Some Limiting Laws in Non-Commutative Probability

Abstract

In this article, we provide some new limiting laws related to the free multiplicative law of large numbers and involving free and Boolean additive convolutions. Some examples of these limiting laws are presented within the framework of non-commutative probability theory.

1. Introduction

In recent decades, a number of articles have studied limit theorems with respect to the free convolution of probability measures. The concept of freeness is the key notion. It may be viewed (for non-commutative random variables) as a type of independence. Like in classical probability, in which the notion of independence leads to classical convolution, the notion of freeness gives rise to a binary operation for real measures: the free convolution. A lot of classic results for the addition and multiplication of independent random variables possess analog properties for this new theory, for example, the central limit theorem, the Lévy–Khintchine formula, the law of large numbers and others. For an introduction to these subjects, we refer to [1]. In [2], the authors provide the distributional model behavior of the sum of free random variables distributed in an identical manner. They explicitly describe the relation between the limiting laws for classical and free additive convolutions. On the other hand, for measures with bounded support, Tucci [3] proved the limiting distribution for the free multiplicative law of large numbers. This result was extended in [4] to measures with unbounded support. Continuing the study of limiting distributions in non-commutative probability, we provide in this article some new limiting laws related to the free multiplicative law of large numbers and involving free and Boolean additive convolutions. For the clarity of the presentation of our results, we need to first recall some concepts of importance in non-commutative probability.

Denote by $\mathcal{P}$ (respectively, ${\mathcal{P}}_{+}$) the set of probability measures on $\mathbb{R}$ (respectively, ${\mathbb{R}}_{+}$).

The Cauchy–Stieltjes transform ${\mathcal{G}}_{\mu }(.)$ of $\mu \in \mathcal{P}$ is defined, for $y\in \mathbb{C}\backslash \mathrm{supp}\left(\mu \right)$, as

$${\mathcal{G}}_{\mu }\left(y\right)=\int \frac{1}{y-\xi }\mu \left(d\xi \right),$$

(1)

where $\mathrm{supp}\left(\mu \right)$ denotes the support of the measure $\mu$.

The free additive convolution of $\mu$ and $\nu$$\in \mathcal{P}$, denoted by $\mu ⊞\nu$, is defined by

$${\mathcal{R}}_{\mu ⊞\nu }\left(\xi \right)={\mathcal{R}}_{\mu }\left(\xi \right)+{\mathcal{R}}_{\nu }\left(\xi \right),$$

(2)

where the free cumulant transform, ${\mathcal{R}}_{\mu }$, of $\mu$ is given by

$${\mathcal{R}}_{\mu }\left({\mathcal{G}}_{\mu }\left(\xi \right)\right)=\xi -\frac{1}{{\mathcal{G}}_{\mu }\left(\xi \right)},\mathrm{for} \mathrm{all}\xi \mathrm{in} \mathrm{an} \mathrm{appropriate} \mathrm{domain}.$$

(3)

See [5] for more details about the free cumulant transform.

A measure $\mu \in \mathcal{P}$ is ⊞-infinitely divisible if for each $q\in \mathbb{N}$, there exists ${\mu }_q\in \mathcal{P}$ such that

$$\mu =\underset{q\mathrm{times}}{\underbrace{{\mu }_q⊞.....⊞{\mu }_q}}.$$

Denote by ${\mu }^{⊞t}$ the t-fold free additive convolution of $\mu$ with itself. This operation is well defined for all $t\ge 1$, (see [6]) and

$${\mathcal{R}}_{{\mu }^{⊞t}}\left(\xi \right)=t{\mathcal{R}}_{\mu }\left(\xi \right).$$

(4)

A measure $\mu \in \mathcal{P}$ is ⊞-infinitely divisible if ${\mu }^{⊞t}$ is well defined for all $t>0$.

The Boolean additive convolution is another interesting convolution in the theory of non-commutative probability, see [7]. For $\mu$, $\nu \in \mathcal{P}$, the Boolean additive convolution $\mu \uplus \nu$ is the probability measure defined by

$$E_{\mu \uplus \nu }\left(\xi \right)=E_{\mu }\left(\xi \right)+E_{\nu }\left(\xi \right),\mathrm{for}\xi \in {\mathbb{C}}^{+},$$

(5)

where

$$E_{\mu }\left(\xi \right)=\xi -\frac{1}{{\mathcal{G}}_{\mu }\left(\xi \right)},$$

(6)

denotes the Boolean cumulant transform of the measure $\mu$.

A measure $\mu \in \mathcal{P}$ is ⊎-infinitely divisible if for each $q\in \mathbb{N}$, there exists ${\mu }_q\in \mathcal{P}$ such that

$$\mu =\underset{q\mathrm{times}}{\underbrace{{\mu }_q\uplus .....\uplus {\mu }_q}}.$$

Note that all measures $\mu \in \mathcal{P}$ are ⊎-infinitely divisible, see [7], Theorem 3.6.

We come now to the concept of free multiplicative convolution. For $\mu \in {\mathcal{P}}_{+}$, $(\mu \ne {\delta }_0)$, the $\mathcal{S}$-transform is given by

$${\mathcal{R}}_{\mu }\left(\zeta {\mathcal{S}}_{\mu }\left(\zeta \right)\right)=\frac{1}{{\mathcal{S}}_{\mu }\left(\zeta \right)}\mathrm{for} \mathrm{all}\zeta \mathrm{in} a \mathrm{neighborhood} \mathrm{of} 0.$$

The multiplication of $\mathcal{S}$-transforms is also an $\mathcal{S}$-transform. For ${\nu }_1$, ${\nu }_2\in {\mathcal{P}}_{+}$, the multiplicative free convolution ${\nu }_1⊠{\nu }_2$ is defined by ${\mathcal{S}}_{{\nu }_1⊠{\nu }_2}\left(\zeta \right)={\mathcal{S}}_{{\nu }_1}\left(\zeta \right){\mathcal{S}}_{{\nu }_2}\left(\zeta \right)$. Multiplicative-free convolution powers ${\mu }^{⊠p}$ are well defined at least for all $p\ge 1$ (see [8], Theorem 2.17) by ${\mathcal{S}}_{{\mu }^{⊠p}}\left(\zeta \right)={\mathcal{S}}_{\mu }{\left(\zeta \right)}^p$.

