2
$\begingroup$

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 :)

$\endgroup$
4
  • $\begingroup$ 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) ? $\endgroup$ Apr 27, 2021 at 8:42
  • $\begingroup$ 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). $\endgroup$ Apr 27, 2021 at 8:46
  • 1
    $\begingroup$ @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? $\endgroup$
    – Ethan Mark
    Apr 27, 2021 at 8:47
  • 1
    $\begingroup$ 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). $\endgroup$ Apr 27, 2021 at 8:51

1 Answer 1

2
$\begingroup$

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 $$ c= \frac1{\int_{-\infty}^{\infty} e^{Ax^2}dx}. $$

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 27, 2021 at 8:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .