Facets of High-Dimensional Gaussian Polytopes

Abstract

We study the number of facets of the convex hull of n independent standard Gaussian points in ${\mathbb {R}}^d$. In particular, we are interested in the expected number of facets when the dimension is allowed to grow with the sample size. We establish an explicit asymptotic formula that is valid whenever $d/n\rightarrow 0$. We also obtain the asymptotic value when d is close to n.

1 Introduction

The convex hull $[X_1,\ldots ,X_n]$ of n independent standard Gaussian samples $X_1,\ldots ,X_n$ from ${\mathbb {R}}^d$ is the Gaussian polytope $P^{(d)}_n$. For fixed dimension d, the face numbers and intrinsic volumes of $P_n^{(d)}$ as n tends to infinity are well understood by now. For $i=0\ldots ,d$ and polytope Q, let $f_i(Q)$ denote the number of i-faces of Q and let $V_i(Q)$ denote the ith intrinsic volume of Q. The asymptotic behavior of the expected value of the number of facets $f_{d-1}(P^{(d)}_n)$ as $n \rightarrow \infty$ was provided by Rényi and Sulanke [22] if $d=2$, and by Raynaud [21] if $d\ge 3$. Namely, they proved that, for any fixed d,

$$\begin{aligned} {\mathbb {E}}f_{d-1}(P^{(d)}_n) = 2^d \pi ^{\frac{d-1}{2}} d^{- \frac{1}{2}} ( \ln n)^{ \frac{d-1}{2} } (1+o(1)) \end{aligned}$$

(1)

as $n \rightarrow \infty$. For $i=0,\ldots ,d$, expected value of $V_i(P^{(d)}_n)$ as $n \rightarrow \infty$ was computed by Affentranger [1], and that of $f_i(P^{(d)}_n)$ was determined Affentranger and Schneider [2] and Baryshnikov and Vitale [3], see Hug et al. [15] and Fleury [12] for a different approach. More recently, Kabluchko and Zaporozhets [18, 19] proved explicit expressions for the expected value of $V_d(P^{(d)}_n)$ and the number of k-faces $f_k(P^{(d)}_n)$. Yet these formulas are complicated and it is not immediate how to deduce asymptotic results for large n high dimensions d.

After various partial results, including the variance estimates of Calka and Yukich [6] and Hug and Reitzner [16], central limit theorems were proved for $f_i(P^{(d)}_n)$ and $V_d(P^{(d)}_n)$ by Bárány and Vu [5], and for $V_i(P^{(d)}_n)$ by Bárány and Thäle [4]. These results have been strengthened considerably by Grote and Thäle [14]. The interesting question whether ${\mathbb {E}}f_{d-1}(P^{(d)}_n)$ is an increasing function in n was answered in the positive by Kabluchko and Thäle [17]. It would be interesting to investigate the monotonicity behavior of the facet number if n and d increases simultaneously.

The “high-dimensional” regime, that is, when d is allowed to grow with n, is of interest in numerous applications in statistics, signal processing, and information theory. The combinatorial structure of $P^{(d)}_n$, when d tends to infinity and n grows proportionally with d, was first investigated by Vershik and Sporyshev [23], and later Donoho and Tanner [11] provided a satisfactory description. For any $t>1$, Donoho and Tanner [11] determined the optimal $\varrho (t)\in (0,1)$ such that if n/d tends to t, then $P^{(d)}_n$ is essentially $\varrho (t)d$-neighbourly (if $0<\eta <\varrho (t)$ and $0\le k\le \eta d$, then $f_k(P^{(d)}_n)$ is asymptotically ${n \atopwithdelims ()k+1}$). See Donoho [10], Candés et al. [7], Candés and Tao [8, 9], Mendoza-Smith et al. [20].

In this note, we consider $f_{d-1}(P^{(d)}_n)$, the number of facets, when both d and n tend to infinity. Our main result is the following estimate for the expected number of facets of the Gaussian polytope. The implied constant in $O(\cdot )$ is always some absolute constant. We write ${{\,\mathrm{{lln}}\,}}x$ for $\ln (\ln x)$.

Theorem 1.1

Assume $P^{(d)}_n$ is a Gaussian polytope. Then for $d \ge 78$ and $n \ge e^e d$, we have

$$\begin{aligned} {\mathbb {E}}f_{d-1}(P^{(d)}_n) = 2^d \pi ^{\frac{d-1}{2}} d^{- \frac{1}{2}} e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\frac{n}{d} - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\frac{n}{d} }{\ln \frac{n}{d} } +(d-1) \frac{\theta }{\ln \frac{n}{d}} \ +O\left( \sqrt{d} e^{ - \frac{1}{10} d}\right) } \end{aligned}$$

with $\theta =\theta (n,d) \in [-34,2]$.

When n/d tends to infinity as $d \rightarrow \infty$, Theorem 1.1 provides the asymptotic formula

$$\begin{aligned} {\mathbb {E}}f_{d-1}(P^{(d)}_n) = \left( (4\pi +o(1)) \ln \frac{n}{d}\right) ^{\frac{d-1}{2}}~. \end{aligned}$$

If $n/(d e^d)\rightarrow \infty$, then we have $\frac{d}{\ln \frac{n}{d}} \rightarrow 0$ and hence

$$\begin{aligned} {\mathbb {E}}f_{d-1}(P^{(d)}_n) = 2^d \pi ^{\frac{d-1}{2}} d^{- \frac{1}{2}} e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\frac{n}{d} - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\frac{n}{d} }{\ln \frac{n}{d} } +o(1)} \end{aligned}$$

as $d \rightarrow \infty$. In the case when n grows even faster such that $(\ln n)/(d\ln d) \rightarrow \infty$, the asymptotic formula simplifies to the result (1) of Rényi and Sulanke [22] and Raynaud [21] for fixed dimension.

Corollary 1.2

Assume $P^{(d)}_n$ is a Gaussian polytope. If $(\ln n)/(d\ln d) \rightarrow \infty$, we have

$$\begin{aligned} {\mathbb {E}}f_{d-1}(P^{(d)}_n) = 2^d \pi ^{\frac{d-1}{2}} d^{- \frac{1}{2}} ( \ln n)^{ \frac{d-1}{2} } (1+o(1))~. \end{aligned}$$

There is a (simpler) counterpart of our main results stating the asymptotic behavior of the expected number of facets of $P_n^{(d)}$, if $n-d$ is small compared to d, that is, if n/d tends to one.

