\in\mathbb{R}\right\\}.$

One can show that $\Lambda_{\mu}^{\ast}$ has finite moments of all orders.

We say that $\mu$ is admissible if it is non-degenerate, i.e. ${\rm
Var}(X)>0$, and satisfies the following conditions:

  1. (i)

There exists $r>0$ such that ${\mathbb{E}}\bigl{(}e^{tX}\bigr{)}<\infty$ for
all $t\in(-r,r)$; in particular, $X$ has finite moments of all orders.

  2. (ii)

If $x^{\ast}<\infty$ then $P(X=x^{\ast})=0$.

  3. (iii)

One of the following holds: (1) $x^{\ast}<+\infty$ or (2) $x^{\ast}=+\infty$
and $\\{\Lambda_{\mu}<\infty\\}={\mathbb{R}}$, or (3) $x^{\ast}=+\infty$,
$\\{\Lambda_{\mu}<\infty\\}$ is bounded and $\mu$ is log-concave.

Finally, we say that $\mu$ satisfies the $\Lambda^{\ast}$-condition if

$\lim\limits_{x\uparrow
x^{\ast}}\dfrac{-\ln\mu([x,\infty))}{\Lambda_{\mu}^{\ast}(x)}=1.$

With these definitions, our main result is the following.

######  Theorem 1.1.

Let $\mu$ be an admissible even probability measure on ${\mathbb{R}}$ that
satisfies the $\Lambda^{\ast}$-condition. Then, for any
$\delta\in\left(0,\tfrac{1}{2}\right)$ we have that

$\lim_{n\to\infty}\varrho(\mu_{n},\delta)=0.$

In Section 4 we give an application of Theorem 1.1 to the case of the product
$p$-measure $\nu_{p}^{n}:=\nu_{p}^{\otimes n}$. For any $p\geqslant 1$ we
denote by $\nu_{p}$ the probability distribution on ${\mathbb{R}}$ with
density $(2\gamma_{p})^{-1}\exp(-|x|^{p})$, where $\gamma_{p}=\Gamma(1+1/p)$.
We show that $\nu_{p}$ satisfies the $\Lambda^{\ast}$-condition.

######  Theorem 1.2.

For any $p\geqslant 1$ we have that

$\lim_{x\to\infty}\frac{-\ln(\nu_{p}[x,\infty))}{\Lambda_{\nu_{p}}^{\ast}(x)}=1.$

Note that the measure $\nu_{p}$ is admissible for all $1\leqslant p<\infty$;
it satisfies condition (iii-3) if $p=1$ and condition (iii-2) for all
$1<p<\infty$. Therefore, Theorem 1.2 implies that if $K_{N}$ is the convex
hull of $N>n$ independent random vectors $\vec{X}_{1},\ldots,\vec{X}_{N}$ with
distribution $\nu_{p}^{n}$ then the expected measure
${\mathbb{E}}_{(\nu_{p}^{n})^{N}}(\nu_{p}^{n}(K_{N}))$ exhibits a sharp
threshold at
$N=\exp((1\pm\varepsilon){\mathbb{E}}_{\nu_{p}}(\Lambda_{\nu_{p}}^{\ast})n)$;
for any $\delta\in\left(0,\tfrac{1}{2}\right)$ we have that
$\lim\limits_{n\to\infty}\varrho(\nu_{p}^{n},\delta)=0$.

We close this introductory section with a brief review of the history of the
problem that we study and related results. A variant of the question, in which
$\mu_{n}(K_{N})$ is replaced by the volume of $K_{N}$, has been studied in the
case where $\mu$ is compactly supported. Define

$\kappa=\kappa(\mu):=\frac{1}{2x^{\ast}}\int_{-x^{\ast}}^{x^{\ast}}\Lambda_{\mu}^{\ast}(x)dx.$

In [14] the following threshold for the expected volume of $K_{N}$ was
established for a large class of compactly supported distributions $\mu$: For
every $\varepsilon>0$,

(1.3)
$\lim_{n\rightarrow\infty}\sup\left\\{(2x^{\ast})^{-n}{\mathbb{E}}(\left\lvert
K_{N}\right\rvert)\colon N\leqslant\exp((\kappa-\varepsilon)n)\right\\}=0$

and

(1.4)
$\lim_{n\rightarrow\infty}\inf\left\\{(2x^{\ast})^{-n}{\mathbb{E}}(\left\lvert
K_{N}\right\rvert)\colon N\geqslant\exp((\kappa+\varepsilon)n)\right\\}=1.$

This result generalized the work of Dyer, Füredi and McDiarmid [10] who
studied the following two cases:

  1. (i)

If $\mu(\\{1\\})=\mu(\\{-1\\})=\tfrac{1}{2}$ then $\Lambda_{\mu}(t)=\ln(\cosh
t)$ and $\Lambda_{\mu}^{\ast}:(-1,1)\rightarrow{\mathbb{R}}$ is given by

$\Lambda_{\mu}^{\ast}(x)=\tfrac{1}{2}(1+x)\ln(1+x)+\tfrac{1}{2}(1-x)\ln(1-x),$

and the result holds with $\kappa=\ln 2-\tfrac{1}{2}$. This is the case of
$\pm 1$ polytopes.

  2. (ii)

If $\mu$ is the uniform distribution on $[-1,1]$, then
$\Lambda_{\mu}(t)=\ln(\sinh t/t)$, and the result holds with

$\kappa=\int_{0}^{\infty}\left(\frac{1}{u}-\frac{1}{e^{u}-1}\right)^{2}du.$

The generalization from [14] states that if $\mu$ is an even, compactly
supported, Borel probability measure on the real line and
$0<\kappa(\mu)<\infty$, then (1.3) holds for every $\varepsilon>0$, and (1.4)
holds for every $\varepsilon>0$ provided that the distribution $\mu$ satisfies
the $\Lambda^{\ast}$-condition.

Further sharp thresholds for the volume of various classes of random polytopes
appear in [20] and [2], [3] where the same question is addressed for a number
of cases where $\vec{X}_{i}$ have rotationally invariant densities.
Exponential in the dimension upper and lower thresholds are obtained in [12]
for the case where $\vec{X}_{i}$ are uniformly distributed in a simplex.
General upper and lower thresholds have been obtained by Chakraborti, Tkocz
and Vritsiou in [7] for some general families of distributions; see also [4].

##  2 Background and auxiliary results

As stated in the introduction, we consider an even Borel probability measure
$\mu$ on the real line and a random variable $X$, on some probability space
$(\varOmega,\mathcal{F},P),$ with distribution $\mu$. In order to avoid
trivialities we assume that ${\rm Var}_{\mu}(X)>0$, and in particular that
$p:=\max\\{P(X=x):x\in{\mathbb{R}}\\}<1$. Recall that $\mu$ is even if
$\mu(-B)=\mu(B)$ for every Borel subset $B$ of ${\mathbb{R}}$. The assumption
that $\mu$ is even merely serves to simplify exposition.

For the proof of our main result we have to make a number of additional
assumptions on $\mu$. The first one is that there exists $r>0$ such that

(2.1)
${\mathbb{E}}\bigl{(}e^{tX}\bigr{)}:=\int_{\mathbb{R}}e^{tx}\,d\mu(x)<\infty$

for all $t\in(-r,r)$. This assumption ensures that $X$ has finite moments of
all orders.

We define
$x^{\ast}:=\sup\left\\{x\in\mathbb{R}\colon\mu([x,\infty))>0\right\\}$ and
$I:=(-x^{\ast},x^{\ast})$. Note that we may have $x^{\ast}=\infty$. Our second
assumption is that if $x^{\ast}<\infty$ then

(2.2) $P(X=x^{\ast})=\mu(\\{x^{\ast}\\})=0.$

Let $g(t):={\mathbb{E}}\bigl{(}e^{tX}\bigr{)}$ for $t\in{\mathbb{R}}$ and
$\Lambda_{\mu}(t):=\ln g(t)$. One can easily check that $\Lambda_{\mu}$ is an
even convex function and $\Lambda_{\mu}(0)=0$, therefore, $\Lambda_{\mu}$ is a
non-negative function. The assumption (2.1) implies that the interval
$J:=\\{\Lambda_{\mu}<\infty\\}$ is a non-degenerate symmetric interval,
possibly the whole real line. We define $t^{\ast}=\sup J$. Then,
$\Lambda_{\mu}$ is $C^{\infty}$ and strictly convex on $(-t^{\ast},t^{\ast})$.

For every $t\in(-t^{\ast},t^{\ast})$ we define the probability measure $P_{t}$
on $(\varOmega,\mathcal{F})$ by

$P_{t}(A):={\mathbb{E}}\bigl{(}e^{tX-\Lambda_{\mu}(t)}\mathds{1}_{A}\bigr{)},\qquad
A\in\mathcal{F}.$

Define also $\mu_{t}(B):=P_{t}(X\in B)$ for any Borel subset $B$ of
$\mathbb{R}$. Then, $\mu_{t}$ has finite moments of all orders, and

(2.3)
${\mathbb{E}}_{t}(X)=\Lambda_{\mu}^{\prime}(t)\quad\hbox{and}\quad\operatorname{Var}_{t}(X)=\Lambda_{\mu}^{\prime\prime}(t),$

where ${\mathbb{E}}_{t}$ and $\operatorname{Var}_{t}$ denote expectation and
variance with respect to $P_{t}$. Notice that $P_{0}=P$ and $\mu_{0}=\mu$.
Since $X$ is non-degenerate with respect to $\mu_{t}$,
$t\in(-t^{\ast},t^{\ast})$, we have that $\Lambda_{\mu}^{\prime\prime}(t)>0$
for all $t\in(-t^{\ast},t^{\ast})$. It follows that $\Lambda_{\mu}^{\prime}$
is strictly increasing and since $\Lambda_{\mu}^{\prime}(0)=0$ we conclude
that $\Lambda_{\mu}$ is strictly increasing on $[0,t^{\ast})$.

