Find pdf of $Y = \frac{1}{X}$ Let $X$ be a uniform random variable on $(-1 , 1)$ and $Y = \frac{1}{X}$. Find pdf of random variable $Y$. My problem is interval of interest. We can solve this problem on interval $(a , b)$ if either $a\gt 0 $ or $b\lt 0$ but I don't know how to solve when interval includes positive and negative numbers.
 A: Starting from cdf, \begin{align}F_Y(\alpha)&=P\{Y\le \alpha\}=P\left\{\frac1X\le \alpha\right\}=P\left\{\frac{1-\alpha X}X\le 0\right\}=\\&=P\{(\alpha X-1\le 0\land X<0)\lor(\alpha X-1\ge 0\land X>0)\}=\\&=\begin{cases}P\left\{\left(X\le \frac1\alpha\land X<0\right)\lor\left(X\ge \frac1\alpha\land X>0\right)\right\}&\text{if }\alpha>0\\P\left\{\left(X\ge \frac1\alpha\land X<0\right)\lor\left(X\le \frac1\alpha\land X>0\right)\right\}&\text{if }\alpha<0\\ P\{X<0\}&\text{if }\alpha=0\end{cases}=\\&=\begin{cases}P\left\{X<0\lor X\ge \frac1\alpha\right\}&\text{if }\alpha>0\\P\left\{ \frac1\alpha\le X<0\right\}&\text{if }\alpha<0\\ P\{X<0\}&\text{if }\alpha=0\end{cases}=\begin{cases}\frac12&\text{if }0<\alpha<1\\ 1-\frac1{2\alpha}&\text{if } 1\le\alpha<\infty\\ \frac12&\text{if }-1<\alpha<0\\ -\frac1{2\alpha}&\text{if }\alpha\le-1\\ \frac12&\text{if }\alpha=0\end{cases}=\\&=\begin{cases}-\frac1{2\alpha}&\text{if }\alpha\le-1\\ \frac12&\text{if }-1<\alpha<1\\ 1-\frac1{2\alpha}&\text{if }\alpha\ge1\end{cases}\end{align}
Therefore it is apparent that any function $f$ such that $f(x)=\frac12x^{-2}$ for all $x\in(-\infty,-1)\cup(1,\infty)$ and $f(x)=0$ for all $x\in (-1,1)$ is a pdf for $Y$.
A: For a possible value $a\in[1,\,\infty)$ of $Y$, $$P(|Y|\le a)=P(Y\in[-a,\,a])=2P(Y\in[1,\,a])=2P\left(X\in\left[\frac1a,\,1\right]\right)=1-\frac1a.$$So $A:=|Y|$ has cdf $\left(1-\frac1a\right)^\prime=\frac{1}{a^2}$. Since $X$ has a symmetric distribution, so does $Y$, with CDF $\frac{1}{2y^2}$.
A: For $y\ge1$ we have
\begin{align}
f_Y(y) = {} & \frac d {dy} \Pr(Y\le y) \\[10pt]
= {} & \frac d {dy} \Pr(Y<0 \text{ or } 0<Y\le y) \\[10pt]
= {} & \frac d {dy} \Pr(0<Y\le y) \\
& \quad \text{since} \tfrac d{dy} \Pr(Y<0)=0 \text{ because} \\
& \quad {\Pr(Y<0)} \text{ does not change as $y$ changes.} \\[12pt]
= {} & \frac d {dy} \Pr\left( \frac 1 y \le X \right) = \frac d {dy} \left( 1- \frac 1 y \right) \\[10pt]
= {} & \frac 1 {y^2}.
\end{align}
And then symmetry of the distribution does the rest.