Theorem 1.3

Assume $P^{(d)}_n$ is a Gaussian polytope. Then for $n - d = o( d)$, we have

$$\begin{aligned} {\mathbb {E}}f_{d-1}(P^{(d)}_n) = {n \atopwithdelims ()d} 2^{-(n-d)+1} e^{\frac{1}{\pi }\frac{(n-d)^2}{d} + O\left( \frac{(n-d)^3}{d^2}\right) +o(1) } \end{aligned}$$

as $d \rightarrow \infty$.

This complements a result of Affentranger and Schneider [2] stating the number of k-dimensional faces for $k \le n-d$ and $n-d$ fixed,

$$\begin{aligned} {\mathbb {E}}f_{k}(P^{(d)}_n) = {n \atopwithdelims ()k+1} (1 +o(1))~, \end{aligned}$$

as $d \rightarrow \infty$.

In the next section we sketch the basic idea of our approach, leaving the technical details to later sections. In Sect. 3 we provide asymptotic approximations for the tail of the normal distribution. In Sect. 4 concentration inequalities are derived for the $\beta$-distribution. Finally, in Sects. 5 and 6, Corollary 1.2 and Theorem 1.3 are proven.

2 Outline of the Argument

For $z\in {\mathbb {R}}$, let

$$\begin{aligned} \Phi (y)=\frac{1}{\sqrt{\pi }}\int \limits _{-\infty }^y e^{- s^2}\,ds, \ \text{ and } {\phi }(y)=\Phi '(y)=\frac{1}{\sqrt{\pi }}e^{- y^2}~. \end{aligned}$$

Our proof is based on the approach of Hug, Munsonius, and Reitzner [15]. In particular, [15, Theorem 3.2] states that if $n\ge d+1$ and $X_1,\ldots ,X_n$ are independent standard Gaussian points in ${\mathbb {R}}^d$, then

$$\begin{aligned} {\mathbb {E}}f_{d-1}([X_1,\ldots ,X_n]) = {n \atopwithdelims ()d} {\mathbb {P}}(Y \notin [Y_1, \dots , Y_{n-d}])~, \end{aligned}$$

where $Y,Y_{1},\dots ,Y_{n-d}$ are independent real-valued random variables with $Y {\mathop {=}\limits ^{d}} N\left( 0,\frac{1}{2d}\right)$ and $Y_{i} {\mathop {=}\limits ^{d}} N\left( 0,\frac{1}{2}\right)$ for $i=1,\ldots ,n-d$. This gives

$$\begin{aligned} {\mathbb {E}}f_{d-1}([X_1,\ldots ,X_n])= & {} 2 {n \atopwithdelims ()d} \frac{\sqrt{d}}{\sqrt{\pi }} \int \limits _{- \infty }^\infty \Phi (y)^{n-d} e^{-dy^2}\,dy\end{aligned}$$

(2)

$$\begin{aligned}= & {} 2 {n \atopwithdelims ()d} \sqrt{d} \, \pi ^{\frac{d-1}{2}} \int \limits _{- \infty }^\infty \Phi (y)^{n-d} {\phi }(y)^d \,dy~. \end{aligned}$$

(3)

Note that similar integrals appear in the analysis of the expected number of k-faces for values of k in the entire range $k=0,\ldots ,d-1$. In our case, the analysis boils down to understanding the integral of $\Phi (y)^{n-d} {\phi }(y)^d$ over the real line. By substituting $(1-u) = \Phi (y)$, we obtain

$$\begin{aligned} \int \limits _{- \infty }^\infty \Phi (y)^{n-d} {\phi }(y)^d \,dy = \int \limits _{0}^1 (1-u)^{n-d} {\phi }(\Phi ^{-1}(1-u))^{d-1}\,du~. \end{aligned}$$

Clearly, $n \ge d+2$ is the nontrivial range. When $n/d \rightarrow \infty$, $(1-u)^{n-d}$ is dominating, and we need to investigate the asymptotic behavior of ${\phi }(\Phi ^{-1}(1-u))$ as $u \rightarrow 0$. We show that the essential term is precisely 2u. Hence, it makes sense to rewrite the integral as

$$\begin{aligned} 2^{d-1} \int \limits _{0}^1 (1-u)^{n-d} u^{d-1} \underbrace{\left( (2u)^{-1} {\phi }(\Phi ^{-1}(1-u))\right) ^{d-1}} _{=: g_d(u)} \,du~. \end{aligned}$$

For $x,y>0$, the Beta-function is given by ${\varvec{B}}(x,y)= \int _0^1 (1-u)^{x-1} u^{y-1} du$. It is well known that for $k,l \in {\mathbb {N}}$ we have ${\varvec{B}}(k,l)= \frac{(k-1)!(l-1)!}{(k+l-1)!}$. A random variable U is ${\varvec{B}}_{(x,y)}$ distributed if its density is given by ${\varvec{B}}(x,y)^{-1} (1-u)^{x-1} u^{y-1}$. With this, we have established the following identity:

Proposition 2.1

$$\begin{aligned} {\mathbb {E}}f_{d-1}([X_1,\ldots ,X_n])= & {} 2^d \pi ^{\frac{d-1}{2}} d^{- \frac{1}{2}} {\mathbb {E}}g_d(U) \end{aligned}$$

(4)

where

$$\begin{aligned} g_d(u) = \left( (2u)^{-1} {\phi }(\Phi ^{-1}(1-u))\right) ^{d-1} \end{aligned}$$

and U is a ${\varvec{B}}(n-d+1,d)$ random variable.

In Lemma 3.3 below we show that

$$\begin{aligned} g_d(u) = (\ln u^{-1})^{- \frac{d-1}{2} } e^{- \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}u^{-1} }{ \ln u^{-1}} - (d-1) \frac{O(1)}{\ln u^{-1}} } \end{aligned}$$

as $u \rightarrow 0$. Because the Beta function is concentrated around $\frac{d}{n}$, see Lemma 4.1 and Lemma 4.2, this yields

$$\begin{aligned} {\mathbb {E}}g_d(U) \approx \left( \ln \frac{n}{d}\right) ^{\frac{d-1}{2}} e^{ - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\frac{n}{d} }{ \ln \frac{n}{d}} - (d-1) \frac{O(1)}{ \ln \frac{n}{d}} } \end{aligned}$$

which implies our main result.

