#  Optimal sub-Gaussian variance proxy
for truncated Gaussian and exponential random variables

Mathias Barreto111Higher School of Economics University, Moscow, Russia,
<EMAIL_ADDRESS>Olivier Marchal222Université Jean Monnet Saint-
Étienne, CNRS, Institut Camille Jordan UMR 5208, Institut Universitaire de
France, F-42023, Saint-Étienne, France<EMAIL_ADDRESS>and Julyan Arbel333Univ. Grenoble Alpes, Inria, CNRS, Grenoble INP, LJK, 38000
Grenoble, France<EMAIL_ADDRESS>
Abstract

This paper establishes the optimal sub-Gaussian variance proxy for truncated
Gaussian and truncated exponential random variables. The proofs rely on first
characterizing the optimal variance proxy as the unique solution to a set of
two equations and then observing that for these two truncated distributions,
one may find explicit solutions to this set of equations. Moreover, we
establish the conditions under which the optimal variance proxy coincides with
the variance, thereby characterizing the strict sub-Gaussianity of the
truncated random variables. Specifically, we demonstrate that truncated
Gaussian variables exhibit strict sub-Gaussian behavior if and only if they
are symmetric, meaning their truncation is symmetric with respect to the mean.
Conversely, truncated exponential variables are shown to never exhibit strict
sub-Gaussian properties. These findings contribute to the understanding of
these prevalent probability distributions in statistics and machine learning,
providing a valuable foundation for improved and optimal modeling and
decision-making processes.

##  1 Introduction

The sub-Gaussian property, a fundamental characteristic extensively explored
in seminal works such as Buldygin and Kozachenko, (1980); Boucheron et al.,
(2013); Pisier, (2016), has played a pivotal role in shaping the landscape of
probability distributions. This property, defined by the tail behavior of
random variables, has garnered significant attention for its implications in
various mathematical disciplines such as concentration inequalities and large
deviation estimates (Hoeffding,, 1963; Kearns and Saul,, 1998; Ledoux,, 1999;
Götze,, 1999; Raginsky and Sason,, 2013; Boucheron et al.,, 2013; Berend and
Kontorovich,, 2013; Perry et al.,, 2020; Ben-Hamou et al.,, 2017), random
series in relation to the geometry of Banach spaces (Pisier,, 1986; Chow,,
2013), spectral properties of random matrices (Litvak et al.,, 2005; Rudelson
and Vershynin,, 2009), or Bayesian statistics (Elder,, 2016; Catoni and
Giulini,, 2018; Vladimirova et al.,, 2019; Lee et al.,, 2021; Vladimirova et
al.,, 2021). More broadly, the sub-Gaussian property holds paramount
importance in machine learning and artificial intelligence applications
(Devroye et al.,, 2016; Cherapanamjeri et al.,, 2019; Genise et al.,, 2019;
Metelli et al.,, 2021; Depersin and Lecué,, 2022; Choi et al.,, 2023; Xie et
al.,, 2023; Michal,, 2023; Cole and Lu,, 2024).

The sub-Gaussianity of a random variable is a key determinant of its
concentration properties which can be defined as follows.

######  Definition 1.1.

A scalar random variable $X$ is called sub-Gaussian if there exists some
$s\geq 0$ such that for all $\theta\in\mathbb{R}$:

$\displaystyle\mathbb{E}[\exp(\theta(X-\mathbb{E}[X]))]\leq\exp\left(\frac{s^{2}\theta^{2}}{2}\right).$
(1)

Any $s^{2}$ satisfying Equation (1) is called a variance proxy, and the
smallest such $s^{2}$ is called the optimal variance proxy, which shall be
denoted as $\|X\|_{{\mathrm{vp}}}^{2}$. It is well known that

${\operatorname{Var}}[X]\leq\|X\|_{{\mathrm{vp}}}^{2},$

and a random variable satisfying
${\operatorname{Var}}[X]=\|X\|_{{\mathrm{vp}}}^{2}$ is called _strictly_ sub-
Gaussian.

Note that there exist many equivalent ways for defining this sub-Gaussian
property, each can be more useful depending on the sought application (see
Proposition 2.5.2 in Vershynin,, 2018). The one recalled in Equation (1) is
often referred to as the Laplace condition. It is equivalent to the following
condition on the tails of $X$:

$\exists\,C>0\,\,\text{such that}\,\,\forall\,t\geq 0,\,\,\mathrm{P}(|X|\geq
t)\leq 2e^{-\frac{t^{2}}{C^{2}}},$

as well as the following condition on the moments of $X$:

$\exists\,C>0\,\,\text{such that}\,\,\forall\,p\geq
1,\,\,\mathbb{E}[|X|^{p}]\leq C^{p}p^{\frac{p}{2}}.$

This shows that a sub-Gaussian random variable has finite moments of any order
larger than one.

Truncated random variables. The focus of this paper lies on the relevance of
the sub-Gaussian property to truncated Gaussian and truncated exponential
distributions. Truncated distributions emerge as natural models when data
exhibits inherent constraints or boundaries. These distributions find
pervasive use in statistics and machine learning, where understanding and
modeling the statistical properties of real-world phenomena are paramount.

Notable fields include survival analysis, for handling censoring (Balakrishnan
and Aggarwala,, 2000), reinforcement learning and bandit problems, for
representing action probabilities or rewards that are bounded within certain
limits (Bubeck et al.,, 2012; Lattimore and Szepesvári,, 2020; Szepesvári,,
2010), Bayesian statistics, to represent prior knowledge or beliefs about
parameters that are restricted to certain intervals (Gelman et al.,, 2013), or
sampling procedures, to reduce Monte Carlo error (Ionides,, 2008; Wawrzynski
and Pacut,, 2007). More broadly, handling truncation in data is useful in
countless applications, ranging from clinical trials, financial modeling, to
environmental modeling, underscoring the ubiquitous nature of truncated
distributions in contemporary research and applications.

Contributions. In general, establishing that a random variable is sub-Gaussian
might be easy, but complication stems from obtaining the optimal variance
proxy. This has been achieved for the most commonly used distributions,
including the beta and multinomial distributions (Marchal and Arbel,, 2017),
Bernoulli, binomial, Kumaraswamy, and triangular distributions (Arbel et al.,,
2020). In the present work, we focus on establishing the optimal variance
proxy for truncated Gaussian and truncated exponential distributions. The
proofs are based initially on defining the optimal variance proxy as the
unique solutions to a pair of equations. Subsequently, it is observed that
explicit solutions to this set of equations can be found for the two truncated
distributions in question.

Furthermore, we establish the conditions under which the optimal variance
proxy matches the variance, thereby identifying the strict sub-Gaussianity of
the truncated random variables. In detail, we illustrate that truncated
Gaussian variables display strict sub-Gaussian behavior if and only if they
exhibit symmetry, meaning their truncation is symmetric relative to the mean.
Conversely, truncated exponential variables are demonstrated to lack strict
sub-Gaussian properties consistently. These results enhance our comprehension
of these common probability distributions in statistics and machine learning,
laying a solid groundwork for enhanced and optimal modeling and decision-
making processes.

Outline. We present the optimal sub-Gaussian variance proxy and the strict
sub-Gaussianity results for truncated Gaussian and truncated exponential
random variables respectively in Section 2 and in Section 3, along with the
main proofs. Future research directions are proposed in Section 4. Technical
details on the proofs are deferred to Appendix A and Appendix B, respectively
for truncated Gaussian and truncated exponential random variables.

##  2 Truncated normal random variables

This section establishes the optimal variance proxy for a truncated normal
variable. Observe that we _a priori_ know already that any truncated random
variable is sub-Gaussian by Hoeffding’s Lemma (when it is truncated along a
finite interval).

In general, if $X$ is a random variable with density $f_{X}$ and cumulative
distribution function $F_{X}$ then its truncated version (which we shall
denote by $X_{\mathrm{T}}$) inside the interval $(a,b)$ (for $a<b$) has the
form:

$\displaystyle
f_{X_{\mathrm{T}}}(x)=\begin{cases}\frac{f_{X}(x)}{F_{X}(b)-F_{X}(a)}\quad&\text{for
$x\in(a,b)$},\\\ 0\quad&\text{otherwise.}\end{cases}$

Let $X\sim\mathcal{N}(\mu,\sigma^{2})$ and $\phi(\cdot),\Phi(\cdot)$ be the
density and cumulative density of a standard normal variable. If we truncate
$X$ on $(a,b)$ then it is known that its moment generating function, mean and
variance are given by:

$\begin{split}\mathbb{E}\left[e^{\theta
X_{\mathrm{T}}}\right]&=\exp\left\\{\theta\mu+\theta^{2}\frac{\sigma^{2}}{2}\right\\}\left[\frac{\Phi(\beta-\sigma\theta)-\Phi(\alpha-\sigma\theta)}{\Phi(\beta)-\Phi(\alpha)}\right],\\\
\mathbb{E}[X_{\mathrm{T}}]&=\mu+\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}\sigma,\\\
{\operatorname{Var}}[X_{\mathrm{T}}]&=\sigma^{2}\left[1-\frac{\beta\phi(\beta)-\alpha\phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)}-\left(\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}\right)^{2}\right],\end{split}$
(2)

where

$\alpha\coloneqq\frac{a-\mu}{\sigma}\,\,,\,\,\beta\coloneqq\frac{b-\mu}{\sigma}.$

Our main result establishes the optimal variance proxy for this density, which
turns out to have a closed-form expression:

######  Theorem 2.1.

Let $X_{\mathrm{T}}$ be a normal variable with mean $\mu\in\mathbb{R}$ and
variance $\sigma^{2}\in\mathbb{R}_{>0}$ truncated along the interval $(a,b)$
with $a<b$. Then its optimal variance proxy is given by:

$\displaystyle\|X_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}=\sigma^{2}\times\begin{cases}1-\frac{2\sigma}{b+a-2\mu}\frac{\phi(\frac{a-\mu}{\sigma})-\phi(\frac{b-\mu}{\sigma})}{\Phi(\frac{b-\mu}{\sigma})-\Phi(\frac{a-\mu}{\sigma})}\quad&\text{if}\,\,-\infty<a<b<+\infty\,\,\text{and}\,\,a+b\neq
2\mu,\\\
1-\frac{2(b-\mu)}{\sigma}\frac{\phi(\frac{b-\mu}{\sigma})}{2\Phi(\frac{b-\mu}{\sigma})-1}\quad&\text{if}\,\,-\infty<a<b<+\infty\,\,\text{and}\,\,a+b=2\mu,\\\
1&\text{if}\,\,a=-\infty\,\,\text{and/or}\,\,b=+\infty.\end{cases}$

In particular, $X_{\mathrm{T}}$ is _strictly_ sub-Gaussian if and only if
$a+b=2\mu$, i.e. if and only if the truncation is symmetric relative to the
mean $\mu$ of the Gaussian variable.

In order to prove Theorem 2.1, let us first note that we can reduce the
problem to the one of truncating a standard Gaussian random variable. Consider
the transformation $Y\coloneqq\frac{X-\mu}{\sigma}$ and let $Y_{\mathrm{T}}$
be the truncated standard normal along the interval $(\alpha,\beta)$. Then, we
have the relation:

$\displaystyle\mathbb{E}\big{[}e^{\theta\left(X_{\mathrm{T}}-\mathbb{E}[X_{\mathrm{T}}]\right)}\big{]}$
$\displaystyle=\int_{a}^{b}e^{\theta(x-\mathbb{E}[X_{\mathrm{T}}])}\frac{\phi(\frac{x-\mu}{\sigma})}{\Phi(\frac{b-\mu}{\sigma})-\Phi(\frac{a-\mu}{\sigma})}\frac{dx}{\sigma}$
$\displaystyle\overset{y=\frac{x-\mu}{\sigma}}{=}\int_{\alpha}^{\beta}\exp\left[\theta(y\sigma+\mu-\mathbb{E}[X_{\mathrm{T}}])\right]\frac{\phi(y)}{\Phi(\beta)-\Phi(\alpha)}dy$
$\displaystyle=\mathbb{E}\big{[}e^{\sigma\theta\left(Y_{\mathrm{T}}-\mathbb{E}[Y_{\mathrm{T}}]\right)}\big{]}.$
(3)

That is, to optimally bound the centered moment generating function of
$X_{\mathrm{T}}$, we can restrict to optimally bounding that of
$Y_{\mathrm{T}}$, as (3) implies
$\|X_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}=\sigma^{2}\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}$.
Hence, an equivalent reformulation of Theorem 2.1 is the following.

######  Theorem 2.2.

Let $Y_{\mathrm{T}}$ be a standard normal variable truncated in the interval
$(\alpha,\beta)$ with $\alpha<\beta$. Then its optimal variance proxy is given
by:

$\displaystyle\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}=\begin{cases}1-\frac{2\left(\phi(\alpha)-\phi(\beta)\right)}{(\alpha+\beta)\left(\Phi(\beta)-\Phi(\alpha)\right)}\quad&\text{if}\,\,-\infty<\alpha<\beta<+\infty\,\,\text{and}\,\,\alpha\neq-\beta,\\\
1-\frac{2\beta\phi(\beta)}{2\Phi(\beta)-1}\quad&\text{if}\,\,-\infty<\alpha<\beta<+\infty\,\,\text{and}\,\,\alpha=-\beta,\\\
1&\text{if}\,\,\alpha=-\infty\,\,\text{and/or}\,\,\beta=+\infty.\end{cases}$

Theorem 2.1 and Theorem 2.2 are illustrated in Figure 1. Note that only the
case $\beta=-\alpha$ (or, equivalently, $a+b=2\mu$), yields strictly sub-
Gaussian random variables. In this case, strict sub-Gaussianity is equivalent
to symmetry (with respect to the mode/mean of the original Gaussian
distribution). The relationship between symmetry and strict sub-Gaussianity is
studied in Arbel et al., (2020).

Figure 1: Illustration of Theorem 2.1 and Theorem 2.2. The top panel
represents the variance (black curve) and variance proxy (orange curve) of
truncated standard Gaussian variables on intervals $(\alpha,\beta)$ with a
fixed value of $\alpha=-2$ and varying values for
$\beta\in\\{-0.5,0,0.5,2\\}$, with colors from red to blue. The corresponding
distributions are depicted on the bottom panel. Note that only the case
$\beta=2$ yields a symmetric distribution, which turns out to be strictly sub-
Gaussian. See the blue dot on the right of the top panel where the variance
and variance proxy are equal.

###### Proof of Theorem 2.2.

Recall by Definition 1.1 that the optimal variance proxy of $Y_{\mathrm{T}}$
corresponds to the smallest possible $s^{2}\geq 0$ such that

$\ln\left[\exp\left(-\theta\left(\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}\right)+\frac{\theta^{2}}{2}\right)\left(\frac{\Phi(\beta-\theta)-\Phi(\alpha-\theta)}{\Phi(\beta)-\Phi(\alpha)}\right)\right]\leq\frac{s^{2}\theta^{2}}{2}\,\,,\,\,\forall\,\theta\in\mathbb{R}.$

By defining:

$\displaystyle
c_{\alpha,\beta}\coloneqq\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)},\quad\theta_{0}\coloneqq\frac{\alpha+\beta}{2}$
$\displaystyle
F_{\alpha,\beta}(\theta)\coloneqq\Phi(\beta-\theta)-\Phi(\alpha-\theta),\quad
f_{\alpha,\beta}(\theta)\coloneqq\ln\left[\frac{F_{\alpha,\beta}(\theta)}{\Phi(\beta)-\Phi(\alpha)}\right]$
$\displaystyle h_{\alpha,\beta}(\theta)\coloneqq
f_{\alpha,\beta}^{\prime}(\theta)=\frac{\phi(\alpha-\theta)-\phi(\beta-\theta)}{\Phi(\beta-\theta)-\Phi(\alpha-\theta)}$
$\displaystyle
h_{\alpha,\beta}^{\prime}(\theta)=\frac{\phi(\alpha-\theta)(\alpha-\theta)-\phi(\beta-\theta)(\beta-\theta)}{\Phi(\beta-\theta)-\Phi(\alpha-\theta)}-\left[\frac{\phi(\alpha-\theta)-\phi(\beta-\theta)}{\Phi(\beta-\theta)-\Phi(\alpha-\theta)}\right]^{2},$

this is equivalent to finding the smallest
$w\coloneqq\frac{s^{2}-1}{2}\geq-\frac{1}{2}$ such that:

$p_{\alpha,\beta;w}(\theta)\coloneqq w\theta^{2}+c_{\alpha,\beta}\theta\geq
f_{\alpha,\beta}(\theta)\,\,,\,\,\forall\,\theta\in\mathbb{R}.$ (4)

The delicate thing is understanding the right-hand side’s function behavior.
Note that this function is independent of the value of $w$. Let us first start
with a lemma regarding the symmetry of the polynomial $p_{\alpha,\beta;w}$ and
functions $f_{\alpha,\beta}$ and $h_{\alpha,\beta}$.

