I was trying to understand the basic ideas of the operator-stable distributions. I found the papers by [Hudson](https://ac.els-cdn.com/0047259X80900792/1-s2.0-0047259X80900792-main.pdf?_tid=2953aa8d-c0a2-4a76-91f6-3f41bfdf0994&acdnat=1533563860_f86a125d48c19655037145541942ca02) and [Sato](https://ac.els-cdn.com/0047259X87900911/1-s2.0-0047259X87900911-main.pdf?_tid=6317fe1b-4153-4d73-b9c5-a8e36fdb790d&acdnat=1533564079_c9c2ff34136b6d94052444a56e4adcbf). However, unfortunately, I am being unable to understand the mathematical expressions, the authors have used, to define the same. So, to make the definition understandable, I wrote: 



> 
> If $X\_1 \stackrel{D}{=} X\_2 \stackrel{D}{=} X$, then $X$ is operator-stable if there exists a real matrix $C$ and a vector $d$ such that $AX\_1 + BX\_2 \stackrel{D}{=} CX + d$, where $A$, and $B$ are real, square matrices and $X\_1$, $X\_2$, and $X$ are real vectors.
> 
> 
> 


Have I understood right (only for simple scenarios)? Also, I have not found any examples (like Gaussian, Cauchy etc.) of the operator-stable distributions. Some examples will be really helpful. If there are no such examples available in closed form, then can I find the density numerically at a given point?



Definition and examples of operator-stable distributions
$\newcommand{\R}{\mathbb{R}}
\newcommand{\D}{\overset{\text{D}}=}
\newcommand{\E}{\operatorname{\mathsf E}}$ 


The fundamental paper in this area is the one by [Sharpe](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=2ahUKEwiO5KTD2NjcAhVIQ6wKHfyQD3kQFjAAegQIABAB&url=http%3A%2F%2Fwww.ams.org%2Ftran%2F1969-136-00%2FS0002-9947-1969-0238365-3%2FS0002-9947-1969-0238365-3.pdf&usg=AOvVaw10G_cIEAPk6JZJ3Yu2wS4l). By Theorem 1 in that paper, a full probability measure $\mu$ on $\R^d$ is (operator) stable if and only if for each natural $n$ there are a $d\times d$ real matrix $B\_n$ and a vector $b\_n\in\R^d$ such that $X\_1+\cdots+X\_n$ equals $B\_nX+b\_n$ in distribution, where $X,X\_1,X\_2,\ldots$ are independent random vectors in $\R^d$ each with distribution $\mu$. (A measure $\mu$ on $\R^d$ is called full if the linear span of its support is $\R^d$.)


If a random vector $X$ in $\R^d$ has a stable distribution, we may say that $X$ itself is stable. Examples of stable random vectors include random vectors with independent stable coordinates. Also, if a a random vector $X$ in $\R^d$ is stable, then any nonsingular affine transformation $BX+b$ of $X$ is also obviously stable (here $B$ is any nonsingular $d\times d$ real matrix and $b\in\R^n$). In particular, any Gaussian distribution on $\R^d$ is stable. 


**Added:** In particular, it follows that the understanding of the notion of operator-stable distributions highlighted in the OP's answer is incorrect, even with the additional assumption that $X\_1$ and $X\_2$ be independent. Indeed, consider, e.g., the case when $X=\begin{bmatrix}\xi\\ \eta\end{bmatrix}$, where $\xi$ and $\eta$ are independent real-valued random variables with the standard Cauchy and Gaussian distributions, respectively. Then the distribution of $X$ is operator-stable. 
Let $X\_1=\begin{bmatrix}\xi\_1\\ \eta\_1\end{bmatrix}$ and $X\_2=\begin{bmatrix}\xi\_2\\ \eta\_2\end{bmatrix}$ be independent copies of $X$. Let $A:=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $B:=\begin{bmatrix}0&1\\1&0\end{bmatrix}$, and suppose that 
\begin{equation}
 Y:=AX\_1+BX\_2\D CX+d=:Z \tag{1}
\end{equation}
for some real matrix $C=\begin{bmatrix}c\_{11}&c\_{12}\\ c\_{21}&c\_{22}\end{bmatrix}$ and a vector $d$. Then $d=0$, since the distributions of the random vectors $Y$ and $CX$ are symmetric about $0$.
We have 
$Y=\begin{bmatrix}\xi\_1+\eta\_2\\ \eta\_1+\xi\_2\end{bmatrix}$, and so, for any $t=\begin{bmatrix}r\\ s\end{bmatrix}\in\R^{2\times1}$,
\begin{equation}
 \E e^{it\cdot Y}=\exp\{-|r|-|s|-r^2/2-s^2/2\},
\end{equation}
whereas 
\begin{equation}
 \E e^{it\cdot Z}=\exp\{-|rc\_{11}+sc\_{21}|-(rc\_{21}+sc\_{22})^2/2\},
\end{equation}
so that 
\begin{equation}
 |r|+|s|+(r^2+s^2)/2=|rc\_{11}+sc\_{21}|+(rc\_{21}+sc\_{22})^2/2
\end{equation}
for all real $r,s$. Letting here $r\to\infty$ and $s=0$, we have $c\_{21}=\pm1$. Similarly, $c\_{22}=\pm1$. On the other hand, letting $r=s\to\infty$, we have $(c\_{21}+c\_{22})^2/2=1$, which contradicts the conditions $c\_{21}=\pm1$ and $c\_{22}=\pm1$. This contradiction shows that (1) cannot hold here for any $C,d$. 