Let $m:[0,x^{\ast})\to[0,\infty)$ be defined by

$m(x)=-\ln\mu([x,\infty)).$

It is clear that $m$ is non-decreasing. Observe that, from Markov’s
inequality, for any $x\in(0,x^{\ast})$ and any $t\geqslant 0$, we have
${\mathbb{E}}\bigl{(}e^{tX}\bigr{)}\geqslant e^{tx}\mu([x,\infty))$, and
hence,

(2.4) $\Lambda_{\mu}(t)\geqslant tx-m(x).$

An important case where (2.1) is satisfied is when $\mu$ is log-concave.
Recall that a Borel measure $\mu$ on $\mathbb{R}$ is called log-concave if
$\mu(\lambda A+(1-\lambda)B)\geqslant\mu(A)^{\lambda}\mu(B)^{1-\lambda}$ for
all compact subsets $A$ and $B$ of ${\mathbb{R}}$ and any $\lambda\in(0,1)$. A
function $f:\mathbb{R}\rightarrow[0,\infty)$ is called log-concave if its
support $\\{f>0\\}$ is an interval in ${\mathbb{R}}$ and the restriction of
$\ln{f}$ to it is concave. Any non-degenerate log-concave probability measure
$\mu$ on ${\mathbb{R}}$ has a log-concave density $f:=f_{{\mu}}$. Since $f$
has finite positive integral, one can check that there exist constants $A,B>0$
such that $f(x)\leqslant Ae^{-B|x|}$ for all $x\in{\mathbb{R}}$ (see [6, Lemma
2.2.1]). In particular, $f$ has finite moments of all orders. We refer to [6]
for more information on log-concave probability measures.

The next lemma describes the behavior of $\Lambda_{\mu}$ at the endpoints of
$J$ for a log-concave probability measure with unbounded support on
${\mathbb{R}}$ .

######  Lemma 2.1.

Let $\mu$ be an even log-concave probability measure on ${\mathbb{R}}$ with

$x^{\ast}=\sup\left\\{x\in\mathbb{R}\colon\mu([x,\infty))>0\right\\}=+\infty.$

If $J$ is a bounded interval, then $J=(-t^{\ast},t^{\ast})$ for some
$t^{\ast}>0$ and $\lim_{t\uparrow t^{\ast}}\Lambda_{\mu}(t)=+\infty$.

###### Proof.

Let $f$ denote the density of $\mu$. Then $f$ can be written as $f=e^{-q}$,
where $q:{\mathbb{R}}\to{\mathbb{R}}$ is an even convex function. By symmetry,
it is enough to consider the convergence of $\Lambda_{\mu}(t)$ for $t>0$.

Note that $\lim_{x\to+\infty}q(x)=+\infty$ and that the function
$u(x)=\frac{q(x)-q(0)}{x}$ is increasing on $(0,\infty)$. First we observe
that we cannot have $\lim_{x\to\infty}u(x)=\infty$. If this was the case then
we would have $\lim_{x\to\infty}\frac{q(x)}{x}=\infty$, and hence

$\int_{0}^{\infty}e^{tx}f(x)dx=\int_{0}^{\infty}e^{x\left(t-\frac{q(x)}{x}\right)}dx<\infty$

for all $t>0$, i.e. $\Lambda_{\mu}(t)<\infty$ for all $t>0$, which is not our
case.

Therefore, since $u$ is increasing, there exists $t^{\ast}>0$ such that

$\lim_{x\to\infty}u(x)=\lim_{x\to\infty}\frac{q(x)-q(0)}{x}=t^{\ast}.$

Assume that $0<t<t^{\ast}$. If $\varepsilon>0$ satisfies
$t+\varepsilon<t^{\ast}$ then there exists $M>0$ such that
$u(x)-t>\varepsilon$ for all $x\geqslant M$ and then

$\int_{0}^{\infty}e^{tx}f(x)dx=e^{-q(0)}\int_{0}^{\infty}e^{-x(u(x)-t)}dx<\infty,$

which shows that $t\in J$, and hence $(-t^{\ast},t^{\ast})\subseteq J$.

On the other hand, if $t=t^{\ast}$ then using the fact that $u(x)\leqslant
t^{\ast}$ for all $x>0$ we get

$\int_{0}^{\infty}e^{t^{\ast}x}f(x)dx=e^{-q(0)}\int_{0}^{\infty}e^{x(t^{\ast}-u(x))}dx=+\infty.$

This shows that $J=(-t^{\ast},t^{\ast})$.

Finally, if we consider a strictly increasing sequence $t_{n}\to t^{\ast}$
then by the monotone convergence theorem we get

$e^{\Lambda_{\mu}(t_{n})}=\int_{0}^{\infty}e^{t_{n}x}f(x)dx\longrightarrow\int_{0}^{\infty}e^{t^{\ast}x}f(x)dx=+\infty,$

which shows that $\lim_{t\uparrow t^{\ast}}\Lambda_{\mu}(t)=+\infty$. ∎

######  Definition 2.2.

Let $\mu$ be an even probability measure on ${\mathbb{R}}$. We will call $\mu$
admissible if it satisfies (2.1) and (2.2), as well as one of the following
conditions:

  1. (i)

$\mu$ is compactly supported, i.e. $x^{\ast}<+\infty$.

  2. (ii)

$x^{\ast}=+\infty$ and $\\{\Lambda_{\mu}<\infty\\}={\mathbb{R}}$.

  3. (iii)

$x^{\ast}=+\infty$, $\\{\Lambda_{\mu}<\infty\\}$ is bounded and $\mu$ is log-
concave.

Note that if $x^{\ast}<+\infty$ then
$\\{\Lambda_{\mu}<\infty\\}={\mathbb{R}}$. Taking also into account Lemma 2.1
we see that, in all the cases that we consider, the interval
$J=\\{\Lambda_{\mu}<\infty\\}$ is open, i.e. $J=(-t^{\ast},t^{\ast})$ where
$t^{\ast}=\sup\,J$.

The next lemma describes the behavior of $\Lambda_{\mu}^{\prime}$ for an
admissible measure $\mu$. The first case was treated in [14].

######  Lemma 2.3.

Let $\mu$ be an admissible even Borel probability measure on the real line.
Then, $\Lambda_{\mu}^{\prime}:J\to I$ is strictly increasing and surjective.
In particular,

$\lim_{t\to\pm t^{\ast}}\Lambda_{\mu}^{\prime}(t)=\pm x^{\ast}.$

###### Proof.

We have already explained that, since
$(\Lambda_{\mu}^{\prime})^{\prime}(t)=\Lambda_{\mu}^{\prime\prime}(t)=\operatorname{Var}_{t}(X)>0$,
the function $\Lambda_{\mu}^{\prime}$ is strictly increasing. Now, we consider
the three cases of the lemma separately.

(i) From the inequality $-x^{\ast}e^{tX}\leqslant Xe^{tX}\leqslant
x^{\ast}e^{tX},$ which holds with probability $1$ for each fixed $t,$ and the
formula
$\Lambda_{\mu}^{\prime}(t)={\mathbb{E}}\bigl{(}Xe^{tX}\bigr{)}/{\mathbb{E}}\bigl{(}e^{tX}\bigr{)}$,
we easily check that $\Lambda_{\mu}^{\prime}(t)\in(-x^{\ast},x^{\ast})$ for
every $t\in{\mathbb{R}}$.

It remains to show that $\Lambda_{\mu}^{\prime}$ is onto $I$. Let
$x\in(0,x^{\ast})$ and $y\in(x,x^{\ast})$. Since $\Lambda_{\mu}(t)\geqslant
ty-m(y)$ for all $t\geqslant 0$, we have that
$\Lambda_{\mu}\bigl{(}m(y)/(y-x)\bigr{)}\geqslant xm(y)/(y-x)$. It follows
that if we consider the function $q_{x}(t):=tx-\Lambda_{\mu}(t)$, then
$q_{x}(0)=0$ and $q_{x}\bigl{(}m(y)/(y-x)\bigr{)}\leqslant 0$. Since $q_{x}$
is concave and $q_{x}^{\prime}(0)=x>0$, this shows that $q_{x}$ attains its
maximum at some point in the open interval $\bigl{(}0,m(y)/(y-x)\bigr{)}$, and
hence, $\Lambda_{\mu}^{\prime}(t)=x$ for some $t$ in this interval. The same
argument applies for all $x\in(-x^{\ast},0)$. Finally, for $x=0$ we have that
$\Lambda_{\mu}^{\prime}(0)=x$.

(ii) We apply the same argument as in (i).

(iii) Assume that $\Lambda_{\mu}^{\prime}$ is bounded from above. Then, there
exists $x>0$ such that $\Lambda_{\mu}^{\prime}(t)<x$ for all $t\in J$. We
consider the function $q_{x}:J\to{\mathbb{R}}$ with
$q_{x}(t)=tx-\Lambda_{\mu}(t)$. Then, $q_{x}$ is strictly increasing. However,
$\lim_{t\uparrow t^{\ast}}q_{x}(t)=-\infty$ because $\lim_{t\uparrow
t^{\ast}}\Lambda_{\mu}(t)=+\infty$, which leads to a contradiction.∎

Let $\mu$ be an admissible even Borel probability measure on the real line.
Lemma 2.3 allows us to define $h\colon I\to J$ by
$h:=(\Lambda_{\mu}^{\prime})^{-1}$. Observe that $h$ is a strictly increasing
$C^{\infty}$ function and

(2.5) $h^{\prime}(x)=\frac{1}{\Lambda_{\mu}^{\prime\prime}(h(x))}.$

Next, consider the Legendre transform of $\Lambda_{\mu}$. This is the function

$\Lambda_{\mu}^{\ast}(x):=\sup\left\\{tx-\Lambda_{\mu}(t)\colon
t\in\mathbb{R}\right\\},\qquad x\in\mathbb{R}.$

In fact, since $tx-\Lambda_{\mu}(t)<0$ for $t<0$ when $x\in[0,x^{\ast})$, we
have that $\Lambda_{\mu}^{\ast}(x)=\sup\\{tx-\Lambda_{\mu}(t)\colon t\geqslant
0\\}$ in this case, and similarly
$\Lambda_{\mu}^{\ast}(x):=\sup\\{tx-\Lambda_{\mu}(t)\colon t\leqslant 0\\}$
when $x\in(-x^{\ast},0]$.

