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

Marcos López-García (Instituto de Matemáticas, Unidad Cuernavaca, Universidad Nacional Autonoma de Mexico, Cuernavaca, Mexico)

Arab Journal of Mathematical Sciences

ISSN: 1319-5166

Article publication date: 6 August 2021

Issue publication date: 29 June 2022

362

Abstract

Purpose

In this work the author gathers several methods and techniques to construct systematically Stieltjes classes for densities defined on R+.

Design/methodology/approach

The author uses complex integration to obtain integrable functions with vanishing moments sequence, and then the author considers some operators defined on the vanishing moments subspace.

Findings

The author gather several methods and techniques to construct systematically Stieltjes classes for densities defined on R+. The author constructs explicitly Stieltjes classes with center at well-known probability 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 |lnx| to construct Stieltjes classes by using a recent result connecting the Krein condition and the Hilbert transform.

Keywords

Citation

López-García, M. (2022), "Operators on the vanishing moments subspace and Stieltjes classes for M-indeterminate probability distributions", Arab Journal of Mathematical Sciences, Vol. 28 No. 2, pp. 229-242. https://doi.org/10.1108/AJMS-04-2021-0083

Publisher

:

Emerald Publishing Limited

Copyright © 2021, Marcos López-García

License

Published in Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) licence. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this licence may be seen at http://creativecommons.org/licences/by/4.0/legalcode


1. Introduction

Consider the subspace M of all functions fL1(R+) with finite moment sequence, i.e.

0xn|f(x)|dx<for all nN0={0,1,}.

The vanishing moments subspace M0 is given as follows

M0=fM:0xnf(x)dx=0for all nN0.

We also consider the subspace M¯ of all functions fL1(R+) with strong finite moment sequence, i.e.

0xn|f(x)|dx<for all nZ={0,±1,±2,},
and the corresponding strong vanishing moments subspace M¯0.

First, we introduce a method to get functions in M0 or M¯0: assume that g is an analytic function on a region containing the sector

Sα={zC*:0argzπα},0<α<1,C*=C\{0}.

We use complex integration to show that

(1)Ig(eiπαxα)M0or M¯0,
provided that g satisfies suitable conditions on the boundary Sα of Sα. As usual, Rz,Iz denote the real and imaginary parts of zC.

Then we introduce some operators mapping the subspace M0 into itself. For instance, we prove that M0 is invariant under the operator

gμ*g,
provided that μ is a positive bounded measure on (R+,B+) with finite moments sequence, see (18). Here μ * g is the convolution of the measure μ and the function g on R+ given by
μ*g(x)=0xg(xs)dμ(s),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., mkE[Xk]=0xkf(x)dx< for all kN0, hence m0 = 1. This means that fM. It is well-known that the moment sequence {mk}k=1 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 Xf. 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 Cramér'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+. 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 hL(R+) such that ‖h = 1, fhM0 and fh is not identically zero. Then the Stieltjes class S(f, h) with center at f and perturbation h is given by

S(f,h)={f(x)[1+εh(x)]:xR+,ε[1,1]}.

Clearly, S(f, h) is a family of densities all having the same moment sequence as f.

If Xf is M-determinate, then the perturbation h = 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 gM0 such that h = g/f is bounded on R+, 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 Xf 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

(2)0lnf(x2)1+x2dx<,
then S(f,sin(Helnf)) is a Stieltjes class, where Helnf is the Hilbert transform of u = ln f:
(3)Heu(t)=2t1/2πP0u(x2)tx2dx,t>0.

In particular, here we compute He(|lnx|m), mN, 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 M0. In Section 3 we introduce some operators defined on M0 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 He(|lnx|m), mN={1,2,}.

2. Functions with vanishing moments

In this section we use complex integration to obtain functions in M0 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(−axα) 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 SC, then hol(S) denotes the space of analytic functions on a region containing S.

Lemma 1.

Let 0 < α < 1, γC. Suppose that g ∈ hol(Sα) satisfies the following conditions

(4) t1+Rγ|g(t)|L1(R+),
(5) limAARγ0πα|g(Aeit)|dt=0,
(6) limε0+εRγ0πα|g(εeit)|dt=0,

Then

(7)0tγ1g(teiπα)dt=eiπαγ0tγ1g(t)dt.
Proof.

