# On the first hitting time of a high-dimensional orthant

Emmanuel Humbert, Kilian Raschel

► To cite this version:

Emmanuel Humbert, Kilian Raschel. On the first hitting time of a high-dimensional orthant. Electronic Communications in Probability, 2025, 30, pp.39. 10.1214/25-ECP688. hal-05220909

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# On the first hitting time of a high-dimensional orthant*

Emmanuel Humbert†

Kilian Raschel‡

## Abstract

We consider a collection of independent standard Brownian particles (or random walks), starting from a configuration where at least one particle is negative, and study the first time they all become positive. This is clearly equivalent to studying the first hitting time of the positive orthant or the first exit time from the complement of the positive orthant. While it turns out to be possible to compute the distribution of these hitting times for one and two particles, the distribution (and even its tail asymptotics) is not known in closed form for three or more particles. In this paper we study the tail asymptotics of the distribution as the number of particles tends to infinity. Our main techniques come from spectral geometry: we prove new asymptotic estimates for the principal eigenvalue of the complement of a high-dimensional orthant, which we believe are of independent interest.

**Keywords:** survival probability; Brownian motion in cones; orthant; eigenvalue problem on the sphere; high-dimensional sphere.

**MSC2020 subject classifications:** 60J65; 60G40; 58C40.

Submitted to ECP on November 26, 2024, final version accepted on April 23, 2025.

## 1 Introduction and main results

Consider $d \ge 1$ standard Brownian particles $B_1(t), \dots, B_d(t)$ (with zero drift and unit variance) which at $t = 0$ are not all non-negative (i.e., at least one of them starts with a negative value). Our main quantity of interest is the first hitting time of the positive orthant, or equivalently the first exit time from the complement of the positive orthant:

$$ \tau_d = \inf\{t > 0 : B_1(t) > 0, \dots, B_d(t) > 0\}. \quad (1.1) $$

In the probabilistic literature, this problem can be interpreted as a capture problem [9] or an absorption problem [17]. This first passage time problem has also attracted some attention in the physics community; see [3, 4, 10], and is inspired by classical experiments in thermodynamics. In fact, if we interpret the negative half-line as an almost confined space with a small hole through which particles can escape, we can imagine the following

<footnote>*EH is supported by the project Einstein-PPF (ANR-23-CE40-0010), funded by the French National Research Agency. KR is supported by the project RAWABRANCH (ANR-23-CE40-0008), funded by the French National Research Agency.</footnote>

<footnote>†Université de Tours, Institut Denis Poisson, Tours, France.</footnote>

<footnote>E-mail: emmanuel.humbert@univ-tours.fr http://www.lmpt.univ-tours.fr/humbert/</footnote>

<footnote>‡CNRS, International Research Laboratory France-Vietnam in mathematics and its applications, Vietnam Institute for Advanced Study in Mathematics, Hanoi, Vietnam.</footnote>

<footnote>E-mail: raschel@math.cnrs.fr https://raschel.perso.math.cnrs.fr/</footnote>
<figure><img src="image_2.png" /><figcaption>Figure 1: Random walks staying in the complement of the d-dimensional orthant, for $d = 1, 2, 3$.</figcaption></figure>

experiment. Let $d$ particles start from the confined space, with a ballistic motion to the right. Then the particles eventually all leave the space and go to the positive half-line. If the same particles follow Brownian motions, how likely is it that they will all have left the room at a given time, especially in the regime of infinitely many particles?