3 Asymptotics of the $\Phi$-Function

To estimate $\Phi (z)$, we need a version of Gordon’s inequality [13] for the Mill’s ratio:

Lemma 3.1

For any $z>1$ there exists $\theta \in (0,1)$, such that

$$\begin{aligned} \Phi (z)= 1-\frac{e^{-z^2}}{2\sqrt{\pi }z} \left( 1 - \frac{\theta }{2z^2}\right) \end{aligned}$$

Proof

It follows by partial integration that

$$\begin{aligned} \int \limits _z^{\infty }e^{-t^2}\,dt= \int \limits _z^{\infty }2t e^{-t^2}\, \frac{1}{2t} \,dt= \frac{e^{-z^2}}{2z} - \int \limits _z^{\infty } \frac{e^{-t^2}}{2t^2} \,dt= \frac{e^{-z^2}}{2z} - \frac{\theta e^{-z^2}}{4z^3} \end{aligned}$$

which yields the lemma. $\square$

Lemma 3.2

For any $u \in (0,e^{-1}]$ there is a ${\delta }={\delta }(u) \in (0,16)$ such that

$$\begin{aligned} \Phi ^{-1}(1-u) = \sqrt{ \ln u^{-1} - \frac{1}{2} {{\,\mathrm{{lln}}\,}}u^{-1} - \ln (2\sqrt{\pi }) + \frac{1}{4} \frac{{{\,\mathrm{{lln}}\,}}u^{-1} }{ \ln u^{-1}} + \frac{\delta }{\ln u^{-1}} }. \end{aligned}$$

(5)

Proof

It is useful to prove (5) for the transformed variable $u=e^{-t}$. We define

$$\begin{aligned} z (t) = \sqrt{t - \frac{1}{2} \ln t - \ln (2\sqrt{\pi }) + \frac{1}{4} \frac{\ln t}{t} +\frac{\delta (t)}{t} } \end{aligned}$$

(6)

which exists for $t > 0$. In a first step we prove that this is the asymptotic expansion of $z=\Phi ^{-1}(1-e^{-t})$ as $z,t \rightarrow \infty$ with a suitable function ${\delta }={\delta }(t)=O(1)$. In a second step we show the bound on $\delta$. Observe that $z \ge 1$ implies $t \ge \ln \Phi (-1))=-2,54\dots$. By Lemma 3.1, for $z \ge 1$

$$\begin{aligned} e^{-t}= 1- \Phi (z) = \frac{1}{2\sqrt{\pi }\, z} e^{- z^2} \left( 1 - \frac{\theta (z)}{2 z^2}\right) \end{aligned}$$

(7)

as $z \rightarrow \infty$ with some $\theta (z) \in (0,1)$, which immediately implies that $z=z(t) \rightarrow \infty$ as $t \rightarrow \infty$. Equation (7) shows that $e^t \ge 2 \sqrt{\pi }z e^{z^2}$ and thus

$$\begin{aligned} t \ge \ln (2 \sqrt{\pi }) + \ln z(t) + z(t)^2 \ge z(t)^2 \end{aligned}$$

for $z \ge 1$. The function $z=z(t)$ is the inverse function we are looking for, if it satisfies

$$\begin{aligned} 4 \pi z(t)^2 e^{-2t} = e^{- 2z(t)^2} \left( 1 - \frac{\theta (z)}{2 z^2}\right) ^2. \end{aligned}$$

(8)

We plug (6) into this equation. This leads to

$$\begin{aligned} t - \frac{1}{2} \ln t - \ln (2\sqrt{\pi }) + \frac{1}{4} \frac{\ln t }{t} + \frac{\delta (t)}{ t}&= t e^{ - \frac{1}{2} \frac{\ln t}{t} - 2\frac{\delta (t)}{t} } \left( 1 - O(t^{-1})\right) \\&= t - \frac{1}{2} \ln t - 2 \delta (t) - O(1) \end{aligned}$$

and shows $- \ln (2\sqrt{\pi }) + o(1)=- 2 \delta (t) - O(1)$. Thus the function z(t) given by (6) in fact satisfies (7) and therefore it is the asymptotic expansion of the inverse function.

The desired estimate for ${\delta }$ follows from some more elaborate but elementary calculations. First we prove that ${\delta }\ge 0$. By (8) and because $e^x \ge 1+x$,

$$\begin{aligned} t - \frac{1}{2} \ln t - \ln (2\sqrt{\pi }) + \frac{1}{4} \frac{\ln t }{t} + \frac{\delta (t)}{ t}&\ge t \left( 1 - \frac{1}{2} \frac{\ln t}{t} - 2\frac{\delta (t)}{t} \right) \left( 1 - \frac{\theta }{2t} \right) ^2\\&\ge (t - \frac{1}{2} \ln t - 2 \delta (t) ) \left( 1 - \frac{\theta }{t} \right) \end{aligned}$$

which is equivalent to

$$\begin{aligned} \delta (t)&\ge \frac{ \ln (2\sqrt{\pi }) - \theta - \frac{1-2\theta \ln t }{4t} }{\left( 2 + \frac{1-2\theta }{t}\right) } >0 \end{aligned}$$

for $t \ge 1$. On the other hand, again by (8),

$$\begin{aligned} t&\ge \left( t - \frac{1}{2} \ln t - \ln (2\sqrt{\pi }) + \frac{1}{4} \frac{\ln t }{t} + \frac{\delta (t)}{ t} \right) e^{\frac{1}{2} \frac{\ln t}{t} + 2\frac{\delta (t)}{t} } \end{aligned}$$

and using $e^x \ge 1+x$ implies

$$\begin{aligned} \delta (t) \le \frac{ \ln (2\sqrt{\pi }) + \frac{2\ln (2\sqrt{\pi }) -1}{4} \frac{\ln t}{t} + \frac{1}{4} \frac{(\ln t)^2}{t} + \frac{1}{8} \frac{(\ln t )^2}{t^2} }{2 - (2\ln (2\sqrt{\pi })-1) \frac{1}{t} - \frac{\ln t }{t} } \le 16. \end{aligned}$$