Now, we present the notion of the free multiplicative law of large numbers. More precisely, we have the following.

Theorem 1

([4], Theorem 2). Let $\sigma \in {\mathcal{P}}_{+}$ and let ${\varphi }_n:{\mathbb{R}}_{+}⟶{\mathbb{R}}_{+}$ be the map ${\varphi }_n\left(y\right)=y^{1/n}$. Set $\alpha =\sigma \left(\right\{0\left\}\right)$. If we consider

$${\sigma }_n={\varphi }_n(\underset{ntimes}{\underbrace{\sigma ⊠.....⊠\sigma }})$$

and thus ${\sigma }_n$ converges weakly to a probability measure denoted $\Phi \left(\sigma \right)$ on ${\mathbb{R}}_{+}$. If σ is a Dirac measure on ${\mathbb{R}}_{+}$, then $\Phi \left(\sigma \right)=\sigma$. Otherwise, $\Phi \left(\sigma \right)$ is the unique probability measure on ${\mathbb{R}}_{+}$ characterized by

$$\Phi \left(\sigma \right)\left(\left[0,\frac{1}{{\mathcal{S}}_{\sigma }(x-1)}\right]\right)=x$$

(7)

for all $x\in (\alpha ,1)$ and $\Phi \left(\sigma \right)\left(\right\{0\left\}\right)=\alpha$. The support of $\Phi \left(\sigma \right)$ is the closure of the interval

$$(a,b)=\left({\left({\int }_0^{\infty }{y}^{-1}\sigma \left(dy\right)\right)}^{-1},{\int }_0^{\infty }y\sigma \left(dy\right)\right)$$

where $0\le a<b\le \infty$.

In [9] (Theorem 3.1), an interesting description is given for the free multiplicative law of large numbers $\Phi \left(\sigma \right)$ in terms of the pseudo-variance function ${\mathbb{V}}_{\sigma }(.)$ of the Cauchy–Stieltjes kernel (CSK) family generated by $\sigma$ (see the next section for CSK families and the corresponding pseudo-variance functions). A number of explicit examples are given for $\Phi \left(\sigma \right)$, see [9].

In this paper, we are interested in explicitly giving the limiting laws $\Phi \left(D_{1/t}\left({\sigma }^{⊞t}\right)\right)$, $\Phi \left(D_{1/r}\left({\sigma }^{\uplus r}\right)\right)$ and $\Phi \left({\mathbb{B}}_r\left(\sigma \right)\right)$ (for $t>1$ and $r>0$) by means of ${\mathbb{V}}_{\sigma }(.)$. Here, $D_c\left(\sigma \right)$ denotes the dilation of measure $\sigma$ by a number $c\ne 0$ and ${\mathbb{B}}_r\left(\sigma \right)={\left({\sigma }^{⊞1+r}\right)}^{\uplus \frac{1}{1+r}}$. Some calculations of $\Phi \left(D_{1/t}\left({\sigma }^{⊞t}\right)\right)$, $\Phi \left(D_{1/r}\left({\sigma }^{\uplus r}\right)\right)$ and $\Phi \left({\mathbb{B}}_r\left(\sigma \right)\right)$ are provided for measures $\sigma$ of importance in non-commutative probability. Section 2 will describe some basic concepts about CSK families and their pseudo-variance functions. Section 3 is devoted to the main result of this article and concludes with some examples.

2. Cauchy–Stieltjes Kernel Families

We introduce some preliminaries about CSK families and their corresponding pseudo-variance functions, see [10] for more details.

Let $\mu$ be a probability measure which is non-degenerate with support bounded from above. The transform

$${\mathcal{M}}_{\mu }\left(\vartheta \right)=\int \frac{1}{1-\vartheta x}\mu \left(dx\right)$$

(8)

converges to ∀$\vartheta \in [0,{\vartheta }_{+}^{\mu })$ with $\frac{1}{{\vartheta }_{+}^{\mu }}=max\{0,sup\mathrm{supp}\left(\mu \right)\}$. For $\vartheta \in [0,{\vartheta }_{+}^{\mu })$, let

$$P_{\vartheta }^{\mu }\left(dx\right)=\frac{1}{{\mathcal{M}}_{\mu }\left(\vartheta \right)(1-\vartheta x)}\mu \left(dx\right).$$

The one-sided CSK family generated by $\mu$ is the set

$${\mathcal{K}}_{+}\left(\mu \right)=\{P_{\vartheta }^{\mu }\left(dx\right);\vartheta \in (0,{\vartheta }_{+}^{\mu })\}.$$

Denote by $k_{\mu }\left(\vartheta \right)=\int x{P}_{\vartheta }^{\mu }\left(dx\right)$ the mean of $P_{\vartheta }^{\mu }\left(dx\right)$. From [11], we know that the function $\vartheta \mapsto k_{\mu }\left(\vartheta \right)$ is strictly increasing on $(0,{\vartheta }_{+}^{\mu })$. Denote by $(m_0^{\mu },m_{+}^{\mu })$ the image of $(0,{\vartheta }_{+}^{\mu })$ by $k_{\mu }(.)$. It is called the (one-sided) mean domain of ${\mathcal{K}}_{+}\left(\mu \right)$. This gives a re-parametrization (by the mean) of ${\mathcal{K}}_{+}\left(\mu \right)$. Consider ${\psi }_{\mu }(.)$, the inverse of $k_{\mu }(.)$, and for $m\in (m_0^{\mu },m_{+}^{\mu })$, write $Q_m^{\mu }\left(dx\right)=P_{{\psi }_{\mu }\left(m\right)}^{\mu }\left(dx\right)$. Then, we obtain

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

Consider

$$B=B\left(\mu \right)=max\{0,sup\mathrm{supp}\left(\mu \right)\}=\frac{1}{{\vartheta }_{+}^{\mu }}\in [0,\infty ).$$

(9)

