2. Proof outlines

Overall, the proof ideas are based on the ones in [Reference Fey, van der Hofstad and Klok8] which were used to study $\lambda_{\max}$ and $\lambda_{\min}$ individually and jointly. More precisely, for the lower bounds in (2) and (3), we relate the condition number probability to that involving n i.i.d. random variables so that the classical Cramér’s theorem can be used.

Proof of Lemma 1. If $\omega\in\{\omega:k^2(p,n)\geq c\}$ , then let us take x = the eigenvector of $\lambda_{\max}$ and y = the eigenvector of $\lambda_{\min}$ . We can think of x and y as normalized vectors so that $||x||=||y||=1$ and $x\cdot y=0$ . Furthermore, $(x\cdot\mathbf{W}x)/(y\cdot\mathbf{W}y)=k^2(p,n)\geq c$ . To see the other direction, let us take $\omega\in\{(x\cdot\mathbf{W}x)/(y\cdot\mathbf{W}y)\geq c\}$ for some $||x||=||y||=1$ and $x\cdot y=0$ . Since $\lambda_{\max}\geq z\cdot\mathbf{W}z$ and $\lambda_{\min}\leq z\cdot\mathbf{W}z$ for all $z\in \mathbb{R}^p$ with $||z||=1$ , it follows that $k^2(p,n)\geq (x\cdot\mathbf{W}x)/$ $(y\cdot\mathbf{W}y)\geq c$ .

With the help of Lemma 1, we can take any two fixed points $x,y\in\mathbb{R}^p$ with $||x||=||y||=1$ and $x\cdot y=0$ , and obtain

(5) \begin{equation}\begin{aligned}\mathbb{P}\left(k^2(p,n)\geq c\right)\geq \mathbb{P}\left((x\cdot\mathbf{W}x)/(y\cdot\mathbf{W}y)\geq c\right)=\mathbb{P}\left(\sum_{i=1}^n(S_{x,i}^2-cS_{y,i}^2)/n\geq0\right).\end{aligned}\end{equation}

Now the classical Cramér’s theorem applied to the n i.i.d. random variables $(S_{x,i}^2-cS_{y,i}^2)$ , $1\leq i\leq n$ , gives the lower bounds.

The upper bounds in (2) and (3) are more complicated. In order to be able to still make use of Lemma 1, we need to divide the surface $S:=\{x\in \mathbb{R}^p:||x||=1\}$ of the p-dimensional sphere of unit radius into smaller pieces, and then take approximations.

The proof of Lemma 2 is given in [Reference Fey, van der Hofstad and Klok8, p. 1056], and the total number of pairs $\{x^{(i)},y^{(i,j)}\}_{i,j\geq1}$ is bounded by $N_d^2$ . Thanks to Lemma 2, the probability $\mathbb{P}\left(k^2(p,n)\geq c\right)$ essentially has an upper bound

\begin{equation*}\sum_{1\leq i,j\leq N_d}\mathbb{P}\big((x^{(i)}\cdot\mathbf{W}x^{(i)})/(y^{(i,j)}\cdot\mathbf{W}y^{(i,j)})\geq c_d\big)\end{equation*}

in an appropriate form for some $c_d$ depending on d (see estimates (9) below for a precise formulation of such an upper bound). However, estimating this upper bound requires subtle relations among $N_d, p$ , and n, and it turns out that the desired upper bounds in (2) and (3) can be derived in this way only for fixed p or $p=p(n)\rightarrow\infty$ with $p(n)=o(n)$ ; see Section 4 for the proof details.

In order to achieve an application in statistics using Wishart matrices, some non-asymptotic estimates will be derived for the distribution function of the condition number using ratios of two independent $\chi^2$ random variables; see Lemma 3 for a precise formulation. To this end, we decompose the joint probability density function of the ordered p eigenvalues.

The non-trivial proofs of the positivity $I_{p,0}(c)>0$ and $I_{\infty,0}(c)>0$ with $c>1$ and of the finiteness $I_{p,0}(c)<\infty$ and $I_{\infty,0}(c)<\infty$ with $c\geq1$ are given in Lemma 7.

3. Wishart matrices

In this section we restrict ourselves to real central Wishart matrices $\mathbf{W}_{p\times p}=\mathbf{X}\mathbf{X}^\top/n$ . The condition number k(p,n) of $\mathbf{W}$ does not depend on the scaling parameter $1/n$ , so throughout this section for simplicity the non-negative real eigenvalues $\lambda_i\geq0$ , $1\leq i\leq p$ , of $\mathbf{X}\mathbf{X}^\top$ are considered. In this setting, let us denote $\lambda=(\lambda_1,\ldots,\lambda_p)$ ; then, the probability density function of $\lambda$ can be written as (cf. [Reference James10])

\begin{equation*}f_{p,n}(\lambda)=c(p,n)\cdot \exp\left\{-\sum_{i=1}^p\lambda_i/2\right\}\cdot\prod_{1\leq i<j\leq p}|\lambda_i-\lambda_j|\cdot \prod_{i=1}^p\lambda_i^{(n-p-1)/2},\end{equation*}

where $c(p,n)=(\pi^{p/2}/2^{(n+2)p/2})\prod_{i=1}^p\left(\Gamma(1+i/2)\Gamma((n-p+i)/2)\right)^{-1}$ . Hence, the probability density function $g_{p,n}(\lambda_\mathrm{ord})$ of the ordered eigenvalues $\lambda_\mathrm{ord}=(\lambda_{(1)},\ldots,\lambda_{(p)})$ with $\lambda_{(1)}\geq \lambda_{(2)}\geq\cdots\geq \lambda_{(p)}\geq0$ is $g_{p,n}(\lambda_\mathrm{ord})=p!f_{p,n}(\lambda_\mathrm{ord})$ on $\lambda_{(1)}\geq \lambda_{(2)}\geq\cdots\geq \lambda_{(p)}\geq0$ . The main result of this section is to control the condition number using ratios of two independent $\chi^2$ random variables.

