3

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?

Rivers McForge
  • 5,787
yoshi
  • 3,509

2 Answers2

3

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.

Pasriv
  • 1,062
0

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

Henry
  • 157,058