$\square$

An asymptotic expansion for ${\phi }(\Phi ^{-1} (1-u))$ follows immediately:

Lemma 3.3

For any $u \in (0,e^{-1}]$ there is a ${\delta }={\delta }(u) \in (0,16)$ such that

$$\begin{aligned} g_d(u)= \left( (2u)^{-1}{\phi }( \Phi ^{-1}(1-u) ) \right) ^{d-1}= (\ln u^{-1})^{ \frac{d-1}{2} } e^{- \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}u^{-1} }{ \ln u^{-1}} - (d-1)\frac{\delta }{\ln u^{-1}} }~. \end{aligned}$$

4 Concentration of the $\beta$-Distribution

A basic integral for us is the Beta-integral

$$\begin{aligned} {\varvec{B}}({\alpha },{\beta }) = \int \limits _{0}^{1}(1- x)^{{\alpha }-1} x^{{\beta }-1}\,dx =\frac{({\alpha }-1)!({\beta }-1)!}{({\alpha }+{\beta }-1)!}. \end{aligned}$$

(9)

Let $U \sim {\varvec{B}}({\alpha }, {\beta })$ distributed. Then ${\mathbb {E}}U=\frac{{\beta }}{{\alpha }+{\beta }}$ and $\textrm{var}(U)= \frac{{\alpha }{\beta }}{({\alpha }+{\beta })^2 ({\alpha }+{\beta }+1)}$ Next we establish concentration inequalities for a Beta-distributed random variable around its mean. Observe that if $U \sim \textbf{B}({\alpha },{\beta })$, then $1-U \sim \textbf{B}({\beta }, {\alpha })$. Hence we may concentrate on the case ${\alpha }\ge {\beta }$.

Lemma 4.1

Let $U \sim {\varvec{B}}(a+1, b+1)$ distributed with $a \ge b$ and set $n=a+b$. Then

$$\begin{aligned} {\mathbb {P}}\left( U \le \frac{b}{n} - s \frac{a^{\frac{1}{2}} b^{\frac{1}{2}}}{n^{\frac{3}{2}}}\right) \le \frac{3e^3}{ \pi } \frac{1}{s} \left( e^{- \frac{1}{6} s^2 } - e^{- \frac{1}{6} \frac{nb}{a} } \right) _+. \end{aligned}$$

Proof

We have to estimate the integral

$$\begin{aligned} \frac{1}{{\varvec{B}}(a+1, b+1)} \int \limits _{0}^{\frac{b-s \sqrt{\frac{ab}{n}}}{n}} (1- x)^a x^b \,dx \end{aligned}$$

For an estimate from above we substitute $x= \frac{b}{n} - \frac{y}{n} \sqrt{\frac{ab}{n}}$.

$$\begin{aligned} J_-= & {} \int \limits _{0}^{\frac{b-s \sqrt{\frac{ab}{n}}}{n}}(1- x)^a x^b\,dx\\= & {} \frac{ a^{a+ \frac{1}{2}} b^{b+ \frac{1}{2}} }{n^{n+\frac{3}{2}}} \ \int \limits _{s }^{\sqrt{\frac{nb}{a}}} \left( 1+ y \sqrt{\frac{b}{an}}\right) ^a \left( 1 - y \sqrt{\frac{a}{bn}}\right) ^b\,dy \end{aligned}$$

It is well known that

$$\begin{aligned} \ln (1+x) = \sum _{k=1}^\infty (-1)^{k-1} \frac{x^k}{k} \le x - \frac{x^2}{6}, \end{aligned}$$

(10)

for $x \in (-1,1]$. Since $a \ge b$, we have

$$\begin{aligned} \left( 1+ y \sqrt{\frac{b}{an}}\right) ^a \left( 1 - y \sqrt{\frac{a}{bn}}\right) ^b \le e^{- \frac{1}{6} y^2 }~, \end{aligned}$$

which implies

$$\begin{aligned} J_-\le & {} \frac{ a^{a+ \frac{1}{2}} b^{b+ \frac{1}{2}} }{n^{n+\frac{3}{2}}} \ \int \limits _{s }^{\sqrt{\frac{nb}{a}}} e^{- \frac{1}{6} y^2 } \,dy\\\le & {} \frac{3 a^{a+ \frac{1}{2}} b^{b+ \frac{1}{2}} }{n^{n+\frac{3}{2}}} \ \frac{1}{s} \left( e^{- \frac{1}{6} s^2 } - e^{- \frac{1}{6} \frac{nb}{a} } \right) ~. \end{aligned}$$

In the last step we use Stirling’s formula,

$$\begin{aligned} \sqrt{2\pi }\, n^{n+\frac{1}{2}} e^{-n} \le n! \le e \, n^{n+\frac{1}{2}} e^{-n}, \end{aligned}$$

to see that

$$\begin{aligned} \frac{ a^{a+ \frac{1}{2}} b^{b+ \frac{1}{2}} }{n^{n+\frac{3}{2}}} \ \le \frac{e^3}{\pi } {\varvec{B}}(a+1,b+1). \end{aligned}$$

(11)

$\square$

Lemma 4.2

Let $U \sim {\varvec{B}}(a+1, b+1)$ distributed with $a \ge b$ and set $n=a+b$. Then for ${\lambda }\ge 2$,

$$\begin{aligned} {\mathbb {P}}\left( U \ge {\lambda }\frac{b}{n}\right) \le \frac{e^3}{\pi } {\lambda }^b b^{\frac{1}{2} } e^{b + \frac{3}{2}} e^{- {\lambda }\frac{ab}{n}}. \end{aligned}$$

Proof

We assume that $a \ge b$ and thus $a \ge \frac{n}{2}$. We have to estimate the probability

$$\begin{aligned} {\mathbb {P}}\left( U \ge {\lambda }\frac{b}{n}\right)\le & {} \frac{1}{{\varvec{B}}(a+1, b+1)} \int \limits _{{\lambda }\frac{b}{n}}^{1}(1- x)^a x^b \,dx \end{aligned}$$

