Cumulative distribution function of Cauchy distribution 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.
 A: $Y$ 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.
A: An intuitive hint
One interpretation of the Cauchy distribution is related to the distribution of bullet holes on a wall caused by a machine gun which is rotating with uniform angular speed. Picture another machine gun set perpendicular to the first one. Let the two machine guns rotate together. The distribution of the bullet (red and blue) holes will not be distinguishable. 

(Think of the similarity of the triangles depicted.)
A: The question can be asked in general: What is the condition that the distribution of a random variable $X$ be the same as that of its reciprocal $\frac{1}{X}$.
The answer is easy if we assume that the distribution function is continuous. 
$$F_{\frac{1}{X}}(x)=P\left(\frac{1}{X}<x\right)=P\left(\frac1X<x,X<0\right)+P\left(\frac1X<x,X>0\right).$$
( We did not have to care about the $X= 0$  event for $P(X=0)=0$.)
Now,
$$F_{\frac1X}(x)=\begin{cases}
P\left(\frac1X<x,X<0\right),&  \text{ if } x<0 \\
P\left(X<0\right)+P\left(\frac1X<x,X>0\right),&  \text{ if } x>0  \\
\end{cases} .$$ 
If $x<0$ then $\{\frac1X<x,X<0\}=\{X>\frac1x,X<0\}=\{0>X>\frac1x\}$. 
If $x>0$ then $\{\frac1X<x,X>0\}=\{X>\frac1x,X>0\}=\{X>\frac1x\}$. 
With this
$$F_{\frac1X}(x)=\begin{cases}
F_X(0)-F_X\left(\frac1x\right),&  \text{ if } x<0 \\
F_X(0)+1-F_X\left(\frac1x\right),&  \text{ if } x>0  \\
\end{cases} .$$ 
For the densities, if they exist,  we have then
$$f_{\frac{1}{X}}(x)=\frac{1}{x^2}f_X\left(\frac{1}{x}\right), \ x \not =0.$$
And the condition we are talking about is
$$f_X(x)=f_{\frac{1}{X}}(x)=\frac{1}{x^2}f_X\left(\frac{1}{x}\right), x\not =0.$$
The Cauchy distribution is such a distribution. The pdf is $f_X(x)=\frac{1}{\pi(1+x^2)}$. So, for $x\not =0$
$$f_{\frac{1}{X}}(x)=\frac{1}{x^2}\frac{1}{\pi\left(1+\frac{1}{x^2}\right)}=\frac{1}{\pi(1+x^2)}=f_X(x).$$
And, I assume,  we can agree that 
$$f_{\frac{1}{X}}(0)=\frac{1}{\pi}=f_X(0).$$
A: You shouldn't say "Let $X$ be a Cauchy distribution", but rather, "Let $X$ have a Cauchy distribution". $X$ is a random variable, not a distribution; $X$ has a distribution.
If $x>0$ then the event $\left[ \dfrac 1 X \le x \right]$ is the same as $\left[ X<0 \text{ or } X> \dfrac 1 x \right].$
The probability of that is $\dfrac 1 2 + \big( 1 - F_X(1/x) \big).$
\begin{align}
& \frac 3 2 - \left( \frac 1 2 + \frac 1 \pi\arctan \frac 1 x \right) \\[8pt]
= {} & 1 - \frac 1 \pi \left( \frac \pi 2 - \arctan x \right) \\[8pt]
= {} & \frac 1 2 + \frac 1 \pi \arctan x.
\end{align}
If $x<0$ then the event $\left[ \dfrac 1 X \le x \right]$ is the same as $\left[ \dfrac 1 x < X <0 \right].$
The probability of that is $F_X(0) - F_X(1/x).$
\begin{align}
& \frac 1 2 - \left( \frac 1 2 + \frac 1 \pi \arctan \frac 1 x \right) \\[8pt]
= {} & -\frac 1 \pi \left( \frac \pi 2 - \arctan x \right) \\[8pt]
= {} & \frac 1 2 + \frac 1 \pi \arctan x.
\end{align}
If $x=0$ then the event $\left[ \dfrac 1 X \le x\right]$ is the same as $\Big[ X<0 \Big].$
