Question

Let $$X_1, \ldots, X_n$$ be independent and identically distributed continuous random variables with a positive continuous joint density function $$f(x_1, \dots, x_n)$$. Suppose that the distribution of $$X_1, \ldots, X_n$$ is radially symmetric about the origin, which means that the joint probability density function $$f$$ satisfies $$f(x_1, \ldots, x_n) = f(y_1, \ldots, y_n)\quad \mathrm{whenever}\quad x_1^2 + \ldots + x_n^2 = y_1^2 + \ldots + y_n^2.$$ What are all possible distributions of $$X_1$$?

My working

The motivation here is to find a function that turns multiplication into addition, so the exponential function comes to mind. However, as the functions represent probability densities, they must also have finite area over the interval $$(-\infty, \infty)$$. Thus, the inverse exponential function is required.

$$\implies f_{X_1}(x_1) = ce^{-x_1^2}$$, where $$c$$ is a constant.

Is my reasoning for deducing the possible distributions of $$X_1$$ correct? If not, how should I approach the question and what should the possible distributions be?

This is my first time encountering radially symmetric distributions, so any intuitive explanations will be greatly appreciated :)

• When you write "a continuous joint density", do you mean that the function $(x_1,\ldots,x_n)\mapsto f(x_1,\ldots,x_n)$ is continuous (in the usual topological sense) ? Apr 27, 2021 at 8:42
• It's a very important piece of information. The answer below uses Cauchy's functional equation for $F$, which is solvable only under some regularity assumptions on $F$ (continuity typically). Apr 27, 2021 at 8:46
• @GabrielRomon I mean, I would suppose that what you said in your first comment is what the question is implying... What other possible ways to interpret "a continuous joint density" are there? Apr 27, 2021 at 8:47
• You're right, I see no other interpretation either. I think the problem can be solved without this continuity assumption on $f$ (this seems to be a well-known result due to Kac). Apr 27, 2021 at 8:51

Let $$r^2 = (x^2+y^2)+ (n-1) 0^2 = x^2 + y^2 + (n-2) 0^2$$. By the radial symmetry, $$f_X(\sqrt{x^2+y^2}) f_X(0)^{n-1}=f_X(x)f_X(y) f_X(0)^{n-2}.$$ Taking logarithm, we have $$\log f_X(\sqrt{x^2+y^2})+\log f_X(0) = \log f_X(x) +\log f_X(y).$$ Thus, $$\log f_X(\sqrt{x^2+y^2})-\log f_X(x) = \log f_X(y)- \log f_X(0).$$ Let $$F(x)=\log f_X(\sqrt x)$$. Then we have $$F(x+y)=F(x)+F(y)-F(0).$$ Let $$G(x)=F(x)-F(0)$$. We have $$G(x+y)=G(x)+G(y)$$ for all positive $$x$$, $$y$$.
Then the usual ways to solve Cauchy's functional equation works. We obtain $$G(x)=Ax$$ for some constant $$A$$, it is $$A=G(1)$$.
Tracing back to $$f_X$$, we obtain $$\log f_X(\sqrt x) = Ax+\log f_X(0).$$ Hence, $$f_X(x)=c e^{Ax^2}.$$ To have a legitimate pdf, we must have $$A<0$$. The constant $$c$$ must be obtained by $$c= \frac1{\int_{-\infty}^{\infty} e^{Ax^2}dx}.$$
• I am using $$x_1=\sqrt{x^2+y^2}, x_2=x_3=\cdots=x_n=0$$ on the LHS and $$x_1=x, x_2=y, x_3=\cdots=x_n=0$$ on the RHS. Apr 27, 2021 at 8:39