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

López-García, Marcos

doi:10.1108/AJMS-04-2021-0083

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 $| \ln x |$ to construct Stieltjes classes by using a recent result connecting the Krein condition and the Hilbert transform.

1. Introduction

Consider the subspace $M$ of all functions $f \in L^{1} (R^{+})$ with finite moment sequence, i.e.

\int_{0}^{\infty} x^{n} | f (x) | d x < \infty for all n \in N_{0} = {0,1, \dots} .

The vanishing moments subspace $M_{0}$ is given as follows

M_{0} = \{f \in M : \int_{0}^{\infty} x^{n} f (x) d x = 0 for all n \in N_{0}\} .

We also consider the subspace $\bar{M}$ of all functions $f \in L^{1} (R^{+})$ with strong finite moment sequence, i.e.

\int_{0}^{\infty} x^{n} | f (x) | d x < \infty for all n \in Z = {0, \pm 1, \pm 2, \dots},

and the corresponding strong vanishing moments subspace ${\bar{M}}_{0}$ ⁠.

First, we introduce a method to get functions in $M_{0}$ or ${\bar{M}}_{0}$ ⁠: assume that g is an analytic function on a region containing the sector

S_{α} = {z \in C^{*} : 0 \leq \arg z \leq π α}, 0 < α < 1, C^{*} = C \ {0} .

We use complex integration to show that

I (g (e^{i π α} x^{α})) \in M_{0} or {\bar{M}}_{0},

(1)

provided that g satisfies suitable conditions on the boundary $\partial$ S_α of S_α. As usual, $R z, I z$ denote the real and imaginary parts of $z \in C$ ⁠.

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 \mapsto μ * 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) = \int_{0}^{x} g (x - s) d μ (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., $m_{k} ≔ E [X^{k}] = \int_{0}^{\infty} x^{k} f (x) d x < \infty$ for all $k \in N_{0}$ ⁠, hence m₀ = 1. This means that $f \in M$ ⁠. It is well-known that the moment sequence ${m_{k}}_{k = 1}^{\infty}$ 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 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 $h \in L^{\infty} (R^{+})$ such that ‖h‖_∞ = 1, $f h \in M_{0}$ and fh is not identically zero. Then the Stieltjes class S(f, h) with center at f and perturbation h is given by

S (f, h) = {f (x) [1 + ε h (x)] : x \in R^{+}, ε \in [- 1,1]} .

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 = 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 $g \in M_{0}$ 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 X ∼ f with a density f in $\bar{M}$ ⁠, 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

\int_{0}^{\infty} - \frac{\ln f (x^{2})}{1 + x^{2}} d x < \infty,

(2)

then $S (f, \sin (H_{e} \ln f))$ is a Stieltjes class, where $H_{e} \ln f$ is the Hilbert transform of u = ln f:

H_{e} u (t) = \frac{2 t^{1 / 2}}{π} P \int_{0}^{\infty} \frac{u (x^{2})}{t - x^{2}} d x, t > 0 .

(3)

In particular, here we compute $H_{e} (| \ln x |^{m})$ ⁠, $m \in N$ ⁠, 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 $H_{e} (| \ln x |^{m})$ ⁠, $m \in N = {1,2, \dots}$ ⁠.

2. Functions with vanishing moments

In this section we use complex integration to obtain functions in $M_{0}$ or ${\bar{M}}_{0}$ ⁠. 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(−ax^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 $S \subset C$ ⁠, then hol(S) denotes the space of analytic functions on a region containing S.

Lemma 1.

Let 0 < α < 1, $γ \in C$ . Suppose that g ∈ hol(S_α) satisfies the following conditions

t^{- 1 + R γ} | g (t) | \in L^{1} (R^{+}),

(4)

\lim_{A \to \infty} A^{R γ} \int_{0}^{π α} | g (A e^{i t}) | d t = 0,

(5)

\lim_{ε \to 0^{+}} ε^{R γ} \int_{0}^{π α} | g (ε e^{i t}) | d t = 0,

(6)

Then

\int_{0}^{\infty} t^{γ - 1} g (t e^{i π α}) d t = e^{- i π α γ} \int_{0}^{\infty} t^{γ - 1} g (t) d t .

(7)

Proof.

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

\oint_{C_{ε, A}}^{} z^{γ - 1} g (z) d z = 0,

where the contour C_ɛ,A consists of the real axis from ɛ to A, the arc of the circle z = Ae^it from t = 0 to t = πα, the straight line from Ae^iπα to ɛe^iπα and the arc of the circle z = ɛe^i(πα−t) from t = 0 to t = πα. Thus,

0 = I_{1} + I_{2} + I_{3} + I_{4} = \int_{ε}^{A} t^{γ - 1} g (t) d t + i \int_{0}^{π α} {(A e^{i t})}^{γ} g (A e^{i t}) d t - e^{i π α} \int_{- A}^{- ε} {(- t e^{i π α})}^{γ - 1} g (- t e^{i π α}) d t - i \int_{0}^{π α} {(ε e^{i (π α - t)})}^{γ} g (ε e^{i (π α - t)}) d t .

