Stochastic Integral and Covariation Representations for Rectangular Lévy Process Ensembles

Abstract

A Bercovici-Pata bijection $\Lambda _{c}$ from the set of symmetric infinitely divisible distributions to the set of ⊞  _c -free infinitely divisible distributions, for certain free convolution ⊞  _c is introduced in Benaych-Georges (Random matrices, related convolutions. Probab Theory Relat Fields 144:471–515, 2009. Revised version of F. Benaych-Georges: Random matrices, related convolutions. arXiv, 2005). This bijection is explained in terms of complex rectangular matrix ensembles whose singular distributions are ⊞  _c -free infinitely divisible. We investigate the rectangular matrix Lévy processes with jumps of rank one associated to these rectangular matrix ensembles. First as general result, a sample path representation by covariation processes for rectangular matrix Lévy processes of rank one jumps is obtained. Second, rectangular matrix ensembles for ⊞  _c -free infinitely divisible distributions are built consisting of matrix stochastic integrals when the corresponding symmetric infinitely divisible distributions under $\Lambda _{c}$ admit stochastic integral representations. These models are realizations of stochastic integrals of nonrandom functions with respect to rectangular matrix Lévy processes. In particular, any ⊞  _c -free selfdecomposable infinitely divisible distribution has a random matrix model of Ornstein-Uhlenbeck type $\int _{0}^{\infty }e^{-t}\mathrm{d}\Psi (t)$, where $\left \{\Psi (t): t \geq 0\right \}$ is a rectangular matrix Lévy process.

Keywords

Mathematics Subject Classification (2000).

Appendix

We prove Lemma 3.2 of Sect. 3.

Lemma 1

Let u,v be independent random vectors, uniformly distributed on the unit sphere of respectively $\mathbb{C}^{d}$ , $\mathbb{C}^{d^{\prime} }.$ Then for any $A \in \mathbb{M}_{d,d^{\prime}}$,

• (a)
  $$\displaystyle{\mathbb{E}_{u,v}\left (\left \langle u,Av\right \rangle \right ) = 0\text{,}}$$
• (b)
  $$\displaystyle{\mathbb{E}_{u,v}\left (\mathfrak{R}\left \langle u,Av\right \rangle \right )^{2} = \frac{1} {2dd^{\prime}}\mathrm{tr}\left (AA^{{\ast}}\right ).}$$

Proof

The assertion (a) clearly follows from independence of u and v. Let us prove the assertion (b). By noting that $\left (vu^{{\ast}}\right )_{ji} = v_{j}\bar{u}_{i}$ we get

$$\displaystyle{\left \langle u,Av\right \rangle =\mathrm{ tr}\left (Avu^{{\ast}}\right ) =\sum _{ i=1}^{d}\sum _{ j=1}^{d^{\prime} }a_{ij}\bar{u}_{i}v_{j},}$$

then

$$\displaystyle{\mathfrak{R}\left \langle u,Av\right \rangle =\sum _{ i=1}^{d}\sum _{ j=1}^{d^{\prime} }\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right ).}$$

If $a_{ij} = a_{ij1} + ia_{ij2},u_{i} = u_{i1} + iu_{i2},$ $v_{j} = v_{j1} + iv_{j2}$ then

$$\displaystyle\begin{array}{rcl} \mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )& =& \mathfrak{R}\left [\left (a_{ij1} + ia_{ij2}\right )\left (u_{i1} - iu_{i2}\right )\left (v_{j1} + iv_{j2}\right )\right ] \\ & =& \left (u_{i1}v_{j1} + u_{i2}v_{j2}\right )a_{ij1} + \left (u_{i2}v_{j1} - u_{i1}v_{j2}\right )a_{ij2}.{}\end{array}$$

(1)

Now

$$\displaystyle\begin{array}{rcl} \left (\mathfrak{R}\left \langle u,Av\right \rangle \right )^{2}& =& \left [\sum _{ i=1}^{d}\sum _{ j=1}^{d^{\prime} }\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )\right ]\left [\sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )\right ] {}\\ & =& \sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }\sum _{k=1}^{d}\sum _{ l=1}^{d^{\prime} }\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )\mathfrak{R}\left (a_{kl}\bar{u}_{k}v_{l}\right ) {}\\ & =& \sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }\left (\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )\right )^{2} +\sum _{ i=1}^{d}\sum _{ j=1}^{d^{\prime} }\sum _{k\neq i}^{d}\sum _{ l\neq j}^{d^{\prime} }\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )\mathfrak{R}\left (a_{kl}\bar{u}_{k}v_{l}\right ). {}\\ \end{array}$$

