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.
 A: 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
$$\mathbb{E}[Z] \leq \sigma \sqrt{ 2 \log n} $$ 
A: 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.
A: 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 $f(z)<e^z$ (and hence $\Phi(z)<e^{z}$ 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
$$
\mathbb{E}Z(n)=\sqrt{2\ln n}+o(1).
$$