To formulate the problem precisely, let $x \in \mathbb{R}^d$ be an arbitrary (starting) point and let $P_x(\cdot)$ be the probability conditional on the event $x = (B_1(0), \dots, B_d(0))$. It is proved in [12, Corollary 1.3] and in [2, Corollary 1] that the survival probability $P_x(\tau_d > t)$ behaves asymptotically when $t \to \infty$ as

$$P_x(\tau_d > t) \sim \frac{V_d(x)}{t^{p_d/2}}, \qquad (1.2)$$

where $V_d(x)$ is the unique (up to multiplicative scaling) harmonic function in the cone $C = \mathbb{R}^d \setminus \mathbb{R}_+^d$ (the complement of the positive orthant) and the exponent $p_d$ is given by

$$p_d = \sqrt{\lambda_1 + \left(\frac{d}{2} - 1\right)^2 - \left(\frac{d}{2} - 1\right)}, \qquad (1.3)$$

with $\lambda_1 = \lambda_1(d)$ defined as follows. It is (see e.g. [2, Equation (1.3)]) the smallest (positive) eigenvalue $\lambda$ of the Dirichlet problem for the Laplace-Beltrami operator $\Delta_{S^{d-1}}$ on the sphere $S^{d-1} := \{(x_1, \dots, x_d) \in \mathbb{R}^d : x_1^2 + \dots + x_d^2 = 1\} \subset \mathbb{R}^d$:

$$\left\{ \begin{array}{ll} -\Delta_{S^{d-1}} m = \lambda m & \text{in } C, \\ m = 0 & \text{in } \partial C, \end{array} \right. \qquad (1.4)$$

where $C$ is, as above, the complement of the positive orthant. Remember that in general the quantity $\lambda_1$ cannot be computed in closed form. An asymptotic statement similar to (1.2) is proved in the discrete setting (random walks) in [13, Theorem 1].

It is well known that $p_1 = 1$ (in dimension 1, $\tau_1$ in (1.1)) is the first passage time at 0 of a standard Brownian motion started at a negative value, whose distribution is classical); see Figure 1. It is also known that $p_2 = \frac{2}{3}$. In dimension 2, we should mention that the enumeration of walks in the complement of a quadrant has recently attracted some attention; see [6–8, 20]. In particular, the exponent $\frac{p_2}{2} = \frac{1}{3}$ is obtained in [6]; see the remarks following theorems 1 and 4 in that paper; see also [20, Theorem 3.1] and [8, Corollary 4.3]. As pointed out in [20, Section 4], the value of $p_3$ (and a fortiori of $p_d$ for $d > 3$) is unknown. However, the approximation $p_3 = 0.4542$ is derived in [5, Section 7.1]; see also [3, Table 1]. It would be interesting to certify the previous approximation, as was done in [11] for linear transformations of three-dimensional orthonants.

Arguing on the number of particles and on (1.2), it is intuitively clear that $p_d$ is decreasing and goes to 0 as $d \to \infty$. In this paper we explore the asymptotic behavior of $p_d$, or equivalently of $\lambda_1$ in the high dimension regime.
**Theorem 1.1.** We have $\lim_{d \to \infty} \frac{\log \lambda_1}{d} = \log(\frac{1}{2})$. More precisely, there exist constants $c, C > 0$ such that for all $d \ge 1$, $\lambda_1 \in [\frac{cd}{2d}, \frac{Cd^3}{2d}]$.

Obviously, the quantity $\frac{1}{2d}$ can be interpreted as the fraction of the volume of an arbitrary orthant in $d$ dimensions. However, to our knowledge there is no easy way to translate this elementary observation into a direct estimate of $\lambda_1$. As an immediate consequence of Theorem 1.1, we have:

**Corollary 1.2.** The critical exponent $p_d$ of the survival probability (1.2) goes to 0 and is equivalent to

$$p_d \sim \frac{\lambda_1}{d}.$$

Let us conclude this introduction by mentioning five challenging open problems. First, it would be interesting to study the joint asymptotics of the survival probability $P_x(\tau_d > t)$ when $d, t \to \infty$ simultaneously. This seems to be a difficult problem, since the asymptotics (1.2) only holds in a regime where $d$ is fixed. However, there are explicit expressions for the survival probability as infinite sums of eigenfunctions; see [2, Theorem 1]. Note that when $t = \frac{T}{d}$, the joint asymptotics of the survival probability when $d, T \to \infty$ is studied in [3] and leads to interesting conjectures based on an approximation of cones with spherical caps [4].

