Notes on Cauchy–Stieltjes Kernel Families

Abstract

The free Meixner family ($\mathbb{FMF}$) is the family of measures that produces quadratic Cauchy–Stieltjes Kernel (CSK) families (i.e., meaning that the associated variance function (VF) is a polynomial with degree $\le 2$ in the mean). Furthermore, a cubic class is introduced in the context of CSK families and is connected to the quadratic class via a reciprocity relation. The associated probability measures are the so-called free analog of the Letac–Mora class (with VF of degree 3). In free probability theory, these two classes of probabilities are crucial. However, a novel transformation of measures is introduced in the setting of free probability, known as the $T_a$-transformation of probability measures. Denote by $\mathcal{P}$ the set of (non-degenerate) real probabilities. For $\nu \in \mathcal{P}$ and $a\in \mathbb{R}$, consider the transformation of measure $\nu$, denoted $T_a\left(\nu \right)$, defined by ${\mathcal{F}}_{T_a\left(\nu \right)}\left(w\right)={\mathcal{F}}_{\nu }(w-a)+a,$ where ${\mathcal{F}}_{\nu }(·)$ is the inverse of the Cauchy–Stieltjes transformation of $\nu$. In this study, we provide important insights into the notion of the $T_a$-transformation of probabilities. We demonstrate that the $\mathbb{FMF}$ (respectively, the free counterpart of the Letac–Mora class of measures) is invariant under the $T_a$-transformation. Furthermore, we develop additional characteristics of the $T_a$-transformation, which yield intriguing findings for significant free probability distributions such as the free Poisson and free Gamma distributions.

1. Introduction

Free probability theory, defined by Dan Voiculescu in the 1980s [1], is a noncommutative probability framework that provides a powerful tool for studying random matrices, operator algebras, and related fields. Unlike classical probability, which relies on independence between random variables, free probability is built on the notion of freeness, a concept analogous to independence but formulated in the context of noncommutative algebras. One of the main motivations for free probability arises from random matrix theory. As the size of random matrices grows, their spectral distributions exhibit asymptotic behaviors that can be effectively described using free probability techniques. This connection has led to significant applications in areas such as statistical physics, wireless communications, and machine learning.

One of the primary goals of probability theory is to understand how the distribution of random variables changes as a result of various operations and transformations, see [2]. This study investigates the transformation of probability measures, a fundamental concept that is critical to these types of inquiries. This concept is heavily used in a variety of applications, including stochastic processes, statistical inference, and more. The transformation of probability measures refers to how the probability distribution of a random variable changes after being subjected to a function, which is often measured. The derivation of distributions of sums of independent variables and applications such as image processing and signal transformation, in which random variables are adjusted to fit certain models or domains, are just two of the many sectors where transformation techniques are widely used.

A number of papers have studied and expanded on the topic of transforming probabilities in the setting of free probability theory, see [3,4,5,6,7,8,9]. Let $\mathcal{P}$ denote the set of real probabilities. The Cauchy–Stieltjes transformation ${\mathcal{G}}_{\nu }(·)$ of $\nu \in \mathcal{P}$ is defined by

$${\mathcal{G}}_{\nu }\left(w\right)=\int {\displaystyle \frac{1}{w-s}}\nu \left(ds\right),w\in \mathbb{C}\setminus \mathrm{supp}\left(\nu \right).$$

For $\nu \in \mathcal{P}$ and $a\in \mathbb{R}$, consider the transformation of $\nu$, denoted $T_a\left(\nu \right)$, defined by

$${\mathcal{F}}_{T_a\left(\nu \right)}\left(w\right)={\mathcal{F}}_{\nu }(w-a)+a,$$

(1)

where

$${\mathcal{F}}_{\nu }\left(w\right)=1/{\mathcal{G}}_{\nu }\left(w\right).$$

(2)

This transformation of measures was investigated in [10] (p. 245). It is a specific instance of the $T_{a,b}$-transformation of measures presented in [11] (where $b=0$ is taken into consideration). The $T_a$-transformation of any measure $\nu$ (with finite moments of all orders) may be viewed in terms of the continued fraction representation of the Cauchy–Stieltjes transformation. That is, for

$${\displaystyle {\mathcal{F}}_{\nu }\left(w\right)=w-{\iota }_0-\frac{{\upsilon }_0}{{\displaystyle w-{\iota }_1-\frac{{\upsilon }_1}{{\displaystyle w-{\iota }_2-\frac{{\upsilon }_2}{{\displaystyle w-{\iota }_3-\frac{{\upsilon }_3}{w-{\iota }_4-...}}}}}}},}$$

we have

$${\displaystyle {\mathcal{F}}_{T_a\left(\nu \right)}\left(w\right)=w-{\iota }_0-\frac{{\upsilon }_0}{{\displaystyle w-{\iota }_1-a-\frac{{\upsilon }_1}{{\displaystyle w-{\iota }_2-a-\frac{{\upsilon }_2}{{\displaystyle w-{\iota }_3-a-\frac{{\upsilon }_3}{w-{\iota }_4-a-...}}}}}}}.}$$

See [10] (pp. 244–245) for more details. Note that $m_0^{T_a\left(\nu \right)}=m_0^{\nu }$ and $Var\left(T_a\left(\nu \right)\right)=Var\left(\nu \right)$.

On the other hand, in free probability, the idea of Cauchy–Stieltjes Kernel (CSK) families (of probabilities) is a new notion that uses the Cauchy–Stieltjes kernel $1/(1-\vartheta x)$. CSK families were studied in [12] for compactly supported probabilities. It has been demonstrated that the mean can be used to parameterize these families. In this parametrization, the variance function (VF) and the mean $m_0^{\nu }$ of $\nu$ uniquely specify the family (and the generating measure $\nu$). The quadratic class of CSK families is investigated, where the VF is a polynomial in the mean m with degree $\le 2$. The free Meixner family ($\mathbb{FMF}$) of probabilities is shown to be the generating measures (for this quadratic class of CSK families). The findings on CSK families are expanded in [13] to include measures $\nu$ with one-sided support boundary. A way is offered to identify the domain of means and the idea of the “pseudo-variance” function (PVF) is presented, which is comparable to the VF but does not have a direct probabilistic interpretation. Furthermore, by establishing a relationship between the free cumulants transformations of the respective generating probabilities, the idea of reciprocity between two CSK families is developed. As a result, a cubic class of CSK families (with support bounded from one side) is characterized. This class is connected to the quadratic class via a reciprocity relation. The so-called free analog of the Letac–Mora class is the generating probability measures for this class of cubic CSK families.

From the standpoint of CSK families and their associated VFs, this paper explores the idea of $T_a$-transformation of measures. In Section 2, we provide preliminary details on the VF of a CSK family to clarify the results to be addressed. This notion will be critical in supporting the numerous outcomes reported in this research. In Section 3, using the VF, we prove that the $\mathbb{FMF}$ (respectively, the free analog of the Letac–Mora class) remains invariant under $T_a$-transformation. In Section 4, we use the VF notion to develop new properties for the $T_a$-transformation, leading to intriguing characteristics for important free probability laws, including the free Poisson and free Gamma laws.

