Probability density function invariant under $x\mapsto 1/x$

Question

Let $f(x)$ be a probability density function (pdf) over $(0,\infty)$, i.e., $f(x)\geq 0$ and $\int_0^\infty dx\ f(x)=1$. Is it possible to characterise the set of such pdfs for which the following holds $$ f(x)=\frac{1}{x^2}f\left(\frac{1}{x}\right)\,? $$ Is this set empty?

Answer 1

The set is not empty since the function $f(x)=\frac{2}{\pi(1+x^2)}$ is in it.

Answer 2

In general, $f:(0,\infty)\to\mathbb{R}$ satisfies $$f\left(\frac{1}{x}\right)=x^2\,f(x)\text{ for every }x>0$$ if and only if there exists $g:[2,\infty)\to\mathbb{R}$ such that $$f(x)=\frac{1}{x}\,g\left(x+\frac{1}{x}\right)\text{ for all }x>0\,.$$ If you want $f$ to be a probability distribution function, then $g:[2,\infty)\to\mathbb{R}_{\geq 0}$ and $$\int_0^\infty\,\frac{1}{x}\,g\left(x+\frac{1}{x}\right)\,\text{d}x=1\,.$$ In Marcel's example, $g(t):=\dfrac{2}{\pi t}$ for all $t\geq 2$. There are other examples such as $g(t):=\dfrac{2}{t^2}$ for all $t\geq 2$, which yields $$f(x)=\frac{2x}{\left(x^2+1\right)^2}\text{ for all }x>0\,.$$ Ergo, you can simply start with any probability density $p:\mathbb{R}\to\mathbb{R}_{\geq 0}$ which is an even function (i.e., $p(-s)=p(+s)$ for all $s\in\mathbb{R}$), and then define $$g(t):=p\Biggl(\text{arccosh}\left(\frac{t}{2}\right)\Biggr)\text{ for all }t\geq 0\,,$$ so that $$f(x)=\frac{1}{x}\,g\left(x+\frac{1}{x}\right)=\frac{1}{x}\,p\big(\ln(x)\big)\text{ for every }x>0\,.$$ Take $p(s):=\dfrac{1}{\sqrt{2\pi}}\exp\left(-\dfrac{s^2}{2}\right)$ for all $s\in\mathbb{R}$, for instance. This gives $$f(x)=\frac{1}{\sqrt{2\pi}\,x}\,\exp\left(-\dfrac{\big(\ln(x)\big)^2}{2}\right)\text{ for }x>0\,.$$ You can also start with a probability density $q:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$, and then set $$f(x):=\frac{1}{2x}\,q\big(|\ln x|\big)\text{ for all }x>0\,.$$ Here is one last example. Let $q(s):=\exp(-s)$ for all $s\geq0$. This gives $$f(x)=\begin{cases}\frac12&\text{if }0<x\leq 1\,,\\ \frac1{2x^2}&\text{if }x>1\,.\end{cases}$$

Answer 3

If the distribution of the positive continuous random variable $X$ is invariant under $X\leftrightarrow X^{-1}$, then the continuous random variable $T=\ln X$ is invariant under $T\leftrightarrow -T$ and viceversa. Therefore any symmetrically distributed continuous random variable furnishes an example.