(Here, $1/0$ is interpreted as ∞.) It was proven in [11] that

$$m_0^{\mu }=\underset{\vartheta \to 0^{+}}{lim}{k}_{\mu }\left(\vartheta \right)\text{and}{m}_{+}^{\mu }=B-\underset{z\to B^{+}}{lim}\frac{1}{{\mathcal{G}}_{\mu }\left(z\right)}.$$

(10)

If the support of $\mu$ is bounded from below, one may similarly introduce the one-sided CSK family. Denote this family by ${\mathcal{K}}_{-}\left(\mu \right)$. We have ${\vartheta }_{-}^{\mu }<\vartheta <0$, where ${\vartheta }_{-}^{\mu }$ is either $1/A\left(\mu \right)$ or $-\infty$ with $A=A\left(\mu \right)=min\{0,inf\mathrm{supp}(\mu \left)\right\}$. The interval $(m_{-}^{\mu },m_0^{\mu })$ is the mean domain for ${\mathcal{K}}_{-}\left(\mu \right)$ with $m_{-}^{\mu }=A-1/{\mathcal{G}}_{\mu }\left(A\right)$. If $\mu$ is compactly supported, then ${\vartheta }_{-}^{\mu }<\vartheta <{\vartheta }_{+}^{\mu }$ and the two-sided CSK family is $\mathcal{K}\left(\mu \right)={\mathcal{K}}_{+}\left(\mu \right)\cup {\mathcal{K}}_{-}\left(\mu \right)\cup \left\{\mu \right\}$.

The variance function is (see [12])

$$x\mapsto {\mathcal{V}}_{\mu }\left(x\right)=\int {(\xi -x)}^2{Q}_x^{\mu }\left(d\xi \right).$$

(11)

If $\mu$ does not have the first moment, all the laws that belong to the CSK family generated by $\mu$ have infinite variance. The concept of the pseudo-variance function ${\mathbb{V}}_{\mu }(.)$ was introduced in [11]. It is defined by

$${\mathbb{V}}_{\mu }\left(x\right)=x\left(\frac{1}{{\psi }_{\mu }\left(x\right)}-x\right),\forall x\in (m_0^{\mu },m_{+}^{\mu }).$$

(12)

If $m_0^{\mu }=\int \xi \mu \left(d\xi \right)$ is finite, then ${\mathcal{V}}_{\mu }(.)$ exists and (see [11])

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

(13)

Throughout the following two remarks, we recall some facts that will be used in the proof of the main result of the paper given by Theorem 2.

Remark 1.

(i)

Consider $\phi :\xi ⟼\lambda \xi +\beta$, where $\lambda \ne 0$ and $\beta \in \mathbb{R}$, and let $\phi \left(\mu \right)$ be the image of μ by φ. Then, for all x close enough to $m_0^{\phi \left(\mu \right)}=\phi \left(m_0^{\mu }\right)=\lambda m_0^{\mu }+\beta$, we have (see [11])

$${\mathbb{V}}_{\phi \left(\mu \right)}\left(x\right)=\frac{{\lambda }^{2}x}{x-\beta }{\mathbb{V}}_{\mu }\left(\frac{x-\beta }{\lambda }\right).$$

(14)

(ii)

According to [11] (Proposition 3.10), for all $t>0$ such that ${\mu }^{⊞t}$ is defined and for all x close enough to $m_0^{{\mu }^{⊞t}}=t{m}_0^{\mu }$,

$${\mathbb{V}}_{{\mu }^{⊞t}}\left(x\right)=t{\mathbb{V}}_{\mu }(x/t).$$

(15)

(iii)

We know from [13] that for all $r>0$ and for all x close enough to $m_0^{{\mu }^{\uplus r}}=r{m}_0^{\mu }$,

$${\mathbb{V}}_{{\mu }^{\uplus r}}\left(x\right)=r{\mathbb{V}}_{\mu }(x/r)+x^2(1/r-1),$$

(16)

Furthermore, for all $s\ge 0$ and for all x close enough to $m_0^{{\mathbb{B}}_s\left(\mu \right)}=m_0^{\mu }$, we have

$${\mathbb{V}}_{{\mathbb{B}}_s\left(\mu \right)}\left(x\right)={\mathbb{V}}_{\mu }\left(x\right)+s{x}^2.$$

(17)

Remark 2.

For $t>0$ such that ${\mu }^{⊞t}$ is defined, there exists an injective analytic map $w_t:{\mathbb{C}}^{+}⟶{\mathbb{C}}^{+}$, called subordination, such that ${\mathcal{G}}_{{\mu }^{⊞t}}\left(\xi \right)={\mathcal{G}}_{\mu }\left(w_t\left(\xi \right)\right)$, for $\xi \in {\mathbb{C}}^{+}$. Furthermore,

$$w_t\left(\xi \right)=\xi /t+\frac{(1-1/t)}{{\mathcal{G}}_{{\mu }^{⊞t}}\left(\xi \right)}$$

(18)

and $H_t\left(w_t\left(\xi \right)\right)=\xi$, where

$$H_t\left(\xi \right)=t\xi +\frac{(1-t)}{{\mathcal{G}}_{\mu }\left(\xi \right)}.$$

(19)

For more details about the subordination function, see [14] (Theorem 2.5).

3. Main Result

In this section, we state and demonstrate the main result of this paper. Some new limiting distributions are provided in relation to the free multiplicative law of large numbers involving free and Boolean additive convolutions.

Theorem 2.

Let $\nu \in {\mathcal{P}}_{+}$ be non-degenerate. Set $\alpha =\nu \left(\right\{0\left\}\right)$. With the notation introduced above, we have

(i)

For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)={(t\alpha -(t-1))}^{+}{\delta }_0+{\left(\frac{t{x}^2}{{\mathbb{V}}_{\nu }\left(x\right)}\right)}^{\prime }{\mathbf{1}}_{\left(-\frac{w_t\left(0\right)}{t-1},m_0^{\nu }\right)}\left(x\right)dx.$$

(20)

(ii)

For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{\alpha }{r-\alpha (r-1)}{\delta }_0+{\left(\frac{r{x}^2}{{\mathbb{V}}_{\nu }\left(x\right)+(1-r)x^2}\right)}^{\prime }{\mathbf{1}}_{(m_{-}^{\nu },m_0^{\nu })}\left(x\right)dx.$$

(21)

(iii)

For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{(1+r){((1+r)\alpha -r)}^{+}}{1+r{((1+r)\alpha -r)}^{+}}{\delta }_0+{\left(\frac{x^2}{{\mathbb{V}}_{\nu }\left(x\right)+r{x}^2}\right)}^{\prime }{\mathbf{1}}_{\left(-\frac{w_{1+r}\left(0\right)}{r},m_0^{\nu }\right)}\left(x\right)dx.$$

(22)

Proof.