######  Lemma 2.3.

For all $\theta\in\mathbb{R}$ it holds that:

$\displaystyle
p_{\alpha,\beta;w}\left(-\frac{c_{\alpha,\beta}}{2w}+\theta\right)=p_{\alpha,\beta;w}\left(-\frac{c_{\alpha,\beta}}{2w}-\theta\right),\quad
f_{\alpha,\beta}\left(\theta_{0}-\theta\right)=f_{\alpha,\beta}\left(\theta_{0}+\theta\right),$
$\displaystyle
h_{\alpha,\beta}\left(\theta_{0}-\theta\right)=-h_{\alpha,\beta}\left(\theta_{0}+\theta\right).$

###### Proof of Lemma 2.3.

The proof is immediate by direct computations. ∎

In particular observe that $p_{\alpha,\beta:w}$ and $f_{\alpha,\beta}$ share
the symmetry line $\theta=\theta_{0}$ only when
$w=-\frac{c_{\alpha,\beta}}{2\theta_{0}}$ or when $c_{\alpha,\beta}=0$. A
second important point is to note that:

$f_{\alpha,\beta}(0)=0=p_{\alpha,\beta;w}(0)\,\,,\,\,f_{\alpha,\beta}(2\theta_{0})=0\,,\,f^{\prime}_{\alpha,\beta}(0)=p^{\prime}_{\alpha,\beta;w}(0).$

Then, the crucial technical point of the proof of the theorem is the following
lemma.

######  Lemma 2.4.

The function $h_{\alpha,\beta}$ is strictly convex on $(-\infty,\theta_{0})$
and strictly concave on $(\theta_{0},+\infty)$ for all $\alpha<\beta$.

###### Proof of Lemma 2.4.

The proof follows from direct and technical computations and is detailed in
Appendix A. ∎

Using Lemma 2.3 and Lemma 2.4, we obtain the optimal variance proxy for a
truncated standard normal random variable given in Theorem 2.2.

Case $\alpha$ and $\beta$ finite. Observe that $f_{\alpha,\beta}$ is strictly
increasing on $(-\infty,\theta_{0})$ and strictly decreasing on
$(\theta_{0},+\infty)$. Thus, it achieves its maximum at $\theta=\theta_{0}$.
Using the strict concavity of $h_{\alpha,\beta}$ inside the interval
$(\theta_{0},+\infty)$ we propose a geometric proof where the optimal variance
proxy case corresponds to fitting the parabola $\theta\mapsto
p_{\alpha,\beta;w}(\theta)$ inside the function $f_{\alpha,\beta}$ as
illustrated in Figure 2.

Figure 2: Illustration of the proof of Theorem 2.2 for $\alpha=-1$ and
$\beta=4$, such that $\theta_{0}=(\alpha+\beta)/2=3/2$. The function
$f_{\alpha,\beta}$ is plotted in dashed black line, while three polynomials
$p_{\alpha,\beta;w}$ are plotted in color: purple corresponds to the use of
the optimal variance proxy, where we can observe that $f_{\alpha,\beta}$ and
$p_{\alpha,\beta;w_{c}}$ are tangent at the purple dot with coordinates
$(\alpha+\beta,0)$. The red curve corresponds to a variance proxy that is not
optimal, while the value used for the blue curve is not a variance proxy.

We now need to split into three cases.

Case $\alpha$ and $\beta$ finite, $\alpha\neq-\beta$. Let us consider the
value
$w_{c}=\frac{\phi(\beta)-\phi(\alpha)}{(\alpha+\beta)(\Phi(\beta)-\Phi(\alpha))}=-\frac{c_{\alpha,\beta}}{2\theta_{0}}$
corresponding to a parabola that is tangent to $f_{\alpha,\beta}$ at
$(0,c_{\alpha,\beta})$ and $(2\theta_{0},-c_{\alpha,\beta})$ and with a
maximum at $\theta=\theta_{0}$. Indeed, we have:

$\begin{split}f_{\alpha,\beta}(0)&=0=p_{\alpha,\beta;w_{c}}(0),\\\
f_{\alpha,\beta}(2\theta_{0})&=0=4w_{c}\theta_{0}^{2}+c_{\alpha,\beta}\theta_{0}=p_{\alpha,\beta;w_{c}}(2\theta_{0}),\\\
f^{\prime}_{\alpha,\beta}(\theta_{0})&=0=2\theta_{0}w_{c}+c_{\alpha,\beta}=p^{\prime}_{\alpha,\beta;w_{c}}(\theta_{0}),\\\
f^{\prime}_{\alpha,\beta}(2\theta_{0})&=-c_{\alpha,\beta}=4\theta_{0}w_{c}+c_{\alpha,\beta}=p^{\prime}_{\alpha,\beta;w_{c}}(2\theta_{0}),\\\
f^{\prime}_{\alpha,\beta}(0)&=c_{\alpha,\beta}=p^{\prime}_{\alpha,\beta;w_{c}}(0).\end{split}$
(5)

Let us define $f_{\alpha,\beta;w}\coloneqq
f_{\alpha,\beta}-p_{\alpha,\beta;w}$ and $h_{\alpha,\beta;w}\coloneqq
h_{\alpha,\beta}-p^{\prime}_{\alpha,\beta;w}$ so that
$h^{\prime\prime}_{\alpha,\beta;w}=h^{\prime\prime}_{\alpha,\beta}$ because
$p_{\alpha,\beta;w}$ is a parabola. We will now prove that inequality (4)
holds for $w=w_{c}$, i.e., we will check that
$f_{\alpha,\beta;w_{c}}(\theta)\leq 0$ for all $\theta\in\mathbb{R}$. From
Lemma 2.4, $h^{\prime}_{\alpha,\beta;w}$ is strictly increasing on
$(-\infty,\theta_{0})$ and strictly decreasing on $(\theta_{0},+\infty)$ and
thus reaches its maximum at $\theta=\theta_{0}$. Thus,
$h^{\prime}_{\alpha,\beta;w}$ may have at most $2$ zeros and by Rolle’s
theorem, $h_{\alpha,\beta;w}$ may have at most $3$ zeros on $\mathbb{R}$. For
$w=w_{c}$, (5) implies that $h_{\alpha,\beta;w_{c}}$ has three distinct zeros
at $\theta\in\\{0,\theta_{0},2\theta_{0}\\}$. This implies that
$h^{\prime}_{\alpha,\beta;w_{c}}(\theta_{0})>0$ and that there exists
$\theta_{1}<\theta_{0}<\theta_{2}$ such that
$h^{\prime}_{\alpha,\beta;w_{c}}(\theta_{1})=h^{\prime}_{\alpha,\beta;w_{c}}(\theta_{2})=0$
and $h^{\prime}_{\alpha,\beta;w_{c}}$ is strictly positive on
$(\theta_{1},\theta_{2})$ and strictly negative on
$(-\infty,\theta_{1})\cup(\theta_{2},+\infty)$ because of Lemma 2.4. Let us
assume that $\theta_{0}>0$ (the case $\theta_{0}<0$ is similar but the set of
zeros $\\{0,\theta_{0},2\theta_{0}\\}$ is ordered in the opposite way) then
variations of $h_{\alpha,\beta;w_{c}}$ imply that
$0<\theta_{1}<\theta_{0}<\theta_{2}<2\theta_{0}$ and thus that
$h_{\alpha,\beta;w_{c}}$ is strictly positive on
$(-\infty,0)\cup(\theta_{0},2\theta_{0})$ and strictly negative on
$(0,\theta_{0})\cup(2\theta_{0},+\infty)$. Hence $f_{\alpha,\beta;w_{c}}$ is
strictly increasing on $(-\infty,0)\cup(\theta_{0},2\theta_{0})$ and strictly
decreasing on $(0,\theta_{0})\cup(2\theta_{0},+\infty)$. Since,
$f_{\alpha,\beta;w_{c}}(2\theta_{0})=f_{\alpha,\beta;w_{c}}(0)=0$, we conclude
that $f_{\alpha,\beta;w_{c}}$ is non-positive and hence that inequality (4)
holds for $w=w_{c}$.

Let us now prove that $w_{c}$ is optimal. Indeed, if we take by contradiction
that $w<w_{c}$, then $f_{\alpha,\beta}(2\theta_{0})=0$ while
$p_{\alpha,\beta;w}(2\theta_{0})=4w\theta_{0}^{2}+c_{\alpha,\beta}\theta_{0}<4w_{c}\theta_{0}^{2}+c_{\alpha,\beta}\theta_{0}=0$.
Therefore inequality (4) is not realized in a neighborhood of $2\theta_{0}$.
So $w_{c}$ is optimal.

Case $\alpha$ and $\beta$ finite, $\alpha=-\beta$. In this case, we have
$c_{-\beta,\beta}=0$ and $\theta_{0}=0=2\theta_{0}$. Moreover, the two curves
are always tangent at $\theta=0$ for any value of $w$:

$\displaystyle f_{-\beta,\beta}(0)=0=p_{-\beta,\beta;w}(0)\quad\text{and}\quad
f^{\prime}_{\alpha,\beta}(0)=0=p^{\prime}_{\alpha,\beta;w}(0).$ (6)

Consider $w_{c}=-\frac{\beta\phi(\beta)}{2\Phi(\beta)-1}$ and let us verify
that it is a variance proxy. In this case, the two curves are tangent at
$\theta=0$ but the second derivatives are also the same:

$f^{\prime\prime}_{\alpha,\beta}(0)=-\frac{2\beta\phi(\beta)}{2\Phi(\beta)-1}=2w_{c}=p^{\prime\prime}_{\alpha,\beta;w_{c}}(0).$
(7)

Let us define $f_{\alpha,\beta;w}\coloneqq
f_{\alpha,\beta}-p_{\alpha,\beta;w}$ and $h_{\alpha,\beta;w}\coloneqq
h_{\alpha,\beta}-p^{\prime}_{\alpha,\beta;w}$ as in the previous case. From
Lemma 2.4, $h^{\prime}_{\alpha,\beta;w_{c}}$ is strictly increasing on
$(-\infty,0)$ and strictly decreasing on $(0,+\infty)$. It achieves its
maximum at $\theta=0$ and $h^{\prime}_{\alpha,\beta;w_{c}}(0)=0$ from (7).
Thus $h^{\prime}_{\alpha,\beta;w_{c}}$ is strictly negative on
$\mathbb{R}\setminus\\{0\\}$ and $h_{\alpha,\beta;w_{c}}$ is a strictly
decreasing function vanishing at $\theta=0$. Hence $f_{\alpha,\beta;w_{c}}$ is
strictly increasing on $\mathbb{R}_{\leq 0}$ and strictly decreasing on
$\mathbb{R}_{\geq 0}$. Its maximum is thus achieved at $\theta=0$ and is null
from (6). Thus, it remains negative on $\mathbb{R}$ and we conclude that
inequality (4) holds for $w=w_{c}$.

Let us now prove that $w_{c}$ is optimal. Indeed, if we take by contradiction
that $w<w_{c}$, then the two curves are still tangent at $\theta=0$ from (6)
but the second derivatives satisfy
$f^{\prime\prime}_{\alpha,\beta}(0)=-\frac{2\beta\phi(\beta)}{2\Phi(\beta)-1}=2w_{c}>2w=p^{\prime\prime}_{\alpha,\beta;w}(0)$.
Hence the parabola $p_{\alpha,\beta;w}$ is locally below the function
$f_{\alpha,\beta}$ so that inequality (4) is not realized in a neighborhood of
$\theta=0$. So $w_{c}$ is optimal. In the end, using the fact that
$s^{2}=2w+1$, we obtain that $s_{c}^{2}\coloneqq 2w_{c}+1$ is the optimal
variance proxy.

Case $\alpha=-\infty$ and/or $\beta=+\infty$. Note first that the case when
$\beta=-\alpha=+\infty$ boils down to not truncating the original standard
Gaussian, which trivially implies strict sub-Gaussianity, while the case
$\alpha$ finite and $\beta=+\infty$ is equivalent to the $\alpha=-\infty$ and
$\beta$ finite by symmetry. Let us thus focus on the latter. Most of the
results proved for arbitrary finite values of $\alpha$ extend by taking the
limit $\alpha\to-\infty$. In particular, $h_{-\infty,\beta}$ is a strictly
concave function on $\mathbb{R}$. Moreover, one can take the limit
$\alpha\to-\infty$ in (4) since all quantities, including $w_{c}(\alpha)$, are
continuous functions of $\alpha$. This provides that $s=1$ is a variance proxy
(i.e. $w=0$) for the truncated Gaussian on $(-\infty,\beta)$. Let us now prove
that this value is optimal. We observe that for $w<0$ we have:

$f_{-\infty,\beta;w}(\theta)\coloneqq
f_{-\infty,\beta}(\theta)-p_{-\infty,\beta;w}(\theta)=\ln\left(\frac{\Phi(\beta-\theta)}{\Phi(\beta)}\right)-w\theta^{2}-c_{-\infty,\beta}\theta\overset{\theta\to-\infty}{\to}+\infty,$
(8)

so that (4) is obviously not verified in a neighborhood of $-\infty$. Hence
$s=1$ is the optimal variance proxy when $\alpha=-\infty$. Similarly, $s=1$ is
the optimal variance proxy for $\beta=+\infty$. This concludes the proof of
Theorem 2.2. ∎

##  3 Truncated exponential random variables

This section considers the truncated version of the classical exponential
distribution. Let $X\sim\text{Exp}(\lambda)$ be an exponential random variable
with rate $\lambda$ (thus with mean $1/\lambda$), and let $X_{\mathrm{T}}$
denote the truncation of $X$ along the interval $(a,b)$ with $0\leq
a<b\leq+\infty$ (recall that the untruncated version is not sub-Gaussian). Its
density, mean, and variance are of the form:

$\displaystyle f_{X_{\mathrm{T}}}(t)$ $\displaystyle=\lambda\frac{e^{-\lambda
t}}{e^{-\lambda a}-e^{-\lambda b}}\boldsymbol{1}_{(a,b)}(t),$
$\displaystyle\mathbb{E}[X_{\mathrm{T}}]$
$\displaystyle=\frac{1}{\lambda}+\frac{be^{\lambda a}-ae^{\lambda
b}}{e^{\lambda a}-e^{\lambda b}}=\frac{1}{\lambda}+\frac{ae^{-\lambda
a}-be^{-\lambda b}}{e^{-\lambda a}-e^{-\lambda b}},$
$\displaystyle{\operatorname{Var}}[X_{\mathrm{T}}]$
$\displaystyle=\frac{1}{\lambda^{2}}-\frac{(b-a)^{2}e^{\lambda(a+b)}}{\left(e^{\lambda
b}-e^{\lambda a}\right)^{2}}.$

It is easy to see that $b$ must be finite in order for $X_{\mathrm{T}}$ to be
sub-Gaussian. Indeed, we have:

$\mathbb{E}[\exp(\theta(X_{\mathrm{T}}-\mathbb{E}[X_{\mathrm{T}}]))]\big{|}_{\theta=\lambda}=\frac{\lambda(b-a)e^{-\lambda\mathbb{E}[X_{\mathrm{T}}]}}{e^{-\lambda
a}-e^{-\lambda
b}}\overset{\lambda\to+\infty}{\sim}e^{\lambda(a-\mathbb{E}[X_{\mathrm{T}}])}\lambda
b,$ (9)

so that the inequality required in Definition 1.1 at $\theta=\lambda$ cannot
be realized for any $s^{2}\geq 0$ when $b\to+\infty$.

######  Theorem 3.1.

Let $X_{\mathrm{T}}$ be an exponential variable with scale $\lambda$ truncated
along the interval $(a,b)$ with $0\leq a<b<\infty$. Then its optimal variance
proxy is given by:

$\|X_{\mathrm{T}}\|_{\mathrm{vp}}^{2}=\frac{(b-a)\left(e^{\lambda
b}+e^{\lambda a}\right)}{2\lambda\left(e^{\lambda b}-e^{\lambda
a}\right)}-\frac{1}{\lambda^{2}}.$

In particular, $X_{\mathrm{T}}$ is never strictly sub-Gaussian.