We substitute $x \rightarrow \frac{1}{a} x + {\lambda }\frac{b}{n}$ and obtain

$$\begin{aligned} \int \limits _{{\lambda }\frac{b}{n}}^{1}(1- x)^a x^b \,dx\le & {} \int \limits _{0}^{\infty }e^{-x - {\lambda }\frac{ab}{n}} \left( \frac{1}{a} x + {\lambda }\frac{b}{n}\right) ^b \ \frac{1}{a} \,dx\\\le & {} a^{-(b+1)} e^{- {\lambda }\frac{ab}{n}} \int \limits _{0}^{\infty }e^{-x} \left( x + {\lambda }\frac{ab}{n}\right) ^b \,dx. \end{aligned}$$

The use of the binomial formula and the Gamma functions yields

$$\begin{aligned} \int \limits _{0}^{\infty }e^{-x }\left( x + {\lambda }\frac{ab}{n}\right) ^b \,dx= & {} \sum _{k=0}^b {b \atopwithdelims ()k} \int \limits _{0}^{\infty }e^{-x } x^{b-k} \left( {\lambda }\frac{ab}{n}\right) ^k \,dx\\= & {} \sum _{k=0}^b {b \atopwithdelims ()k} (b-k)! \left( {\lambda }\frac{ab}{n}\right) ^k\\\le & {} b \left( {\lambda }\frac{ab}{n}\right) ^b \end{aligned}$$

because $b \le {\lambda }\frac{ab}{n}$ for $a \ge \frac{n}{2} \ge b$ and ${\lambda }\ge 2$, and $\frac{1}{k!} \left( {\lambda }\frac{ab}{n}\right) ^k$ is increasing for $k\le \left( {\lambda }\frac{ab}{n}\right)$. Using (11) this gives

$$\begin{aligned} {\mathbb {P}}\left( U \ge {\lambda }\frac{b}{n}\right)\le & {} \frac{e^3}{\pi } \left( 1+ \frac{b}{a} \right) ^{a+ \frac{3}{2}} b^{\frac{1}{2} } {\lambda }^b e^{- {\lambda }\frac{ab}{n}} \end{aligned}$$

and with $(1+x) \le e^x$ the lemma. $\square$

5 The Case $n-d$ Large

In this section we combine Lemma 3.3 which gives the asymptotic behavior of $g_d(u)$ as $u \rightarrow 0$, with the concentration properties of the Beta function just obtained. We split our proof in two Lemmata.

Lemma 5.1

For $d \ge d_0=78$ and $n \ge e^e d$ we have

$$\begin{aligned} {\mathbb {E}}g_d(U) \le e^{\frac{d-1}{2} {{\,\mathrm{{lln}}\,}}(\frac{n}{d} ) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}(\frac{n}{d}) }{ \ln (\frac{n}{d})} + (d-1) \frac{2}{ \ln (\frac{n}{d})}} e^{\frac{e^6}{\pi } \sqrt{d} e^{-\frac{1}{10} d} }. \end{aligned}$$

Lemma 5.2

For $d \ge d_0=78$ and $n \ge e^e d$ we have

$$\begin{aligned} {\mathbb {E}}g_d(U) \ge e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}(\frac{n}{d}) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\frac{n}{d} }{\ln \frac{n}{d} } -(d-1) \frac{34}{\ln \frac{n}{d}} } \ e^{- \frac{2e^6}{\pi } \sqrt{d} e^{ - \frac{1}{10} d}}. \end{aligned}$$

These two bounds prove Theorem 1.1. The idea is to split the expectation into the main term close to $\frac{d}{n}$ and two error terms,

$$\begin{aligned} {\mathbb {E}}g_d(U)= & {} {\mathbb {E}}g_d(U) \, {\mathbbm {1}}\left( U \le e^{-2} \frac{d}{n} \right) \\{} & {} +\,{\mathbb {E}}g_d(U) \, {\mathbbm {1}} \left( U \in \left[ e^{-2} \frac{d}{n},2 \frac{d}{n}\right] \right) \\{} & {} +\,{\mathbb {E}}g_d(U) \, {\mathbbm {1}} \left( U \ge 2 \frac{d}{n} \right) ~. \end{aligned}$$

Proof of Lemma 5.2

Recall that U is ${\varvec{B}}(n-d+1,d)$-distributed. Lemma 4.2 with $a=n -d$ and $b= d-1$ shows that

$$\begin{aligned} {\mathbb {P}}\left( U \ge {\lambda }\frac{d}{n}\right) \le {\mathbb {P}}\left( U \ge {\lambda }\frac{d-1}{n-1}\right) \le \frac{e^3}{\pi } {\lambda }^{d-1} (d-1)^{\frac{1}{2} } e^{(d-1) + \frac{3}{2}} e^{- {\lambda }\frac{(n-d)(d-1)}{n-1}} \end{aligned}$$

because $\frac{d-1}{n-1}< \frac{d}{n}$. For ${\lambda }=2$ this gives

$$\begin{aligned} {\mathbb {P}}\left( U \ge 2 \frac{d}{n}\right) \le \frac{e^6}{2\pi } \sqrt{d} e^{ ( \ln 2- 1 + 2 \frac{d}{n}) d} \le \frac{e^6}{2\pi } \sqrt{d} e^{ - \frac{1}{10} d} \end{aligned}$$

(12)

for $n \ge 10 d$. The probability that U is small is estimated by Lemma 4.1 with $s= (1- e^{-2})\sqrt{\frac{(d-1)(n-1)}{n-d}}$,

$$\begin{aligned} {\mathbb {P}}\left( U \le e^{-2} \frac{d-1}{n-1} \right)\le & {} \frac{3e^3}{ \pi } (1- e^{-2})^{-1} \sqrt{\frac{n-d}{(d-1)(n-1)}} e^{- \frac{1}{6} (1- e^{-2})^2 \frac{(d-1)(n-1)}{n-d} }\\\le & {} \frac{e^6}{ 2 \pi } e^{- \frac{1}{10}d } \end{aligned}$$

for $d \ge 6$. Combining both estimates and using

$$\begin{aligned} \ln (1-x) \ge - 2x \end{aligned}$$

(13)

for $x \in [0, \frac{1}{2}]$, we have

$$\begin{aligned} {\mathbb {P}}\left( U \in \left[ \frac{1}{2} \frac{d}{n},2 \frac{d}{n} \right] \right) \ge 1 - \frac{e^6}{2\pi } \sqrt{d} e^{ - \frac{1}{10} d} - \frac{e^6}{ 2 \pi } e^{- \frac{1}{10}d } \ge e^{ - \frac{2e^6}{\pi } \sqrt{d} e^{ - \frac{1}{10} d}} \end{aligned}$$