Since $| z^{λ} | = | z |^{R λ} \exp (- \arg z I λ)$ for all $λ \in C$ ⁠, $z \in C^{*}$ ⁠, conditions (5) and (6) imply that $\lim_{A \to \infty} I_{2} = \lim_{ε \to 0^{+}} I_{4} = 0$ ⁠. Therefore from condition (4) we get

\lim_{ε \to 0^{+}, A \to \infty} (I_{1} + I_{3}) = 0,

□

and the result follows.

Theorem 2.

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

x^{n} | g (x^{α}) | \in L^{1} (R^{+}) for all n \in N_{0} or n \in Z,

(8)

then relationholds.

Proof.

By setting γ = (n + 1)/α for all $n \in N_{0}$ or $n \in Z$ ⁠, t = x^α, 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 e^x ≥ x for all x > 0 we have

e^{- x} \leq 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.

Example 3.

For all b₁, b₂ > 0, 0 < c₁, c₂ < 1/2 and $a \in R$ we have

x^{a - 1} \sin (π a + b_{2} \tan (π c_{2}) x^{- c_{2}} - b_{1} \tan (π c_{1}) x^{c_{1}}) \exp (- b_{1} x^{c_{1}} - b_{2} x^{- c_{2}}) \in M_{0}^{s} .

(10)

Indeed, we just apply Theorem 2 with g(z) = z^β exp(−ρ₁z^λ − ρ₂z⁻¹) for any $β \in R$ ⁠, λ, ρ₁, ρ₂ > 0. Clearly g is an analytic function on ${z \in C^{*} : | \arg z | < π}$ ⁠.

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

0 < \cos (π α) \leq \cos t \leq 1, 0 < \cos (π α λ) \leq \cos (λ t) \leq 1 for all t \in [0, π α],

(11)

together the inequality in (9) imply that

\lim_{A \to \infty} A^{(n + 1) / α + β} \int_{0}^{π α} \exp (- A^{λ} ρ_{1} \cos (λ t) - A^{- 1} ρ_{2} \cos t) d t = 0 for all n \in Z,

and, by making A = 1/ɛ, the last case implies

\lim_{ε \to 0^{+}} ε^{(n + 1) / α + β} \int_{0}^{π α} \exp (- ε^{λ} ρ_{1} \cos (λ t) - ε^{- 1} ρ_{2} \cos t) d t = 0 for all n \in Z .

From Theorem 2 we have

x^{α β} I (\exp (i π α β - ρ_{1} e^{i π α λ} x^{α λ} - ρ_{2} e^{- i π α} x^{- α})) \in M_{0}^{s},

and the result follows by setting α = c₂, λ = c₁/c₂, β = (a − 1)/c₂, ρ_i = b_i/cos(πc_i), i = 1, 2. Thus $h (x) = \sin (π a + b_{2} \tan (π c_{2}) x^{- c_{2}} - b_{1} \tan (π c_{1}) x^{c_{1}})$ is a perturbation for the strong Stieltjes class with center at $f (x) = K x^{a - 1} \exp (- b_{1} x^{c_{1}} - b_{2} x^{- c_{2}})$ ⁠, x > 0. This Stieltjes class was also founded in [6].

Example 4.

For all 0 < α < 1/2, a, b > 0 we have

x^{a - 1} \sin (π a - b \tan (π α) x^{α}) \exp (- b x^{α}) \in M_{0} .

(12)

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

\lim_{ε \to 0^{+}} ε^{(n + 1) / α + β} \int_{0}^{π α} \exp (- ε ρ \cos t) d t \leq π α \lim_{ε \to 0^{+}} ε^{(n + 1) / α + β} = 0 for all n \in N_{0},

Theorem 2 implies that

x^{α β} I (e^{i π α β} \exp (- ρ e^{i π α} x^{α})) \in M_{0},

and the result follows by setting β = (a − 1)/α and ρ = b/cos(πα). Thus h(x) = sin(πa − b tan(πα)x^α) is a perturbation for the Stieltjes class with center at f(x) = Kx^a−1 exp(−bx^α), 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 $x, y \in R$ ⁠, $m \in N$ ⁠, we have

{(x + i y)}^{m} = \sum_{j = 0}^{⌊ m / 2 ⌋} (\binom{m}{2 j}) {(- 1)}^{j} x^{m - 2 j} y^{2 j} + i \sum_{j = 0}^{⌈ m / 2 ⌉ - 1} (\binom{m}{2 j + 1}) {(- 1)}^{j} x^{m - 2 j - 1} y^{2 j + 1} .

(13)

Example 5.