2. The Variance Function

This section is devoted to describe the concept of VF of a CSK family. In fact, the CSK families of probability measures plays a fundamental role in free probability, particularly in characterizing spectral distributions and solving distributional equations in the noncommutative setting. This family is defined through the Cauchy–Stieltjes transform (also known as the Stieltjes transform or resolvent function), which encodes essential structural properties of probability measures in free probability. Denote by ${\mathcal{P}}_{ba}$ (respectively, by ${\mathcal{P}}_c$) the family of (non-degenerate) probabilities having one sided support boundary from above (respectively, with compact support). Let $\mu \in {\mathcal{P}}_{ba}$, then

$${\mathcal{M}}_{\mu }\left(\vartheta \right)=\int {\displaystyle \frac{\mu \left(ds\right)}{1-\vartheta s}}$$

converges for $\vartheta \in [0,{\vartheta }_{+}^{\mu })$ with $1/{\vartheta }_{+}^{\mu }=max\{sup\mathrm{supp}\left(\mu \right),0\}$. The family

$${\mathcal{K}}_{+}\left(\mu \right)=\left\{P_{\vartheta }^{\mu }\left(ds\right)={\displaystyle \frac{\mu \left(ds\right)}{{\mathcal{M}}_{\mu }\left(\vartheta \right)(1-\vartheta s)}}:0<\vartheta <{\vartheta }_{+}^{\mu }\right\}$$

is said the CSK family induced by $\mu$.

The function of means $\vartheta \mapsto k_{\mu }\left(\vartheta \right)=\int s{P}_{\vartheta }^{\mu }\left(ds\right)$ realizes a bijection from $(0,{\vartheta }_{+}^{\mu })$ into the interval $(m_0^{\mu },m_{+}^{\mu })=k_{\mu }\left((0,{\vartheta }_{+}^{\mu })\right)$, which is said to be the (one-sided) mean domain of ${\mathcal{K}}_{+}\left(\mu \right)$; see [13]. The inverse of $k_{\mu }(·)$ is denoted by ${\psi }_{\mu }(·)$ and for $m\in (m_0^{\mu },m_{+}^{\mu })$, write $Q_m^{\mu }\left(ds\right)=P_{{\psi }_{\mu }\left(m\right)}^{\mu }\left(ds\right)$. We obtain the mean parametrization of ${\mathcal{K}}_{+}\left(\mu \right)$:

$${\mathcal{K}}_{+}\left(\mu \right)=\{Q_m^{\mu }\left(ds\right):m\in (m_0^{\mu },m_{+}^{\mu })\}.$$

With $B=B\left(\mu \right)={\displaystyle \frac{1}{{\vartheta }_{+}^{\mu }}}$, it is proved in [13] that $m_0^{\mu }={lim}_{\vartheta \to 0^{+}}{k}_{\mu }\left(\vartheta \right)$ and $m_{+}^{\mu }=B-{lim}_{z\to B^{+}}{\displaystyle \frac{1}{{\mathcal{G}}_{\mu }\left(z\right)}}.$

The CSK family is represented by ${\mathcal{K}}_{-}\left(\mu \right)$ in the case where the support of $\mu$ is bounded from below. We have ${\vartheta }_{-}^{\mu }<\vartheta <0$, where ${\vartheta }_{-}^{\mu }$ can be either $1/A\left(\mu \right)$ or $-\infty$ with $A=A\left(\mu \right)=min\{0,infsupp(\mu \left)\right\}$. The mean domain for ${\mathcal{K}}_{-}\left(\mu \right)$ is $(m_{-}^{\mu },m_0^{\mu })$ with $m_{-}^{\mu }=A-1/{\mathcal{G}}_{\mu }\left(A\right)$. If $\mu \in {\mathcal{P}}_c$, then ${\vartheta }_{-}^{\mu }<\vartheta <{\vartheta }_{+}^{\mu }$ and $\mathcal{K}\left(\mu \right)={\mathcal{K}}_{-}\left(\mu \right)\cup \left\{\mu \right\}\cup {\mathcal{K}}_{+}\left(\mu \right)$ represent the two-sided CSK family.

The VF is (see [12])

$$m\mapsto {\mathcal{V}}_{\mu }\left(m\right)=\int {(s-m)}^2{Q}_m^{\mu }\left(ds\right).$$

In the case where $\mu$ has not a moment of order 1, all measures from ${\mathcal{K}}_{+}\left(\mu \right)$ have infinite variance. The notion of PVF, denoted ${\mathbb{V}}_{\mu }(·)$, is presented in [13] as ${\mathbb{V}}_{\mu }\left(m\right)=m\left(1/{\psi }_{\mu }\left(m\right)-m\right).$

If $m_0^{\mu }=\int s\mu \left(ds\right)$ is finite, the VF exists and we have (see [13])

$${\mathbb{V}}_{\mu }\left(m\right)={\displaystyle \frac{m{\mathcal{V}}_{\mu }\left(m\right)}{m-m_0^{\mu }}}.$$

(3)

To conclude this part, some relevant facts are offered in the following remark to aid in the proof of the article’s primary findings.

Remark 1.

(i) A measure $\mu \in {\mathcal{P}}_{ba}$ is determined by ${\mathbb{V}}_{\mu }(·)$. If we set $\Sigma =\Sigma \left(m\right)=m+{\displaystyle \frac{{\mathbb{V}}_{\mu }\left(m\right)}{m}},$ then

$${\mathcal{G}}_{\mu }\left(\Sigma \right)={\displaystyle \frac{m}{{\mathbb{V}}_{\mu }\left(m\right)}}.$$

(4)

If $m_0^{\mu }$ is finite, then

$${\mathcal{G}}_{\mu }\left(\Sigma \right)={\displaystyle \frac{m-m_0^{\mu }}{{\mathcal{V}}_{\mu }\left(m\right)}}.$$

(5)

Thus, ${\mathcal{V}}_{\mu }(·)$ and $m_0^{\mu }$ characterizes μ.

