Showing 2 Distributions are the Same 
Let $X_1, X_2, \dots$ be i.i.d. exponentially distributed RVs. For $n = 1,2,\dots$ consider:
$Y_n := \max(X_1, X_2, \dots, X_n)$
$U_n := \sum_{i=1}^{n}\frac{X_i}{i}$
Show that $Y_n$ and $U_n$ have the same distribution


What I've tried:
$P(Y_n<y)= P(X_i<y)^n = (1- e^{\lambda y})^n => P(Y_n=y) = n(1- e^{\lambda y})^{n-1}$
But I get stuck with $U_n$. I tried an MGF, $U_n$ evaluates nicely but then $Y_n$ gets messy. Any thoughts?
 A: Let $\phi_X(u)$ indicate the characteristic function of the r.v. $X$ in the following. 
Moreover let $\mathbf{X}=[X_1,X_2,...,X_n]$. Let $\mathbf{A}=\{A_j\}$, where $A_j=\frac{n!}{j},$ $1\le j\le n$.
Using independence of the $X_j$ and the scaling property of the characteristic function you have:
$$
\begin{align}
\phi_{U_n}(u)&=\phi_{\sum_{j=1}^n \frac{X_j}{j}}(u)\\
&=\phi_{\frac{\mathbf{A}^T\mathbf{X}}{n!}}(u)\\
&=\phi_{\mathbf{A}^T\mathbf{X}}\left(\frac{u}{n!}\right)\\
&=\phi_{\mathbf{X}}\left(\frac{\mathbf{A}^Tu}{n!}\right)\\
&=\left(\phi_{X_1}\left(\frac{A_1}{n!}u\right)\right)^n\\
&=\phi_{Y_n}\left(u\right),\\
\end{align}$$
and the characteristic function characterizes the probability.
A: Repeating an answer given to a duplicate later question:
There is an intuitive argument that can be made rigorous, using a couple of properties of iid exponential distributions:

*

*the minimum, i.e. first event, of $n$ has an exponential distribution with $n$ times the rate, the same distribution as $\frac{X_n}{n}$


*exponential distributions are memoryless, so if their values exceed $k$ then the distribution of the excess above $k$ is the same exponential distribution as the original
So the minimum has the same distribution as $\frac{X_n}{n}$.  The additional amount to the second lowest is similar but now with $n-1$ remaining variables so the same distribution as $\frac{X_{n-1}}{n-1}$, independently of the distribution of the minimum.  Inductively this continues so the next additional amount has the same distribution as $\frac{X_{n-2}}{n-2}$  and so on up to $\frac{X_{1}}{1}$.
Adding up all the additional amounts gives the distribution of the maximum values as being the same as  $\frac{X_n}{n}+ \frac{X_{n-1}}{n-1}+\frac{X_{n-2}}{n-2}  + \cdots +\frac{X_{1}}{1}$, as required