Lemma 3. For any $c\geq1$ and $2\leq p\leq n$ ,

\begin{align*} \mathbb{P}(U_1/U_2\geq c)\leq \mathbb{P}(k^2(p,n)\geq c)\leq a(p,n)\cdot \mathbb{P}(U_3/U_4\geq c), \end{align*}

where $U_1\sim \chi^2(n)$ , $U_2\sim \chi^2(n)$ , $U_3\sim \chi^2(n+3p-5)$ , and $U_4\sim \chi^2(n-p+1)$ are four independent $\chi^2$ random variables, and

\begin{equation*}a(p,n)=\frac{\pi\Gamma((n+3p-5)/2)\Gamma((n-p+1)/2)}{\Gamma(p/2)\Gamma(n/2)\Gamma((p-1)/2)\Gamma((n-1)/2)}.\end{equation*}

Proof. The lower bound follows from (5). More specifically, in the summation $\sum_{i=1}^n(S_{x,i}^2-cS_{y,i}^2)$ , each $S_{x,i}=\sum_{k=1}^px_k X_{ki}\sim N(0,1)$ and $S_{y,i}=\sum_{k=1}^py_k X_{ki}\sim N(0,1)$ . Furthermore, $S_{x,i}$ and $S_{y,j}$ are independent since $\mathbb{E}(S_{x,i}S_{y,j})=x\cdot y=0$ . Therefore, $\sum_{i=1}^nS_{x,i}^2\sim \chi^2(n)$ , which is independent of $\sum_{i=1}^nS_{y,i}^2\sim \chi^2(n)$ .

For the upper bound, we decompose the density function $g_{p,n}(\lambda_\mathrm{ord})$ of the ordered eigenvalues $\lambda_{(1)}\geq \lambda_{(2)}\geq\cdots\geq \lambda_{(p)}\geq0$ as follows:

\begin{align*}g_{p,n}(\lambda_\mathrm{ord}) & = p!f_{p,n}(\lambda_\mathrm{ord}) \\ & = p!c(p,n)\cdot \exp\left\{-\sum_{i=1}^p\lambda_{(i)}/2\right\}\cdot\prod_{1\leq i<j\leq p}(\lambda_{(i)}-\lambda_{(j)})\cdot \prod_{i=1}^p\lambda_{(i)}^{(n-p-1)/2} \\ & = C(p,n)\cdot \bigg[(\lambda_{(1)}\lambda_{(p)})^{(n-p-1)/2}\mathrm{e}^{-(\lambda_{(1)}+\lambda_{(p)})/2}\prod_{2\leq i\leq p}(\lambda_{(1)}-\lambda_{(i)}) \\ & \qquad\qquad\qquad\prod_{2\leq i\leq p-1}(\lambda_{(i)}-\lambda_{(p)})\bigg]\cdot h_{p,n}(\lambda_{(2)},\ldots,\lambda_{(p-1)}),\end{align*}

where $C(p,n)=(p!c(p,n))/((p-2)!c(p-2,n-2))$ , and $h_{p,n}$ is another density function given by $h_{p,n}(\lambda_{(2)},\ldots,\lambda_{(p-1)})=(p-2)!c(p-2,n-2)\cdot \exp\big\{{-}\sum_{i=2}^{p-1}\lambda_{(i)}/2\big\}\cdot\prod_{2\leq i<j\leq p-1}(\lambda_{(i)}-\lambda_{(j)})\cdot \prod_{i=2}^{p-1}\lambda_{(i)}^{(n-p-1)/2}$ . If one applies the bound $\prod_{2\leq i\leq p}(\lambda_{(1)}-\lambda_{(i)})\prod_{2\leq i\leq p-1}(\lambda_{(i)}-\lambda_{(p)})\leq \lambda_{(1)}^{(p-1)+(p-2)}$ on $\lambda_{(1)}\geq \lambda_{(2)}\geq\cdots\geq \lambda_{(p)}\geq0$ , then $g_{p,n}(\lambda_\mathrm{ord})\leq C(p,n)\cdot \lambda_{(1)}^{(n+3p-7)/2}\cdot \lambda_{(p)}^{(n-p-1)/2}\cdot \mathrm{e}^{-(\lambda_{(1)}+\lambda_{(p)})/2}\cdot h_{p,n}(\lambda_{(2)},\ldots,\lambda_{(p-1)})$ . Hence,

\begin{align*} \mathbb{P}(k^2(p,n)\geq c) & = \int_{\lambda_{(1)}/\lambda_{(p)}\geq c, \lambda_{(1)}\geq\cdots\geq\lambda_{(p)}\geq0} g_{p,n}(\lambda_\mathrm{ord}) \, \mathrm{d} \lambda_{(1)}\cdots \mathrm{d} \lambda_{(p)} \\ & \leq \int_{\lambda_{(1)}/\lambda_{(p)}\geq c}\int_{\lambda_{(2)}\geq\cdots\geq\lambda_{(p-1)}\geq0} g_{p,n}(\lambda_\mathrm{ord}) \, \mathrm{d}\lambda_{(1)}\cdots \mathrm{d}\lambda_{(p)} \\ & \leq C(p,n)\int_{\lambda_{(1)}/\lambda_{(p)}\geq c} \!\!\! \lambda_{(1)}^{(n+3p-7)/2}\cdot \lambda_{(p)}^{(n-p-1)/2}\cdot \mathrm{e}^{-(\lambda_{(1)}+\lambda_{(p)})/2} \, \mathrm{d}\lambda_{(1)}\mathrm{d}\lambda_{(p)} \\ & = a(p,n)\cdot \mathbb{P}(U_3/U_4\geq c),\end{align*}

where the last equality comes from identifying the density functions of the corresponding two independent $\chi^2$ random variables. We remark that similar decompositions were used for $\lambda_{\max}$ and $\lambda_{\min}$ individually in [Reference Jiang and Li11].

As mentioned earlier, the upper tail controls specified in Lemma 3 yield an independent concise proof of Corollary 1 (with $p=o(n/\ln n)$ ), which is presented here for the sake of completeness.

Proof of Corollary 1. It suffices to prove the upper bound for $c>1$ . Lemma 3 yields

\begin{align*}\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}\left(k^2(p,n)\geq c\right)\leq \limsup_{n\rightarrow\infty}n^{-1}\ln a(p,n)+\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}(U_3/U_4\geq c).\end{align*}

We shall first prove $\lim_{n\rightarrow\infty}n^{-1}\ln a(p,n)=0$ under $p=o(n/\ln n)$ . If p is fixed, then, as $n\rightarrow\infty$ ,

\begin{align*}\ln a(p,n)&=\ln\left[\frac{\pi\Gamma((n+3p-5)/2)\Gamma((n-p+1)/2)}{\Gamma(p/2)\Gamma(n/2)\Gamma((p-1)/2)\Gamma((n-1)/2)}\right]\\&=\ln\Gamma((n+3p-5)/2)+\ln\Gamma((n-p+1)/2)\\&\,\,\,\,\,\,-\ln\Gamma(n/2)-\ln\Gamma((n-1)/2)+o(n).\end{align*}

It follows from Stirling’s approximation that $\ln \Gamma(x)=x\ln x-x+o(x)$ as $x\rightarrow\infty$ , so $\ln a(p,n)$ becomes $(n/2)\ln((n+3p-5)/n)+(n/2)\ln((n-p+1)/(n-1))+o(n)=o(n)$ , which proves the limit $\lim_{n\rightarrow\infty}n^{-1}\ln a(p,n)=0$ . When $p=p(n)\rightarrow\infty$ with $p(n)=o(n/\ln n)$ as $n\rightarrow\infty$ , similar arguments to those above yield $\ln a(p,n)=(n/2)\ln((n+3p-5)/n)+(n/2)\ln((n-p+1)/(n-1))+O(1)p\ln n+o(n)$ , and this again implies the limit $\lim_{n\rightarrow\infty}n^{-1}\ln a(p,n)=0$ .

