Given a random vector

\begin{equation} x=(x_1, \ldots, x_n) \end{equation}

with independent and identically distributed entries $x_i \sim \mathcal{N}(0,\sigma^2)$, I would like to find a lower bound $f(n)$

\begin{equation} \mathbb{E}[||x||^2_{\infty}] \geq f(n) \end{equation}

which is reasonably tight. I know that the following equality for the non squared norm holds when $\sigma^2 =1$:

\begin{equation} E(\|x||_\infty)=\int_0^\infty(1-(2\Phi(x)-1)^n)dx, \end{equation}

where $\Phi$ is the CDF of $\mathcal{N}(0,1)$, see the comment to this question by @Did here. Unfortunately I am not even sure on how to (tightly) lower bound the right integral for this special case.

Any help on solving the general case is much appreciated.


Let $Z_i:=x_i$ and $M:=M_n:=\|x\|_\infty=\max_1^n|Z_i|$. By rescaling, without loss of generality $\sigma=1$. So, for real $u>0$ \begin{multline} P(M^2>u)=P(M>\sqrt u)=1-P(\max_1^n|Z_i|\le\sqrt u)=1-P(|Z_1|\le\sqrt u)^n \\ =1-(1-2G(\sqrt u))^n=1-e^{-ng(u)}, \tag{1} \end{multline} where $$G(x):=P(Z_1>x)\sim\frac1{x\sqrt{2\pi}}e^{-x^2/2} $$ as $x\to\infty$ and $$g(u):=-\ln(1-2G(\sqrt u))\sim2G(\sqrt u)\sim\frac2{\sqrt{2\pi u}}e^{-u/2} =e^{-u/(2+o(1))} $$ as $u\to\infty$.

Also, $g(u)$ decreases from $\infty$ to $0$ as $u$ increases from $0$ to $\infty$. So, for each natural $n\ge3$ there are unique positive real numbers $u_n$ and $v_n$ such that $$ng(u_n)=\ln n,\quad ng(v_n)=1. $$ Clearly, $0<u_n<v_n<\infty$. Also, $$\frac{\ln n}n=g(u_n)=e^{-u_n/(2+o(1))}\quad\text{and}\quad \frac{\ln n}n=e^{-(1+o(1))\ln n}, $$ whence $$u_n\sim2\ln n\quad\text{and, similarly,}\quad v_n\sim2\ln n. $$

Next, \begin{equation} EM^2=\int_0^\infty P(M^2>u)\,du=\int_0^\infty (1-e^{-ng(u)})\,du=I_1+I_2+I_3, \tag{2} \end{equation} where $$I_1:=\int_0^{u_n}(1-e^{-ng(u)})\,du,\quad I_2:=\int_{u_n}^{v_n}(1-e^{-ng(u)})\,du,\quad I_3:=\int_{v_n}^\infty (1-e^{-ng(u)})\,du. $$ If $0<u<u_n$, then $0<e^{-ng(u)}<e^{-ng(u_n)}=1/n$. So, $$I_1\sim u_n.$$ Next, $$I_2\le v_n-u_n=o(u_n), $$ $$I_3<\int_{v_n}^\infty ng(u)\,du\sim \int_{v_n}^\infty n\frac2{\sqrt{2\pi u}}e^{-u/2}\,du \sim 2n\frac2{\sqrt{2\pi v_n}}e^{-v_n/2} \sim 2ng(v_n)=2=o(u_n). $$ We conclude that, for $\sigma=1$, $$E\|x\|_\infty^2=EM^2\sim u_n\sim2\ln n. $$ So, for any real $\sigma>0$, $$E\|x\|_\infty^2\sim2\sigma^2\ln n. $$