(ii) Let $f\left(\mu \right)$ denote the image of μ by $f:s\mapsto \zeta s+\gamma$, where $\zeta \ne 0$ and $\gamma \in \mathbb{R}$. Then, ∀m close sufficiently to $m_0^{f\left(\mu \right)}=f\left(m_0^{\mu }\right)=\zeta m_0^{\mu }+\gamma$,

$${\mathbb{V}}_{f\left(\mu \right)}\left(m\right)={\displaystyle \frac{{\zeta }^{2}m}{m-\gamma }}{\mathbb{V}}_{\mu }\left({\displaystyle \frac{m-\gamma }{\zeta }}\right).$$

(6)

If ${\mathcal{V}}_{\mu }(·)$ exists, then

$${\mathcal{V}}_{f\left(\mu \right)}\left(m\right)={\zeta }^2{\mathcal{V}}_{\mu }\left({\displaystyle \frac{m-\gamma }{\zeta }}\right).$$

(7)

(iii) From [13] (Corollary 3.6), one has $m/{\mathbb{V}}_{\mu }\left(m\right)\stackrel{m\searrow m_0^{\mu }}{\to }0$ and $m^2/{\mathbb{V}}_{\mu }\left(m\right)\stackrel{m\searrow m_0^{\mu }}{\to }0.$

(iv) The free cumulant transformation is the function defined by the relation [14]

$${\mathcal{R}}_{\mu }\left({\mathcal{G}}_{\mu }\left(w\right)\right)=w-{\displaystyle \frac{1}{{\mathcal{G}}_{\mu }\left(w\right)}},wclose to\infty .$$

Furthermore, if the variance of μ is finite, then according to [15], we have

$${\mathcal{R}}_{\mu }\left(\xi \right)=m_0^{\mu }+Var\left(\mu \right)\xi +\xi \epsilon \left(\xi \right),with\epsilon \left(\xi \right)\stackrel{\xi \to 0}{\to }0.$$

(8)

3. Notes on CSK Families with Quadratic VF

The class of quadratic CSK families with VF

$${\mathcal{V}}_{\mu }\left(m\right)=1+\gamma m+\lambda m^2,\gamma \in \mathbb{R},\lambda \ge -1,$$

(9)

with $m_0^{\mu }=0$, was described in [12]. The relative laws correspond to the $\mathbb{FMF}$:

$$\mu \left(d\zeta \right)={\displaystyle \frac{\sqrt{4(1+\lambda )-{(\zeta -\gamma )}^2}}{2\pi (\lambda {\zeta }^2+\gamma \zeta +1)}}{1}_{(\gamma -2\sqrt{1+\lambda },\gamma +2\sqrt{1+\lambda })}\left(\zeta \right)d\zeta +k_1{\delta }_{{\zeta }_1}+k_2{\delta }_{{\zeta }_2}.$$

We have the following:

(i)

if $\lambda =0,{\gamma }^2>1$, then $k_1=1-1/{\gamma }^2,{\zeta }_1=-1/\gamma ,k_2=0$.

(ii)

If $\lambda >0$ and ${\gamma }^2>4\lambda$, then $k_1=max\{0,1-{\displaystyle \frac{\left|\gamma \right|-\sqrt{{\gamma }^2-4\lambda }}{2\lambda \sqrt{{\gamma }^2-4\lambda }}}\},k_2=0$, and ${\zeta }_1=\pm {\displaystyle \frac{\left|\gamma \right|-\sqrt{{\gamma }^2-4\lambda }}{2\lambda }}$ with the sign opposite to the sign of $\gamma$.

(iii)

If $-1\le \gamma <0$, then

$${\zeta }_{1,2}={\displaystyle \frac{-\gamma \pm \sqrt{{\gamma }^2-4\lambda }}{2\lambda }},k_{1,2}=1+{\displaystyle \frac{\sqrt{{\gamma }^2-4\lambda }\mp \gamma }{2\lambda \sqrt{{\gamma }^2-4\lambda }}}.$$

Important laws are covered by this finding. Up to a dilation and convolution, measure $\mu$ is as follows:

(i)

The semicircle (free Gaussian) law if $\gamma =\lambda =0$.

(ii)

The Marchenko–Pastur (free Poisson) type law if $\lambda =0$ and $\gamma \ne 0$.

(iii)

The free Pascal (free negative binomial) type law if $\lambda >0$ and ${\gamma }^2>4\lambda$.

(iv)

The free Gamma type law if $\lambda >0$ and ${\gamma }^2=4\lambda$.

(v)

The free analog of hyperbolic type law if $\lambda >0$ and ${\gamma }^2<4\lambda$.

(vi)

The free binomial type law if $-1\le \lambda <0$.

The following result gives how the VF of a CSK family changes when considering $T_a$-transformation of the generating measure.

Proposition 1.

Let $\nu \in {\mathcal{P}}_{ba}$. For $a\in \mathbb{R}$, consider the $T_a$-transformation defined by (1). Then, for u close to $m_0^{T_a\left(\nu \right)}=m_0^{\nu }$,

$${\mathbb{V}}_{T_a\left(\nu \right)}\left(u\right)={\mathbb{V}}_{\nu }\left(u\right)+au.$$

(10)

If $m_0^{\nu }$ is finite,

$${\mathcal{V}}_{T_a\left(\nu \right)}\left(u\right)={\mathcal{V}}_{\nu }\left(u\right)+a(u-m_0^{\nu }).$$

(11)

Proof. 

Combining (2) with (4), we obtain

$${\mathcal{F}}_{\nu }\left(u+{\displaystyle \frac{{\mathbb{V}}_{\nu }\left(u\right)}{u}}\right)={\displaystyle \frac{{\mathbb{V}}_{\nu }\left(u\right)}{u}}.$$

(12)

Using (1) and (12), we get

$${\mathcal{F}}_{T_a\left(\nu \right)}\left(u+{\displaystyle \frac{{\mathbb{V}}_{\nu }\left(u\right)+au}{u}}\right)={\mathcal{F}}_{\nu }\left(u+{\displaystyle \frac{{\mathbb{V}}_{\nu }\left(u\right)+au}{u}}-a\right)+a={\mathcal{F}}_{\nu }\left(u+{\displaystyle \frac{{\mathbb{V}}_{\nu }\left(u\right)}{u}}\right)+a={\displaystyle \frac{{\mathbb{V}}_{\nu }\left(u\right)+au}{u}}.$$

Then, relation (10) can be obtained from (12) by the use of measure $T_a\left(\nu \right)$ instead of $\nu$. Furthermore, if $m_0^{\nu }$ is finite, then relation (11) is obtained by combining (10) and (3). □

Next, we demonstrate the invariance property of the $\mathbb{FMF}$ when applying $T_a$-transformation.

Theorem 1.

Suppose $\nu \in \mathbb{FMF}$. Then, for each $a\in \mathbb{R}$, $T_a\left(\nu \right)\in \mathbb{FMF}$.

Proof. 

Assume that $\nu \in \mathbb{FMF}$. Then, for for u close to 0, the VF of $\mathcal{K}\left(\nu \right)$ is of the form

$${\mathcal{V}}_{\nu }\left(u\right)=1+{\gamma }^{\prime }u+{\lambda }^{\prime }{u}^2,{\gamma }^{\prime }\in \mathbb{R},{\lambda }^{\prime }\ge -1.$$

(13)

Combining (13) and (11), for u close to $m_0^{T_a\left(\nu \right)}=0$, we obtain

$${\mathcal{V}}_{T_a\left(\nu \right)}\left(u\right)=1+({\gamma }^{\prime }+a)u+{\lambda }^{\prime }{u}^2,$$

(14)

which is a VF of the form (9). So, $T_a\left(\nu \right)\in \mathbb{FMF}$. □

We will use specific laws to demonstrate the significance of Theorem 1.

