I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (3rd Ed, Exercise 15.1.3):
Let $n \in \mathbb N$ and let $X_1, \ldots, X_n$ be i.i.d. exponentially distributed random variables with parameter $1$. Let $Y_1, \ldots, Y_n$ be independent exponentially distributed random variables with $\mathbf P_{Y_k} = \exp_k$. That is, $$(Y_1, \ldots, Y_n) \stackrel{\mathcal D}{=} (X_1, X_2/2, X_3/3, \ldots, X_n/n)$$ where $\stackrel{\mathcal D}=$ denotes equivalently distributed. Finally, sort the values of $X_i$ by size $X_{(1)} > X_{(2)} > \cdots > X_{(n)}$. Show that $$ \left( X_{(n)}, X_{(n-1)}, \ldots, X_{(1)}\right) \stackrel{\mathcal D}= \left( Y_n, Y_{n-1} + Y_n, \ldots, Y_1 + Y_2 + \cdots + Y_n\right). $$ Hint: First check that $X_{(n)} \stackrel{\mathcal D}= Y_n$. Show that the conditional distribution $\mathcal L\left[\left(X_{(1)} - X_{(n)}, \ldots, X_{(n-1)} - X_{(n)}\right) | X_{(n)}\right]$ does not depend on $X_{(n)}$ and that it equals the (unconditional) distribution of the ordered values of $X_1, \ldots, X_{n-1}$.
Here Distribution of ordered independent exponential random variables there is a theoretical solution.
Since I am not able to finish it, I think I am doing something wrong in the conditional law's density calculation.
Indeed, for the distribution of the ordered values $X_{(1)},...,X_{(n-1)}$ I get, by integrating out the minumim (please, mind that $X_{(1)}$ is the maximum in the notation used by prof. Klenke) from the joint distribution of the ordered statistics, this density:
$f_{X_{(1)},...,X_{(n-1)}}(x_1,...,x_{(n-1)})= n! f_{X_1}(x_1)...f_{X_1}(x_{n-1})F_{X_1}(x_{n-1})=n!e^{-(x_1+...+x_{n-1})}(1-e^{x_{n-1}})$.
By applying the change of variables theorem I get for the conditional density above:
$$\frac{f_{X_{(1)}-x_n,...,X_{(n-1)}-x_n,X_{(n)}}(x_1,...,x_{n-1},x_n)}{f_{X_{(n)}}(x_n)}= \\\quad \frac{f_{X_{(1)},...,X_{(n-1)},X_{(n)}}(x_1+x_n,...,x_{n-1}+x_n,x_n)}{f_{X_{(n)}}(x_n)}= \\\quad \frac{n!f_{X_1}(x_1+x_n)...f_{X_1}(x_{n-1}+x_n)f_{X_1}(x_{n})}{{f_{X_{(n)}}(x_n)}}= \\\quad \frac{n!e^{-(x_1+...+x_{n-1})}e^{-nx_n}}{ne^{-nx_n}}$$
This latter should be equal to the unconditional density that I calculated above. Any suggestions?
Edit
Could it be something like the following equations based on the iid and “memoryless” properties of the sample?
$P[X_{(1)}-X_{(n)}<y_1,…,X_{(n-1)}-X_{(n)}<y_{n-1}| X_{(n)}=x_n]= \\= n! P[X_{1}-X_{n}<y_1,…,X_{n-1}-X_{n}<y_{n-1}| X_{n}=x_n]= \\=n!P[X_{1}<y_1+x_n,…,X_{n-1}<y_{n-1}+x_n| X_{n}=x_n]=\\=n!P[X_1<y_1]…P[X_{n-1}<y_{n-1}]$
Basically I am considering in the $n!$ permutation also the fact that the lowest observed value could come from any observation in the sample. Thus I condition on the (arbitrarily) first observation to be the lowest and no more on the minimum $X_{(n)}$. Before I tried to divide the joint density of the ordered statistics by the density of the minimum. Is my reasoning right?
Thank you.