As a second problem, we could consider Brownian motions $B_1(t), \dots, B_d(t)$ with their own variances $\sigma_1^2, \dots, \sigma_d^2$ and ask the same asymptotic question as in (1.2). Technically, this reduces to estimating the principal eigenvalue $\lambda_1$ of the complement of the linear transformation of an orthant. The quantity $\lambda_1$ should also decay exponentially fast to zero.

Third, we could even tackle a more general problem by adding a drift to each Brownian particle. In this case, the problem becomes even more complicated because the associated multidimensional random walk has a drift and, to the best of our knowledge, there is no general expression for the asymptotics of the survival probability in the case of a random walk with drift, except for partial results [14, 15].

Fourth, there is a deep connection between the exponent $p_d$ in (1.3) and the following occupation time problem. Considering a Brownian motion $B(t) = (B_1(t), \dots, B_d(t))$ starting from the origin, one can define the occupation time of the orthant $\mathbb{R}_+^d$ as

$$T = \int_0^1 1_{\{B(t) \in \mathbb{R}_+^d\}} dt.$$

It is well known that in dimension $d = 1$, $T$ follows the arcsine law. On the other hand, the distribution of $T$ is unknown as soon as $d \ge 2$. However, it was proved in [19, Theorem 2] that as $t \to 0$, $P(T \le t)/t^{p_d/2}$ is bounded from below and above by two positive constants. (This result actually holds for a much larger family of cones.) In this context, Theorem 1.1 provides a precise approximation to the short-time behaviour of the occupation time of a large-dimensional orthant, meaning the study of $P(T \le t)$ as $t \to 0$.

Fifth, in a broader setting, this work initiates the study of random walks in cones in the high-dimensional regime. When there is no cone constraint, high-dimensional random walks are quite well understood; see e.g. [16] for a study of a random walk when time and dimension simultaneously go to infinity. It would be interesting to understand how a high-dimensional cone structure affects this joint limit behaviour.

## 2 Proof of Theorem 1.1

In the whole proof, if $r \in \mathbb{N}^* = \{1, 2, 3, \dots\}$ and $A \subset \mathbb{S}^r$, $|A|_r$ will denote the volume of $A$ with respect to the standard metric $\sigma_r$ of $\mathbb{S}^r$ induced by the Euclidean metric of $\mathbb{R}^{r+1}$. In particular, $|\mathbb{S}^r|_r$ is denoted by $\omega_r$.
On the first hitting time of a high-dimensional orthant

<figure><img src="image_3.png" /><figcaption>Figure 2: Illustration of $U_d$ in (2.1) for $d = 2, 3$.</figcaption></figure>

Let $d \in \mathbb{N}$ and set

$$U_d = (\mathbb{R}^d \setminus \mathbb{R}_+^d) \cap \mathbb{S}^{d-1}; \qquad (2.1)$$

see Figure 2. Theorem 1.1 aims to estimate the first Dirichlet eigenvalue $\lambda_1$ of $U_d$ for the Dirichlet problem (1.4).

## 2.1 A preliminary lemma

For the upper bound part of the proof, we will need the following crucial lemma, to be proved in Section 2.4.

**Lemma 2.1.** Define for $d \in \mathbb{N}^*$ and $a \in [0, 1]$

$$\Sigma_d(a) = [-a, 1]^d \cap \mathbb{S}^{d-1}. \qquad (2.2)$$

Let $(a_d)_{d \ge 1}$ be any sequence such that $a_d \ll d^{-3/2}$. Then there exists $C > 0$ such that for all $d \ge 1$,

$$|\Sigma_d(a_d)|_{d-1} \le C \frac{\omega_{d-1}}{2^d}.$$

To the best of our knowledge, the volume of $\Sigma_d(a)$ does not admit a closed-form expression.