Corollary 1.

Consider $\nu \left(d\xi \right)={\displaystyle \frac{1}{2}}{\delta }_{-1}+{\displaystyle \frac{1}{2}}{\delta }_1$. Then, $T_a\left(\nu \right)$ is the free binomial law with $\gamma =a$ and $\lambda =-1$.

Proof. 

The VF of the symmetric Bernoulli CSK family is ${\mathcal{V}}_{\nu }\left(m\right)=1-m^2$. From (14), we have that

$${\mathcal{V}}_{T_a\left(\nu \right)}\left(u\right)=1+au-u^2.$$

(15)

By the use of (5), an identification between (15) and (9) gives that $T_a\left(\nu \right)$ is a the free binomial law with $\gamma =a$ and $\lambda =-1$. □

Corollary 2.

Consider the semicircle measure

$${\displaystyle \nu \left(d\xi \right)=\frac{\sqrt{4-{\xi }^2}}{2\pi }{\mathbf{1}}_{(-2,2)}\left(\xi \right)d\xi .}$$

(16)

So, $T_a\left(\nu \right)$ is

(i) the semicircle measure if $a=0$.

(ii) the free Poisson measure with $\gamma =a$ and $\lambda =0$ if $a\ne 0$.

Proof. 

We know that ${\gamma }^{\prime }={\lambda }^{\prime }=0$ for the VF of the semicircle CSK family provided by (13). Relation (14) gives

$${\mathcal{V}}_{T_a\left(\nu \right)}\left(u\right)=1+au.$$

(17)

An identification between (17) and (9) gives the following:

(i) If $a=0$, then $T_a\left(\nu \right)=\nu$.

(ii) If $a\ne 0$, then $T_a\left(\nu \right)$ is the free Poisson measure with $\gamma =a$ and $\lambda =0$.

□

Corollary 3.

For ${\gamma }^{\prime }\ne 0$ and ${\lambda }^{\prime }=0$, consider the free Poisson measure

$$\nu \left(d\xi \right)={\displaystyle \frac{\sqrt{4-{(\xi -{\gamma }^{\prime })}^2}}{2\pi ({\gamma }^{\prime }\xi +1)}}{\mathbf{1}}_{({\gamma }^{\prime }-2,{\gamma }^{\prime }+2)}\left(\xi \right)d\xi +{(1-1/{\gamma }^{\prime 2})}^{+}{\delta }_{-1/{\gamma }^{\prime }}.$$

(18)

So, $T_a\left(\nu \right)$ is a free Poisson measure with $\gamma ={\gamma }^{\prime }+a$ and $\lambda =0$.

Proof. 

We know that ${\gamma }^{\prime }\ne 0$ and ${\lambda }^{\prime }=0$ for the VF of the free Poisson CSK family provided by (13). Relation (14) gives

$${\mathcal{V}}_{T_a\left(\nu \right)}\left(u\right)=1+({\gamma }^{\prime }+a)u.$$

(19)

An identification between (19) and (9) implies that $T_a\left(\nu \right)$ is the Marchenko–Pastur measure with $\gamma ={\gamma }^{\prime }+a$ and $\lambda =0$. □

Corollary 4.

For ${\gamma }^{\prime }\ne 0$ and ${\lambda }^{\prime }={\gamma }^{\prime 2}/4$, consider the free Gamma measure

$$\nu \left(d\xi \right)={\displaystyle \frac{\sqrt{4(1+{\lambda }^{\prime })-{(\xi -{\gamma }^{\prime })}^2}}{2\pi ({\lambda }^{\prime }{\xi }^2+{\gamma }^{\prime }\xi +1)}}{\mathbf{1}}_{({\gamma }^{\prime }-2\sqrt{{\lambda }^{\prime }+1},{\gamma }^{\prime }+2\sqrt{{\lambda }^{\prime }+1})}\left(\xi \right)d\xi .$$

(20)

So, $T_a\left(\nu \right)$ is

(i) The free Gamma measure with $\gamma ={\gamma }^{\prime }$ and $\lambda ={\lambda }^{\prime }={\gamma }^{\prime 2}/4$ if $a=0$.

(ii) The free analog of hyperbolic measure with $\gamma ={\gamma }^{\prime }+a$ and $\lambda ={\lambda }^{\prime }={\gamma }^{\prime 2}/4$ if $a\ne 0$ and $a^2+2a{\gamma }^{\prime }<0$.

(iii) The free Pascal measure with $\gamma ={\gamma }^{\prime }+a$ and $\lambda ={\lambda }^{\prime }={\gamma }^{\prime 2}/4$ if $a\ne 0$ and $a^2+2a{\gamma }^{\prime }>0$.

Proof. 

We know that ${\gamma }^{\prime }\ne 0$ and ${\lambda }^{\prime }={\gamma }^{\prime 2}/4$ for the VF of the free Gamma CSK family provided by (13). Relation (14) gives

$${\mathcal{V}}_{T_a\left(\nu \right)}\left(u\right)=1+({\gamma }^{\prime }+a)u+{\displaystyle \frac{{\gamma }^{\prime 2}}{4}}{u}^2.$$

(21)

An identification between (21) and (9) gives the following:

(i) If $a=0$, then $T_a\left(\nu \right)=\nu$.

(ii) If $a\ne 0$ and $a^2+2a{\gamma }^{\prime }<0$, then $T_a\left(\nu \right)$ is of the free analog of hyperbolic measure with $\gamma ={\gamma }^{\prime }+a$ and $\lambda ={\lambda }^{\prime }={\gamma }^{\prime 2}/4$.

(iii) If $a\ne 0$ and $a^2+2a{\gamma }^{\prime }>0$, then $T_a\left(\nu \right)$ is of the free Pascal measure with $\gamma ={\gamma }^{\prime }+a$ and $\lambda ={\lambda }^{\prime }={\gamma }^{\prime 2}/4$.

□

4. Notes on CSK Families with Cubic PVF

In [13], an identification is given for a set of cubic CSK families with PVF

$${\mathbb{V}}_{\mu }\left(m\right)=m(\gamma m^2+\lambda m+\rho ),$$

(22)

with $\gamma >0$. The corresponding measures are of the form

$$\mu \left(d\zeta \right)={\displaystyle \frac{\sqrt{4\gamma \rho -{(\lambda +1)}^2-4\gamma \zeta }}{2\pi (\rho +\lambda \zeta +\gamma {\zeta }^2)}}{\mathbf{1}}_{\left(-\infty ,\rho -{\displaystyle \frac{{(\lambda +1)}^2}{\left(4\gamma \right)}}\right)}\left(\zeta \right)d\zeta +p(\gamma ,\lambda ,\rho ){\delta }_{-(\lambda +\sqrt{{\lambda }^2-4\gamma \rho })/\left(2\gamma \right)},$$

(23)

where $p(\gamma ,\lambda ,\rho )=1-1/\sqrt{{\lambda }^2-4\gamma \rho }$ if ${\lambda }^2>4\gamma \rho +1$, and 0 otherwise.