(14)

for $d \ge d_0=78$. (Observe that $\frac{2e^6}{\pi } \sqrt{d}_0 e^{ - \frac{1}{10} d_0} \le \frac{1}{2}$.) In the last step we compute

$$\begin{aligned} \min _{u \in \left[ e^{-2} \frac{d}{n},2 \frac{d}{n}\right] } g_d(u)= & {} \min _{u \in \left[ e^{-2} \frac{d}{n},2 \frac{d}{n}\right] } e^{\frac{d-1}{2} {{\,\mathrm{{lln}}\,}}u^{-1} - \frac{d-1}{4} \frac{\ln \ln u^{-1} }{ \ln u^{-1}} - (d-1) \frac{ \delta }{ \ln u^{-1}} }\\\ge & {} e^{\frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( \frac{1}{2} \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( \frac{1}{2} \frac{n}{d}\right) }{ \ln \left( \frac{1}{2} \frac{n}{d}\right) } - (d-1) \frac{\max \delta }{ \ln \left( \frac{1}{2} \frac{n}{d}\right) } } \end{aligned}$$

for $n \ge e^e d$. Here, note that $\frac{{{\,\mathrm{{lln}}\,}}x}{\ln x}$ is decreasing for $x \ge e^e$. Now using

$$\begin{aligned} {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d} \right) \ge {{\,\mathrm{{lln}}\,}}\left( \frac{1}{2} \frac{n}{d} \right) = {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d} \right) +\ln \left( 1- \frac{\ln 2 }{ \ln \left( \frac{n}{d}\right) } \right) \ge {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d} \right) - \frac{2\ln 2 }{ \ln \left( \frac{n}{d}\right) }, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\ln \left( \frac{1}{2} \frac{n}{d}\right) } = \frac{1}{\ln \left( \frac{n}{d}\right) - \ln 2 } \le \frac{1}{\ln \left( \frac{n}{d}\right) } \left( 1 + 2 \frac{\ln 2}{\ln \left( \frac{n}{d}\right) }\right) \le 2 \frac{1}{\ln \left( \frac{n}{d}\right) } \end{aligned}$$

for $n \ge e^{e} d$, we have

$$\begin{aligned} \min _{u \in \left[ e^{-2} \frac{d}{n},2 \frac{d}{n}\right] } g_d(u)\ge & {} e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\frac{n}{d} - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\frac{n}{d} }{\ln \frac{n}{d}} -(d-1) \frac{\delta '}{\ln \frac{n}{d}}} \end{aligned}$$

with $\delta '=\frac{3 \ln 2}{2}+2 \max \delta \in [0, 34]$. Combining this estimate with (14) we obtain

$$\begin{aligned} {\mathbb {E}}g_d(U)\ge & {} \min _{u \in \left[ e^{-2} \frac{d}{n},2 \frac{d}{n}\right] } g_d(u)\ {\mathbb {E}}{\mathbbm {1}}\left( U \in \left[ e^{-2} \frac{d}{n},2 \frac{d}{n}\right] \right) \\\ge & {} e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\frac{n}{d} - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\frac{n}{d} }{\ln \frac{n}{d} } -(d-1) \frac{\delta '}{\ln \frac{n}{d}} } \ e^{- \frac{2e^6}{\pi } \sqrt{d} e^{ - \frac{1}{10} d}} \end{aligned}$$

for $d \ge d_0$ and $n \ge e^e d$.

Proof of Lemma 5.1

As an upper bound we have

$$\begin{aligned} {\mathbb {E}}g_d(U)\le & {} {\mathbb {E}}g_d(U){\mathbbm {1}}\left( U \le e^{-2} \frac{d}{n} \right) \\{} & {} + \max _{u \in [e^{-2} \frac{d}{n},2 \frac{d}{n}]} g_d(u) \ {\mathbb {P}}\left( U \in \left[ e^{-2} \frac{d}{n},2 \frac{d}{n} \right] \right) \\{} & {} + \underbrace{ \max _{u \in \left[ 2 \frac{d}{n},1\right] } g_d(u)}_{\le \max _{u \in \left[ \frac{d}{n},1\right] } g_d(u)} {\mathbb {P}}\left( U \ge 2 \frac{d}{n} \right) \\\le & {} {\mathbb {E}}g_d(U){\mathbbm {1}}\left( U \le e^{-2} \frac{d}{n} \right) \\{} & {} + e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) }{ \ln \left( e^2 \frac{n}{d}\right) } }\\{} & {} + e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( \frac{n}{d}\right) }{ \ln \left( \frac{n}{d}\right) } } \frac{e^6}{2\pi } \sqrt{d} e^{ - \frac{1}{10} d} \end{aligned}$$

since $\delta \ge 0$, and where the last term follows from (12). For the first term we use that ${\phi }(\Phi ^{-1}(\cdot ))$ is a symmetric and concave function and thus increasing on $[0, e^{-2}\frac{d}{ n}]$, and that ${\delta }\ge 0$.

$$\begin{aligned}{} & {} {\mathbb {E}}g_d(U){\mathbbm {1}} \left( U \le e^{-2}\frac{d}{n} \right) \\{} & {} \quad \le \frac{1}{{\varvec{B}}(n-d+1, d)} \int \limits _0^{e^{-2}\frac{d}{n}} e^{\frac{d-1}{2} {{\,\mathrm{{lln}}\,}}x^{-1} - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}x^{-1} }{ \ln x^{-1}} } (1-x)^{n-d} x^{d-1} dx\\{} & {} \quad \le \frac{1}{{\varvec{B}}(n-d+1, d)} e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) }{ \ln \left( e^2 \frac{n}{d}\right) } } \left( e^{-2} \frac{d}{n}\right) ^{d-1} \int \limits _0^{\infty } e^{-(n-d)x} dx \end{aligned}$$