Similarly as in the Gaussian case, in order to prove Theorem 3.1 we restrict
ourselves to the case of a standard exponential random variable
$Y\sim\text{Exp}(1)$. Indeed, let $Y\sim\text{Exp}(1)$ and $Y_{\mathrm{T}}$ be
its truncation on $[\alpha,\beta]$ with $0\leq\alpha<\beta<+\infty$, then
defining $X=\frac{Y}{\lambda}$,
$(a,b)=\left(\frac{\alpha}{\lambda},\frac{\beta}{\lambda}\right)$ and
$X_{\mathrm{T}}$ the truncation of $X$ on $[a,b]$, we have that
$X_{\mathrm{T}}=\frac{Y_{\mathrm{T}}}{\lambda}$. Thus, we get:

$\displaystyle\mathbb{E}\left[\text{exp}\Big{(}\theta(X_{\mathrm{T}}-\mathbb{E}[X_{\mathrm{T}}]))\right]=\mathbb{E}\left[\text{exp}\left(\frac{\theta}{\lambda}(Y_{\mathrm{T}}-\mathbb{E}[Y_{\mathrm{T}}])\right)\right]\,\,,\,\,\forall\,\theta\in\mathbb{R},$

from where it follows that
$\lambda^{2}\|X_{\mathrm{T}}\|_{\mathrm{vp}}^{2}=\|Y_{\mathrm{T}}\|_{\mathrm{vp}}^{2}$.
Hence an equivalent formulation is:

######  Theorem 3.2.

Let $Y_{\mathrm{T}}$ be an exponential variable with mean $1$ truncated along
the interval $(\alpha,\beta)$ with $0\leq\alpha<\beta<+\infty$. Then its
variance proxy is given by:

$\|Y_{\mathrm{T}}\|_{\mathrm{vp}}^{2}=\frac{(\beta-\alpha)\left(e^{\alpha}+e^{\beta}\right)}{2\left(e^{\beta}-e^{\alpha}\right)}-1.$

In particular, $Y_{\mathrm{T}}$ is _never strictly_ sub-Gaussian.

As discussed above, the exponential random variable obtained with
$\beta=+\infty$ and any $\alpha$ is not sub-Gaussian. As a result, the
variance proxy given in Theorem 3.2 needs to compensate for this lack of sub-
Gaussianity by diverging to $+\infty$ as $\beta\to+\infty$. More specifically,
we have that $\|Y_{\mathrm{T}}\|_{\mathrm{vp}}^{2}$ is equivalent to $\beta/2$
as $\beta\to+\infty$, for fixed $\alpha$. Theorem 3.1 and Theorem 3.2 are
illustrated in Figure 3.

Figure 3: Illustration of Theorem 3.1 and Theorem 3.2. The top panel
represents the variance (black curve) and variance proxy (orange curve) of
truncated standard exponential random variables on intervals $(\alpha,\beta)$
with a fixed value of $\alpha=1/2$ and varying values for $\beta\in\\{2,4\\}$
(in violet and blue). The corresponding distributions are depicted on the
bottom panel. Note that truncated standard exponential random variables are
never strictly sub-Gaussian (the black and orange curves on the top panel
never coincide).

The proof of Theorem 3.2 relies on a direct study of the difference

$g_{\alpha,\beta,s}(\theta)\coloneqq\mathbb{E}[\exp(\theta(Y_{\mathrm{T}}-\mathbb{E}[Y_{\mathrm{T}}]))]-\exp\left(\frac{s^{2}\theta^{2}}{2}\right).$
(10)

The most important step is to verify that the optimal variance proxy is
characterized by the unique pair
$(\theta_{c},\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}})$ that solves the following
system of equations:

$g_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}}(\theta_{c})=0\,\,\text{
and
}g^{\prime}_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}}(\theta_{c})=0\,\text{
with }s_{\text{inf}}(\alpha,\beta)<\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}\leq
s_{1}(\alpha,\beta)\,\text{ and }\,\theta_{c}\neq 0,$ (11)

where:

$\displaystyle
g_{\alpha,\beta,s}(\theta)=\frac{e^{\frac{1}{2}s^{2}\theta^{2}}}{e^{\beta}-e^{\alpha}}G_{\alpha,\beta,s}(\theta),$
$\displaystyle
G_{\alpha,\beta,s}(\theta)=e^{\alpha}-e^{\beta}+\frac{1}{\theta-1}\left(e^{-\frac{1}{2}s^{2}\theta^{2}+(\beta-\mathbb{E}[Y_{\mathrm{T}}])\theta+\alpha}-e^{-\frac{1}{2}s^{2}\theta^{2}+(\alpha-\mathbb{E}[Y_{\mathrm{T}}])\theta+\beta}\right),$
$\displaystyle
s_{\text{inf}}(\alpha,\beta)\coloneqq\sqrt{1-\frac{(\beta-\alpha)^{2}e^{\alpha+\beta}}{\left(e^{\beta}-e^{\alpha}\right)^{2}}}\,,\,\,s_{1}(\alpha,\beta)\coloneqq\sqrt{\frac{(\beta-\alpha)^{2}(\beta-\mathbb{E}[Y_{\mathrm{T}}]+2)-\delta_{\alpha,\beta}}{(\beta-\alpha)^{2}+12}},$
$\displaystyle\delta_{\alpha,\beta}=2(\beta-\alpha)\left(-3\mathbb{E}[Y_{\mathrm{T}}]^{2}+(6\beta-(\beta-\alpha)^{2})\mathbb{E}[Y_{\mathrm{T}}]+\beta^{3}-2\alpha\beta^{2}+\alpha^{2}\beta+\alpha^{2}-2\alpha\beta-2\beta^{2}\right)^{\frac{1}{2}}.$
(12)

Furthermore, by uniqueness and observing that:

$\theta_{c}=2\,\,,\,\,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}=\frac{{(\beta-\alpha+2)e^{\alpha}+(\beta-\alpha-2)e^{\beta}}}{{2}(e^{\beta}-e^{\alpha})}$

is a solution of (11) we conclude the computation of
$\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}$ . Eventually, the statement that
$Y_{\mathrm{T}}$ is never strictly sub-Gaussian follows from a direct analysis
of the difference between $\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}$ and
${\operatorname{Var}}[Y_{\mathrm{T}}]$ that is detailed in Appendix B.1.

###### Proof of Theorem 3.2.

We will prove that $\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}$ is the optimal
variance proxy if and only if it is the solution to (11) and that such
solution is unique. First note that $g_{\alpha,\beta,s}$ and
$G_{\alpha,\beta,s}$ have the same sign and that they are smooth functions on
$\mathbb{R}$444Although the function $g_{\alpha,\beta,s}$ defined in Equation
(10) itself is undefined at $\theta=1$, setting it equal to $-e^{s^{2}/2}$
makes it smooth. In the proof of Theorem 3.2, we may have to avoid $\theta=1$
at some places. Still, it is obvious by continuity of $g_{\alpha,\beta,s}$
that the inequality of Definition 1.1 extends at $\theta=1$ if it is proved
valid in a neighborhood of this point.. Clearly $s^{2}$ is a variance proxy if
$G_{\alpha,\beta,s}(\theta)\leq 0$ for all $\theta\in\mathbb{R}$. In
particular, we have:

$\displaystyle
g_{\alpha,\beta,s}(0)=0\,\,,\,\,g_{\alpha,\beta,s}^{\prime}(0)=0\,\,,\,\,\forall
s>0,$ $\displaystyle
G_{\alpha,\beta,s}(0)=G^{\prime}_{\alpha,\beta,s}(0)=0\,\,,\,\,\underset{\theta\to\pm\infty}{\lim}G_{\alpha,\beta,s}(\theta)=e^{\alpha}-e^{\beta}<0$
(13)

Thus, another necessary condition for $s^{2}$ to be a variance proxy is that
$G^{\prime\prime}_{\alpha,\beta,s}(0)\leq 0$, i.e., we get the claimed lower
bound for the optimal variance proxy:

$\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}\geq
s_{\text{inf}}(\alpha,\beta)\coloneqq\sqrt{1-\frac{(\beta-\alpha)^{2}e^{\alpha+\beta}}{\left(e^{\beta}-e^{\alpha}\right)^{2}}}.$
(14)

In the following, we shall thus assume that $s\geq
s_{\text{inf}}(\alpha,\beta)$. A simple computation yields that
$G^{\prime}_{\alpha,\beta,s}$ is given by

$G^{\prime}_{\alpha,\beta,s}(\theta)=\frac{1}{(\theta-1)^{2}}e^{\alpha\theta+\beta}e^{-\frac{1}{2}s^{2}\theta^{2}-\mathbb{E}[Y_{\mathrm{T}}]\theta}h_{\alpha,\beta,s}(\theta),$
(15)

where:

$\displaystyle
h_{\alpha,\beta,s}(\theta)\coloneqq\left(-s^{2}\theta^{2}+(s^{2}+\beta-\mathbb{E}[Y_{\mathrm{T}}])\theta+\mathbb{E}[Y_{\mathrm{T}}]-\beta-1\right)e^{(\theta-1)(\beta-\alpha)}$
$\displaystyle+s^{2}\theta^{2}-(s^{2}+\alpha-\mathbb{E}[Y_{\mathrm{T}}])\theta-\mathbb{E}[Y_{\mathrm{T}}]+\alpha+1$
$\displaystyle
h^{\prime\prime\prime}_{\alpha,\beta,s}(\theta)=\frac{e^{-(\theta-1)(\beta-\alpha)}}{(\beta-\alpha)}P_{\alpha,\beta,s}(\theta),$

with:

$\displaystyle P_{\alpha,\beta,s}(\theta)\coloneqq
A_{\alpha,\beta,s}\theta^{2}+B_{\alpha,\beta,s}\theta+C_{\alpha,\beta,s},$
$\displaystyle A_{\alpha,\beta,s}\coloneqq-s^{2}(\beta-\alpha)^{2},$
$\displaystyle
B_{\alpha,\beta,s}\coloneqq(\alpha-\beta)(\alpha-\beta+6)s^{2}+(\beta-\alpha)^{2}(\beta-\mathbb{E}[Y_{\mathrm{T}}]),$
$\displaystyle C_{\alpha,\beta,s}\coloneqq
3(\beta-\alpha-2)s^{2}+(\beta-\alpha)\left((\beta-\alpha-3)\mathbb{E}[Y_{\mathrm{T}}]+\alpha\beta-\beta^{2}+\alpha+2\beta\right).$

The discriminant of the second-degree polynomial $P_{\alpha,\beta,s}$ is
crucial in establishing the previously mentioned characterization of the
optimal variance proxy, as shown in Lemma 3.3. It is given by:

$\displaystyle\Delta_{\alpha,\beta,s}\coloneqq$ $\displaystyle
B_{\alpha,\beta,s}^{2}-4A_{\alpha,\beta,s}C_{\alpha,\beta,s}$ $\displaystyle=$
$\displaystyle(\beta-\alpha)^{2}\left((\beta-\alpha)^{2}+12)s^{4}-2(\beta-\alpha)^{2}(\beta-\mathbb{E}[Y_{\mathrm{T}}]+2)s^{2}+(\beta-\alpha)^{2}(\beta-\mathbb{E}[Y_{\mathrm{T}}])^{2}\right).$

######  Lemma 3.3.

If $\Delta_{\alpha,\beta,s}\leq 0$, then $s^{2}$ is always a variance proxy
because $G_{\alpha,\beta,s}$ has a unique maximum at $\theta=0$ which is
vanishing. On the contrary, if $\Delta_{\alpha,\beta,s}>0$ then
$G_{\alpha,\beta,s}$ has exactly two local maximums. One at $\theta=0$ which
is vanishing and also another one denoted by $\theta_{c}(\alpha,\beta,s)$
which is nonzero.

###### Proof of Lemma 3.3.

For the proof we will need the following immediate results:

$\displaystyle\underset{\theta\to-\infty}{\lim}h^{\prime\prime}_{\alpha,\beta,s}(\theta)=2s^{2}>0,\quad\underset{\theta\to+\infty}{\lim}h^{\prime\prime}_{\alpha,\beta,s}(\theta)=-\infty,$
$\displaystyle\underset{\theta\to\pm\infty}{\lim}h^{\prime}_{\alpha,\beta,s}(\theta)=-\infty,\quad
h^{\prime}_{\alpha,\beta,s}(1)=0,$
$\displaystyle\underset{\theta\to-\infty}{\lim}h_{\alpha,\beta,s}(\theta)=+\infty,\quad
h_{\alpha,\beta,s}(1)=0,$
$\displaystyle\underset{\theta\to+\infty}{\lim}h_{\alpha,\beta,s}(\theta)=-\infty.$
(16)

Case $\Delta_{\alpha,\beta,s}\leq 0$. Let us assume that
$\Delta_{\alpha,\beta,s}\leq 0$. Since $A_{\alpha,\beta,s}<0$, we get that
$h^{\prime\prime\prime}_{\alpha,\beta,s}$ is a strictly negative function on
$\mathbb{R}$ except at one point where it vanishes when
$\Delta_{\alpha,\beta,s}=0$. This implies that
$h^{\prime\prime}_{\alpha,\beta,s}$ is a strictly decreasing function on
$\mathbb{R}$. From the first two equations of (16), we conclude that there
exists a unique value $\theta_{1}\in\mathbb{R}$ such that
$h^{\prime\prime}_{\alpha,\beta,s}(\theta_{1})=0$. Moreover,
$h^{\prime\prime}_{\alpha,\beta,s}$ is strictly positive on
$(-\infty,\theta_{1})$ and strictly negative on $(\theta_{1},+\infty)$. Thus,
$h^{\prime}_{\alpha,\beta,s}$ is a strictly increasing function on
$(-\infty,\theta_{1})$ and a strictly decreasing function on
$(\theta_{1},+\infty)$. Since we know that $h^{\prime}_{\alpha,\beta,s}(1)=0$
and
$\underset{\theta\to\pm\infty}{\lim}h^{\prime}_{\alpha,\beta,s}(\theta)=-\infty$,
we conclude that $h^{\prime}_{\alpha,\beta,s}$ may have at most two zeros. We
have two subcases:

  * •

$h^{\prime}_{\alpha,\beta,s}$ has a double zero at $\theta_{1}=1$. In this
case, from the limit at infinity given by (16), $h^{\prime}_{\alpha,\beta,s}$
is strictly negative on $\mathbb{R}\setminus\\{1\\}$ and thus
$h_{\alpha,\beta,s}$ is a strictly decreasing function on $\mathbb{R}$. From
(16), we have $h_{\alpha,\beta,s}(1)=0$ so that it implies that
$h_{\alpha,\beta,s}$ is strictly positive on $(-\infty,1)$ and strictly
negative on $(1,+\infty)$ and so is $G^{\prime}_{\alpha,\beta,s}$ from (15).
But this is a contradiction to the fact that
$G^{\prime}_{\alpha,\beta,s}(0)=0$ in (13) so that this subcase may be
discarded.

  * •

$h^{\prime}_{\alpha,\beta,s}$ has two distinct zeros, i.e. there exists a
unique $\theta_{2}\neq 1$ such that
$h^{\prime}_{\alpha,\beta,s}(\theta_{2})=0$. Let us denote
$\tilde{\theta}_{1}=\min(1,\theta_{1})$ and
$\tilde{\theta}_{2}=\max(1,\theta_{2})$. Then $h^{\prime}_{\alpha,\beta,s}$ is
strictly negative on
$(-\infty,\tilde{\theta}_{1})\cup(\tilde{\theta}_{2},+\infty)$ and strictly
positive on $(\tilde{\theta}_{1},\tilde{\theta}_{2})$. Thus,
$h_{\alpha,\beta,s}$ is strictly decreasing on $(-\infty,\tilde{\theta}_{1})$,
strictly increasing on $(\tilde{\theta}_{1},\tilde{\theta}_{2})$ and strictly
decreasing on $(\tilde{\theta}_{2},+\infty)$. But from (16) we have
$h_{\alpha,\beta,s}(1)=0$ so that $h_{\alpha,\beta,s}$ has a local extremum at
$\theta=1$ that is null. Hence $h_{\alpha,\beta,s}$ has constant sign locally
around $\theta=1$. Thus, there exists a unique value $\theta_{3}\neq 1$ such
that $h_{\alpha,\beta,s}(\theta_{3})=0$. Moreover, $h_{\alpha,\beta,s}$ is
strictly positive on $(-\infty,\theta_{3})$ and strictly negative on
$(\theta_{3},+\infty)$ and so is $G^{\prime}_{\alpha,\beta,s}$ from (15).
However, from (13) and the fact that $s\geq s_{\text{inf}}(\alpha,\beta)$, we
know that $\theta=0$ is a local maximum of $G_{\alpha,\beta,s}$ so that we
necessarily have $\theta_{3}=0$. Hence $G_{\alpha,\beta,s}$ achieves a unique
maximum at $\theta=0$ and this maximum is null from (13) so that we conclude
that $G_{\alpha,\beta,s}$ is negative on $\mathbb{R}$, i.e. that $s^{2}$ is a
variance proxy.

