Finite sequences representing expected order statistics

Characterizations of finite sequences $\beta_{1}<\cdots<\beta_{n}$ representing expected values of order statistics from a random sample of size $n$ are given. As a by-product, a characterization of binomial mixtures, when the mixing random variable is supported in the open interval $(0,1)$, is presented; this enables the exact description of the convex hull of the open binomial curve, as well as the open moment curve.


Introduction
In the present note we consider the following problem: Given n real numbers under what conditions on β's is there an integrable random variable (r.v.) X such that IEX i:n = β i , 1 ≤ i ≤ n?
[Here, X 1:n ≤ • • • ≤ X n:n are the order statistics of independent, identically distributed r.v.'s X 1 , . . ., X n , each with distribution like X.] Notice that the number n is held fixed; the question for infinite sequences is closely connected to the Hausdorff (1921) moment problem, and its answer is well-known from the works of Huang (1998), Kadane (1971Kadane ( , 1974)), Kolodynski (2000) and Papadatos (2017).Some relative results for the finite case can be found in Mallows (1973).order statistics Without loss of generality we may consider the numbers instead of β i .Clearly these numbers will be the expected order statistics (=EOS) from (X − c)/λ if and only if the β's are the EOS from X. First, we seek for a necessary condition.Assume that X is a non-degenerate random variable with distribution function (d.f.) F and IE|X| < ∞.Let X 1 , . . ., X n be independent, identically distributed (i.i.d.) random variables with d.f.F , and denote by X 1:n ≤ • • • ≤ X n:n the corresponding order statistics.It is known that where µ j:n := IEX j:n , j = 1, . . ., n; this follows by a trivial application of Newton's formula to the expression , where F −1 (u) := inf{x : F (x) ≥ u}, 0 < u < 1, is the left-continuous inverse of F .From (1) with k = 1, 2, (j − 1)µ j:n . ( On the other hand, it is well-known (see Jones and Balakrishnan (2002)) that where α < ω are the endpoints of the support of X; actually this formula goes back to Karl Pearson (1902).Notice that −∞ ≤ α < ω ≤ ∞, α < ω because F is nondegenerate, and the integral in (3) is finite since X is integrable.From ( 1) and (3) (applied to n = 2), yields Choosing the numbers µ j:n = (µ j:n − c)/λ are the EOS from (X − c)/λ.Therefore, by considering (X − c)/λ in place of X, we may further assume that µ 2 − µ 1 = 1.Then, Since F (y)(1 − F (y)) > 0 for y ∈ (α, ω), and zero outside [α, ω), it follows that f Y (y) := F (y)(1 − F (y)) defines a Lebesgue density of a random variable, say Y , supported in the (finite or infinite) interval (α, ω).From (4), the numbers p j := j(n − j)(µ j+1:n − µ j:n )/(n(n − 1)), j = 1, . . ., n − 1, are strictly positive probabilities (summing to 1).Also, the integral in (3) can be rewritten as where T := F (Y ) is a random variable taking values in the interval (0, 1) w.p. 1, because, by definition, IP(α < Y < ω) = 1.Hence, (3) reads as and we have shown the following Proposition 1 If X 1 , . . ., X n are i.i.d.integrable non-degenerate r.v.'s, then there exists an r.v.T , with IP(0 < T < 1) = 1, such that It is of interest to observe that the binomial moments of T appear in the r.h.s. of (5).Clearly, the r.v.T in this representation need not be unique; any other r.v.T ′ with IP(0 < T ′ < 1) = 1, possessing identical moments up to order n − 2 with T , will fulfill the same relationship.
Remark 1 For any integrable non-degenerate r.v.X with d.f.F we may define the r.v.T as in the proof of Proposition 1, that is, It can be shown, using Lemma 4.1 in Papadatos (2001), that the d.f. of T is specified by Notice that λ = 1 0 (2t−1)F −1 (t)dt and, hence, the function F −1 determines uniquely the d.f. of T .Moreover, (6) shows that the entire location-scale family of X, {c+λX : c ∈ R, λ > 0}, is mapped to a single r.v.T ∈ (0, 1).Provided that X has (finite or infinite) interval support, non-vanishing density f and differentiable inverse d.f.F −1 , we conclude from (6) that a density of T is given by Next, we proceed to verify that the preceding procedure can be inverted, showing sufficiency of (5).To this end, we shall make use of the following lemma, which is of independent interest in itself.A detailed proof is postponed to the appendix.
Lemma 1 Let T be an r.v. with d.f.F T such that IP(0 < T < 1) = 1.Then, there exists a unique, non-degenerate, integrable, r.v.X, satisfying where X k:k = max{X 1 , . . ., X k } with X 1 , X 2 , . . .being i.i.d.copies of X.The inverse distribution function of X is given by and I denotes an indicator function.
Remark 3 Suppose that the r.v.T of Lemma 1 is absolutely continuous with density f T .Assume also that the corresponding r.v.X (with IEX = 0, IEX 2:2 = 1, inverse d.f.F −1 0 as in ( 9)) is absolutely continuous, admitting a non-vanishing density f 0 in the (finite or infinite) interval support of X, and that The characterization for finite n reads as follows.
Remark 4 The r.h.s. of (11) corresponds to a Binomial Mixture (of a particular form, since IP(T = 0) = IP(T = 1) = 0).Obviously, the necessary and sufficient condition (12) is always satisfied for n = 2 and n = 3.Hence, the true problem begins at n = 4.

