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

Mathias Barreto, Olivier Marchal, Julyan Arbel

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Mathias Barreto, Olivier Marchal, Julyan Arbel. Optimal sub-Gaussian variance proxy for truncated Gaussian and exponential random variables. Statistics and Probability Letters, 2026, 228, pp.110555. 10.1016/j.spl.2025.110555. hal-04503494v2

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

Mathias Barreto<sup>a</sup>, Olivier Marchal<sup>b</sup>, Julyan Arbel<sup>c</sup>

<sup>a</sup>University of British Columbia, Department of Mathematics, Vancouver, Canada

<sup>b</sup>Université Jean Monnet, Institut Camille Jordan CNRS UMR 5208, Institut Universitaire de France, 23 Rue du Dr. Michelon, 42000 Saint-Etienne, France,

<sup>c</sup>Univ. Grenoble Alpes, Inria, CNRS, Grenoble INP, LJK, 38000 Grenoble, France,

## Abstract

This paper establishes the optimal sub-Gaussian variance proxy for truncated Gaussian and truncated exponential random variables. The proofs are based initially on reducing each distribution to their standarized versions. Geometrically, for the normal distribution, our argument consists of fitting a parabola to another parabola-looking function, which emerges from its moment generating function. For the exponential case, we show that the optimal variance proxy is the unique solution to a pair of equations and then provide this solution explicitly. Moreover, 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-Gaussianity.

**Keywords:** Sub-Gaussian random variables, Optimal variance proxy, Truncated distributions

## 1. Introduction

The sub-Gaussian property, a fundamental characteristic extensively explored in seminal works such as [7, 5, 33], 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 [22, 24, 26, 21, 34, 5, 4, 31, 3], random series in relation to the geometry of Banach spaces [32, 12], spectral properties of random matrices [28, 35], or Bayesian statistics [18, 8, 40, 27, 38]. For instance, the optimal variance proxy (see Definition 1.1) appears explicitly in convergence results of the Langevin Monte Carlo algorithm [see Theorem 6.2.8 in 10], or in bandit algorithms [14], just to mention a few. More broadly, the sub-Gaussian property holds paramount importance in machine learning and artificial intelligence applications [17, 9, 20, 30, 15, 11, 42, 16, 13].

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 $s \ge 0$ such that for all $\theta \in \mathbb{R}$:

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

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 $\|X\|_{\text{vp}}^2$. A Taylor expansion of the left hand side of Definition 1.1 at $\theta = 0$ immediately shows that $\text{Var}[X] \le \|X\|_{\text{vp}}^2$, and a random variable satisfying $\text{Var}[X] = \|X\|_{\text{vp}}^2$ is called *strictly sub-Gaussian*, a terminology that seems to have been used firstly by [7].
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 37]. 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:

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

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

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

This shows that a sub-Gaussian random variable has finite moments of any order larger than or equal to 1.

**Truncated random variables.** The focus of this paper lies on the relevance of the sub-Gaussian property to truncated Gaussian and truncated exponential distributions. 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 $X_T$) inside the interval $(a, b)$ (for $a < b$, and possibly $a = -\infty$ or $b = +\infty$) has the form:

$$ f_{X_T}(x) = \begin{cases} \frac{f_X(x)}{F_X(b) - F_X(a)} & \text{for } x \in (a, b), \\ 0 & \text{otherwise.} \end{cases} $$

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 [2], reinforcement learning and bandit problems, for representing action probabilities or rewards that are bounded within certain limits [6, 25, 36], Bayesian statistics, to represent prior knowledge or beliefs about parameters that are restricted to certain intervals [19], or sampling procedures, to reduce Monte Carlo error [23, 41]. 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.

**Hoeffding's Lemma.** Any finitely supported random variable $X$ (satisfying $a \le X \le b$ almost surely) is sub-Gaussian. If we denote the half-interval by $\ell = \frac{b-a}{2}$, we also know that the optimal variance proxy satisfies $\|X\|_{\text{vp}}^2 \le s_H^2 := \ell^2$, where the upper bound $s_H^2$ directly follows from Hoeffding's Lemma:

**Lemma 1.2 (Hoeffding's Lemma).** Let $X$ be a real-valued random variable such that $a \le X \le b$ almost surely. Then for all $\theta \in \mathbb{R}$ it holds that: $\mathbb{E}[\exp(\theta(X - \mathbb{E}[X]))] \le \exp(\frac{\theta^2(b-a)^2}{8})$.

However, this bound is known to be potentially very loose, as can be seen in the two cases studied here. In order to compare the optimal variance proxy values derived in the next two sections against Hoeffding's upper bound $s_H^2 = \ell^2$, we studied their limiting behavior in the cases of small or large truncation range $\beta - \alpha = 2\ell$. In both limits, the behavior does not depend anymore on the location of the truncation, but only on $\ell$. The limiting behavior (in the sense of asymptotical equivalence) is summarized in Table 1. It shows that our optimal variance proxy is superior to $s_H^2 = \ell^2$ by a multiplicative factor of $1/3$ when $\ell \to 0$. It is even more superior in the $\ell \to \infty$ limit.

<table><caption>Table 1: Comparison of the optimal variance proxy (vp) for the Gaussian case (first column, derived in Section 2) and the exponential case (second column, derived in Section 3) with Hoeffding's bound of Lemma 1.2 (third column). Limiting cases of small truncation range $\beta - \alpha = 2\ell \to 0^+$ (first row) and large range $\ell \to \infty$ are compared. We remark that these are the results for the standarized densities ($\mu = 0, \sigma = 1$ for the Gaussian case, and $\lambda = 1$ for the exponential case).</caption><thead><tr><th></th><th>Gaussian optimal vp</th><th>Exponential optimal vp</th><th>Hoeffding's bound $s_H^2$</th></tr></thead><tbody><tr><td>$\ell \to 0^+$</td><td>$\frac{\ell^2}{3}$</td><td>$\frac{\ell^2}{3}$</td><td>$\ell^2$</td></tr><tr><td>$\ell \to \infty$</td><td>1</td><td>$\ell$</td><td>$\ell^2$</td></tr></tbody></table>
**Contributions.** In general, establishing that a random variable is sub-Gaussian might be easy, but com- plication stems from obtaining the optimal variance proxy. This has been achieved for the most commonly used distributions, including the beta and Dirichlet distributions [29], Bernoulli, binomial, Kumaraswamy, and triangular distributions [1]. In the present work, we focus on establishing the optimal variance proxy for truncated Gaussian (Theorem 2.1) and truncated exponential distributions (Theorem 3.1).

The proofs are based initially on showing that it suffices to consider the standardized cases or $\mu = 0, \sigma^2 = 1$ for the truncated normal case, and $\lambda = 1$ for the truncated exponential case. Subsequently, for the truncated normal, the argument to obtain a closed-form expression of the variance proxy relies on proving the concavity of a function inside a region and then separating the cases where the truncation intervals are finite and/or symmetric towards the mean. For the exponential case, we show that the optimal variance proxy is the unique solution to a pair of equations and then provide this solution explicitly.

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 random variable truncated along a finite interval is sub-Gaussian by Hoeffding's Lemma, see Lemma 1.2.

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 the moment generating function is $\mathbb{E}[e^{\theta X_T}] = \exp\{\theta\mu + \theta^2 \frac{\sigma^2}{2}\} \left[\frac{\Phi(\beta - \sigma\theta) - \Phi(\alpha - \sigma\theta)}{\Phi(\beta) - \Phi(\alpha)}\right]$, while mean and variance are given by:

$$ \mathbb{E}[X_T] = \mu + \frac{\phi(\alpha) - \phi(\beta)}{\Phi(\beta) - \Phi(\alpha)}\sigma, \quad \mathrm{Var}[X_T] = \sigma^2 \left[1 - \frac{\beta\phi(\beta) - \alpha\phi(\alpha)}{\Phi(\beta) - \Phi(\alpha)} - \left(\frac{\phi(\alpha) - \phi(\beta)}{\Phi(\beta) - \Phi(\alpha)}\right)^2\right], $$

where $\alpha := \frac{a-\mu}{\sigma}$ and $\beta := \frac{b-\mu}{\sigma}$. The main result of this section establishes the optimal variance proxy for this density, which turns out to have a closed-form expression:

**Theorem 2.1.** Let $X_T$ 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:

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

In particular, $X_T$ 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.
<figure><img src="image_2.png" /><figcaption>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.</figcaption></figure>

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 $Y := \frac{X-\mu}{\sigma}$ and let $Y_T$ be the truncated standard normal along the interval $(\alpha, \beta)$. Then, we have the relation $\mathbb{E}[e^{\theta(X_T - \mathbb{E}[X_T])}] = \mathbb{E}[e^{\sigma\theta(Y_T - \mathbb{E}[Y_T])}]$. That is, to optimally bound the centered moment generating function of $X_T$, we can restrict to optimally bounding that of $Y_T$, since the previous equation implies $\|X_T\|_{vp}^2 = \sigma^2 \|Y_T\|_{vp}^2$. Hence, an equivalent reformulation of Theorem 2.1 is the following.

**Theorem 2.2.** Let $Y_T$ be a standard normal variable truncated in the interval $(\alpha, \beta)$ with $\alpha < \beta$. Then its optimal variance proxy is given by:

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

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 [1]. The limiting behavior of the optimal variance proxy (when $\ell \to 0$ and $\ell \to \infty$ where $2\ell = \beta - \alpha$) can be obtained by using fourth order Taylor expansions of the Gaussian pdf $\phi$ and cdf $\Phi$, combined with identities $\phi'(x) = -x\phi(x)$, $\phi''(x) = (x^2 - 1)\phi(x)$, and $\phi'''(x) = -x(x^2 - 3)\phi(x)$. It shows that our optimal variance proxy is superior to $s_H^2 = \ell^2$ by a multiplicative factor of $1/3$ when $\ell \to 0$. It is even more superior in the $\ell \to \infty$ limit, where $\|Y_T\|_{vp}^2 \to 1$, while $s_H^2$ is quadratic in $\ell$. See Table 1.