Next, we shall prove the following estimate:

(6) \begin{equation} \limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}(U_3/U_4\geq c)\leq-2^{-1}\ln\left[(c+1)^2/(4c)\right]. \end{equation}

To this end, let us first rewrite the ratio of the two independent $\chi^2$ random variables as summations of squares of independent standard normal random variables,

\begin{align*}\mathbb{P}(U_3/U_4\geq c)=\mathbb{P}\left(\sum_{i=1}^{n+3p-5}\xi_i^2-c\sum_{i=1}^{n-p+1}\eta_i^2\geq0\right),\end{align*}

where $\xi_i$ and $\eta_i$ are independent standard normal random variables. However, the two degrees of freedom $n+3p-5$ and $n-p+1$ are different, so Cramér’s theorem cannot be readily applied. To achieve the large-deviation asymptotics, we make use of the Gärtner–Ellis theorem [Reference Dembo and Zeitouni5, Section 2.3]. To this end, let us define

\begin{equation*}Z_n=\left(\sum_{i=1}^{n+3p-5}\xi_i^2-c\sum_{i=1}^{n-p+1}\eta_i^2\right)/n,\end{equation*}

and consider the logarithmic moment-generating function of $Z_n$ defined as $\Lambda_n(\theta)=\ln \mathbb{E}\exp\{\theta Z_n\}.$ By using the fact that $\mathbb{E}\exp\{\theta\xi_i^2\}=(1-2\theta)^{-1/2}$ for $\theta<1/2$ , one can establish the limit, for $-1/(2c)<\theta<1/2$ and $p=o(n)$ ,

\begin{align*}\Lambda(\theta)&=\lim_{n\rightarrow\infty}n^{-1}\Lambda_n(n\theta)=\lim_{n\rightarrow\infty}n^{-1}\ln \mathbb{E}\exp\{n\theta Z_n\}=-2^{-1}\ln\left[(1-2\theta)(1+2c\theta)\right].\end{align*}

If one defines the Fenchel–Legendre transform of $\Lambda(\theta)$ as $\Lambda^*(\alpha)=\sup_{\theta \in \mathbb{R}}[\theta\cdot \alpha-\Lambda(\theta)]$ , $\alpha\in \mathbb{R}$ , then the Gärtner–Ellis theorem says that, for any closed set F,

(7) \begin{align}\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}\left(Z_n \in F\right)\leq -\inf_{\alpha\in F}\Lambda^*(\alpha).\end{align}

Now we take $F=[0,\infty)$ , and we claim that, for any $\alpha\geq0$ , $\Lambda^*(\alpha)=\sup_{\theta \geq 0}[\theta\cdot \alpha-\Lambda(\theta)].$ This comes from the fact that $\Lambda(\theta)=\ln \mathbb{E}\exp\{\theta\xi_i^2-c\theta\eta_i^2\}\geq\theta(1-c),$ thus implying $\theta\cdot \alpha-\Lambda(\theta)\leq \theta(\alpha+(c-1))\leq0$ for any $\theta\leq0.$ Therefore, $\Lambda^*(\alpha)$ is non-decreasing in $\alpha$ for $\alpha\geq0$ . Hence, in (7) we have $\inf_{\alpha\in F}\Lambda^*(\alpha)=\Lambda^*(0)=2^{-1}\ln\left[(c+1)^2/(4c)\right]$ , so (6) now follows from (7).

4. Random matrices with sub-Gaussian entries

In this section we consider sub-Gaussian random matrices $\mathbf{X}$ with i.i.d. entries $X_{ij}$ being sub-Gaussian satisfying (1). To establish large-deviation asymptotics for the condition numbers in Theorem 1, we handle the two cases p being fixed and p(n) being dependent on n separately since there are more subtle relations among $N_d$ (which appeared in Lemma 2), d, p, and n when $p(n)\rightarrow\infty$ . To avoid triviality let us focus on the case $c>1$ .

Since the lower bounds have already been proved in Section 2, we focus here on the upper bounds. In the spirit of Lemmas 1 and 2, the set $\{k^2(p,n)\geq c\}$ can be rewritten and estimated as follows (for notational simplicity, all points x, y, $x^{(i)}$ , and $y^{(i,j)}$ below will be on S, and we will not write this explicitly):

(8) \begin{equation} \begin{aligned} \{k^2(p,n)\geq c\} & = \{ x,y \text{ s.t. }x\cdot y=0 \text{ and }(x\cdot\mathbf{W}x)/(y\cdot\mathbf{W}y)\geq c\}\\ & \subseteq \{ x^{(i)},y^{(i,j)} \text{ s.t. }x^{(i)}\cdot y^{(i,j)}=0 \text{ and }\\ & \qquad x^{(i)}\cdot\mathbf{W}x^{(i)}-cy^{(i,j)}\cdot\mathbf{W}y^{(i,j)}\geq -2\lambda_{\max}d(2c+1)\},\end{aligned}\end{equation}

where the inclusion $\subseteq$ comes from Lemma 2 and the fact $|x\cdot\mathbf{W}x-x^{(i)}\cdot\mathbf{W}x^{(i)}|\leq (||x||+||x^{(i)}||)||W||||x-x^{(i)}||\leq 2 \lambda_{\max}d.$

4.1. Fixed dimension p

When p is fixed, the number $N_d$ of spherical caps of chord $2d\sqrt{1-d^2/4}$ needed to cover S can be chosen fixed as well, and the exact expression of $N_d$ is not important. To analyze the probability of the latter set in (8), we further consider two disjoint sets $\{\lambda_{\max}\leq p K\}$ and $\{\lambda_{\max}> p K\}$ for some large K which will be specified later. Since $\lambda_{\max}\leq \mathrm{tr}(\mathbf{W})$ , with $\mathrm{tr}(\mathbf{W})$ denoting the trace of $\mathbf{W}$ , it then follows from (8) that

