What is the variance of the maximum of a sample? I'm looking for bounds on the variance of the maximum of a set of random variables. In other words, I'm looking for closed-form formulas for $B$, such that
$$
\mbox{Var}(\max_i X_i) \leq B \enspace,
$$
where $X = \{ X_1, \ldots, X_M \}$ is a fixed set of $M$ random variables with finite means $\mu_1, \ldots, \mu_M$ and variances $\sigma_1^2, \ldots, \sigma_M^2$.
I can deduce that
$$
\mbox{Var}(\max_i X_i) \leq \sum_i \sigma_i^2 \enspace,
$$
but this bound seems very loose. A numerical test seems to indicate that $B = \max_i \sigma_i^2$ might be a possibility, but I have not been able to prove this. Any help is appreciated.
 A: A question on MathOverflow is related to this question.
For IID random variables, the $k$th highest is called an order statistic. 
Even for IID Bernoulli random variables, the variance of any order statistic other than the median can be greater than the variance of the population. For example, if $X_i$ is $1$ with probability $1/10$ and $0$ with probability $9/10$ and $M=10$, then the maximum is $1$ with probability $\approx 1- 1/e$, so the variance of the population is $0.09$ while the variance of the maximum is about $0.23$. 
Here are two papers on the variances of order statistics:
Yang, H. (1982) "On the variances of median and some other order statistics." Bull. Inst. Math. Acad. Sinica, 10(2) pp. 197-204
Papadatos, N. (1995) "Maximum variance of order statistics." Ann. Inst. Statist. Math., 47(1) pp. 185-193
I believe the upper bound on the variance of the maximum in the second paper is $M\sigma^2$. They point out that equality can't occur, but any lower value can occur for IID Bernoulli random variables.
A: For any $n$ random variables $X_i$ , the best general bound is 
$\newcommand{\Var}{\mathrm{Var}}\Var(\max X_i) \le \sum_i  \Var(X_i)$ as stated in the original question.
Here is a proof sketch: If X,Y are IID then $E[(X-Y)^2] =2\Var(X)$.  Given a vector of possibly dependent variables $(X_1,\ldots ,X_n)$, let $(Y_1,\ldots ,Y_n)$ be an independent vector with the same joint distribution. For any $r>0$, we have by the union bound that $P[ |\max_i X_i-\max_i Y_i|^2 >r] \le \sum_i P[ | X_i-Y_i|^2 >r]$, and integrating this $dr$ from $0$ to $\infty$ yields the claimed inequality.
If $X_i$ are IID indicators of events of probability $\epsilon$,
then $\max X_i$ is an indicator of an event of probability $n\epsilon+O(n^2 \epsilon^2)$.  Fixing $n$ and letting $\epsilon$ tend to zero, we get $\Var(X_i)=\epsilon-\epsilon^2$ and $\Var(\max_i X_i)= n\epsilon +O(n^2\epsilon^2)$.
