Let $X = \{X_1, ..., X_k\}$ be a set of $k$ iid variables drawn from a binomial distribution: $X_i \sim B(n, p)$. How to calculate the upper bound of the expected value of $max(X_i)$?
Several related question (such as: Bounds for the maximum of binomial random variables or Maximum of Binomial Random Variables) give such estimates for cases when $n = k$. I am, however, interested in the general case.
$\newcommand\P{\mathbb{P}}\newcommand\E{\mathbb{E}}\newcommand\ol{\overline}$Write $\ol X_{\max} = \max\{\ol X_i\} = \max\{\tfrac{1}{n} X_i\}$. We can compute $$ \P(\ol X_{\max} > p + t) = \P(\ol X_i > p + t \text{ for some } i=1,\ldots,k) \leq k\P(\ol X_1 > p + t) \leq k\,e^{-2nt^2} $$ by the union bound and Hoeffding's inequality. Denoting $(x)_+ = \max\{x,0\}$, this implies $$ \E \ol X_{\max} \leq p + \E(\ol X_{\max} - p)_+ = p + \int_0^\infty \P(X_{\max} > p + t)dt \leq p + k\sqrt{\frac{\pi}{8n}} $$ In terms of the original variables $X_{\max} = n\ol X_{\max}$, we have $$ \E X_{\max} \leq np + k\sqrt{\frac{n\pi}{8}} \leq np + \tfrac{2}{3}k\sqrt{n}. $$ Note that this is much weaker in terms of $k$ than the asymptotically correct bound $$ \E X_{\max} \asymp np + \sqrt{2p(1-p)} \sqrt{n\log k} $$ based on a normal approximation and the fact that if $Z_i\sim\mathcal{N}(0,1)$, $\E Z_{\max} \asymp \sqrt{2\log k}$, but it does give the right dependence on $n$.