(9) \begin{equation} \begin{aligned} \mathbb{P}(k^2(p,n)\geq c) & \leq \mathbb{P}\big(\text{there exists } x^{(i)},y^{(i,j)} \text{ s.t. }x^{(i)}\cdot y^{(i,j)}=0 \text{ and }\\ &\qquad\,\,\, x^{(i)}\cdot\mathbf{W}x^{(i)}-cy^{(i,j)}\cdot\mathbf{W}y^{(i,j)}\geq -2pKd(2c+1)\big)\\ &\quad+\mathbb{P}\big(\mathrm{tr}(\mathbf{W})>pK\big)\\ &\leq \sum_{1\leq i,j\leq N_d}\mathbb{P}\big(x^{(i)}\cdot\mathbf{W}x^{(i)}-cy^{(i,j)}\cdot\mathbf{W}y^{(i,j)}\geq -2pKd(2c+1)\big)\\ &\quad+\mathbb{P}\big(\mathrm{tr}(\mathbf{W})>pK\big).\end{aligned}\end{equation}

Therefore, denoting $\varepsilon:=2pKd(2c+1)$ with $-\varepsilon>1-c$ for small d, Cramér’s theorem yields

\begin{align*}\limsup_{n\rightarrow\infty}n^{-1}\ln & \sum_{1\leq i,j\leq N_d}\mathbb{P}\big(x^{(i)}\cdot\mathbf{W}x^{(i)}-cy^{(i,j)}\cdot\mathbf{W}y^{(i,j)}\geq -\varepsilon\big)\\&=\max_{1\leq i,j\leq N_d}\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}\big(x^{(i)}\cdot\mathbf{W}x^{(i)}-cy^{(i,j)}\cdot\mathbf{W}y^{(i,j)}\geq -\varepsilon\big)\\&\leq \max_{1\leq i,j\leq N_d} -\sup_{\theta\in\mathbb{R}}\big[-\varepsilon\theta-\ln \mathbb{E}\exp\big\{\theta\big(S_{x^{(i)},1}^2-cS_{y^{(i,j)},1}^2\big)\big\}\big]\\&\leq -I_{p,-\varepsilon}(c).\end{align*}

On the other hand, it follows from Cramér’s theorem again that, with $K>1$ ,

\begin{align*}\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}\big(\mathrm{tr}(\mathbf{W})>pK\big) &=\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}\left(\sum_{i=1}^p\sum_{j=1}^nX_{ij}^2/(np)>K\right)\\&\leq -p I_{X^2}(K),\end{align*}

where $I_{X^2}(K)=\sup_{\theta\in\mathbb{R}}\big[\theta K-\ln \mathbb{E}\exp\big\{\theta X_{11}^2\big\}\big]$ . We now apply the following fact, whose proof will be presented in Section 6.

Lemma 4. Under the assumptions of Theorem 1 with a fixed p, for any $c>1$ we have $\lim_{\varepsilon\rightarrow0^-}I_{p,\varepsilon}(c)=I_{p,0}(c)<\infty$ and $\lim_{K\rightarrow\infty}I_{X^2}(K)=\infty$ .

Taking into account all these observations, we obtain, from (9),

\begin{equation*}\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}(k^2(p,n)\geq c)\leq \max\left\{-I_{p,-\varepsilon}(c), -p I_{X^2}(K)\right\}.\end{equation*}

By sending $d\rightarrow0^+$ (equivalently $\varepsilon\rightarrow0^+$ ), it follows that $I_{p,-\varepsilon}(c)\rightarrow I_{p,0}(c)$ . Furthermore, by taking large K, it is clear that $I_{X^2}(K)>I_{p,0}(c)$ . Therefore,

\begin{align*}\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}(k^2(p,n)\geq c)\leq -I_{p,0}(c).\end{align*}

4.2. High dimension $${\bf{p}}\; = \;{\bf{p}}({\bf{n}})$$

We note that the arguments in Section 4.1 do not go through when $p=p(n)\rightarrow\infty$ as $n\rightarrow\infty$ since, when we take the limit $n\rightarrow\infty$ , the parameter $\varepsilon\rightarrow\infty$ as well. Furthermore, the number $N_d$ of spherical caps of chord $2\tilde{d}:=2d\sqrt{1-d^2/4}$ needed to cover S is important and an explicit expression for this in terms of d is needed. According to [Reference Rogers16], $N_d=4p(n)^{3/2}\tilde{d}^{-p(n)}(\ln p(n)+\ln\ln p(n)-\ln \tilde{d})(1+O(1/\ln p(n)))$ for all $d<1/2$ and large p(n). The main ingredient of the proof in the high dimension setting is the following concentration inequality for the maximal singular value of the matrix $\mathbf{X}_{p\times n}$ .

Lemma 5. [Reference Vershynin24, Theorem 5.39] With $p\leq n$ , suppose that the entries $X_{ij}$ , $1\leq i\leq p$ , $1\leq j\leq n$ are i.i.d. sub-Gaussian. Then, for any $\gamma\geq0$ , $\mathbb{P}(\lambda_{\max}>(1+\kappa_1+\gamma)^2)\leq 2\exp\{-\kappa_2\gamma^2n\}$ , where $\kappa_1,\kappa_2>0$ are two constants depending only on the sub-Gaussian norm of $X_{ij}$ .

With such a concentration inequality, we estimate as follows: $\mathbb{P}(k^2(p,n)\geq c)\leq \mathbb{P}(k^2(p,n)\geq c,\lambda_{\max}\leq (1+\kappa_1+\gamma)^2)+\mathbb{P}(\lambda_{\max}>(1+\kappa_1+\gamma)^2)$ . For the term $\mathbb{P}(k^2(p,n)\geq c,\lambda_{\max}\leq (1+\kappa_1+\gamma)^2)$ we can employ the idea used to prove (9) to obtain

\begin{multline*} \mathbb{P}\big(k^2(p,n)\geq c,\lambda_{\max}\leq (1+\kappa_1+\gamma)^2\big)\\ \leq N_d^2\max_{1\leq i,j\leq N_d}\mathbb{P}\big(x^{(i)}\cdot\mathbf{W}x^{(i)}-cy^{(i,j)}\cdot\mathbf{W}y^{(i,j)}\geq -2(1+\kappa_1+\gamma)^2d(2c+1)\big).\end{multline*}