Case $\Delta_{\alpha,\beta,s}>0$. Let us assume that
$\Delta_{\alpha,\beta,s}>0$. We get that there exists two distinct values
$\theta_{0}\neq\theta_{1}$ such that
$h^{\prime\prime\prime}_{\alpha,\beta,s}(\theta_{0})=h^{\prime\prime\prime}_{\alpha,\beta,s}(\theta_{1})=0$.
Moreover, since $A_{\alpha,\beta,s}<0$, we get that
$h^{\prime\prime\prime}_{\alpha,\beta,s}$ is strictly negative on
$(-\infty,\theta_{1})\cup(\theta_{2},+\infty)$ and strictly positive on
$(\theta_{1},\theta_{2})$. Thus, $h^{\prime\prime}_{\alpha,\beta,s}$ is
strictly decreasing on $(-\infty,\theta_{1})$, strictly increasing on
$(\theta_{1},\theta_{2})$ and strictly decreasing on $(\theta_{2},+\infty)$.
Since
$\underset{\theta\to-\infty}{\lim}h^{\prime\prime}_{\alpha,\beta,s}(\theta)>0$
and
$\underset{\theta\to+\infty}{\lim}h^{\prime\prime}_{\alpha,\beta,s}(\theta)=-\infty$
we conclude that $h^{\prime\prime}_{\alpha,\beta,s}$ may have one or two (one
simple and one double zero) or three distinct zeros depending on the sign of
$h^{\prime}{\alpha,\beta,s}(\theta_{1})$ and
$h^{\prime}{\alpha,\beta,s}(\theta_{2})$.

The case when $h^{\prime\prime}_{\alpha,\beta,s}$ has only one or two zeros is
identical to the previous case (because in both cases
$h^{\prime\prime}_{\alpha,\beta,s}$ changes sign only at the unique simple
zero) and we refer to its conclusion. Let us thus assume that
$h^{\prime\prime}_{\alpha,\beta,s}$ has three distinct zeros
$\hat{\theta}_{1}<\theta_{0}<\hat{\theta}_{2}<\theta_{1}<\hat{\theta}_{3}$.
Then $h^{\prime\prime}_{\alpha,\beta,s}$ is strictly positive on
$(-\infty,\hat{\theta}_{1})\cup(\hat{\theta}_{2},\hat{\theta}_{3})$ and is
strictly negative on
$(\hat{\theta}_{1},\hat{\theta}_{2})\cup(\hat{\theta}_{3},+\infty)$. Thus,
since
$\underset{\theta\to\pm\infty}{\lim}h^{\prime}_{\alpha,\beta,s}(\theta)=-\infty$,
$h^{\prime}_{\alpha,\beta,s}$ may have two distinct zeros or three distinct
zeros (in this case, one double zero and two simple zeros) or four distinct
zeros.

Cases corresponding to two or three distinct zeros are identical to the
previous case (note that in both cases, $h^{\prime}_{\alpha,\beta,s}$ only
changes sign twice) and we refer to its conclusion. Let us thus assume that
$h^{\prime}_{\alpha,\beta,s}$ has four distinct zeros. These zeros are
necessarily simple zeros and from Rolle’s Theorem, we conclude that
$h_{\alpha,\beta,s}$ may have at most $5$ distinct zeros. However, we have
$h_{\alpha,\beta,s}(1)=h^{\prime}_{\alpha,\beta,s}(1)=0$ so that $\theta=1$ is
always a double zero of $h_{\alpha,\beta,s}$. Hence, $h_{\alpha,\beta,s}$ may
only change sign at most thrice. However since
$\underset{\theta\to-\infty}{\lim}h_{\alpha,\beta,s}(\theta)=\infty$ and
$\underset{\theta\to+\infty}{\lim}h_{\alpha,\beta,s}(\theta)=-\infty$, it
cannot change sign twice so that $h_{\alpha,\beta,s}$ may only change sign at
one or three distinct locations and so is $G^{\prime}_{\alpha,\beta,s}$ from
(15).

The case when $h_{\alpha,\beta,s}$ changes sign once is identical to the
previous one and we refer it its conclusion. Let us thus assume that
$G^{\prime}_{\alpha,\beta,s}$ changes sign at three distinct locations denoted
$\tilde{\theta_{1}}<\tilde{\theta}_{2}<\tilde{\theta}_{3}$. From the limit at
infinity, $G^{\prime}_{\alpha,\beta,s}$ is necessarily strictly positive on
$(-\infty,\tilde{\theta_{1}})\cup(\tilde{\theta_{2}},\tilde{\theta_{3}})$ and
strictly negative on
$(\tilde{\theta_{1}},\tilde{\theta_{2}})\cup(\tilde{\theta_{3}},+\infty)$.
This implies that $G_{\alpha,\beta,s}$ is strictly increasing on
$(-\infty,\tilde{\theta_{1}})$ and then strictly decreasing on
$(\tilde{\theta}_{1},\tilde{\theta}_{2})$, then strictly increasing on
$(\tilde{\theta}_{2},\tilde{\theta}_{3})$ and finally strictly decreasing
$(\tilde{\theta}_{3},+\infty)$. Hence, $G_{\alpha,\beta,s}$ has two local
maxima and one local minimum. However, from (13), we know that $\theta=0$ is a
local extremum of $G_{\alpha,\beta,s}$ and from (14) we have assumed $s\geq
s_{\text{inf}}(\alpha,\beta)$ so that
$G^{\prime\prime}_{\alpha,\beta,s}(0)<0$. Hence $\theta=0$ is necessarily a
local maximum of $G_{\alpha,\beta,s}$ and its value is null from (13). Hence
we conclude that $G_{\alpha,\beta,s}$ has a vanishing local maximum at
$\theta=0$, a local minimum with a strictly negative value, and a second local
maximum at $\theta_{3}\neq 0$. The sign of this second local maximum
determines the sign of $G_{\alpha,\beta,s}$ and hence is equivalent to
deciding if $s^{2}$ is a variance proxy. This concludes the proof of Lemma
3.3. ∎

By the condition on $\Delta_{\alpha,\beta,s}$, we also get the claimed upper
bound of the optimal variance proxy from this lemma. Indeed,
$\Delta_{\alpha,\beta,s}\leq 0$ is equivalent to have
$s\in[s_{1}(\alpha,\beta),s_{2}(\alpha,\beta)]$ while
$\Delta_{\alpha,\beta,s}>0$ is equivalent to have
$s\in(0,s_{1}(\alpha,\beta))\cup(s_{2}(\alpha,\beta),+\infty)$. Therefore
$\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}\leq s_{1}(\alpha,\beta)$, with:

$\displaystyle
s_{1}(\alpha,\beta)\coloneqq\sqrt{\frac{(\beta-\alpha)^{2}(\beta-\mathbb{E}[Y_{\mathrm{T}}]+2)-\delta_{\alpha,\beta}}{(\beta-\alpha)^{2}+12}},\quad
s_{2}(\alpha,\beta)\coloneqq\sqrt{\frac{(\beta-\alpha)^{2}(\beta-\mathbb{E}[Y_{\mathrm{T}}]+2)+\delta_{\alpha,\beta}}{(\beta-\alpha)^{2}+12}},$

and where $\delta_{\alpha,\beta}$ was defined in (12). It is obvious that the
position of the second maximum $\theta_{c}(\alpha,\beta,s)$ depends smoothly
on $s$. Moreover, we know that the optimal variance proxy exists and is non-
zero because of (14). The previous analysis implies that for
$s<\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}$ we must have
$\Delta_{\alpha,\beta,s}>0$ and the existence of
$\theta_{c}(\alpha,\beta,s)\neq 0$ with
$G_{\alpha,\beta,s}(\theta_{c}(\alpha,\beta,s))>0$. Thus, taking the limit
$s\to\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}$ with
$s<\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}$ implies that
$G_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{v}p}(\theta_{c}(\alpha,\beta,\|Y_{\mathrm{T}}\|_{v}p))=0$.
Notice that $\theta_{c}(\alpha,\beta,\|Y_{\mathrm{T}}\|_{v}p)=\pm\infty$ is
not possible because we have
$\underset{\theta\to\pm\infty}{\lim}G_{\alpha,\beta,s}(\theta)=e^{\alpha}-e^{\beta}<0$
independent of $s$. Hence we must have

$G_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}}(\theta_{c}(\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}))=0.$

To finish it remains to prove that
$\theta_{c}(\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}})\neq 0$. Indeed,
let us assume by contradiction that
$\theta_{c}(\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}})=0$. The previous
analysis shows that for $s<\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}$,
$G^{\prime}_{\alpha,\beta,s}$ has two distinct zeros inside
$(0,\theta_{c}(\alpha,\beta,s))$ and thus by Rolle’s theorem that
$G^{\prime\prime}_{\alpha,\beta,s}$ has at least one zero in
$(0,\theta_{c}(\alpha,\beta,s))$. Thus at the limit, we would get
$G^{\prime\prime}_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}}(0)=0$,
i.e. $\theta=0$ would be at least a triple zero of
$G_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}}$. In fact, in order to
remain locally negative around $\theta=0$, we would necessarily have a zero of
order four, i.e.
$G^{\prime\prime\prime}_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}}(0)=0$.
Hence $\theta=0$ would be a zero of order four of
$g_{\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}}$. This would imply
$\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}=s_{\text{inf}}(\alpha,\beta)$ (triple zero
using the definition of $s_{\text{inf}}(\alpha,\beta)$ and
$g^{\prime\prime\prime}_{\alpha,\beta,s_{\text{inf}}(\alpha,\beta)}(0)$ (zero
of order four). However we have:

$g^{\prime\prime\prime}_{\alpha,\alpha+\epsilon,s_{\text{inf}}(\alpha,\beta)}(0)=\frac{2e^{3\epsilon}-(\epsilon^{3}+6)e^{2\epsilon}+(6-\epsilon^{3})e^{\epsilon}-2}{(e^{\epsilon}-1)^{3}}=\frac{P(\epsilon)}{(e^{\epsilon}-1)^{3}}.$

From Lemma B.1 in Appendix B we have that
$g^{\prime\prime\prime}_{\alpha,\alpha+\epsilon,s_{\text{inf}}(\alpha,\alpha+\epsilon)}(0)>0$
for all $\epsilon>0$ so that
$g^{\prime\prime\prime}_{\alpha,\beta,s_{\text{inf}}(\alpha,\beta)}(0)>0$ for
any $\alpha<\beta$ i.e., we cannot have
$\theta_{c}(\alpha,\beta,\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}})=0$. We conclude
that the optimal variance proxy is characterized by the unique solution of the
below system of equations, which is coherent with the illustration of Figure
4. This concludes the proof of Theorem 3.2. ∎

Figure 4: Illustration of the proof of Theorem 3.2. Function
$g_{\alpha,\beta,s}$ defined in Equation (10) is represented with $\alpha=1$
and $\beta=4$ and for different values of $s$. Non-positivity of the
$g_{\alpha,\beta,s}$ function is equivalent to $s^{2}$ being a variance proxy.
In purple is the optimal variance proxy case
$s^{2}=s_{c}^{2}=\|Y_{\mathrm{T}}\|_{\mathrm{vp}}^{2}$ ($s=s_{c}\approx
0.8107$). Red represents $s=0.812>s_{c}$ (with $s^{2}$ being a variance proxy)
and blue represents $s=0.8095<s_{c}$ (with $s^{2}$ not being a variance
proxy).

##  4 Future research directions

In summary, our work has identified the optimal sub-Gaussian variance proxy
for truncated Gaussian and truncated exponential random variables, while also
delineating the conditions under which strict sub-Gaussianity may or may not
be observed in these truncated distributions.

Moving forward, there are several avenues for extending this work. Firstly,
exploring additional commonly encountered distributions beyond Gaussian and
exponential would broaden the scope of our findings. Additionally,
investigating the truncation of multivariate distributions, such as the
multivariate Gaussian distribution, could provide valuable insights into more
complex scenarios.

Furthermore, considering the sub-Weibull property of truncated random
variables represents a promising direction for future research (Vladimirova et
al.,, 2020). This concept, which serves as a generalization of the sub-
Gaussian property, offers an intriguing framework for further understanding
the behavior of truncated distributions across various contexts.

## Acknowledgments

The first author was partially supported by Huawei’s Math Modelling Lab in
Moscow, Russia. The second author used part of his IUF junior grant G752IUFMAR
for this research. The third author was partially supported by
ANR-21-JSTM-0001 grant.

##  Appendix A Proofs for truncated Gaussian random variables

This entire Appendix A is devoted to proving Lemma 2.4, showing that the
function $h_{\alpha,\beta}$ is strictly concave on $[\theta_{0},+\infty)$ for
any finite $\alpha<\beta$.

###### Proof of Lemma 2.4.

To begin with, observe that it is sufficient to check the concavity of the
function only for the case when $\alpha=-\beta$ (i.e.
$\theta_{0}:=\frac{\alpha+\beta}{2}=0$). Indeed, suppose that
$h^{\prime\prime}_{-\beta,\beta}(\theta)<0$ for all $\beta\in\mathbb{R}_{>0}$
and $\theta>0$. Then, this implies that
$h_{-\frac{\beta-\alpha}{2},\frac{\beta-\alpha}{2}}(\theta)=h_{\alpha,\beta}(\theta+\theta_{0})$
is strictly concave for $\theta\geq 0$, i.e. that $x\mapsto
h_{\alpha,\beta}(x)$ is strictly concave for $x\geq\theta_{0}$.

###  A.1 Some notations and preliminary results

Let us recall that we have:

$\begin{split}&\Phi(x)\coloneqq\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}s^{2}}ds,\\\
&F_{-\beta,\beta}(\theta)\coloneqq\Phi(\beta-\theta)-\Phi(-\beta-\theta),\\\
&f_{-\beta,\beta}(\theta)\coloneqq\ln\left(\frac{F_{-\beta,\beta}(\theta)}{2\Phi(\beta)-1}\right),\\\
&F^{\prime}_{-\beta,\beta}(\theta)=\phi(\theta+\beta)-\phi(\theta-\beta)=\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{1}{2}(\theta+\beta)^{2}}-e^{-\frac{1}{2}(\theta-\beta)^{2}}\right),\\\
&F^{\prime\prime}_{-\beta,\beta}(\theta)=\frac{1}{\sqrt{2\pi}}\left(-(\beta+\theta)e^{-\frac{1}{2}(\theta+\beta)^{2}}+(\theta-\beta)e^{-\frac{1}{2}(\theta-\beta)^{2}}\right),\\\
&F^{\prime\prime\prime}_{-\beta,\beta}(\theta)=\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{1}{2}(\theta+\beta)^{2}}\left[(\beta+\theta)^{2}-1\right]+e^{-\frac{1}{2}(\theta-\beta)^{2}}\left[1-(\beta-\theta)^{2}\right]\right),\\\
&h_{-\beta,\beta}(\theta)\coloneqq
f^{\prime}_{-\beta,\beta}(\theta)=\frac{F^{\prime}_{-\beta,\beta}(\theta)}{F_{-\beta,\beta}(\theta)},\\\
&h^{\prime}_{-\beta,\beta}(\theta)=\frac{F_{-\beta,\beta}(\theta)F^{\prime\prime}_{-\beta,\beta}(\theta)-(F^{\prime}_{-\beta,\beta}(\theta))^{2}}{F_{-\beta,\beta}(\theta)^{2}},\\\
&h^{\prime\prime}_{-\beta,\beta}(\theta)=\frac{(F_{-\beta,\beta}(\theta))^{2}F^{\prime\prime\prime}_{-\beta,\beta}(\theta)-3F_{-\beta,\beta}(\theta)F^{\prime}_{-\beta,\beta}(\theta)F^{\prime\prime}_{-\beta,\beta}(\theta)+2(F^{\prime}_{-\beta,\beta}(\theta))^{3}}{F_{-\beta,\beta}(\theta)^{3}},\\\
&\underset{\theta\to+\infty}{\lim}F^{\prime}_{-\beta,\beta}(\theta)=\underset{\theta\to+\infty}{\lim}F^{\prime\prime}_{-\beta,\beta}(\theta)=\underset{\theta\to+\infty}{\lim}F^{\prime\prime\prime}_{-\beta,\beta}(\theta)=0,\\\
&\underset{{\tiny{\theta\to+\infty}}}{\lim}F_{-\beta,\beta}(\theta)=0.\end{split}$
(A.17)

It is obvious by definition that $F_{-\beta,\beta}$ is strictly positive on
$\mathbb{R}$. It is also straightforward to obtain the important linear
relation

$F^{\prime\prime\prime}_{-\beta,\beta}(\theta)=-P_{2}^{\prime}(\theta)F^{\prime\prime}_{-\beta,\beta}(\theta)-P_{2}(\theta)F^{\prime}_{-\beta,\beta}(\theta),$
(A.18)

where

$P_{2}(\theta)\coloneqq\theta^{2}+1-\beta^{2}.$ (A.19)

In order to prove the strict concavity of $h_{-\beta,\beta}$ on
$\mathbb{R}_{\geq 0}$, we shall need information on the function
$F_{-\beta,\beta}$ and its derivatives. Thus, we shall first prove the
following lemma.

######  Lemma A.1.

We have the following results:

  * •

We have $F^{\prime}_{-\beta,\beta}(\theta)<0$ for any $\beta>0$ and any
$\theta\in\mathbb{R}_{>0}$.