We pick 0 < ɛ < A < , Cauchy's theorem implies that

Cε,Azγ1g(z)dz=0,
where the contour Cɛ,A consists of the real axis from ɛ to A, the arc of the circle z = Aeit from t = 0 to t = πα, the straight line from Aeiπα to ɛeiπα and the arc of the circle z = ɛei(παt) from t = 0 to t = πα. Thus,
0=I1+I2+I3+I4=εAtγ1g(t)dt+i0πα(Aeit)γg(Aeit)dteiπαAε(teiπα)γ1g(teiπα)dti0πα(εei(παt))γg(εei(παt))dt.

Since |zλ|=|z|Rλexp(argzIλ) for all λC, zC*, conditions (5) and (6) imply that limAI2=limε0+I4=0. Therefore from condition (4) we get

limε0+,A(I1+I3)=0,
and the result follows.
Theorem 2.

Let 0 < α < 1. Suppose that g ∈ hol(Sα) satisfies conditions (5), (6) with γ = (n + 1)/α, g(x)R for all x > 0 and

(8) xn|g(xα)|L1(R+)for all nN0or nZ,
then relation (1) holds.

Proof.

By setting γ = (n + 1)/α for all nN0 or nZ, t = xα, t−1dt = αx−1dx in (7) and taking the imaginary part, the result follows.□

We recall an inequality that will be useful to get our estimates: since ex ≥ x for all x > 0 we have

(9)exssxsfor all x,s>0.

Throughout this work the constant K will be a normalizing constant to produce a density function in each case.

Example 3.

For all b1, b2 > 0, 0 < c1, c2 < 1/2 and aR we have

(10) xa1sinπa+b2tan(πc2)xc2b1tan(πc1)xc1exp(b1xc1b2xc2)M0s.

Indeed, we just apply Theorem 2 with g(z) = zβ  exp(−ρ1zλ − ρ2z−1) for any βR, λ, ρ1, ρ2 > 0. Clearly g is an analytic function on {zC*:|argz|<π}.

From (9) we have that g satisfies condition (8) for all 0 < α < 1, nZ. Assume that 0 < α, αλ < 1/2, then the inequalities

(11)0<cos(πα)cost1,0<cos(παλ)cos(λt)1for all t[0,πα],
together the inequality in (9) imply that
limAA(n+1)/α+β0παexpAλρ1cos(λt)A1ρ2costdt=0for all nZ,
and, by making A = 1/ɛ, the last case implies
limε0+ε(n+1)/α+β0παexpελρ1cos(λt)ε1ρ2costdt=0for all nZ.

From Theorem 2 we have

xαβIexp(iπαβρ1eiπαλxαλρ2eiπαxα)M0s,
and the result follows by setting α = c2, λ = c1/c2, β = (a − 1)/c2, ρi = bi/cos(πci), i = 1, 2. Thus h(x)=sinπa+b2tan(πc2)xc2b1tan(πc1)xc1 is a perturbation for the strong Stieltjes class with center at f(x)=Kxa1exp(b1xc1b2xc2), x > 0. This Stieltjes class was also founded in [6].
Example 4.

For all 0 < α < 1/2, a, b > 0 we have

(12) xa1sin(πabtan(πα)xα)exp(bxα)M0.

Indeed, consider g(z) = zβ  exp(−ρz) for any β > − 1/α, ρ > 0. From (9) we have that g satisfies condition (8) for all nN0. The inequality in (11) implies condition (5) holds for all μ = (n + 1)/α, nN0. Since

limε0+ε(n+1)/α+β0παexp(ερcost)dtπαlimε0+ε(n+1)/α+β=0for all nN0,

Theorem 2 implies that

xαβIeiπαβexp(ρeiπαxα)M0,
and the result follows by setting β = (a − 1)/α and ρ = b/cos(πα). Thus h(x) = sin(πab tan(πα)xα) is a perturbation for the Stieltjes class with center at f(x) = Kxa−1  exp(−bxα), x > 0. This Stieltjes class was also founded in [7, Example 3.2].