Now the remaining integration is trivial. We use Stirling’s formula (11) to estimate the Beta-function and obtain

$$\begin{aligned}{} & {} {\mathbb {E}}g_d(U){\mathbbm {1}} \left( U \le e^{-2}\frac{d}{n} \right) \\{} & {} \quad \le \frac{e^3}{\pi } \frac{(n-1)^{n +\frac{1}{2} }}{(n-d)^{n-d+\frac{3}{2}} (d-1)^{d- \frac{1}{2}}} e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) }{\ln \left( e^2 \frac{n}{d}\right) } } \left( e^{-2} \frac{d}{n}\right) ^{d-1}\\{} & {} \quad \le e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) }{ \ln \left( e^2 \frac{n}{d}\right) } } \frac{e^5}{\pi } e^{(d-1) + \frac{(d-1)}{(n-d) }\left( \frac{3}{2}\right) + 1+ \frac{1}{(d-1)}\frac{1}{2} -2d}\\{} & {} \quad \le e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) }{ \ln \left( e^2 \frac{n}{d}\right) } } \frac{e^5}{\pi } e^{-\frac{1}{10} d} \end{aligned}$$

e.g. for $n \ge e^e d$ and $d \ge 78$. Combining our results gives

$$\begin{aligned} {\mathbb {E}}g_d(U)\le & {} e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) }{ \ln \left( e^2 \frac{n}{d}\right) } } \frac{e^5}{\pi } e^{-\frac{1}{10} d}\\{} & {} + e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d}\right) }{ \ln \left( e^2 \frac{n}{d}\right) } }\\{} & {} + e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d}\right) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}\left( \frac{n}{d}\right) }{ \ln \left( \frac{n}{d}\right) } } \frac{e^6}{2\pi } \sqrt{d} e^{ - \frac{1}{10} d} \end{aligned}$$

In a similar way as above, we get rid of the involved constant $e^2$ by using

$$\begin{aligned} {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d} \right) \le {{\,\mathrm{{lln}}\,}}\left( e^2 \frac{n}{d} \right) = {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d} \right) +\ln \left( 1 + \frac{ 2 }{ \ln (\frac{n}{d})} \right) \le {{\,\mathrm{{lln}}\,}}\left( \frac{n}{d} \right) + \frac{2}{ \ln (\frac{n}{d})}, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\ln \left( e^2 \frac{n}{d}\right) } = \frac{1}{\ln (\frac{n}{d})} \left( 1+ \frac{2}{\ln (\frac{n}{d})} \right) ^{-1} \ge \frac{1}{\ln (\frac{n}{d} ) } \left( 1 - \frac{2}{\ln (\frac{n}{d}) } \right) . \end{aligned}$$

This yields

$$\begin{aligned} {\mathbb {E}}g_d(U) \le e^{ \frac{d-1}{2} {{\,\mathrm{{lln}}\,}}(\frac{n}{d} ) - \frac{d-1}{4} \frac{{{\,\mathrm{{lln}}\,}}(\frac{n}{d}) }{ \ln (\frac{n}{d})} + (d-1) \frac{\frac{3}{2}}{ \ln (\frac{n}{d})} } \left( 1+ \frac{e^6}{\pi } \sqrt{d} e^{-\frac{1}{10} d} \right) \end{aligned}$$

(15)

$\square$

6 The Case $n-d$ Small

Finally, it remains to prove Theorem 1.3. The starting point here is again formula (2), together with the substitution $y \rightarrow \frac{y}{\sqrt{d}}$.

$$\begin{aligned} {\mathbb {E}}f_{d-1}([X_1,\ldots ,X_n])= & {} 2 {n \atopwithdelims ()d} \frac{\sqrt{d}}{\sqrt{\pi }} \int \limits _{- \infty }^\infty \Phi (y)^{n-d} e^{-dy^2}\,dy\nonumber \\= & {} 2 {n \atopwithdelims ()d} \frac{1}{\sqrt{\pi }} \int \limits _{- \infty }^\infty \Phi \left( \frac{y}{\sqrt{d}}\right) ^{n-d} e^{-y^2}\,dy \end{aligned}$$

(16)

The Taylor expansion of $\Phi$ at $y=0$ is given by

$$\begin{aligned} \Phi (y)= & {} \frac{1}{2} + \frac{1}{\sqrt{\pi }} y + \frac{1}{\sqrt{\pi }} (-\theta _1) e^{-\theta _1^2} \, y^2 = \frac{1}{2} + \frac{1}{\sqrt{\pi }} y (1-\theta _2 y) \end{aligned}$$

with some $\theta _1, \theta _2 \in {\mathbb {R}}$ depending on y. Since $\Phi (y)$ is above its tangent at 0 for $y>0$ and below it for $y<0$, we have $0 \le 1- \theta _2 y \le 1$. Further,

$$\begin{aligned} |\theta _2| \le \max _{\theta _1} \theta _1 e^{-\theta _1^2} = \frac{1}{\sqrt{2e}}. \end{aligned}$$

Hence an expression for $\ln \Phi$ at $y=0$ is given by

$$\begin{aligned} \ln \Phi (y)= & {} - \ln 2 + \ln \left( 1 + \frac{2}{\sqrt{\pi }} y (1- \theta _2 y) \right) . \end{aligned}$$

