Ratios of order statistics of Pareto distributed random variables

Question

Let $X_1, X_2, \dots$ be i.i.d. with $X_i \sim 1-1/x$. It's well known that $$ \mathbb{P} \left[ \frac{X}{t} \leq a \: \middle| \: X \geq t \right] = \mathbb{P}\left[X \leq a \right]. $$ Let $X_{1,n} \leq X_{2,n} \leq \cdots \leq X_{n,n}$ be the order statistics of the first $n$ random variables $X_i$. Is it true that for any $n \geq 2$, $1 \leq k \leq n$ $$ \sum \limits _{i=0}^{k-1}\frac{X_{n-i,n}}{X_{n-k,n}} =_D \sum \limits_{i=1}^kX_i $$ where $=_D$ denotes equal in distribution.

This would seem to be the case, heuristically, since $X_{n-i,n}$ are "conditioned" to be at least $X_{n-k,n}$, however, I'm not sure how to show this rigorously. Note that I'm not interested in the order statiscis $X_{n-i,n}$ themselves, only the sum.

Answer 1

I fear that the property is false in its additive form, but is true in a muliplicative form which has the same intuitive derivation, namely $$ \prod_{i=0}^{k-1} \frac{X_{n-i,n}}{X_{n-k,n}} =_{D} \prod_{i=1}^{k} X_i. $$ This is a consequence of Rényi's representation which gives the joint distribution of the order statistics for the standard exponential. Indeed, if $X_i$ is a sample of the standard Pareto, the r.vs $Y_i:= \log X_i$ form a sample of the standard exponential. It is known that the spacing $Z_{n-i} := Y_{n-i,n} -Y_{n-i-1,n}$ has an exponential distribution with mean $1/(i + 1)$ for $i=0$ to $i=n$, where $Y_{0,n}:=0$. Moreover the r.vs $Z_{n-i}$ are independent. Then $$ \sum_{i=0}^{k-1} [Y_{n-i,n} -Y_{n-k,n}] = \sum_{i=0}^{k-1} \sum_{j=i}^{k-1} [Y_{n-j,n} -Y_{n-j-1,n}] = \sum_{i=0}^{k-1} \sum_{j=i}^{k-1} Z_{n-j} = \sum_{j=0}^{k-1} (j+1) Z_{n-j}. $$ But the $k$ r.vs $(j+1) Z_{n-j}$ for $j=0$ to $k-1$ form a sample of size $k$ of the standard exponential, as do the $k$ r.vs $Y_j$ for $j=1$ to $k$. Hence $$ \sum_{i=0}^{k-1} [Y_{n-i,n} -Y_{n-k,n}] =_D \sum_{j=1}^{k} Y_j. $$ The result follows by taking the exponential.

See the book by Embrechts et al. cited by @djohnson, example 4.1.5.