It now follows again from Cramér’s theorem that, with $\varepsilon=2(1+\kappa_1+\gamma)^2d(2c+1)$ ,

\begin{align*}n^{-1}\ln \mathbb{P}&(k^2(p,n)\geq c,\lambda_{\max}\leq (1+\kappa_1+\gamma)^2)\\&\leq (2/n)\ln N_d-\min_{1\leq i,j\leq N_d}\sup_{\theta\in\mathbb{R}}\big[-\varepsilon\theta-\ln \mathbb{E}\exp\big\{\theta\big(S_{x^{(i)},1}^2-cS_{y^{(i,j)},1}^2\big)\big\}\big]\\&\leq (2/n)\ln N_d- I_{p(n),-\varepsilon}(c)\\&\leq (2/n)\ln N_d- I_{\infty,-\varepsilon}(c),\end{align*}

where the last inequality comes from the fact that $I_{p,-\varepsilon}\geq I_{\infty,-\varepsilon}(c)$ for small enough $\varepsilon.$ The assumption $p=p(n)=o(n)$ implies that $\lim_{n\rightarrow\infty} (2/n)\ln N_d=0$ for each fixed d. Therefore, $\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}(k^2(p,n)\geq c,\lambda_{\max}\leq (1+\kappa_1+\gamma)^2)\leq - I_{\infty,-\varepsilon}(c)$ . Taking the limit $d\rightarrow0^+$ implies that $\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}(k^2(p,n)\geq c,\lambda_{\max}\leq (1+\kappa_1+\gamma)^2)\leq - I_{\infty,0}(c)$ . The limit $\lim_{\varepsilon\rightarrow0^-}I_{\infty,\varepsilon}(c)=I_{\infty,0}(c)$ is established in Lemma 6. In summary, $\limsup_{n\rightarrow\infty}n^{-1}\ln \mathbb{P}(k^2(p,n)\geq c)\leq \max(-I_{\infty,0}(c),-\kappa_2 \gamma^2)$ , and the proof is completed by sending $\gamma\rightarrow\infty$ .

Lemma 6. Under the assumptions of Theorem 1 with $p=p(n)\rightarrow\infty$ and $p(n)=o(n)$ , we have, for any $c>1$ , $\lim_{\varepsilon\rightarrow0^-}I_{\infty,\varepsilon}(c)=I_{\infty,0}(c)<\infty$ .

Proof. Since $I_{p,\varepsilon}(c)$ is non-increasing in p, the arguments leading to (10) still work in this case. Therefore, by taking $p(n)\rightarrow\infty$ we obtain $2(N+1)\varepsilon/\varepsilon_0\leq I_{\infty,\varepsilon}(c)-I_{\infty,0}(c)\leq 0$ , and complete the proof by sending $\varepsilon\rightarrow0^-$ .

5. Application: The union-intersection test method

In this section, suppose that a population is p-variate normal with a zero mean vector and covariance matrix $\Sigma_{p\times p}$ . The sphericity test deals with the hypotheses $H_0: \Sigma_{p\times p}=\sigma^2 I_{p\times p}$ for some $\sigma>0$ , $H_1: \Sigma_{p\times p}\neq\sigma^2 I_{p\times p}$ . Among others, the union-intersection test method [Reference Srivastava and Khatri20, Section 7.4] suggests that $H_0$ is rejected if $k^2(p,n) \geq c$ , where c is determined from $\mathbb{P}(k^2(p,n) \geq c)=\alpha$ with a given significance level $\alpha$ . Unfortunately, so far in the literature there is no efficient way to evaluate the probability $\mathbb{P}(k^2(p,n) \geq c)$ under the hypothesis $H_0$ . We remark here (again) that an exact expression for $\mathbb{P}(k^2(p,n) \geq c)$ was derived in [Reference Anderson and Wells1] with $\Sigma_{p\times p}=I_{p\times p}$ for all $2\leq p\leq n$ involving complicated zonal polynomials which prevents us efficiently obtaining the probability $\mathbb{P}(k^2(p,n) \geq c)$ (especially for large n). In this section we aim to control the probability $\mathbb{P}(k^2(p,n) \geq c)$ (under $\Sigma_{p\times p}=I_{p\times p}$ in which we choose, without loss of generality, $\sigma=1$ since the condition number does not depend on $\sigma$ ) using the upper bound in Lemma 3 which is easy and efficient, and then apply it to test the above hypotheses using the union-intersection test method.

Recall the upper bound of $\mathbb{P}(k^2(p,n) \geq c)$ in Lemma 3: $\mathbb{P}(k^2(p,n)\geq c)\leq a(p,n)\cdot \mathbb{P}(U_3/U_4\geq c)$ , where $U_3\sim \chi^2(n+3p-5)$ and $U_4\sim \chi^2(n-p+1)$ are two independent $\chi^2$ random variables. With such an upper bound, we are able to control the p-value for the above hypotheses test. More specifically, let $k^2_\mathrm{obs}(p,n)$ denote the observed condition number for data simulated with $\Sigma_{p\times p}=I_{p\times p}$ ; then, the p-value has an upper bound $p_\mathrm{upp}$ defined as $p_\mathrm{upp}= a(p,n)\cdot \mathbb{P}(U_3/U_4\geq k^2_\mathrm{obs}(p,n))$ . The null hypothesis $H_0$ is then rejected if $p_\mathrm{upp}<\alpha$ . Based on these, the simulated powers $\hat{h}$ of the hypotheses are obtained for various $\Sigma_{p\times p}$ through computer simulations (performed $N=10\,000$ times); see Tables 1 and 2.

Table 1.

Simulated powers when $p=2$ , for $\Sigma_{p\times p}= \begin{bmatrix} 1\;\;\;\; & 0.5\\ 0.5\;\;\;\; & 1 \end{bmatrix} $ , $p=2$ , and $\alpha=0.05$ .

Table 2.

Simulated powers when $p=4$ , for $\Sigma_{p\times p}= \begin{bmatrix} 1\;\;\;\;\;\; & 0.25\;\;\;\;\;\; & 0.25\;\;\;\;\;\; & 0.4 \\ 0.25\;\;\;\;\;\; & 1\;\;\;\;\;\; & 0.1\;\;\;\;\;\; & 0.2 \\ 0.25\;\;\;\;\;\; & 0.1\;\;\;\;\;\; & 1\;\;\;\;\;\; & 0.3 \\ 0.4\;\;\;\;\;\; & 0.2\;\;\;\;\;\; & 0.3\;\;\;\;\;\; & 1 \end{bmatrix} $ , $p=4$ , and $\alpha=0.05$ .