Along the same lines, one can give an explicit non-asymptotic lower bound on $E\|x\|_\infty^2$ which will be asymptotically equivalent to $E\|x\|_\infty^2$ as $n\to\infty$. Indeed, one can use the inequality $$G(t)\ge B(t):=\frac{f(t)}{\sqrt{t^2+2}} $$ for real $t\ge0$, where $f$ is the standard normal pdf, so that $f(t)=\frac1{\sqrt{2\pi}}\,e^{-t^2/2}$ for real $t$. The latter lower bound on $G(t)$ is a simpler, even if a bit less accurate, version of Birnbaum's lower bound $B_1(t):=f(t)(\sqrt{t^2+4}-t)/2$ on $G(t)$; we have $B_1(t)>B(t)$ for all real $t\ge0$. So, assuming that $\sigma=1$ and $n\ge16$, letting \begin{equation*} w_n:=2\ln n-2\ln\ln n, \end{equation*} and recalling (1), for $u\in[0,w_n]$ we have $\ln\ln n>1$ and \begin{align*} P(M^2>u)&=1-(1-2G(\sqrt u))^n \\ &\ge1-\exp\big\{-2nG(\sqrt u)\big\} \\ &\ge1-\exp\big\{-2nG(\sqrt w_n)\big\} \\ &\ge1-\exp\Big\{-2n\frac{f(\sqrt w_n)}{\sqrt{w_n+2}}\Big\} \\ &=1-\exp\Big\{-\frac1{\sqrt\pi}\frac{\ln n}{\sqrt{1+\ln n-\ln\ln n}}\Big\} \\ &\ge1-\delta_n, \end{align*} where \begin{equation*} \delta_n:=\exp\Big\{-\sqrt{\frac{\ln n}\pi}\,\Big\}\to0. \end{equation*} Now it follows from (2) that, for $\sigma=1$, \begin{equation*} E\|x\|_\infty^2=EM^2\ge\int_0^{w_n} P(M^2>u)\,du \ge(1-\delta_n)w_n=(1-\delta_n)(1-\epsilon_n)2\ln n, \end{equation*} where \begin{equation*} \epsilon_n:=\frac{\ln\ln n}{\ln n}\to0. \end{equation*} So, for any real $\sigma>0$ and any $n\ge16$, $$E\|x\|_\infty^2\ge2(1-\delta_n)(1-\epsilon_n)\sigma^2\ln n\sim2\sigma^2\ln n, $$ as claimed.

For $n=1,\dots,15$, the values of $E\|x\|_\infty^2$ can be easily computed numerically, using (2) and (1), with any degree of accuracy, say to get these $15$ approximate values for $E\|x\|_\infty^2/\sigma^2$: $1, 1.63662, 2.10266, 2.47021, 2.77375, 3.03236, 3.25771, 3.45743, 3.6368, 3.79962, 3.9487, 4.08621, 4.21382, 4.33288, 4.44447$.

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  • $\begingroup$ Can you add a few sentences about the structure of the argument here? It’s not easy to dive in to all the formulas. $\endgroup$ – Matt F. Nov 8 '19 at 18:38
  • $\begingroup$ @MattF. : I am not sure what you mean. This whole argument is a straightforward calculation, consisting of 10 sentences, if I counted them correctly. If anything in particular is unclear or hard to comprehend, please let me know. $\endgroup$ – Iosif Pinelis Nov 8 '19 at 20:47
  • $\begingroup$ I have added an explicit non-asymptotic lower bound on $E\|x\|_\infty^2$ which is asymptotically equivalent to $E\|x\|_\infty^2$ as $n\to\infty$. $\endgroup$ – Iosif Pinelis Nov 10 '19 at 2:40

It is known that the max of i.i.d. subgaussian random variables with variance $\sigma^2$ is on the order of $\sigma \sqrt{\log n}$ so you can expect the squared max of the random variables to be roughly $\sigma^2 \log n$. A reference for this result is 'High Dimensional Probability' by Vershynin. In this case, Jensen's inequality immediately gives you that $$\mathbf{E}[||x||_{\infty}^2] \ge (\mathbf{E}[||x||_{\infty}])^2 = \Omega(\sigma^2 \log n).$$

We can also show that this is the right order of magnitude. Consider an arbitrary $\lambda > 0$ ($\lambda$ will have to satisfy a condition that we will address later). Then

\begin{align*} \exp(\lambda \mathbf{E}(\max_i |x_i|)^2) &\le \mathbf{E}\exp(\lambda (\max_i |x_i|)^2) \\ &= \mathbf{E} \max_i \exp(\lambda |x_i|^2) \\ &\le \sum_{i=1}^n \mathbf{E}\exp(\lambda x_i^2). \end{align*} Now the last quantity is just the MGF of a chi-squared distribution which has an explicit form which is $1/(\sqrt{1-2\lambda \sigma^2})$. Then by taking the logs we have $$ \mathbf{E}(\max_i |x_i|)^2 \le \frac{\log n}{\lambda} + \frac{1}{\lambda \sqrt{1-2\lambda \sigma^2}}. $$ Optimizing this quantity in $\lambda$, we let $\lambda$ be such that $$\lambda \sigma^2 = \frac{1}2 - \frac{1}{2 \log(n)^2}.$$ (Note in the conference of the MGF above, we needed $\lambda \sigma^2 < 1/2$ which is satisfied here). Then plugging back in, we see that $$ \mathbf{E}(\max_i |x_i|)^2 \le \frac{4 \sigma^2 \log(n)}{1-1/\log(n)^2}$$ so $\sigma^2 \log n$ is the right order.