The inverse semicircle law

$$\mu \left(d\zeta \right)={\displaystyle \frac{\sqrt{-1-4\zeta }}{2\pi {\zeta }^2}}{\mathbf{1}}_{(-\infty ,-1/4)}\left(\zeta \right)d\zeta$$

(24)

is the most interesting example in this class. We have ${\mathbb{V}}_{\mu }\left(m\right)=m^3$∀$m\in (m_0^{\mu },m_{+}^{\mu })=(-\infty ,-1)$. It corresponds to the case $\gamma =1$, $\lambda =0$ and $\rho =0$. Similarly, the detailed descriptions of the free equivalent of the Letac–Mora class is provided in [13] (Section 4.2).

Denote by $\mathbb{CPVF}$ the class of measures that generates CSK families with PFV of the form (23). We have the following result.

Theorem 2.

Suppose that $\nu \in \mathbb{CPVF}$, then for each $a\in \mathbb{R}$, $T_a\left(\nu \right)\in \mathbb{CPVF}$.

Proof. 

Suppose that $\nu \in \mathbb{CPVF}$. The PVF of ${\mathcal{K}}_{+}\left(\nu \right)$ is of the form

$${\mathbb{V}}_{\nu }\left(m\right)=m({\gamma }^{\prime }{m}^2+{\lambda }^{\prime }m+{\rho }^{\prime }),{\gamma }^{\prime }>0,{\lambda }^{\prime },{\rho }^{\prime }\in \mathbb{R}.$$

(25)

Combining (25) and (10), ∀m close to $m_0^{T_a\left(\nu \right)}=-\infty$, one has

$${\mathbb{V}}_{T_a\left(\nu \right)}\left(m\right)=m({\gamma }^{\prime }{m}^2+{\lambda }^{\prime }m+{\rho }^{\prime }+a),$$

(26)

which is a cubic PVF in the mean m of the form (22). Thus, $T_a\left(\nu \right)$ belongs to $\mathbb{CPVF}$. □

Corollary 5.

Let ν be the inverse semicircle law (24) with $m_0^{\nu }=-\infty$ and ${\gamma }^{\prime }=1$, ${\lambda }^{\prime }={\rho }^{\prime }=0$. Then, $T_1\left(\nu \right)$ is the free strict arcsine measure with $\gamma =1$, $\lambda =0$ and $\rho =1$.

Proof. 

We know that ${\gamma }^{\prime }=1$ and ${\lambda }^{\prime }={\rho }^{\prime }=0$ for the PVF of the inverse semicircle CSK family provided by (25). From (28), we have that ${\mathbb{V}}_{T_1\left(\nu \right)}\left(m\right)=m(m^2+1),$ which is nothing but the PVF of the free strict arcsine CSK family, see [13] (p. 590). This achieves the proof by (4). □

5. Some Properties of ${\mathit{T}}_{\mathit{a}}$-Transformation

In this section, we provide some results related the Marchenko–Pastur and free Gamma distributions. These results involves the notion of $T_a$-transformation of measures. For $\alpha \in \mathbb{R}\setminus \left\{0\right\}$, consider the function ${\mathbf{H}}_{\alpha }\left(x\right)=\alpha x$. Before presenting and proving the results of this section, we first clarify some facts on the domain of means.

Remark 2.

For $\rho \in {\mathcal{P}}_{ba}$, it is well know that $m_0^{f\left(\rho \right)}=f\left(m_0^{\rho }\right)$ and for $s>0$, in order for ${\rho }^{⊞s}$ to be well defined, we have $m_0^{{\rho }^{⊞s}}=s{m}_0^{\rho }$. However, we do not have a general formula for $m_{+}^{\rho }$ when considering powers of free additive convolution or affine transformation for ρ.

This fact has led the authors of [16] to extend the domain of means preserving the PVF (or preserving the VF in case of existence). The upper end of the extended mean domain is defined as

$${\mathbf{M}}_{+}^{\rho }=inf\left\{m>m_0^{\rho }:{\mathbb{V}}_{\rho }\left(m\right)/m<0\right\}.$$

Note that, for $a\in \mathbb{R}$, we have ${\mathbf{M}}_{+}^{{\mathbf{H}}_a\left(\rho \right)}=a{\mathbf{M}}_{+}^{\rho }$ and ${\mathbf{M}}_{+}^{{\rho }^{⊞s}}=s{\mathbf{M}}_{+}^{\rho }$. In the rest of this section, for $\rho \in {\mathcal{P}}_{ba}$, we will consider the domain of means of the form $(m_0^{\rho },{\mathbf{M}}_{+}^{\rho })$.

Next, we present and prove the section’s primary findings.

Theorem 3.

Let $\nu \in {\mathcal{P}}_{ba}$ with finite first moment $m_0^{\nu }$. Introduce the family of probability measures $T_a\left({\mathcal{K}}_{+}\left(\nu \right)\right)=\left\{T_a\left(Q_m^{\nu }\right)\left(ds\right)\right\}$. Then:

(i) For $a>0$, if $T_a\left({\mathcal{K}}_{+}\left(\nu \right)\right)={\mathcal{K}}_{+}\left({\mathbf{H}}_a\left(\nu \right)\right)$, then (up to affinity) ν is of the free Gamma measure.

(ii) For $a>1$, if $T_a\left({\mathcal{K}}_{+}\left(\nu \right)\right)={\mathcal{K}}_{+}\left({\nu }^{⊞a}\right)$, then (up to affinity) ν is of the free Poisson measure.

Proof. 

(i) For $a>0$, suppose that $T_a\left({\mathcal{K}}_{+}\left(\nu \right)\right)={\mathcal{K}}_{+}\left({\mathbf{H}}_t\left(\nu \right)\right)$. Then, ∀$m\in (m_0^{\nu },{\mathbf{M}}_{+}^{\nu })$, there is $r\in (m_0^{{\mathbf{H}}_a\left(\nu \right)},{\mathbf{M}}_{+}^{{\mathbf{H}}_a\left(\nu \right)})=(a{m}_0^{\nu },a{\mathbf{M}}_{+}^{\nu })$ so that

$$T_a\left(Q_m^{\nu }\right)=Q_r^{{\mathbf{H}}_a\left(\nu \right)}.$$

(27)

In terms of the free cumulant transform this may be written as

$${\mathcal{R}}_{T_a\left(Q_m^{\nu }\right)}\left(\zeta \right)={\mathcal{R}}_{Q_r^{{\mathbf{H}}_a\left(\nu \right)}}\left(\zeta \right),\forall \zeta \mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e} \mathrm{t}\mathrm{o}0.$$

(28)

