What's asymptotics of $E(\max_{0I've encounterd the following problem: let $x_i,\,i=1,...,n$ be i.i.d. random variables with distribution $N(0,1)$. I'm interested in the fowllowing random variable $z_n:=\max_{0<j<k<n}(\sum_{t=j}^k x_t)$, namely $z_n$ is the maxim of sum of all intervals.
What can we say about $z_n$, for example asymtotics of $E(z_n)$? And another question is what's the distribution of $(j/n,k/n)$ when $\sum_{t=j}^k x_t$ is reaches maxium and $n\rightarrow\infty$? I guess not evenly distributed, any helps are appreciated.
 A: Denote $S_{n}=\sum_{i=1}^{n}X_{i}$, by Donsker's theorem, the scaled partial sum process converges (under Skorokhod topology) to a standard Brownian motion $S_{\lfloor nt\rfloor}/\sqrt{n}\Rightarrow B_{t}$. Therefore, by continuous mapping theorem
$$\frac{Z_{n}}{\sqrt{n}}=\frac{1}{\sqrt{n}}\max_{1\leq j\leq k\leq n}\sum_{i=j}^{k}X_{i}=\max_{1\leq j\leq k\leq n}\biggl(\frac{1}{\sqrt{n}}\sum_{i=1}^{k}X_{i}-\frac{1}{\sqrt{n}}\sum_{i=1}^{j-1}X_{i}\biggr)\overset{d}{\to} \sup_{0\leq s\leq t\leq 1}(B_{t}-B_{s}).$$
This shows $Z_{n}/\sqrt{n}$ behaves like the range of Brownian motion as $n\to\infty$, whose distribution is known
$$\begin{align*}
\mathbb{P}\Bigl(\sup_{0\leq s\leq t\leq 1}(B_{t}-B_{s})\leq v\Bigr)=\sum_{k=1}^{\infty}&(-1)^{k+1}k\Bigl\{\mathrm{erfc}\Bigl(\frac{(k+1)v}{\sqrt{2}}\Bigr)\\&-2\mathrm{erfc}\Bigl(\frac{kv}{\sqrt{2}}\Bigr)+\mathrm{erfc}\Bigl(\frac{(k-1)v}{\sqrt{2}}\Bigr)\Bigr\}.\label{eq:v}\tag{1}\end{align*}$$
Above convergence doesn't imply convergence of $\mathbb{E}[(Z_{n}/\sqrt{n})^{p}]$, maybe the Gaussianity of $X_{i}$ can give some uniformly integrable conditions that sufficient for the convergence of moments. The assiciated moments are
$$\mathbb{E}\Bigl[\Bigl(\sup_{0\leq s\leq t\leq1}(B_{t}-B_{s})\Bigr)^{p}\Bigr]=\frac{2^{p/2+2}}{\sqrt{\pi}}\Bigl(1-\frac{4}{2^{p}}\Bigr)\Gamma\Bigl(\frac{p+1}{2}\Bigr)\zeta(p-1).$$
When $p=1$ the result is $\sqrt{8/\pi}$ and $p=2$ the result is $4\log 2$.
For $(j/n,k/n)$, they play the role of $(s,t)$ the time interval of $B$ reaches the supremum range on $[0,1]$. The derivation of their distribution is more complicated, since they are not stopping times. However, given a range $v$, the first range time $\tau_{v}:=\inf\{t:\sup_{0\leq s\leq t}B_{s}-\inf_{0\leq s\leq t}B_{s}=v\}$ is a stopping time, with the density given by
$$\mathbb{P}(\tau_{v}\in dt)=\frac{4v}{t\sqrt{2\pi t}}\sum_{k=1}^{\infty}(-1)^{k+1}k^{2}e^{-k^{2}v^{2}/(2t)}dt.\label{eq:t}\tag{2}$$
Conditional on the range $v$ and $W_{\tau_{v}}$, the left point $s$ of the interval is the first hitting time of $W_{\tau_{v}}\mp v$, depending on whether $W_{\tau_{v}}$ is positive. The density of first hitting time $H_{y}:=\inf\{s:W_{s}=y\}$ is
$$\mathbb{P}(H_{y}\in dt)=\frac{|y|}{\sqrt{2\pi}t^{3/2}}e^{-y^{2}/(2t)}dt.\label{eq:s}\tag{3}$$
The joint distribution of $(s,t)$ can be derived using (\ref{eq:v})-(\ref{eq:s}) and
$$\mathbb{P}(\tau_{v}\in dt,W_{\tau_{v}}\in dz)=1_{\{|z|\leq v\}}\frac{\sqrt{2}}{t\sqrt{\pi t}}\sum_{k=-\infty}^{\infty}k\bigl(1-(|z|+2kv)^{2}/t\bigr)e^{-(|z|+2kv)^{2}/(2t)}dtdz.$$
Reference: Borodin, A. N., & Salminen, P. (2015). Handbook of Brownian motion-facts and formulae. Springer Science & Business Media.