Recall that ⌊⋅⌋ is the floor function and ⌈⋅⌉ is the ceiling function. For x,yR, mN, we have

(13)(x+iy)m=j=0m/2m2j(1)jxm2jy2j+ij=0m/21m2j+1(1)jxm2j1y2j+1.
Example 5.

For all 0 < α < 1/2, b > 0, mN, we have

ebxα(θm(x)cos(πα)ψm(x)sin(πα))sinbxα(ψm(x)cos(πα)+θm(x)sin(πα))M0,
where
(14) θm(x)=j=0m/2m2j(1)jπ2j(lnx)m2j,ψm(x)=j=0m/21m2j+1(1)jπ2j+1(lnx)m2j1.

To see this, consider g(z) = exp(−ρz(Log z)m) for any ρ > 0, here Log z stands for the principal branch of the logarithm function. For nN0 we write

0xng(xα)dx=01+1e+exng(xα)dx=I1+I2+I3.

Clearly I2 < , and (9) implies that

I3eCxn(ραmxα)sdx<for s>0 big enough.

Clearly I1 <  when m is even. Assume that m is odd, thus

I1=1yn2exp(ραmyα(lny)m)dyC1dyyn+2<.

Hence g satisfies condition (8) for all nN0, mN.

Since the real part is an additive function, we have for A > e and t ∈ [0, πα] that

(15)Reit(lnA+it)m=(lnA)mcost+j=0m1mj(lnA)jR(eit(it)mj)(lnA)mcos(πα)C(lnA)m1
for some constant C > 0. Therefore,
limAA(n+1)/α0πα|g(Aeit)|dtπαlimAA(n+1)/αexpcos(πα)ρA(lnA)m+CρA(lnA)m1=0
for all nN0.

On the other hand, there is C > 0 such that Reit(lnε+it)mC(|lnε|m+1) for all t ∈ [0, πα], thus

limε0+ε(n+1)/α0πα|g(εeit)|dtπαlimε0+ε(n+1)/αexp(Cε|lnε|m)=0for all nN0.

Theorem 2 implies that

Iexp(ραmeiπαxα(iπ+lnx)m)M0,
the result follows by setting ρ = b/αm and using (13). As before, we can consider the corresponding Stieltjes class for the density
f(x)=Kebxα(θm(x)cos(πα)ψm(x)sin(πα)),x>0.
Example 6.

For all aR, b > 0, mN, we have

xaebθ2m(x)sin(πabψ2m(x))M¯0.

Consider g(z) = zβ  exp(−ρ(Log z)2m) for any βR, ρ > 0. We make the change of variable y = ln x to get

0xng(xα)dx=expρα2my2m+(n+αβ+1)ydy<for all nZ.

As in (15) and using (13) we can see that

limAA(n+1)/α0πα|g(Aeit)|dtπαlimAA(n+1)/α+βexpρ(lnA)2m+Cρ(lnA)2m2=0
for all nZ. The function g also satisfies condition (6) for μ = (n + 1)/α, nZ, we just set A = 1/ɛ and apply the last case.

Theorem 2 implies that

xαβIeiπαβexp(α2mρ(lnx+iπ)2m)M¯0,
and the result follows by setting ρ = b/α2m, β = a/α and using (13). Thus h(x) = sin(πa2m(x)) is a perturbation for the strong Stieltjes class with center at f(x)=Kxaebθ2m(x).
Example 7.

For all a > 0, 0 < α < 1/2 we have

xa1eθ(x)sin(πaψ(x))+2eθ(x)sin(πa)+e2θ(x)sin(πa+ψ(x))1+2cos(ψ(x))eθ(x)+e2θ(x)2M0,
where θ(x) = xα  cos(πα), ψ(x) = xα  sin(πα).