  * •

For $0<\beta\leq\sqrt{3}$: The function
$F^{\prime\prime\prime}_{-\beta,\beta}$ has only one zero on $\mathbb{R}_{>0}$
that we shall denote $\theta_{1}$. Moreover, it is strictly positive on
$(0,\theta_{1})$ and strictly negative on $(\theta_{1},+\infty)$.
Consequently, the function $F^{\prime\prime}_{-\beta,\beta}$ is strictly
increasing on $[0,\theta_{1}]$ from
$F^{\prime\prime}_{-\beta,\beta}(0)=-\frac{2\beta}{\sqrt{2\pi}}e^{-\frac{1}{2}\beta^{2}}<0$
to $F^{\prime\prime}_{-\beta,\beta}(\theta_{1})>0$ and strictly decreasing on
$(\theta_{1},+\infty)$ from $F^{\prime\prime}_{-\beta,\beta}(\theta_{1})>0$ to
$0$. In particular $F^{\prime\prime}_{-\beta,\beta}$ has only one zero on
$\mathbb{R}_{>0}$ denoted $\alpha_{1}$ satisfying $\alpha_{1}>\theta_{1}$.

  * •

For $\beta>\sqrt{3}$: The function $F^{\prime\prime\prime}_{-\beta,\beta}$ has
two distinct zeros on $\mathbb{R}_{>0}$ denoted $\theta_{1}<\theta_{2}$ on
$\mathbb{R}_{>0}$. Moreover, it is strictly negative on
$(0,\theta_{1})\cup(\theta_{2},+\infty)$ and strictly positive on
$(\theta_{1},\theta_{2})$. Consequently, the function
$F^{\prime\prime}_{-\beta,\beta}$ is decreasing on $(0,\theta_{1})$ from
$F^{\prime\prime}_{-\beta,\beta}(0)=-\frac{2\beta}{\sqrt{2\pi}}e^{-\frac{1}{2}\beta^{2}}<0$
to
$F^{\prime\prime}_{-\beta,\beta}(\theta_{1})<F^{\prime\prime}_{-\beta,\beta}(0)<0$.
$F^{\prime\prime}_{-\beta,\beta}$ is then increasing on
$(\theta_{1},\theta_{2})$ up to $F^{\prime\prime}_{-\beta,\beta}(\theta_{2})$.
It finally decreases from $F^{\prime\prime}_{-\beta,\beta}(\theta_{2})$ to its
limit $0$ at $\theta\to+\infty$. In particular, we must have
$F^{\prime\prime}_{-\beta,\beta}(\theta_{2})>0$ and
$F^{\prime\prime}_{-\beta,\beta}$ has only one zero on $\mathbb{R}_{>0}$
denoted $\alpha_{1}$ that verifies $\alpha_{2}\in(\theta_{1},\theta_{2})$.

###### Proof of Lemma A.1.

The first point is obvious from (A.17). Let us observe that

$F^{\prime\prime}_{-\beta,\beta}(\theta)=\frac{2}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}(\beta^{2}+\theta^{2})\right)\left(\theta\sinh(\beta\theta)-\beta\cosh(\beta\theta)\right).$
(A.20)

In particular
$F^{\prime\prime}_{-\beta,\beta}(0)=-\frac{2\beta}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\beta^{2}\right)<0$
and
$\underset{\theta\to+\infty}{\lim}F^{\prime\prime}_{-\beta,\beta}(\theta)=0$.
Let us assume by contradiction that $F^{\prime\prime\prime}_{\beta}$ does not
vanish on $\mathbb{R}_{>0}$. This implies that
$F_{-\beta,\beta}^{\prime\prime}$ would be a strictly monotonous function and
therefore would be a strictly increasing function because of the values of
$F^{\prime\prime}_{-\beta,\beta}(0)$ and
$F^{\prime\prime}_{-\beta,\beta}(+\infty)$. Since
$\underset{\theta\to+\infty}{\lim}F^{\prime\prime}_{-\beta,\beta}(\theta)=0$,
it implies that $F^{\prime\prime}_{-\beta,\beta}$ would be a strictly negative
function on $\mathbb{R}_{>0}$ so that $F^{\prime}_{-\beta,\beta}$ would be a
strictly decreasing function on $\mathbb{R}_{>0}$. This is a contradiction
with the fact that
$\underset{\theta\to+\infty}{\lim}F^{\prime}_{-\beta,\beta}(\theta)=0$ and
$F^{\prime}_{-\beta,\beta}$ is strictly negative on $\mathbb{R}_{>0}$.
Therefore, we conclude that there is at least one value of
$\theta_{c}\in\mathbb{R}_{>0}$ for which
$F^{\prime\prime\prime}_{-\beta,\beta}(\theta_{c})=0$. This point shall be
used later to determine the sign of some quantities by contradiction555In
fact, the reasoning implies that there exists a non-trivial interval on which
$F^{\prime\prime\prime}_{-\beta,\beta}$ is strictly positive.. Let us now
compute

$F^{\prime\prime\prime}_{-\beta,\beta}\left(\theta=\frac{s}{\beta}\right)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\beta^{2}+\frac{s^{2}}{\beta^{2}}\right)\right)e^{-s}\left(\left(1-\left(\frac{s}{\beta}-\beta\right)^{2}\right)e^{2s}-\left(1-\left(\frac{s}{\beta}+\beta\right)^{2}\right)\right).$
(A.21)

Therefore, the sign and zeros of $F^{\prime\prime\prime}_{-\beta,\beta}$ are
equivalent to the sign and zeros of $g_{\beta}(s)$ defined by

$g_{\beta}(s)\coloneqq\left(1-\left(\frac{s}{\beta}-\beta\right)^{2}\right)e^{2s}-1+\left(\frac{s}{\beta}+\beta\right)^{2}.$
(A.22)

We have

$\displaystyle g_{\beta}^{\prime}(s)$
$\displaystyle=-\frac{2(\beta^{4}-2(s+1)\beta^{2}+s^{2}+s)}{\beta^{2}}e^{2s}+\frac{2(\beta^{2}+s)}{\beta^{2}},$
$\displaystyle g^{\prime\prime}_{\beta}(s)$
$\displaystyle=-\frac{2(2\beta^{4}-4\beta^{2}s-6\beta^{2}+2s^{2}+4s+1)}{\beta^{2}}e^{2s}+\frac{2}{\beta^{2}},$
$\displaystyle g^{\prime\prime\prime}_{\beta}(s)$
$\displaystyle=-\frac{4(2s^{2}-(4\beta^{2}-6)s+2\beta^{4}-8\beta^{2}+3)}{\beta^{2}}e^{2s}.$

The discriminant of the numerator of $g^{\prime\prime\prime}_{\beta}$ is given
by $\Delta=4(4\beta^{2}+3)>0$. Hence $g^{\prime\prime\prime}_{\beta}$ has two
distinct zeros on $\mathbb{R}$ and we need to study the sign of these zeros.
Let us denote

$P(\beta)\coloneqq 2\beta^{4}-8\beta^{2}+3\,\,,\,\,S(\beta)\coloneqq
4\beta^{2}-6,$ (A.23)

so that the product (resp. the sum) of the two zeros of
$g^{\prime\prime\prime}_{\beta}$ are given by $\frac{1}{2}P(\beta)$ (resp.
$\frac{1}{2}S(\beta)$). $P(\beta)$ is strictly positive on
$\left(0,\frac{\sqrt{8-2\sqrt{10}}}{2}\right)\cup\left(\frac{\sqrt{8+2\sqrt{10}}}{2},+\infty\right)$
and strictly negative on
$\left(\frac{\sqrt{8-2\sqrt{10}}}{2},\frac{\sqrt{8+2\sqrt{10}}}{2}\right)$.
$S(\beta)$ is strictly positive on $\left(\frac{\sqrt{6}}{2},+\infty\right)$
and strictly negative on $\left(0,\frac{\sqrt{6}}{2}\right)$. We shall also
observe:

$\displaystyle\lim_{s\to+\infty}g_{\beta}(s)=\lim_{s\to+\infty}g^{\prime}_{\beta}(s)=\lim_{s\to+\infty}g^{\prime\prime}_{\beta}(s)=\lim_{s\to+\infty}g^{\prime\prime\prime}_{\beta}(s)=-\infty,$
$\displaystyle
g^{\prime\prime\prime}_{\beta}(0)=-\frac{4P(\beta)}{\beta^{2}},$
$\displaystyle g^{\prime\prime}_{\beta}(0)=4(3-\beta^{2}),$ $\displaystyle
g^{\prime}_{\beta}(0)=2(3-\beta^{2}),$ $\displaystyle g_{\beta}(0)=0.$

Let us now study the following four cases.

Case $\beta>\frac{\sqrt{8+2\sqrt{10}}}{2}$. Here,
$g^{\prime\prime\prime}_{\beta}$ has two roots on $\mathbb{R}$ and their
product in strictly positive and their sum is strictly positive. Thus,
$g^{\prime\prime\prime}_{\beta}$ has two distinct roots on $\mathbb{R}_{>0}$
denoted $0<s_{1}<s_{2}$. It is strictly negative on
$(0,s_{1})\cup(s_{2},+\infty)$ and strictly positive in $(s_{1},s_{2})$.
Therefore $g^{\prime\prime}_{\beta}$ strictly decreases on $(0,s_{1})$ and
since $g^{\prime\prime}_{\beta}(0)<0$ it remains strictly negative on
$(0,s_{1})$. $g^{\prime\prime}_{\beta}$ is then strictly increasing from
$g^{\prime\prime}_{\beta}(s_{1})$ up to $g^{\prime\prime}_{\beta}(s_{2})$. We
have
$g^{\prime\prime}_{\beta}(s_{2})=\frac{2}{\beta^{2}}\left(e^{2\beta^{2}+\sqrt{4\beta^{2}+3}-3}(\sqrt{4\beta^{2}+3}-1)+1\right)>0$.
Since $g^{\prime\prime}_{\beta}$ diverges towards $-\infty$ when
$s\to+\infty$, we end up with fact that there exist exactly two values
$(s_{3},s_{4})\in(\mathbb{R}_{>0})^{2}$ such that
$g^{\prime\prime}_{\beta}(s_{3})=g^{\prime\prime}_{\beta}(s_{4})=0$ and they
satisfy $s_{3}<s_{2}<s_{4}$. Moreover $g^{\prime\prime}_{\beta}$ is strictly
positive on $\left(s_{3},s_{4}\right)$ and strictly negative on
$\left(0,s_{3}\right)\cup\left(s_{4},+\infty\right)$. Therefore
$g^{\prime}_{\beta}$ strictly decreases on $\left(0,s_{3}\right)$ and since
$g^{\prime}_{\beta}(0)<0$ it remains strictly negative on $(0,s_{3})$. Note
that we necessarily have $g^{\prime}_{\beta}(s_{4})>0$. Indeed, if
$g^{\prime}_{\beta}(s_{4})\leq 0$ then $g^{\prime}_{\beta}$ would be strictly
negative on $\mathbb{R}_{>0}$ so that $g_{\beta}$ would be strictly decreasing
and since $g_{\beta}(0)=0$ it would not vanish on $\mathbb{R}_{>0}$ which is a
contradiction (because we have proved that
$F^{\prime\prime\prime}_{-\beta,\beta}$ must vanish at least once). Since
$g^{\prime}_{\beta}(s_{4})>0$, we obtain that there exists exactly two values
$(s_{5},s_{6})\in(\mathbb{R}_{>0})^{2}$ such that
$g^{\prime}_{\beta}(s_{5})=g^{\prime}_{\beta}(s_{6})=0$ and they satisfy
$s_{3}<s_{5}<s_{4}<s_{6}$. $g_{\beta}$ is thus strictly decreasing on
$(0,s_{5})$ and since $g_{\beta}(0)=0$ it remains strictly negative. Again
$g_{\beta}(s_{6})$ must be strictly positive otherwise $g_{\beta}$ would not
vanish and nor would $F^{\prime\prime\prime}_{-\beta,\beta}$ which is a
contradiction. Therefore, the variations of $g_{\beta}$ implies that there
exists exactly two values $(s_{7},s_{8})\in(\mathbb{R}_{>0})^{2}$ such that
$g_{\beta}(s_{7})=g_{\beta}(s_{8})=0$ and they satisfy
$s_{5}<s_{7}<s_{6}<s_{8}$.

Case $\beta\in\Big{(}\sqrt{3},\frac{\sqrt{8+2\sqrt{10}}}{2}\,\Big{]}$. Here,
$g^{\prime\prime\prime}_{\beta}$ has two roots on $\mathbb{R}$ and their
product in negative or null and their sum is strictly positive. This means
that $g^{\prime\prime\prime}_{\beta}$ has exactly one root on
$\mathbb{R}_{>0}$ that we shall denote $s_{2}>0$.
$g^{\prime\prime\prime}_{\beta}$ is strictly positive on $(0,s_{2})$ and
strictly negative on $(s_{2},+\infty)$. Since $g^{\prime\prime}_{\beta}(0)<0$
and
$g^{\prime\prime}_{\beta}(s_{2})=\frac{2}{\beta^{2}}\left(e^{2\beta^{2}+\sqrt{4\beta^{2}+3}-3}(\sqrt{4\beta^{2}+3}-1)+1\right)>0$
and $\underset{s\to+\infty}{\lim}g^{\prime\prime}_{\beta}(s)=-\infty$, we get
that there exists exactly two distinct values
$(s_{3},s_{4})\in(\mathbb{R}_{>0})^{2}$ such that
$g^{\prime\prime}_{\beta}(s_{3})=g^{\prime\prime}_{\beta}(s_{4})=0$ and they
satisfy $s_{3}<s_{2}<s_{4}$. Moreover, $g^{\prime}_{\beta}$ is strictly
decreasing on $(0,s_{3})$, strictly increasing on $(s_{3},s_{4})$ and strictly
decreasing on $(s_{4},+\infty)$. We have $g^{\prime}_{\beta}(0)<0$ so that
$g^{\prime}_{\beta}(s_{3})<0$. Similarly to the previous case, we must have
$g^{\prime}_{\beta}(s_{4})>0$ otherwise $g_{\beta}$ would not vanish on
$\mathbb{R}_{>0}$ (it would be strictly decreasing on $\mathbb{R}_{>0}$ with
$g_{\beta}(0)=0$). We obtain that there exists exactly two values
$(s_{5},s_{6})\in(\mathbb{R}_{>0})^{2}$ such that
$g^{\prime}_{\beta}(s_{5})=g^{\prime}_{\beta}(s_{6})=0$ and they satisfy
$s_{3}<s_{5}<s_{4}<s_{6}$. $g_{\beta}$ is then strictly decreasing from
$g_{\beta}(0)=0$ to $g_{\beta}(s_{5})<0$. We must have $g_{\beta}(s_{6})>0$
otherwise $g_{\beta}$ would not vanish on $\mathbb{R}_{>0}$ (which is a
contradiction since we know that $F^{\prime\prime\prime}_{-\beta,\beta}$ has
at least one zero on $\mathbb{R}_{>0}$). Therefore, the variations of
$g_{\beta}$ implies that there exists exactly two values
$(s_{7},s_{8})\in(\mathbb{R}_{>0})^{2}$ such that
$g_{\beta}(s_{7})=g_{\beta}(s_{8})=0$ and they satisfy
$s_{5}<s_{7}<s_{6}<s_{8}$.

Case $\beta\in\Big{(}\frac{\sqrt{8-2\sqrt{10}}}{2},\sqrt{3}\Big{]}$. Here,
$g^{\prime\prime\prime}_{\beta}$ has two roots on $\mathbb{R}$ and their
product is strictly negative. This means that $g^{\prime\prime\prime}_{\beta}$
has exactly one root on $\mathbb{R}_{>0}$ that we shall denote $s_{2}>0$.
$g^{\prime\prime\prime}_{\beta}$ is strictly positive on $(0,s_{2})$ and
strictly negative on $(s_{2},+\infty)$. We have
$g^{\prime\prime}_{\beta}(0)\geq 0$ and
$\underset{s\to+\infty}{\lim}g^{\prime\prime}_{\beta}(s)=-\infty$. Thus, since
$g_{\beta}^{\prime\prime}$ is strictly increasing and then strictly
decreasing, we get that there exists exactly one value
$s_{4}\in\mathbb{R}_{>0}$ such that $g^{\prime\prime}_{\beta}(s_{4})=0$ and it
satisfies $s_{2}<s_{4}$. Moreover, $g^{\prime}_{\beta}$ is strictly increasing
on $(0,s_{4})$ with $g_{\beta}^{\prime}(0)\geq 0$. It then strictly decreases
from $g^{\prime}_{\beta}(s_{4})>0$ towards $-\infty$. Hence, there exists
exactly one value $s_{6}>s_{4}$ such that $g^{\prime}_{\beta}(s_{6})=0$ and
$g^{\prime}_{\beta}$ is strictly positive on $(0,s_{6})$ and strictly negative
on $(s_{6},+\infty)$. Thus, $g_{\beta}$ is strictly increasing from
$g_{\beta}(0)=0$ to $g_{\beta}(s_{6})>0$ and then strictly decreasing towards
$-\infty$. We conclude that there exists only one value $s_{8}$ on
$\mathbb{R}_{>0}$ such that $g_{\beta}(s_{8})=0$.