Using (8), we have

$${\mathcal{R}}_{Q_r^{{\mathbf{H}}_a\left(\nu \right)}}\left(\zeta \right)=m_0^{Q_r^{{\mathbf{H}}_a\left(\nu \right)}}+Var\left(Q_r^{{\mathbf{H}}_a\left(\nu \right)}\right)\zeta +\zeta \epsilon \left(\zeta \right)=r+{\mathcal{V}}_{{\mathbf{H}}_a\left(\nu \right)}\left(r\right)\zeta +\zeta \epsilon \left(\zeta \right),with\epsilon \left(\zeta \right)\stackrel{\zeta \to 0}{\to }0.$$

(29)

We also have, knowing that $Var\left(T_a\left(Q_m^{\nu }\right)\right)=Var\left(Q_m^{\nu }\right)={\mathcal{V}}_{\nu }\left(m\right)$,

$${\mathcal{R}}_{T_a\left(Q_m^{\nu }\right)}\left(\zeta \right)=m_0^{T_a\left(Q_m^{\nu }\right)}+Var\left(T_a\left(Q_m^{\nu }\right)\right)\zeta +\zeta {\epsilon }_1\left(\zeta \right)=m+{\mathcal{V}}_{\nu }\left(m\right)\zeta +\zeta {\epsilon }_1\left(\zeta \right),with{\epsilon }_1\left(\zeta \right)\stackrel{\zeta \to 0}{\to }0.$$

(30)

Combining (28)–(30), we obtain

$$m+{\mathcal{V}}_{\nu }\left(m\right)\zeta +\zeta {\epsilon }_1\left(\zeta \right)=r+{\mathcal{V}}_{{\mathbf{H}}_a\left(\nu \right)}\left(r\right)\zeta +\zeta \epsilon \left(\zeta \right).$$

This implies that $r=m$, and then ${\mathcal{V}}_{{\mathbf{H}}_a\left(\nu \right)}\left(m\right)={\mathcal{V}}_{\nu }\left(m\right)$. This together with (7) implies that

$$a^2{\mathcal{V}}_{\nu }(m/a)={\mathcal{V}}_{\nu }\left(m\right)\forall m\in (m_0^{\nu },{\mathbf{M}}_{+}^{\nu })and\forall a>0.$$

(31)

As $\nu$ is non degenerate by assumption, then ${\mathcal{V}}_{\nu }(·)\ne 0$. Thus, relation (31) gives that ${\mathcal{V}}_{\nu }\left(m\right)=\kappa m^2$ for some $\kappa >0$.

• If $m_0^{\nu }=0$, there cannot exist a VF $\mathcal{V}(·)$ of the type $\mathcal{V}\left(m\right)=\kappa m^2$, where $\kappa >0$. By contraposition, we demonstrate this fact. Assume that the VF of ${\mathcal{K}}_{+}\left(\nu \right)$ with $m_0^{\nu }=0$ is $\mathbb{V}\left(m\right)=\mathcal{V}\left(m\right)=\kappa m^2$ with $\kappa >0$. In contrast to Remark 1(iv), we have that

$${\displaystyle \frac{m^2}{\mathcal{V}\left(m\right)}}={\displaystyle \frac{m^2}{\mathbb{V}\left(m\right)}}={\displaystyle \frac{1}{\kappa }}\stackrel{m\to 0}{\to }{\displaystyle \frac{1}{\kappa }}\ne 0.$$

• If $m_0^{\nu }\ne 0$, then $\nu$ is the image by $s\mapsto m_0^{\nu }({\displaystyle \frac{{\gamma }^{\prime }}{2}}s+1)$ of the free Gamma law (20). In this case, $\kappa ={\displaystyle \frac{{\gamma }^{\prime 2}}{4}}={\lambda }^{\prime }$.

Remark 3.

In the case of the free Gamma law, we have ${\mathbf{M}}_{+}^{\nu }=+\infty .$ Suppose that $m_0^{\nu }>0$ and $a\in (0,1]$. Then, ∀$m\in (m_0^{\nu },+\infty )$, we have $r=m\in (m_0^{\nu },+\infty )\subset (a{m}_0^{\nu },+\infty )$. Thus, relation (27) is well defined. If $m_0^{\nu }>0$ and $a>1$, relation $T_a\left({\mathcal{K}}_{+}\left(\nu \right)\right)={\mathcal{K}}_{+}\left({\mathbf{H}}_a\left(\nu \right)\right)$ may be interpreted as: ∀$r\in (m_0^{{\mathbf{H}}_a\left(\nu \right)},{\mathbf{M}}_{+}^{{\mathbf{H}}_a\left(\nu \right)})=(a{m}_0^{\nu },a{\mathbf{M}}_{+}^{\nu })=(a{m}_0^{\nu },+\infty )$, where $m\in (m_0^{\nu },{\mathbf{M}}_{+}^{\nu })=(m_0^{\nu },+\infty )$ such that relation (27) holds. This implies that $r=m$. Thus, ∀$r\in (a{m}_0^{\nu },+\infty )$ we have $m=r\in (a{m}_0^{\nu },+\infty )\subset (m_0^{\nu },+\infty )$. Thus, relation (27) is well defined. If $m_0^{\nu }<0$, the same arguments implies that relation (27) is well defined.

(ii) For $a>1$, suppose that $T_a\left({\mathcal{K}}_{+}\left(\nu \right)\right)={\mathcal{K}}_{+}\left({\nu }^{⊞a}\right)$. Then, ∀$m\in (m_0^{\nu },{\mathbf{M}}_{+}^{\nu })$, there is $r\in (m_0^{{\nu }^{⊞a}},{\mathbf{M}}_{+}^{{\nu }^{⊞a}})=(a{m}_0^{\nu },a{\mathbf{M}}_{+}^{\nu })$ so that

$$T_a\left(Q_m^{\nu }\right)=Q_r^{{\nu }^{⊞a}}.$$

(32)

In terms of the free cumulant transform, this means that

$${\mathcal{R}}_{T_a\left(Q_m^{\nu }\right)}\left(\zeta \right)={\mathcal{R}}_{Q_r^{{\nu }^{⊞a}}}\left(\zeta \right),\forall \zeta \mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e} \mathrm{t}\mathrm{o}0.$$

(33)

Or equivalently,

$$m+{\mathcal{V}}_{\nu }\left(m\right)\zeta +\zeta {\epsilon }_1\left(\zeta \right)=r+{\mathcal{V}}_{{\nu }^{⊞a}}\left(r\right)\zeta +\zeta {\epsilon }_3\left(\zeta \right)with{\epsilon }_3\left(\zeta \right)\stackrel{\zeta \to 0}{\to }0.$$

(34)

This implies that $r=m$ and then ${\mathcal{V}}_{{\nu }^{⊞a}}\left(m\right)={\mathcal{V}}_{\nu }\left(m\right)$. Together with the fact that ${\mathcal{V}}_{{\nu }^{⊞a}}\left(m\right)=a{\mathcal{V}}_{\nu }(m/a)$ (see [13] (Proposition 3.10, Equation (3.17))), this implies that

$$a{\mathcal{V}}_{\nu }(m/a)={\mathcal{V}}_{\nu }\left(m\right),\forall m\in (m_0^{\nu },{\mathbf{M}}_{+}^{\nu })and\forall a>1.$$

(35)

As $\nu$ is non degenerate by assumption, then ${\mathcal{V}}_{\nu }(·)\ne 0$. Thus, relation (35) gives that ${\mathcal{V}}_{\nu }\left(m\right)=\eta m$ for some $\eta \ne 0$.

