# Expectation of the maximum of gaussian random variables

Is there an exact or good approximate expression for the expectation, variance or other moments of the maximum of $n$ independent, identically distributed gaussian random variables where $n$ is large?

If $F$ is the cumulative distribution function for a standard gaussian and $f$ is the probability density function, then the CDF for the maximum is (from the study of order statistics) given by

$$F_{\rm max}(x) = F(x)^n$$

and the PDF is

$$f_{\rm max}(x) = n F(x)^{n-1} f(x)$$

so it's certainly possible to write down integrals which evaluate to the expectation and other moments, but it's not pretty. My intuition tells me that the expectation of the maximum would be proportional to $\log n$, although I don't see how to go about proving this.

• I presume you are interested in the large $n$ regime ? Dec 6, 2011 at 21:26
• @Sasha yes, I'll edit to include that Dec 6, 2011 at 21:38
• You might be interested in this related question: Does exceptionalism persist as sample size gets large? Dec 7, 2011 at 5:28
• Note: the answers to this related question on cstheory.stackexchange are useful in answering your question. Dec 4, 2012 at 23:31
• More generally, the expectation and variance of the range depends on how fat the tail of your distribution is. For the variance, it is $O(n^{-B})$ where $B$ depends on your distribution ($B = 2$ for uniform, $B = 1$ for Gaussian, and $B = 0$ for exponential.) May 24, 2019 at 23:30

How precise an answer are you looking for? Giving (upper) bounds on the maximum of i.i.d Gaussians is easier than precisely characterizing its moments. Here is one way to go about this (another would be to combine a tail bound on Gaussian RVs with a union bound).

Let $X_i$ for $i = 1,\ldots,n$ be i.i.d $\mathcal{N}(0,\sigma^2)$.

Defining, $$Z = [\max_{i} X_i]$$

By Jensen's inequality,

$$\exp \{t\mathbb{E}[ Z] \} \leq \mathbb{E} \exp \{tZ\} = \mathbb{E} \max_i \exp \{tX_i\} \leq \sum_{i = 1}^n \mathbb{E} [\exp \{tX_i\}] = n \exp \{t^2 \sigma^2/2 \}$$

where the last equality follows from the definition of the Gaussian moment generating function (a bound for sub-Gaussian random variables also follows by this same argument).

Rewriting this,

$$\mathbb{E}[Z] \leq \frac{\log n}{t} + \frac{t \sigma^2}{2}$$

Now, set $t = \frac{\sqrt{2 \log n}}{\sigma}$ to get

$$\mathbb{E}[Z] \leq \sigma \sqrt{ 2 \log n}$$

• The reason Sivaraman set t = \sqrt{2\log{n}}/\sigma is because that is the point at which the upper bound is at a minimum. You can see this by taking the derivative of the bound with respect to t and setting it to zero. Nov 2, 2014 at 17:15
• I find it interesting that this doesn't need the independence assumption.
– Arun
Dec 10, 2014 at 18:12
• Can we similarly prove the lower bound? I've trying to use this hint in one of my exercises that $P(Z\geq t) = 1- P(X_1 \leq t)^n$. Mar 25, 2016 at 15:16
• This uses the Cramer-Chernoff method. For completeness and reference, the proof provided above appears as a special case in Pascal Massart: "Concentration inequalities and model selection", p. 17f, link.springer.com/10.1007/978-3-540-48503-2 Jul 7, 2017 at 15:56
• Here's a proof of a lower bound: gautamkamath.com/writings/gaussian_max.pdf Jun 26, 2019 at 2:10

The $$\max$$-central limit theorem (Fisher-Tippet-Gnedenko theorem) can be used to provide a decent approximation when $$n$$ is large. See this example at reference page for extreme value distribution in Mathematica.

The $$\max$$-central limit theorem states that $$F_\max(x) = \left(\Phi(x)\right)^n \approx F_{\text{EV}}\left(\frac{x-\mu_n}{\sigma_n}\right)$$, where $$F_{EV} = \exp(-\exp(-x))$$ is the cumulative distribution function for the extreme value distribution, and $$\mu_n = \Phi^{-1}\left(1-\frac{1}{n} \right) \qquad \qquad \sigma_n = \Phi^{-1}\left(1-\frac{1}{n} \cdot \mathrm{e}^{-1}\right)- \Phi^{-1}\left(1-\frac{1}{n} \right)$$ Here $$\Phi^{-1}(q)$$ denotes the inverse cdf of the standard normal distribution.