Case $\beta\in\Big{(}0,\frac{\sqrt{8-2\sqrt{10}}}{2}\,\Big{]}$. Here,
$g^{\prime\prime\prime}_{\beta}$ has two roots on $\mathbb{R}$, and their
product is positive or null, while their sum is strictly negative. This
implies that $g^{\prime\prime\prime}_{\beta}$ does not vanish on
$\mathbb{R}_{>0}$ and thus is strictly negative. $g^{\prime\prime}_{\beta}$ is
thus strictly decreasing on $\mathbb{R}_{>0}$ with
$g^{\prime\prime}_{\beta}(0)>0$ and
$g_{\beta}^{\prime\prime}(+\infty)=-\infty$. We get that there exists exactly
one value $s_{4}\in\mathbb{R}_{>0}$ such that
$g^{\prime\prime}_{\beta}(s_{4})=0$. Moreover, $g^{\prime}_{\beta}$ is
strictly increasing on $(0,s_{4})$ with $g_{\beta}^{\prime}(0)>0$. It then
strictly decreases from $g^{\prime}_{\beta}(s_{4})>0$ towards $-\infty$.
Hence, there exists exactly one value $s_{6}>s_{4}$ such that
$g^{\prime}_{\beta}(s_{6})=0$ and $g^{\prime}_{\beta}$ is strictly positive on
$(0,s_{6})$ and strictly negative on $(s_{6},+\infty)$. Thus, $g_{\beta}$ is
strictly increasing from $g_{\beta}(0)=0$ to $g_{\beta}(s_{6})>0$ and then
strictly decreasing towards $-\infty$. We conclude that there exists only one
value $s_{8}$ on $\mathbb{R}_{>0}$ such that $g_{\beta}(s_{8})=0$.

Summarizing the four different cases and using $\theta=\frac{s}{\beta}$, we
conclude that for $\beta\in\Big{(}0,\sqrt{3}\Big{]}$,
$F^{\prime\prime\prime}_{-\beta,\beta}$ has exactly one root $\theta_{1}$ on
$\mathbb{R}_{>0}$ and it is strictly positive on $\left(0,\theta_{1}\right)$
and it is strictly negative on $\left(\theta_{1},+\infty\right)$. On the
contrary, for $\beta>\sqrt{3}$, $F^{\prime\prime\prime}_{-\beta,\beta}$ has
exactly two distinct roots on $\mathbb{R}_{>0}$ denoted
$\theta_{1}<\theta_{2}$ and it is strictly positive on
$\left(\theta_{1},\theta_{2}\right)$ and strictly negative on
$\left(0,\theta_{1}\right)\cup\left(\theta_{2},+\infty\right)$. The rest of
the lemma is then obvious, which concludes the proof of Lemma A.1. ∎

###  A.2 A sufficient condition for strict concavity

Let us first prove that for any $\beta>0$, the function $h_{-\beta,\beta}$ is
strictly concave in a positive neighborhood of $\theta=0$. It is a
straightforward computation by taking Taylor series around $\theta=0$ to
observe that

$h^{\prime\prime}_{-\beta,\beta}(\theta)\overset{\theta\to
0}{=}-\frac{\beta\exp\left(-\frac{\beta^{2}}{2}\right)}{(2\Phi(\beta)-1)^{2}}\left(\frac{6\beta}{\pi}\exp\left(-\frac{\beta^{2}}{2}\right)+\frac{2(\beta^{2}-3)}{\sqrt{2\pi}}(2\Phi(\beta)-1)\right)\theta+O(\theta^{2}).$
(A.24)

The leading order is strictly negative for $\beta\geq\sqrt{3}$ because both
terms are negative or null. For $\beta\in[0,\sqrt{3})$ we define

$f_{1}(\beta)\coloneqq\frac{6\beta}{\pi(\beta^{2}-3)}\exp\left(-\frac{\beta^{2}}{2}\right)+\frac{2}{\sqrt{2\pi}}(2\Phi(\beta)-1),$
(A.25)

whose derivative is
$f_{1}^{\prime}(\beta)=-\frac{4\beta^{4}\exp\left(-\frac{\beta^{2}}{2}\right)}{\pi(\beta^{2}-3)^{2}}<0$.
Therefore $f_{1}$ is strictly decreasing on $[0,\sqrt{3})$ and since
$f_{1}(0)=0$ we get that $f_{1}$ is strictly negative on $[0,\sqrt{3})$. This
implies that the leading order of $h^{\prime\prime}_{-\beta,\beta}(\theta)$ as
$\theta\to 0$ is strictly negative on $\mathbb{R}_{>0}$. In particular, we get
that for any $\beta>0$, there exists a positive neighborhood of $\theta=0$ on
which $h_{-\beta,\beta}$ is strictly concave.

Let us now reformulate the problem of strict concavity more simply. From
(A.17), the equation $h_{-\beta,\beta}^{\prime\prime}(\theta)=0$ is equivalent
to

$Z_{\beta}(\theta)\coloneqq(F_{-\beta,\beta}(\theta))^{2}F^{\prime\prime\prime}_{-\beta,\beta}(\theta)-3F_{-\beta,\beta}(\theta)F^{\prime}_{-\beta,\beta}(\theta)F^{\prime\prime}_{-\beta,\beta}(\theta)+2(F^{\prime}_{-\beta,\beta}(\theta))^{3}=0.$
(A.26)

We shall denote

$\displaystyle a_{\beta}(\theta)$ $\displaystyle\coloneqq
F^{\prime\prime\prime}_{-\beta,\beta}(\theta),$ $\displaystyle
b_{\beta}(\theta)$
$\displaystyle\coloneqq-3F^{\prime}_{-\beta,\beta}(\theta)F^{\prime\prime}_{-\beta,\beta}(\theta),$
$\displaystyle c_{\beta}(\theta)$ $\displaystyle\coloneqq
2(F^{\prime}_{-\beta,\beta}(\theta))^{3},$

so that

$Z_{\beta}(\theta)=a_{\beta}(\theta)(F_{-\beta,\beta}(\theta))^{2}+b_{\beta}(\theta)F_{-\beta,\beta}(\theta)+c_{\beta}(\theta).$
(A.27)

Let us observe that if we can prove that $Z_{\beta}(\theta)$ does not vanish
on $\mathbb{R}_{>0}$, then it proves that $h_{-\beta,\beta}$ is strictly
concave on $\mathbb{R}_{>0}$. Indeed, the function $Z_{\beta}(\theta)$ is
continuous in $\theta$. Moreover, we have proved that it is strictly negative
in a positive neighborhood of $\theta=0$ so that if it does not vanish on
$\mathbb{R}_{>0}$, then the intermediate value theorem implies that it must
remain strictly negative on $\mathbb{R}_{>0}$. Therefore, we obtain the
following sufficient condition to prove strict concavity of $h_{-\beta,\beta}$
on $\mathbb{R}_{>0}$.

######  Proposition A.1.

Let $\beta>0$. Proving that $Z_{\beta}$ does not vanish on $\mathbb{R}_{>0}$
is a sufficient condition to proving the strict concavity of
$h_{-\beta,\beta}$ on $\mathbb{R}_{>0}$.

The next step is to use the fact that the r.h.s. of (A.27) may be seen as a
polynomial of degree $2$ in $F_{-\beta,\beta}(\theta)$. In particular, zeros
$\theta_{c}$ of $Z_{\beta}$ are either zeros of
$F^{\prime\prime\prime}_{\beta}$ or solutions of the system

$\begin{split}&9(F^{\prime}_{-\beta,\beta}(\theta_{c}))^{2}(F^{\prime\prime}_{-\beta,\beta}(\theta_{c}))^{2}-8(F^{\prime}_{-\beta,\beta}(\theta_{c}))^{3}F^{\prime\prime\prime}_{-\beta,\beta}(\theta_{c})\geq
0,\\\
&F_{-\beta,\beta}(\theta_{c})=\frac{3}{2}\frac{F^{\prime}_{-\beta,\beta}(\theta_{c})F^{\prime\prime}_{-\beta,\beta}(\theta_{c})}{F^{\prime\prime\prime}_{-\beta,\beta}(\theta_{c})}\\\
&\quad\pm\frac{1}{2F^{\prime\prime\prime}_{-\beta,\beta}(\theta_{c})}\sqrt{9(F^{\prime}_{-\beta,\beta}(\theta_{c}))^{2}(F^{\prime\prime}_{-\beta,\beta}(\theta_{c}))^{2}-8(F^{\prime}_{-\beta,\beta}(\theta_{c}))^{3}F^{\prime\prime\prime}_{-\beta,\beta}(\theta_{c})}\\\
&=-\frac{b_{\beta}(\theta_{c})}{2a_{\beta}(\theta_{c})}\pm\frac{1}{2a_{\beta}(\theta_{c})}\sqrt{b_{\beta}(\theta_{c})^{2}-4a_{\beta}(\theta_{c})c_{\beta}(\theta_{c})}.\end{split}$
(A.28)

The first inequality in (A.28) is necessary otherwise (A.26) which is
polynomial of degree $2$ in $F_{-\beta,\beta}(\theta_{c})$ would have no real
roots and thus (A.26) would not have solutions on $\mathbb{R}_{>0}$ ending the
proof. Therefore, we shall define

$\displaystyle T_{\beta,\pm}(\theta)$ $\displaystyle\coloneqq-
F_{-\beta,\beta}(\theta)+\frac{3}{2}\frac{F^{\prime}_{-\beta,\beta}(\theta)F^{\prime\prime}_{-\beta,\beta}(\theta)}{F^{\prime\prime\prime}_{-\beta,\beta}(\theta)}$
$\displaystyle\quad\pm\frac{1}{2F^{\prime\prime\prime}_{-\beta,\beta}(\theta)}\sqrt{9(F^{\prime}_{-\beta,\beta}(\theta))^{2}(F^{\prime\prime}_{-\beta,\beta}(\theta))^{2}-8(F^{\prime}_{-\beta,\beta}(\theta))^{3}F^{\prime\prime\prime}_{-\beta,\beta}(\theta)}$
$\displaystyle=-\frac{b_{\beta}(\theta)}{2a_{\beta}(\theta)}\pm\frac{1}{2a_{\beta}(\theta)}\sqrt{b_{\beta}(\theta)^{2}-4a_{\beta}(\theta)c_{\beta}(\theta)}-F_{-\beta,\beta}(\theta),$

and we have the following proposition.

######  Proposition A.2.

Let $\beta>0$. A sufficient condition to prove strict concavity of
$h_{-\beta,\beta}$ on $\mathbb{R}_{>0}$ is to prove that the function
$S_{\beta}$ is strictly positive on $\mathbb{R}_{>0}$ where

$\begin{split}S_{\beta}(\theta)&\coloneqq
9\theta(F_{-\beta,\beta}^{\prime\prime}(\theta))^{5}\\\
&\quad+\left(42\theta^{2}-9\right)F_{-\beta,\beta}^{\prime}(\theta)(F_{-\beta,\beta}^{\prime\prime}(\theta))^{4}\\\
&\quad-15\left(\beta^{2}-\frac{79}{15}\theta^{2}+2\right)\theta(F_{-\beta,\beta}^{\prime}(\theta))^{2}(F_{-\beta,\beta}^{\prime\prime}(\theta))^{3}\\\
&\quad+\left(75\theta^{4}-(42\beta^{2}+36)\theta^{2}-\beta^{4}+12\beta^{2}-3\right)(F_{-\beta,\beta}^{\prime}(\theta))^{3}(F_{-\beta,\beta}^{\prime\prime}(\theta))^{2}\\\
&\quad+4\left(9\theta^{4}-\left(10\beta^{2}+\frac{9}{2}\right)\theta^{2}+\beta^{4}+\frac{9}{2}\beta^{2}-\frac{3}{2}\right)\theta(F_{-\beta,\beta}^{\prime}(\theta))^{4}F_{-\beta,\beta}^{\prime\prime}(\theta)\\\
&\quad+\left(7\theta^{6}-(13\beta^{2}+3)\theta^{4}+(5\beta^{4}+6\beta^{2}-3)\theta^{2}+(\beta^{2}-1)^{3}\right)(F_{-\beta,\beta}^{\prime}(\theta))^{5}.\end{split}$
(A.29)

###### Proof of Proposition A.2.

As explained in Proposition A.1, a sufficient condition to prove the strict
concavity of $h_{-\beta,\beta}$ on $\mathbb{R}_{>0}$ is to show that
$Z_{\beta}$ does not vanish on $\mathbb{R}_{>0}$. Moreover, the previous
discussion implies that zeros of $Z_{\beta}$ are either zeros of
$F^{\prime\prime\prime}_{-\beta,\beta}$ or zeros of $T_{\pm,\beta}$. Let us
first prove that the functions $T_{\beta,+}$ and $T_{\beta,-}$ do not vanish
on $\mathbb{R}_{>0}\setminus\mathcal{R}_{\beta}$ where $\mathcal{R}_{\beta}$
is the set of zeros of $F^{\prime\prime\prime}_{-\beta,\beta}$. Let us observe
that for any $\theta\in\mathbb{R}_{>0}\setminus\mathcal{R}_{\beta}$:

$\begin{split}T_{\beta,\pm}^{\prime}(\theta)&=\left(\frac{3F^{\prime}_{-\beta,\beta}(\theta)F^{\prime\prime}_{-\beta,\beta}(\theta)\pm\sqrt{9(F^{\prime}_{-\beta,\beta}(\theta))^{2}(F^{\prime\prime}_{-\beta,\beta}(\theta))^{2}-8(F^{\prime}_{-\beta,\beta}(\theta))^{3}F^{\prime\prime\prime}_{-\beta,\beta}(\theta)}}{2F^{\prime\prime\prime}_{-\beta,\beta}(\theta)}\right)^{\prime}-F^{\prime}_{-\beta,\beta}(\theta)\\\
&=\frac{1}{2\sqrt{b_{\beta}(\theta)^{2}-4a_{\beta}(\theta)c_{\beta}(\theta)}}\Big{(}\left(a_{\beta}^{\prime}(\theta)b_{\beta}(\theta)-a_{\beta}(\theta)b^{\prime}_{\beta}(\theta)-2F^{\prime\prime}_{-\beta,\beta}(\theta)\right)\sqrt{b_{\beta}(\theta)^{2}-4a_{\beta}(\theta)c_{\beta}(\theta)}\\\
&\qquad\qquad\mp\left(b_{\beta}(\theta)^{2}a_{\beta}^{\prime}(\theta)-b_{\beta}(\theta)a_{\beta}(\theta)b_{\beta}^{\prime}(\theta)+2a_{\beta}(\theta)c^{\prime}_{\beta}(\theta)-2a_{\beta}(\theta)c_{\beta}(\theta)a^{\prime}_{\beta}(\theta)\right)\Big{)}\end{split}$
(A.30)

is only expressed in terms of $F_{-\beta,\beta}^{\prime}$ and its derivatives
that are classical functions. Zeros of $T^{\prime}_{\beta,\pm}$ must satisfy
(taking the square of the numerator of (A.30) to remove the $\pm$ sign):

$\left(b_{\beta}(\theta)^{2}a_{\beta}^{\prime}(\theta)-b_{\beta}(\theta)a_{\beta}(\theta)b_{\beta}^{\prime}(\theta)+2a_{\beta}(\theta)c^{\prime}_{\beta}(\theta)-2a_{\beta}(\theta)c_{\beta}(\theta)a^{\prime}_{\beta}(\theta)\right)^{2}\\\
-\left(a_{\beta}^{\prime}(\theta)b_{\beta}(\theta)-a_{\beta}(\theta)b^{\prime}_{\beta}(\theta)-2F^{\prime\prime}_{-\beta,\beta}(\theta)\right)^{2}\left(b_{\beta}(\theta)^{2}-4a_{\beta}(\theta)c_{\beta}(\theta)\right)=0.$

Replacing $F_{-\beta,\beta}^{\prime\prime\prime}$ and $F_{-\beta,\beta}^{(4)}$
in terms of $F_{-\beta,\beta}^{\prime}$ and $F_{-\beta,\beta}^{\prime\prime}$
using (A.18) gives after a tedious computation that the r.h.s. is of the form
$8F^{\prime\prime\prime}_{-\beta,\beta}(\theta)^{2}F^{\prime}_{-\beta,\beta}(\theta)S_{\beta}(\theta)$.
Therefore zeros of $T^{\prime}_{\beta,\pm}$ are among those of $S_{\beta}$.
Simple asymptotic expansions around $\theta=0$ and $\theta\to+\infty$ provide
the following results:

