Operators on the vanishing moments subspace and Stieltjes classes for M-indeterminate probability distributions

Purpose – In this work the author gathers several methods and techniques to construct systematically Stieltjes classes for densities defined on R þ . Design/methodology/approach – The author uses complex integration to obtain integrable functions with vanishingmomentssequence,andthentheauthorconsiderssomeoperatorsdefinedonthevanishingmomentssubspace. Findings – TheauthorgatherseveralmethodsandtechniquestoconstructsystematicallyStieltjesclassesfor densitiesdefinedon R þ .TheauthorconstructsexplicitlyStieltjesclasseswithcenteratwell-knownprobability densities. The author gives a lot of examples, including old cases and new ones. Originality/value – The author computes the Hilbert transform of powers of j ln x j to construct Stieltjes classes by using a recent result connecting the Krein condition and the Hilbert transform.


Introduction
Consider the subspace M of all functions f ∈ L 1 ðR þ Þ with finite moment sequence, i.e. Z ∞ 0 x n jf ðxÞjdx < ∞ for all n ∈ N 0 ¼ f0; 1; . . .g: The vanishing moments subspace M 0 is given as follows We also consider the subspace M of all functions f ∈ L 1 ðR þ Þ with strong finite moment sequence, i.e. Z ∞ 0 x n jf ðxÞjdx < ∞ for all n ∈ Z ¼ f0; ±1; ±2; . . .g; and the corresponding strong vanishing moments subspace M 0 .

Mindeterminate probability distributions
First, we introduce a method to get functions in M 0 or M 0 : assume that g is an analytic function on a region containing the sector S α ¼ fz ∈ C * : 0 ≤ arg z ≤ παg; 0 < α < 1; C * ¼ Cn f0g: We use complex integration to show that provided that g satisfies suitable conditions on the boundary vS α of S α . As usual, Rz; Iz denote the real and imaginary parts of z ∈ C.
Then we introduce some operators mapping the subspace M 0 into itself. For instance, we prove that M 0 is invariant under the operator g ↦ μ * g; provided that μ is a positive bounded measure on ðR þ ; B þ Þ with finite moments sequence, see (18). Here μ * g is the convolution of the measure μ and the function g on R þ given by μ * gðxÞ ¼ Z x 0 gðx À sÞdμðsÞ; x > 0: Suppose now that f is a probability density function (we use further just density) of a random variable X such that all moments are finite, i.e., m k dE½X k ¼ R ∞ 0 x k f ðxÞdx < ∞ for all k ∈ N 0 , hence m 0 5 1. This means that f ∈ M. It is well-known that the moment sequence fm k g ∞ k¼1 either determines X and f uniquely, and we say that X, and also f, is M-determinate, or that f is M-indeterminate. In the latter case there are infinitely many continuous and infinitely many discrete distributions all sharing the same moments as X ∼ f. This is a fundamental qualitative result, see [1,2].
In the survey [3] the author revisited recent developments on the checkable moment-(in) determinacy criteria including Cram er's condition, Carleman's condition, Hardy's condition, Krein's condition and the growth rate of moments. In this survey the author analyzes Hamburger and Stieltjes cases.
In this work we only focus in the Stieltjes case, i.e we consider distributions supported on R þ . Recall that in [4] was introduced the concept of Stieltjes class for M-indeterminate absolutely continuous distribution function. Let f be a density in M. Assume that there exists a function h ∈ L ∞ ðR þ Þ such that khk ∞ 5 1, fh ∈ M 0 and fh is not identically zero. Then the Stieltjes class S(f, h) with center at f and perturbation h is given by Clearly, S(f, h) is a family of densities all having the same moment sequence as f.
If X ∼ f is M-determinate, then the perturbation h 5 0, and the Stieltjes class consists of a single element, the center f.
The main aim of this work is to find perturbations for Stieltjes classes with center at a density f > 0. To do this, the basic idea is take a function g ∈ M 0 such that h 5 g/f is bounded on R þ , therefore h will be a perturbation (up to scaling by a constant) for a Stieltjes class with center at f. Thus, in this paper all the densities f are M-indeterminate.
When X ∼ f with a density f in M, we make the obvious changes to define the strong Stieltjes class with center at f.
In [5,Theorem 1.2] the author proved that if f is a density in M satisfying the Krein condition AJMS then Sðf ; sinðH e ln f ÞÞ is a Stieltjes class, where H e ln f is the Hilbert transform of u 5 ln f: In particular, here we compute H e ðjln xj m Þ, m ∈ N, to obtain new Stieltjes classes corresponding to M-indeterminate generalized log-normal random variables. In order to test our approach we apply the developed methods to the generalized gamma (GG) distribution (see Examples 4,8,11 and 12), powers of the generalized inverse gaussian (GIG) distribution (see Examples 3, 14 and 16), powers of the half-logistic distribution (see Example 7) and to the generalized lognormal (GLN) distribution (see Examples 5,6,19 and 21).
This work is organized as follows. In Section 2 we give the precise conditions on g to prove (1), and we apply this result to get functions in M 0 . In Section 3 we introduce some operators defined on M 0 and we use the functions obtained in Section 2 to get new perturbations in Examples 5,6,8,11,19 and 21, hence we give new Stieltjes classes. In the last section we compute H e ðjln xj m Þ, m ∈ N ¼ f1; 2; . . .g.

