Can we give any upper bound on $E[\max_{n \leq N} X_n]$ in terms of $\max_{n \leq N} E[X_n]$ Consider a sequence $\{X_n\}$ of $N$ random variables. Can we give any upper bound on $E[\max_{n \leq N} X_n]$ in terms of $\max_{n \leq N} E[X_n]$. I think in general it is not possible.
If $\{X_n\}$ is a zero mean martingale can we say something. 
 A: Indeed you can't get an upper bound, not even for martingales of bounded support. 
Doob's $L^{p}$ inequality gives that if $(Y_{n})$ is a martingale, then $E\max_{m\leq n} |Y_{m}|^{p} \leq  q E(|Y_{n}|^{p})$ for $p>1$ and $q$ such that $p^{-1}+q^{-1}=1$, where $|\cdot|$ denotes absolute value. 
However, this last conclusion is $\textit{not}$ true if $p=1$.  To see this, let $(S_{n})$ be the SRW (simple random walk) with $S_{0}= 1$. (That is, $S_{n}$ is the sum of $n$ independent random variables that take values $\{-1, 1\}$ with probability 1/2). Considering the stopped process $X_{n} = S_{\min\{n,K\}}$ for $K= \inf\{n: S_{n} = 0\}$, we have that $E(\max_{m\leq n} X_{m}) \rightarrow +\infty$ as $n \rightarrow \infty$. (To check this remember that the probability that a SRW started at the origin hits $b$ before $-a$ is $\frac{a}{a+b}$ to calculate the tail of the distribution of the $\max X_{n}$ and show that their sum corresponds to the harmonic series, so that $E(\max X_{n}) = \infty$, and the claim follows by the monotone convergence theorem). However, $\max E(X_{n})=1$, as we have that $E(X_{n}) = ES_{0} = 1$, for all $n$ (this is a property of stopped martingales, see for instance Williams (1991), Probability with Martingales, section 10.9). Hope this helps.