The basic properties of $\Lambda_{\mu}^{\ast}$ are described in the next lemma
(for a proof, see e.g. [13, Prposition 2.12]).

######  Lemma 2.4.

Let $\mu$ be an admissible even probability measure on ${\mathbb{R}}$. Then,

  1. (i)

$\Lambda_{\mu}^{\ast}\geqslant 0$, $\Lambda_{\mu}^{\ast}(0)=0$ and
$\Lambda_{\mu}^{\ast}(x)=\infty$ for
$x\in\mathbb{R}\setminus[-x^{\ast},x^{\ast}]$.

  2. (ii)

For every $x\in I$ we have $\Lambda_{\mu}^{\ast}(x)=tx-\Lambda_{\mu}(t)$ if
and only if $\Lambda_{\mu}^{\prime}(t)=x;$ hence

$\Lambda_{\mu}^{\ast}(x)=xh(x)-\Lambda_{\mu}(h(x))\qquad\text{for}\ x\in I.$

  3. (iii)

$\Lambda_{\mu}^{\ast}$ is a strictly convex $C^{\infty}$ function on $I,$ and

$(\Lambda_{\mu}^{\ast})^{\prime}(x)=h(x).$

  4. (iv)

$\Lambda_{\mu}^{\ast}$ attains its unique minimum on $I$ at $x=0$.

  5. (v)

$\Lambda_{\mu}^{\ast}(x)\leqslant m(x)$ for all $x\in[0,x^{\ast})$; this is a
direct consequence of (2.4).

######  Corollary 2.5.

We have that $\lim_{x\uparrow x^{\ast}}\Lambda_{\mu}^{\ast}(x)=+\infty$.

###### Proof.

If $x^{\ast}=+\infty$ then the convexity of $\Lambda_{\mu}^{\ast}$ and the
fact that $(\Lambda_{\mu}^{\ast})^{\prime}(x)>0$ for all $x>0$ imply that
$\lim_{x\uparrow x^{\ast}}\Lambda_{\mu}^{\ast}(x)=+\infty$.

Next, assume that $x^{\ast}<+\infty$. Since
$\Lambda_{\mu}^{\prime}(t)\leqslant x^{\ast}$ for all $t$, the function
$t\mapsto tx^{\ast}-\Lambda_{\mu}(t)$ is non-decreasing. Therefore,

$\Lambda_{\mu}^{\ast}(x^{\ast})=\sup\limits_{t\in{\mathbb{R}}}[tx^{\ast}-\Lambda_{\mu}(t)]=\lim\limits_{t\uparrow\infty}[tx^{\ast}-\Lambda_{\mu}(t)].$

However,

$\lim\limits_{t\uparrow\infty}e^{-(tx^{\ast}-\Lambda_{\mu}(t))}=\lim\limits_{t\uparrow\infty}e^{-tx^{\ast}}g(t)=\lim\limits_{t\uparrow\infty}{\mathbb{E}}\bigl{(}e^{t(X-x^{\ast})}\bigr{)}={\mathbb{E}}\\!\left(\lim\limits_{t\uparrow\infty}e^{t(X-x^{\ast})}\right)=P(X=x^{\ast}),$

the third equality being a consequence of the dominated convergence theorem.
It follows that $\Lambda_{\mu}^{\ast}(x^{\ast})=-\ln P(X=x^{\ast})=+\infty$.
Since $\Lambda_{\mu}^{\ast}$ is lower semi-continuous on ${\mathbb{R}}$ as the
pointwise supremum of the continuous functions $x\mapsto tx-\Lambda_{\mu}(t)$,
$t\in{\mathbb{R}}$, it follows that $\lim_{x\uparrow
x^{\ast}}\Lambda_{\mu}^{\ast}(x)=+\infty$.∎

The next result generalizes an observation from [5] which states that
$\Lambda_{\mu}^{\ast}$ has finite moments of all orders in the case where
$\mu$ is absolutely continuous with respect to Lebesgue measure. The more
general statement of the next proposition can be found as an exercise in [9].

######  Proposition 2.6.

Let $\mu$ be an even probability measure on ${\mathbb{R}}$. Then,

$\int_{I}e^{\Lambda_{\mu}^{\ast}(x)/2}d\mu(x)\leqslant 4.$

In particular, for all $p\geqslant 1$ we have that
$\int_{I}(\Lambda_{\mu}^{\ast}(x))^{p}\,d\mu(x)<+\infty$.

###### Sketch of the proof.

We define $F(x)=\mu((-\infty,x])$ and for any $z>0$ we set
$\alpha(x)=\mu((x,z])$ and $\beta(x)=\exp(I(x)/2)$ where $I(x)=0$ if
$x\leqslant 0$ and $I(x)=\Lambda_{\mu}^{\ast}(x)$ if $x>0$. Note that $\alpha$
is right continuous and decreasing, and $\beta$ is increasing. Applying [15,
Theorem 21.67 (iv)] for any $z>0$ we write

$\int_{0}^{z}\beta(x)d\alpha(x)+\int_{0}^{z}\alpha(x-)d\beta(x)=\alpha(z)e^{I(z+)/2}-\alpha(0-)e^{I(0-)/2}.$

It follows that

$\displaystyle\int_{0}^{z}e^{\Lambda_{\mu}^{\ast}(x)/2}d\mu(x)$
$\displaystyle=-\int_{0}^{z}\beta(x)d\alpha(x)=\int_{0}^{z}\alpha(x-)d\beta(x)-\alpha(z)e^{I(z+)/2}+\alpha(0-)e^{I(0-)/2}$
$\displaystyle\leqslant\int_{0}^{z}e^{-I(x)}d\beta(x)+1,$

where we have used the fact that $\alpha(x-)=\mu([x,z])\leqslant
e^{-\Lambda_{\mu}^{\ast}(x)}$ and $I(0-)=0$, $\alpha(0-)\leqslant 1$. Finally,
we note that

$\int_{0}^{z}e^{-I(x)}d\beta(x)+1=\int_{0}^{z}\beta(x)^{-2}d\beta(x)+1\leqslant\int_{1}^{\infty}t^{-2}dt+1=2,$

because $\beta$ is strictly increasing and continuous on $[0,z]$ and
$\beta(0)=1$. The result follows by symmetry.∎

######  Definition 2.7.

Let $\mu$ be an admissible even Borel probability measure on the real line.
Recall that $\Lambda_{\mu}^{\ast}(x)\leqslant m(x)$ for all
$x\in[0,x^{\ast})$. We shall say that $\mu$ satisfies the
$\Lambda^{\ast}$-condition if

$\lim\limits_{x\uparrow x^{\ast}}\dfrac{m(x)}{\Lambda_{\mu}^{\ast}(x)}=1.$

##  3 Proof of the main theorem

Let $\mu$ be an admissible even Borel probability measure on the real line.
Recall that $\mu_{n}=\mu\otimes\cdots\otimes\mu$ ($n$ times), and hence the
support of $\mu_{n}$ is $I_{\mu_{n}}=I^{n}$. The logarithmic Laplace transform
of $\mu_{n}$ is defined by

$\Lambda_{\mu_{n}}(\xi)=\ln\Big{(}\int_{{\mathbb{R}}^{n}}e^{\langle\xi,z\rangle}d\mu_{n}(z)\Big{)},\qquad\xi\in{\mathbb{R}}^{n}$

and the Cramer transform of $\mu_{n}$ is the Legendre transform of
$\Lambda_{\mu_{n}}$, defined by

$\Lambda_{\mu_{n}}^{\ast}(x)=\sup_{\xi\in{\mathbb{R}}^{n}}\left\\{\langle
x,\xi\rangle-\Lambda_{\mu_{n}}(\xi)\right\\},\qquad x\in{\mathbb{R}}^{n}.$

Since $\mu_{n}$ is a product measure, we can easily check that
$\Lambda_{\mu_{n}}^{\ast}(x)=\sum_{i=1}^{n}\Lambda_{\mu}^{\ast}(x_{i})$ for
all $x=(x_{1},\ldots,x_{n})\in I_{\mu_{n}}$, which implies that

$\int_{I_{\mu_{n}}}e^{\Lambda_{\mu_{n}}^{\ast}(x)/2}d\mu_{n}(x)=\prod_{i=1}^{n}\left(\int_{I}e^{\Lambda_{\mu}^{\ast}(x_{i})/2}d\mu(x_{i})\right)<+\infty.$

In particular, for all $p\geqslant 1$ we have that
$\int_{I_{\mu_{n}}}(\Lambda_{\mu_{n}}^{\ast}(x))^{p}\,d\mu_{n}(x)<+\infty$. We
also define the parameter

(3.1) $\beta(\mu_{n})=\frac{{\rm
Var}_{\mu_{n}}(\Lambda_{\mu_{n}}^{\ast})}{({\mathbb{E}}_{\mu_{n}}(\Lambda_{\mu_{n}}^{\ast}))^{2}}.$

Since $\mu_{n}=\mu\otimes\cdots\otimes\mu$, we have ${\rm
Var}_{\mu_{n}}(\Lambda_{\mu_{n}}^{\ast})=n{\rm
Var}_{\mu}(\Lambda_{\mu}^{\ast})$ and
${\mathbb{E}}_{\mu_{n}}(\Lambda_{\mu}^{\ast})=n{\mathbb{E}}_{\mu}(\Lambda_{\mu}^{\ast})$.
Therefore,

$\beta(\mu_{n})=\frac{{\rm
Var}_{\mu_{n}}(\Lambda_{\mu_{n}}^{\ast})}{({\mathbb{E}}_{\mu_{n}}(\Lambda_{\mu_{n}}^{\ast}))^{2}}=\frac{\beta(\mu)}{n},$

where $\beta(\mu)$ is a finite positive constant which is independent from
$n$. In particular, $\beta(\mu_{n})\to 0$ as $n\to\infty$.