For all 0 < α < 1/2, b > 0, $m \in N$ , we have

e^{- b x^{α} (θ_{m} (x) \cos (π α) - ψ_{m} (x) \sin (π α))} \sin (b x^{α} (ψ_{m} (x) \cos (π α) + θ_{m} (x) \sin (π α))) \in M_{0},

where

\begin{matrix} θ_{m} (x) & = & \sum_{j = 0}^{⌊ m / 2 ⌋} (\binom{m}{2 j}) {(- 1)}^{j} π^{2 j} {(\ln x)}^{m - 2 j}, \\ ψ_{m} (x) & = & \sum_{j = 0}^{⌈ m / 2 ⌉ - 1} (\binom{m}{2 j + 1}) {(- 1)}^{j} π^{2 j + 1} {(\ln x)}^{m - 2 j - 1} . \end{matrix}

(14)

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 $n \in N_{0}$ we write

\int_{0}^{\infty} x^{n} g (x^{α}) d x = (\int_{0}^{1} + \int_{1}^{e} + \int_{e}^{\infty}) x^{n} g (x^{α}) d x = I_{1} + I_{2} + I_{3} .

Clearly I₂ < ∞, and (9) implies that

I_{3} \leq \int_{e}^{\infty} \frac{C x^{n}}{{(ρ α^{m} x^{α})}^{s}} d x < \infty for s > 0 big enough .

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

I_{1} = \int_{1}^{\infty} y^{- n - 2} \exp (ρ α^{m} y^{- α} {(\ln y)}^{m}) d y \leq C \int_{1}^{\infty} \frac{d y}{y^{n + 2}} < \infty .

Hence g satisfies condition (8) for all $n \in N_{0}$ ⁠, $m \in N$ ⁠.

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

\begin{matrix} R (e^{i t} {(\ln A + i t)}^{m}) & = & {(\ln A)}^{m} \cos t + \sum_{j = 0}^{m - 1} (\binom{m}{j}) {(\ln A)}^{j} R (e^{i t} {(i t)}^{m - j}) \\ \geq & {(\ln A)}^{m} \cos (π α) - C {(\ln A)}^{m - 1} \end{matrix}

(15)

for some constant C > 0. Therefore,

\lim_{A \to \infty} A^{(n + 1) / α} \int_{0}^{π α} | g (A e^{i t}) | d t \leq π α \lim_{A \to \infty} A^{(n + 1) / α} \exp (- \cos (π α) ρ A {(\ln A)}^{m} + C ρ A {(\ln A)}^{m - 1}) = 0

for all $n \in N_{0}$ ⁠.

On the other hand, there is C > 0 such that $- R (e^{i t} {(\ln ε + i t)}^{m}) \leq C (| \ln ε |^{m} + 1)$ for all t ∈ [0, πα], thus

\lim_{ε \to 0^{+}} ε^{(n + 1) / α} \int_{0}^{π α} | g (ε e^{i t}) | d t \leq π α \lim_{ε \to 0^{+}} ε^{(n + 1) / α} \exp (C ε | \ln ε |^{m}) = 0 for all n \in N_{0} .

Theorem 2 implies that

I (\exp (- ρ α^{m} e^{i π α} x^{α} {(i π + \ln x)}^{m})) \in M_{0},

the result follows by setting ρ = b/α^m and using (13). As before, we can consider the corresponding Stieltjes class for the density

f (x) = K e^{- b x^{α} (θ_{m} (x) \cos (π α) - ψ_{m} (x) \sin (π α))}, x > 0 .

Example 6.

For all $a \in R$ , b > 0, $m \in N$ , we have

x^{a} e^{- b θ_{2 m} (x)} \sin (π a - b ψ_{2 m} (x)) \in {\bar{M}}_{0} .

Consider g(z) = z^β exp(−ρ(Log z)^2m) for any $β \in R$ ⁠, ρ > 0. We make the change of variable y = ln x to get

\int_{0}^{\infty} x^{n} g (x^{α}) d x = \int_{- \infty}^{\infty} \exp (- ρ α^{2 m} y^{2 m} + (n + α β + 1) y) d y < \infty for all n \in Z .

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

\lim_{A \to \infty} A^{(n + 1) / α} \int_{0}^{π α} | g (A e^{i t}) | d t \leq π α \lim_{A \to \infty} A^{(n + 1) / α + β} \exp (- ρ {(\ln A)}^{2 m} + C ρ {(\ln A)}^{2 m - 2}) = 0

for all $n \in Z$ ⁠. The function g also satisfies condition (6) for μ = (n + 1)/α, $n \in Z$ ⁠, we just set A = 1/ɛ and apply the last case.

Theorem 2 implies that

x^{α β} I (e^{i π α β} \exp (- α^{2 m} ρ {(\ln x + i π)}^{2 m})) \in {\bar{M}}_{0},

and the result follows by setting ρ = b/α^2m, β = a/α and using (13). Thus h(x) = sin(πa − bψ_2m(x)) is a perturbation for the strong Stieltjes class with center at $f (x) = K x^{a} e^{- b θ_{2 m} (x)}$ ⁠.

Example 7.