These simulated powers suggest that when n is large, it is easy to make a correct conclusion that the population covariance matrix is not an identity matrix. Here we also note that if one uses the above procedures to simulate the type-I error $\alpha$ (under the setting that the population covariance matrix is an identity matrix), it can happen that the simulated type-I error is much smaller than $\alpha$ since an upper bound of the p-value is used during the procedures. For instance, with $p=4$ , $n=100$ , $N=10\,000$ , and $\alpha=0.05$ , the simulated type-I error is around $10^{-4}$ .

6. Other detailed proofs

This section contains proofs of the facts and the auxiliary lemmas used in previous sections.

Proof of Lemma 4. We first prove that $\lim_{K\rightarrow\infty}I_{X^2}(K)=\infty$ . Since $X_{ij}$ are sub-Gaussian, it follows from the super-exponential moment that $\mathbb{E}\exp\{t X_{11}^2\}<\infty$ for some $t>0$ (actually, one can take $t=1/K_3^2$ ). Therefore,

\begin{equation*} I_{X^2}(K) = \sup_{\theta\in\mathbb{R}}\left[\theta K-\ln \mathbb{E}\exp\{\theta X_{11}^2\}\right] \geq t K-\ln \mathbb{E}\exp\{t X_{11}^2\} \rightarrow\infty \text{ as }K\rightarrow\infty.\end{equation*}

We then prove that $I_{p,0}(c)<\infty$ for all $p\geq2$ and $c\geq1$ . It suffices to just prove the case $p=2$ since $0\leq I_{p,0}(c)\leq I_{2,0}(c)$ . To show this, we take $x^*=(x_1,x_2)=(1/\sqrt{2},1/\sqrt{2})$ and $y^*=(y_1,y_2)=(1/\sqrt{2},-1/\sqrt{2})$ . Hence,

\begin{align*}S_{x^*,1}^2-cS_{y^*,1}^2&=\big(X_{11}/\sqrt{2}+X_{21}/\sqrt{2}\big)^2-c\big(X_{11}/\sqrt{2}-X_{21}/\sqrt{2}\big)^2\\&=(1-c)\big(X_{11}^2+X_{21}^2\big)/2+(1+c)X_{11}X_{21}.\end{align*}

If everything is restricted to the set $\{X_{21}+\varepsilon_1\geq X_{11}\geq X_{21}\geq \varepsilon_2\geq0\}$ for some $\varepsilon_1,\varepsilon_2\geq0$ , then

\begin{align*}S_{x^*,1}^2-cS_{y^*,1}^2&\geq(1-c)X_{11}^2+(1+c)X_{11}(X_{11}-\varepsilon_1)\\&=X_{11}(2X_{11}-(1+c)\varepsilon_1)\\&\geq X_{11}(2\varepsilon_2-(1+c)\varepsilon_1)\\&\geq 0 \quad \text{ if }(1+c)\varepsilon_1\leq 2\varepsilon_2.\end{align*}

Therefore, for $(1+c)\varepsilon_1\leq 2\varepsilon_2$ ,

\begin{align*}I_{2,0}(c)&=\inf_{x,y\in \mathbb{R}^2,||x||=||y||=1,x\cdot y=0} \, \sup_{\theta\in\mathbb{R}}\left[-\ln \mathbb{E}\exp\{\theta(S_{x,1}^2-cS_{y,1}^2)\}\right]\\&=\inf_{x,y\in \mathbb{R}^2,||x||=||y||=1,x\cdot y=0}\, \sup_{\theta\geq0}\left[-\ln \mathbb{E}\exp\{\theta(S_{x,1}^2-cS_{y,1}^2)\}\right]\\&\leq \sup_{\theta\geq0}\left[-\ln \mathbb{E}\exp\{\theta(S_{x^*,1}^2-cS_{y^*,1}^2)\}\right]\\&\leq -\ln \mathbb{P}(X_{21}+\varepsilon_1\geq X_{11}\geq X_{21}\geq \varepsilon_2\geq0).\end{align*}

We now claim that, for any entries $X_{ij}$ satisfying the assumptions of Theorem 1, there always exist $\varepsilon_1$ and $\varepsilon_2$ with $(1+c)\varepsilon_1\leq 2\varepsilon_2$ such that $\mathbb{P}(X_{21}+\varepsilon_1\geq X_{11}\geq X_{21}\geq \varepsilon_2\geq0)>0$ . Here we chose the right side of zero without loss of generality because $\mathbb{E}(X_{ij}=0)$ , but if needed the left side of zero can be chosen. To see this claim, let us first look at the case when the distribution of $X_{ij}$ has a pure discrete part on $\mathbb{R}^+$ , and in this case we can simply take $\varepsilon_1=\varepsilon_2=0$ since $\mathbb{P}(X_{11}=X_{21}\geq0)>0$ . If the distribution of $X_{ij}$ does not contain a pure discrete part, then there must exist $0<c_1<c_2<c_3$ with $c_3=c_1+2c_1/(1+c)$ (here it is stressed again that, without loss of generality, we only consider the distribution on $\mathbb{R}^+$ ) such that $\mathbb{P}(X_{ij}\in(c_1,c_2))>0$ and $\mathbb{P}(X_{ij}\in(c_2,c_3))>0$ . Setting $\varepsilon_1=c_3-c_1$ and $\varepsilon_2=c_1$ , it follows that $(1+c)\varepsilon_1= 2\varepsilon_2$ and

\begin{align*}\mathbb{P}(X_{21}+\varepsilon_1\geq X_{11}\geq X_{21}\geq \varepsilon_2\geq0)&\geq \mathbb{P}(X_{21}\in (c_1,c_2),X_{11}\in (c_2,c_3)) \\&\geq \mathbb{P}(X_{21}\in (c_1,c_2))\mathbb{P}(X_{11}\in (c_2,c_3))>0.\end{align*}