Consider g(z)=zβ(1+ez)2ez for arbitrary β > − 1/α. For all nN0 we have that

0xng(xα)dx0xn+αβexαdx<.
When Rz>0 we have that |1+ez|1eRz, therefore
limAA(n+1)/α0πα|g(Aeit)|dtπαlimAA(n+1)/α+βeAcos(πα)(1eAcos(πα))2=0for all nN0.
For 0 < ɛ < 1 we have
0παdt|1+eεit|2=0παdt2+2cos(εt)πα2,
Hence
limε0+ε(n+1)/α0πα|g(εeit)|dtπα2limε0+ε(n+1)/α+βeεcos(πα)=0.

Theorem 2 implies that

xαβIeiπαβ(1+exp(xαeiπα))2exp(xαeiπα)M0,
and the result follows by setting β = (a − 1)/α.

Notice that

h̃(x)=sin(πaψ(x))+2eθ(x)sin(πa)+e2θ(x)sin(πa+ψ(x))(1+eθ(x))21+2cos(ψ(x))eθ(x)+e2θ(x)2
is a bounded continuous function on R+, and it can be used to construct a Stieltjes class with center at f̃(x)=Kxa1(1+eθ(x))2eθ(x). Now we set δ = (cos(πα))−1/α, and a change of variable implies that h(x)=h̃(δx) can be used to get a Stieltjes class with center at
f(x)=δf̃(δx)=Kxa1exα(1+exα)2.
For 0 < α < 1/2 the last densities are the densities of M-indeterminate powers of random variables following a half-logistic distribution, see [8, Section 6].

3. Operators on the vanishing moment subspace

For mN, s > 0 we introduce the operator Tm,s as follows

Tm,sg(x)=x1/m1g(x1/ms)χ(sm,)(x).

The binomial formula implies that Tm,sM0M0:

(16)smxng(x1/ms)dxx11/m=m0(x+s)mng(x)dx=mj=0mnmnjsj0xmnjg(x)dx=0.

The case m = 1 was considered in [8, Lemma 1].

For a, b > 0 and 0 < α < 1 we have

(17)(a+b)αaα+bα.
Example 8.

Let mN,s>0,0<α<1/2 fixed. From (12) we have

x1/m1sin(tan(πα)((x1/ms)α))e(x1/ms)αχ(sm,)(x)M0,
thus (17) implies that
h(x)=sin(tan(πα)((x1/ms)α))e(xα/m(x1/ms)α)χ(sm,)(x)
is a bounded continuous function on R+ that can be used to obtain a Stieltjess class with center at f(x) = Kx1/m−1  exp(−xα/m), x > 0.

If g1,gmM0 and a1,amR then iaigiM0, hence the following result is a generalization of the last observation.

Proposition 9.

Let (J, μ) be a measure space. Assume that G:R+×JR is a measurable function such that xnG(x,ω)L1(R+×J,dxdμ) for all nN0 or nZ and

G(,ω)M0or M¯0for all ωΩ,
therefore ΩG(,ω)dμ(ω)M0or M¯0.

Proof.

Fubini's theorem implies that

0xnΩG(x,ω)dμ(ω)dx=Ω0xnG(x,ω)dxdμ(ω)=0for all nN0or nZ.

Corollary 10.

Let μ be a positive bounded measure on (R+,B+) such that

(18) 0xndμ<for all nN0.
If gM0, then μ*gM0.

Proof.

We consider the function G:R+×R+R given by

G(x,s)=g(xs)χ(s,)(x).

Clearly G is a measurable function and satisfies

00xn|G(x,s)|dxdμ(s)=00(x+s)n|g(x)|dxdμ(s)=j=0nnj0xj|g(x)|dx0snjdμ(s)<
for all nN0. Since gM0 we have that G(,s)M0 for all s > 0, and the last result implies that 0G(,s)dsM0. On the other hand, we have
0G(x,s)ds=0g(xs)χ(0,x)(s)dμ(s)=μ*g(x),x>0.

We apply the last result to obtain new Stieltjes classes with center at f(x) = K exp(−xα), x > 0, as follows.

Example 11.