For all a > 0, 0 < α < 1/2 we have

x^{a - 1} e^{- θ (x)} \frac{\sin (π a - ψ (x)) + 2 e^{- θ (x)} \sin (π a) + e^{- 2 θ (x)} \sin (π a + ψ (x))}{{(1 + 2 \cos (ψ (x)) e^{- θ (x)} + e^{- 2 θ (x)})}^{2}} \in M_{0},

whereθ(x) = x^α cos(πα), ψ(x) = x^α sin(πα).

Consider $g (z) = z^{β} {(1 + e^{- z})}^{- 2} e^{- z}$ for arbitrary β > − 1/α. For all $n \in N_{0}$ we have that

\int_{0}^{\infty} x^{n} g (x^{α}) d x \leq \int_{0}^{\infty} x^{n + α β} e^{- x^{α}} d x < \infty .

When $R z > 0$ we have that $| 1 + e^{- z} | \geq 1 - e^{- R z}$ ⁠, therefore

\lim_{A \to \infty} A^{(n + 1) / α} \int_{0}^{π α} | g (A e^{i t}) | d t \leq π α \lim_{A \to \infty} A^{(n + 1) / α + β} \frac{e^{- A \cos (π α)}}{{(1 - e^{- A \cos (π α)})}^{2}} = 0 for all n \in N_{0} .

For 0 < ɛ < 1 we have

\int_{0}^{π α} \frac{d t}{| 1 + e^{- ε i t} |^{2}} = \int_{0}^{π α} \frac{d t}{2 + 2 \cos (ε t)} \leq \frac{π α}{2},

Hence

\lim_{ε \to 0^{+}} ε^{(n + 1) / α} \int_{0}^{π α} | g (ε e^{i t}) | d t \leq \frac{π α}{2} \lim_{ε \to 0^{+}} ε^{(n + 1) / α + β} e^{- ε \cos (π α)} = 0 .

Theorem 2 implies that

x^{α β} I (e^{i π α β} {(1 + \exp (- x^{α} e^{i π α}))}^{- 2} \exp (- x^{α} e^{i π α})) \in M_{0},

and the result follows by setting β = (a − 1)/α.

Notice that

\tilde{h} (x) = \frac{\sin (π a - ψ (x)) + 2 e^{- θ (x)} \sin (π a) + e^{- 2 θ (x)} \sin (π a + ψ (x))}{{(1 + e^{- θ (x)})}^{- 2} {(1 + 2 \cos (ψ (x)) e^{- θ (x)} + e^{- 2 θ (x)})}^{2}}

is a bounded continuous function on $R^{+}$ ⁠, and it can be used to construct a Stieltjes class with center at $\tilde{f} (x) = K x^{a - 1} {(1 + e^{- θ (x)})}^{- 2} e^{- θ (x)}$ ⁠. Now we set δ = (cos(πα))^−1/α, and a change of variable implies that $h (x) = \tilde{h} (δ x)$ can be used to get a Stieltjes class with center at

f (x) = δ \tilde{f} (δ x) = K^{'} x^{a - 1} \frac{e^{- x^{α}}}{{(1 + e^{- x^{α}})}^{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 $m \in N$ ⁠, s > 0 we introduce the operator T_m,s as follows

T_{m, s} g (x) = x^{1 / m - 1} g (x^{1 / m} - s) χ_{(s^{m}, \infty)} (x) .

The binomial formula implies that $T_{m, s} M_{0} \subset M_{0}$ ⁠:

\int_{s^{m}}^{\infty} x^{n} g (x^{1 / m} - s) \frac{d x}{x^{1 - 1 / m}} = m \int_{0}^{\infty} {(x + s)}^{m n} g (x) d x = m \sum_{j = 0}^{m n} (\binom{m n}{j}) s^{j} \int_{0}^{\infty} x^{m n - j} g (x) d x = 0 .

(16)

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

For a, b > 0 and 0 < α < 1 we have

{(a + b)}^{α} \leq a^{α} + b^{α} .

(17)

Example 8.

Let $m \in N, s > 0,0 < α < 1 / 2$ fixed. From (12) we have

x^{1 / m - 1} \sin (\tan (π α) ({(x^{1 / m} - s)}^{α})) e^{- {(x^{1 / m} - s)}^{α}} χ_{(s^{m}, \infty)} (x) \in M_{0},

thusimplies that

h (x) = \sin (\tan (π α) ({(x^{1 / m} - s)}^{α})) e^{(x^{α / m} - {(x^{1 / m} - s)}^{α})} χ_{(s^{m}, \infty)} (x)

is a bounded continuous function on

R^{+}

that can be used to obtain a Stieltjess class with center at f(x) = Kx^1/m−1 exp(−x^α/m),x > 0.

If $g_{1}, \dots g_{m} \in M_{0}$ and $a_{1}, \dots a_{m} \in R$ then $\sum_{i} a_{i} g_{i} \in M_{0}$ ⁠, hence the following result is a generalization of the last observation.

Proposition 9.