In order to estimate $\varrho_{i}(\mu_{n},\delta)$, $i=1,2$, we shall follow
the approach of [5]. For every $r>0$ we define

$B_{r}(\mu_{n}):=\\{x\in{\mathbb{R}}^{n}:\Lambda^{\ast}_{\mu_{n}}(x)\leqslant
r\\}.$

Note that, since
$\Lambda_{\mu_{n}}^{\ast}(x)=\sum_{i=1}^{n}\Lambda_{\mu}^{\ast}(x_{i})$ for
all $x=(x_{1},\ldots,x_{n})$ and $\Lambda_{\mu}^{\ast}(y)$ increases to
$+\infty$ as $y\uparrow x^{\ast}$, for every $r>0$ there exists
$0<M_{r}<x^{\ast}$ such that
$B_{r}(\mu_{n})\subseteq[-M_{r},M_{r}]^{n}\subseteq I^{n}$, and hence
$B_{r}(\mu_{n})$ is a compact subset of $I^{n}$.

For any $x\in{\mathbb{R}}^{n}$ we denote by ${\cal H}(x)$ the set of all half-
spaces $H$ of ${\mathbb{R}}^{n}$ containing $x$. Then we define

$\varphi_{\mu_{n}}(x)=\inf\\{\mu_{n}(H):H\in{\cal H}(x)\\}.$

The function $\varphi_{\mu_{n}}$ is called Tukey’s half-space depth. We refer
the reader to the survey article of Nagy, Schütt and Werner [18] for a
comprehensive account and references. We start with the upper threshold. Note
that the $\Lambda^{\ast}$-condition is not required for this result.

######  Theorem 3.1.

Let $\mu$ be an even probability measure on ${\mathbb{R}}$. Then, for any
$\delta\in\left(0,\tfrac{1}{2}\right)$ there exist $c(\mu,\delta)>0$ and
$n_{0}(\mu,\delta)\in{\mathbb{N}}$ such that

$\varrho_{1}(\mu_{n},\delta)\geqslant\left(1-\frac{c(\mu,\delta)}{\sqrt{n}}\right)\mathbb{E}_{\mu}(\Lambda_{\mu}^{*})$

for all $n\geqslant n_{0}(\mu)$.

###### Proof.

The standard approach towards an upper threshold is based on the next fact
which holds true in general, for any Borel probability measure on
${\mathbb{R}}^{n}$. For every $r>0$ and every $N>n$ we have

(3.2)
${\mathbb{E}}_{\mu_{n}^{N}}(\mu_{n}(K_{N}))\leqslant\mu_{n}(B_{r}(\mu_{n}))+N\exp(-r).$

This estimate appeared originally in [10] and follows from the observation
that (by the definition of $\varphi_{\mu_{n}}$, Markov’s inequality and the
definition of $\Lambda_{\mu_{n}}^{\ast}$) for every $x\in{\mathbb{R}}^{n}$ we
have

(3.3) $\varphi_{\mu_{n}}(x)\leqslant\exp(-\Lambda_{\mu_{n}}^{\ast}(x)).$

We use (3.2) in the following way. Let
$T_{1}:={\mathbb{E}}_{\mu}(\Lambda_{\mu}^{\ast})$ and
$T_{n}:=\mathbb{E}_{\mu_{n}}(\Lambda_{\mu_{n}}^{*})=T_{1}n$. Then, for all
$\zeta\in(0,1)$, from Chebyshev’s inequality we have that

$\displaystyle\mu_{n}(\\{\Lambda_{\mu_{n}}^{*}\leqslant T_{n}-\zeta T_{n}\\})$
$\displaystyle\leqslant\mu_{n}(\\{|\Lambda_{\mu_{n}}^{*}-T_{n}|\geqslant\zeta
T_{n}\\})\leqslant\frac{\mathbb{E}_{\mu_{n}}|\Lambda_{\mu_{n}}^{*}-T_{n}|^{2}}{\zeta^{2}T_{n}^{2}}=\frac{\beta(\mu_{n})}{\zeta^{2}}=\frac{\beta(\mu)}{\zeta^{2}n}.$

Equivalently,

$\mu_{n}(B_{(1-\zeta)T_{n}}(\mu_{n}))\leqslant\frac{\beta(\mu)}{\zeta^{2}n}.$

Let $\delta\in\left(0,\tfrac{1}{2}\right)$. We may find $n_{0}(\mu,\delta)$
such that if $n\geqslant n_{0}(\mu,\delta)$ then $8\beta(\mu)/n<\delta<1/2$.
We choose $\zeta=\sqrt{2\beta(\mu)/n\delta}$, which implies that

$\mu(B_{(1-\zeta)T_{n}}(\mu_{n}))\leqslant\frac{\delta}{2}.$

From (3.2) we see that

$\displaystyle\sup\\{{\mathbb{E}}_{\mu_{n}^{N}}(\mu_{n}(K_{N})):N\leqslant
e^{(1-2\zeta)T_{n}}\\}$
$\displaystyle\leqslant\mu_{n}(B_{(1-\zeta)T_{n}}(\mu_{n}))+e^{(1-2\zeta)T_{n}}e^{-(1-\zeta)T_{n}}$
$\displaystyle\leqslant\frac{\delta}{2}+e^{-\zeta T_{n}}\leqslant\delta,$

provided that $\zeta T_{n}\geqslant\ln(2/\delta)$. Since $T_{n}=T_{1}n$, the
last condition takes the form $T_{1}n\geqslant c_{1}\ln(2/\delta)\sqrt{\delta
n/\beta(\mu)}$ and it is certainly satisfied if $n\geqslant n_{0}(\mu)$, where
$n_{0}(\mu)$ depends only on $\beta(\mu)$ because $\sqrt{\delta}\ln(2/\delta)$
is bounded on $\left(0,\tfrac{1}{2}\right)$. By the choice of $\zeta$ we
conclude that

$\varrho_{1}(\mu_{n},\delta)\geqslant\left(1-\sqrt{8\beta(\mu)/n\delta}\right)\mathbb{E}_{\mu}(\Lambda_{\mu}^{*})$

as claimed.∎

For the proof of the lower threshold we need a basic fact that plays a main
role in the proof of all the lower thresholds that have been obtained so far.
For a proof see [14, Lemma 4.1].

######  Lemma 3.2.

For every Borel subset $A$ of ${\mathbb{R}}^{n}$ we have that

$1-\mu_{n}^{N}(K_{N}\supseteq
A)\leqslant\binom{N}{n}p^{N-n}+2\binom{N}{n}\left(1-\inf_{x\in
A}\varphi_{\mu_{n}}(x)\right)^{N-n}.$

where $p=\max\\{P(X=x):x\in{\mathbb{R}}\\}<1$. Therefore,

(3.4)
${\mathbb{E}}_{\mu_{n}^{N}}\,[\mu_{n}(K_{N})]\geqslant\mu_{n}(A)\left(1-\binom{N}{n}p^{N-n}-2\binom{N}{n}\left(1-\inf_{x\in
A}\varphi_{\mu_{n}}(x)\right)^{N-n}\right).$

We are going to apply Lemma 3.2 with $A=B_{(1+\varepsilon)T_{n}}(\mu_{n})$,
using Chebyshev’s inequality exactly as in the proof of Theorem 3.1. From
(3.4) it is clear that we will also need a lower bound for $\inf_{x\in
B_{(1+\varepsilon)T_{n}}(\mu_{n})}\varphi_{\mu_{n}}(x)$ which will imply that

$2\binom{N}{n}\left(1-\inf_{x\in
B_{(1+\varepsilon)T_{n}}(\mu_{n})}\varphi_{\mu_{n}}(x)\right)^{N-n}=o_{n}(1).$

Therefore, the main technical step is to obtain the next inequality.