By (12) we have g(x)=sin(tan(πα)xα)exp(xα)M0, with 0 < α < 1/2 fixed. Consider the measures 1(s) = χ(0,1)ds and 2(s) = esds on R+. Thus,

μ1*g(x)=0x1sin(tan(πα)(xs)α)e(xs)αdsM0,
where x ∧ 1 = min{x, 1} and
μ2*g(x)=0xsin(tan(πα)(xs)α)e(xs)αsdsM0.

From (17) we get

|μ1*g(x)|01esαdsexαfor all x>0,
hence the bounded function h(x) = μ1 * g(x) exp(xα) can be used to construct a Stieltjes class with center at f(x) = K exp(−xα), x > 0.

Since (x/e)α ≤ x/e for all x ≥ e, there exist a constant 0 < C < 1 such that xα − x ≤ −Cx for all x ≥ e, thus

|μ2*g(x)|0xesαsdsexαC~+0eCsdsexαfor all x>0,
and we proceed as before to construct the corresponding Stieltjes class.

Now, let p be a polynomial with real coefficients, with p(0) = 0 and p′ > 0 on R+. We introduce the operator Rp as follows

Rpg(x)=g(p1(x))p(p1(x)),x>0,
where p−1 is the inverse function of p on R+. As in (16), a change of variable and the binomial formula implies that RpM0M0.
Example 12.

Let 0 < α < 1/2, a, b > 0 be fixed and 1 ≤ n < (2α)−1, nN. From (12) we have that g(x)=xna1sin(πnabtan(nπα)xnα)exp(bxnα)M0. We set pn (x) = xn, x ≥ 0, to get that

Rpng(x)=n1xa1sin(πnabtan(πnα)xα)exp(bxα)M0,
therefore hn(x) = sin(πnab tan(nπα)xα), x > 0, is a perturbation for a Stieltjes class with center at f(x) = Kxa−1  exp(−bxα), x > 0. As far as we know, these are new Stieltjes classes when 2 ≤ n < (2α)−1, nN.

Remark 13.

Let Λ ≠ ∅. Assume that {fλ}λΛM0. If x1fλ0(x1){fλ}λΛ, then fλ0M¯0.

Example 14.

Let 0 < c1, c2 < 1/2, b1, b2 > 0, aR be fixed and 1n<21(c11c21), nN. From (10) we have that

xna1sinπna+b2tan(πnc2)xnc2b1tan(πnc1)xnc1exp(b1xnc1b2xnc2)M¯0,

we proceed as in Example 12 and use Remark 13 to get that

xa1sinπna+b2tan(πnc2)xc2b1tan(πnc1)xc1exp(b1xc1b2xc2)M¯0.

Once again, we obtain new perturbations for strong Stieltjes classes with center at f(x)=Kxa1exp(b1xc1b2xc2), x > 0.

4. Krein criterion and the Hilbert transform

In this section we use a different technique to construct Stieltjes classes. This method involves the computation of the Hilbert transform of ln f, where f is a density fM satisfying the Krein criterion (2). In [5, Theorem 1.2] was proved that S(f,sin(Helnf)) is a Stieltjes class with center at f, where the Hilbert transform He is defined in (3). The following result can be found in [5, Remark 2.1, Lemmas 2.2 and 2.3] and provides the computation of the Hilbert transform of power functions, constant functions and the logarithm function.

Proposition 15.

a) For any constant cR we have He(c)0.

b) Let 0 < |γ| < 1. Then

He(xγ)(t)=tan(γπ)tγ,t>0.

c) He(lnx)π.

As a consequence we obtain a Stieltjes class with center at a generalized inverse Gaussian density.

Example 16.

Let aR, b1, b2 > 0, 0 < c1, c2 < 1/2. Consider the density f(x)=Kxa1exp(b1xc1b2xc2), x > 0. Proposition 15 implies that

He(lnf)(t)=π(a1)b2tan(πc2)tc2+b1tan(πc1)tc1,t>0.

Thus h(x)=sinπa+b2tan(πc2)xc2b1tan(πc1)xc1, x > 0, is a perturbation for the Stieltjes class with center at f. This is the case n = 1 in Example 14.

Remark 17.