(i) From [9] (Theorem 3.1), we have

$$\Phi \left(\nu \right)\left(dx\right)=\alpha {\delta }_0+{\left(\frac{x^2}{{\mathbb{V}}_{\nu }\left(x\right)}\right)}^{\prime }{\mathbf{1}}_{(m_{-}^{\nu },m_0^{\nu })}\left(x\right)dx.$$

(23)

Furthermore, we know from [14] (hl Theorem 3.1) that ${\nu }^{⊞t}$ has an atom at 0 for $t>1$ if and only if $\nu \left(\right\{0\left\}\right)=\alpha >1-1/t$. In this case,

$${\nu }^{⊞t}\left(\left\{0\right\}\right)=t\alpha -(t-1).$$

(24)

Since $\nu \in {\mathcal{P}}_{+}$ is non-degenerate, this is the same for ${\nu }^{⊞t}$. Furthermore, $A\left({\nu }^{⊞t}\right)=min\{0,infsupp\left({\nu }^{⊞t}\right)\}=0$. Then,

$$m_{-}^{{\nu }^{⊞t}}=A\left({\nu }^{⊞t}\right)-1/{\mathcal{G}}_{{\nu }^{⊞t}}\left(A\left({\nu }^{⊞t}\right)\right)=-1/{\mathcal{G}}_{{\nu }^{⊞t}}\left(0\right).$$

(25)

Combining (25) and (18), we obtain for all $t>1$

$$m_{-}^{{\nu }^{⊞t}}=\frac{t{w}_t\left(0\right)}{1-t}.$$

(26)

Combining (15), (23), (24) and (26), we get for all $t>1$ that

$$\begin{array}{l}\Phi \left({\nu }^{⊞t}\right)\left(dx\right)={\nu }^{⊞t}\left(\left\{0\right\}\right){\delta }_0+{\left(\frac{x^2}{{\mathbb{V}}_{{\nu }^{⊞t}}\left(x\right)}\right)}^{\prime }{\mathbf{1}}_{\left(m_{-}^{{\nu }^{⊞t}},m_0^{{\nu }^{⊞t}}\right)}\left(x\right)dx\\ ={(t\alpha -(t-1))}^{+}{\delta }_0+{\left(\frac{x^2}{t{\mathbb{V}}_{\nu }(x/t)}\right)}^{\prime }{\mathbf{1}}_{\left(-\frac{t{w}_t\left(0\right)}{t-1},t{m}_0^{\nu }\right)}\left(x\right)dx.\end{array}$$

(27)

From [15] (Lemma 2.7), for all $c>0$ and for all $\mu \in {\mathcal{P}}_{+}$, we have

$$D_c(\Phi \left(\mu \right))=\Phi \left(D_c\left(\mu \right)\right).$$

(28)

Then,

$$D_{1/t}(\Phi \left({\nu }^{⊞t}\right))=\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right).$$

(29)

Equation (20) follows from (27) and (29).

(ii) From [15] (Corollary 2.3), we known that ${\nu }^{\uplus r}$ has an atom at 0 if and only if $\nu$ has an atom at 0. In this case, we have

$${\nu }^{\uplus r}\left(\left\{0\right\}\right)=\frac{\alpha }{r-\alpha (r-1)},r>0.$$

(30)

Since $\nu \in {\mathcal{P}}_{+}$ is non-degenerate, this is the same for ${\nu }^{\uplus r}$. Then, $A\left({\nu }^{\uplus r}\right)=min\{0,infsupp\left({\nu }^{\uplus r}\right)\}=0$. Thus,

$$m_{-}^{{\nu }^{\uplus r}}=-1/{\mathcal{G}}_{{\nu }^{\uplus r}}\left(0\right)=-r/{\mathcal{G}}_{\nu }\left(0\right)=r{m}_{-}^{\nu }.$$

(31)

Combining (16), (23), (30) and (31), we get for all $r>0$

$$\begin{array}{l}\Phi \left({\nu }^{\uplus r}\right)\left(dx\right)={\nu }^{\uplus r}\left(\left\{0\right\}\right){\delta }_0+{\left(\frac{x^2}{{\mathbb{V}}_{{\nu }^{\uplus r}}\left(x\right)}\right)}^{\prime }{\mathbf{1}}_{(m_{-}^{{\nu }^{\uplus r}},m_0^{{\nu }^{\uplus r}})}\left(x\right)dx\\ =\frac{\alpha }{r-\alpha (r-1)}{\delta }_0+{\left(\frac{x^2}{r{\mathbb{V}}_{\nu }(x/r)+x^2(1/r-1)}\right)}^{\prime }{\mathbf{1}}_{(r{m}_{-}^{\nu },r{m}_0^{\nu })}\left(x\right)dx.\end{array}$$

(32)

Relation (21) follows from (28) and (32).