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Here's a general way how to obtain lower bounds in this case. The infinity norm $||\vec{X}||_\infty $ is defined on $\mathbb{R}^n$ as

$$ ||\vec{X}||_\infty =\max(|X_1|, ..., |X_n|)$$

Consequently, we also have

$$ ||\vec{X}||_\infty^2 = \max(X_1^2, ..., X_n^2)$$

as $X_i \leq X_j \implies X_i^2 \leq X_j^2$. Now, the maximum of several i.i.d variables, also known as the largest order statistic, has the distribution

$$ \max(X_1^2, ..., X_n^2) \sim \frac{d}{dx} F(x)^{n} = n \left(F(x)\right)^{n-1} f(x) $$

where $f(x)$ is the distribution of $X_i^2$ and $F(x)$ is its CDF. Finally, we therefore have

$$ \mathbb{E}[||\vec{X}||_\infty^2] = n \int_{-\infty}^{\infty} x F(x)^{n-1} f(x) \ dx = \int_{-\infty}^{\infty} x \frac{d}{dx} F(x)^{n} \ dx$$

Use the substitution $u = F(x)$, and we obtain

$$ \mathbb{E}[||\vec{X}||_\infty^2] = n \int_0^1 F^{-1}(u) u^{n-1} du $$

Now, for the random variables $X_i^2$ specifically, we have

$$ f(x) = \frac{1}{\sqrt{2 \pi x}\ \sigma} e^{- \frac{x}{2 \sigma^2}} \theta(x), \;\;\; F(x) = \text{erf}\left(\frac{\sqrt{x}}{\sqrt{2} \sigma}\right) $$

and therefore

$$F^{-1}(u) = 2 \sigma^2 \text{erf}^{-1}(u)^2 $$ .

Now, remarkably in this case, the function $\text{erf}^{-1}(u)$ has a Taylor series around $u = 0$ where all of the coefficients are positive. This means that any truncation of the series is strictly less than $\text{erf}^{-1}(u)$.

The first few terms are

$$ \text{erf}^{-1}(u) = \frac{\sqrt{\pi}}{2} u + \frac{\pi^{\frac{3}{2}}}{24} u^3 + O(u^5)$$

Therefore, if you replace $\text{erf}^{-1}(u)$ with the truncated series in the expression for $F^{-1}(u)$ and integrate the resulting polynomial expression, you obtain a lower bound.

As an example, take the lowest order term in the above expansion. We then must have

$$ \mathbb{E}[||\vec{X}||_\infty^2] \geq n \int_0^1 \sigma^2 \frac{\pi}{2} u^{n+1} du = \sigma^2 \frac{\pi}{2} \frac{n}{n+2}$$

As you take more terms in the Taylor expansion and evaluate the integral, you will obtain increasingly tighter lower bounds.

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  • 1
    $\begingroup$ Each of your lower bounds will be asymptotically constant for large $n$, whereas the correct asymptotics is $\sim2\sigma^2\ln n$. $\endgroup$ – Iosif Pinelis Nov 8 '19 at 17:16
  • $\begingroup$ @IosifPinelis That's right for a fixed number of terms. If you want to reproduce the asymptotics, I suspect you'd also have to make the number of terms you include in the approximation a function of $n$. Admittedly, I wouldn't know how to connect this up at this point, and I'm also unsure which provides a better bound for various regions of the value of $n$. $\endgroup$ – bursneh Nov 8 '19 at 19:15
  • $\begingroup$ Ah, as it happens, the Taylor series approach misses the fact that $\text{erf}^{-1}(u)$ has a logarithmic singularity at $u = 1$. This is likely the source of the logarithmic dependence on $n$. I'll edit the answer to include that when I have more time. $\endgroup$ – bursneh Nov 8 '19 at 20:30

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