Let (J, μ) be a measure space. Assume that $G : R^{+} \times J \to R$ is a measurable function such that $x^{n} G (x, ω) \in L^{1} (R^{+} \times J, d x \otimes d μ)$ for all $n \in N_{0}$ or $n \in Z$ and

G (\cdot, ω) \in M_{0} or {\bar{M}}_{0} for all ω \in Ω,

therefore

\int_{Ω} G (\cdot, ω) d μ (ω) \in M_{0} or {\bar{M}}_{0}

⁠.

Proof.

Fubini's theorem implies that

\int_{0}^{\infty} x^{n} \int_{Ω} G (x, ω) d μ (ω) d x = \int_{Ω} \int_{0}^{\infty} x^{n} G (x, ω) d x d μ (ω) = 0 for all n \in N_{0} or n \in Z .

□

Corollary 10.

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

\int_{0}^{\infty} x^{n} d μ < \infty for all n \in N_{0} .

(18)

If

g \in M_{0}

⁠, then

μ * g \in M_{0}

⁠.

Proof.

We consider the function $G : R^{+} \times R^{+} \to R$ given by

G (x, s) = g (x - s) χ_{(s, \infty)} (x) .

Clearly $G$ is a measurable function and satisfies

\int_{0}^{\infty} \int_{0}^{\infty} x^{n} | G (x, s) | d x \otimes d μ (s) = \int_{0}^{\infty} \int_{0}^{\infty} {(x + s)}^{n} | g (x) | d x d μ (s) = \sum_{j = 0}^{n} (\binom{n}{j}) \int_{0}^{\infty} x^{j} | g (x) | d x \int_{0}^{\infty} s^{n - j} d μ (s) < \infty

for all

n \in N_{0}

⁠. Since

g \in M_{0}

we have that

G (\cdot, s) \in M_{0}

for all s > 0, and the last result implies that

\int_{0}^{\infty} G (\cdot, s) d s \in M_{0}

⁠. On the other hand, we have

\int_{0}^{\infty} G (x, s) d s = \int_{0}^{\infty} g (x - s) χ_{(0, x)} (s) d μ (s) = μ * g (x), x > 0 .

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

Example 11.

By (12) we have $g (x) = \sin (\tan (π α) x^{α}) \exp (- x^{α}) \in M_{0}$ , with 0 < α < 1/2 fixed. Consider the measures dμ₁(s) = χ_(0,1)ds and dμ₂(s) = e^−sds on $R^{+}$ . Thus,

μ_{1} * g (x) = \int_{0}^{x \land 1} \sin (\tan (π α) {(x - s)}^{α}) e^{- {(x - s)}^{α}} d s \in M_{0},

wherex ∧ 1 = min{x, 1} and

μ_{2} * g (x) = \int_{0}^{x} \sin (\tan (π α) {(x - s)}^{α}) e^{- {(x - s)}^{α} - s} d s \in M_{0} .

From (17) we get

| μ_{1} * g (x) | \leq \int_{0}^{1} e^{s^{α}} d s e^{- x^{α}} for all x > 0,

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

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

| μ_{2} * g (x) | \leq \int_{0}^{x} e^{s^{α} - s} d s e^{- x^{α}} \leq (\tilde{C} + \int_{0}^{\infty} e^{- C s} d s) e^{- x^{α}} for 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 R_p as follows

R_{p} g (x) = \frac{g (p^{- 1} (x))}{p^{'} (p^{- 1} (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 $R_{p} M_{0} \subset M_{0}$ ⁠.

Example 12.

Let 0 < α < 1/2, a, b > 0 be fixed and 1 ≤ n < (2α)⁻¹, $n \in N$ . From (12) we have that $g (x) = x^{n a - 1} \sin (π n a - b \tan (n π α) x^{n α}) \exp (- b x^{n α}) \in M_{0}$ . We set p_n (x) = xⁿ, x ≥ 0, to get that

R_{p_{n}} g (x) = n^{- 1} x^{a - 1} \sin (π n a - b \tan (π n α) x^{α}) \exp (- b x^{α}) \in M_{0},

thereforeh_n(x) = sin(πna − b tan(nπα)x^α),x > 0, is a perturbation for a Stieltjes class with center atf(x) = Kx^a−1 exp(−bx^α),x > 0. As far as we know, these are new Stieltjes classes when 2 ≤ n < (2α)⁻¹,

n \in N

.

Remark 13.

Let Λ ≠ ∅. Assume that ${f_{λ}}_{λ \in Λ} \subset M_{0}$ . If $x^{- 1} f_{λ_{0}} (x^{- 1}) \in {f_{λ}}_{λ \in Λ}$ , then $f_{λ_{0}} \in {\bar{M}}_{0}$ .

Example 14.