$\begin{split}&\lim_{\theta\to+\infty}T_{\beta,\pm}(\theta)=0,\\\
&T_{\beta,+}(0)=-\frac{\left(\sqrt{2\pi}\left(2\phi(\beta)-1\right)(\beta^{2}-3)\exp\left(\frac{\beta^{2}}{2}\right)+6\beta\right)\exp\left(-\frac{\beta^{2}}{2}\right)\sqrt{2}}{2\sqrt{\pi}(\beta^{2}-3)}\,\,,\forall\,\beta\in\mathbb{R}_{>0}\setminus\\{\sqrt{3}\\},\\\
&T^{\prime}_{\beta,+}(\theta)=-\frac{2\beta}{3\sqrt{2\pi}}e^{-\frac{1}{2}\beta^{2}}\theta+O(\theta^{2})\,\,,\,\,\forall\,\beta\in\mathbb{R}_{>0}\setminus\\{\sqrt{3}\\},\\\
&T_{\beta,-}(0)=-2\Phi(\beta)+1+O(\theta^{2})\,\,,\,\,\forall\,\beta\in\mathbb{R}_{>0}\setminus\\{\sqrt{3}\\},\\\
&T^{\prime}_{\beta,-}(\theta)=-\frac{4\beta^{5}}{3\sqrt{2\pi}(\beta^{2}-3)^{2}}e^{-\frac{1}{2}\beta^{2}}\theta+O(\theta^{2})\,\,,\,\,\forall\,\beta\in\mathbb{R}_{>0}\setminus\\{\sqrt{3}\\},\\\
&T_{\sqrt{3},+}(\theta)=\frac{3e^{-\frac{3}{2}}\sqrt{6}}{\sqrt{\pi}\theta^{2}}+O(1),\\\
&T^{\prime}_{\sqrt{3},+}(\theta)=-\frac{6e^{-\frac{3}{2}}\sqrt{6}}{\sqrt{\pi}\theta^{3}}+O(\theta),\\\
&T_{\sqrt{3},-}(\theta)=-2\Phi(\sqrt{3})+1+O(\theta^{2}),\\\
&T^{\prime}_{\sqrt{3},-}(\theta)=-\frac{1}{3\sqrt{\pi}}e^{-\frac{3}{2}}\sqrt{6}\theta+O(\theta^{2}).\end{split}$
(A.31)

In particular, in all cases, we get that the functions
$\left(T^{\prime}_{\beta,-},T^{\prime}_{\beta,+}\right)$ are always strictly
negative in a positive neighborhood of $\theta=0$ for any value of $\beta>0$.
Let us show that proving that $S_{\beta}$ is strictly positive on
$\mathbb{R}_{>0}$ is a sufficient condition to get that both functions
$(T_{\beta,-},T_{\beta,+})$ do not vanish on $\mathbb{R}_{>0}$. Indeed, if we
assume that $S_{\beta}(\theta)$ is strictly positive on $\mathbb{R}_{>0}$ then
it implies that $T^{\prime}_{\beta,\pm}$ cannot vanish. Depending on the value
of $\beta$, we have three cases:

  * •

For $\beta<\sqrt{3}$: we have that $T^{\prime}_{\beta,\pm}$ is strictly
negative on $(0,\theta_{1})$ so that $T_{\beta,\pm}$ is a strictly decreasing
function on $(0,\theta_{1})$. Note that $T_{\beta,-}$ diverges at $\theta_{1}$
and changes sign (because
$3F^{\prime}_{-\beta,\beta}(\theta_{1})F^{\prime\prime}_{-\beta,\beta}(\theta_{1})<0$
from Lemma A.1). Moreover, from (A.31), we have $T_{\beta,-}(0)<0$. This
implies that $\underset{\theta\to\theta_{1,-}}{\lim}T_{\beta,-}=-\infty$ and
$\underset{\theta\to\theta_{1,+}}{\lim}T_{\beta,-}=+\infty$. Therefore, the
sign of $T^{\prime}_{\beta,-}$ is necessarily negative on
$(\theta_{1},+\infty)$ so that $T_{\beta,-}$ is strictly decreasing on
$(\theta_{1},+\infty)$. Since its limit is zero at infinity, it remains
strictly positive on $(\theta_{1},+\infty)$. Thus, we conclude that
$T_{\beta,-}$ never vanishes on $\mathbb{R}_{>0}$. The situation for
$T_{\beta,+}$ is simpler. Indeed, $T_{\beta,+}$ is a smooth function at
$\theta=\theta_{1}$ (because
$3F^{\prime}_{-\beta,\beta}(\theta_{1})F^{\prime\prime}_{-\beta,\beta}(\theta_{1})<0$
from Lemma A.1). Therefore the sign of $T_{\beta,+}^{\prime}$ remains constant
on $\mathbb{R}_{>0}$ and thus $T_{\beta,+}$ is a decreasing function on
$\mathbb{R}_{>0}$. Since its limit at infinity is null, we get that it remains
strictly positive on $\mathbb{R}_{>0}$. In both cases, $T_{\beta,\pm}$ does
not vanish on $\mathbb{R}_{>0}$.

  * •

For $\beta=\sqrt{3}$: we have that $T^{\prime}_{\sqrt{3},\pm}$ is strictly
negative in $(0,\theta_{1})$ so that $T_{\sqrt{3},\pm}$ is a strictly
decreasing function on $(0,\theta_{1})$. Note that $T_{\sqrt{3},-}$ diverges
at $\theta_{1}$ and changes sign (because
$3F^{\prime}_{-\sqrt{3},\sqrt{3}}(\theta_{1})F^{\prime\prime}_{-\sqrt{3},\sqrt{3}}(\theta_{1})<0$
from Lemma A.1). Moreover, from (A.31), we have $T_{\sqrt{3},-}(0)<0$. This
implies that $\underset{\theta\to\theta_{1,-}}{\lim}T_{\sqrt{3},-}=-\infty$
and $\underset{\theta\to\theta_{1,+}}{\lim}T_{\sqrt{3},-}=+\infty$. Therefore,
the sign of $T^{\prime}_{\sqrt{3},-}$ is necessarily negative on
$(\theta_{1},+\infty)$ so that $T_{\sqrt{3},-}$ is strictly decreasing on
$(\theta_{1},+\infty)$. Since its limit is zero at infinity, it remains
strictly positive on $(\theta_{1},+\infty)$. Thus, we conclude that
$T_{\sqrt{3},-}$ never vanishes on $\mathbb{R}_{>0}$. The situation for
$T_{\sqrt{3},+}$ is simpler. We have from (A.31) that $T_{\sqrt{3},+}(0)>0$
and $T_{\sqrt{3},+}$ is a smooth function at $\theta=\theta_{1}$ (because
$3F^{\prime}_{-\sqrt{3},\sqrt{3}}(\theta_{1})F^{\prime\prime}_{-\sqrt{3},\sqrt{3}}(\theta_{1})<0$
from Lemma A.1). Therefore the sign of $T_{\sqrt{3},+}^{\prime}$ remains
constant on $\mathbb{R}_{>0}$ and $T_{\sqrt{3},+}$ is a decreasing function on
$\mathbb{R}_{>0}$. Since its limit is null, we get that it remains strictly
positive on $\mathbb{R}_{>0}$. In both cases, $T_{\sqrt{3},\pm}$ does not
vanish on $\mathbb{R}_{>0}$.

  * •

For $\beta>\sqrt{3}$, we get that $T^{\prime}_{\beta,\pm}$ is strictly
negative in $(0,\theta_{1})$. Note that $T_{\beta,+}$ is smooth at
$\theta_{2}$ but diverges and changes sign at $\theta=\theta_{1}$ (because
$3F^{\prime}_{-\beta,\beta}(\theta_{1})F^{\prime\prime}_{-\beta,\beta}(\theta_{1})<0$
and
$3F^{\prime}_{-\beta,\beta}(\theta_{2})F^{\prime\prime}_{-\beta,\beta}(\theta_{2})>0$
from Lemma A.1). Moreover, we have from (A.31) $T_{\beta,+}(0)<0$ so that
$\underset{\theta\to\theta_{1,-}}{\lim}T_{\beta,+}=-\infty$ and
$\underset{\theta\to\theta_{1,+}}{\lim}T_{\beta,+}=+\infty$. Therefore
$T_{\beta,+}$ must be decreasing on $(\theta_{1},+\infty)$ and since its limit
is $0$ at infinity we conclude that it is strictly positive on
$(\theta_{1},+\infty)$. The situation for $T_{\beta,-}$ is similar. It is
smooth at $\theta_{1}$ but diverges and changes sign at $\theta_{2}$ (because
$3F^{\prime}_{-\beta,\beta}(\theta_{1})F^{\prime\prime}_{-\beta,\beta}(\theta_{1})<0$
and
$3F^{\prime}_{-\beta,\beta}(\theta_{2})F^{\prime\prime}_{-\beta,\beta}(\theta_{2})>0$
from Lemma A.1). We have $T_{\beta,-}(0)<0$ so that it is strictly negative on
$(0,\theta_{2})$ and
$\underset{\theta\to\theta_{2,-}}{\lim}T_{\beta,-}=-\infty$ and
$\underset{\theta\to\theta_{2,+}}{\lim}T_{\beta,-}=+\infty$. Therefore it must
strictly decrease on $(\theta_{2},+\infty)$ and since its limit is $0$ at
infinity, we end up with the fact that it is strictly positive on
$(\theta_{2},+\infty)$. In both cases, $T_{\beta,\pm}$ does not vanish on
$\mathbb{R}_{>0}$.

Thus, we conclude that proving that $S_{\beta}$ is strictly positive on
$\mathbb{R}_{>0}$ is a sufficient condition to get that $T_{\beta,\pm}$ does
not vanish on $\mathbb{R}_{>0}\setminus\mathcal{R}_{\beta}$. Let us prove that
the fact that $S_{\beta}$ is strictly positive on $\mathbb{R}_{>0}$ also
excludes the zeros of $F^{\prime\prime\prime}_{-\beta,\beta}$ as potential
zeros of $Z_{\beta}$. Under the assumption that $S_{\beta}$ is strictly
positive on $\mathbb{R}_{>0}$ we have

  * •

For $\beta>\sqrt{3}$, $F^{\prime\prime\prime}_{-\beta,\beta}$ has exactly two
zeros denoted $\theta_{1}<\theta_{2}$ on $\mathbb{R}_{>0}$. As explained
above, $T_{\beta,+}$ is a smooth function at $\theta_{2}$ and we have
$T_{\beta,+}(\theta_{2})=-\frac{c_{\beta}(\theta_{2})}{b_{\beta}(\theta_{2})}-F_{-\beta,\beta}(\theta_{2})$.
We have proved above that $T_{\beta,+}$ is strictly positive on
$(\theta_{1},+\infty)$ so that
$T_{\beta,+}(\theta_{2})=-\frac{c_{\beta}(\theta_{2})}{b_{\beta}(\theta_{2})}-F_{-\beta,\beta}(\theta_{2})>0$.
However, if $\theta_{2}$ was a zero of $Z_{\beta}$, then we would have from
(A.26)
$F_{-\beta,\beta}(\theta_{2})+\frac{c_{\beta}(\theta_{2})}{b_{\beta}(\theta_{2})}=0$
leading to a contradiction. Therefore $\theta_{2}$ is not a zero of
$Z_{\beta}$. Similarly, we have proved that $T_{\beta,-}$ is a smooth function
at $\theta_{1}$ and we have
$T_{\beta,-}(\theta_{1})=-\frac{c_{\beta}(\theta_{1})}{b_{\beta}(\theta_{1})}-F_{-\beta,\beta}(\theta_{1})$.
Moreover, we have proved above that $T_{\beta,-}$ is strictly negative on
$(0,\theta_{2})$ so that
$T_{\beta,-}(\theta_{1})=-\frac{c_{\beta}(\theta_{1})}{b_{\beta}(\theta_{1})}-F_{-\beta,\beta}(\theta_{1})<0$.
However, if $\theta_{1}$ was a zero of $Z_{\beta}$, then we would have from
(A.26)
$F_{-\beta,\beta}(\theta_{1})+\frac{c_{\beta}(\theta_{1})}{b_{\beta}(\theta_{1})}=0$
leading to a contradiction. Therefore $\theta_{1}$ is not a zero of
$Z_{\beta}$.

  * •

For $\beta\leq\sqrt{3}$: $F^{\prime\prime\prime}_{-\beta,\beta}$ has exactly
one zero denoted $\theta_{1}$ on $\mathbb{R}_{>0}$. As explained above,
$T_{\beta,+}$ is a smooth function at $\theta=\theta_{1}$ and it is strictly
positive on $\mathbb{R}_{>0}$. In particular we have
$T_{\beta,+}(\theta_{1})=-\frac{c_{\beta}(\theta_{1})}{b_{\beta}(\theta_{1})}-F_{-\beta,\beta}(\theta_{1})>0$.
However, if $\theta_{1}$ was a zero of $Z_{\beta}$, then we would have from
(A.26)
$F_{-\beta,\beta}(\theta_{1})+\frac{c_{\beta}(\theta_{1})}{b_{\beta}(\theta_{1})}=0$
leading to a contradiction. Therefore $\theta_{1}$ is not a zero of
$Z_{\beta}$.

This concludes the proof of Proposition A.2. ∎

###  A.3 Proof of the sufficient condition

From Proposition A.2, a sufficient condition to have strict concavity of
$h_{-\beta,\beta}$ on $\mathbb{R}_{>0}$ is to prove that $S_{\beta}$ is
strictly positive on $\mathbb{R}_{>0}$. In this section, we shall propose a
sufficient condition to obtain this result. Let us first observe that:

$\displaystyle F^{\prime}(\theta)$
$\displaystyle=-\frac{2}{\sqrt{2\pi}}e^{-\frac{1}{2}\theta^{2}-\frac{1}{2}\beta^{2}}\sinh(\beta\theta),$
$\displaystyle F^{\prime\prime}(\theta)$
$\displaystyle=\frac{2}{\sqrt{2\pi}}e^{-\frac{1}{2}\theta^{2}-\frac{1}{2}\beta^{2}}\left(\theta\sinh(\beta\theta)-\beta\cosh(\beta\theta)\right).$

The term
$\frac{2}{\sqrt{2\pi}}e^{-\frac{1}{2}\theta^{2}-\frac{1}{2}\beta^{2}}$ factors
out of $S_{\beta}$ because we have homogeneous powers. Therefore, let us thus
rewrite
$S_{\beta}(\theta)=2\left(\frac{2}{\sqrt{2\pi}}e^{-\frac{1}{2}\theta^{2}-\frac{1}{2}\beta^{2}}\right)^{5}\tilde{S}_{\beta}(\theta)$.
We have:

$\displaystyle\tilde{S}_{\beta}(\theta)$
$\displaystyle=\sinh(5\beta\theta)+\left(4\beta^{6}+4(3\theta^{2}+6)\beta^{4}+12\beta^{2}-5\right)\sinh(3\beta\theta)$
$\displaystyle+\left(-12\beta^{6}+4(3\theta^{2}+18)\beta^{4}-36\beta^{2}+10)\right)\sinh(\beta\theta)$
$\displaystyle-4\left(3\beta^{2}+\theta^{2}+6\right)\beta^{3}\theta\cosh(3\beta\theta)$
$\displaystyle+4\left(-33\beta^{2}+\theta^{2}+6\right)\beta^{3}\theta\cosh(\beta\theta).$

Proving that $S_{\beta}$ is strictly positive on $\mathbb{R}_{>0}$ is
equivalent to prove that $\tilde{S}_{\beta}$ is strictly positive on
$\mathbb{R}_{>0}$. Let us perform the following change of variables:
$s=\beta\theta$ and define
$A_{\beta}(s)\coloneqq\tilde{S}_{\beta}\left(\frac{s}{\beta}\right)$. Since
$\beta>0$, it is obvious that proving that $\tilde{S}_{\beta}$ is strictly
positive on $\mathbb{R}_{>0}$ is equivalent to proving that $A_{\beta}$ is
also strictly positive on $\mathbb{R}_{>0}$. We obtain:

$\displaystyle A_{\beta}(s)$
$\displaystyle=\sinh(5s)+\left(4\beta^{6}+24\beta^{4}+12s^{2}\beta^{2}+12\beta^{2}-5\right)\sinh(3s)$
$\displaystyle+\left(-12\beta^{6}+72\beta^{4}+12s^{2}\beta^{2}-36\beta^{2}+10\right)\sinh(s)$
$\displaystyle-4(3\beta^{4}+6\beta^{2}+s^{2})s\cosh(3s)+4(-33\beta^{4}+6\beta^{2}+s^{2})s\cosh(s).$

Let us observe that $B_{s}(\mu)\coloneqq A_{\beta}(s)$ is a polynomial of
degree $3$ in $\mu\coloneqq\beta^{2}$:

$\displaystyle B_{s}(\mu)$ $\displaystyle\coloneqq A_{\mu^{2}}(s)$
$\displaystyle=4\left(\sinh(3s)-3\sinh(s)\right)\mu^{3}$
$\displaystyle\quad+4\left(6\sinh(3s)+18\sinh(s)-33s\cosh(s)-3s\cosh(3s)\right)\mu^{2}$
$\displaystyle\quad+4\left(3(s^{2}+1)\sinh(3s)+3(s^{2}-3)\sinh(s)+6s\cosh(s)-6s\cosh(3s)\right)\mu$
$\displaystyle\quad+\sinh(5s)-5\sinh(3s)+10\sinh(s)+4s^{3}\cosh(s)-4s^{3}\cosh(3s).$

Let us first notice that
$B_{s}(0)=\sinh(5s)-5\sinh(3s)+10\sinh(s)+4s^{3}\cosh(s)-4s^{3}\cosh(3s)>0$
for any $s>0$ so that $B_{s}$ is strictly positive in a positive neighborhood
of $s=0$. Indeed, we have the following lemma.

