An identity on order statistics of a set of random variables

Abstract

When an n × 1 random vector X = (X₁, …, X_n)^T has a sign-invariant distribution, Strait [J. Multivariate Anal. 4 (1974) 494–496] proved that the expectations of max(0, X₁, X₁ + X₂, …, X₁ + X_n) and max(0, X₁, …, X_n) are equal. In this note we assume a weaker condition that (X₁, X₂, …, X_n) and (−X₁, X₂, …, X_n) are equal in distribution and prove a more general result that the expectations of L_r(0, X₁, X₁ + X₂, …, X₁ + X_n) and L_r(0, X₁, …, X_n) are equal, where L_r(0, X₁, …, X_n) is the rth order statistic of 0, X₁, …, X_n for r = 1, …, n + 1.

Let X = (X₁, …, X_n)^T be an n × 1 random vector, and let H be an n × n diagonal matrix that belongs to ℋ(n) = {diag(h₁, …, h_n), $h_i^2=1$, i = 1, …, n}. The distribution of X is said to be sign-invariant if $\mathbf{X}\stackrel{d}{=}\mathbf{\text{HX}}$ for every H ∈ ℋ(n), where the notation $\stackrel{d}{=}$ means that both sides of the equality have the same distribution. The properties of the sign-invariant distribution can be found in Berman [1,2], while the problem of estimating its location parameters was investigated by Xu [4]. When X has a sign-invariant distribution with E(X₁) = 0, Strait [3] proved that

$$E[\text{max}(0,X_1,X_1+X_2,\dots ,X_1+X_n)]=E[\text{max}(0,X_1,\dots ,X_n)].$$ (1)

His method appeared to be overly lengthy. Below we assume that

$$(X_1,X_2,\dots ,X_n)\stackrel{\mathrm{d}}{=}(-X_1,X_2,\dots ,X_n)$$ (2)

with E(X₁) = 0. Assumption (2) is clearly weaker than assumption of sign-invariance. Under assumption (2), we prove a more general result,

$$E[L_r(0,X_1,X_1+X_2,\dots ,X_1+X_n)]=E[L_r(0,X_1,\dots ,X_n)],$$ (3)

provided that expectations exist, where L_r(0, X₁, …, X_n) is the rth order statistic of 0, X₁, …, X_n for r = 1, …, n + 1. Clearly, when r = n + 1, (3) becomes Strait’s result (1). To show (3), we use (2) to obtain

$$\left(\begin{array}{c}0\\ X_1\\ X_1+X_2\\ ⋮\\ X_1+X_n\end{array}\right)=\left(\begin{array}{cccccc}0& 0& 0& 0& \cdots & 0\\ 1& 0& 0& 0& \cdots & 0\\ 1& 1& 0& 0& \cdots & 0\\ ⋮& ⋮& ⋮& ⋮& \cdots & ⋮\\ 1& 0& 0& 0& \cdots & 1\end{array}\right)\left(\begin{array}{c}{X}_1\\ X_2\\ X_3\\ ⋮\\ X_n\end{array}\right)\stackrel{\mathrm{d}}{=}\left(\begin{array}{cccccc}0& 0& 0& 0& \cdots & 0\\ 1& 0& 0& 0& \cdots & 0\\ 1& 1& 0& 0& \cdots & 0\\ ⋮& ⋮& ⋮& ⋮& \cdots & ⋮\\ 1& 0& 0& 0& \cdots & 1\end{array}\right)\left(\begin{array}{c}-X_1\\ X_2\\ X_3\\ ⋮\\ X_n\end{array}\right)=\left(\begin{array}{c}0\\ -X_1\\ X_2-X_1\\ ⋮\\ X_n-X_1\end{array}\right).$$ (4)

Let U and V denote the vector of the left-hand side and the right-hand side of (4), respectively. Then, (4) states $\mathbf{U}\stackrel{\mathrm{d}}{=}\mathbf{V}$, which implies $f(\mathbf{U})\stackrel{\mathrm{d}}{=}f(\mathbf{V})$ for any function f. Taking f = L_r will conclude that $L_r(\mathbf{U})\stackrel{\mathrm{d}}{=}{L}_r(\mathbf{V})$. Furthermore, using the property of the order statistics yields that

$$L_r(\mathbf{V})=L_r(0,-X_1,X_2-X_1,\dots ,X_n-X_1)=L_r(X_1-X_1,0-X_1,X_2-X_1,\dots ,X_n-X_1)=L_r(X_1,0,X_2,\dots ,X_n)-X_1=L_r(0,X_1,X_2,\dots ,X_n)-X_1.$$ (5)

Noting that L_r(U) = L_r(0, X₁, X₁ + X₂, …, X₁ + X_n) and using (5) will conclude that $L_r(\mathbf{U})\stackrel{\mathrm{d}}{=}{L}_r(\mathbf{V})$ is equivalent to

$$L_r(0,X_1,X_1+X_2,\dots ,X_1+X_n)\stackrel{\mathrm{d}}{=}{L}_r(0,X_1,X_2,\dots ,X_n)-X_1.$$ (6)

Taking the expectation on both sides of (6) and using the assumption that E(X₁) = 0 will immediately yield (3). It is worth mentioning that it appears to be almost impossible to prove (3) if one wants to employ Strait’s [3] method.