Let 0 < c₁, c₂ < 1/2, b₁, b₂ > 0, $a \in R$ be fixed and $1 \leq n < 2^{- 1} (c_{1}^{- 1} \land c_{2}^{- 1})$ , $n \in N$ . From (10) we have that

x^{n a - 1} \sin (π n a + b_{2} \tan (π n c_{2}) x^{- n c_{2}} - b_{1} \tan (π n c_{1}) x^{n c_{1}}) \exp (- b_{1} x^{n c_{1}} - b_{2} x^{- n c_{2}}) \in {\bar{M}}_{0},

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

x^{a - 1} \sin (π n a + b_{2} \tan (π n c_{2}) x^{- c_{2}} - b_{1} \tan (π n c_{1}) x^{c_{1}}) \exp (- b_{1} x^{c_{1}} - b_{2} x^{- c_{2}}) \in {\bar{M}}_{0} .

Once again, we obtain new perturbations for strong Stieltjes classes with center at $f (x) = K x^{a - 1} \exp (- b_{1} x^{c_{1}} - b_{2} x^{- c_{2}})$ , 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 $f \in M$ satisfying the Krein criterion (2). In [5, Theorem 1.2] was proved that $S (f, \sin (H_{e} \ln f))$ is a Stieltjes class with center at f, where the Hilbert transform $H_{e}$ 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 $c \in R$ we have $H_{e} (c) \equiv 0$ .

b) Let 0 < |γ| < 1. Then

H_{e} (x^{γ}) (t) = - \tan (γ π) t^{γ}, t > 0 .

c) $H_{e} (\ln x) \equiv - π$ ⁠.

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

Example 16.

Let $a \in R$ , b₁, b₂ > 0, 0 < c₁, c₂ < 1/2. Consider the density $f (x) = K x^{a - 1} \exp (- b_{1} x^{c_{1}} - b_{2} x^{- c_{2}})$ , x > 0. Proposition 15 implies that

H_{e} (\ln f) (t) = - π (a - 1) - b_{2} \tan (π c_{2}) t^{- c_{2}} + b_{1} \tan (π c_{1}) t^{c_{1}}, t > 0 .

Thus $h (x) = \sin (π a + b_{2} \tan (π c_{2}) x^{- c_{2}} - b_{1} \tan (π c_{1}) x^{c_{1}})$ , 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(πa − b tan(πα)x^α), x > 0, is a perturbation for the Stieltjes class with center at the density f(x) = Kx^a−1 exp(−bx^α), 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, $n \in N$ ⁠. Thus, we need to compute the principal value of the singular integral in (3) with u = | ln x|ⁿ, $n \in N$ ⁠.

For all $k, n \in N_{0}$ we have the identity, see [10, p. 69, eq. 4.1.51],

\int x^{n} {(\ln x)}^{k} d x = x^{n + 1} \sum_{j = 0}^{k} \frac{{(- 1)}^{j} k!}{{(n + 1)}^{j + 1} (k - j)!} {(\ln x)}^{k - j} .

(19)

We introduce the following constants

γ_{j} ≔ \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{j}} = (1 - 2^{- j}) ζ (j), j > 1, j \in N,

where ζ(z) is the zeta function.

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

Lemma 18.

For $m \in N$ we have

H_{e} (| \ln x |^{2 m}) (t) = - \frac{2 (2 m)!}{π} \sum_{ℓ = 1}^{m} \frac{2^{2 ℓ} γ_{2 ℓ}}{(2 m - 2 ℓ + 1)!} {(\ln t)}^{2 m - 2 ℓ + 1}, t > 0 .

(20)

Proof.

By (3) it follows that

H_{e} (| \ln x |^{2 m}) (t^{2}) = \frac{2^{2 m + 1} t}{π} P \int_{0}^{\infty} \frac{| \ln x |^{2 m}}{t^{2} - x^{2}} d x, 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, t −ɛ], and by using (19), we get that

\begin{matrix} \frac{1}{t^{2}} \int_{0}^{t - ε} \frac{| \ln x |^{2 m}}{1 - {(x / t)}^{2}} d x & = & \sum_{n = 0}^{\infty} \frac{1}{t^{2 n + 2}} \int_{0}^{t - ε} x^{2 n} | \ln x |^{2 m} d x \\ = & \sum_{j = 0}^{2 m} \frac{{(- 1)}^{j} (2 m)!}{(2 m - j)!} \sum_{n = 0}^{\infty} \frac{1}{t^{2 n + 2}} [{\frac{x^{2 n + 1} {(\ln x)}^{2 m - j}}{{(2 n + 1)}^{j + 1}}|}_{x = 0}^{x = t - ε}] \\ = & \sum_{j = 0}^{2 m} \frac{{(- 1)}^{j} (2 m)!}{(2 m - j)!} {(\ln (t - ε))}^{2 m - j} \sum_{n = 0}^{\infty} \frac{{(t - ε)}^{2 n + 1}}{{(2 n + 1)}^{j + 1} t^{2 n + 2}} . \end{matrix}