######  Theorem 3.3.

Let $\mu$ be an admissible even probability measure on ${\mathbb{R}}$ that
satisfies the $\Lambda^{\ast}$-condition, i.e.
$m(x)\sim\Lambda_{\mu}^{\ast}(x)$ as $x\uparrow x^{\ast}$. Then, for every
$\varepsilon>0,$ there exists $n_{0}(\mu,\varepsilon)\in\mathbb{N},$ depending
only on $\varepsilon$ and $\mu$, such that for all $r>0$ and all $n\geqslant
n_{0}(\mu,\varepsilon)$ we have that

$\inf_{x\in
B_{r}(\mu_{n})}\varphi_{\mu_{n}}(x)\geqslant\exp(-(1+\varepsilon)r-2\varepsilon
n).$

###### Proof.

Let $x\in B_{r}(\mu_{n})$ and $H_{1}$ be a closed half-space with
$x\in\partial{H_{1}}$. There exists $v\in{\mathbb{R}}^{n}\setminus\\{0\\}$
such that $H_{1}=\\{y\in{\mathbb{R}}^{n}:\langle v,y-x\rangle\geqslant 0\\}$.
Consider the function $q:B_{r}(\mu_{n})\to{\mathbb{R}}$, $q(w)=\langle
v,w\rangle$. Since $q$ is continuous and $B_{r}(\mu_{n})$ is compact, $q$
attains its maximum at some point $z\in B_{r}(\mu_{n})$. Define
$H=\\{y\in{\mathbb{R}}^{n}:\langle v,y-z\rangle\geqslant 0\\}$. Then,
$z\in\partial(H)$ and for every $w\in B_{r}(\mu_{n})$ we have $\langle
v,w\rangle\leqslant\langle v,z\rangle$, which shows that $\partial(H)$
supports $B_{r}(\mu_{n})$ at $z$ (see Schneider’s book [21] for basic facts
from the theory of convex bodies). Moreover, $H\subseteq H_{1}$ and hence
$P^{n}(\vec{X}\in H)\leqslant P^{n}(\vec{X}\in H_{1})$. This shows that
$\inf\\{\varphi_{\mu_{n}}(x):x\in B_{r}(\mu_{n})\\}$ is attained for some
closed half-space $H$ whose bounding hyperplane supports $B_{r}(\mu_{n})$.
Therefore, for the proof of the theorem it suffices to show that given
$\varepsilon>0$ we may find $n_{0}(\mu,\varepsilon)$ so that if $n\geqslant
n_{0}(\mu,\varepsilon)$ then

(3.5) $P^{n}\bigl{(}\vec{X}\in
H\bigr{)}\geqslant\exp(-(1+\varepsilon)r-2\varepsilon n)$

for any closed half-space $H$ whose bounding hyperplane supports
$B_{r}(\mu_{n})$.

Let $H$ be such a half-space. Then, there exists
$x\in\partial(B_{r}(\mu_{n}))$ such that

$P^{n}\bigl{(}\vec{X}\in
H\bigr{)}=P^{n}\\!\left(\sum\limits_{i=1}^{n}t_{i}(X_{i}-x_{i})\geqslant
0\right),$

where $t_{i}=h(x_{i})$, because the normal vector to $H$ is
$\nabla\Lambda_{\mu_{n}}^{\ast}(x)$ and $(\Lambda_{\mu}^{\ast})^{\prime}=h$ by
Lemma 2.4 (iii). We fix this $x$ for the rest of the proof. By symmetry and
independence we may assume that $x_{i}\geqslant 0$ for all $1\leqslant
i\leqslant n$. Recall that $\Lambda_{\mu}^{\ast}(0)=0$ and that $\mu$
satisfies the $\Lambda^{\ast}$-condition: we have
$m(x)\sim\Lambda_{\mu}^{\ast}(x)$ as $x\uparrow x^{\ast}$. Therefore, we can
find $M>\tau>0$ with the following properties:

  1. (i)

If $0\leqslant x\leqslant\tau$ then
$0\leqslant\Lambda_{\mu}^{\ast}(x)\leqslant\varepsilon$.

  2. (ii)

If $M<x<x^{\ast}$ then $P(X\geqslant
x)\geqslant\exp(-\Lambda_{\mu}^{\ast}(x)(1+\varepsilon))$.

Set $[n]=\\{1,\ldots,n\\}$. We consider the sets of indices

$\displaystyle A_{1}=A_{1}(x)$
$\displaystyle:=\left\\{i\in[n]:x_{i}<\tau\right\\}$ $\displaystyle
A_{2}=A_{2}(x)$ $\displaystyle:=\left\\{i\in[n]:\tau\leqslant x_{i}\leqslant
M\right\\},$ $\displaystyle A_{3}=A_{3}(x)$
$\displaystyle:=\left\\{i\in[n]:x_{i}>M\right\\}$

and the probabilities

$P_{j}=P_{j}(x):=P^{n}\\!\left(\sum\limits_{i\in
A_{j}}t_{i}(X_{i}-x_{i})\geqslant 0\right)\qquad j=1,2,3.$

By independence we have that

$P^{n}\bigl{(}\vec{X}\in
H\bigr{)}=P^{n}\\!\left(\sum\limits_{i=1}^{n}t_{i}(X_{i}-x_{i})\geqslant
0\right)\geqslant P_{1}P_{2}P_{3}.$

We will give lower bounds for $P_{1}$, $P_{2}$ and $P_{3}$ separately.

######  Lemma 3.4.

We have that

$P_{1}\geqslant\exp\\!\left(-\sum_{i\in
A_{1}}(\Lambda_{\mu}^{\ast}(x_{i})+\varepsilon)-c_{1}\ln\left\lvert
A_{1}\right\rvert-c_{2}\right),$

where $c_{1},c_{2}>0$ depend only on $\varepsilon$ and $\mu$.

###### Proof.

We write

(3.6) $P_{1}=P^{n}\\!\left(\sum\limits_{i\in A_{1}}t_{i}(X_{i}-x_{i})\geqslant
0\right)\geqslant P^{n}\\!\left(\sum_{i\in A_{1}}t_{i}(X_{i}-\tau)\geqslant
0\right),$

and use the following fact (see [14, Lemma 4.3]): For every
$\tau\in(0,x^{\ast}),$ there exists $c(\tau)>0$ depending only on $\tau$ and
$\mu,$ such that for any $k\in\mathbb{N}$ and any
$v_{1},\ldots,v_{k}\in\mathbb{R}$ with $\sum_{i=1}^{k}v_{i}>0$ we have that

$P^{k}\\!\left(\sum_{i=1}^{k}v_{i}(X_{i}-\tau)\geqslant 0\right)\geqslant
c(\tau)\,k^{-3/2}\,e^{-k\Lambda_{\mu}^{\ast}(\tau)}.$

Combining the above with (3.6) and using the simple bound
$\Lambda_{\mu}^{\ast}(\tau)\leqslant\varepsilon\leqslant\Lambda_{\mu}^{\ast}(x)+\varepsilon$
for $x$ in $[0,\tau]$, we conclude the proof of the lemma.∎

######  Lemma 3.5.

We have that

$P_{3}\geqslant\exp\\!\left(-(1+\varepsilon)\sum_{i\in
A_{3}}\Lambda_{\mu}^{\ast}(x_{i})\right).$

###### Proof.

By independence, we can write

$P_{3}=P^{n}\\!\left(\sum\limits_{i\in A_{3}}t_{i}(X_{i}-x_{i})\geqslant
0\right)\geqslant\prod_{i\in A_{3}}P(X_{i}\geqslant x_{i}).$

By the choice of $M$ we see that

$P(X_{i}\geqslant x_{i})\geqslant
e^{-\Lambda_{\mu}^{\ast}(x_{i})(1+\varepsilon)}$

for all $i\in A_{3}$, and this immediately gives the lemma.∎

######  Lemma 3.6.

There exist $c_{3},c_{4}>0$ depending only on $\varepsilon,M$ and $\mu,$ such
that

$P^{n}\\!\left(\sum\limits_{i\in A_{2}}t_{i}(X_{i}-x_{i})\geqslant
0\right)\geqslant\exp\\!\left(-\sum_{i\in
A_{2}}\Lambda_{\mu}^{\ast}(x_{i})-c_{3}\sqrt{|A_{2}|}-c_{4}\right).$

The proof of this estimate requires some preparation. Without loss of
generality, we may assume that $A_{2}=\left\\{1,\ldots,k\right\\}$ for some
$k\leqslant n$. Recall that $t_{i}=h(x_{i})$ for each $i$, and that this is
equivalent to having $x_{i}=\Lambda_{\mu}^{\prime}(t_{i})$ for each $i$.
Define the probability measure $P_{x_{1},\ldots,x_{k}}$ on
$(\varOmega^{k},\mathcal{F}^{\otimes k})$, by

$P_{x_{1},\ldots,x_{k}}(A):={\mathbb{E}}^{k}\\!\left[\mathds{1}_{A}\cdot\exp\\!\left(\sum\limits_{i=1}^{k}(t_{i}X_{i}-\Lambda_{\mu}(t_{i}))\right)\right]$

for $A\in\mathcal{F}^{\otimes k}$ (${\mathbb{E}}^{k}$ denotes expectation with
respect to the product measure $P^{k}$ on $\mathcal{F}^{\otimes k}$). Direct
computation shows that, under $P_{x_{1},\ldots,x_{k}},$ the random variables
$t_{1}X_{1},\ldots,t_{k}X_{k}$ are independent, with mean, variance and
absolute central third moment given by

$\displaystyle{\mathbb{E}}_{x_{1},\ldots,x_{k}}(t_{i}X_{i})$
$\displaystyle=t_{i}\Lambda_{\mu}^{\prime}(t_{i})=t_{i}x_{i},$
$\displaystyle{\mathbb{E}}_{x_{1},\ldots,x_{k}}\bigl{(}\left\lvert
t_{i}(X_{i}-x_{i})\right\rvert^{2}\bigr{)}$
$\displaystyle=t_{i}^{2}\Lambda_{\mu}^{\prime\prime}(t_{i}),$
$\displaystyle{\mathbb{E}}_{x_{1},\ldots,x_{k}}\bigl{(}\left\lvert
t_{i}(X_{i}-x_{i})\right\rvert^{3}\bigr{)}$ $\displaystyle=\left\lvert
t_{i}\right\rvert^{3}{\mathbb{E}}_{t_{i}}\bigl{(}\left\lvert
X-\Lambda_{\mu}^{\prime}(t_{i})\right\rvert^{3}\bigr{)},$

respectively. Set
$\sigma_{i}^{2}:=t_{i}^{2}\Lambda_{\mu}^{\prime\prime}(t_{i})$,