The mean of the maximum of the size $$n$$ normal sample, for large $$n$$, is well approximated by $$\begin{eqnarray} m_n &=& \sqrt{2} \left((\gamma -1) \Phi^{-1}\left(2-\frac{2}{n}\right)-\gamma \Phi^{-1}\left(2-\frac{2}{e n}\right)\right) \\ &=& \sqrt{\log \left(\frac{n^2}{2 \pi \log \left(\frac{n^2}{2\pi} \right)}\right)} \cdot \left(1 + \frac{\gamma}{\log (n)} + \mathcal{o} \left(\frac{1}{\log (n)} \right) \right) \end{eqnarray}$$ where $$\gamma$$ is the Euler-Mascheroni constant.

• +1. See also Section 10.5 ("The Asymptotic Distribution of the Extreme") in David and Nagaraja's Order Statistics. They explicitly discuss the normal distribution on page 302. Dec 6, 2011 at 22:35
• Doesn't the inverse cdf have domain $[0,1]$? Dec 30, 2012 at 7:01
• (+1) Two comments: (1) The somewhat nonstandard use of $Q$ for the inverse normal is a little unfortunate given that it is a standard notation in some contexts for the upper-tail distribution of the standard normal $\mathbb P(Z \geq z)$. I would suggest $\Phi^{-1}$ instead. (2) As you know, convergence in distribution doesn't imply convergence of moments, in general; but, in the case of extreme values of iid random variables it does (curiously enough). This was proved in Pickands (1968). Dec 30, 2012 at 16:19
• Unless I misunderstood something, the first line of your expression for $m_n$ is negative Jan 16, 2017 at 2:29
• The first expression for $m_n$ should be $(1-\gamma)*\Phi^{-1}(1-1/n) + \gamma\Phi^{-1}(1-1/(en))$, which is the mean of the extreme value distribution with the given parameters $\mu_n$ and $\sigma_n$. Jan 9, 2018 at 16:46

Here is a good asymptotics for this:

Let $$f(x)=\frac{e^{-x^2/2}}{\sqrt{2\pi}}$$ be the density and $$\Phi(x)=\int_{-\infty}^x f(u)du$$ be the CDF of the standard normal random variable $$X$$. Let $$Z=Z(n)=\max(X_1,\dots,X_n)$$ where $$X_i$$ are i.i.d.\ normal $${\cal N}(0,1)$$. We are interested in $$\mathbb{E}Z(n)$$.

Since the CDF of $$Z$$ is given by $$F_Z(z)=\mathbb{P}(Z\le z)=\mathbb{P}(X_1\le z,\dots,X_n\le z)=\mathbb{P}(X\le z)^n=\Phi(z)^n$$, hence its density is $$\frac{d}{dz}\left[\Phi(z)^n\right]=n f(z)\Phi(z)^{n-1}$$ and \begin{align*} \mathbb{E}Z(n)&=\int_{-\infty}^\infty z\cdot n f(z)\Phi(z)^{n-1} dz =\int_{-\infty}^0 z\cdot \frac{d}{dz}\left[\Phi(z)^n\right] dz -\int_0^{\infty} z\cdot \frac{d}{dz}\left[1-\Phi(z)^n\right] dz \\ &=-\int_{-\infty}^0 \Phi(z)^n dz +\int_0^{\infty} 1-\Phi(z)^n dz \end{align*} using the integration parts and noting that $$\lim_{z\to-\infty}z\cdot \Phi(z)^n=0$$ and $$\lim_{z\to\infty}z\cdot \left[1-\Phi(z)^n\right]=0$$ by e.g.\ L'Hôpital's rule. Next, \begin{align*} 0\le \int_{-\infty}^0 \Phi(z)^n dz \le \int_{-\infty}^{-1} e^{zn} dz +\int_{-1}^{0} \Phi(0)^n dz = \frac{1}{n}e^{-n} + 2^{-n}\to 0 \end{align*} since $$f(z) (and hence $$\Phi(z) too) for $$x\le -1$$, and $$\Phi(z)\le \Phi(0)=1/2$$ for $$z\le 0$$.