**Proof of Theorem 2.2.** Recall by Definition 1.1 that the optimal variance proxy of $Y_T$ corresponds to the smallest possible $s^2 \ge 0$ such that

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

$$c_{\alpha, \beta} := \frac{\phi(\alpha) - \phi(\beta)}{\Phi(\beta) - \Phi(\alpha)}, \quad \theta_0 := \frac{\alpha + \beta}{2}$$

$$F_{\alpha,\beta}(\theta) := \Phi(\beta - \theta) - \Phi(\alpha - \theta), \quad f_{\alpha,\beta}(\theta) := \ln \left[ \frac{F_{\alpha,\beta}(\theta)}{\Phi(\beta) - \Phi(\alpha)} \right]$$

$$h_{\alpha,\beta}(\theta) := f'_{\alpha,\beta}(\theta) = \frac{\phi(\alpha - \theta) - \phi(\beta - \theta)}{\Phi(\beta - \theta) - \Phi(\alpha - \theta)}$$

$$h'_{\alpha,\beta}(\theta) = \frac{\phi(\alpha - \theta)(\alpha - \theta) - \phi(\beta - \theta)(\beta - \theta)}{\Phi(\beta - \theta) - \Phi(\alpha - \theta)} - \left[ \frac{\phi(\alpha - \theta) - \phi(\beta - \theta)}{\Phi(\beta - \theta) - \Phi(\alpha - \theta)} \right]^2,$$

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

$$p_{\alpha,\beta;w}(\theta) := w\theta^2 + c_{\alpha,\beta}\theta \ge f_{\alpha,\beta}(\theta), \quad \forall \theta \in \mathbb{R}. \qquad (2)$$

The delicate thing is understanding the right-hand side's function behavior. Note that this function is independent of the value of $w$. Observe also 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 $c_{\alpha,\beta} = 0$. Another important point is to note that:

$$f_{\alpha,\beta}(0) = 0 = p_{\alpha,\beta;w}(0), \quad f_{\alpha,\beta}(2\theta_0) = 0, \quad f'_{\alpha,\beta}(0) = p'_{\alpha,\beta;w}(0).$$

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

**Lemma 2.3.** Let $\alpha < \beta$. Function $h_{\alpha,\beta}$ is strictly convex on $(-\infty, \theta_0)$ and strictly concave on $(\theta_0, +\infty)$.

*Proof of Lemma 2.3.* The proof follows from direct and technical computations detailed in Appendix A. $\square$

Using Lemma 2.3, we obtain the optimal variance proxy for a truncated standard normal random variable given in Theorem 2.2. We now need to split into three cases.

**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.

**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:

$$\begin{align*}
f_{\alpha,\beta}(0) &= 0 = p_{\alpha,\beta;w_c}(0), \\
f_{\alpha,\beta}(2\theta_0) &= 0 = 4w_c\theta_0^2 + c_{\alpha,\beta}\theta_0 = p_{\alpha,\beta;w_c}(2\theta_0), \\
f'_{\alpha,\beta}(\theta_0) &= 0 = 2\theta_0w_c + c_{\alpha,\beta} = p'_{\alpha,\beta;w_c}(\theta_0), \tag{3} \\
f'_{\alpha,\beta}(2\theta_0) &= -c_{\alpha,\beta} = 4\theta_0w_c + c_{\alpha,\beta} = p'_{\alpha,\beta;w_c}(2\theta_0), \\
f'_{\alpha,\beta}(0) &= c_{\alpha,\beta} = p'_{\alpha,\beta;w_c}(0).
\end{align*}$$

Let us define $f_{\alpha,\beta;w} := f_{\alpha,\beta} - p_{\alpha,\beta;w}$ and $h_{\alpha,\beta;w} := h_{\alpha,\beta} - p'_{\alpha,\beta;w}$ so that $h''_{\alpha,\beta;w} = h''_{\alpha,\beta}$ because $p_{\alpha,\beta;w}$ is a parabola. We will now prove that inequality (2) holds for $w = w_c$, i.e., we will check that $f_{\alpha,\beta;w_c}(\theta) \le 0$ for all $\theta \in \mathbb{R}$. From Lemma 2.3, $h'_{\alpha,\beta;w}$ is strictly increasing on $(-\infty, \theta_0)$ and strictly decreasing on $(\theta_0, +\infty)$ and thus reaches its maximum at $\theta = \theta_0$. Thus, $h'_{\alpha,\beta;w}$ 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 $w = w_c$, (3) implies that $h_{\alpha,\beta;w_c}$ has three distinct zeros at $\theta \in \{0, \theta_0, 2\theta_0\}$. This implies that $h'_{\alpha,\beta;w_c}(\theta_0) > 0$ and that there exists $\theta_1 < \theta_0 < \theta_2$ such that $h'_{\alpha,\beta;w_c}(\theta_1) = h'_{\alpha,\beta;w_c}(\theta_2) = 0$ and $h'_{\alpha,\beta;w_c}$ is strictly positive on $(\theta_1, \theta_2)$ and strictly negative on $(-\infty, \theta_1) \cup (\theta_2, +\infty)$ because of Lemma 2.3. Let us assume that $\theta_0 > 0$ (the case $\theta_0 < 0$ is similar
<figure><img src="image_3.png" /><figcaption>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.</figcaption></figure>

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 (2) 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 (2) is not realized in a neighborhood of $2\theta_0$. So $w_c$ is optimal.

**Case α and β 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 $w$:

$$f_{-\beta,\beta}(0) = 0 = p_{-\beta,\beta;w}(0) \quad \text{and} \quad f'_{\alpha,\beta}(0) = 0 = p'_{\alpha,\beta;w}(0). \qquad (4)$$

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:

$$f''_{\alpha,\beta}(0) = -\frac{2\beta\phi(\beta)}{2\Phi(\beta) - 1} = 2w_c = p''_{\alpha,\beta;w_c}(0). \qquad (5)$$

Let us define $f_{\alpha,\beta;w} := f_{\alpha,\beta} - p_{\alpha,\beta;w}$ and $h_{\alpha,\beta;w} := h_{\alpha,\beta} - p'_{\alpha,\beta;w}$ as in the previous case. From Lemma 2.3, $h'_{\alpha,\beta;w_c}$ is strictly increasing on $(-\infty, 0)$ and strictly decreasing on $(0, +\infty)$. It achieves its maximum at $\theta = 0$ and $h'_{\alpha,\beta;w_c}(0) = 0$ from (5). Thus $h'_{\alpha,\beta;w_c}$ 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 $\mathbb{R}_{\le 0}$ and strictly decreasing on $\mathbb{R}_{\ge 0}$. Its maximum is thus achieved at $\theta = 0$ and is null from (4). Thus, it remains negative on $\mathbb{R}$ and we conclude that inequality (2) 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 (4) but the second derivatives satisfy $f''_{\alpha,\beta}(0) = -\frac{2\beta\phi(\beta)}{2\Phi(\beta)-1} = 2w_c > 2w = p''_{\alpha,\beta;w}(0)$. Hence the parabola $p_{\alpha,\beta;w}$ is locally below the function $f_{\alpha,\beta}$ so that inequality (2) 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 $s_c^2 := 2w_c + 1$ 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 (2) 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:

$$f_{-\infty,\beta;w}(\theta) := f_{-\infty,\beta}(\theta) - p_{-\infty,\beta;w}(\theta) = \ln\left(\frac{\Phi(\beta - \theta)}{\Phi(\beta)}\right) - w\theta^2 - c_{-\infty,\beta}\theta^{\theta \to -\infty} + \infty, \quad (6)$$

so that (2) 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 $X_T$ denote the truncation of $X$ along the interval $(a, b)$ with $0 \le a < b \le +\infty$ (recall that the untruncated version is not sub-Gaussian). Its density function $f_{X_T}(t) = \lambda \frac{e^{-\lambda t}}{e^{-\lambda a} - e^{-\lambda b}} \mathbf{1}_{(a,b)}(t)$ directly provides expressions for the mean and variance:

$$\mathbb{E}[X_T] = \frac{1}{\lambda} + \frac{be^{\lambda a} - ae^{\lambda b}}{e^{\lambda a} - e^{\lambda b}} = \frac{1}{\lambda} + \frac{ae^{-\lambda a} - be^{-\lambda b}}{e^{-\lambda a} - e^{-\lambda b}}, \quad \text{Var}[X_T] = \frac{1}{\lambda^2} - \frac{(b-a)^2 e^{\lambda(a+b)}}{(e^{\lambda b} - e^{\lambda a})^2}.$$

It is easy to see that $b$ must be finite in order for $X_T$ to be sub-Gaussian. Indeed, we have:

$$\mathbb{E}[\exp(\theta(X_T - \mathbb{E}[X_T]))] \Big|_{\theta=\lambda} = \frac{\lambda(b-a)e^{-\lambda\mathbb{E}[X_T]}}{e^{-\lambda a} - e^{-\lambda b}} \stackrel{\lambda \to +\infty}{\sim} e^{\lambda(a - \mathbb{E}[X_T])} \lambda b,$$

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

**Theorem 3.1.** Let $X_T$ be an exponential variable with scale $\lambda$ truncated along the interval $(a, b)$ with $0 \le a < b < \infty$. Then $X_T$ is never strictly sub-Gaussian and its optimal variance proxy is given by:

$$\|X_T\|_{\text{vp}}^2 = \frac{(b-a)(e^{\lambda b} + e^{\lambda a})}{2\lambda(e^{\lambda b} - e^{\lambda a})} - \frac{1}{\lambda^2}.$$

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 $Y_T$ be its truncation on $[\alpha, \beta]$ with $0 \le \alpha < \beta < +\infty$, then defining $X = \frac{Y}{\lambda}$, $(a, b) = (\frac{\alpha}{\lambda}, \frac{\beta}{\lambda})$ and $X_T$ the truncation of $X$ on $[a, b]$, we have that $X_T = \frac{Y_T}{\lambda}$ and $\lambda^2 \|X_T\|_{\text{vp}}^2 = \|Y_T\|_{\text{vp}}^2$. Hence an equivalent formulation is:

**Theorem 3.2.** Let $Y_T$ be an exponential variable with mean 1 truncated along the interval $(\alpha, \beta)$ with $0 \le \alpha < \beta < +\infty$, and denote the half-length by $\ell = \frac{\beta - \alpha}{2}$. Then $Y_T$ is never strictly sub-Gaussian; its optimal variance proxy depends on $\alpha$ and $\beta$ only through $\ell$ and is given by:

$$\|Y_T\|_{\text{vp}}^2 = \ell \frac{e^{2\ell} + 1}{e^{2\ell} - 1} - 1.$$
<figure><img src="image_4.png" /><figcaption>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).</figcaption></figure>

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 $\ell \to +\infty$. More specifically, we have that $\|Y_T\|_{\text{vp}}^2$ is equivalent to $\ell$ as $\ell \to +\infty$, for any $\alpha$ and $\beta = \alpha + 2\ell$. Theorem 3.1 and Theorem 3.2 are illustrated in Figure 3.

