Let X be a Cauchy distribution with $X\sim\mathrm{Cauchy(1)}$ (so $a=1$).
Prove that $Y=\frac1X$ has the same cumulative distrubtion as $X$.
Now I've tried taking $F_X(x)$ for $a=1$ combined with the identity $\arctan(x)+\arctan(1/x)=\pm\frac\pi2$
But this forces $F_Y(y)=\frac12 \pm \frac12-\frac1{\pi\arctan(x)}\ldots$
I don't see how this is equal.
Best Answer
$Y$ is a transformation of the original Cauchy distribution. There are a couple of ways of dealing with these, see for example http://www2.econ.iastate.edu/classes/econ671/hallam/documents/Transformations.pdf.
Talking through it we find the inverse transformation, so if $Y=\frac{1}{X}$ then we have $X=\frac{1}{Y}=w(Y)$, say. We take the derivaitive of this with respect to $Y$ and $\frac{d}{dy}w(y)=\frac{-1}{y^2}$. Then we follow the formula to get $$f_Y(y)=f_X(w^{-1}(y)).\left|\frac{d}{dy}w^{-1}(y)\right|= \frac{1}{\pi(1+(\frac{1}{y})^2)}.\frac{1}{y^2} = \frac{1}{\pi} \frac{y^2}{y^2+1}.\frac{1}{y^2} = \frac{1}{\pi(y^2+1)}$$ which is a Cauchy distribution. So they must have the same CDF.
It makes sense as we can think of a Cauchy as the quotient of two normal distributions, and this transformation would be like swapping them over.