Multiplying the last equality by t and using that $arctanh (x) = \sum_{n = 0}^{\infty} x^{2 n + 1} / (2 n + 1)$ for |x| < 1, we have

\begin{matrix} t \int_{0}^{t - ε} \frac{| \ln x |^{2 m}}{t^{2} - x^{2}} d x & = & \sum_{n = 0}^{\infty} \frac{(2 m)! {(t - ε)}^{2 n + 1}}{{(2 n + 1)}^{2 m + 1} t^{2 n + 1}} + {(\ln (t - ε))}^{2 m} arctanh (\frac{t - ε}{t}) \\ + \sum_{j = 1}^{2 m - 1} \frac{{(- 1)}^{j} (2 m)!}{(2 m - j)!} {(\ln (t - ε))}^{2 m - j} \sum_{n = 0}^{\infty} \frac{{(t - ε)}^{2 n + 1}}{{(2 n + 1)}^{j + 1} t^{2 n + 1}} \\ = & I_{1} (ε) + I_{2} (ε) + I_{3} (ε) . \end{matrix}

(21)

Similarly, we can obtain that

- \int_{t + ε}^{\infty} \frac{1}{x^{2}} \frac{| \ln x |^{2 m}}{1 - {(t / x)}^{2}} d x = - \sum_{j = 0}^{2 m} \frac{(2 m)!}{(2 m - j)!} {(\ln (t + ε))}^{2 m - j} \sum_{n = 0}^{\infty} \frac{t^{2 n}}{{(2 n + 1)}^{j + 1} {(t + ε)}^{2 n + 1}} .

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

\begin{matrix} t \int_{t + ε}^{\infty} \frac{| \ln x |^{2 m}}{t^{2} - x^{2}} d x & = & - \sum_{n = 0}^{\infty} \frac{(2 m)! t^{2 n + 1}}{{(2 n + 1)}^{2 m + 1} {(t + ε)}^{2 n + 1}} - {(\ln (t + ε))}^{2 m} arctanh (\frac{t}{t + ε}) \\ - \sum_{j = 1}^{2 m - 1} \frac{(2 m)!}{(2 m - j)!} {(\ln (t + ε))}^{2 m - j} \sum_{n = 0}^{\infty} \frac{t^{2 n + 1}}{{(2 n + 1)}^{j + 1} {(t + ε)}^{2 n + 1}} \\ = & J_{1} (ε) + J_{2} (ε) + J_{3} (ε) . \end{matrix}

(22)

By the other hand,

\lim_{ε \to 0^{+}} I_{1} (ε) + J_{1} (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

\lim_{ε \to 0^{+}} I_{3} (ε) + J_{3} (ε) = - \sum_{ℓ = 1}^{m} \frac{2 (2 m)! γ_{2 ℓ}}{(2 m - 2 ℓ + 1)!} {(\ln t)}^{2 m - 2 ℓ + 1} .

Finally, we use that $arctanh (x) = 2^{- 1} \ln \frac{1 + x}{1 - x}$ ⁠, |x| < 1, to obtain

\begin{matrix} H_{e} (| \ln x |^{2 m}) (t^{2}) & = & - \frac{2^{2 m + 2}}{π} \sum_{ℓ = 1}^{m} \frac{(2 m)! γ_{2 ℓ}}{(2 m - 2 ℓ + 1)!} {(\ln t)}^{2 m - 2 ℓ + 1} \\ + \frac{2^{m}}{π} \lim_{ε \to 0^{+}} [{(\ln (t - ε))}^{2 m} \ln \frac{2 t - ε}{ε} - {(\ln (t + ε))}^{2 m} \ln \frac{2 t + ε}{ε}] . \end{matrix}

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 x^s ≤ exp(x^s) for all x, s > 0, we have

\ln x \leq \frac{x^{s}}{s} for all x, s > 0 .

(23)

Example 19.

For $m \in N$ consider the density f(x) = K exp(−| ln x|^2m), x > 0. Clearly

\int_{1}^{\infty} x^{n} f (x) d x = K \int_{0}^{\infty} e^{- x^{2 m} + (n + 1) x} d x < \infty,

for alln

\in Z

, then

f \in \bar{M}

. Fromwe get

\int_{0}^{\infty} \frac{| \ln x |^{2 m}}{1 + x^{2}} d x = 2 \int_{1}^{\infty} \frac{| \ln x |^{2 m}}{1 + x^{2}} d x < \infty,

thusfsatisfies, therefore

\sin (H_{e} (| \ln x |^{2 m}) (t))

is a perturbation for the Stieltjes class with center atf, where

H_{e} (| \ln x |^{2 m})

is given in.

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 $m \in N$ we have

H_{e} (| \ln x |^{2 m - 1}) (t) = \frac{2 (2 m - 1)!}{π} [\sum_{ℓ = 1}^{m} \frac{2^{2 ℓ} γ_{2 ℓ}}{(2 m - 2 ℓ)!} {(\ln t)}^{2 m - 2 ℓ} - \sum_{n = 0}^{\infty} \frac{2^{2 m} t^{n + \frac{1}{2}}}{{(2 n + 1)}^{2 m}}], 0 < t < 1 .

(24)

For t > 1 we have $H_{e} (| \ln x |^{2 m - 1}) (t) = H_{e} (| \ln x |^{2 m - 1}) (t^{- 1})$ .

Proof.