As in the previous section, we compared our optimal variance proxy against Hoeffding's upper bound $s_H^2 = \ell^2$ in the limiting case of small and large truncation range. See the results in Table 1.

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

$$g_{\alpha, \beta, s}(\theta) := \mathbb{E}[\exp(\theta(Y_T - \mathbb{E}[Y_T]))] - \exp\left(\frac{s^2\theta^2}{2}\right). \qquad (7)$$

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

$$g_{\alpha, \beta, \|Y_T\|_{\text{vp}}}(\theta_c) = 0 \quad \text{and} \quad g'_{\alpha, \beta, \|Y_T\|_{\text{vp}}}(\theta_c) = 0 \quad \text{with} \quad s_{\text{inf}}(\alpha, \beta) < \|Y_T\|_{\text{vp}} \le s_1(\alpha, \beta) \quad \text{and} \quad \theta_c \ne 0, \qquad (8)$$

where:

$$g_{\alpha, \beta, s}(\theta) = \frac{e^{\frac{1}{2}s^2\theta^2}}{e^{\beta} - e^{\alpha}} G_{\alpha, \beta, s}(\theta),$$

$$G_{\alpha, \beta, s}(\theta) = e^{\alpha} - e^{\beta} + \frac{1}{\theta - 1} \left( e^{-\frac{1}{2}s^2\theta^2 + (\beta - \mathbb{E}[Y_T])\theta + \alpha} - e^{-\frac{1}{2}s^2\theta^2 + (\alpha - \mathbb{E}[Y_T])\theta + \beta} \right),$$

$$s_{\text{inf}}(\alpha, \beta) := \sqrt{1 - \frac{(\beta - \alpha)^2 e^{\alpha + \beta}}{(e^{\beta} - e^{\alpha})^2}}, \quad s_1(\alpha, \beta) := \sqrt{\frac{(\beta - \alpha)^2 (\beta - \mathbb{E}[Y_T] + 2) - \delta_{\alpha, \beta}}{(\beta - \alpha)^2 + 12}},$$

$$\delta_{\alpha, \beta} = 2(\beta - \alpha) \left(-3\mathbb{E}[Y_T]^2 + (6\beta - (\beta - \alpha)^2)\mathbb{E}[Y_T] + \beta^3 - 2\alpha\beta^2 + \alpha^2\beta + \alpha^2 - 2\alpha\beta - 2\beta^2\right)^{\frac{1}{2}}. \qquad (9)$$

Furthermore, by uniqueness and observing that:
$$ \theta_c = 2, \quad \|Y_T\|_{\text{vp}}^2 = \frac{(\beta - \alpha + 2)e^{\alpha} + (\beta - \alpha - 2)e^{\beta}}{2(e^{\beta} - e^{\alpha})} $$

is a solution of (8) we conclude the computation of $\|Y_T\|_{\text{vp}}^2$. Eventually, the statement that $Y_T$ is never strictly sub-Gaussian follows from a direct analysis of the difference between $\|Y_T\|_{\text{vp}}^2$ and $\text{Var}[Y_T]$ that is detailed in Appendix B.4.

## 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 [39]. 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.

## Funding

OM was supported by his IUF junior grant G752IUFMAR, and JA by grant ANR-21-JSTM-0001.

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# Appendix A. Proofs for truncated Gaussian random variables

This entire Appendix A is devoted to proving Lemma 2.3, showing that the function $h_{\alpha,\beta}$ is strictly concave on $[\theta_0, +\infty)$ for any finite $\alpha < \beta$.

**Proof of Lemma 2.3.** 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 $h''_{-\beta,\beta}(\theta) < 0$ 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 $\theta \ge 0$, i.e. that $x \mapsto h_{\alpha,\beta}(x)$ is strictly concave for $x \ge \theta_0$.

## Appendix A.1. Some notations and preliminary results

Let us recall that we have:

$$
\begin{align*}
\Phi(x) &:= \int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}s^2} ds, \\
F_{-\beta,\beta}(\theta) &:= \Phi(\beta - \theta) - \Phi(-\beta - \theta), \\
f_{-\beta,\beta}(\theta) &:= \ln \left( \frac{F_{-\beta,\beta}(\theta)}{2\Phi(\beta) - 1} \right), \\
F'_{-\beta,\beta}(\theta) &:= \phi(\theta + \beta) - \phi(\theta - \beta) = \frac{1}{\sqrt{2\pi}} \left( e^{-\frac{1}{2}(\theta+\beta)^2} - e^{-\frac{1}{2}(\theta-\beta)^2} \right), \\
F''_{-\beta,\beta}(\theta) &:= \frac{1}{\sqrt{2\pi}} \left( -(\beta + \theta)e^{-\frac{1}{2}(\theta+\beta)^2} + (\theta - \beta)e^{-\frac{1}{2}(\theta-\beta)^2} \right), \\
F'''_{-\beta,\beta}(\theta) &:= \frac{1}{\sqrt{2\pi}} \left( e^{-\frac{1}{2}(\theta+\beta)^2} [(\beta + \theta)^2 - 1] + e^{-\frac{1}{2}(\theta-\beta)^2} [1 - (\beta - \theta)^2] \right), \tag{A.1} \\
h_{-\beta,\beta}(\theta) &:= f'_{-\beta,\beta}(\theta) = \frac{F'_{-\beta,\beta}(\theta)}{F_{-\beta,\beta}(\theta)}, \\
h'_{-\beta,\beta}(\theta) &:= \frac{F_{-\beta,\beta}(\theta)F''_{-\beta,\beta}(\theta) - (F'_{-\beta,\beta}(\theta))^2}{F_{-\beta,\beta}(\theta)^2}, \\
h''_{-\beta,\beta}(\theta) &:= \frac{(F_{-\beta,\beta}(\theta))^2 F'''_{-\beta,\beta}(\theta) - 3F_{-\beta,\beta}(\theta)F'_{-\beta,\beta}(\theta)F''_{-\beta,\beta}(\theta) + 2(F'_{-\beta,\beta}(\theta))^3}{F_{-\beta,\beta}(\theta)^3}, \\
\lim_{\theta \to +\infty} F'_{-\beta,\beta}(\theta) &= \lim_{\theta \to +\infty} F''_{-\beta,\beta}(\theta) = \lim_{\theta \to +\infty} F'''_{-\beta,\beta}(\theta) = 0, \\
\lim_{\theta \to +\infty} F_{-\beta,\beta}(\theta) &= 0.
\end{align*}
$$

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

$$F'''_{-\beta,\beta}(\theta) = -P'_{2}(\theta)F''_{-\beta,\beta}(\theta) - P_{2}(\theta)F'_{-\beta,\beta}(\theta), \quad (A.2)$$

where

$$P_2(\theta) := \theta^2 + 1 - \beta^2. \quad (A.3)$$

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

## Lemma Appendix A.1. We have the following results:

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

* For $0 < \beta \le \sqrt{3}$: The function $F'''_{-\beta,\beta}$ 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 $F''_{-\beta,\beta}$ is strictly increasing on $[0, \theta_1]$ from $F''_{-\beta,\beta}(0) = -\frac{2\beta}{\sqrt{2\pi}}e^{-\frac{1}{2}\beta^2} < 0$ to $F''_{-\beta,\beta}(\theta_1) > 0$ and strictly decreasing on $(\theta_1, +\infty)$ from $F''_{-\beta,\beta}(\theta_1) > 0$ to 0. In particular $F''_{-\beta,\beta}$ has only one zero on $\mathbb{R}_{>0}$ denoted $\alpha_1$ satisfying $\alpha_1 > \theta_1$.
• For $\beta > \sqrt{3}$: The function $F''''_{-\beta,\beta}$ 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 $F''_{-\beta,\beta}$ is decreasing on $(0, \theta_1)$ from $F''_{-\beta,\beta}(0) = -\frac{2\beta}{\sqrt{2\pi}}e^{-\frac{1}{2}\beta^2} < 0$ to $F''_{-\beta,\beta}(\theta_1) < F''_{-\beta,\beta}(0) < 0$. $F''_{-\beta,\beta}$ is then increasing on $(\theta_1, \theta_2)$ up to $F''_{-\beta,\beta}(\theta_2)$. It finally decreases from $F''_{-\beta,\beta}(\theta_2)$ to its limit 0 at $\theta \to +\infty$. In particular, we must have $F''_{-\beta,\beta}(\theta_2) > 0$ and $F''_{-\beta,\beta}$ has only one zero on $\mathbb{R}_{>0}$ denoted $\alpha_1$ that verifies $\alpha_2 \in (\theta_1, \theta_2)$.

**Proof of Lemma Appendix A.1.** The first point is obvious from (A.1). Let us observe that

$$F''_{-\beta,\beta}(\theta) = \frac{2}{\sqrt{2\pi}} \exp\left(-\frac{1}{2}(\beta^2 + \theta^2)\right) (\theta \sinh(\beta\theta) - \beta \cosh(\beta\theta)). \quad (A.4)$$

In particular $F''_{-\beta,\beta}(0) = -\frac{2\beta}{\sqrt{2\pi}} \exp(-\frac{1}{2}\beta^2) < 0$ and $\lim_{\theta \to +\infty} F''_{-\beta,\beta}(\theta) = 0$. Let us assume by contradiction that $F'''_{-\beta,\beta}$ does not vanish on $\mathbb{R}_{>0}$. This implies that $F''_{-\beta,\beta}$ would be a strictly monotonous function and therefore would be a strictly increasing function because of the values of $F''_{-\beta,\beta}(0)$ and $F''_{-\beta,\beta}(+\infty)$. Since $\lim_{\theta \to +\infty} F''_{-\beta,\beta}(\theta) = 0$, it implies that $F''_{-\beta,\beta}$ would be a strictly negative function on $\mathbb{R}_{>0}$ so that $F'_{-\beta,\beta}$ would be a strictly decreasing function on $\mathbb{R}_{>0}$. This is a contradiction with the fact that $\lim_{\theta \to +\infty} F'_{-\beta,\beta}(\theta) = 0$ and $F'_{-\beta,\beta}$ 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 $F'''_{-\beta,\beta}(\theta_c) = 0$. This point shall be used later to determine the sign of some quantities by contradiction<fnref n="1" />. Let us now compute

$$F'''_{-\beta,\beta} \left( \theta = \frac{s}{\beta} \right) = \frac{1}{\sqrt{2\pi}} \exp \left( -\frac{1}{2} \left( \beta^2 + \frac{s^2}{\beta^2} \right) \right) e^{-s} \left( \left( 1 - \left( \frac{s}{\beta} - \beta \right)^2 \right) e^{2s} - \left( 1 - \left( \frac{s}{\beta} + \beta \right)^2 \right) \right). \quad (A.5)$$