$s_{k}^{2}:=\sum\limits_{i=1}^{k}{\mathbb{E}}_{x_{1},\ldots,x_{k}}\bigl{(}\left\lvert
t_{i}(X_{i}-x_{i})\right\rvert^{2}\bigr{)}=\sum\limits_{i=1}^{k}t_{i}^{2}\Lambda_{\mu}^{\prime\prime}(t_{i})=\sum\limits_{i=1}^{k}\sigma_{i}^{2}$

and

$S_{k}:=\sum\limits_{i=1}^{k}t_{i}(X_{i}-x_{i}),$

and let $F_{k}\colon\mathbb{R}\rightarrow\mathbb{R}$ denote the cumulative
distribution function of the random variable $S_{k}/s_{k}$ under the
probability law $P_{x_{1},\ldots,x_{k}}$:
$F_{k}(x):=P_{x_{1},\ldots,x_{k}}(S_{k}\leqslant xs_{k})$ $(x\in\mathbb{R})$.
Write also $\nu_{k}$ for the probability measure on $\mathbb{R}$ defined by
$\nu_{k}(-\infty,x]:=F_{k}(x)$ $(x\in\mathbb{R})$. Notice that
${\mathbb{E}}_{x_{1},\ldots,x_{k}}(S_{k}/s_{k})=0$ and
$\operatorname{Var}_{x_{1},\ldots,x_{k}}(S_{k}/s_{k})=1$.

######  Lemma 3.7.

The following identity holds:

$P^{k}\\!\left(\sum\limits_{i=1}^{k}t_{i}(X_{i}-x_{i})\geqslant
0\right)=\left(\int_{[0,\infty)}e^{-s_{k}u}\,d\nu_{k}(u)\right)\,\exp\\!\left(-\sum_{i=1}^{k}\Lambda_{\mu}^{\ast}(x_{i})\right).$

###### Proof.

By definition of the measure $P_{x_{1},\ldots,x_{k}}$, we have that

$P^{k}\\!\left(\sum\limits_{i=1}^{k}t_{i}(X_{i}-x_{i})\geqslant
0\right)=P^{k}(S_{k}\geqslant
0)={\mathbb{E}}_{x_{1},\ldots,x_{k}}\\!\left[\mathds{1}_{[0,\infty)}(S_{k})\cdot\exp\\!\left(-\sum_{i=1}^{k}(t_{i}X_{i}-\Lambda_{\mu}(t_{i}))\right)\right].$

It follows that

$P^{k}\\!\left(\sum\limits_{i=1}^{k}t_{i}(X_{i}-x_{i})\geqslant
0\right)=\int_{[0,\infty)}e^{-s_{k}u}\,d\nu_{k}(u)\cdot\exp\\!\left(\sum_{i=1}^{k}(\Lambda_{\mu}(t_{i})-t_{i}x_{i})\right),$

and the lemma now follows from Lemma 2.4 (ii). ∎

We will also use the following consequence of the Berry-Esseen theorem (cf.
[11], p. 544).

######  Lemma 3.8.

For any $a,b>0,$ there exist $k_{0}\in{\mathbb{N}}$ and $\zeta>0$ with the
following property: If $k\geqslant k_{0},$ and if $Y_{1},\ldots,Y_{k}$ are
independent random variables with

$\mathbb{E}(Y_{i})=0,\quad\sigma_{i}^{2}:=\mathbb{E}(Y_{i}^{2})\geqslant
a,\quad\mathbb{E}(|Y_{i}|^{3})\leqslant b,$

then

$\mathbb{P}\left(0\leqslant\sum_{i=1}^{k}Y_{i}\leqslant\sigma\right)\geqslant\zeta,$

where $\sigma^{2}=\sigma_{1}^{2}+\cdots+\sigma_{k}^{2}$.

###### Proof of Lemma 3.6.

Consider the random variables $Y_{i}:=t_{i}(X_{i}-x_{i})$, $i\in
A_{2}=\\{1,\ldots,k\\}$, which are independent with respect to
$P_{x_{1},\ldots,x_{k}}$ and satisfy
${\mathbb{E}}_{x_{1},\ldots,x_{k}}(Y_{i})=0$ for all $1\leqslant i\leqslant
k$. Set $J^{\ast}=(\Lambda_{\mu}^{\prime})^{-1}([\tau,M])$. Since
$\tau\leqslant x_{i}\leqslant M$ for all $1\leqslant i\leqslant k$, we see
that

$\sigma_{i}^{2}={\mathbb{E}}_{x_{1},\ldots,x_{k}}\,(Y_{i}^{2})=t_{i}^{2}\Lambda_{\mu}^{\prime\prime}(t_{i})\geqslant\min_{t\in
J^{\ast}}t^{2}\Lambda_{\mu}^{\prime\prime}(t)=:a_{1}>0$

and

${\mathbb{E}}_{x_{1},\ldots,x_{k}}\,(|Y_{i}|^{3})=\left\lvert
t_{i}\right\rvert^{3}{\mathbb{E}}_{t_{i}}\bigl{(}\left\lvert
X-\Lambda_{\mu}^{\prime}(t_{i})\right\rvert^{3}\bigr{)}\leqslant\max_{t\in
J^{\ast}}\left\lvert t\right\rvert^{3}{\mathbb{E}}_{t}\bigl{(}\left\lvert
X-\Lambda_{\mu}^{\prime}(t)\right\rvert^{3}\bigr{)}=:b_{1}<+\infty$

for all $1\leqslant i\leqslant k$. Applying Lemma 3.8 we find $\zeta>0$ and
$k_{0}\in{\mathbb{N}}$ such that if $k\geqslant k_{0}$ then

(3.7)
$\mathbb{P}_{x_{1},\ldots,x_{k}}\left(0\leqslant\sum_{i=1}^{k}Y_{i}\leqslant
s_{k}\right)\geqslant\zeta.$

Now, we distinguish two cases:

Case 1: Assume that $k<k_{0}$. Then, working as for $A_{3}$, we see that

$P^{n}\\!\left(\sum\limits_{i\in A_{2}}t_{i}(X_{i}-x_{i})\geqslant
0\right)\geqslant\prod\limits_{i\in A_{2}}P(X_{i}\geqslant
x_{i})\geqslant\prod\limits_{i\in A_{2}}P(X_{i}\geqslant
M)=e^{-m(M)k}\geqslant e^{-m(M)k_{0}}.$

Case 2: Assume that $k\geqslant k_{0}$. From Lemma 3.7 we have

(3.8) $\displaystyle P^{n}\\!\left(\sum\limits_{i\in
A_{2}}t_{i}(X_{i}-x_{i})\geqslant 0\right)$
$\displaystyle=\left(\int_{[0,\infty)}e^{-s_{k}u}\,d\nu_{k}(u)\right)\,\exp\\!\left(-\sum_{i=1}^{k}\Lambda_{\mu}^{\ast}(x_{i})\right)$
$\displaystyle\geqslant e^{-s_{k}}\nu_{k}([0,1])\exp\\!\left(-\sum_{i\in
A_{2}}\Lambda_{\mu}^{\ast}(x_{i})\right).$

From (3.7) we see that

$\nu_{k}([0,1])=P_{x_{1},\ldots,x_{k}}(0\leqslant S_{k}\leqslant
s_{k})=\mathbb{P}\left(0\leqslant\sum_{i=1}^{k}Y_{i}\leqslant
s_{k}\right)\geqslant\zeta.$

Moreover, $s_{k}\leqslant c\sqrt{k}$. Combining the two cases we get the
estimate of Lemma 3.6 for $P_{2}$.∎

We can now complete the proof of Theorem 3.3. Collecting the estimates from
Lemma 3.4, Lemma 3.5 and Lemma 3.6, we may write

$\displaystyle P^{n}\\!\left(\sum\limits_{i=1}^{n}t_{i}(X_{i}-x_{i})\geqslant
0\right)$
$\displaystyle\geqslant\prod_{j=1}^{3}P^{n}\\!\left(\sum\limits_{i\in
A_{j}}t_{i}(X_{i}-x_{i})\geqslant 0\right)$
$\displaystyle\geqslant\exp\\!\left(-\sum_{i=1}^{n}\Lambda_{\mu}^{\ast}(x_{i})\right)$
$\displaystyle\quad\times\exp\\!\left(-\varepsilon\left\lvert
A_{1}\right\rvert-c_{1}\ln\left\lvert A_{1}\right\rvert-
c_{2}-\varepsilon\sum_{i\in
A_{3}}\Lambda_{\mu}^{\ast}(x_{i})-c_{3}\sqrt{\left\lvert
A_{2}\right\rvert}-c_{4}\right)$
$\displaystyle\geqslant\exp\\!\left(-\sum_{i=1}^{n}\Lambda_{\mu}^{\ast}(x_{i})-\varepsilon\sum_{i=1}^{n}\Lambda_{\mu}^{\ast}(x_{i})-2\varepsilon
n\right),$

provided $n\geqslant n(\mu,\varepsilon)$ for an appropriate
$n(\mu,\varepsilon)\in\mathbb{N}$ depending only on $\varepsilon$ and $\mu$.
This proves (3.5). ∎

We are now able to provide an upper bound for $\varrho_{2}(\mu_{n},\delta)$.

######  Theorem 3.9.

Let $\mu$ be an admissible even probability measure on ${\mathbb{R}}$ that
satisfies the $\Lambda^{\ast}$-condition, i.e.
$m(x)\sim\Lambda_{\mu}^{\ast}(x)$ as $x\uparrow x^{\ast}$. Then, for any
$\delta\in\left(0,\tfrac{1}{2}\right)$ and $\varepsilon\in(0,1)$ we can find
$n_{0}(\mu,\delta,\varepsilon)$ such that

$\varrho_{2}(\mu_{n},\delta)\leqslant\left(1+\varepsilon\right)\mathbb{E}_{\mu}(\Lambda_{\mu}^{*})$

for all $n\geqslant n_{0}(\mu,\delta,\varepsilon)$.

