I am supposed to find a general condition for a PDF of a random variable X, so that the distributions of X and 1/X are the same. I showed this for standard Cauchy distribution using the formula $ g(y) = \frac{ 1 }{ y^2 } f\left( \frac{ 1 }{ y } \right)$ , where g is PDF of 1/X and f PDF of X. I thought this formula would be a good start for finding a general condition, but I didn't come far with it. I'd appreciate some hints.
It also came to my mind that the only possible condition could be that the PDF should satisfy $ f(y) = \frac{ 1 }{ y^2 } f\left( \frac{ 1 }{ y } \right)$, but that seems a bit too easy.
The condition looks correct to me as a consequence of the change of variable formula and that the function that $x \rightarrow 1/x $ is one to one, apart from sets of measure 0.
Note that if you fix any positive function $z(x) :[-1,1] \rightarrow R^+$, with integral over [-1,1] equal to 1/2 , you can extend it over the whole line using $z(1/x)=x^2z(x)$, when x in [-1,1], to a normalized function $f$. The check is a change of variable in the integral of the density.
A random variable $Z$ with the resulting density $f(x)$ than would have the same distribution as its inverse $1/Z$. In particular, there are infinite such functions. As a particular case starting with $z(x)=1/(1+x^2)$ (appropriately normalized) you recover the Cauchy distribution.