## 2.2 The lower bound

We denote by $H^1(\mathbb{S}^d)$ the Sobolev space of $L^2$-functions on $\mathbb{S}^d$ whose all derivatives of order one are also in $L^2$. Assume that $d \ge 3$. For any $u \in H^1(\mathbb{S}^d)$, $u \ne 0$, define

$$Q(u) = \frac{\int_{\mathbb{S}^d} |\nabla u|^2 d\sigma_d + d(d-1) \int_{\mathbb{S}^d} u^2 d\sigma_d}{\left( \int_{\mathbb{S}^d} |u|^{\frac{2d}{d-2}} d\sigma_d \right)^{\frac{d-2}{d}}}.$$

The function $Q$ is called the Yamabe functional and appeared naturally in the solution of the famous Yamabe problem (see e.g. [18] for more information).

It is well known (see for instance [18, Theorem 3.2]) that the infimum of $Q$ over the set $H^1(\mathbb{S}^d) \setminus \{0\}$ is attained for the constant function $u = 1$. This implies that for all $u \in H^1(\mathbb{S}^d)$, $u \ne 0$, $Q(u) \ge Q(1)$ and thus

$$\left( \int_{\mathbb{S}^d} |u|^{\frac{2d}{d-2}} d\sigma_d \right)^{\frac{d-2}{d}} \le \frac{1}{d(d-1)\omega_d^2} \left( \frac{4(d-1)}{d-2} \int_{\mathbb{S}^d} |\nabla u|^2 d\sigma_d + d(d-1) \int_{\mathbb{S}^d} u^2 d\sigma_d \right). \quad (2.3)$$

Let $u_{d+1}$ be an eigenfunction of $U_{d+1}$ associated with $\lambda_1$, the smallest eigenvalue for the Dirichlet problem; see (2.1). Notice that in this paragraph, $\lambda_1$ is the first Dirichlet
eigenvalue of $U_{d+1} \subset \mathbb{S}^d$ while in the next section, it denotes the first Dirichlet eigenvalue of $U_d \subset \mathbb{S}^{d-1}$. We extend $u_{d+1}$ on $\mathbb{S}^d$ by 0. Without loss of generality, we can assume that

$$ \int_{\mathbb{S}^d} u_{d+1}^2 d\sigma_d = 1 $$

and by Green's first identity, we get

$$ \int_{\mathbb{S}^d} |\nabla u_{d+1}|^2 d\sigma_d = \lambda_1. $$

Plugging these equalities into the inequality (2.3) and using Hölder's inequality, we get

$$ \begin{aligned} 1 = \int_{\mathbb{S}^d} u_{d+1}^2 d\sigma_d &\le \mathrm{Vol}_{\sigma_d}(U_{d+1})^{\frac{2}{d}} \left( \int_{\mathbb{S}^d} |u_d|^{\frac{2d}{d-2}} d\sigma_d \right)^{\frac{d-2}{d}} \\ &\le \mathrm{Vol}_{\sigma_d}(U_{d+1})^{\frac{2}{d}} \left( \frac{4\lambda_1}{d(d-2)\omega_d^{\frac{2}{d}}} + \frac{1}{\omega_d^{\frac{2}{d}}} \right). \end{aligned} $$

We now notice that $\mathrm{Vol}_{\sigma_d}(U_{d+1}) = (1 - \frac{1}{2^d}) \omega_d$. If we set $x_d = (1 - \frac{1}{2^d})^{\frac{2}{d}}$, the last inequality then reads as

$$ \lambda_1 \ge \frac{d(d-2)(1-x_d)}{4x_d}. $$

We finally note that when $d \to \infty$, $1 - x_d \sim \frac{2}{d} \frac{1}{2^d}$ and we obtain

$$ \lambda_1 \ge c \frac{d}{2^d} $$

for some $c > 0$, which gives the desired lower bound in Theorem 1.1.