(iii) Using (24) and (30), we have for all $r>0$

$$\begin{array}{l}\left({\mathbb{B}}_r\left(\nu \right)\right)\left\{0\right\}={\left({\nu }^{⊞1+r}\right)}^{\uplus \frac{1}{1+r}}\left\{0\right\}=\frac{{\nu }^{⊞1+r}\left\{0\right\}}{\frac{1}{1+r}-\left(\frac{1}{1+r}-1\right){\nu }^{⊞1+r}\left\{0\right\}}\\ =\frac{{((1+r)\alpha -r)}^{+}}{\frac{1}{1+r}-\left(\frac{1}{1+r}-1\right){((1+r)\alpha -r)}^{+}}=\frac{(1+r){((1+r)\alpha -r)}^{+}}{1+r{((1+r)\alpha -r)}^{+}}.\end{array}$$

(33)

Furthermore, from (26) and (31), one see that

$$m_{-}^{{\mathbb{B}}_r\left(\nu \right)}=m_{-}^{{\left({\nu }^{⊞1+r}\right)}^{\uplus \frac{1}{1+r}}}=\frac{1}{1+r}{m}_{-}^{{\nu }^{⊞1+r}}=-\frac{w_{1+r}\left(0\right)}{r}.$$

(34)

Combining (23), (33), (34) and (17), we get for all $r>0$

$$\begin{array}{l}\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)={\mathbb{B}}_r\left(\nu \right)\left(\left\{0\right\}\right){\delta }_0+{\left(\frac{x^2}{{\mathbb{V}}_{{\mathbb{B}}_r\left(\nu \right)}\left(x\right)}\right)}^{\prime }{\mathbf{1}}_{(m_{-}^{{\mathbb{B}}_r\left(\nu \right)},m_0^{{\mathbb{B}}_r\left(\nu \right)})}\left(x\right)dx\\ =\frac{(1+r){((1+r)\alpha -r)}^{+}}{1+r{((1+r)\alpha -r)}^{+}}{\delta }_0+{\left(\frac{x^2}{{\mathbb{V}}_{\nu }\left(x\right)+r{x}^2}\right)}^{\prime }{\mathbf{1}}_{\left(-\frac{w_{1+r}\left(0\right)}{r},m_0^{\nu }\right)}\left(x\right)dx.\end{array}$$

(35)

□

Next, some examples are provided for the limiting laws $\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)$, $\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)$ and $\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)$ using measures $\nu$ of importance in non-commutative probability.

Example 1.

Let $\gamma =\frac{1}{2}{\delta }_{-1}+\frac{1}{2}{\delta }_1$ be the symmetric Bernoulli distribution, with $m_0^{\gamma }=0$. We have ${\mathcal{V}}_{\gamma }\left(x\right)=1-x^2={\mathbb{V}}_{\gamma }\left(x\right)$, $\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-1,0)$. Consider the translation $\phi :\zeta ⟼\zeta +1$. Then, $\nu =\phi \left(\gamma \right)=\frac{1}{2}{\delta }_0+\frac{1}{2}{\delta }_2$, with $m_0^{\nu }=1$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=\frac{x^2(2-x)}{x-1},\forall x\in (m_{-}^{\nu },m_0^{\nu })=(0,1)and{\mathcal{G}}_{\nu }\left(\xi \right)=\frac{\xi -1}{\xi (\xi -2)}.$$

From (19), we get

$$H_t\left(\xi \right)=\frac{{\xi }^2+\xi (t-2)}{\xi -1}$$

and therefore

$$w_t\left(u\right)=\frac{1}{2}\left(2+u-t-\sqrt{u^2-2tu+{(t-2)}^2}\right),$$

for $u\in \mathbb{C}\backslash (t-2\sqrt{t-1},t+2\sqrt{t-1})$. Then,

$$w_t\left(0\right)=\frac{2-t-|t-2|}{2}.$$

We have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)={(1-t/2)}^{+}{\delta }_0+\frac{t}{{(2-x)}^2}{\mathbf{1}}_{\left(\frac{t-2+|t-2|}{2(t-1)},1\right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{1}{1+r}{\delta }_0+\frac{r}{{(1+r(1-x))}^2}{\mathbf{1}}_{(0,1)}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{\frac{1}{2}(1+r){(1-r)}^{+}}{1+\frac{1}{2}r{(1-r)}^{+}}{\delta }_0+\frac{1}{{((r-1)x+2-r)}^2}{\mathbf{1}}_{\left(\frac{r-1+|r-1|}{2r},1\right)}\left(x\right)dx.$$

Example 2.

Consider Wigner’s law

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

with $m_0^{\gamma }=0$. We have

$${\mathcal{V}}_{\gamma }\left(x\right)=1={\mathbb{V}}_{\gamma }\left(x\right),\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-1,0)and{\mathcal{G}}_{\gamma }\left(\xi \right)=\frac{1}{2}\left(\xi -\sqrt{{\xi }^2-4}\right).$$

Consider the translation $\phi :\zeta ⟼\zeta +2$. Then,

$${\displaystyle \nu \left(d\zeta \right)=f\left(\gamma \right)\left(d\zeta \right)=\frac{\sqrt{\zeta (4-\zeta )}}{2\pi }{\mathbf{1}}_{(0,4)}\left(\zeta \right)d\zeta ,}$$

with $m_0^{\nu }=2$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=\frac{x}{x-2},\forall x\in (m_{-}^{\nu },m_0^{\nu })=(1,2)and{\mathcal{G}}_{\nu }\left(\xi \right)=\frac{1}{2}\left(\xi -2-\sqrt{\xi (\xi -4)}\right).$$

From (19), we have that

$$H_t\left(\xi \right)=t\xi -\frac{2(t-1)}{\xi -2-\sqrt{\xi (\xi -4)}}$$

and consequently

$$w_t\left(u\right)=\frac{2t-2t^2+u+tu-(t-1)\sqrt{4t(t-1)-4tu+u^2}}{2t}$$

for all $u\in \mathbb{C}\backslash (2t-2\sqrt{t},2t+2\sqrt{t})$. Then,

$$w_t\left(0\right)=(t-1)\left(-1-\sqrt{\frac{t-1}{t}}\right).$$

Then, we have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)=2t(x-1){\mathbf{1}}_{\left(1+\sqrt{\frac{t-1}{t}},2\right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{2r(x-1)}{{(1+(1-r)(x^2-2x))}^2}{\mathbf{1}}_{(1,2)}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{2(x-1)}{{(1+r(x^2-2x))}^2}{\mathbf{1}}_{\left(1+\sqrt{\frac{r}{1+r}},2\right)}\left(x\right)dx.$$

Example 3.