• If $m_0^{\nu }=0$, then $\mathcal{V}\left(m\right)=\eta m$ with $\eta \ne 0$ cannot be a VF (see [17] (p. 6)).

• If $m_0^{\nu }\ne 0$, then $\nu$ is the image by $s\mapsto m_0^{\nu }(1+{\gamma }^{\prime }s)$ of the free Poisson law provided by (18). In this case, $\eta ={\gamma }^{\prime 2}{m}_0^{\nu }$.

Remark 4.

In the case of the free Poisson law, we have ${\mathbf{M}}_{+}^{\nu }=+\infty .$ Suppose that $m_0^{\nu }<0$. Then, ∀$m\in (m_0^{\nu },+\infty )$, we have $r=m\in (m_0^{\nu },+\infty )\subset (a{m}_0^{\nu },+\infty )$. Thus, relation (32) is well defined.

If $m_0^{\nu }>0$, relation $T_a\left({\mathcal{K}}_{+}\left(\nu \right)\right)={\mathcal{K}}_{+}\left({\nu }^{⊞a}\right)$ may be interpreted as: ∀$r\in (m_0^{{\nu }^{⊞a}},{\mathbf{M}}_{+}^{{\nu }^{⊞a}})=(a{m}_0^{\nu },a{\mathbf{M}}_{+}^{\nu })=(a{m}_0^{\nu },+\infty )$, there is $m\in (m_0^{\nu },{\mathbf{M}}_{+}^{\nu })=(m_0^{\nu },+\infty )$ such that relation (32) hold. This implies that $r=m$. Thus, ∀$r\in (a{m}_0^{\nu },+\infty )$ we have $m=r\in (a{m}_0^{\nu },+\infty )\subset (m_0^{\nu },+\infty )$. Thus, relation (32) is well defined.

Next, we demonstrate that the inverse implication in Theorem 3(i) is invalid. Assume that $m_0^{\nu }\ne 0$ and $\nu$ is the image by $s\mapsto m_0^{\nu }({\displaystyle \frac{{\gamma }^{\prime }}{2}}s+1)$ of the free Gamma law given by (20). We show that

$$T_a\left(Q_m^{\nu }\right)\ne Q_m^{{\mathbf{H}}_a\left(\nu \right)}.$$

(36)

We have that $m_0^{T_a\left(Q_m^{\nu }\right)}=m=m_0^{Q_m^{{\mathbf{H}}_a\left(\nu \right)}}$. Then, $\varsigma >0$ exists so that ${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}(·)$ and ${\mathbb{V}}_{Q_m^{{\mathbf{H}}_a\left(\nu \right)}}(·)$ are well defined on $(m,m+\varsigma )$. For $v\in (m_0^{\nu },+\infty )$, we have

$${\mathbb{V}}_{\nu }\left(v\right)={\displaystyle \frac{{\gamma }^{\prime 2}{v}^3}{v-m_0^{\nu }}}.$$

(37)

We calculate ${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}(·)$. In [16], (Theorem 3.2), a relationship is established between ${\mathbb{V}}_{Q_m^{\nu }}(·)$ and ${\mathbb{V}}_{\nu }(·)$. Using [16] (Theorem 3.2, Equation (2.9)), we get

$$y={\displaystyle \frac{v^2\left(\frac{{\gamma }^{\prime 2}{m}^3}{m-m_0^{\nu }}\right)-m^2\left(\frac{{\gamma }^{\prime 2}{v}^3}{v-m_0^{\nu }}\right)}{v\left(\frac{{\gamma }^{\prime 2}{m}^3}{m-m_0^{\nu }}\right)-m\left(\frac{{\gamma }^{\prime 2}{v}^3}{v-m_0^{\nu }}\right)}}={\displaystyle \frac{vm{m}_0^{\nu }}{m_0^{\nu }(v+m)-vm}}.$$

(38)

Using [16] (Theorem 3.2, Equation (2.10)), we obtain for $y\ne m$

$${\mathbb{V}}_{Q_m^{\nu }}\left(y\right)=y\left({\displaystyle \frac{{\mathbb{V}}_{\nu }\left(v\right)}{v}}+v-y\right).$$

(39)

From (38), we have

$$v={\displaystyle \frac{ym{m}_0^{\nu }}{ym+m{m}_0^{\nu }-y{m}_0^{\nu }}}$$

(40)

For $y\ne m$, Equations (37), (39), and (40) give

$${\mathbb{V}}_{Q_m^{\nu }}\left(y\right)={\displaystyle \frac{y^3({\gamma }^{\prime 2}{m}^2-(y-m)(m-m_0^{\nu }))}{(y-m)(y(m-m_0^{\nu })+m{m}_0^{\nu })}}$$

(41)

Combining (41) and (10), we get

$${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}\left(y\right)={\mathbb{V}}_{Q_m^{\nu }}\left(y\right)+ay={\displaystyle \frac{y^3({\gamma }^{\prime 2}{m}^2-(y-m)(m-m_0^{\nu }))+ay(y-m)(y(m-m_0^{\nu })+m{m}_0^{\nu })}{(y-m)(y(m-m_0^{\nu })+m{m}_0^{\nu })}}.$$

(42)

Now, we calculate ${\mathbb{V}}_{Q_m^{{\mathbf{H}}_a\left(\nu \right)}}(·)$. For $u\in (t{m}_0^{\nu },+\infty )$, we have

$${\mathbb{V}}_{{\mathbf{H}}_a\left(\nu \right)}\left(u\right)=a^2{\mathbb{V}}_{\nu }(u/a)={\displaystyle \frac{{\gamma }^{\prime 2}{u}^3}{u-a{m}_0^{\nu }}}.$$

(43)

Using [16] (Theorem 3.2, Equation (2.9)), we get

$$y={\displaystyle \frac{u^2\left(\frac{{\gamma }^{\prime 2}{m}^3}{m-a{m}_0^{\nu }}\right)-m^2\left(\frac{{\gamma }^{\prime 2}{u}^3}{u-a{m}_0^{\nu }}\right)}{u\left(\frac{{\gamma }^{\prime 2}{m}^3}{m-a{m}_0^{\nu }}\right)-m\left(\frac{{\gamma }^{\prime 2}{u}^3}{u-a{m}_0^{\nu }}\right)}}={\displaystyle \frac{uma{m}_0^{\nu }}{a{m}_0^{\nu }(u+m)-um}}.$$

(44)