We need again estimates for the logarithm, namely $\ln (1+x) = x - \theta _3 x^2 < x$ with some $\theta _3 = \theta _3 (x) \ge 0$. In addition, there exists $c_3 \in {\mathbb {R}}$ such that $\theta _3 < c_3$ if x is bounded away from $-1$, for example, for $x \ge 2 \Phi (-1)-1$. This gives

$$\begin{aligned} \ln \Phi (y)\le & {} - \ln 2 + \frac{2}{\sqrt{\pi }} y - \frac{2}{\sqrt{\pi }} \theta _2 y^2 \end{aligned}$$

and

$$\begin{aligned} \ln \Phi (y)= & {} - \ln 2 + \frac{2}{\sqrt{\pi }} y (1- \theta _2 y) - \theta _3 \frac{4}{\pi } y^2 \underbrace{(1- \theta _2 y)^2}_{\le 1}\\\ge & {} - \ln 2 + \frac{2}{\sqrt{\pi }} y - \frac{2}{\sqrt{\pi }} \theta _2 y^2 - \theta _3 \frac{4}{\pi } y^2 \end{aligned}$$

with $\theta _3 < c_3$ for $y \ge -1$. Thus the Taylor expansion of $\ln \Phi$ at $y=0$ is given by

$$\begin{aligned} \ln \Phi (y)= & {} - \ln 2 + \frac{2}{\sqrt{\pi }} y - \theta _4 y^2 \end{aligned}$$

with some $\theta _4= \theta _4(y) > - \frac{1}{2}$, and there exists a $c_4 \in {\mathbb {R}}$ with $\theta _4 \le c_4$ for $y \ge -1$. We plug this into (16) and obtain

$$\begin{aligned} \int \limits _{- \infty }^\infty \Phi \left( \frac{y}{\sqrt{d}}\right) ^{n-d} e^{-y^2}\,dy= & {} e^{- (n-d)\ln 2} \int \limits _{- \infty }^\infty e^{ \frac{2}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} y - \theta _4 \frac{n-d}{d} y^2 -y^2 } \,dy~. \end{aligned}$$

Since $\frac{n-d}{d} \rightarrow 0$ we assume that $1+\theta _4 \frac{n-d}{d} \ge 1- \frac{1}{2} \frac{n-d}{d} > 0.$ As an estimate from above we have

$$\begin{aligned} \int \limits _{- \infty }^\infty e^{ \frac{2}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} y - (1+\theta _4 \frac{n-d}{d}) y^2 } \,dy\le & {} \int \limits _{- \infty }^\infty e^{ \frac{2}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} y - (1- \frac{1}{2} \frac{n-d}{d}) y^2 } \,dy \nonumber \\= & {} e^{\frac{\frac{4}{\pi }\frac{(n-d)^2}{d}}{4 (1- \frac{1}{2} \frac{n-d}{d})}} \int \limits _{- \infty }^\infty e^{ -\left( \frac{\frac{2}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}}}{2 \sqrt{(1- \frac{1}{2} \frac{n-d}{d})}} - \sqrt{(1- \frac{1}{2} \frac{n-d}{d})}y\right) ^2 } \,dy \nonumber \\= & {} e^{ \frac{1}{\pi }\frac{(n-d)^2}{d} \left( 1+ O\left( \frac{n-d}{d}\right) \right) } \frac{\sqrt{\pi }}{\sqrt{\left( 1- \frac{1}{2} \frac{n-d}{d}\right) } }\nonumber \\= & {} \sqrt{\pi }e^{ \frac{1}{\pi }\frac{(n-d)^2}{d} + O\left( \frac{(n-d)^3}{d^2}\right) + O\left( \frac{n-d}{d}\right) }. \end{aligned}$$

(17)

The estimate from below is slightly more complicated. For $y \ge -\sqrt{d}$ there is an upper bound $c_4$ for $\theta _4$. Using this we have

$$\begin{aligned} \int \limits _{- \infty }^\infty e^{ \frac{2}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} y - \theta _4 \frac{n-d}{d} y^2 -y^2 } \,dy\ge & {} e^{\frac{1}{\pi }\frac{(n-d)^2}{d}} \int \limits _{\frac{1}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}}- \sqrt{d}}^\infty e^{-\left( \frac{1}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} - y \right) ^2 - c_4 \frac{n-d}{d} y^2 } \,dy\\\ge & {} e^{\frac{1}{\pi }\frac{(n-d)^2}{d}} \int \limits _{- \infty }^{\sqrt{d} } e^{-y^2 - c_4 \frac{n-d}{d} \left( \frac{1}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} - y \right) ^2 } \,dy~. \end{aligned}$$

Now we use $(a-b)^2 \le 2a^2 + 2b^2$ which shows that

$$\begin{aligned} \int \limits _{- \infty }^\infty e^{ \frac{2}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} y - \theta _4 \frac{n-d}{d} y^2 -y^2 } \,dy\ge & {} e^{\frac{1}{\pi }\frac{(n-d)^2}{d} + O(\frac{(n-d)^3}{d^2}) } \int \limits _{- \infty }^{\sqrt{d} } e^{- (1+ 2 c_4 \frac{n-d}{d}) y^2 } \,dy\nonumber \\= & {} e^{\frac{1}{\pi }\frac{(n-d)^2}{d} + O\left( \frac{(n-d)^3}{d^2}\right) } \frac{1}{\sqrt{(1+ 2 c_4 \frac{n-d}{d})}} \int \limits _{- \infty }^{\sqrt{d (1+ 2 c_4 \frac{n-d}{d})}} e^{- y^2 } \,dy \nonumber \\\ge & {} e^{\frac{1}{\pi }\frac{(n-d)^2}{d} + O\left( \frac{(n-d)^3}{d^2}\right) +O\left( \frac{n-d}{d}\right) } \int \limits _{- \infty }^{\sqrt{d} } e^{- y^2 } \,dy. \end{aligned}$$

(18)

Recall the estimate for $\Phi (z)$ from Lemma 3.1,

$$\begin{aligned} \int \limits _{- \infty }^{\sqrt{d} } e^{- y^2 } \,dy = \sqrt{\pi }\, \Phi (\sqrt{d}) \ge \sqrt{\pi }(1- e^{-d}) = \sqrt{\pi }e^{O(e^{-d})}. \end{aligned}$$

(19)

We combine Eqs. (17), (18) and (19) and obtain

$$\begin{aligned} \int \limits _{- \infty }^\infty e^{ \frac{2}{\sqrt{\pi }} \frac{n-d}{\sqrt{d}} y - \theta _4 \frac{n-d}{d} y^2 -y^2 } \,dy = \sqrt{\pi }e^{\frac{1}{\pi }\frac{(n-d)^2}{d} + O\big (\frac{(n-d)^3}{d^2}\big ) +O\big (\frac{n-d}{d}\big ) + O(e^{-d}) } \end{aligned}$$

which yields Theorem 1.3.