######  Lemma A.2.

For any $s>0$, we have

$\sinh(5s)-5\sinh(3s)+10\sinh(s)+4s^{3}\cosh(s)-4s^{3}\cosh(3s)>0.$ (A.32)

###### Proof of Lemma A.2.

One may rewrite

$\displaystyle\sinh(5s)-5\sinh(3s)+10\sinh(s)+4s^{3}\cosh(s)-4s^{3}\cosh(3s)=16\sinh(s)^{2}\left(\sinh(s)^{3}-s^{3}\cosh(s)\right)$
$\displaystyle=16\sinh(s)^{2}\cosh(s)\left(\frac{\sinh(s)^{3}}{\cosh(s)}-s^{3}\right).$

Let us then define $a(s)\coloneqq\frac{\sinh(s)^{3}}{\cosh(s)}-s^{3}$ for
$s>0$. We have

$\displaystyle a^{\prime}(s)$
$\displaystyle=\frac{2\cosh(s)^{4}+(-3s^{2}-1)\cosh(s)^{2}-1}{\cosh(s)^{2}},$
$\displaystyle a^{\prime\prime}(s)$
$\displaystyle=\frac{(4\sinh(s)\cosh(s)^{4}-6s\cosh(s)^{3}+2\sinh(s))}{\cosh(s)^{3}},$
$\displaystyle a^{(3)}(s)$
$\displaystyle=\frac{2\sinh(s)^{4}(4\cosh(s)^{2}+3)}{\cosh(s)^{4}}>0,$

so that $a^{\prime\prime}$ is strictly increasing on $\mathbb{R}_{>0}$ and
since $a^{\prime\prime}(0)=0$, we get that $a^{\prime\prime}$ is strictly
positive on $\mathbb{R}_{>0}$. This implies that $a^{\prime}$ is strictly
increasing on $\mathbb{R}_{>0}$ and since $a^{\prime}(0)=0$ we end up with
$a^{\prime}$ strictly positive on $\mathbb{R}_{>0}$. In the end, $a$ is
strictly increasing on $\mathbb{R}_{>0}$ and $a(0)=0$ so that $a$ is strictly
positive on $\mathbb{R}_{>0}$, ending the proof of Lemma A.2. ∎

Then, we observe that the leading coefficient of the polynomial $\mu\mapsto
B_{s}(\mu)$ is given by $\sinh(3s)-3\sinh(s)$ and is obviously strictly
positive for any $s>0$. We want to prove that for any $s>0$, $\mu\mapsto
B_{s}(\mu)$ is a strictly positive function on $\mathbb{R}_{>0}$. We have:

$\displaystyle B_{s}^{\prime}(\mu)$
$\displaystyle=12\Big{[}\left(\sinh(3s)-3\sinh(s)\right)\mu^{2}$
$\displaystyle+2\left(2\sinh(3s)+6\sinh(s)-11s\cosh(s)-s\cosh(3s)\right)\mu$
$\displaystyle+(s^{2}+1)\sinh(3s)+(s^{2}-3)\sinh(s)+2s\cosh(s)-2s\cosh(3s)\Big{]}.$

The discriminant $144\Delta(s)$ of this polynomial of degree two is given by
$\Delta(s)\coloneqq-64\cosh(s)\tilde{\Delta}(s)$ with

$\displaystyle\tilde{\Delta}(s)$
$\displaystyle=-3\cosh(s)^{5}+2s\sinh(s)\cosh(s)^{4}-6s^{2}\cosh(s)^{3}+14s\sinh(s)\cosh(s)^{2}$
$\displaystyle-3\cosh(s)^{3}-3s^{2}\cosh(s)+2s\sinh(s)+6\cosh(s).$

Let us assume that $\tilde{\Delta}$ is strictly positive on $\mathbb{R}_{>0}$
so that $\Delta(s)$ is strictly negative, i.e. $B_{s}^{\prime}$ does not
vanish on $\mathbb{R}$ and thus remains strictly positive on $\mathbb{R}_{>0}$
(because its leading coefficient is strictly positive). Since we have proved
in Lemma A.2 that $B_{s}(0)>0$, we obtain that $B_{s}$ is strictly positive on
$\mathbb{R}_{>0}$ which is equivalent to say that for any
$(\beta,s)\in\mathbb{R}_{>0}\times\mathbb{R}_{>0}$: $A_{\beta}(s)$ is strictly
positive. Therefore, a sufficient condition to obtain that $S_{\beta}$ is
strictly positive on $\mathbb{R}_{>0}$ is that $\tilde{\Delta}$ is strictly
positive on $\mathbb{R}_{>0}$. In order to prove this sufficient condition,
let us observe that

$\displaystyle\tilde{\Delta}^{\prime}(s)$
$\displaystyle=\sinh(s)\Big{[}10s\cosh(s)^{3}\sinh(s)-13\cosh(s)^{4}-18s^{2}\cosh(s)^{2}$
$\displaystyle+32s\sinh(s)\cosh(s)+5\cosh(s)^{2}-3s^{2}+8\Big{]}$
$\displaystyle\coloneqq\sinh(s)Q(s),$

with

$\displaystyle Q^{\prime}(s)$
$\displaystyle=2\sinh(s)\Big{[}20s\sinh(s)\cosh(s)^{2}-21\cosh(s)^{3}-18s^{2}\cosh(s)+19s\sinh(s)+21\cosh(s)\Big{]}$
$\displaystyle\coloneqq 2\sinh(s)R(s),$

with

$R(s)\coloneqq
20s\sinh(s)\cosh(s)^{2}-21\cosh(s)^{3}-18s^{2}\cosh(s)+19s\sinh(s)+21\cosh(s).$

Let us assume that $R(s)>0$ for any $s>0$, then $Q$ is a strictly increasing
function on $\mathbb{R}_{>0}$ with $Q(0)=0$ so that it is strictly positive on
$\mathbb{R}_{>0}$. This implies that $\tilde{\Delta}$ is a strictly increasing
function on $\mathbb{R}_{>0}$ so that since $\tilde{\Delta}(0)=0$ we get that
$\tilde{\Delta}$ is also a strictly positive function on $\mathbb{R}_{>0}$.
Therefore, we have the following result.

######  Proposition A.3.

Let $\beta>0$. A sufficient condition to obtain that $S_{\beta}$ is strictly
positive on $\mathbb{R}_{>0}$ is to prove that $s\mapsto R(s)\coloneqq
20s\sinh(s)\cosh(s)^{2}-21\cosh(s)^{3}-18s^{2}\cosh(s)+19s\sinh(s)+21\cosh(s)$
is strictly positive on $\mathbb{R}_{>0}$.

Finally we may prove this sufficient condition using the following
proposition.

######  Proposition A.4.

Let $\beta>0$. The function $s\mapsto R(s)\coloneqq
20s\sinh(s)\cosh(s)^{2}-21\cosh(s)^{3}-18s^{2}\cosh(s)+19s\sinh(s)+21\cosh(s)$
is strictly positive on $\mathbb{R}_{>0}$.

###### Proof of Proposition A.4.

Let us first rewrite:

$\displaystyle R(s)$
$\displaystyle=20s\sinh(s)\cosh(s)^{2}-21\cosh(s)^{3}-18s^{2}\cosh(s)+19s\sinh(s)+21\cosh(s)$
$\displaystyle=20s\sinh(s)\cosh(s)^{2}-21\cosh(s)\sinh(s)^{2}-18s^{2}\cosh(s)+19s\sinh(s)$
$\displaystyle=-\left(18\cosh(s)s^{2}-(20\sinh(s)\cosh(s)^{2}+19\sinh(s))s+21\cosh(s)\sinh(s)^{2}\right),$

and observe that

$R(s)=\frac{64}{15}s^{6}+O(s^{7}),$

so that $R$ is strictly positive in a positive neighborhood of zero. Moreover,
equation $R(s)=0$ may be seen as a polynomial of degree two in $s$. In other
words $R(s)=0$ with $s>0$ is equivalent to

$s=\frac{20\sinh(s)\cosh(s)^{2}+19\sinh(s)}{36\cosh(s)}\pm\frac{\sinh(s)}{36\cosh(s)}\sqrt{400\cosh(s)^{4}-752\cosh(s)^{2}+361}.$
(A.33)

Let us define

$R_{\pm}(s)\coloneqq
s-\frac{20\sinh(s)\cosh(s)^{2}+19\sinh(s)}{36\cosh(s)}-\pm\frac{\sinh(s)}{36\cosh(s)}\sqrt{400\cosh(s)^{4}-752\cosh(s)^{2}+361}.$

Note in particular that

$\begin{split}R_{+}(s)&=\frac{64}{45}s^{5}+O(s^{6}),\\\
R_{-}(s)&=-\frac{1}{6}s-\frac{7}{18}s^{3}+O(s^{4}).\end{split}$ (A.34)

Moreover, we have

$R_{\pm}^{\prime}(s)=\frac{(-40\cosh(s)^{4}+56\cosh(s)^{2}-19)\sqrt{400\cosh(s)^{4}-752\cosh(s)^{2}+361}\pm(800\cosh(s)^{6}-1152\cosh(s)^{4}+361)}{36\cosh(s)^{2}\sqrt{400\cosh(s)^{4}-752\cosh(s)^{2}+361}}.$

Therefore, zeros of $R^{\prime}_{\pm}$ must satisfy

$\left(-40\cosh(s)^{4}+56\cosh(s)^{2}-19\right)^{2}\left(400\cosh(s)^{4}-752\cosh(s)^{2}+361\right)-\left(800\cosh(s)^{6}-1152\cosh(s)^{4}+361\right)^{2}=0,$

which is equivalent to

$-576\sinh(s)^{4}\cosh(s)^{2}\left(2000\cosh(s)^{4}-3781\cosh(s)^{2}+1805\right)=0.$

Since the discriminant of $2000X^{2}-3781X+1805$ is negative (equal to
$-144039$), we conclude that $R_{\pm}^{\prime}$ does not vanish on
$\mathbb{R}_{>0}$ and thus has a constant sign on $\mathbb{R}_{>0}$. In
particular, from (A.34), we get that $R^{\prime}_{+}$ is strictly positive on
$\mathbb{R}_{>0}$ so that $R_{+}$ is strictly increasing and since
$R_{+}(0)=0$ we conclude that $R_{+}$ is strictly positive on
$\mathbb{R}_{>0}$. Similarly, from (A.34) $R^{\prime}_{-}$ is strictly
negative on $\mathbb{R}_{>0}$ so that $R_{-}$ is strictly decreasing and since
$R_{-}(0)=0$ we conclude that $R_{-}$ is strictly negative on
$\mathbb{R}_{>0}$. In both cases, the function $R_{\pm}$ do not vanish on
$\mathbb{R}_{>0}$ so that (A.33) cannot be satisfied on $\mathbb{R}_{>0}$ and
eventually $R$ does not vanish on $\mathbb{R}_{>0}$. Since we have shown that
it is strictly positive in a positive neighborhood of zero and since it is a
smooth function, we conclude that function $R$ is strictly positive on
$\mathbb{R}_{>0}$. This concludes the proof of Proposition A.4. ∎

Finally, we conclude from Proposition A.3 and Proposition A.4 that for any
$\beta>0$, $S_{\beta}$ is a strictly positive function on $\mathbb{R}_{>0}$ so
that from Proposition A.2 we obtain that $h_{-\beta,\beta}$ is a strictly
concave function on $\mathbb{R}_{>0}$. This concludes the proof of Lemma 2.4
stated in the main part of the paper. ∎

##  Appendix B Proofs for truncated exponential random variables

######  Lemma B.1.

For any $\epsilon>0$, we have

$P(\epsilon)\coloneqq
2e^{3\epsilon}-(\epsilon^{3}+6)e^{2\epsilon}+(6-\epsilon^{3})e^{\epsilon}-2>0.$

###### Proof of Lemma B.1.

We have:

$\displaystyle P^{\prime}(\epsilon)$
$\displaystyle=e^{3\epsilon}\left((6-3\epsilon^{2}-\epsilon^{3})e^{-2\epsilon}-(12+3\epsilon^{2}+2\epsilon^{3})e^{-\epsilon}+6\right)\coloneqq
e^{3\epsilon}Q(\epsilon),$ $\displaystyle Q^{\prime}(\epsilon)$
$\displaystyle=e^{-\epsilon}\left[\left(2\epsilon^{3}+3\epsilon^{2}-6\epsilon-12\right)e^{-\epsilon}+2\epsilon^{3}-3\epsilon^{2}-6\epsilon+12\right]\coloneqq
e^{-\epsilon}R(\epsilon).$

It is then straightforward to compute
$R^{(4)}(\epsilon)=e^{-\epsilon}\epsilon(2\epsilon^{2}-21\epsilon+42)$ whose
roots are $\epsilon_{\pm}=\frac{21}{4}\pm\sqrt{105}{4}$. Thus
$R^{\prime\prime\prime}$ is strictly increasing on $(0,\epsilon_{-})$ and then
strictly decreasing on $(\epsilon_{-},\epsilon_{+})$ and finally increasing on
$(\epsilon_{+},+\infty)$. Since $R^{\prime\prime\prime}(0)=0$ and
$R^{\prime\prime\prime}(\epsilon_{+})>0$ we conclude that
$R^{\prime\prime\prime}$ is strictly positive on $(0,+\infty)$. Thus,
$R^{\prime\prime}$ is strictly increasing on $\mathbb{R}_{\geq 0}$ and since
$R^{\prime\prime}(0)=0$ it is strictly positive on $\mathbb{R}_{>0}$. Hence
$R^{\prime}$ is strictly increasing on $\mathbb{R}_{\geq 0}$ and since
$R^{\prime}(0)=0$ it is strictly positive on $\mathbb{R}_{>0}$. Eventually,
$R$ is strictly increasing on $\mathbb{R}_{\geq 0}$ and since $R(0)=0$, we get
that $R$ is strictly positive on $\mathbb{R}_{>0}$ so that $Q$ is strictly
increasing on $\mathbb{R}_{\geq 0}$. In the end, since $Q(0)=0$, $Q$ is
strictly positive on $\mathbb{R}_{>0}$ so that $P$ is strictly increasing on
$\mathbb{R}_{\geq 0}$. Since $P(0)=0$, we conclude that $P$ is strictly
positive on $\mathbb{R}_{>0}$, which concludes the proof of Lemma B.1. ∎

###  B.1 Proving that the truncated exponential is never strictly sub-Gaussian

Let us study the sign of

$\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}-\text{Var}[Y_{\mathrm{T}}]=\frac{e^{\alpha+\beta}}{2\left(e^{\beta}-e^{\alpha}\right)}\left((\alpha-\beta-4)e^{\alpha-\beta}+(\beta-\alpha-4)e^{\beta-\alpha}+2(\beta-\alpha)^{2}+8\right).$

Observe that the last term is only a function of $\epsilon=\beta-\alpha>0$.
Thus, the sign of
$\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}-\text{Var}[Y_{\mathrm{T}}]$ is the
same as the sign of the function $K$ on $\mathbb{R}_{>0}$ defined by

$K(\epsilon)\coloneqq(\epsilon-4)e^{\epsilon}-(\epsilon+4)e^{-\epsilon}+2\epsilon^{2}+8=2\epsilon\sinh\epsilon-8\cosh\epsilon+2\epsilon^{2}+8.$

We have:

$\displaystyle K^{\prime}(\epsilon)$
$\displaystyle=2\epsilon\cosh\epsilon-6\sinh\epsilon+4\epsilon,$
$\displaystyle K^{\prime\prime}(\epsilon)$
$\displaystyle=2\epsilon\sinh\epsilon-4\cosh\epsilon+4,$ $\displaystyle
K^{\prime\prime\prime}(\epsilon)$
$\displaystyle=2\left(\epsilon\cosh\epsilon-\sinh\epsilon\right).$

It is obvious that $K^{\prime\prime\prime}$ is strictly positive on
$\mathbb{R}_{>0}$. Since $K^{\prime\prime}(0)=0$, $K^{\prime}(0)=0$ and
$K(0)=0$ we get successively that $K^{\prime\prime}$, $K^{\prime}$ and $K$ are
strictly increasing and strictly positive on $\mathbb{R}_{>0}$. Thus, we
conclude that for all $\alpha<\beta$, we have
$\|Y_{\mathrm{T}}\|_{{\mathrm{vp}}}^{2}>\text{Var}[Y_{\mathrm{T}}]$ so that
the truncated exponential is never strictly sub-Gaussian.

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