Functions with vanishing moments
In this section we use complex integration to obtain functions in M 0 or M 0 . We follow the technique introduced in [6], also in [7], where a similar result appears. In fact, in [6] the author asks the condition g(x) ≥ A exp(Àax α ) for some A > 0, a > 0 and some α ∈ (0, 1/2), which is replaced with our integrability conditions (4), (5) and (6) below.
Let S ⊂ C, then hol(S) denotes the space of analytic functions on a region containing S. Lemma 1. Let 0 < α < 1, γ ∈ C. Suppose that g ∈ hol(S α ) satisfies the following conditions Then Proof. We pick 0 < « < A < ∞, Cauchy's theorem implies that I where the contour C «,A consists of the real axis from « to A, the arc of the circle z 5 Ae it from t 5 0 to t 5 πα, the straight line from Ae iπα to «e iπα and the arc of the circle z 5 «e i(παÀt) from t 5 0 to t 5 πα. Thus, Mindeterminate probability distributions Since jz λ j ¼ jzj Rλ expð−arg z IλÞ for all λ ∈ C, z ∈ C * , conditions (5) and (6) imply that lim A→∞ I 2 ¼ lim ε→0 þ I 4 ¼ 0. Therefore from condition (4) we get lim and the result follows.
Proof. By setting γ 5 (n þ 1)/α for all n ∈ N 0 or n ∈ Z, t 5 x α , t À1 dt 5 αx À1 dx in (7) and taking the imaginary part, the result follows. , We recall an inequality that will be useful to get our estimates: since e x ≥ x for all x > 0 we have e −x ≤ s s x −s for all x; s > 0: (9) Throughout this work the constant K will be a normalizing constant to produce a density function in each case.
Recall that b$c is the floor function and ⌈$⌉ is the ceiling function. For x; y ∈ R, m ∈ N, we have ðÀ1Þ j π 2j ðln xÞ m−2j ; To see this, consider g(z) 5 exp(Àρz(Log z) m ) for any ρ > 0, here Log z stands for the principal branch of the logarithm function. For n ∈ N 0 we write Clearly I 2 < ∞, and (9) implies that Cx n ðρα m x α Þ s dx < ∞ for s > 0 big enough:

Mindeterminate probability distributions
Clearly I 1 < ∞ when m is even. Assume that m is odd, thus Hence g satisfies condition (8) for all n ∈ N 0 , m ∈ N.
Since the real part is an additive function, we have for A > e and t ∈ [0, πα] that for some constant C > 0. Therefore, On the other hand, there is Consider g(z) 5 z β exp(Àρ(Log z) 2m ) for any β ∈ R, ρ > 0. We make the change of variable As in (15) and using (13) we can see that for all n ∈ Z. The function g also satisfies condition (6) for μ 5 (n þ 1)/α, n ∈ Z, we just set A 5 1/« and apply the last case.

Operators on the vanishing moment subspace
For m ∈ N, s > 0 we introduce the operator T m,s as follows T m;s gðxÞ ¼ x 1=m−1 gðx 1=m À sÞχ ðs m ;∞Þ ðxÞ: The binomial formula implies that T m; The case m 5 1 was considered in [8, Lemma 1].
If g 1 ; . . . g m ∈ M 0 and a 1 ; . . . a m ∈ R then P i a i g i ∈ M 0 , hence the following result is a generalization of the last observation.

AJMS
Clearly G is a measurable function and satisfies for all n ∈ N 0 . Since g ∈ M 0 we have that Gð$; sÞ ∈ M 0 for all s > 0, and the last result implies that R ∞ 0 Gð$; sÞds ∈ M 0 . On the other hand, we have gðx À sÞχ ð0;xÞ ðsÞdμðsÞ ¼ μ * gðxÞ; x > 0: We apply the last result to obtain new Stieltjes classes with center at f(x) 5 K exp(Àx α ), x > 0, as follows.
Since (x/e) α ≤ x/e for all x ≥ e, there exist a constant 0 < C < 1 such that x α À x ≤ ÀCx for all x ≥ e, thus and we proceed as before to construct the corresponding Stieltjes class. Now, let p be a polynomial with real coefficients, with p(0) 5 0 and p 0 > 0 on R þ . We introduce the operator R p as follows where p À1 is the inverse function of p on R þ . As in (16), a change of variable and the binomial formula implies that R p M 0 ⊂ M 0 .
Example 14. Let 0 < c 1 , c 2 < 1/2, b 1 , b 2 > 0, a ∈ R be fixed and 1 ≤ n < 2 −1 ðc −1 1 ∧c −1 2 Þ, n ∈ N. From (10) we have that we proceed as in Example 12 and use Remark 13 to get that Once again, we obtain new perturbations for strong Stieltjes classes with center at As a consequence we obtain a Stieltjes class with center at a generalized inverse Gaussian density.

Krein criterion and the Hilbert transform
Þ , x > 0, is a perturbation for the Stieltjes class with center at f. This is the case n 5 1 in Example 14.
Finally, in the last examples we get two Stieltjes classes that we could not obtain by the method of complex integration given in Section 2. The densities involved are special cases of generalized log-normal densities, see [9]. In order to construct these examples we need to find out the Hilbert transform of j ln xj n , x > 0, n ∈ N. Thus, we need to compute the principal value of the singular integral in (3) with u 5 j ln xj n , n ∈ N.
For all k; n ∈ N 0 we have the identity, see [10, p.
We introduce the following constants where ζ(z) is the zeta function. First we compute the Hilbert transform of even powers of j ln xj.