As before, we can see that h(x) = sin(πab tan(πα)xα), x > 0, is a perturbation for the Stieltjes class with center at the density f(x) = Kxa−1  exp(−bxα), x > 0, provided that 0 < α < 1/2, a, b > 0. This is the case n = 1 in Example 12.

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 | ln x|n, x > 0, nN. Thus, we need to compute the principal value of the singular integral in (3) with u = | ln x|n, nN.

For all k,nN0 we have the identity, see [10, p. 69, eq. 4.1.51],

(19)xn(lnx)kdx=xn+1j=0k(1)jk!(n+1)j+1(kj)!(lnx)kj.

We introduce the following constants

γjn=01(2n+1)j=(12j)ζ(j),j>1,jN,
where ζ(z) is the zeta function.

First we compute the Hilbert transform of even powers of | ln x|.

Lemma 18.

For mN we have

(20) He(|lnx|2m)(t)=2(2m)!π=1m22γ2(2m2+1)!(lnt)2m2+1,t>0.

Proof.

By (3) it follows that

He(|lnx|2m)(t2)=22m+1tπP0|lnx|2mt2x2dx,t>0.

Let t > 0 fixed and ɛ > 0 small enough. Since the geometric series with ratio r = x2/t2 converges uniformly for x ∈ [0, tɛ], and by using (19), we get that

1t20tε|lnx|2m1(x/t)2dx=n=01t2n+20tεx2n|lnx|2mdx=j=02m(1)j(2m)!(2mj)!n=01t2n+2[x2n+1(lnx)2mj(2n+1)j+1x=0x=tε]=j=02m(1)j(2m)!(2mj)!(ln(tε))2mjn=0(tε)2n+1(2n+1)j+1t2n+2.

Multiplying the last equality by t and using that arctanh(x)=n=0x2n+1/(2n+1) for |x| < 1, we have

(21)t0tε|lnx|2mt2x2dx=n=0(2m)!(tε)2n+1(2n+1)2m+1t2n+1+(ln(tε))2marctanhtεt+j=12m1(1)j(2m)!(2mj)!(ln(tε))2mjn=0(tε)2n+1(2n+1)j+1t2n+1=I1(ε)+I2(ε)+I3(ε).

Similarly, we can obtain that

t+ε1x2|lnx|2m1(t/x)2dx=j=02m(2m)!(2mj)!(ln(t+ε))2mjn=0t2n(2n+1)j+1(t+ε)2n+1.

As before, we multiply the last equality by t to get

(22)tt+ε|lnx|2mt2x2dx=n=0(2m)!t2n+1(2n+1)2m+1(t+ε)2n+1(ln(t+ε))2marctanhtt+εj=12m1(2m)!(2mj)!(ln(t+ε))2mjn=0t2n+1(2n+1)j+1(t+ε)2n+1=J1(ε)+J2(ε)+J3(ε).

By the other hand,

limε0+I1(ε)+J1(ε)=0,
and we apply the Weierstrass M-test to the third terms in (21) and (22), considering ɛ ∈ [0, ɛ0) with ɛ0 small enough, to obtain
limε0+I3(ε)+J3(ε)==1m2(2m)!γ2(2m2+1)!(lnt)2m2+1.

Finally, we use that arctanh(x)=21ln1+x1x, |x| < 1, to obtain

He(|lnx|2m)(t2)=22m+2π=1m(2m)!γ2(2m2+1)!(lnt)2m2+1+2mπlimε0+(ln(tε))2mln2tεε(ln(t+ε))2mln2t+εε.

L'Hôpital's rule implies that the last limit is equal to zero, and the result follows.□

Similar to (9), now we give a basic estimate for the logarithm function: since xs ≤ exp(xs) for all x, s > 0, we have

(23)lnxxssfor all x,s>0.
Example 19.

For mN consider the density f(x) = K exp(−| ln x|2m), x > 0. Clearly

1xnf(x)dx=K0ex2m+(n+1)xdx<,
for all n Z, then fM¯. From (23) we get
0|lnx|2m1+x2dx=21|lnx|2m1+x2dx<,
thus f satisfies (2), therefore sin(He(|lnx|2m)(t)) is a perturbation for the Stieltjes class with center at f, where He(|lnx|2m) is given in (20).