Using [16] (Theorem 3.2, Equation (2.10)), we obtain for $y\ne m$

$${\mathbb{V}}_{Q_m^{{\mathbf{H}}_a\left(\nu \right)}}\left(y\right)=y\left({\displaystyle \frac{{\mathbb{V}}_{{\mathbf{H}}_a\left(\nu \right)}\left(u\right)}{u}}+u-y\right).$$

(45)

From (44), we have

$$u={\displaystyle \frac{yma{m}_0^{\nu }}{ym+ma{m}_0^{\nu }-ya{m}_0^{\nu }}}$$

(46)

For $y\ne m$, Equations (43), (45), and (46) give

$${\mathbb{V}}_{Q_m^{{\mathbf{H}}_a\left(\nu \right)}}\left(y\right)={\displaystyle \frac{y^3({\gamma }^{\prime 2}{m}^2-(y-m)(m-a{m}_0^{\nu }))}{(y-m)(y(m-a{m}_0^{\nu })+ma{m}_0^{\nu })}}.$$

(47)

It is clear from (42) and (47) that ${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}\left(y\right)\ne {\mathbb{V}}_{Q_m^{{\mathbf{H}}_a\left(\nu \right)}}\left(y\right)$, ∀$y\in (m,m+\varsigma )$. This achieve the proof of (36) by (4).

Next, we demonstrate that the inverse implication in Theorem 3(ii) is also invalid. Assume that $m_0^{\nu }\ne 0$, and $\nu$ is the image by $s\mapsto m_0^{\nu }(1+a^{\prime }s)$ of the free Poisson law provided by (18). We show that

$$T_a\left(Q_m^{\nu }\right)\ne Q_m^{{\nu }^{⊞a}}.$$

(48)

We have that $m_0^{T_a\left(Q_m^{\nu }\right)}=m=m_0^{Q_m^{{\nu }^{⊞a}}}$. Then, $\varsigma >0$ exists so that ${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}(·)$ and ${\mathbb{V}}_{Q_m^{{\nu }^{⊞a}}}(·)$ are well defined on $(m,m+\varsigma )$.

We first determine ${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}(·)$. For $x\in (m_0^{\nu },+\infty )$, we have

$${\mathbb{V}}_{\nu }\left(x\right)={\displaystyle \frac{a^{\prime 2}{m}_0^{\nu }{x}^2}{x-m_0^{\nu }}}.$$

(49)

Using [16] (Theorem 3.2, Equation (2.9)), we get

$$y={\displaystyle \frac{x^2\left(\frac{a^{\prime 2}{m}_0^{\nu }{m}^2}{m-m_0^{\nu }}\right)-m^2\left(\frac{a^{\prime 2}{m}_0^{\nu }{x}^2}{x-m_0^{\nu }}\right)}{x\left(\frac{a^{\prime 2}{m}_0^{\nu }{m}^2}{m-m_0^{\nu }}\right)-m\left(\frac{a^{\prime 2}{m}_0^{\nu }{x}^2}{x-m_0^{\nu }}\right)}}={\displaystyle \frac{mx}{m_0^{\nu }}}.$$

(50)

Using [16] (Theorem 3.2, Equation (2.10)), we obtain for $y\ne m$

$${\mathbb{V}}_{{\mathbb{Q}}_m^{\nu }}\left(y\right)=y\left({\displaystyle \frac{{\mathbb{V}}_{\nu }\left(x\right)}{x}}+x-y\right).$$

(51)

From (50), we have

$$x={\displaystyle \frac{y{m}_0^{\nu }}{m}}$$

(52)

For $y\ne m$, Equations (49), (51), and (52) give

$${\mathbb{V}}_{Q_m^{\nu }}\left(y\right)=y^2\left({\displaystyle \frac{a^{\prime 2}{m}_0^{\nu }}{y-m}}+{\displaystyle \frac{m_0^{\nu }}{m}}-1\right).$$

(53)

Thus,

$${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}\left(y\right)={\mathbb{V}}_{Q_m^{\nu }}\left(y\right)+ay=y^2\left({\displaystyle \frac{a^{\prime 2}{m}_0^{\nu }}{y-m}}+{\displaystyle \frac{m_0^{\nu }}{m}}-1\right)+ay.$$

(54)

Now, we calculate ${\mathbb{V}}_{Q_m^{{\nu }^{⊞a}}}(·)$. From the fact that ${\mathbb{V}}_{{\nu }^{⊞a}}\left(m\right)=a{\mathbb{V}}_{\nu }(m/a)$, (see [13] (Proposition 3.10, Equation (3.17))), we get

$${\mathbb{V}}_{{\nu }^{⊞a}}\left(x\right)={\displaystyle \frac{a^{\prime 2}{m}_0^{\nu }{x}^2}{x-a{m}_0^{\nu }}}.$$

(55)

Using [16] (Theorem 3.2, Equation (2.9)), we get

$$y={\displaystyle \frac{x^2\left(\frac{a^{\prime 2}{m}_0^{\nu }{m}^2}{m-a{m}_0^{\nu }}\right)-m^2\left(\frac{a^{\prime 2}{m}_0^{\nu }{x}^2}{x-a{m}_0^{\nu }}\right)}{x\left(\frac{a^{\prime 2}{m}_0^{\nu }{m}^2}{m-a{m}_0^{\nu }}\right)-m\left(\frac{a^{\prime 2}{m}_0^{\nu }{x}^2}{x-a{m}_0^{\nu }}\right)}}={\displaystyle \frac{xm}{a{m}_0^{\nu }}}.$$

(56)

Using [16] (Theorem 3.2, Equation (2.10)), we obtain for $y\ne m$

$${\mathbb{V}}_{Q_m^{{\nu }^{⊞a}}}\left(y\right)=y\left({\displaystyle \frac{{\mathbb{V}}_{{\nu }^{⊞a}}\left(x\right)}{x}}+x-y\right).$$

(57)

From (56), we have

$$x={\displaystyle \frac{a{m}_0^{\nu }y}{m}}.$$

(58)

For $y\ne m$, Equations (55), (57), and (58) give

$${\mathbb{V}}_{Q_m^{{\nu }^{⊞a}}}\left(y\right)=y^2\left({\displaystyle \frac{a^{\prime 2}{m}_0^{\nu }}{y-m}}+{\displaystyle \frac{a{m}_0^{\nu }}{m}}-1\right).$$

(59)

It is clear from (54) and (59) that ${\mathbb{V}}_{T_a\left(Q_m^{\nu }\right)}\left(y\right)\ne {\mathbb{V}}_{Q_m^{{\nu }^{⊞a}}}\left(y\right)$, ∀$y\in (m,m+\varsigma )$. This achieves the proof of (48) by (4).

□

6. Conclusions

In this paper, we have proved that the $\mathbb{FMF}$ (respectively, the free analog of the Letac–Mora class) is invariant when applying $T_a$-transformation of measures. Moreover, some new characteristic properties related to the $T_a$-transformation are established, which lead to interesting results for important distributions in free probability such as the free Poisson and free Gamma distributions. These works deepen our understanding of the $T_a$-transformation in the free probability context.