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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.

kjetil b halvorsen
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    I believe I answered this question at https://stats.stackexchange.com/a/185709/919, at least for variables with positive support (regardless of whether they have densities). The generalization to variables with positive and negative support should be clear. A broader generalization is discussed in remarks at the end of https://stats.stackexchange.com/a/29010/919. BTW, if you need the backslash to represent "not zero," why not use the $\TeX$ markup `\ne`? – whuber Jan 16 '19 at 12:56
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    Ah... I see. I typed this "\" by copying from another post and pasting it. Some (many?) keyboards [accept ASCII character code input,](https://tools.oratory.com/altcodes.html) but I haven't been successful with this one: try it, but YMMV. – whuber Jan 16 '19 at 13:15

1 Answers1

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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 $$\int_{-\infty}^\infty g(x) \, dx =1$$

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.

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

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)

Sextus Empiricus
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    This is a special case of the general construction I described at https://stats.stackexchange.com/a/29010/919. The group $G$ in this case is the pair of functions $e(x)=x$ and $f(x)=1/x$ defined on the extended real line $\mathbb{R}\cup\{\infty\}.$ The general construction provides a comprehensive answer. – whuber Jan 16 '19 at 14:15
  • The function for the moments is derived by $$\int_1^\infty x^n g(x) dx = \int_0^1 t^{-n} g(1/t)/t^2 dt = \int_0^1 t^{-n} g(t) dt$$ – Sextus Empiricus May 28 '20 at 15:18
  • If you instead consider integrands of the form $f(x)\mathrm{d}x/x,$ your formulas will become symmetrical: $f(x)$ will be interchanged with $f(1/x),$ *but that's all.* No longer does that annoying factor of $1/t^2$ appear in the calculations. – whuber May 28 '20 at 17:10