Method of CDF for Y = 1/X I am trying to solve this question:
Let X be a standard cauchy variable. Define Y to be 1/X. I want to find the CDF of Y. 
My problem: I am finding the CDF to be:
https://arachnoid.com/latex/?equ=%5Cfrac%7B1%7D%7B2%20%7D-%5Cfrac%7B1%7D%7B%5Cpi%20%7Darctan(%5Cfrac%7B1%7D%7By%20%7D)
But as a take the limit to infinity, it doesn't equal 1, it equals 1/2. I searched online for answers but I couldn't fix my problem.

A standard trigonometric identity says
$$
\arctan \frac 1 y = \frac \pi 2 - \arctan y \text{ if } y>0
$$
so we get
$$
\frac 1 2 -\frac 1 \pi \arctan \frac 1 y = \frac 1 2 - \frac 1 \pi\left( \frac \pi 2 - \arctan y \right) \text{ if } y>0
$$
and this simplifies to
$$
\frac 1 \pi \arctan y
$$
Somewhere you should have had $\text{“} + \text{constant''}.$ With the right constant, the limit as $y\to\infty$ is $1.$
And so the density is
$$
\frac 1 {\pi(1+y^2)}.
$$
Then the case where $y<0$ needs to be similarly treated.
Thus you see that if $X$ has a standard Cauchy distribution, then $1/X$ has that same standard Cauchy distribution.
