What can we say about distributions of random variables $X$ such that $X$ and its inverse $1/X$ have the same distribution?

Question

What can we say about random variables such that it and its inverse have the same distribution? One example is Cauchy distributed random variables, easily proved via the fact that if $X, Y$ are IID standard normal then both $X/Y$ and $Y/X$ are Cauchy distributed.

If we restrict to random variables having a density, we can show that the density function $f$ must satisfy the functional equation $$ u(1/t)=t^2 u(t) $$ for $t\not =0$. This math SE post and this one have information on solutions.

I'm interested both in general solutions and solutions when restricted to symmetry, and in references.

Answer 1

Any positive function $f(x)$ defined for $-1 \leq x \leq 1$ and for which you have $$\int_{-1}^1 f(x) \, dx =0.5$$ can be extended to a function $$g(x) = \begin{cases} f(x) & \quad \text{if } -1 \leq x \leq 1 \\ f(1/x)/x^2 &\quad \text{otherwise}\end{cases}$$ such that $$\int_{-\infty}^\infty g(x) \, dx =1$$

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More to say is that the moments can be computed by:

$$\mu_n = \int_{-1}^1 \left(x^n+\frac 1 {x^n}\right)f(x) \, dx$$

and $\lim_{x \to 0 } f(x) \neq 0 $ is a sufficient condition for the moments to be undefined or infinite.

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That function for the moments can be derived by using $$\int_1^\infty x^n g(x) dx = \int_0^1 t^{-n} g(1/t)/t^2 dt = \int_0^1 t^{-n} g(t) dt$$

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Also, a ratio of two the same distributions has the property that 1/X and X have the same distribution. (https://stats.stackexchange.com/a/350537/164061)