## 2.3 The upper bound

Assume that $z_d := \frac{2^d \lambda_1}{d^3}$ is not bounded. Up to extracting a subsequence, we can assume that

$$ \lim_{d \to \infty} z_d = \infty. \qquad (2.4) $$

Choose a sequence $\alpha_d$ tending to 0 such that

$$ z_d \gg \alpha_d^{-2} \qquad (2.5) $$

and set $a_d = \alpha_d d^{-\frac{3}{2}}$. Let $\theta_d : \mathbb{R} \to \mathbb{R}$ be a smooth function satisfying $0 \le \theta_d \le 1$, $\theta_d(t) = 0$ if $t \ge 0$, $\theta_d(t) = 1$ if $t \le -a_d$ and $|\theta'| \le 2a_d^{-1}$. For all $x \in \mathbb{R}^d$, set

$$ \eta_d(x) = \max\{\theta_d(x_1), \dots, \theta_d(x_d)\}. $$

Then the restriction of $\eta_d$ (still denoted by $\eta_d$) to $\mathbb{S}^{d-1}$ has the following properties:

* the function $\eta_d$ is supported in $U_d$ since when $x \notin U_d$ then $x_i > 0$ for all $i$ and thus $\theta_d(x_i) = 0$ for all $i$;

* $\nabla \eta_d$ exists almost everywhere and satisfies $|\nabla_{\mathbb{S}^{d-1}} \eta_d| \le |\nabla_{\mathbb{R}^d} \eta_d| \le 2a_d^{-1}$.

Introduce the domains

$$ \Sigma_1 := \{x \in \mathbb{S}^{d-1} : \eta_d(x) = 1\} \quad \text{and} \quad \Sigma_{01} := \{x \in \mathbb{S}^{d-1} : 0 < \eta_d(x) < 1\}. $$

**Step 1.** It holds that $|\Sigma_{01}|_{d-1} \le C (\frac{1}{2})^d \omega_{d-1}$.
**Proof.** Using our notation (2.2), we notice that

$$ \Sigma_{01} \subset \Sigma_{d-1}(a_d) = \{x \in \mathbb{S}^{d-1} : \forall i, x_i \ge -a_d\}. $$

Indeed, as soon as $x_i \le -a_d$ for some $i$, it holds that $\eta_d(x) = 1$. Since $a_d = \alpha_d d^{-\frac{3}{2}} \ll d^{-\frac{3}{2}}$, we get from Lemma 2.1 that

$$ |\Sigma_{d-1}(a_d)|_{d-1} \le C \left(\frac{1}{2}\right)^d \omega_{d-1}, \qquad (2.6) $$

which proves Step 1. □

**Step 2.** There exists $C > 0$ such that $|\Sigma_1|_{d-1} \ge C\omega_{d-1}$.

**Proof.** We first show that

$$ \mathbb{S}^{d-1} \setminus \Sigma_{d-1}(a_d) \subset \Sigma_1. \qquad (2.7) $$

Indeed, if $x \notin \Sigma_{d-1}(a_d)$, there exists $i$ such that $x_i \le -a_d$, so that $\theta_d(x_i) = 1$, and finally $\eta_d(x) = 1$. As a consequence, $x \in \Sigma_1$. Since by (2.6)

$$ |\Sigma_{d-1}(a_d)|_{d-1} \le o(1)\omega_{d-1}, $$

we obtain from (2.7) that

$$ |\Sigma_1|_{d-1} \ge \omega_{d-1}(1 - o(1)), $$

which implies Step 2. □

**Step 3.** Variational characterization.

**Proof.** Recall from [1, Theorem 4.4 p.103] that $\lambda_1$ has the following variational characterization:

$$ \lambda_1 = \inf_{u \in H_0^1(U_d) \setminus \{0\}} Y(u), $$

where $H_0^1(U_d)$ is the set of $L^2(U_d)$-functions vanishing on $\partial U_d$ and whose derivatives of order one are also $L^2$, and where

$$ Y(u) = \frac{\int_{U_d} |\nabla u|^2 d\sigma_{d-1}}{\int_{U_d} u^2 d\sigma_{d-1}}. $$

