Problem about partial sum of exponential random variable Let $X_1, X_2, \dots,X_n, X_{n+1}$ be independent random variable of exponential distribution, and the mean is 1.
Let $S_i = X_1 + \dots + X_i$
I want to know $\mathbb{E}\left[\max_{k=1}^n\left(\frac{S_k}{S_{n+1}} - \mathbb{E}\left(\frac{S_k}{S_{n+1}}\right)\right)\right]$. Approximated solution is OK.
Here, $\left(\frac{S_1}{S_{n+1}}, \frac{S_2}{S_{n+1}}, \dots, \frac{S_n}{S_{n+1}}\right) $ has the same distribution with the order statistics of $n$ uniform $U(0,1)$ random variables. Please see Didier's answer to Question about order statistics.
 A: By exchangeability, $\mathrm E\left(\frac{S_k}{S_n}\right)=\frac{k}n$ for every $0\leqslant k\leqslant n$. 
Heuristics based on the functional central limit theorem suggest that, for every $0\leqslant t\leqslant 1$,  when $k\approx tn$, $S_k\approx nt+\sqrt{n}\cdot B_t$ for a standard Brownian motion $(B_t)_{0\leqslant t\leqslant 1}$. Thus, 
$$
\frac{S_k}{S_n}-\mathrm E\left(\frac{S_k}{S_n}\right)\approx\frac{nt+\sqrt{n}\cdot B_t}{n+\sqrt{n}\cdot B_1}-t\approx\frac1{\sqrt{n}}(B_t-tB_1).
$$
The process $(B_t-tB_1)_{0\leqslant t\leqslant 1}$ is a standard Brownian bridge. 
This suggests that the expectation $\mathfrak e_n$ you are asking for behaves like $\mathrm E(M)/\sqrt{n}$, where
$$
M=\max\limits_{0\leqslant t\leqslant 1}(B_t-tB_1).
$$
Finally, since $\mathrm P(M\geqslant x)=\mathrm e^{-2x^2}$ for every $x\geqslant0$, $\mathrm E(M)=\sqrt{\pi/8}$ and all this would yield the asymptotics
$$
\lim\limits_{n\to\infty}\sqrt{n}\cdot\mathfrak e_n=\sqrt{\pi/8}=0.626657\ldots
$$
Edit The Brownian bridge is distributed like $(B_t)_{0\leqslant t\leqslant 1}$ conditioned on $B_1=0$, hence, for every $x\gt0$,
$$
\mathrm P(M\geqslant x)=\mathrm P(M_1\geqslant x\mid B_1=0),\qquad
M_1=\max\limits_{0\leqslant t\leqslant 1}B_t.
$$ 
Informally, intyroducing $T_x=\inf\{t\geqslant0\mid B_t\geqslant x\}$,
$[M_1\geqslant x]=[T_x\leqslant1]$, hence
$$
\mathrm P(M_1\geqslant x\mid B_1=0)\approx\left.\frac{Q(\mathrm dz)}{\mathrm P(B_1\in\mathrm dz)}\right|_{z=0}
$$
where
$$
Q(\mathrm dz)=\mathrm P(T_x\leqslant1,B_1\in\mathrm dz)=\mathrm P(T_x\leqslant1)\mathrm P(B_1\in\mathrm dz\mid T_x\leqslant1).
$$
By the reflection principle, for $z\lt x$,
$$
\mathrm P(B_1\in\mathrm dz\mid T_x\leqslant1)=\mathrm P(B_1\in\mathrm 2x-dz\mid T_x\leqslant1),
$$ 
hence $Q(\mathrm dz)=\mathrm P(B_1\in\mathrm 2x-dz, T_x\leqslant1)=\mathrm P(B_1\in\mathrm 2x-dz)$. Introducing the density $\varphi$ of the distribution of $B_1$ and applying this to $z=0$, this yields $\mathrm P(M\geqslant x)=\varphi(2x)/\varphi(0)$, which is the formula used above.
A: Let $U_{k:n}$ be $k$-th order statistics in a uniform sample of size $n$. $U_{k:n}$ is equal in distribution to a beta random variables with parameters $\alpha = k$ and $\beta=n+1-k$, so that
$$
    \mathbb{E}\left( \frac{S_k}{S_{n+1}} \right) = \mathbb{E}(U_{k:n}) = \frac{k}{n+1}
$$
Thus we need to find the expectation of random variable $V_n = \max_{k=1}^n \left(U_{k:n} - \frac{k}{n+1} \right)$. Clearly $ \mathbb{P}\left(-\frac{1}{n+1} \leqslant V_n \leqslant \frac{n}{n+1} \right) = 1$.
Let $-\frac{1}{n+1} <z<\frac{n}{n+1}$, and consider
$$ \begin{eqnarray}
   \mathbb{P}(V_n \leqslant z) &=& \mathbb{P}\left(\land_{k=1}^n  \left(U_{k:n} \leqslant z+\frac{k}{n+1}\right)\right) = F_{U_{1:n}, \ldots, U_{n:n}}\left(z + \frac{1}{n+1},\ldots,z + \frac{n}{n+1} \right) \\
   &=& \sum_{\begin{array}{c} s_1 \leqslant s_2 \leqslant \cdots \leqslant s_n \\ s_1+s_2 + \cdots+s_n+s_{n+1} = n \\ s_i \geqslant i \end{array}} \binom{n}{s_1,s_2,\ldots,s_{n+1}}\prod_{k=1}^{n+1} \left(F_U(z_k) - F_U(z_{k-1})\right)^{s_k}
\end{eqnarray}
$$
where $z_i = z + \frac{i}{n+1}$ and $z_0 = 0$ and $z_{n+1} = 1$. This gives an exact law for variable $V_n$.
Computation of the mean can be done as 
$$
   \mathbb{E}(V_n) = -\frac{1}{n+1} + \int\limits_{-\frac{1}{n+1}}^{\frac{n}{n+1}} \left( 1- \mathbb{P}(V_n \leqslant z) \right) \mathrm{d} z = \frac{n}{n+1} - \int\limits_{-\frac{1}{n+1}}^{\frac{n}{n+1}} \mathbb{P}(V_n \leqslant z)  \mathrm{d} z
$$
With this it is easy to evaluate moments for low values of $n$:
In[76]:= Table[n/(n + 1) - 
  Integrate[(
      CDF[OrderDistribution[{UniformDistribution[], n}, Range[n]], 
       z + Range[n]/(n + 1)]) // Simplify, {z, -1/(n+1), n/(
    n+1)}], {n, 1, 10}]

Out[76]= {0, 8/81, 129/1024, 2104/15625, 38275/279936, 784356/5764801, 
 18009033/134217728, 459423728/3486784401, 12913657911/100000000000, 
 396907517500/3138428376721}

Denominators of the expectation $\mathbb{E}(V_n)$ equal to $(n+1)^{n+2}$. 
Exact computations of $\mathbb{E}(V_n)$ for $n$ much higher than $n=10$ is difficult due to bad complexity. 
However, simulation is always at hand. 
