# Ratios of order statistics of Pareto distributed random variables

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.

• This isn't a direct answer to your query but it may be useful nonetheless. Embrechts, et al's, 1996 book Modelling Extremal Events remains, even after 20 years, the single best text reference for questions like yours. – DJohnson Dec 4 '17 at 13:30

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.