Show that random variables $max(X_1, ..., X_n)$ and $X_1+\dfrac{X_2}{2}+...+\dfrac{X_n}{n}$ have the same distribution, if $X_i \sim Exp(\lambda)$

Question

Let's define $\xi = max(X_1, ..., X_n)$ and $\eta = X_1+\dfrac{X_2}{2}+...+\dfrac{X_n}{n}$. I tried to show that these random variables have same cumulative distribution functions and I got following: $$\phi_{\xi}(t) = \prod\limits_{k=1}^{n}(1-\dfrac{it}{\lambda k})^{-1}$$ $$\phi_{\eta}(t) = \sum\limits_{k=1}^{n}(-1)^{k+1}C_n^k(1-\dfrac{it}{\lambda k})^{-1}$$ Wolfram says these functions are the same function, but I don't know, how to prove it.

How to show that functions are the same?
May be there is another solution of original problem?

EDIT: Random variables $X_i$ are independently distributed

Hello :) Are the $X_i$ indenpendently distributed? Try to compute the distribution explicitely. Let $Y:=\max_iX_i$, then $ F_Y(y)=P(Y \leq y) = P(X_1 \leq y, \ldots, X_n \leq y) = \prod_{i=1}^{n} P(X_i \leq y) = F_{X_1}(y)^n$ — Jochen, Oct 14 '21 at 10:38
@Jochen That leaves finding the convolution for $X_1+\frac{X_2}{2}+\ldots+$\frac{X_{n}}${n}$ — Henry, Oct 14 '21 at 10:54
Previously asked: https://math.stackexchange.com/q/1794095/321264, https://math.stackexchange.com/q/3979952/321264, https://math.stackexchange.com/q/364691/321264 — StubbornAtom, Oct 14 '21 at 14:21
Does this answer your question? Showing 2 Distributions are the Same — Henry, Oct 16 '21 at 10:01

score 0 · Accepted Answer · answered Oct 14 '21 at 10:46

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

Show that random variables $max(X_1, ..., X_n)$ and $X_1+\dfrac{X_2}{2}+...+\dfrac{X_n}{n}$ have the same distribution, if $X_i \sim Exp(\lambda)$

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