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

- \int_{0}^{t - ε} \frac{{(\ln x)}^{2 m - 1}}{t^{2} - x^{2}} d x = - \sum_{j = 0}^{2 m - 1} \frac{{(- 1)}^{j} (2 m - 1)!}{(2 m - 1 - j)!} {(\ln (t - ε))}^{2 m - 1 - j} \sum_{n = 0}^{\infty} \frac{{(t - ε)}^{2 n + 1}}{{(2 n + 1)}^{j + 1} t^{n + 2}},

- \int_{t + ε}^{1} \frac{{(\ln x)}^{2 m - 1}}{t^{2} - x^{2}} d x = - \sum_{n = 0}^{\infty} \frac{(2 m - 1)! t^{2 n}}{{(2 n + 1)}^{2 m}} + \sum_{j = 0}^{2 m - 1} \frac{(2 m - 1)!}{(2 m - 1 - j)!} \sum_{n = 0}^{\infty} \frac{{(\ln (t + ε))}^{2 m - 1 - j} t^{2 n}}{{(2 n + 1)}^{j + 1} {(t + ε)}^{2 n + 1}},

and \int_{1}^{\infty} \frac{{(\ln x)}^{2 m - 1}}{t^{2} - x^{2}} d x = - (2 m - 1)! \sum_{n = 0}^{\infty} \frac{t^{2 n}}{{(2 n + 1)}^{2 m}} .

Therefore we obtain

\begin{matrix} \frac{π}{2^{2 m + 1}} H_{e} (| \ln x |^{2 m - 1}) (t^{2}) & = & - \sum_{n = 0}^{\infty} \frac{(2 m - 1)! t^{2 n + 1}}{{(2 n + 1)}^{2 m}} + \sum_{ℓ = 1}^{m} \frac{(2 m - 1)! γ_{2 ℓ}}{(2 m - 2 ℓ)!} {(\ln t)}^{2 m - 2 ℓ} \\ + \frac{1}{4} \lim_{ε \to 0^{+}} {(\ln (t + ε))}^{2 m - 1} \ln \frac{2 t + ε}{ε} - {(\ln (t - ε))}^{2 m - 1} \ln \frac{2 t - ε}{ε} . \end{matrix}

The last limit is equal to zero and the result follows. When t > 1 a change of variables shows that $H_{e} (| \ln x |^{2 m - 1}) (t^{2}) = H_{e} (| \ln x |^{2 m - 1}) (t^{- 2})$ ⁠.□

Example 21.

For $m \in N$ consider the density f(x) = K exp(−| ln x|^2m−1), x > 0. Then $\sin (H_{e} (| \ln x |^{2 m - 1}) (t))$ is a perturbation for the Stieltjes class with center at f, where $H_{e} (| \ln x |^{2 m - 1})$ is given in (24).

In this setting, we also can use the functions in $M_{0}$ 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.

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

References

1.

Berg

Ch

.

From discrete to absolutely continuous solutions of indeterminate moment problems

.

Arab J Math Sci

.

1998

;

4

(

2

):

1

-

18

.

Google Scholar

2.

Berg

Ch

and

Christensen

JPR

.

Density questions in the classical theory of moments

.

Ann Inst Fourier (Grenoble)

.

1981

;

31

(

3

):

99

-

114

.

Google Scholar

Crossref

3.

Lin

GD

.

Recent developments on the moment problem

.

J Stat Dists Appl

.

2017

;

4

(

1

):

5

.

Google Scholar

Crossref

4.

Stoyanov

J

.

Stieltjes classes for moment-indeterminate probability distributions

.

J Appl Probab

.

2004

;

41A

:

281

-

94

.

[Stochastic methods and their applications]

.

Google Scholar

5.

López-García

M

.

Krein condition and the Hilbert transform

.

Electron Commun Probab

,

2020

;

25

(

71

):

7

.

Google Scholar

6.

Ostrovska

S

,

Stoyanov

J

.

Stieltjes classes for M-indeterminate powers of inverse Gaussian distributions

.

Statist Probab Lett

.

2005

;

71

(

2

):

165

-

71

.

Google Scholar

Crossref

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

.

Google Scholar

8.

Stoyanov

J

,

Tolmatz

L

.

Method for constructing Stieltjes classes for M-indeterminate probability distributions

.

Appl Math Comput

.

2005

;

165

(

3

):

669

-

85

.

Google Scholar

9.

Kleiber

C.

The generalized lognormal distribution and the Stieltjes moment problem

.

J Theoret Probab

.

2014

;

27

(

4

):

1167

-

77

.

Google Scholar

Crossref

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

.

Google Scholar

2021

Marcos López-García

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

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

1. Introduction

2. Functions with vanishing moments

3. Operators on the vanishing moment subspace

4. Krein criterion and the Hilbert transform

5. Conclusion

References

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

1. Introduction

2. Functions with vanishing moments

3. Operators on the vanishing moment subspace

4. Krein criterion and the Hilbert transform

5. Conclusion

References

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