For $0<a^2<1$, consider the (absolutely continuous) Marchenko–Pastur law

$${\displaystyle \gamma \left(d\zeta \right)=\frac{\sqrt{4-{(\zeta -a)}^2}}{2\pi (1+a\zeta )}{\mathbf{1}}_{(a-2,a+2)}\left(\zeta \right)d\zeta ,}$$

with $m_0^{\gamma }=0$. We have

$${\mathcal{V}}_{\gamma }\left(x\right)=1+ax={\mathbb{V}}_{\gamma }\left(x\right),\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-1,0)and{\mathcal{G}}_{\gamma }\left(\xi \right)=\frac{\left(a+\xi -\sqrt{{(a-\xi )}^2-4}\right)}{2(1+a\xi )}.$$

Consider the affine transformation $\phi :\zeta ⟼a\zeta +1$. Then,

$$\nu \left(d\zeta \right)=\phi \left(\gamma \right)\left(d\zeta \right)=\frac{\sqrt{\left({(a+1)}^2-\zeta \right)\left(\zeta -{(a-1)}^2\right)}}{2\pi a^2\zeta }{\mathbf{1}}_{({(a-1)}^2,{(a+1)}^2)}\left(\zeta \right)d\zeta ,$$

with $m_0^{\nu }=1$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=\frac{a^2{x}^2}{x-1},\forall x\in (m_{-}^{\nu },m_0^{\nu })=(1-a^2,1)$$

and

$${\mathcal{G}}_{\nu }\left(\xi \right)=\frac{1}{2\xi a}\left(a+\frac{(\xi -1)}{a}-\sqrt{{\left(a-\frac{(\xi -1)}{a}\right)}^2-4}\right).$$

(36)

We have that

$$H_t\left(\xi \right)=t\xi -\frac{2\xi a(t-1)}{\left(a+\frac{(\xi -1)}{a}-\sqrt{{\left(a-\frac{(\xi -1)}{a}\right)}^2-4}\right)}$$

(37)

and

$$w_t\left(u\right)=\frac{-a^2+t+a^{2}t-t^2+u+tu-(t-1)\sqrt{{(t-a^2)}^2-2u(a^2+t)+u^2}}{2t},$$

(38)

for all $u\in \mathbb{C}\backslash (a^2+t-2\sqrt{t{a}^2},a^2+t+2\sqrt{t{a}^2})$. Then,

$$w_t\left(0\right)=-\frac{(t-1)(t-a^2)}{t}.$$

Then, we have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)=\frac{t}{a^2}{\mathbf{1}}_{(1-a^2/t,1)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{a^{2}r}{{(a^2+(1-r)(x-1))}^2}{\mathbf{1}}_{(1-a^2,1)}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{a^2}{{(a^2+r(x-1))}^2}{\mathbf{1}}_{\left(1-\frac{a^2}{r+1},1\right)}\left(x\right)dx.$$

Example 4.

For $a^2>1$, consider the Marchenko–Pastur law

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

with $m_0^{\gamma }=0$. We have

$${\mathcal{V}}_{\gamma }\left(x\right)=1+ax={\mathbb{V}}_{\gamma }\left(x\right),\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-1,0).$$

Consider the affine transformation $\phi :\zeta ⟼a\zeta +1$. Then,

$$\nu \left(d\zeta \right)=f\left(\gamma \right)\left(d\zeta \right)=\frac{\sqrt{\left({(a+1)}^2-\zeta \right)\left(\zeta -{(a-1)}^2\right)}}{2\pi a^2\zeta }{\mathbf{1}}_{\left({(a-1)}^2,{(a+1)}^2\right)}\left(\zeta \right)d\zeta +(1-1/a^2){\delta }_0$$

with $m_0^{\nu }=1$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=\frac{a^2{x}^2}{x-1},\forall x\in (m_{-}^{\nu },m_0^{\nu })=(0,1).$$

The Cauchy transform of ν is given by (36). The measure ν has a Dirac mass at 0. This implies that ${\int }_0^{\infty }{y}^{-1}\nu \left(dy\right)=+\infty$ and so $m_{-}^{\nu }=-1/{\mathcal{G}}_{\nu }\left(0\right)=0$.

The functions $H_t(.)$ and $w_t(.)$ are given, respectively, by (37) and (38). We have that

$$w_t\left(0\right)=\frac{(t-1)(a^2-t-|t-a^2\left|\right)}{2t}.$$

Then, we have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)={(1-t/a^2)}^{+}{\delta }_0+\frac{t}{a^2}{\mathbf{1}}_{\left(\frac{t-a^2+|t-a^2|}{2t},1\right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\left(\frac{a^2-1}{a^2+r-1}\right){\delta }_0+\frac{a^{2}r}{{(a^2+(1-r)(x-1))}^2}{\mathbf{1}}_{(0,1)}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\left(\frac{(1+r){\left(1-\frac{1+r}{a^2}\right)}^{+}}{1+r{\left(1-\frac{1+r}{a^2}\right)}^{+}}\right){\delta }_0+\frac{a^2}{{(a^2+r(x-1))}^2}{\mathbf{1}}_{\left(\frac{r+1-a^2+|r+1-a^2|}{2(r+1)},1\right)}\left(x\right)dx.$$

Example 5.

For $a\ne 0$, consider the free Gamma law,

$$\gamma \left(d\zeta \right)=\frac{\sqrt{4(1+a^2)-{(\zeta -2a)}^2}}{2\pi (a^2{\zeta }^2+2a\zeta +1)}{\mathbf{1}}_{(2a-2\sqrt{1+a^2},2a+2\sqrt{1+a^2})}\left(\zeta \right)\left(d\zeta \right),$$

with $m_0^{\gamma }=0$. We have

$${\mathbb{V}}_{\gamma }\left(x\right)={\mathcal{V}}_{\gamma }\left(x\right)={(1+ax)}^{2}and{\mathcal{G}}_{\gamma }\left(\xi \right)=\frac{2a+\xi +2a^2\xi -\sqrt{{(2a-\xi )}^2-4(1+a^2)}}{2{(1+a\xi )}^2}.$$