The function $\eta_d$ is supported in $U_d$, and can thus be used as a test function in the above functional. In particular, it holds that $\lambda_1 \le Y(\eta_d)$. The support of $\nabla\eta_d$ is included in $\Sigma_{01}$, since it is a constant outside. Accordingly, it holds from Step 1 that

$$ \int_{U_d} |\nabla \eta_d|^2 d\sigma_{d-1} \le \|\nabla \eta_d\|_{L^\infty}^2 |\Sigma_{01}|_{d-1} \le C a_d^{-2} |\Sigma_{01}|_{d-1} \le C d^3 \alpha_d^{-2} \left(\frac{1}{2}\right)^d \omega_{d-1}, $$

while by Step 2

$$ \int_{U_d} \eta_d^2 d\sigma_{d-1} \ge |\Sigma_1|_{d-1} \ge C\omega_{d-1}. $$

This implies that $\lambda_1 \le Y(\eta_d) \le C d^3 \alpha_d^{-2} \left(\frac{1}{2}\right)^d$, which contradicts (2.4) and (2.5). We thus obtain that

$$ \lambda_1 \le C d^3 \left(\frac{1}{2}\right)^d, $$

which proves the desired upper bound. □
## 2.4 Proof of Lemma 2.1

Introduce for any $d \ge 2$ and $0 \le k \le d$ the set

$$V_{k,d}(a) = ([-a, 1]^k \times [0, 1]^{d-k}) \cap \mathbb{S}^{d-1}.$$

The objective of Lemma 2.1 is to estimate the volume of $V_{d,d}$. We shall decompose the proof into two steps.

**Step 1.** For all $a \in [0, 1]$, all $d \ge 1$ and $0 \le k \le d$,

$$|V_{k+1,d+1}(a)|_d \le |V_{k,d+1}(a)|_d + a|V_{k,d}(\frac{a}{\sqrt{1-a^2}})|_{d-1} \quad (2.8)$$

*Proof.* Let us write $V_{k+1,d+1}(a) = V_+ \cup V_-$, where

$$\begin{cases} V_+ = V_{k+1,d+1}(a) \cap \{x_{k+1} \ge 0\}, \\ V_- = V_{k+1,d+1}(a) \cap \{x_{k+1} < 0\}. \end{cases}$$

First, let us note that $V_+ = V_{k,d+1}(a)$. Moreover, write $|V_-|_d = \int_0^a |H_s|_{d-1} ds$, with

$$H_s = V_{k+1,d+1}(a) \cap \{x_{k+1} = s\}.$$

We now set $y_1 = \frac{x_1}{\sqrt{1-s^2}}$, ..., $y_k = \frac{x_k}{\sqrt{1-s^2}}$ and $y_{k+1} = \frac{x_{k+2}}{\sqrt{1-s^2}}$, ..., $y_d = \frac{x_{d+1}}{\sqrt{1-s^2}}$. With this notation, $(x_1, \dots, x_{d+1}) \in H_s$ if and only if $x_{k+1} = s$ and $(y_1, \dots, y_d) \in V_{k,d}(\frac{a}{\sqrt{1-s^2}})$. We thus have

$$|H_s|_{d-1} \le (1-s^2)^{\frac{d-1}{2}} |V_{k,d}(\frac{a}{\sqrt{1-s^2}})|_{d-1} \le (1-s^2)^{\frac{d-1}{2}} |V_{k,d}(\frac{a}{\sqrt{1-a^2}})|_{d-1} \le |V_{k,d}(\frac{a}{\sqrt{1-a^2}})|_{d-1}.$$

We deduce that $|V_-|_d \le a|V_{k,d}(\frac{a}{\sqrt{1-a^2}})|_{d-1}$. The proof of (2.8) is complete. $\square$

**Step 2.** For all $D \ge 2$, there exists a quantity $\varepsilon_D = O(\frac{1}{\sqrt{D}})$ such that

$$\forall 0 \le k \le d, \quad \forall a \le \frac{\varepsilon_D}{\sqrt{d - (d-k)\varepsilon_D^2}}, \quad |V_{k,d}(a)|_{d-1} \le 4 \left(\frac{1}{2} + \varepsilon_D\right)^k \frac{\omega_{d-1}}{2^{d-k}} \quad (2.9)$$