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

$$g_{\beta}(s) := \left(1 - \left(\frac{s}{\beta} - \beta\right)^2\right) e^{2s} - 1 + \left(\frac{s}{\beta} + \beta\right)^2. \quad (A.6)$$

We have

$$g'_{\beta}(s) = -\frac{2(\beta^4 - 2(s+1)\beta^2 + s^2 + s)}{\beta^2} e^{2s} + \frac{2(\beta^2 + s)}{\beta^2},$$

$$g''_{\beta}(s) = -\frac{2(2\beta^4 - 4\beta^2s - 6\beta^2 + 2s^2 + 4s + 1)}{\beta^2} e^{2s} + \frac{2}{\beta^2},$$

$$g'''_{\beta}(s) = -\frac{4(2s^2 - (4\beta^2 - 6)s + 2\beta^4 - 8\beta^2 + 3)}{\beta^2} e^{2s}.$$

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

$$P(\beta) := 2\beta^4 - 8\beta^2 + 3, \quad S(\beta) := 4\beta^2 - 6, \quad (A.7)$$

so that the product (resp. the sum) of the two zeros of $g'''_{\beta}$ are given by $\frac{1}{2}P(\beta)$ (resp. $\frac{1}{2}S(\beta)$). $P(\beta)$ is strictly positive on $(0, \frac{\sqrt{8-2\sqrt{10}}}{2}) \cup (\frac{\sqrt{8+2\sqrt{10}}}{2}, +\infty)$ and strictly negative on $(\frac{\sqrt{8-2\sqrt{10}}}{2}, \frac{\sqrt{8+2\sqrt{10}}}{2})$. $S(\beta)$

<footnote n="1">¹In fact, the reasoning implies that there exists a non-trivial interval on which $F'''_{-\beta,\beta}$ is strictly positive.</footnote>
is strictly positive on $(\frac{\sqrt{6}}{2}, +\infty)$ and strictly negative on $(0, \frac{\sqrt{6}}{2})$. We shall also observe:

$$ 
\begin{align*}
\lim_{s \to +\infty} g_{\beta}(s) &= \lim_{s \to +\infty} g'_{\beta}(s) = \lim_{s \to +\infty} g''_{\beta}(s) = \lim_{s \to +\infty} g'''_{\beta}(s) = -\infty, \\
g''''_{\beta}(0) &= -\frac{4P(\beta)}{\beta^2}, \\
g''_{\beta}(0) &= 4(3 - \beta^2), \\
g'_{\beta}(0) &= 2(3 - \beta^2), \\
g_{\beta}(0) &= 0.
\end{align*}
$$

Let us now study the following four cases.

**Case** $\beta > \frac{\sqrt{8+2\sqrt{10}}}{2}$. Here, $g''''_{\beta}$ has two roots on $\mathbb{R}$ and their product in strictly positive and their sum is strictly positive. Thus, $g''''_{\beta}$ 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 $g''_{\beta}$ strictly decreases on $(0, s_1)$ and since $g''_{\beta}(0) < 0$ it remains strictly negative on $(0, s_1)$. $g''_{\beta}$ is then strictly increasing from $g''_{\beta}(s_1)$ up to $g''_{\beta}(s_2)$. We have $g''_{\beta}(s_2) = \frac{2}{\beta^2} (e^{2\beta^2} + \sqrt{4\beta^2+3-3}(\sqrt{4\beta^2+3}-1)+1) > 0$. Since $g''_{\beta}$ 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 $g''_{\beta}(s_3) = g''_{\beta}(s_4) = 0$ and they satisfy $s_3 < s_2 < s_4$. Moreover $g''_{\beta}$ is strictly positive on $(s_3, s_4)$ and strictly negative on $(0, s_3) \cup (s_4, +\infty)$. Therefore $g'_{\beta}$ strictly decreases on $(0, s_3)$ and since $g'_{\beta}(0) < 0$ it remains strictly negative on $(0, s_3)$. Note that we necessarily have $g'_{\beta}(s_4) > 0$. Indeed, if $g'_{\beta}(s_4) \le 0$ then $g'_{\beta}$ 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 $F'''_{-\beta,\beta}$ must vanish at least once). Since $g'_{\beta}(s_4) > 0$, we obtain that there exists exactly two values $(s_5, s_6) \in (\mathbb{R}_{>0})^2$ such that $g'_{\beta}(s_5) = g'_{\beta}(s_6) = 0$ 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 $F'''_{-\beta,\beta}$ 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** $\beta \in (\sqrt{3}, \frac{\sqrt{8+2\sqrt{10}}}{2}]$. Here, $g'''_{\beta}$ has two roots on $\mathbb{R}$ and their product in negative or null and their sum is strictly positive. This means that $g'''_{\beta}$ has exactly one root on $\mathbb{R}_{>0}$ that we shall denote $s_2 > 0$. $g'''_{\beta}$ is strictly positive on $(0, s_2)$ and strictly negative on $(s_2, +\infty)$. Since $g''_{\beta}(0) < 0$ and $g''_{\beta}(s_2) = \frac{2}{\beta^2} (e^{2\beta^2} + \sqrt{4\beta^2+3-3}(\sqrt{4\beta^2+3}-1)+1) > 0$ and $\lim_{s \to +\infty} g''_{\beta}(s) = -\infty$, we get that there exists exactly two distinct values $(s_3, s_4) \in (\mathbb{R}_{>0})^2$ such that $g''_{\beta}(s_3) = g''_{\beta}(s_4) = 0$ and they satisfy $s_3 < s_2 < s_4$. Moreover, $g'_{\beta}$ is strictly decreasing on $(0, s_3)$, strictly increasing on $(s_3, s_4)$ and strictly decreasing on $(s_4, +\infty)$. We have $g'_{\beta}(0) < 0$ so that $g'_{\beta}(s_3) < 0$. Similarly to the previous case, we must have $g'_{\beta}(s_4) > 0$ 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 $g'_{\beta}(s_5) = g'_{\beta}(s_6) = 0$ 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 $F'''_{-\beta,\beta}$ 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** $\beta \in (\frac{\sqrt{8-2\sqrt{10}}}{2}, \sqrt{3}]$. Here, $g'''_{\beta}$ has two roots on $\mathbb{R}$ and their product is strictly negative. This means that $g'''_{\beta}$ has exactly one root on $\mathbb{R}_{>0}$ that we shall denote $s_2 > 0$. $g'''_{\beta}$ is strictly positive on $(0, s_2)$ and strictly negative on $(s_2, +\infty)$. We have $g''_{\beta}(0) \ge 0$ and $\lim_{s \to +\infty} g''_{\beta}(s) = -\infty$. Thus, since $g''_{\beta}$ is strictly increasing and then strictly decreasing, we get that there exists exactly one value $s_4 \in \mathbb{R}_{>0}$ such that $g''_{\beta}(s_4) = 0$ and it satisfies $s_2 < s_4$. Moreover, $g'_{\beta}$ is strictly increasing on $(0, s_4)$ with $g'_{\beta}(0) \ge 0$. It then strictly decreases from $g'_{\beta}(s_4) > 0$ towards $-\infty$. Hence, there exists exactly one value $s_6 > s_4$ such that $g'_{\beta}(s_6) = 0$ and $g'_{\beta}$ 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 $g_β(s_6) > 0$ and then strictly decreasing towards $-∞$. We conclude that there exists only one value $s_8$ on $R_{>0}$ such that $g_β(s_8) = 0$.

**Case** $β \in (0, \frac{\sqrt{8-2\sqrt{10}}}{2}]$. Here, $g_β'''$ has two roots on $R$, and their product is positive or null, while their sum is strictly negative. This implies that $g_β'''$ does not vanish on $R_{>0}$ and thus is strictly negative. $g_β''$ is thus strictly decreasing on $R_{>0}$ with $g_β''(0) > 0$ and $g_β''(+∞) = -∞$. We get that there exists exactly one value $s_4 \in R_{>0}$ such that $g_β''(s_4) = 0$. Moreover, $g_β'$ is strictly increasing on $(0, s_4)$ with $g_β'(0) > 0$. It then strictly decreases from $g_β'(s_4) > 0$ towards $-∞$. Hence, there exists exactly one value $s_6 > s_4$ such that $g_β'(s_6) = 0$ and $g_β'$ is strictly positive on $(0, s_6)$ and strictly negative on $(s_6, +∞)$. Thus, $g_β$ is strictly increasing from $g_β(0) = 0$ to $g_β(s_6) > 0$ and then strictly decreasing towards $-∞$. We conclude that there exists only one value $s_8$ on $R_{>0}$ such that $g_β(s_8) = 0$.

Summarizing the four different cases and using $θ = \frac{s}{β}$, we conclude that for $β \in (0, \sqrt{3}]$, $F_{-β,β}'''$ has exactly one root $θ_1$ on $R_{>0}$ and it is strictly positive on $(0, θ_1)$ and it is strictly negative on $(θ_1, +∞)$. On the contrary, for $β > \sqrt{3}$, $F_{-β,β}'''$ has exactly two distinct roots on $R_{>0}$ denoted $θ_1 < θ_2$ and it is strictly positive on $(θ_1, θ_2)$ and strictly negative on $(0, θ_1) \cup (θ_2, +∞)$. The rest of the lemma is then obvious, which concludes the proof of Lemma Appendix A.1. □

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

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

$$h''_{-β,β}(θ) \stackrel{θ \to 0}{=} -\frac{\beta \exp\left(-\frac{\beta^2}{2}\right)}{(2\Phi(β) - 1)^2} \left( \frac{6\beta}{\pi} \exp\left(-\frac{\beta^2}{2}\right) + \frac{2(\beta^2 - 3)}{\sqrt{2\pi}}(2\Phi(β) - 1) \right) θ + O(θ^2). \quad (A.8)$$

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

$$f_1(β) := \frac{6β}{\pi(β^2 - 3)} \exp\left(-\frac{β^2}{2}\right) + \frac{2}{\sqrt{2π}}(2\Phi(β) - 1), \quad (A.9)$$

whose derivative is $f'_1(β) = -\frac{4β^4 \exp(-\frac{β^2}{2})}{\pi(β^2-3)^2} < 0$. 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 $h''_{-β,β}(θ)$ as $θ \to 0$ is strictly negative on $R_{>0}$. In particular, we get that for any $β > 0$, there exists a positive neighborhood of $θ = 0$ on which $h_{-β,β}$ is strictly concave.