Finally, we compute the Hilbert transform of odd powers of | ln x|. The computations are very similar to those in the proof of Lemma 18.

Lemma 20.

For mN we have

(24) He(|lnx|2m1)(t)=2(2m1)!π=1m22γ2(2m2)!(lnt)2m2n=022mtn+12(2n+1)2m,0<t<1.

For t > 1 we have He(|lnx|2m1)(t)=He(|lnx|2m1)(t1).

Proof.

We just make a sketch of the proof. Let t ∈ (0, 1) fixed. We have the following equalities

0tε(lnx)2m1t2x2dx=j=02m1(1)j(2m1)!(2m1j)!(ln(tε))2m1jn=0(tε)2n+1(2n+1)j+1tn+2,
t+ε1(lnx)2m1t2x2dx=n=0(2m1)!t2n(2n+1)2m+j=02m1(2m1)!(2m1j)!n=0(ln(t+ε))2m1jt2n(2n+1)j+1(t+ε)2n+1,
and 1(lnx)2m1t2x2dx=(2m1)!n=0t2n(2n+1)2m.

Therefore we obtain

π22m+1He(|lnx|2m1)(t2)=n=0(2m1)!t2n+1(2n+1)2m+=1m(2m1)!γ2(2m2)!(lnt)2m2+14limε0+(ln(t+ε))2m1ln2t+εε(ln(tε))2m1ln2tεε.

The last limit is equal to zero and the result follows. When t > 1 a change of variables shows that He(|lnx|2m1)(t2)=He(|lnx|2m1)(t2).□

Example 21.

For mN consider the density f(x) = K exp(−| ln x|2m−1), x > 0. Then sin(He(|lnx|2m1)(t)) is a perturbation for the Stieltjes class with center at f, where He(|lnx|2m1) is given in (24).

In this setting, we also can use the functions in M0 obtained in Examples 5 and 6 to construct perturbations for Stieltjes classes with center at generalized log-normal densities.

5. Conclusion

We gather several methods and techniques to construct systematically Stieltjes classes for M-indeterminate probability densities defined on R+. We construct explicitly Stieltjes classes with centers at densities of M-indeterminate powers of generalized log-normal random variables.

References

1.Berg Ch. From discrete to absolutely continuous solutions of indeterminate moment problems. Arab J Math Sci. 1998; 4(2): 1-18.

2.Berg Ch and Christensen JPR. Density questions in the classical theory of moments. Ann Inst Fourier (Grenoble). 1981; 31(3): 99-114.

3.Lin GD. Recent developments on the moment problem. J Stat Dists Appl. 2017; 4(1): 5.

4.Stoyanov J. Stieltjes classes for moment-indeterminate probability distributions. J Appl Probab. 2004; 41A: 281-94. [Stochastic methods and their applications].

5.López-García M. Krein condition and the Hilbert transform. Electron Commun Probab, 2020; 25(71): 7.

6.Ostrovska S, Stoyanov J. Stieltjes classes for M-indeterminate powers of inverse Gaussian distributions. Statist Probab Lett. 2005; 71(2): 165-71.

7.Ostrovska S. Constructing Stieltjes classes for M-indeterminate absolutely continuous probability distributions. ALEA Lat Am J Probab Math Stat. 2014; 11(1): 253-58.

8.Stoyanov J, Tolmatz L. Method for constructing Stieltjes classes for M-indeterminate probability distributions. Appl Math Comput. 2005; 165(3): 669-85.

9.Kleiber C. The generalized lognormal distribution and the Stieltjes moment problem. J Theoret Probab. 2014; 27(4): 1167-77.

10.Abramowitz M, Irene A. Stegun. Handbook of mathematical Functions with formulas, graphs, and mathematical tables, New York, NY: Dover, ninth dover printing, tenth gpo printing edition, 1964.

Acknowledgements

Funding: This research received no specific grant from any funding agency.

Corresponding author

Marcos López-García can be contacted at: marcos.lopez@im.unam.mx

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