The standard cauchy distribution was given as follows $f_X(x) = \frac{1}{\pi} \frac{1}{1+x^2} \;\;\;\;\; x\in \mathbb{R}$
I need to find the distribution for $Y= \frac{1}{X}$ and show it is the same as $X$.
I managed to do the problem easily for $y \neq 0$ that $f_Y(y) = (F_X(\frac{1}{y}))'(\frac{1}{y})' = f_X(y)$.
For $y = 0$, can I use the right continuous property of CDFs?
I was hoping to do $F_Y(0) = \lim\limits_{t \to 0+} F_Y(t)$ and then go on to differentiate from there.
Is what I am doing correct? I am unsure because we haven't verified that $F_Y$ is in fact a CDF or something.
![[1]: https://i.sstatic.net/ovBO5.png](https://i.sstatic.net/0ljSI.png)
If the random variable $X$ is standard Cauchy then so is $1/X$ : In this question, they used a theorem called inverse-jacobian which we have not done yet.
Cumulative distribution function of Cauchy distribution : I also saw this and realise that there are correct methods I can compare to.
Best Answer
$f_Y(y)=f_X(y)$ almost everywhere (w.r.t. Lebesgue measure) implies that $X$ and $Y$ have the same distribution. There is no need to worry about $F_Y(0)$.
Some basics: If the densities $f$ and $g$ of $X$ and $Y$ are equal almost everywhere then $P(X \leq x)=\int_{-\infty}^{x} f(t)dt=\int_{-\infty}^{x} g(t)dt=P(Y \leq x)$ for all $x$. Changing the value of an integrand at one point or on a set of measure zero does not change the value of the integral.