Let us now reformulate the problem of strict concavity more simply. From (A.1), the equation $h''_{-β,β}(θ) = 0$ is equivalent to

$$Z_{\beta}(\theta) := (F_{-\beta,\beta}(\theta))^2 F_{-\beta,\beta}'''(\theta) - 3F_{-\beta,\beta}(\theta)F_{-\beta,\beta}'(\theta)F_{-\beta,\beta}''(\theta) + 2(F_{-\beta,\beta}'(\theta))^3 = 0. \quad (A.10)$$

We shall denote

$$\begin{align*} a_{\beta}(\theta) &:= F_{-\beta,\beta}'''(\theta), \\ b_{\beta}(\theta) &:= -3F_{-\beta,\beta}'(\theta)F_{-\beta,\beta}''(\theta), \\ c_{\beta}(\theta) &:= 2(F_{-\beta,\beta}'(\theta))^3, \end{align*}$$

so that

$$Z_{\beta}(\theta) = a_{\beta}(\theta)(F_{-\beta,\beta}(\theta))^2 + b_{\beta}(\theta)F_{-\beta,\beta}(\theta) + c_{\beta}(\theta). \quad (A.11)$$

Let us observe that if we can prove that $Z_β(θ)$ does not vanish on $R_{>0}$, then it proves that $h_{-β,β}$ is strictly concave on $R_{>0}$. Indeed, the function $Z_β(θ)$ is continuous in $θ$. Moreover, we have proved that it is strictly negative in a positive neighborhood of $θ = 0$ so that if it does not vanish on $R_{>0}$, then the intermediate value theorem implies that it must remain strictly negative on $R_{>0}$. Therefore, we obtain the following sufficient condition to prove strict concavity of $h_{-β,β}$ on $R_{>0}$.
**Proposition Appendix 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.11) 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 $F_{\beta}'''$ or solutions of the system

$$ 
\begin{aligned}
& 9(F'_{-\beta,\beta}(\theta_c))^2(F''_{-\beta,\beta}(\theta_c))^2 - 8(F'_{-\beta,\beta}(\theta_c))^3 F'''_{-\beta,\beta}(\theta_c) \ge 0, \\
& F_{-\beta,\beta}(\theta_c) = \frac{3 F'_{-\beta,\beta}(\theta_c) F''_{-\beta,\beta}(\theta_c)}{2 F'''_{-\beta,\beta}(\theta_c)} \\
& \quad \pm \frac{1}{2F'''_{-\beta,\beta}(\theta_c)} \sqrt{9(F'_{-\beta,\beta}(\theta_c))^2(F''_{-\beta,\beta}(\theta_c))^2 - 8(F'_{-\beta,\beta}(\theta_c))^3 F'''_{-\beta,\beta}(\theta_c)} \quad (A.12) \\
& = -\frac{b_\beta(\theta_c)}{2a_\beta(\theta_c)} \pm \frac{1}{2a_\beta(\theta_c)} \sqrt{b_\beta(\theta_c)^2 - 4a_\beta(\theta_c)c_\beta(\theta_c)}.
\end{aligned}
$$

The first inequality in (A.12) is necessary otherwise (A.10) which is polynomial of degree 2 in $F_{-\beta,\beta}(\theta_c)$ would have no real roots and thus (A.10) would not have solutions on $\mathbb{R}_{>0}$ ending the proof. Therefore, we shall define

$$ 
\begin{aligned}
T_{\beta, \pm}(\theta) &:= -F_{-\beta,\beta}(\theta) + \frac{3 F'_{-\beta,\beta}(\theta) F''_{-\beta,\beta}(\theta)}{2 F'''_{-\beta,\beta}(\theta)} \\
& \quad \pm \frac{1}{2F'''_{-\beta,\beta}(\theta)} \sqrt{9(F'_{-\beta,\beta}(\theta))^2(F''_{-\beta,\beta}(\theta))^2 - 8(F'_{-\beta,\beta}(\theta))^3 F'''_{-\beta,\beta}(\theta)} \\
&= -\frac{b_\beta(\theta)}{2a_\beta(\theta)} \pm \frac{1}{2a_\beta(\theta)} \sqrt{b_\beta(\theta)^2 - 4a_\beta(\theta)c_\beta(\theta)} - F_{-\beta,\beta}(\theta),
\end{aligned}
$$

and we have the following proposition.

**Proposition Appendix 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

$$ 
\begin{aligned}
S_\beta(\theta) &:= 9\theta(F''_{-\beta,\beta}(\theta))^5 \\
& \quad + (42\theta^2 - 9) F'_{-\beta,\beta}(\theta) (F''_{-\beta,\beta}(\theta))^4 \\
& \quad - 15 \left(\beta^2 - \frac{79}{15}\theta^2 + 2\right) \theta (F'_{-\beta,\beta}(\theta))^2 (F''_{-\beta,\beta}(\theta))^3 \\
& \quad + (75\theta^4 - (42\beta^2 + 36)\theta^2 - \beta^4 + 12\beta^2 - 3) (F'_{-\beta,\beta}(\theta))^3 (F''_{-\beta,\beta}(\theta))^2 \\
& \quad + 4 \left(9\theta^4 - \left(10\beta^2 + \frac{9}{2}\right)\theta^2 + \beta^4 + \frac{9}{2}\beta^2 - \frac{3}{2}\right) \theta (F'_{-\beta,\beta}(\theta))^4 F''_{-\beta,\beta}(\theta) \\
& \quad + (7\theta^6 - (13\beta^2 + 3)\theta^4 + (5\beta^4 + 6\beta^2 - 3)\theta^2 + (\beta^2 - 1)^3) (F'_{-\beta,\beta}(\theta))^5.
\end{aligned}
\quad (A.13)
$$

*Proof of Proposition Appendix A.2.* As explained in Proposition Appendix 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 $F_{\beta}'''$ 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 $F_{\beta}'''$. Let us
observe that for any $\theta \in \mathbb{R}_{>0} \setminus \mathcal{R}_{\beta}$:

$$ 
\begin{aligned}
T'_{\beta, \pm}(\theta) &= \left( \frac{3F'_{-\beta, \beta}(\theta)F''_{-\beta, \beta}(\theta) \pm \sqrt{9(F'_{-\beta, \beta}(\theta))^2(F''_{-\beta, \beta}(\theta))^2 - 8(F'_{-\beta, \beta}(\theta))^3F'''_{-\beta, \beta}(\theta)}}{2F'''_{-\beta, \beta}(\theta)} \right)' - F'_{-\beta, \beta}(\theta) \\
&= \frac{1}{2\sqrt{b_{\beta}(\theta)^2 - 4a_{\beta}(\theta)c_{\beta}(\theta)}} \left( (a'_{\beta}(\theta)b_{\beta}(\theta) - a_{\beta}(\theta)b'_{\beta}(\theta) - 2F''_{-\beta, \beta}(\theta)) \sqrt{b_{\beta}(\theta)^2 - 4a_{\beta}(\theta)c_{\beta}(\theta)} \right. \\
&\qquad \left. \mp (b_{\beta}(\theta)^2a'_{\beta}(\theta) - b_{\beta}(\theta)a_{\beta}(\theta)b'_{\beta}(\theta) + 2a_{\beta}(\theta)c'_{\beta}(\theta) - 2a_{\beta}(\theta)c_{\beta}(\theta)a'_{\beta}(\theta)) \right) \quad (A.14)
\end{aligned}
$$

is only expressed in terms of $F'_{-\beta, \beta}$ and its derivatives that are classical functions. Zeros of $T'_{\beta, \pm}$ must satisfy (taking the square of the numerator of (A.14) to remove the $\pm$ sign):

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

Replacing $F'''_{-\beta, \beta}$ and $F^{(4)}_{-\beta, \beta}$ in terms of $F'_{-\beta, \beta}$ and $F''_{-\beta, \beta}$ using (A.2) gives after a tedious computation that the r.h.s. is of the form $8F'''_{-\beta, \beta}(\theta)^2F'_{-\beta, \beta}(\theta)S_{\beta}(\theta)$. Therefore zeros of $T'_{\beta, \pm}$ are among those of $S_{\beta}$. Simple asymptotic expansions around $\theta = 0$ and $\theta \to +\infty$ provide the following results:

$$ 
\begin{aligned}
\lim_{\theta \to +\infty} T_{\beta, \pm}(\theta) &= 0, \\
T_{\beta,+}(0) &= -\frac{\left(\sqrt{2\pi}(2\phi(\beta) - 1)(\beta^2 - 3)\exp\left(\frac{\beta^2}{2}\right) + 6\beta\right)\exp\left(-\frac{\beta^2}{2}\right)\sqrt{2}}{2\sqrt{\pi}(\beta^2 - 3)}, \quad \forall \beta \in \mathbb{R}_{>0} \setminus \{\sqrt{3}\}, \\
T'_{\beta,+}(\theta) &= -\frac{2\beta}{3\sqrt{2\pi}}e^{-\frac{1}{2}\beta^2}\theta + O(\theta^2), \quad \forall \beta \in \mathbb{R}_{>0} \setminus \{\sqrt{3}\}, \\
T_{\beta,-}(0) &= -2\Phi(\beta) + 1 + O(\theta^2), \quad \forall \beta \in \mathbb{R}_{>0} \setminus \{\sqrt{3}\}, \\
T'_{\beta,-}(\theta) &= -\frac{4\beta^5}{3\sqrt{2\pi}(\beta^2 - 3)^2}e^{-\frac{1}{2}\beta^2}\theta + O(\theta^2), \quad \forall \beta \in \mathbb{R}_{>0} \setminus \{\sqrt{3}\}, \quad (A.15) \\
T_{\sqrt{3},+}(\theta) &= \frac{3e^{-\frac{3}{2}\sqrt{6}}}{\sqrt{\pi}\theta^2} + O(1), \\
T'_{\sqrt{3},+}(\theta) &= -\frac{6e^{-\frac{3}{2}\sqrt{6}}}{\sqrt{\pi}\theta^3} + O(\theta), \\
T_{\sqrt{3},-}(\theta) &= -2\Phi(\sqrt{3}) + 1 + O(\theta^2), \\
T'_{\sqrt{3},-}(\theta) &= -\frac{1}{3\sqrt{\pi}}e^{-\frac{3}{2}\sqrt{6}\theta} + O(\theta^2).
\end{aligned}
$$