Expanding $\left [\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )\right ]^{2}$ by using (1) we get

$$\displaystyle\begin{array}{rcl} \left (\mathfrak{R}\left \langle u,Av\right \rangle \right )^{2}& =& \sum _{ i=1}^{d}\sum _{ j=1}^{d^{\prime} }\left [\left (u_{i1}^{2}v_{ j1}^{2} + u_{ i2}^{2}v_{ j2}^{2}\right )a_{ ij1}^{2} + \left (u_{ i2}^{2}v_{ j1}^{2} + u_{ i1}^{2}v_{ j2}^{2}\right )a_{ ij2}^{2}\right ] {}\\ & +& \sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }[\left (2u_{i2}^{2}v_{ j1}v_{j2} - 2u_{i1}u_{i2}v_{j2}^{2} + 2u_{ i1}u_{i2}v_{j1}^{2} - 2u_{ i1}^{2}v_{ j1}v_{j2}\right )a_{ij1}a_{ij2} {}\\ & +& 2u_{i1}u_{i2}v_{j1}v_{j2}\left (a_{ij1}^{2} - a_{ ij2}^{2}\right )] {}\\ & +& \sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }\sum _{k\neq i}^{d}\sum _{ l\neq j}^{d^{\prime} }\mathfrak{R}\left (a_{ij}\bar{u}_{i}v_{j}\right )\mathfrak{R}\left (a_{kl}\bar{u}_{k}v_{l}\right ). {}\\ \end{array}$$

Taking expectation of $\left (\mathfrak{R}\left \langle u,Av\right \rangle \right )^{2}$ we obtain that the expectation of the terms in the second and third summand are zero. Thus, using component-wise the Lemma 2 below,

$$\displaystyle\begin{array}{rcl} \mathbb{E}_{u,v}\left (\mathfrak{R}\left \langle u,Av\right \rangle \right )^{2}& =& \sum _{ i=1}^{d}\sum _{ j=1}^{d^{\prime} }\mathbb{E}\left [\left (u_{i1}^{2}v_{ j1}^{2} + u_{ i2}^{2}v_{ j2}^{2}\right )a_{ ij1}^{2} + \left (u_{ i2}^{2}v_{ j1}^{2} + u_{ i1}^{2}v_{ j2}^{2}\right )a_{ ij2}^{2}\right ] {}\\ & =& \sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }\left [\left ( \frac{1} {2d} \frac{1} {2d^{\prime}} + \frac{1} {2d} \frac{1} {2d^{\prime}}\right )a_{ij1}^{2} + \left ( \frac{1} {2d} \frac{1} {2d^{\prime}} + \frac{1} {2d} \frac{1} {2d^{\prime}}\right )a_{ij2}^{2}\right ] {}\\ & =& \frac{1} {2dd^{\prime}}\sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }\left (a_{ij1}^{2} + a_{ ij2}^{2}\right ) {}\\ & =& \frac{1} {2dd^{\prime}}\sum _{i=1}^{d}\sum _{ j=1}^{d^{\prime} }\left \vert a_{ij}\right \vert ^{2} {}\\ & =& \frac{1} {2dd^{\prime}}\mathrm{tr}\left (AA^{{\ast}}\right ). {}\\ \end{array}$$

 □ 

The following lemma is well known in the real vector case, see e.g. [15, eq. (2)]. We give a proof for the completeness of the paper.

Lemma 2

Let $U = \left (U_{1},\ldots,U_{d}\right )^{T}$  be a random vector uniformly distributed on the unit sphere of $\mathbb{C}^{d}$ . Then their components U _k , k = 1,2,…,d are identically distributed with symmetric density function

$$\displaystyle{ \frac{d - 1} {\pi } \left (1 - x^{2} - y^{2}\right )^{d-2},\quad x^{2} + y^{2} \leq 1. }$$

(2)

The components $X_{k},Y _{k}$ of $U_{k} = X_{k} + iY _{k},$ are identically distributed with density

$$\displaystyle{ \frac{1} {\sqrt{\pi }} \frac{\Gamma \left (n\right )} {\Gamma \left (n -\frac{1} {2}\right )}\left (1 - t^{2}\right )^{n-3/2},\ -1 \leq t \leq 1,}$$

and their first two marginal moments are

$$\displaystyle{EX_{k} = EY _{k} = 0}$$

and

$$\displaystyle{EX_{k}^{2} = EY _{ k}^{2} = \frac{1} {2d}.}$$

Proof

Let $u = \left (u_{1},\ldots,u_{d}\right )$ a random vector choosen uniformly on $\{u \in \mathbb{C}^{d}: \left \Vert u\right \Vert = 1\},$ observe that $\left \Vert u\right \Vert ^{2} = uu^{{\ast}} =\sum _{ i=1}^{d}\left \vert u_{i}\right \vert ^{2}.$ By [13, pp 140] the distribution of u _k for 1 ≤ k ≤ d is

$$\displaystyle{\frac{d - 1} {\pi } \left (1 - r^{2}\right )^{d-2}r\mathrm{d}r\mathrm{d}\theta \quad (u_{ k} = re^{i\theta },0 \leq r \leq 1,0 \leq \theta \leq 2\pi ).}$$

