Deduce PDF of $1/X$ from PDF of $X$ Assume that we have a pdf $f(x)$ with random variable $X.$ How can you find the pdf of the random variable $Y = 1/X$?
I checked some examples including Cauchy distribution and I saw Jacobian in Cauchy distribution's exapmle.
Anyway, for or example if we have a cdf $F(x)$ with random variable $X,$ and if we want to find the cdf of r.v. $Y = aX+b,$ we do the following:
$$F(x) = P( X \le x ) \Longrightarrow F(y) = P( Y \le y ) = P( aX+b \le y ) = P\left( X \le \frac{y-b}a \right).$$
How about $Y = 1/X$?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\tt\mbox{The easy way}\ldots}$

\begin{align}
&\pp\pars{y}\,\dd y=\,{\rm P}\pars{x}\,\dd x\ \imp\
\pp\pars{y}\verts{-\,{\dd x \over x^{2}}}=\,{\rm P}\pars{x}\,\verts{\dd x}\ \imp\
\color{#66f}{\large\pp\pars{y}}=\,{\rm P}\pars{x}x^{2}
\\[5mm]&=\color{#66f}{\large{\,{\rm P}\pars{1/y} \over y^{2}}}
\end{align}
A: The univariate transformation formula is this:
For $Y=g(X)$, $f_Y(y)=f_X(g^{-1}(y))|(g^{-1})'(y)|$.
Applied to $Y=\frac{1}{X}$, this gives:
$f_Y(y)=\frac{f_X(\frac{1}{y})}{y^2}$.
A: If $X$ is always positive you can say
\begin{align}
f_Y(y)& =\frac{d}{dy}\Pr\left( Y\le y\right) = \frac d{dy}\Pr\left(X\ge \frac 1 y\right) = \frac{d}{dy} \left(1-F_X\left(\frac 1 y\right)\right) \\[10pt]
& = -f_X\left(\frac 1 y\right)\cdot\frac d {dy}\,\frac 1 y \\[10pt]
& = f_X\left(\frac 1 y\right)\cdot \frac 1 {y^2}
\end{align}
and then do simplifications that depend on which function $f$ is.
But that won't work for the Cauchy distribution.  You have to do it piecewise, considering positive and negative parts separately.
(And if $X\sim\mathrm{Cauchy}$ then $1/X\sim\mathrm{Cauchy}$.  If you're collecting the basic facts about this distribution, then include that one.)