In particular, in all cases, we get that the functions $(T'_{\beta,-}, T'_{\beta,+})$ 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 $T'_{\beta, \pm}$ cannot vanish. Depending on the value of $\beta$, we have three cases:

* For $\beta < \sqrt{3}$: we have that $T'_{\beta, \pm}$ 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 $3F'_{-\beta, \beta}(\theta_1)F''_{-\beta, \beta}(\theta_1) < 0$ from Lemma Appendix A.1). Moreover, from (A.15), we have $T_{\beta, -}(0) < 0$. This implies that
$\lim_{\theta \to \theta_{1,-}} T_{\beta,-} = -\infty$ and $\lim_{\theta \to \theta_{1,+}} T_{\beta,-} = +\infty$. Therefore, the sign of $T'_{\beta,-}$ 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 $3F'_{-\beta,\beta}(\theta_1)F''_{-\beta,\beta}(\theta_1) < 0$ from Lemma Appendix A.1). Therefore the sign of $T'_{\beta,+}$ 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 $T'_{\sqrt{3},\pm}$ 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 $3F'_{-\sqrt{3},\sqrt{3}}(\theta_1)F''_{-\sqrt{3},\sqrt{3}}(\theta_1) < 0$ from Lemma Appendix A.1). Moreover, from (A.15), we have $T_{\sqrt{3},-}(0) < 0$. This implies that $\lim_{\theta \to \theta_{1,-}} T_{\sqrt{3},-} = -\infty$ and $\lim_{\theta \to \theta_{1,+}} T_{\sqrt{3},-} = +\infty$. Therefore, the sign of $T'_{\sqrt{3},-}$ 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.15) that $T_{\sqrt{3},+}(0) > 0$ and $T_{\sqrt{3},+}$ is a smooth function at $\theta = \theta_1$ (because $3F'_{-\sqrt{3},\sqrt{3}}(\theta_1)F''_{-\sqrt{3},\sqrt{3}}(\theta_1) < 0$ from Lemma Appendix A.1). Therefore the sign of $T'_{\sqrt{3},+}$ 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 $T'_{\beta,\pm}$ 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 $3F'_{-\beta,\beta}(\theta_1)F''_{-\beta,\beta}(\theta_1) < 0$ and $3F'_{-\beta,\beta}(\theta_2)F''_{-\beta,\beta}(\theta_2) > 0$ from Lemma Appendix A.1). Moreover, we have from (A.15) $T_{\beta,+}(0) < 0$ so that $\lim_{\theta \to \theta_{1,-}} T_{\beta,+} = -\infty$ and $\lim_{\theta \to \theta_{1,+}} T_{\beta,+} = +\infty$. 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 $3F'_{-\beta,\beta}(\theta_1)F''_{-\beta,\beta}(\theta_1) < 0$ and $3F'_{-\beta,\beta}(\theta_2)F''_{-\beta,\beta}(\theta_2) > 0$ from Lemma Appendix A.1). We have $T_{\beta,-}(0) < 0$ so that it is strictly negative on $(0, \theta_2)$ and $\lim_{\theta \to \theta_{2,-}} T_{\beta,-} = -\infty$ and $\lim_{\theta \to \theta_{2,+}} T_{\beta,-} = +\infty$. 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 $F'''_{-\beta,\beta}$ as potential zeros of $Z_\beta$. Under the assumption that $S_\beta$ is strictly positive on $\mathbb{R}_{>0}$ we have

* For $\beta > \sqrt{3}$, $F'''_{-\beta,\beta}$ 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.10) $F_{-\beta,\beta}(\theta_2) + \frac{c_\beta(\theta_2)}{b_\beta(\theta_2)} = 0$ 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.10) $F_{-\beta,\beta}(\theta_1) + \frac{c_\beta(\theta_1)}{b_\beta(\theta_1)} = 0$ leading to a contradiction. Therefore $\theta_1$ is not a zero of $Z_\beta$.

* For $\beta \le \sqrt{3}$: $F'''_{-\beta,\beta}$ 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.10) $F_{-\beta,\beta}(\theta_1) + \frac{c_\beta(\theta_1)}{b_\beta(\theta_1)} = 0$ leading to a contradiction. Therefore $\theta_1$ is not a zero of $Z_\beta$.
This concludes the proof of Proposition Appendix A.2. □

## Appendix A.3. Proof of the sufficient condition

From Proposition Appendix 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:

$$ 
\begin{aligned}
F'(\theta) &= -\frac{2}{\sqrt{2\pi}} e^{-\frac{1}{2}\theta^2 - \frac{1}{2}\beta^2} \sinh(\beta\theta), \\
F''(\theta) &= \frac{2}{\sqrt{2\pi}} e^{-\frac{1}{2}\theta^2 - \frac{1}{2}\beta^2} (\theta \sinh(\beta\theta) - \beta \cosh(\beta\theta)).
\end{aligned}
$$

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:

$$ 
\begin{aligned}
\tilde{S}_{\beta}(\theta) &= \sinh(5\beta\theta) + (4\beta^6 + 4(3\theta^2 + 6)\beta^4 + 12\beta^2 - 5) \sinh(3\beta\theta) \\
&\quad + (-12\beta^6 + 4(3\theta^2 + 18)\beta^4 - 36\beta^2 + 10)) \sinh(\beta\theta) \\
&\quad - 4(3\beta^2 + \theta^2 + 6) \beta^3\theta \cosh(3\beta\theta) \\
&\quad + 4(-33\beta^2 + \theta^2 + 6) \beta^3\theta \cosh(\beta\theta).
\end{aligned}
$$

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 $A_\beta(s) := \tilde{S}_\beta\left(\frac{s}{\beta}\right)$. 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:

$$ 
\begin{aligned}
A_\beta(s) &= \sinh(5s) + (4\beta^6 + 24\beta^4 + 12s^2\beta^2 + 12\beta^2 - 5) \sinh(3s) \\
&\quad + (-12\beta^6 + 72\beta^4 + 12s^2\beta^2 - 36\beta^2 + 10) \sinh(s) \\
&\quad - 4(3\beta^4 + 6\beta^2 + s^2)s \cosh(3s) + 4(-33\beta^4 + 6\beta^2 + s^2)s \cosh(s).
\end{aligned}
$$

Let us observe that $B_s(\mu) := A_\beta(s)$ is a polynomial of degree 3 in $\mu := \beta^2$:

$$ 
\begin{aligned}
B_s(\mu) &:= A_{\mu^2}(s) \\
&= 4(\sinh(3s) - 3\sinh(s))\mu^3 \\
&\quad + 4(6\sinh(3s) + 18\sinh(s) - 33s\cosh(s) - 3s\cosh(3s))\mu^2 \\
&\quad + 4(3(s^2 + 1)\sinh(3s) + 3(s^2 - 3)\sinh(s) + 6s\cosh(s) - 6s\cosh(3s))\mu \\
&\quad + \sinh(5s) - 5\sinh(3s) + 10\sinh(s) + 4s^3\cosh(s) - 4s^3\cosh(3s).
\end{aligned}
$$

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 Appendix A.2. For any $s > 0$, we have

$$ \sinh(5s) - 5\sinh(3s) + 10\sinh(s) + 4s^3\cosh(s) - 4s^3\cosh(3s) > 0. \quad (A.16) $$

*Proof of Lemma Appendix A.2.* One may rewrite

$$ 
\begin{aligned}
& \sinh(5s) - 5\sinh(3s) + 10\sinh(s) + 4s^3\cosh(s) - 4s^3\cosh(3s) = 16\sinh(s)^2(\sinh(s)^3 - s^3\cosh(s)) \\
& = 16\sinh(s)^2\cosh(s)\left(\frac{\sinh(s)^3}{\cosh(s)} - s^3\right).
\end{aligned}
$$
Let us then define $a(s) := \frac{\sinh(s)^3}{\cosh(s)} - s^3$ for $s > 0$. We have

$$
\begin{align*}
a'(s) &= \frac{2 \cosh(s)^4 + (-3s^2 - 1) \cosh(s)^2 - 1}{\cosh(s)^2}, \\
a''(s) &= \frac{(4 \sinh(s) \cosh(s)^4 - 6s \cosh(s)^3 + 2 \sinh(s))}{\cosh(s)^3}, \\
a^{(3)}(s) &= \frac{2 \sinh(s)^4 (4 \cosh(s)^2 + 3)}{\cosh(s)^4} > 0,
\end{align*}
$$

so that $a''$ is strictly increasing on $\mathbb{R}_{>0}$ and since $a''(0) = 0$, we get that $a''$ is strictly positive on $\mathbb{R}_{>0}$. This implies that $a'$ is strictly increasing on $\mathbb{R}_{>0}$ and since $a'(0) = 0$ we end up with $a'$ 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 Appendix A.2. $\square$

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:

$$
B'_{s}(\mu) = 12 \left[ \begin{aligned}
& (\sinh(3s) - 3\sinh(s))\mu^2 \\
& + 2(2\sinh(3s) + 6\sinh(s) - 11s\cosh(s) - s\cosh(3s))\mu \\
& + (s^2 + 1)\sinh(3s) + (s^2 - 3)\sinh(s) + 2s\cosh(s) - 2s\cosh(3s)
\end{aligned} \right].
$$

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

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

Let us assume that $\tilde{\Delta}$ is strictly positive on $\mathbb{R}_{>0}$ so that $\Delta(s)$ is strictly negative, i.e. $B'_s$ 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 Appendix 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

$$
\begin{align*}
\tilde{\Delta}'(s) &= \sinh(s) [10s \cosh(s)^3 \sinh(s) - 13 \cosh(s)^4 - 18s^2 \cosh(s)^2 \\
&\quad + 32s \sinh(s) \cosh(s) + 5 \cosh(s)^2 - 3s^2 + 8] \\
&:= \sinh(s)Q(s),
\end{align*}
$$

with

$$
\begin{align*}
Q'(s) &= 2 \sinh(s) \left[ 20s \sinh(s) \cosh(s)^2 - 21 \cosh(s)^3 - 18s^2 \cosh(s) + 19s \sinh(s) + 21 \cosh(s) \right] \\
&:= 2 \sinh(s) R(s),
\end{align*}
$$

with

$$
R(s) := 20s \sinh(s) \cosh(s)^2 - 21 \cosh(s)^3 - 18s^2 \cosh(s) + 19s \sinh(s) + 21 \cosh(s).
$$

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

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

**Proposition Appendix A.4.** Let $\beta > 0$. The function $s \mapsto R(s) := 20s \sinh(s) \cosh(s)^2 - 21 \cosh(s)^3 - 18s^2 \cosh(s) + 19s \sinh(s) + 21 \cosh(s)$ is strictly positive on $\mathbb{R}_{>0}$.

