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Question

Let $X_1, \ldots, X_n$ be independent and identically distributed continuous random variables with a positive continuous joint density function $f(x_1, \dots, x_n)$ satisfying the relation $$f(x_1, \ldots, x_n) = f(y_1, \ldots, y_n)\quad \mathrm{whenever}\quad \lvert x_1 \rvert + \ldots + \lvert x_n \rvert = \lvert y_1 \rvert + \ldots + \lvert y_n \rvert.$$ What are all possible distributions of $X_1$?

My working

If $\lvert x_1 \rvert = -x_1$, then all possible distributions of $X_1$ have density of the form $$f_{X_1}(x_1) = \frac {e^{-cx_1}} {\int^{\infty}_{-\infty} e^{-cx}\ \mathrm{d}x}\ \forall\ c \in \mathbb{R}^-.$$

If $\lvert x_1 \rvert = x_1$, then all possible distributions of $X_1$ have density of the form $$f_{X_1}(x_1) = \frac {e^{-cx_1}} {\int^{\infty}_{-\infty} e^{-cx}\ \mathrm{d}x}\ \forall\ c \in \mathbb{R}^+.$$

Thus, combining both cases, all possible distributions of $X_1$ have density of the form $$f_{X_1}(x_1) = \frac {e^{-\lvert cx_1 \rvert}} {\int^{\infty}_{-\infty} e^{-\lvert cx \rvert}\ \mathrm{d}x}\ \forall\ c \in \mathbb{R}.$$

Is my answer correct? I am very new to radially symmetric distributions, so if I have gone wrong anywhere, any intuitive explanations will be greatly appreciated :)

Ethan Mark
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    A remark: let $|\cdot|$ be a norm on $\mathbb R^n$. In both definitions of "symmetry", the joint density $f$ has the form $f(x_1,\ldots,x_n) = g(|(x_1,\ldots,x_n)|)$. In this question, $|\cdot|$ is the $1$-norm, and in your previous question $|\cdot|$ is the $2$-norm. – Gabriel Romon Apr 28 '21 at 07:27
  • You can further simplify this, I think – JungleKing Apr 28 '21 at 16:42

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