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If $X = (X_1,...,X_n)' \in \mathbb{R}^n$ has a multivariate normal distribution, any linear combination of its components has an univariate normal distribution. Especially, $w'X$ is identically distributed for every $w\in \mathbb{R}^n$ up to location and scale, i.e. there exist $a>0$ and $b\in\mathbb{R}$ such that $w'X$ and $a\cdot X_1+b$ have the same distribution. This also holds true, if $X$ has a multivariate elliptical distribution. For every $w\in \mathbb{R}^n$, $w'X$ is symmetric around its mean and has finite variance.

If $X_1,...,X_n$ are i.i.d. stable random variables, $w'X$ is identically distributed for every $w\in\mathbb{R}^n$ with $w>0$ up to location and scale. There exist non-symmetric (i.e. skewed) stable distributions which all have infinite variance.

Question: Does there exist a non-elliptical (skewed) multivariate random variable $X$ with finite variance which has the property that every linear combination $w'X$ of its components with non-negative weights $w > 0$ is identically distributed up to location and scale?

My Idea: The characteristic function of $X$ could look like $$\phi_X(t) = e^{it'\mu} \cdot e^{-\frac{1}{2} \psi(t' \Sigma t) \cdot (1 - i \gamma \cdot \text{sgn}(e' t))}$$ for some constants $\mu \in \mathbb{R}^n$, $\gamma\in\mathbb{R}$ and a continuous function $\psi:\mathbb{R}\to\mathbb{R}$. (Here, $e = (1,...,1)'$ and $sgn$ is the signum function.) If $\gamma = 0$, $X$ has an elliptical distribution. If $X$ has a characteristic function $\phi_X$ of the form above, it fulfills the desired property. But, for which $\mu, \gamma, \psi$ does $\phi_X$ correspond to a characteristic function of a random vector $X$?

Edit: Following the idea of kimchi lover:

A stronger requirement is: $w'X \cong ||w|| X_1$ for every $w\in\mathbb{R}^n$ with $w>0$ (i.e. $w_i > 0$ for all $i$). The characteristic function equivalently then satisfies $$\phi_X(t) = \mathbb{E}[e^{it' X}] = \mathbb{E}[e^{i||t|| X_1}] = \phi_X(||t|| e_1) \text{, and} \\ \phi_X(-t) = \phi_X(- ||t|| e_1)$$ for every $t \in \mathbb{R}^n$ with $t > 0$.

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  • $\begingroup$ Aren't you really asking for multivariate distributions invariant under rotations? So that the characteristic functions restricted to lines through the origin are all equal? $\endgroup$ Jan 2 '18 at 14:28
  • $\begingroup$ @kimchilover I think multivariate distributions invariant under rotations would be scaled spherical (and thus elliptical) ones. They indeed fulfill the property, but are not skewed. However, not all rotations are allowed such that more distributions (and especially skewed ones) might also fit the property. $\endgroup$
    – lunos
    Jan 2 '18 at 14:44
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Here is a ham-fisted partial answer. Let $\mu_1$ and $\mu_2$ be two distinct rotationally invariant probability measures on the plane, for which the origin is not an atom. (Such as standard bivariate gaussian and uniform on the unit disk.) Let $H$ be the union of the open right half plane and the open positive $Y$ axis. Let $\mu(A)=\mu_1(A\cap H)+\mu_2(A\cap H^c)$. (If the $\mu_i$ have densities $f_i$, then the density of $\mu$ is $f_1(x,y)$ if $x>0$ or $x=0, y>0$, and $f_2(x,y)$ otherwise.)

In a higher dimensional setting: let $H$ be such that $H\cap(-H)=\phi$ and $\mathbb R^n= H\cap(-H) \cup \{0\}$. If both $\mu_i$ are rotationally invariant, and assign equal mass to the origin, then I think the above recipe works.

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  • $\begingroup$ Thank you for your useful answer! I think your idea is the way to go. I edited my original post to check whether your proposed measure fulfills the property. I think it does not fulfill the symmetry property derived in the last sentence. Maybe one have to define H as the union of the first and third quadrant? $\endgroup$
    – lunos
    Jan 4 '18 at 10:09
  • $\begingroup$ Al of the 1 dimensional projection measures I constructed are asymmetric. $\endgroup$ Jan 4 '18 at 12:39
  • $\begingroup$ One necessary condition is that the marginals X and Y should have identical distributions up to location and scale. This is not the case in your example. $\endgroup$
    – lunos
    Jan 8 '18 at 8:05

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