*Proof of Proposition Appendix A.4.* Let us first rewrite:

$$ 
\begin{aligned}
R(s) &= 20s \sinh(s) \cosh(s)^2 - 21 \cosh(s)^3 - 18s^2 \cosh(s) + 19s \sinh(s) + 21 \cosh(s) \\
&= 20s \sinh(s) \cosh(s)^2 - 21 \cosh(s) \sinh(s)^2 - 18s^2 \cosh(s) + 19s \sinh(s) \\
&= -(18 \cosh(s)s^2 - (20 \sinh(s) \cosh(s)^2 + 19 \sinh(s))s + 21 \cosh(s) \sinh(s)^2),
\end{aligned}
$$

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

$$s = \frac{20 \sinh(s) \cosh(s)^2 + 19 \sinh(s)}{36 \cosh(s)} \pm \frac{\sinh(s)}{36 \cosh(s)} \sqrt{400 \cosh(s)^4 - 752 \cosh(s)^2 + 361}. \quad (A.17)$$

Let us define

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

Note in particular that

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

Moreover, we have

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

Therefore, zeros of $R'_\pm$ must satisfy

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

which is equivalent to

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

Since the discriminant of $2000X^2 - 3781X + 1805$ is negative (equal to $-144039$), we conclude that $R'_\pm$ does not vanish on $\mathbb{R}_{>0}$ and thus has a constant sign on $\mathbb{R}_{>0}$. In particular, from (A.18), we get that $R'_+$ 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.18) $R'_-$ 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.17) 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 Appendix A.4. $\square$

Finally, we conclude from Proposition Appendix A.3 and Proposition Appendix A.4 that for any $\beta > 0$, $S_\beta$ is a strictly positive function on $\mathbb{R}_{>0}$ so that from Proposition Appendix A.2 we obtain that $h_{-\beta,\beta}$ is a strictly concave function on $\mathbb{R}_{>0}$. This concludes the proof of Lemma 2.3 stated in the main part of the paper. $\square$
# Appendix B. Proofs for truncated exponential random variables

## Appendix B.1. Restriction to $\lambda = 1$

Let $X \sim \text{Exp}(\lambda)$ with $\lambda > 0$ and define $X_T$ the truncated distribution on $(a, b)$ with $a \ge 0$ and $b > 0$. Let $Y = \lambda X \sim \text{Exp}(1)$ and $Y_T$ be the truncated distribution on $[\alpha := \lambda a, \beta := \lambda b]$. Then obviously $Y_T = \lambda X_T$ so that:

$$ \mathbb{E}[\exp(\theta(X_T - \mathbb{E}[X_T]))] = \mathbb{E}\left[\exp\left(\frac{\theta}{\lambda}(Y_T - \mathbb{E}[Y_T])\right)\right], \quad \forall \theta \in \mathbb{R}. \quad (B.1) $$

Thus, we immediately obtain that

$$ \|Y_T\|_{\text{vp}} = \lambda \|X_T\|_{\text{vp}}, \quad (B.2) $$

and, we may restrict the analysis to $Y \sim \text{Exp}(1)$.

## Appendix B.2. Proof of the optimal proxy variance for $Y \sim \text{Exp}(1)$

*Proof of Theorem 3.2.* We will prove that $\|Y_T\|_{\text{vp}}^2$ is the optimal variance proxy if and only if it is the solution to (8) 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 $\mathbb{R}^2$. Clearly $s^2$ is a variance proxy if $G_{\alpha, \beta, s}(\theta) \le 0$ for all $\theta \in \mathbb{R}$. In particular, we have:

$$ \begin{gathered} g_{\alpha, \beta, s}(0) = 0, \quad g'_{\alpha, \beta, s}(0) = 0, \quad \forall s > 0, \\ G_{\alpha, \beta, s}(0) = G'_{\alpha, \beta, s}(0) = 0, \quad \lim_{\theta \to \pm\infty} G_{\alpha, \beta, s}(\theta) = e^{\alpha} - e^{\beta} < 0 \end{gathered} \quad (B.3) $$

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

$$ \|Y_T\|_{\text{vp}} \ge s_{\inf}(\alpha, \beta) := \sqrt{1 - \frac{(\beta - \alpha)^2 e^{\alpha + \beta}}{(e^{\beta} - e^{\alpha})^2}}. \quad (B.4) $$

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

$$ G'_{\alpha, \beta, s}(\theta) = \frac{1}{(\theta - 1)^2} e^{\alpha\theta + \beta} e^{-\frac{1}{2}s^2\theta^2 - \mathbb{E}[Y_T]\theta} h_{\alpha, \beta, s}(\theta), \quad (B.5) $$

where:

$$ \begin{aligned} h_{\alpha, \beta, s}(\theta) &:= (-s^2\theta^2 + (s^2 + \beta - \mathbb{E}[Y_T])\theta + \mathbb{E}[Y_T] - \beta - 1) e^{(\theta-1)(\beta-\alpha)} \\ &\quad + s^2\theta^2 - (s^2 + \alpha - \mathbb{E}[Y_T])\theta - \mathbb{E}[Y_T] + \alpha + 1 \\ h'''_{\alpha, \beta, s}(\theta) &= \frac{e^{-(\theta-1)(\beta-\alpha)}}{(\beta-\alpha)} P_{\alpha, \beta, s}(\theta), \end{aligned} $$

with:

$$ \begin{aligned} P_{\alpha, \beta, s}(\theta) &:= A_{\alpha, \beta, s}\theta^2 + B_{\alpha, \beta, s}\theta + C_{\alpha, \beta, s}, \\ A_{\alpha, \beta, s} &:= -s^2(\beta - \alpha)^2, \\ B_{\alpha, \beta, s} &:= (\alpha - \beta)(\alpha - \beta + 6)s^2 + (\beta - \alpha)^2(\beta - \mathbb{E}[Y_T]), \\ C_{\alpha, \beta, s} &:= 3(\beta - \alpha - 2)s^2 + (\beta - \alpha)((\beta - \alpha - 3)\mathbb{E}[Y_T] + \alpha\beta - \beta^2 + \alpha + 2\beta). \end{aligned} $$

<footnote n="2">²Although the function $g_{\alpha, \beta, s}$ defined in Equation (7) 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.</footnote>
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 Appendix B.1. It is given by:

$$ \Delta_{\alpha,\beta,s} := B_{\alpha,\beta,s}^2 - 4A_{\alpha,\beta,s}C_{\alpha,\beta,s} \\ = (\beta - \alpha)^2 ((\beta - \alpha)^2 + 12)s^4 - 2(\beta - \alpha)^2(\beta - \mathbb{E}[Y_T] + 2)s^2 + (\beta - \alpha)^2(\beta - \mathbb{E}[Y_T])^2). $$

**Lemma Appendix B.1.** If $\Delta_{\alpha,\beta,s} \le 0$, 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 Appendix B.1.* For the proof we will need the following immediate results:

$$ \begin{align*} \lim_{\theta \to -\infty} h''_{\alpha,\beta,s}(\theta) &= 2s^2 > 0, & \lim_{\theta \to +\infty} h''_{\alpha,\beta,s}(\theta) &= -\infty, \\ \lim_{\theta \to \pm\infty} h'_{\alpha,\beta,s}(\theta) &= -\infty, & h'_{\alpha,\beta,s}(1) &= 0, \\ \lim_{\theta \to -\infty} h_{\alpha,\beta,s}(\theta) &= +\infty, & h_{\alpha,\beta,s}(1) &= 0, \\ \lim_{\theta \to +\infty} h_{\alpha,\beta,s}(\theta) &= -\infty. & & \end{align*} \tag{B.6} $$

**Case** $\Delta_{\alpha,\beta,s} \le 0$. Let us assume that $\Delta_{\alpha,\beta,s} \le 0$. Since $A_{\alpha,\beta,s} < 0$, we get that $h'''_{\alpha,\beta,s}$ is a strictly negative function on $\mathbb{R}$ except at one point where it vanishes when $\Delta_{\alpha,\beta,s} = 0$. This implies that $h''_{\alpha,\beta,s}$ is a strictly decreasing function on $\mathbb{R}$. From the first two equations of (B.6), we conclude that there exists a unique value $\theta_1 \in \mathbb{R}$ such that $h''_{\alpha,\beta,s}(\theta_1) = 0$. Moreover, $h''_{\alpha,\beta,s}$ is strictly positive on $(-\infty, \theta_1)$ and strictly negative on $(\theta_1, +\infty)$. Thus, $h'_{\alpha,\beta,s}$ is a strictly increasing function on $(-\infty, \theta_1)$ and a strictly decreasing function on $(\theta_1, +\infty)$. Since we know that $h'_{\alpha,\beta,s}(1) = 0$ and $\lim_{\theta \to \pm\infty} h'_{\alpha,\beta,s}(\theta) = -\infty$, we conclude that $h'_{\alpha,\beta,s}$ may have at most two zeros. We have two subcases:

* $h'_{\alpha,\beta,s}$ has a double zero at $\theta_1 = 1$. In this case, from the limit at infinity given by (B.6), $h'_{\alpha,\beta,s}$ is strictly negative on $\mathbb{R} \setminus \{1\}$ and thus $h_{\alpha,\beta,s}$ is a strictly decreasing function on $\mathbb{R}$. From (B.6), 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 $G'_{\alpha,\beta,s}$ from (B.5). But this is a contradiction to the fact that $G'_{\alpha,\beta,s}(0) = 0$ in (B.3) so that this subcase may be discarded.

