Given a set of random variables $x_1,x_2,...,x_n$, and we know their means and variances $(\mu_1,\sigma_1),(\mu_2,\sigma_2),...,(\mu_n,\sigma_n)$. How to compute mean and variance of the maximum distribution of $x_1,x_2,...,x_n$? We can assume $x_1,x_2,...,x_n$ are gaussian distributions.

$\begingroup$ Maybe this is helpful: ocw.mit.edu/courses/civilandenvironmentalengineering/… $\endgroup$ – Markus Sprecher Apr 28 '18 at 8:55

$\begingroup$ See stats.stackexchange.com/questions/18433/… and references therein, such as Wikipedia. Also again stats.SE for the Gaussian case. $\endgroup$ – Tobias Fritz Apr 28 '18 at 12:20

1$\begingroup$ @TobiasFritz  those references are for the case of identically distributed Gaussian variables; the OP asks for the nonidentical case, which seems quite a bit more complicated. $\endgroup$ – Carlo Beenakker Apr 28 '18 at 12:22

$\begingroup$ @CarloBeenakker: right, I wasn't paying proper attention. Thanks. $\endgroup$ – Tobias Fritz Apr 28 '18 at 12:28

2$\begingroup$ Do you just want an answer and don't care about having an elegant solution? Then do it by stochastic (Monte Carlo) simulation. You can incorporate whatever dependencies or distributions you want. $\endgroup$ – Mark L. Stone Apr 28 '18 at 12:36
For independently distributed $x_i$'s, each with cumulative distribution $$F_i(x_i)=\tfrac{1}{2}+\tfrac{1}{2}\operatorname{Erf}\,[(x_i\mu_i)/(\sigma_i\sqrt 2],$$ the cumulative distribution of the maximum is given by $$P(\max_i \,x_i<X_{\max})=\prod_{i=1}^n P(x_i<X_{\max})=\prod_{i=1}^n F_i(X_{\max}).$$ For small $n$ you can now calculate moments of $X_{\rm max}$ by integration, $$E(X_{\max}^p)=\int_{\infty}^\infty x^p\frac{d}{dx}\left(\prod_{i=1}^n F_i(x)\right)\,dx.$$ There is unlikely to be a closedform answer for arbitrary $n$, in fact, even the $n=2$ integral seems problematic (Mathematica fails to evaluate it). If you take the $\mu_i$'s and $\sigma_i$'s to be the same, then progress can be made, for $n=2$ I find $$E(X_{\max})=\mu+\sigma/\sqrt\pi,\;\operatorname{Var}(X_{\max})=(11/\pi) \sigma^2.$$
Perhaps you are satisfied with a large$n$ approximation. For identical $\mu_i$'s and $\sigma_i$'s it is given by the Fisher–Tippett–Gnedenko theorem, see for example this MSE posting. I have found one paper that generalizes this to arbitrary $\mu_i$'s and $\sigma_i$'s: On the distribution of the maximum of n independent normal random variables: iid and inid cases, but I have difficulty parsing their result (a rescaled Gumbel distribution).
_{There is more in that reference that I do not understand. They give the inequality $$\frac{1}{n}\sum_i\mu_i\leq E(X_{\rm max})\leq \frac{1}{n}\sum_i\mu_i+\frac{n1}{n} \max_i\,\mu_i$$ which contradicts the $n=2$ result given above.}

$\begingroup$ I changed several instances of {\rm max} to \max. That affects spacing: with the latter the amount of space to the right and left depends on the context without any manual adjustments, and you formerly saw ${\rm max}_i$ rather than $\displaystyle\max_i$ near the bottom line. $\endgroup$ – Michael Hardy May 1 '18 at 18:44