# What's asymptotics of $E(\max_{0<j<k\le n}(\sum_{t=j}^k x_t))$ for $x_i\,\sim\,N(0,1)$

I'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 $$z_n:=\max_{0, namely $$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.

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 associated 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 $$\mathbb{P}(\tau_{v}\in dt,W_{\tau_{v}}\in dz)=1_{\{|z|\leq v\}}\frac{\sqrt{2}}{t\sqrt{\pi t}}\sum_{k=-\infty}^{\infty}k\bigl(1-(|z|+2kv)^{2}/t\bigr)e^{-(|z|+2kv)^{2}/(2t)}dtdz.$$