Using polar coordinates, $r = \sqrt{x^{2 } + y^{2}}$ and $\theta =\arctan \left (y/x\right )$ wich implies $\frac{\partial \left (r,\theta \right )} {\partial \left (x,y\right )} = \frac{1} {\sqrt{x^{2 } +y^{2}}}.$ By the change of variable formula if we denote u _k  = x + iy then the distribution of u _k is

$$\displaystyle{\frac{d - 1} {\pi } \left (1 - x^{2} - y^{2}\right )^{d-2},}$$

where $x^{2} + y^{2} \leq 1.$ This proves (2). Hence X _k and Y _k are identically distributed symmetric distribution around zero and therefore $EX_{k} = EY _{k} = 0$. Next we compute the marginal density of X _k

$$\displaystyle\begin{array}{rcl} f_{X_{k}}\left (x\right )& =& \frac{n - 1} {\pi } \int _{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}} }\left (1 - x^{2} - y^{2}\right )^{n-2}\mathrm{d}y {}\\ & =& \frac{2\left (n - 1\right )} {\pi } \int _{0}^{\sqrt{1-x^{2}} }\left (1 - x^{2} - y^{2}\right )^{n-2}\mathrm{d}y\text{,} {}\\ \end{array}$$

by change of variable $s = y^{2}/\left (1 - x^{2}\right )$ we get

$$\displaystyle\begin{array}{rcl} f_{X_{k}}\left (x\right )& =& \frac{2\left (n - 1\right )} {\pi } \int _{0}^{\sqrt{1-x^{2}} }\left (1 - x^{2} - y^{2}\right )^{n-2}\mathrm{d}y {}\\ & =& \frac{\left (n - 1\right )\left (1 - x^{2}\right )^{n-3/2}} {\pi } \int _{0}^{1}s^{-\frac{1} {2} }\left (1 - s\right )^{n-2}\mathrm{d}y {}\\ & =& \frac{\left (n - 1\right )\left (1 - x^{2}\right )^{n-3/2}} {\pi } \mathrm{Beta}\left (\frac{1} {2},n - 1\right ) {}\\ & =& \frac{1} {\sqrt{\pi }} \frac{\Gamma \left (n\right )} {\Gamma \left (n -\frac{1} {2}\right )}\left (1 - x^{2}\right )^{n-3/2}. {}\\ \end{array}$$

Now

$$\displaystyle\begin{array}{rcl} \mathrm{E}X_{k}^{2}& =& \int \int _{ D}x^{2}f\left (x,y\right )\mathrm{d}x\mathrm{d}y = \frac{d - 1} {\pi } \int _{-1}^{1}\int _{ -\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}} }x^{2}\left (1 - x^{2} - y^{2}\right )^{d-2}\mathrm{d}x\mathrm{d}y {}\\ & =& \frac{d - 1} {\pi } \int _{0}^{2\pi }\int _{ 0}^{1}r^{2}\cos ^{2}\theta \left (1 - r^{2}\right )^{d-2}r\mathrm{d}r\mathrm{d}\theta {}\\ & =& \frac{d - 1} {\pi } \frac{\pi } {2d\left (d - 1\right )} {}\\ & =& \frac{1} {2d}, {}\\ \end{array}$$

where we have used the identity $\mathrm{Beta}\left (a,b\right ) =\int _{ 0}^{1}t^{a-1}\left (1 - t\right )^{b-1}\mathrm{d}t = \frac{\Gamma \left (a\right )\Gamma \left (b\right )} {\Gamma \left (a+b\right )}$ to compute $\int _{0}^{1}r^{3}\left (1 - r^{2}\right )^{d-2}\mathrm{d}r = \frac{1} {2}\int _{0}^{1}s\left (1 - s\right )^{d-2}\mathrm{d}s = \frac{1} {2}\mathrm{Beta}\left (2,d - 1\right ) = \frac{1} {2d\left (d-1\right )},$ the change of variable s = r ² with $dr = ds/\left (2\sqrt{s}\right )$ and $\int _{0}^{2\pi }\cos ^{2}\theta \mathrm{d}\theta =\pi$. □