Are $\sum_{i=1}^{n} \xi_i / i$ and $\max\xi_n$ equally distributed? Let the random variables $\xi_1, \xi_2, \ldots, \xi_n$ be independent, have an exponential distribution and expectation of each is 2.
Are $\sum_{i=1}^{n} \xi_i / i$ and $\max\xi_n$ equally distributed?
 A: Claim: the statement is true and may be generalized easily for  any order statistic from exponential distribution with any $\lambda$, not only $\lambda = 2$.
At first, let us note that it's sufficient to consider $\lambda = 1$ since $\frac1{\lambda}$ is a scale parameter.
See proposition 3 p.55 here:1
https://doi.org/10.1214/aoms/1177699058
So, the claim is true and may be showed by direct computation.
Moreover,
$$X_{(j)} = \sum_{i=1}^{j} \frac{\xi_i}{n-i+1}$$  in distribution.
1Chernoff, H.; Gastwirth, Joseph L.; Johns, M. V. jun., Asymptotic distribution of linear combinations of functions of order statistics with applications to estimation, Ann. Math. Stat. 38, 52-72 (1967). ZBL0157.47701.
A: Yes. The argument in this blog post shows that they are equal in mean, but it can be extended to show they are actually equal in distribution.
Let $a(1),\dots,a(n)$ be the (random) ordering of the indices such that $\xi_{a(1)}\leq \xi_{a(2)}\leq \cdots \leq \xi_{a(n)}$. Then
$$\max(\xi_1,\dots,\xi_n)=\xi_{a(n)}=\xi_{a(1)}+(\xi_{a(2)}-\xi_{a(1)})+\cdots+(\xi_{a(n)}-\xi_{a(n-1)}).$$
Since $\xi_{a(1)}$ is the minimum of $n$ independent $\text{Exp}(\lambda)$ random variables, its distribution is $\text{Exp}(n\lambda)$.
Since $\xi_{a(1)}$ is the smallest of the $\xi_i$, we have that $\xi_{a(2)}-\xi_{a(1)}$ is the minimum of $n-1$ independent $\text{Exp}(\lambda)$ random variables, namely
$$\xi_{a(2)}-\xi_{a(1)}=\min_{i \neq a(1)}(\xi_{i}-\xi_{a(1)}).$$
Note that $(\xi_i-\xi_{a(1)})\sim\text{Exp}(\lambda)$ for $i \neq a(1)$ by the memoryless property of the exponential distribution. Thus $\xi_{a(2)}-\xi_{a(1)}$ has the distribution $\text{Exp}((n-1)\lambda)$, and it is independent of $\xi_{a(1)}$ by the memoryless property.
Next we have $\xi_{a(3)}-\xi_{a(2)}$ is the minimum of $n-2$ independent $\text{Exp}(\lambda)$ random variables:
$$\xi_{a(3)}-\xi_{a(2)}=\min_{i \neq a(1),a(2)}(\xi_{i}-\xi_{a(2)})$$
and so has the distribution of $\text{Exp}((n-2)\lambda)$. It is independent of $\xi_{a(1)}$ and $\xi_{a(2)}-\xi_{a(1)}$ by the memoryless property.
Continuing this argument, we have that $\max(\xi_1,\dots,\xi_n)$ is the sum of $n$ independent exponential random variables with rates $n\lambda,(n-1)\lambda,\dots,\lambda$.
On the other hand, $\frac{\xi_i}{i}$ is an exponential random variable with rate $\lambda i$. Since the $\xi_i$ are independent, $\sum_{i=1}^n \frac{\xi_i}{i}$ is the sum of $n$ independent exponential random variables with rates $\lambda, 2\lambda, \dots, n\lambda$. So we conclude that $\max(\xi_1,\dots,\xi_n)$ and $\sum_{i=1}^n\frac{\xi_i}{i}$ are equal in distribution.