###### Proof.

Let $\varepsilon\in(0,1)$ and define $\zeta=T_{1}\varepsilon/(3T_{1}+4)$. Note
that if $T_{n}:=\mathbb{E}_{\mu_{n}}(\Lambda_{\mu_{n}}^{*})$ then, as in the
proof of Theorem 3.1, Chebyshev’s inequality implies that

$\mu_{n}(\\{\Lambda_{\mu_{n}}^{*}\geqslant T_{n}+\zeta
T_{n}\\})\leqslant\mu_{n}(\\{|\Lambda_{\mu_{n}}^{*}-T_{n}|\geqslant\zeta
T_{n}\\})\leqslant\frac{\beta(\mu)}{\zeta^{2}n}.$

Since $\zeta$ depends only on $\varepsilon$ and $\mu$ we can find
$n_{0}(\mu,\delta,\varepsilon)$ such that

$\frac{\beta(\mu)}{\zeta^{2}n}\leqslant\frac{\delta}{2}$

and hence

$\mu_{n}(B_{(1+\zeta)T_{n}}(\mu_{n}))\geqslant 1-\frac{\delta}{2}$

for all $n\geqslant n_{0}(\mu,\delta,\varepsilon)$. Assume that
$N\geqslant\exp((1+\varepsilon)T_{n})=\exp((1+3\zeta)T_{n}+4\zeta n)$.
Applying (3.4) with $A=B_{(1+\zeta)T_{n}}(\mu_{n})$ and using the estimate of
Theorem 3.3 we get

(3.9)
${\mathbb{E}}_{\mu_{n}^{N}}\,[\mu_{n}(K_{N})]\geqslant\mu_{n}(B_{(1+\zeta)T_{n}}(\mu_{n}))\left(1-\binom{N}{n}p^{N-n}-2\binom{N}{n}\left(1-\exp(-(1+\zeta)^{2}T_{n}-2\zeta
n)\right)^{N-n}\right).$

Therefore, taking into account the fact that $(1+\zeta)^{2}<1+3\zeta$ for
$\zeta<1$, we will have that

$\varrho_{2}(\mu_{n},\delta)\leqslant(1+\varepsilon)T_{1}$

if we check that

$\binom{N}{n}\,p^{N-n}+2\binom{N}{n}\left(1-\exp(-(1+3\zeta)T_{1}n-2\zeta
n)\right)^{N-n}\leqslant\frac{\delta}{2}.$

We first claim that there exists $n_{1}(\mu,\delta)$ such that

$\binom{N}{n}p^{N-n}<\frac{\delta}{4}$

for all $n\geqslant n_{1}(\mu,\delta)$. Indeed, since
$\binom{N}{n}\leqslant(eN/n)^{n}$, it suffices to check that

(3.10) $1+\ln\\!\left(\frac{N}{n}\right)+\frac{N-n}{n}\,\ln
p<\frac{1}{n}\ln(\delta/4).$

Set $x:=N/n$. Then, (3.10) is equivalent to

$(x-1)\ln(1/p)-\ln x>1+\frac{1}{n}\ln(4/\delta).$

The claim follows from the facts that the function on the left-hand side
increases to infinity as $x\to\infty$, and
$x=N/n\geqslant\exp((1+3\zeta)T_{1}n+4\zeta n)/n\geqslant e^{4\zeta
n}/n\to\infty$ when $n\to\infty$.

Next we check that there exists $n_{2}(\mu,\delta,\varepsilon)$ such that

$2\,\binom{N}{n}\bigl{[}1-\exp(-(1+3\zeta)T_{1}n-2\zeta
n)\bigr{]}^{N-n}<\frac{\delta}{4}$

for all $n\geqslant n_{2}(\mu,\delta,\varepsilon)$. Since $1-y\leqslant
e^{-y}$, it suffices to check that

(3.11)
$\left(\frac{2eN}{n}\right)^{n}\exp\\!\left(-(N-n)\exp(-(1+3\zeta)T_{1}n-2\zeta
n)\right)<\frac{\delta}{4}$

for all $n\geqslant n_{2}$. Setting $x:=N/n$, we see that this inequality is
equivalent to

$\exp((1+3\zeta)T_{1}n+2\zeta n)<\frac{x-1}{\ln
x+\ln(2e)+n^{-1}\ln(4/\delta)}.$

Since $N\geqslant\exp((1+3\zeta)T_{1}n+4\zeta n)$, we easily check that the
right-hand side exceeds $\exp((1+3\zeta)T_{1}n+3\zeta n)$ when $n\geqslant
n_{2}(\mu,\zeta,\delta)=n_{2}(\mu,\varepsilon,\delta)$, and hence we get
(3.11). Combining the above we conclude that

$\varrho_{2}(\mu_{n},\delta)\leqslant\left(1+\varepsilon\right)T_{1}$

for all $n\geqslant n_{0}$, where $n_{0}=n_{0}(\mu,\delta,\varepsilon)$
depends only on $\mu$, $\delta$ and $\varepsilon$. ∎

###### Proof of Theorem 1.1.

Let $\delta\in\left(0,\tfrac{1}{2}\right)$ and $\varepsilon\in(0,1)$. From the
estimates of Theorem 3.1 and Theorem 3.9 we see that there exists
$n_{0}(\mu,\delta,\varepsilon)$ such that if $n\geqslant n_{0}$ then
$\frac{c(\mu,\delta)}{\sqrt{n}}<\varepsilon$ (where $c(\mu,\delta)$ is the
constant in Theorem 3.1) and

$\varrho_{1}(\mu_{n},\delta)\geqslant\left(1-\frac{c(\mu,\delta)}{\sqrt{n}}\right)\mathbb{E}_{\mu}(\Lambda_{\mu}^{*})$

as well as

$\varrho_{2}(\mu_{n},\delta)\leqslant\left(1+\varepsilon\right)\mathbb{E}_{\mu}(\Lambda_{\mu}^{*}).$

Therefore,

$\varrho(\mu_{n},\delta)\leqslant
2\varepsilon\mathbb{E}_{\mu}(\Lambda_{\mu}^{*})$

for all $n\geqslant n_{0}$. Since $\varepsilon\in(0,1)$ was arbitrary, we see
that $\lim\limits_{n\to\infty}\varrho(\mu_{n},\delta)\to 0$, as claimed in
Theorem 1.1.∎

##  4 Threshold for the $p$-measures

We write $\nu$ for the symmetric exponential distribution on ${\mathbb{R}}$;
thus, $\nu$ is the probability measure with density $\frac{1}{2}\exp(-|x|)$.
More generally, for any $p\geqslant 1$ we denote by $\nu_{p}$ the probability
distribution on ${\mathbb{R}}$ with density
$(2\gamma_{p})^{-1}\exp(-|x|^{p})$, where $\gamma_{p}=\Gamma(1+1/p)$. Note
that $\nu_{1}=\nu$. The product measure $\nu_{p}^{n}=\nu_{p}^{\otimes n}$ has
density $(2\gamma_{p})^{-n}\exp(-\|x\|_{p}^{p})$, where
$\|x\|_{p}=\left(\sum_{i=1}^{n}|x_{i}|^{p}\right)^{1/p}$ is the
$\ell_{p}^{n}$-norm.

Let $g_{p}(x)=x^{2}$ for $0\leqslant x<1$ and $g_{p}(x)=x^{p}$ for $x\geqslant
1$. It is shown in [16] that for any $p\geqslant 1$ and $x\in{\mathbb{R}}$ one
has

$\Lambda_{\nu_{p}}^{\ast}(x/c)\leqslant
g_{p}(|x|)\leqslant\Lambda_{\nu_{p}}^{\ast}(cx)$

where $c>1$ is an absolute constant. Our aim in this section is to show that
$\nu_{p}$ satisfies the $\Lambda^{\ast}$-condition.

######  Theorem 4.1.

For any $p\geqslant 1$ we have that
$-\ln(\nu_{p}[x,\infty))\sim\Lambda_{\nu_{p}}^{\ast}(x)$ as $x\to\infty$. In
other words,

(4.1)
$\lim_{x\to+\infty}\frac{-\ln(\nu_{p}[x,\infty))}{\Lambda_{\nu_{p}}^{\ast}(x)}=1.$

###### Proof of the case $p=1$.

We start with the case $p=1$ which is simple because $\Lambda_{\nu}^{\ast}$
can be computed explicitly. A direct calculation shows that

$\Lambda_{\nu}^{\ast}(x)=\sqrt{1+x^{2}}-1-\ln\left(\frac{\sqrt{1+x^{2}}+1}{2}\right),\qquad
x\in{\mathbb{R}}.$

It follows that $\Lambda_{\nu}^{\ast}(x)\sim x$ as $x\to\infty$. On the other
hand, $\nu([x,\infty))=\frac{1}{2}e^{-x}$ for all $x>0$, which shows that
$-\ln(\nu([x,\infty))=x+\ln 2$, and hence $-\ln(\nu[x,\infty))\sim x$ as
$x\to\infty$. Combining the above we immediately see that (4.1) is satisfied
for $p=1$.∎

For the rest of this section we fix $p>1$. Following [1] we say that a non-
negative function $f:{\mathbb{R}}\to{\mathbb{R}}$ is regularly varying of
index $s\in{\mathbb{R}}$, and write $f\in R_{s}$, if
$\lim\limits_{x\to\infty}f(\lambda x)/f(x)=\lambda^{s}$ for every $\lambda>0$.
It is proved in [1, Theorem 4.12.10] that if $f\in R_{s}$ for some $s>0$ then

$-\ln\left(\int_{x}^{\infty}e^{-f(t)}dt\right)\sim f(x)$

as $x\to\infty$. Let $f_{p}(x)=|x|^{p}$, $x\geqslant 0$. It is clear that
$f_{p}\in R_{p}$, and hence

$-\ln(\nu_{p}[x,\infty))=-\ln\left((2\gamma_{p})^{-1}\int_{x}^{\infty}e^{-f_{p}(t)}dt\right)=\ln(2\gamma_{p})-\ln\left(\int_{x}^{\infty}e^{-f_{p}(t)}dt\right)\sim
f_{p}(x)$

as $x\to\infty$. This proves the following.