This proves $I_{2,0}(c)<\infty$ for any $c\geq1$ . Next, we shall prove that, for $c>1$ , $\lim_{\varepsilon\rightarrow0^+}I_{p,-\varepsilon}(c)=I_{p,0}(c)$ . To achieve this limit, first note that we can actually prove a stronger finiteness result: $I_{2,\varepsilon_0}(c)<\infty$ for some small $\varepsilon_0>0$ . (‘Stronger’ here refers to the fact that $I_{2,\alpha}(c)$ is non-decreasing in $\alpha$ for all $\alpha\geq 1-c$ .) By taking the same $x^*$ and $y^*$ as in the above arguments, we obtain

\begin{align*}\mathbb{E}\exp\big\{\theta&\big(S_{x^*,1}^2-cS_{y^*,1}^2\big)\big\}\geq\\&\begin{cases}\exp\{\theta\varepsilon_2(2\varepsilon_2-(1+c)\varepsilon_1)\}\mathbb{P}(X_{21}+\varepsilon_1\geq X_{11}\geq X_{21}\geq \varepsilon_2\geq0),& \mathrm{(I)} \\\exp\{\theta 2\varepsilon_2^2\}\mathbb{P}(X_{11}=X_{21}\geq \varepsilon_2\geq0),& \mathrm{(II)}\end{cases}\end{align*}

where (I) represents the case when the distribution of $X_{ij}$ has no pure discrete part, and (II) means that the distribution has a pure discrete part. If in the above arguments we take $c_3=c_1+1.5c_1/(1+c)$ (everything else will be kept the same), then $(1+c)\varepsilon_1=1.5c_1\leq 2\varepsilon_2$ . Hence, $I_{2,\varepsilon_0}(c)<\infty$ for any $\varepsilon_0\leq \min\{\varepsilon_2(2\varepsilon_2-(1+c)\varepsilon_1),2\varepsilon_2^2\}$ , implying that $I_{p,\varepsilon_0}(c)<\infty$ for any $p\geq2$ .

If we denote $g_{x,y}(\alpha)=\sup_{\theta\in\mathbb{R}}\big[\theta \alpha-\ln \mathbb{E}\exp\big\{\theta\big(S_{x,1}^2-cS_{y,1}^2\big)\big\}\big]$ , then it is clear that $I_{p,\alpha}(c)=\inf_{x,y\in \mathbb{R}^p,||x||=||y||=1,x\cdot y=0}g_{x,y}(\alpha)$ . Since $0\leq I_{p,\varepsilon}(c)\leq I_{p,\varepsilon_0}(c):=N<\infty$ for all $1-c\leq \varepsilon\leq \varepsilon_0$ , it suffices to consider $A:=\{(x,y)\in \mathbb{R}^{2p}: ||x||=||y||=1,x\cdot y=0, g_{x,y}(\varepsilon_0)\leq N+1\}$ , that is, $I_{p,\varepsilon}(c)=\inf_{x,y\in A}g_{x,y}(\varepsilon)$ for all $1-c\leq \varepsilon\leq \varepsilon_0$ . Then, for any $(x,y)\in A$ , the convexity of $g_{x,y}(\varepsilon)$ in $\varepsilon\in (1-c,0)$ yields

\begin{equation*}(g_{x,y}(0)-g_{x,y}(\varepsilon))/(0-\varepsilon)\leq (g_{x,y}(\varepsilon_0)-g_{x,y}(0))/(\varepsilon_0).\end{equation*}

This gives us the following nice bounds (with $g_{x,y}(0)\geq 0$ in mind):

\begin{equation*}0\leq g_{x,y}(0)-g_{x,y}(\varepsilon)\leq (N+1)(-\varepsilon)/\varepsilon_0.\end{equation*}

Hence,

(10) \begin{equation}\begin{aligned}(N+1)\varepsilon/\varepsilon_0&\leq \inf_{(x,y)\in A}\left(g_{x,y}(\varepsilon)-g_{x,y}(0)\right)\\&\leq \inf_{(x,y)\in A}g_{x,y}(\varepsilon)-\inf_{(x,y)\in A}g_{x,y}(0)\\&=I_{p,\varepsilon}(c)-I_{p,0}(c)\leq 0.\end{aligned}\end{equation}

Sending $\varepsilon\rightarrow0^-$ in (10) completes the proof.

Proof. The proof of the finiteness of $I_{p,0}(c)$ and $I_{\infty,0}(c)$ has been done separately in Lemmas 4 and 6. Here we prove the rest.

Jensen’s inequality yields $\ln \mathbb{E}\exp\{\theta(S_{x,1}^2-S_{y,1}^2)\}\geq \mathbb{E}(\theta(S_{x,1}^2-S_{y,1}^2))=0$ . Therefore, $\sup_{\theta\in\mathbb{R}}\big[-\ln \mathbb{E}\exp\big\{\theta\big(S_{x,1}^2-S_{y,1}^2\big)\big\}\big]\leq0$ , implying that $I_{p,0}(1)\leq 0$ . Taking $\theta=0$ gives $I_{p,0}(1)= 0$ . The term $I_{\infty,0}(1)$ as the limit of $I_{p,0}(1)$ must also satisfy $I_{\infty,0}(1)=0$ .

Now we prove that $I_{p,0}(c)>0$ and $I_{\infty,0}(c)>0$ for $c>1$ ; this will be done in three steps.

Step 1 We claim that, for any $c\geq1,p\geq2$ ,

(11) \begin{equation}\begin{aligned}I_{p,0}(c)\geq 1/2\,\,\cdot\inf_{x,y\in \mathbb{R}^p,||x||=||y||=1,x\cdot y=0} \, \sup_{\theta\geq0}\left[-\ln \left(\mathbb{E}\exp\{\theta S_{x,1}^2\}\mathbb{E}\exp\{-\theta cS_{y,1}^2\}\right)\right].\end{aligned}\end{equation}

To see (11), it suffices to notice that

\begin{equation*}\mathbb{E}\exp\{\theta(S_{x,1}^2-cS_{y,1}^2)\}\leq (\mathbb{E}\exp\{2\theta S_{x,1}^2\})^{1/2}(\mathbb{E}\exp\{-2\theta cS_{y,1}^2\})^{1/2}.\end{equation*}

Step 2 For $c>1$ and fixed $p\geq2$ , it will be proved that $I_{p,0}(c)>0$ . To this end, let us first note that, because of $\mathbb{E}\exp\{tX_{11}^2\}<\infty$ for some $t>0$ (as remarked in the proof of Lemma 4),

(12) \begin{align}\mathbb{E}\exp\{t S_{x,1}^2\}\leq \mathbb{E}\exp\left\{t\sum_{i=1}^pX_{i1}^2\right\}=\left(\mathbb{E}\exp\{tX_{11}^2\}\right)^p<\infty.\end{align}