Now it only remains to estimate $$\int_0^{\infty} 1-\Phi(z)^n dz$$. Observe that $$\frac{1-\Phi(z)}{f(z)/z}\to 1\text{ as }z\to+\infty.$$ since for $$z> 0$$ \begin{align*} 1-\Phi(z)&=\int_z^\infty \frac{e^{-x^2/2}}{\sqrt{2\pi}}dx \le\int_z^\infty \frac{x}{z} \cdot \frac{ e^{-x^2/2}}{\sqrt{2\pi}}dx =\frac 1z\cdot \frac{ e^{-z^2/2}}{\sqrt{2\pi}}; \\ 1-\Phi(z) &\ge \int_z^{z+1} \frac{e^{-x^2/2}}{\sqrt{2\pi}}dx \ge \int_z^{z+1} \frac{x}{z+1} \cdot \frac{ e^{-x^2/2}}{\sqrt{2\pi}}dx = \frac {1-e^{-1/2-z}}{1+1/z}\cdot \frac{ e^{-z^2/2}}{z\sqrt{2\pi}}, \end{align*} so we can write $$1-\Phi(z)=c_z\frac{ e^{-z^2/2}}{z\sqrt{2\pi}}\quad\text{for }z>0$$ where $$c_z\to 1$$ as $$z\to +\infty$$.

Let $$y=y(n)=\sqrt{2\ln n}$$. For $$z\ge y$$ we have %$$1-\Phi(z)\ll \frac 1n$$, \begin{align*} \int_{y}^{\infty} 1-\Phi(z)^n dz&= \int_{y}^{\infty} 1-\left[1- c_z\frac{ e^{-z^2/2}}{z\sqrt{2\pi}}\right]^n dz \le \int_{y}^{\infty} n c_z\frac{ e^{-z^2/2}}{z\sqrt{2\pi}} dz \\ &\le \int_{y}^{\infty} n c_z \frac{z}{y^2} \frac{e^{-z^2/2}}{\sqrt{2\pi}} dz \le \frac{n\max_{z\ge y} c_z}{y^2\sqrt{2\pi}} \int_{y}^{\infty} z e^{-z^2/2} dz =O\left(\frac1{\ln n}\right). \end{align*}

Let us estimate now $$\int_{0}^{y} \Phi(z)^n dz$$. Fix $$\varepsilon>0$$ small, and note that if $$1\le z\le y-\varepsilon$$ then $$1-\Phi(z)\ge \left(\min_{1\le z\le y} c_z\right) \frac{n^{-1}\, e^{\varepsilon y-\frac{\varepsilon^2}2}}{(y-\varepsilon)\sqrt{2\pi}} = \frac 1n\, \frac{c_1}{1-o(1)} \frac{e^{\varepsilon y}}{y}\gg \frac 1n$$ for some $$c_1(\varepsilon)>0$$ and where $$o(1)\to 0$$ as $$n\to\infty$$; note also that $$e^{\varepsilon y}/y\to\infty$$ as $$y\to\infty$$. This yields $$\Phi(z)^n\le \exp\left(-(c_1+o(1))e^{\varepsilon y}/y\right)$$. Hence \begin{align*} \int_0^{y-\varepsilon}\Phi(z)^n dz&\le \int_0^1 \Phi(1)^n dz+ \int_1^{y-\varepsilon} \Phi(z)^n dz \le 0.85^n+\int_1^{y-\varepsilon} \exp\left(-(c_1+o(1))e^{\varepsilon y}/y\right) dz \\ & \le 0.85^n+ y\exp\left(-(c_1+o(1))e^{\varepsilon y}/y\right) \to 0\qquad\text{ as }n,y\to\infty. \end{align*} Consequently, since $$\int_{y-\varepsilon}^y \Phi(z)^n dz\le \varepsilon$$, we conclude $$\limsup_{n\to\infty} |\mathbb{E}Z(n)-y(n)|\le \varepsilon$$ Since $$\varepsilon>0$$ is arbitrary, this means $$\mathbb{E}Z(n)=\sqrt{2\ln n}+o(1).$$