Consider the affine transformation $\phi :x⟼a\zeta +1$. Then,

$$\nu \left(d\zeta \right)=\phi \left(\gamma \right)\left(d\zeta \right)=\frac{\sqrt{({(\sqrt{a^2+1}+a)}^2-\zeta )(\zeta -{(\sqrt{a^2+1}-a)}^2)}}{2\pi a^2{\zeta }^2}{\mathbf{1}}_{\left(\right(\sqrt{a^2+1}-{\left|a\right|)}^2,(\sqrt{a^2+1}+{\left|a\right|)}^2)}\left(\zeta \right)d\zeta ,$$

with $m_0^{\nu }=1$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=\frac{a^2{x}^3}{x-1},\forall x\in (m_{-}^{\nu },m_0^{\nu })=\left(\frac{1}{1+a^2},1\right)$$

and

$${\mathcal{G}}_{\nu }\left(\xi \right)=\frac{\frac{\xi -1}{a}+2a\xi -\sqrt{{(2a-\frac{\xi -1}{a})}^2-4(1+a^2)}}{2a{\xi }^2}.$$

We have that

$$H_t\left(\xi \right)=t\xi -\frac{2(t-1)a{\xi }^2}{\frac{\xi -1}{a}+2a\xi -\sqrt{{(2a-\frac{\xi -1}{a})}^2-4(1+a^2)}}$$

and

$$w_t\left(u\right)=\frac{t-t^2+u(1+2a^2+t)-(t-1)\sqrt{t^2-(4a^2+2t)u+u^2}}{2(a^2+t)}$$

for all $u\in \mathbb{C}\backslash (2a^2+t-\sqrt{{(2a^2+t)}^2-1},2a^2+t+\sqrt{{(2a^2+t)}^2-1})$. Then,

$$w_t\left(0\right)=-\frac{t(t-1)}{a^2+t}.$$

We have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)=\frac{t}{a^2{x}^2}{\mathbf{1}}_{\left(\frac{t}{t+a^2},1\right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{a^{2}r}{{(a^{2}x+(1-r)(x-1))}^2}{\mathbf{1}}_{\left(\frac{1}{1+a^2},1\right)}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{a^2}{{(a^{2}x+r(x-1))}^2}{\mathbf{1}}_{\left(\frac{r+1}{r+1+a^2},1\right)}\left(x\right)dx.$$

Example 6.

The inverse semicircle law is given by

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

with $m_0^{\gamma }=-\infty$. We have

$${\mathbb{V}}_{\gamma }\left(x\right)=x^3,\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-\infty ,-1)and{\mathcal{G}}_{\gamma }\left(\xi \right)=\frac{1+2\xi -\sqrt{1+4\xi }}{2{\xi }^2}.$$

Consider the transformation $\phi :\zeta ⟼-\zeta$. Then,

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

with $m_0^{\nu }=+\infty$. We have that

$${\mathbb{V}}_{\nu }\left(x\right)=-x^3,\forall x\in (m_{-}^{\nu },m_0^{\nu })=(1,+\infty )and{\mathcal{G}}_{\nu }\left(\xi \right)=\frac{2\xi -1+\sqrt{1-4\xi }}{2{\xi }^2}.$$

We have that

$$H_t\left(\xi \right)=t\xi -\frac{2(t-1){\xi }^2}{2\xi -1+\sqrt{1-4\xi }}$$

and

$$w_t\left(u\right)=\frac{1}{2}\left((t-1)(-t-\sqrt{t^2-4u})+2u\right)$$

for all $u\in \mathbb{C}\backslash (t^2/4,+\infty )$. Then,

$$w_t\left(0\right)=-t(t-1).$$

We have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)=\frac{t}{x^2}{\mathbf{1}}_{\left(t,+\infty \right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{r}{{(-x+(1-r))}^2}{\mathbf{1}}_{(1,+\infty )}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{1}{{(x-r)}^2}{\mathbf{1}}_{\left(1+r,+\infty \right)}\left(x\right)dx.$$

Example 7.

The free Ressel law is given by

$${\displaystyle \gamma \left(d\zeta \right)=\frac{-1}{\pi \zeta \sqrt{-1-\zeta }}{\mathbf{1}}_{(-\infty ,-1)}\left(\zeta \right)d\zeta ,}$$

with $m_0^{\gamma }=-\infty$. We have

$${\mathbb{V}}_{\gamma }\left(x\right)=x^2(x+1),\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-\infty ,-2)and{\mathcal{G}}_{\gamma }\left(\xi \right)=\frac{1+\xi -\sqrt{1+\xi }}{\xi (1+\xi )}.$$

Consider the transformation $\phi :\zeta ⟼-\zeta$. Then,

$$\nu \left(d\zeta \right)=\phi \left(\gamma \right)\left(d\zeta \right)=\frac{1}{\pi \zeta \sqrt{\zeta -1}}{\mathbf{1}}_{\left(1,+\infty \right)}\left(\zeta \right)d\zeta ,$$

with $m_0^{\nu }=+\infty$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=x^2(1-x),\forall x\in (m_{-}^{\nu },m_0^{\nu })=(2,+\infty )and{\mathcal{G}}_{\nu }\left(\xi \right)=\frac{1-\xi -\sqrt{1-\xi }}{\xi (1-\xi )}.$$

We have that

$$H_t\left(\xi \right)=t\xi -\frac{(t-1)\xi (1-\xi )}{1-\xi -\sqrt{1-\xi }},$$

and

$$w_t\left(u\right)=\frac{1}{2}\left(1-t^2-(t-1)\sqrt{{(1+t)}^2-4u}+2u\right).$$

for all $u\in \mathbb{C}\backslash ((1+t^2)/4,+\infty )$. Then,

$$w_t\left(0\right)=-(t-1)(1+t).$$

We have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)=\frac{t}{{(1-x)}^2}{\mathbf{1}}_{\left(1+t,+\infty \right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{r}{{(2-r-x)}^2}{\mathbf{1}}_{(2,+\infty )}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{1}{{(1-x+r)}^2}{\mathbf{1}}_{\left(2+r,+\infty \right)}\left(x\right)dx.$$

Example 8.