######  Lemma 4.2.

For every $p\geqslant 1$ we have that $-\ln(\nu_{p}[x,\infty))\sim x^{p}$ as
$x\to\infty$.

Lemma 4.2 shows that in order to complete the proof of the theorem we have to
show that $\Lambda_{\nu_{p}}^{\ast}(x)\sim x^{p}$ as $x\to\infty$. We shall
apply the Laplace method; more precisely, we shall use the next version of
Watson’s lemma (see [17, Section 2.2]).

######  Proposition 4.3.

Let $S<a<T\leqslant\infty$ and $g,h:[S,T]\to{\mathbb{R}}$, where $g$ is
continuous with a Taylor series in a neighborhood of $a$, and $h$ is twice
continuously differentiable and has its maximum at $a$ and satisfies
$h^{\prime}(a)=0$ and $h^{\prime\prime}(a)<0$. Assume also that the integral

$\int_{S}^{T}g(x)e^{th(x)}\,dx$

converges for large values of $t$. Then,

$\int_{S}^{T}g(x)e^{th(x)}\,dx\sim
g(a)\left(-\frac{2\pi}{th^{\prime\prime}(a)}\right)^{1/2}e^{th(a)}+e^{th(a)}O(t^{-1/2})$

as $t\to+\infty$.

We apply Proposition 4.3 to get the next asymptotic estimate.

######  Lemma 4.4.

Let $p>1$ and $q$ be the conjugate exponent of $p$. Then, setting $y=t^{q}$ we
have that

$I(t):=\int_{0}^{\infty}e^{tx-x^{p}}\,dx\sim
y^{\frac{1}{p}}e^{yh(a)}\left[\left(-\frac{2\pi}{yh^{\prime\prime}(a)}\right)^{1/2}+O(y^{-1/2})\right]$

as $t\to+\infty$, where $h(s)=s-s^{p}$ on $[0,\infty)$ and $a=p^{-q/p}$.

###### Proof.

We set $x=\lambda s$ and $t=\lambda^{p-1}$. Then,

$I(t)=I(\lambda^{p-1})=\lambda\int_{0}^{\infty}e^{\lambda^{p}(s-s^{p})}\,ds.$

Now, set $y=\lambda^{p}=t^{q}$. Then,

$I(t)=y^{1/p}\int_{0}^{\infty}e^{y(s-s^{p})}ds.$

We have $h^{\prime}(s)=1-ps^{p-1}$, therefore $h$ attains its maximum at
$a=(1/p)^{\frac{1}{p-1}}=p^{-q/p}$. Now, applying Proposition 4.3 with
$g\equiv 1$ we see that

$\int_{0}^{\infty}e^{yh(s)}ds\sim
e^{yh(a)}\left[\left(-\frac{2\pi}{yh^{\prime\prime}(a)}\right)^{1/2}+O(y^{-1/2})\right],$

and the lemma follows.∎

We proceed to study the asymptotic behavior of $\Lambda_{\nu_{p}}(t)$. Recall
that

$\Lambda_{\nu_{p}}(t)=\ln\left(c_{p}\int_{-\infty}^{\infty}e^{tx-|x|^{p}}\,dx\right),$

where $c_{p}=(2\Gamma(1+1/p))^{-1}$. By the dominated convergence theorem,

$\int_{-\infty}^{0}e^{tx-|x|^{p}}\,dx\longrightarrow 0$

as $t\to+\infty$. Therefore, from Lemma 4.4,

$c_{p}\int_{-\infty}^{\infty}e^{tx-|x|^{p}}\,dx\sim
c_{p}\int_{0}^{\infty}e^{tx-x^{p}}\,dx\sim
c_{p}y^{\frac{1}{p}}e^{yh(a)}\left[\left(-\frac{2\pi}{yh^{\prime\prime}(a)}\right)^{1/2}+O(y^{-1/2})\right],$

where $h(s)=s-s^{p}$ on $[0,\infty)$, $a=p^{-q/p}$ and $y=t^{q}$. Now,

$\displaystyle\ln\left(c_{p}y^{\frac{1}{p}}e^{yh(a)}\left[\left(-\frac{2\pi}{yh^{\prime\prime}(a)}\right)^{1/2}+O(y^{-1/2})\right]\right)$
$\displaystyle\hskip 56.9055pt=\ln c_{p}+\frac{1}{p}\ln y+yh(a)+O(\ln y)\sim
yh(a).$

It follows that $\Lambda_{\nu_{p}}(t)\sim yh(a)\sim(p^{-q/p}-p^{-q})t^{q}$,
where $q$ is the conjugate exponent of $p$. We rewrite this as follows.

######  Lemma 4.5.

Let $p>1$ and $q$ be the conjugate exponent of $p$. Then,

$\Lambda_{\nu_{p}}(t)\sim\frac{p-1}{p^{q}}t^{q}\quad\hbox{as}\;\;t\to+\infty.$

Lemma 4.5 allows us to determine the asymptotic behavior of
$\Lambda_{\nu_{p}}^{\ast}(x)$ as $x\to\infty$. We need a lemma which appears
in [8] and [19].

######  Lemma 4.6.

Let $q\geqslant 1$, $a>0$ and $f:[a,\infty)\to{\mathbb{R}}$ be a continuously
differentiable function such that $f^{\prime}$ is increasing on $[a,\infty)$
and $f(t)\sim t^{q}$ as $t\to+\infty$. Then, $f^{\prime}(t)\sim qt^{q-1}$ as
$t\to+\infty$.

###### Sketch of the proof.

Let $\varepsilon\in(0,1)$. There exists $b>a$ and
$\eta:[b,\infty)\to{\mathbb{R}}$ such that $|\eta(t)|\leqslant\varepsilon$ and
$f(t)=t^{q}(1+\eta(t))$ for all $t>b$. Since $f^{\prime}$ is increasing, for
any $s>0$ we have that

$\displaystyle sf^{\prime}(t)$
$\displaystyle\leqslant\int_{t}^{t+s}f^{\prime}(u)\,du=f(t+s)-f(t)=\big{(}(t+s)^{q}-t^{q}\big{)}+\big{(}(t+s)^{q}\eta(t+s)-t^{q}\eta(t)\big{)}$
$\displaystyle\leqslant sq(t+s)^{q-1}+2\varepsilon(t+s)^{q}.$

We set $s=\sqrt{\varepsilon}t$. Then,

$f^{\prime}(t)\leqslant
qt^{q-1}\big{(}(1+\sqrt{\varepsilon})^{q-1}+2q^{-1}\sqrt{\varepsilon}(1+\sqrt{\varepsilon})^{q}\big{)}$

for all $t>b$. In the same way we see that

$f^{\prime}(t)\geqslant
qt^{q-1}\big{(}(1-\sqrt{\varepsilon})^{q-1}-2q^{-1}\sqrt{\varepsilon}\big{)}$

for all $t>b/(1-\sqrt{\varepsilon})$, and the lemma follows.∎

###### Proof of the case $p>1$ in Theorem 4.1.

Now, we can show that

(4.2) $\Lambda_{\nu_{p}}^{\ast}(x)\sim x^{p}$

as $x\to\infty$. We know that
$\Lambda_{\nu_{p}}^{\ast}(x)=xh(x)-\Lambda_{\nu_{p}}(h(x))$ where
$h(x)=(\Lambda_{\nu_{p}}^{\prime})^{-1}(x)$. From Lemma 4.5 and Lemma 4.6 we
see that $\Lambda_{\nu_{p}}^{\prime}(t)\sim p^{-(q-1)}t^{q-1}$, and hence

$h(x)\sim px^{\frac{1}{q-1}}=px^{p-1},$

since $(p-1)(q-1)=1$. It follows that

$\frac{\Lambda_{\nu_{p}}^{\ast}(x)}{x^{p}}=\frac{h(x)}{x^{p-1}}-\frac{\Lambda_{\nu_{p}}(h(x))}{x^{p}}=\frac{h(x)}{x^{p-1}}-\frac{\Lambda_{\nu_{p}}(h(x))}{h(x)^{\frac{p}{p-1}}}\left(\frac{h(x)^{\frac{1}{p-1}}}{x}\right)^{p}\longrightarrow
p-\frac{p-1}{p^{q}}\cdot p^{q}=1$

as $x\to\infty$. This proves (4.2) and completes the proof of the theorem.∎

Acknowledgement. The author acknowledges support by the Hellenic Foundation
for Research and Innovation (H.F.R.I.) under the “Third Call for H.F.R.I. PhD
Fellowships” (Fellowship Number: 5779).

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Keywords: threshold, random polytopes, convex bodies, half-space depth, Cramer
transform.

2020 MSC: Primary 60D05; Secondary 60E15, 62H05, 52A22, 52A23.

Minas Pafis: Department of Mathematics, National and Kapodistrian University
of Athens, Panepistimioupolis 157-84, Athens, Greece.

E-mail<EMAIL_ADDRESS>