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.

  • $\begingroup$ I presume you are interested in the large $n$ regime ? $\endgroup$
    – Sasha
    Dec 6 '11 at 21:26
  • $\begingroup$ @Sasha yes, I'll edit to include that $\endgroup$ Dec 6 '11 at 21:38
  • $\begingroup$ You might be interested in this related question: Does exceptionalism persist as sample size gets large? $\endgroup$ Dec 7 '11 at 5:28
  • $\begingroup$ Note: the answers to this related question on cstheory.stackexchange are useful in answering your question. $\endgroup$
    – Neal Young
    Dec 4 '12 at 23:31
  • $\begingroup$ 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.) $\endgroup$ May 24 '19 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

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

  • 5
    $\begingroup$ 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. $\endgroup$
    – SigmaX
    Nov 2 '14 at 17:15
  • 22
    $\begingroup$ I find it interesting that this doesn't need the independence assumption. $\endgroup$
    – Arun
    Dec 10 '14 at 18:12
  • 1
    $\begingroup$ Can we similarly prove the lower bound? I've trying to use this hint in one of my exercises that $P(Z\geq t) = 1- P(X_1 \leq t)^n$. $\endgroup$ Mar 25 '16 at 15:16
  • 1
    $\begingroup$ 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 $\endgroup$
    – user32849
    Jul 7 '17 at 15:56
  • 4
    $\begingroup$ Here's a proof of a lower bound: gautamkamath.com/writings/gaussian_max.pdf $\endgroup$ Jun 26 '19 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.

  • 4
    $\begingroup$ +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. $\endgroup$ Dec 6 '11 at 22:35
  • 1
    $\begingroup$ Doesn't the inverse cdf have domain $[0,1]$? $\endgroup$ Dec 30 '12 at 7:01
  • 9
    $\begingroup$ (+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). $\endgroup$
    – cardinal
    Dec 30 '12 at 16:19
  • 2
    $\begingroup$ Unless I misunderstood something, the first line of your expression for $m_n$ is negative $\endgroup$
    – Glen_b
    Jan 16 '17 at 2:29
  • 4
    $\begingroup$ 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 $\sigma_n$. $\endgroup$ Jan 9 '18 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.