The Free Abel law is given by

$${\displaystyle \gamma \left(d\zeta \right)=\frac{1}{\pi (1-\zeta )\sqrt{-\zeta }}{\mathbf{1}}_{(-\infty ,0)}\left(\zeta \right)d\zeta ,}$$

with $m_0^{\gamma }=-\infty$. We have

$${\mathbb{V}}_{\gamma }\left(x\right)=x^2(x-1),\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-\infty ,0).$$

Consider the transformation $\phi :\zeta ⟼-\zeta$. Then,

$$\nu \left(d\zeta \right)=\phi \left(\gamma \right)\left(d\zeta \right)=\frac{1}{\pi (1+\zeta )\sqrt{\zeta }}{\mathbf{1}}_{\left(0,+\infty \right)}\left(\zeta \right)d\zeta ,$$

with $m_0^{\nu }=+\infty$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=-x^2(1+x),\forall x\in (m_{-}^{\nu },m_0^{\nu })=(0,+\infty )and{\mathcal{G}}_{\nu }\left(\xi \right)=\frac{\xi +\sqrt{-\xi }}{\xi (1+\xi )}.$$

We have that

$$H_t\left(\xi \right)=t\xi -\frac{(t-1)\xi (1+\xi )}{\xi +\sqrt{-\xi }}$$

and

$$w_t\left(u\right)=\frac{1}{2}\left(-{(t-1)}^2-(t-1)\sqrt{{(t-1)}^2-4u}+2u\right)$$

for all $u\in \mathbb{C}\backslash ({(t-1)}^2/4,+\infty )$. Then,

$$w_t\left(0\right)=-{(t-1)}^2.$$

We have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)=\frac{t}{{(1+x)}^2}{\mathbf{1}}_{\left(t-1,+\infty \right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{r}{{(r+x)}^2}{\mathbf{1}}_{(0,+\infty )}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left(B_r\left(\nu \right)\right)\left(dx\right)=\frac{1}{{(r-1-x)}^2}{\mathbf{1}}_{\left(r,+\infty \right)}\left(x\right)dx.$$

Example 9.

The free strict arcsine law is given by

$${\displaystyle \gamma \left(d\zeta \right)=\frac{\sqrt{3-4\zeta }}{2\pi (1+{\zeta }^2)}{\mathbf{1}}_{(-\infty ,3/4)}\left(\zeta \right)d\zeta ,}$$

with $m_0^{\gamma }=-\infty$. We have

$${\mathbb{V}}_{\gamma }\left(x\right)=x(1+x^2),\forall x\in (m_{-}^{\gamma },m_0^{\gamma })=(-\infty ,-1/2).$$

Consider the affine transformation $\phi :\zeta ⟼-\zeta +3/4$. Then,

$$\nu \left(d\zeta \right)=\phi \left(\gamma \right)\left(d\zeta \right)=\frac{\sqrt{\zeta }}{\pi (1+{(3/4-\zeta )}^2)}{\mathbf{1}}_{\left(0,+\infty \right)}\left(\zeta \right)d\zeta ,$$

with $m_0^{\nu }=+\infty$. We have

$${\mathbb{V}}_{\nu }\left(x\right)=-x\left(x^2-\frac{3}{2}x+\frac{25}{16}\right),\forall x\in (m_{-}^{\nu },m_0^{\nu })=(5/4,+\infty )and{\mathcal{G}}_{\nu }\left(\xi \right)=\frac{\frac{5}{4}-\xi -\sqrt{-\xi }}{-\frac{25}{16}+\frac{3}{2}\xi -{\xi }^2}.$$

We have that

$$H_t\left(\xi \right)=t\xi -\frac{(t-1)(-\frac{25}{16}+\frac{3}{2}\xi -{\xi }^2)}{\frac{5}{4}-\xi -\sqrt{-\xi }}$$

and

$$w_t\left(u\right)=\frac{1}{4}\left(-(t-1)(2t+3)-2(t-1)\sqrt{(t-1)(t+4)-4u}+4u\right)$$

for all $u\in \mathbb{C}\backslash \left(\right(t-1\left)\right(t+4)/4,+\infty )$. Then,

$$w_t\left(0\right)=\frac{1}{4}(t-1)\left[-(2t+3)-2\sqrt{(t-1)(t+4)}\right].$$

We have the following:

(i) For all $t>1$,

$$\Phi \left(D_{1/t}\left({\nu }^{⊞t}\right)\right)\left(dx\right)=\frac{t\left(x^2-\frac{25}{16}\right)}{{\left(x^2-\frac{3}{2}x+\frac{25}{16}\right)}^2}{\mathbf{1}}_{\left(\frac{2t+3+2\sqrt{(t-1)(t+4)}}{4},+\infty \right)}\left(x\right)dx.$$

(ii) For all $r>0$,

$$\Phi \left(D_{1/r}\left({\nu }^{\uplus r}\right)\right)\left(dx\right)=\frac{r\left(x^2-\frac{25}{16}\right)}{{\left((1-r)x-\left(x^2-\frac{3}{2}x+\frac{25}{16}\right)\right)}^2}{\mathbf{1}}_{(5/4,+\infty )}\left(x\right)dx.$$

(iii) For all $r>0$,

$$\Phi \left({\mathbb{B}}_r\left(\nu \right)\right)\left(dx\right)=\frac{x^2-\frac{25}{16}}{{\left(rx-\left(x^2-\frac{3}{2}x+\frac{25}{16}\right)\right)}^2}{\mathbf{1}}_{\left(\frac{2r+5+2\sqrt{r(r+5)}}{4},+\infty \right)}\left(x\right)dx.$$

4. Conclusions

In classical probability, the law of large numbers for classical multiplicative convolution is deduced from the law for the classical additive convolution. This is not the case in non-commutative probability. The free additive law was demonstrated in [16] for probability measures with bounded support and extended in [17] to all probability measures with a first moment. The free multiplicative law was demonstrated in [3] for measures with bounded support and extended in [4] to measures with unbounded support. Contrary to the case of classical multiplicative convolution, the limiting measure for the free multiplicative law of large numbers is not a Dirac measure, except in the case where the original measure is a Dirac measure. In [9], an interesting description is given for the free multiplicative law of large numbers by means of the pseudo-variance function. In this paper, we have explicitly demonstrated the law of large numbers of three types of non-commutative probability measures involving the free and the Boolean additive convolutions. The results are explained using several probability measures and may be useful for researchers in the field of non-commutative probability.