Then, it follows from $\mathrm{e}^x=1+x+x^2/2+\mathrm{e}^{\alpha x}x^3/6$ for some $\alpha\in[0,1]$ that

\begin{equation*}\mathbb{E}\exp\{sS_{x,1}^2\}=1+s+s^2\mathbb{E}(S_{x,1}^4)/2+s^3\mathbb{E}\left(\exp\{\alpha s S_{x,1}^2\}\cdot S_{x,1}^6\right)/6\end{equation*}

for small $0\leq s\leq \varepsilon_0$ . The term $\mathbb{E}(S_{x,1}^4)$ can be bounded as

\begin{equation*}\mathbb{E}(S_{x,1}^4)=\mathbb{E}(X_{11}^4)\sum_{i=1}^px_i^4+6\mathbb{E}(X_{11}^2)\sum_{1\leq i<j\leq p}x_i^2x_j^2\leq 16K_2^4+12pK_2^2,\end{equation*}

and the term

\begin{equation*}\mathbb{E}\left(\exp\{\alpha s S_{x,1}^2\}\cdot S_{x,1}^6\right)\leq \mathbb{E}\left(\exp\{\alpha s S_{x,1}^2\}\cdot (1+\exp\{\varepsilon S_{x,1}^2\})K\right)<\infty\end{equation*}

for some $K>0$ and $\varepsilon$ with $\varepsilon_0+\varepsilon\leq t$ , where the finiteness comes from (12). Therefore, we can rewrite, for small $0\leq s\leq\varepsilon_0$ ,

\begin{align*}\mathbb{E}\exp\{sS_{x,1}^2\}&=1+s+s^2\cdot c_1+s^3\cdot c_2(s),\\\mathbb{E}\exp\{sS_{y,1}^2\}&=1+s+s^2\cdot c_1^{'}+s^3\cdot c_2^{'}(s),\end{align*}

where $0<c_1,c_1^{'}<\infty$ and $0<c_2(s), c_2^{'}(s)<\infty$ for all $0\leq s\leq \varepsilon_0$ . Now we apply the estimate $\ln(1+x)\leq x-x^2/4$ for all $|x|<1$ and obtain

\begin{align*}\ln \mathbb{E}\exp\{\theta S_{x,1}^2\}&\leq (\theta+\theta^2\cdot c_1+\theta^3\cdot c_2(\theta))-o(\theta),\\\ln \mathbb{E}\exp\{-\theta cS_{y,1}^2\}&\leq(-\theta c+(\theta c)^2\cdot c_1^{'}-(\theta c)^3\cdot c_2^{'}(\theta))-o(\theta) \end{align*}

for small $\theta$ . Hence, for small enough $\theta$ , uniform in x and y,

\begin{align*}\ln \left(\mathbb{E}\exp\{\theta S_{x,1}^2\}\mathbb{E}\exp\{-\theta cS_{y,1}^2\}\right)\leq \theta(1-c)+o(\theta).\end{align*}

Taking this inequality back to (11), we get, for small $\theta\leq \epsilon$ ,

\begin{align*}I_{p,0}(c)&\geq 1/2\,\,\cdot\inf_{x,y\in \mathbb{R}^p,||x||=||y||=1,x\cdot y=0} \, \sup_{\theta\leq \epsilon}\left[-\ln \left(\mathbb{E}\exp\{\theta S_{x,1}^2\}\mathbb{E}\exp\{-\theta cS_{y,1}^2\}\right)\right]\\&\geq \sup_{\theta\leq \epsilon}[\theta(c-1)+o(\theta)]>0.\end{align*}

Step 3 For $c>1$ , we prove that $I_{\infty,0}(c)>0$ . The proof will be identical to the one in Step 2 if we one can show, instead of using (12), that $\mathbb{E}\exp\{t S_{x,1}^2\}\leq C<\infty$ for some $t>0$ and some constant $C>0$ which is independent of p and x. To this end, we use the moment-generating function upper bound in the definition of a sub-Gaussian random variable:

(13) \begin{align}\mathbb{E}\exp\{t X_{11}\}\leq \exp\{t^2K_4^2\} \qquad \text{for all }t\in\mathbb{R}.\end{align}

Now we apply (13) and the fact that $\mathbb{E}\exp\{t Z\}=\exp\{t^2/2\}$ for all $t\in\mathbb{R}$ , where $Z\sim N(0,1)$ , which is independent of the entries, and derive, for any $0\leq t\leq 1/(8K_4^2)$ , that

\begin{align*}\mathbb{E}\exp\{t S_{x,1}^2\} & = \mathbb{E}\big(\mathbb{E}\big(\exp\big\{tS_{x,1}^2\big\} \mid S_{x,1}\big)\big)=\mathbb{E}\big(\mathbb{E}\big(\mathbb{E}\big(\exp\big\{\sqrt{2t}S_{x,1}Z\big\} \mid S_{x,1}\big) \mid S_{x,1}\big)\big) \\&=\mathbb{E}\big(\mathbb{E}\big(\exp\big\{\sqrt{2t}S_{x,1}Z\big\} \mid S_{x,1}\big)\big)=\mathbb{E}\exp\big\{\sqrt{2t}S_{x,1}Z\big\} \\ & =\mathbb{E}\big(\mathbb{E}\big(\exp\big\{\sqrt{2t}S_{x,1}Z\big\} \mid Z\big)\big)\\&=\mathbb{E}\left(\mathbb{E}\left(\prod_{i=1}^p\mathbb{E}\exp\big\{\sqrt{2t}x_iX_{i1}Z\big\} \mid Z\right)\right)\leq \mathbb{E}\left(\prod_{i=1}^p\exp\big\{2tx_i^2Z^2K_4^2\big\}\right)\\&=\mathbb{E}\exp\big\{2tZ^2K_4^2\big\}=\big(1-4tK_4^2\big)^{-1/2}\leq \sqrt{2}.\end{align*}

This completes the proof.

7. Open problems and future work

In this paper we have only considered condition numbers for some special rectangular random matrices (i.e. $p<n$ with $p=o(n)$ ). It would be interesting and challenging to study large-deviation probabilities of condition numbers for square random matrices with $p=n$ (and also for almost square random matrices with $n=p+b$ for b fixed or $b=o(n)$ ). The main difficulty lies in the fact that the various estimates used throughout the paper become imprecise when $p=n$ (or $n=p+b$ ).

Another rectangular case we have not touched on here is when $p/n\rightarrow \kappa\in (0,1)$ , and in this case $k^2(p,n)\rightarrow(1+\kappa^{1/2})^2/(1-\kappa^{1/2})^2$ in probability. With this law of large numbers it is also natural to study the corresponding large-deviation asymptotics of k(p,n) for large n.

The last open problem is about a universality result (mentioned in (4)): under what assumptions on the distribution of the entries does $I_{\infty,0}(c)=2^{-1}\ln\left[(c+1)^2/(4c)\right]$ ? If the assumptions can be specified, then the large-deviation asymptotics of condition numbers for all such distributions coincide with those for Wishart matrices.