Let $X_1, \dots, X_n$ be i.i.d. $\exp(\lambda)$ and denote the order statistics with $Y_1 \leq \dots \leq Y_n$.
I want to show that the distribution of $ \frac{ \sum_{i = 1}^j Y_i } {\sum_{i=1}^n Y_i}$ is independent of $\lambda$.
I know that $Y_i = \sum_{j=1}^i \frac{Z_j}{n-j+1}$, where the $Z_j's$ are i.i.d. $\exp(\lambda)$.
Best Answer
I am supposing the probability density function of $X_1$ is $$ f_{X_1}(x) = \lambda \exp(-\lambda x)1(x>0). $$ Therefore the distribution of $\lambda X_1$ is an exponential distribution with parameter $1$. Now you can write your quantity of interest as
$$ \frac{ \sum_{i = 1}^j\lambda Y_i } {\sum_{i=1}^n \lambda Y_i} = \frac{ \sum_{i = 1}^j Y_i' } {\sum_{i=1}^n Y_i'}, $$, where $Y_i'$s are the order statistics of $\lambda X_1,\ldots,\lambda X_n$. We already observed that $\lambda X_i$'s are exponential distribution with parameter $1$. (of course it is independent of $\lambda$)