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:

$$P(Y_n P(Y_n=y) = n(1- e^{\lambda y})^{n-1}$$

But I get stuck with $$U_n$$. I tried an MGF, $$U_n$$ evaluates nicely but then $$Y_n$$ gets messy. Any thoughts?

• What did you try to compute the MGF of $Y_n$?
– Did
Commented May 21, 2016 at 14:14
• I just wrote the definition down and then didn't recognize the integral as anything I could do. Commented May 21, 2016 at 14:18
• And... can we see these tries?
– Did
Commented May 21, 2016 at 14:21
• @Did MGF $U_n$ => $E[e^{tu}] = E[e^{t\sum{X_i/i}}] = \prod_i E[e^{t*x_i/i}] = \prod_{i} \frac{\lambda}{\lambda-t/i}$ MGF $Y_n$ = $\int n e^{ty}(1-e^{\lambda y})^{n-1}$ Commented May 21, 2016 at 15:08
• math.stackexchange.com/q/364691/321264 Commented Jan 10, 2021 at 17:35

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.

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