Let $\ X_1,X_2,...\ $ be i.i.d. exponential random variables. For $\ n=1,2,...\ $consider the random variables $\ Y_n= \max\{X_1,...,X_n\} \ $ and $\ U_n=\sum_{i=1}^{n}X_i/i \ $. Show that for each n, the random variables $\ Y_n \ $and $\ U_n \ $ have the same distribution function.
I know that the sum of exponentials is a Gamma(n,$\lambda$) distribution, but the indices in the denominator here are making it difficult. In addition, I know the distribution function of $\ Y_n\ $is $\ (1-\exp^{-\lambda x})^n\ $. So I just need to show the distribution of $\ U_n $ is the same.
You are right that the distribution of $\ Y_n= \max\{X_1,...,X_n\} \ $ goes something like this
$$P(Y_n \le x) = F_1(x).F_2(x).... F_n(x) = \left(F_i(x)\right)^n = (1-e^{-\lambda x})^n$$
$$ f_{Y_n}(x) = n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\tag 1$$
Now, let $\frac{X_1}{1} = Z_1$, $\frac{X_2}{2} = Z_2$, ... $\frac{X_n}{n} = Z_n$
$$P(X_1 \le x) = P(Z_1\le x) = (1-e^{-\lambda x})$$
$$f_{Z_1}(x) = \lambda.e^{-\lambda x}$$
Similarly
$$P(X_2 \le 2x) = P(Z_2\le x) = (1-e^{-2\lambda x})$$
$$f_{Z_2}(x) = 2\lambda.e^{-2\lambda x}$$
And thus
$$P(X_n \le nx) = P(Z_n\le x) = (1-e^{-n\lambda x})$$
$$f_{Z_n}(x) = n\lambda.e^{-n\lambda x}$$
So $Z_1$ ~ $Exp(\lambda)$, $Z_2$ ~ $Exp(2\lambda)$, .., $Z_n$ ~ $Exp(n\lambda)$
Refer to the article below to find the distribution of sum of exponentials with different parameters to get $U_n = \sum \frac{X_i}{i} = Z_1+Z_2+\cdots+Z_n$
https://people.maths.bris.ac.uk/~mb13434/sumexp.pdf
$$f_{Z_1+Z_2+\cdots+Z_n}(x) = \left(\prod_{i=1}^{n} \lambda_i\right)\sum_{j=1}^{n}\dfrac{e^{-\lambda_j x}}{\prod_{k\ne j k=1}^{n}(\lambda_k-\lambda_j)}\tag 3$$
$$=n\lambda\left( e^{-\lambda x} - (n-1).e^{-2\lambda x}+ \frac{(n-1)(n-2)}{2}e^{-3\lambda x} + \cdots + {(n-1)\choose(n-1)}e^{-n\lambda x}\right)$$
$$=n\lambda e^{-\lambda x}\left( 1- (n-1).e^{-\lambda x}+ \frac{(n-1)(n-2)}{2}e^{-2\lambda x} + \cdots + {(n-1)\choose(n-1)}e^{-(n-1)\lambda x}\right)$$
$$f_{U_n}(X) = f_{Z_1+Z_2+\cdots+Z_n}(x)= n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\tag 2$$
Hence we have proved that $Y_n$ and $U_n$ have the same distribution. To illustrate the first few terms of $U_n$ in (3),
j = 1,
$$=(1.2.3...n)\lambda^n \dfrac{e^{-\lambda x}}{(2\lambda - 1\lambda) (3\lambda - 1\lambda)(4\lambda - 1\lambda)..(n\lambda - 1\lambda)}$$ $$=(1.2.3...n)\lambda^n \dfrac{e^{-\lambda x}}{1.2.3.4..(n-1)\lambda^{n-1}}$$
$$ = n\lambda.e^{-\lambda x}$$
j = 2
$$=(1.2.3...n)\lambda^n \dfrac{e^{-2\lambda x}}{(1\lambda - 2\lambda) (3\lambda - 2\lambda)(4\lambda - 2\lambda)..(n\lambda - 2\lambda)}$$ $$=(1.2.3...n)\lambda^n \dfrac{e^{-2\lambda x}}{-1.1.2.3.4..(n-2)\lambda^{n-1}}$$
$$ = -n(n-1)\lambda.e^{-2\lambda x}$$
j = 3
$$=(1.2.3...n)\lambda^n \dfrac{e^{-3\lambda x}}{(1\lambda - 3\lambda) (2\lambda - 3\lambda)(4\lambda - 3\lambda)..(n\lambda - 3\lambda)}$$ $$=(1.2.3...n)\lambda^n \dfrac{e^{-3\lambda x}}{-2.-1.1.2.3.4..(n-3)\lambda^{n-1}}$$
$$ = \frac{n(n-1)(n-2)}{2}\lambda.e^{-3\lambda x}$$