*Proof.* We prove (2.9) by induction on $d \in \{2, \dots, D\}$. The property (2.9) is clearly valid for $d = 2$. Let us now assume that (2.9) holds for some $d$. We will prove it now for $d+1$. We will show the latter property by induction on $k$. It is clearly satisfied at $k = 0$, using the following: the value of $V_{k,d+1}$ is easily calculated for $k = 0$, at which

$$|V_{0,d+1}(a)|_d = |[0, 1]^{d+1} \cap \mathbb{S}^d|_d = \frac{\omega_d}{2^{d+1}}.$$

Assuming it holds at a given $k$, we want to prove that

$$\forall a \le \frac{\varepsilon_D}{\sqrt{d+1 - (d-k)\varepsilon_D^2}}, \quad |V_{k+1,d+1}(a)|_d \le 4 \left(\frac{1}{2} + \varepsilon_D\right)^{k+1} \frac{\omega_d}{2^{d-k}} \quad (2.10)$$

To that aim, we use the upper bound (2.8), which is a sum of two terms:

* to obtain an upper bound of the term $|V_{k,d+1}(a)|_d$, we use the induction hypothesis at $k$;

* to obtain an upper bound of $|V_{k,d}(\frac{a}{\sqrt{1-a^2}})|_{d-1}$, we use the induction hypothesis in the index $d$.
To apply the second upper bound, we need to verify that $\frac{a}{\sqrt{1-a^2}}$ satisfies the inequality in (2.9). Starting from the inequality on $a$ in (2.10), we easily obtain

$$ \frac{a}{\sqrt{1-a^2}} \le \frac{\varepsilon_D}{\sqrt{d - (d-k)\varepsilon_D^2 + (1-\varepsilon_D^2)}} \le \frac{\varepsilon_D}{\sqrt{d - (d-k)\varepsilon_D^2}}, $$

since the function $x \mapsto \frac{x}{\sqrt{1-x^2}}$ is increasing on $[0, 1)$. As a consequence, using (2.8), (2.9) and (2.10), we obtain

$$ |V_{k+1,d+1}(a)| \le 4 \left( \frac{1}{2} + \varepsilon_D \right)^k \frac{1}{2^{d-k}} \omega_d \left( \frac{1}{2} + a \frac{\omega_{d-1}}{\omega_d} \right). $$

Using the fact that for all $d \ge 1$, $\frac{\omega_{d-1}}{\omega_d} \le \sqrt{d}$, we easily obtain

$$ a \frac{\omega_{d-1}}{\omega_d} \le \frac{\sqrt{d\varepsilon_D}}{\sqrt{d+1-d\varepsilon_D^2}} \le \varepsilon_D, $$

assuming in the latter inequality that $d\varepsilon_D^2 \in (0, 1]$. $\square$

We are now ready to finish the proof of Lemma 2.1. We apply (2.9) at $d = D$ and deduce that for all $a \le \frac{\varepsilon_D}{\sqrt{D}}$,

$$ |V_{D,D}(a)|_{D-1} \le C \left( \frac{1}{2} + \varepsilon_D \right)^D \omega_{D-1}. $$

As soon as $\frac{\varepsilon_D}{D} \to 0$, we have $(\frac{1}{2} + \varepsilon_D)^D \le C \frac{1}{2D}$ for some $C > 0$. The proof of Lemma 2.1 is completed.

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**Acknowledgments.** We would like to thank Michel Bauer and Françoise Cornu for suggesting the problem. KR would like to thank Nicolas Raymond and Grégory Schehr for interesting discussions. KR thanks the VIASM (Hanoï, Vietnam) for their hospitality and wonderful working conditions. We thank the two reviewers and the editors for their careful reading and suggestions.
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