* $h'_{\alpha,\beta,s}$ has two distinct zeros, i.e. there exists a unique $\theta_2 \ne 1$ such that $h'_{\alpha,\beta,s}(\theta_2) = 0$. Let us denote $\tilde{\theta}_1 = \min(1, \theta_1)$ and $\tilde{\theta}_2 = \max(1, \theta_2)$. Then $h'_{\alpha,\beta,s}$ 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 (B.6) 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 $\theta_3 \ne 1$ 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 $G'_{\alpha,\beta,s}$ from (B.5). However, from (B.3) and the fact that $s \ge s_{\text{inf}}(\alpha, \beta)$, 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 (B.3) 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 $\theta_0 \ne \theta_1$ such that $h'''_{\alpha,\beta,s}(\theta_0) = h'''_{\alpha,\beta,s}(\theta_1) = 0$. Moreover, since $A_{\alpha,\beta,s} < 0$, we get that $h'''_{\alpha,\beta,s}$ is strictly negative on $(-\infty, \theta_1) \cup (\theta_2, +\infty)$ and strictly positive on $(\theta_1, \theta_2)$. Thus, $h''_{\alpha,\beta,s}$ is strictly decreasing on $(-\infty, \theta_1)$, strictly increasing on $(\theta_1, \theta_2)$ and strictly decreasing on $(\theta_2, +\infty)$. Since $\lim_{\theta \to -\infty} h''_{\alpha,\beta,s}(\theta) > 0$ and $\lim_{\theta \to +\infty} h''_{\alpha,\beta,s}(\theta) = 0$, we conclude that $\Delta_{\alpha,\beta,s} > 0$.
$-\infty$ we conclude that $h''_{\alpha,\beta,s}$ may have one or two (one simple and one double zero) or three distinct zeros depending on the sign of $h'_{\alpha, \beta}, s(\theta_1)$ and $h'_{\alpha, \beta}, s(\theta_2)$.

The case when $h''_{\alpha,\beta,s}$ has only one or two zeros is identical to the previous case (because in both cases $h''_{\alpha,\beta,s}$ changes sign only at the unique simple zero) and we refer to its conclusion. Let us thus assume that $h''_{\alpha,\beta,s}$ has three distinct zeros $\hat{\theta}_1 < \theta_0 < \hat{\theta}_2 < \theta_1 < \hat{\theta}_3$. Then $h''_{\alpha,\beta,s}$ 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 $\lim_{\theta \to \pm\infty} h'_{\alpha,\beta,s}(\theta) = -\infty, h'_{\alpha,\beta,s}$ 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, $h'_{\alpha,\beta,s}$ only changes sign twice) and we refer to its conclusion. Let us thus assume that $h'_{\alpha,\beta,s}$ 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 $h_{\alpha,\beta,s}(1) = h'_{\alpha,\beta,s}(1) = 0$ 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 $\lim_{\theta \to +\infty} h_{\alpha,\beta,s}(\theta) = +\infty$ and $\lim_{\theta \to -\infty} h_{\alpha,\beta,s}(\theta) = -\infty$, it cannot change sign twice so that $h_{\alpha,\beta,s}$ may only change sign at one or three distinct locations and so is $G'_{\alpha,\beta,s}$ from (B.5).

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 $G'_{\alpha,\beta,s}$ changes sign at three distinct locations denoted $\tilde{\theta}_1 < \tilde{\theta}_2 < \tilde{\theta}_3$. From the limit at infinity, $G'_{\alpha,\beta,s}$ 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 (B.3), we know that $\theta = 0$ is a local extremum of $G_{\alpha,\beta,s}$ and from (B.4) we have assumed $s \ge s_{\inf}(\alpha, \beta)$ so that $G'''_{\alpha,\beta,s}(0) < 0$. Hence $\theta = 0$ is necessarily a local maximum of $G_{\alpha,\beta,s}$ and its value is null from (B.3). 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 $\theta_3 \ne 0$. 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 Appendix B.1. $\square$

By the condition on $\Delta_{\alpha,\beta,s}$, we also get the claimed upper bound of the optimal variance proxy from this lemma. Indeed, $\Delta_{\alpha,\beta,s} \le 0$ 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 $\|Y_T\|_{vp} \le s_1(\alpha, \beta)$, with:

$$s_1(\alpha, \beta) := \sqrt{\frac{(\beta - \alpha)^2 (\beta - \mathbb{E}[Y_T] + 2) - \delta_{\alpha, \beta}}{(\beta - \alpha)^2 + 12}}, \quad s_2(\alpha, \beta) := \sqrt{\frac{(\beta - \alpha)^2 (\beta - \mathbb{E}[Y_T] + 2) + \delta_{\alpha, \beta}}{(\beta - \alpha)^2 + 12}},$$

and where $\delta_{\alpha,\beta}$ was defined in (9). 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 (B.4). The previous analysis implies that for $s < \|Y_T\|_{vp}$ we must have $\Delta_{\alpha,\beta,s} > 0$ and the existence of $\theta_c(\alpha, \beta, s) \ne 0$ with $G_{\alpha,\beta,s}(\theta_c(\alpha, \beta, s)) > 0$. Thus, taking the limit $s \to \|Y_T\|_{vp}$ with $s < \|Y_T\|_{vp}$ implies that $G_{\alpha,\beta, \|Y_T\|_{vp}}(\theta_c(\alpha, \beta, \|Y_T\|_p)) = 0$. Notice that $\theta_c(\alpha, \beta, \|Y_T\|_p) = \pm\infty$ is not possible because we have $\lim_{\theta \to \pm\infty} G_{\alpha,\beta,s}(\theta) = e^{\alpha} - e^{\beta} < 0$ independent of $s$. Hence we must have

$$G_{\alpha,\beta, \|Y_T\|_{vp}}(\theta_c(\alpha, \beta, \|Y_T\|_{vp})) = 0.$$

To finish it remains to prove that $\theta_c(\alpha, \beta, \|Y_T\|_{vp}) \ne 0$. Indeed, let us assume by contradiction that $\theta_c(\alpha, \beta, \|Y_T\|_{vp}) = 0$. The previous analysis shows that for $s < \|Y_T\|_{vp}$, $G'_{\alpha,\beta,s}$ has two distinct zeros inside $(0, \theta_c(\alpha, \beta, s))$ and thus by Rolle's theorem that $G''_{\alpha,\beta,s}$ has at least one zero in $(0, \theta_c(\alpha, \beta, s))$. Thus at the limit, we would get $G'''_{\alpha,\beta, \|Y_T\|_{vp}}(0) = 0$, i.e. $\theta = 0$ would be at least a triple zero of $G_{\alpha,\beta, \|Y_T\|_{vp}}$. In fact, in order to remain locally negative around $\theta = 0$, we would necessarily have a zero of order four, i.e. $G'''_{\alpha,\beta, \|Y_T\|_{vp}}(0) = 0$. Hence $\theta = 0$ would be a zero of order four of $g_{\alpha,\beta, \|Y_T\|_{vp}}$. This would imply
<figure><img src="image_5.png" /><figcaption>Figure B.4: Illustration of the proof of Theorem 3.2. Function $g_{\alpha,\beta,s}$ defined in Equation (7) 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 $s^2 = s_c^2 = \|Y_T\|_{vp}^2$ ($s = s_c \approx 0.8107$). 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).</figcaption></figure>

$\|Y_T\|_{vp} = s_{inf}(\alpha, \beta)$ (triple zero using the definition of $s_{inf}(\alpha, \beta)$ and $g'''_{\alpha,\beta,s_{inf}(\alpha,\beta)}(0)$ (zero of order four). However we have:

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

We have that $g'''_{\alpha,\alpha+\epsilon,s_{inf}(\alpha,\alpha+\epsilon)}(0) > 0$ for all $\epsilon > 0$ so that $g'''_{\alpha,\beta,s_{inf}(\alpha,\beta)}(0) > 0$ for any $\alpha < \beta$ i.e., we cannot have $\theta_c(\alpha, \beta, \|Y_T\|_{vp}) = 0$. Indeed, the series expansion of $P(\epsilon)$ at $\epsilon = 0$ starts as

$$P(\epsilon) \stackrel{\epsilon \to 0}{\rightleftharpoons} \frac{1}{120}\epsilon^7 + \frac{1}{80}\epsilon^8 + \frac{59}{6048}\epsilon^9 + \frac{53}{10080}\epsilon^{10} + \frac{1327}{604800}\epsilon^{11} + \frac{19}{25200}\epsilon^{12} + \frac{59}{266112}\epsilon^{13} + O(\epsilon^{14})$$

Then, we get that the general Taylor coefficient $c_n$ of $P$ is given by $c_n = \frac{2 \times 3^n - 6 \times 2^n + 6 - n(n-1)(n-2)(2^{n-3}+1)}{n!}$ and because $3^n$ is growing faster than $2^n$ times any cubic polynomial at infinity, it is then easy to see by induction that $(n!)c_n$ is positive and growing for $n \ge 14$. Finally, since $P(\epsilon)$ is analytic and its Taylor series has only non-negative coefficients, it is then positive on $\mathbb{R}_+^*$.

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 B.4. This concludes the proof of Theorem 3.2. □

## Appendix B.3. Exact solution

We have proved that the optimal proxy variance is the unique solution of

$$g_{\alpha,\beta,\|Y_T\|}(\theta_c) = 0 \quad \text{and} \quad g'_{\alpha,\beta,\|Y_T\|}(\theta_c) = 0 \quad \text{with} \quad s_{inf}(\alpha, \beta) < \|Y_T\| \le s_1(\alpha, \beta) \quad \text{and} \quad \theta_c \ne 0. \qquad (B.7)$$

We observe that

$$\theta_c = 2, \quad \|Y_T\| = \frac{\sqrt{(\beta - \alpha + 2)e^{\alpha} + (\beta - \alpha - 2)e^{\beta}}}{\sqrt{2}\sqrt{e^{\beta} - e^{\alpha}}} \qquad (B.8)$$

is a solution of $g_{\alpha,\beta,\|Y_T\|}(\theta_c) = 0$ and $g'_{\alpha,\beta,\|Y_T\|}(\theta_c) = 0$. By uniqueness, we therefore conclude that the optimal variance proxy is given by:
$$ \|Y_T\|_{\text{vp}}^2 = \frac{(\beta - \alpha + 2)e^{\alpha} + (\beta - \alpha - 2)e^{\beta}}{2(e^{\beta} - e^{\alpha})} = \frac{(\beta - \alpha)(e^{\alpha} + e^{\beta})}{2(e^{\beta} - e^{\alpha})} - 1. $$

Appendix B.4. Proving that the truncated exponential is never strictly sub-Gaussian

Let us study the sign of

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

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

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

We have:

$$ K'(\epsilon) = 2\epsilon \cosh \epsilon - 6 \sinh \epsilon + 4\epsilon, $$

$$ K''(\epsilon) = 2\epsilon \sinh \epsilon - 4 \cosh \epsilon + 4, $$

$$ K'''(\epsilon) = 2(\epsilon \cosh \epsilon - \sinh \epsilon). $$

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