Expectation of maximum of n i.i.d random variables I have $n$ i.i.d. random variables, $X_1,..., X_n$ which follow some arbitrary distribution. Based on experiments in Python with various distributions, it seems that $\mathbb{E}(\max(X_1,...,X_n))$ is a linear (or seemingly close to linear) function of $\mathbb{E}(X_i)$. It is indeed linear for some examples where it is possible to get a closed form solution for $\mathbb{E}(\max(X_1,...,X_n))$ or a good approximation.  
Expected value of $\max\{X_1,\ldots,X_n\}$ where $X_i$ are iid uniform.
Expectation of the maximum of i.i.d. geometric random variables
I wonder if this is the case more generally? Is there some way to prove it?
 A: A general technique to get a bound that is often pretty decent is to use the MGF if you have it: for all $t\geq 0$:
\begin{align}
\exp(t\mathbb{E}[\max_i X_i])&\leq \mathbb{E}[\exp(t\max_i X_i)]\\
&\leq\mathbb{E}[\sum_{i=1}^n \exp(t X_i)] \\
&= n\mathbb{E}[\exp(t X)],
\end{align}
so 
\begin{equation}
\mathbb{E}[\max_i X_i]\leq \frac{\log(n\mathbb{E}[\exp(tX)])}{t}.
\end{equation}
You can then optimize in $t\geq 0$ to get a decent upper bound. For instance, doing this with Gaussians with variance $\sigma^2$ would show $\mathbb{E}[\max_i X_i]\leq \sigma\sqrt{2\log n}$, which turns out to be right up to a constant.
A: This is question is related to something called order statistics in probability theory. You can read more about them here. For $n$ iid variables $X_1, …, X_n$ with cumulative density function $F$ and density function $f$, the density function of the maximum is:
$$f_{max}(x) = nf(x)F(x)^{n-1}$$
Then this implies the expected value would be:
$$E[X_{max}] = \int_{-\infty}^{\infty} nxf(x)F(x)^{n-1} dx$$
I don't see any linear relationship here in general between $E[X_{max}]$ and $E[X]$
A: It seems the idea generalizes. Say $E(max(X_1,X_2,...,X_n))=X_j =E(X_j \geq X_1 ,X_j >X_2,...,X_j>X_n)$ Then(By assumed independence of the $X_i$) Let $f_i$ be the pdf. of $X_i$ :$$ P(X_j \geq X_1 ,X_j >X_2,...,X_j>X_n)= P(X_j >X_1)P(X_j > X_2).....P(X_j >X_n) = (\int_{- \infty}^{x_j} f_idx_i)^n$$ and then $f_x= \frac {d}{dt}(F_x)= $ ( By chain rule) $$(d/dxF(x))n( \int_{-\infty}^{\infty}f_idx_i)^{n-1} $$ , so that the expected value is $$n\int_{-\infty}^{\infty} x \int_{x_j}^{\infty} f_idx_i $$
I don't see how this is linear on $E(X_i):=\int _{-\infty}^{\infty}x_if_idx_i$