The truncated moment problem for finite open intervals
In this section, we obtain a precise characterization by invoking results from the truncated moment problem for finite intervals.The existing results are limited to compact intervals and are not applicable to our case, since, according to the characterization of Theorem 1, a suitable T lies in the open interval (0, 1) w.p. 1.
According to Theorem 1, the β's are EOS if and only if the ν's fulfill the truncated moment problem in the interval (0, 1).However, for the truncated moment problem, well-known results exist for a compact interval [a, b]; see, e.g., Theorem IV.1.1 of Karlin and Studden (1966) or Theorems 10.1, 10.2 in Schmüdgen (2017).In order to obtain the corresponding necessary and sufficient conditions for open intervals, we shall make use of the following Theorem 2 (Richter-Tchakaloff Theorem; see Schmüdgen (2017), Theorem 1.24).Let (X , F , µ) be a measure space and V a finite dimensional linear subspace of A symmetric n × n matrix A with real entries is positive definite (denoted by A ≻ 0) if x T Ax > 0 for all x ∈ R n \ {0}, where x T denotes the transpose of a column vector x ∈ R n .Similarly, A is positive semi-definite (or nonnegative definite) if x T Ax ≥ 0 for all x ∈ R n , and this is denoted by A 0.
From Theorem 1(iii) it follows that the β's are EOS.
Next, suppose that n = 2m + 3.Then, A 1 = B 1 and it follows that showing that A 1 (and B 1 ) is singular and positive semi-definite.On the other hand, x k /2 k ) 2 ≥ 0, and (ii) shows that the β's are EOS.Hence, although the numbers ν k (β) (k = 0, . . ., n − 2) do not satisfy the condition A 1 ≻ 0 and B 1 ≻ 0, the corresponding β j are EOS.
It can be checked that the given β's are the EOS from the two-valued r.v.X with IP(X = 0) = IP(X = 2 n ) = 1/2.
Remark 5 (a) Assume that for i = 0 or 1, A i 0, B i 0, and either det A i = 0 or det B i = 0 (or both).Then, the measure µ = µ 0 is [0, 1]-determinate from its moments (ν k ) n−2 k=0 ; see Theorem 10.7 in Schmüdgen (2017).Hence, if this is the case, we can find ε ∈ (0, 1/2) such that A i (ε) 0 and B i (ε) 0 if and only if the support of (the unique) µ 0 does not contain any of the endpoints 0 and 1.
(b) The finite supporting set of the discrete measure µ 0 , constructed in the proof of Theorem 3, can be chosen to contain at most k ≤ n/2 (rather than k ≤ n − 1) points.