Show that $1/T$ has the same distribution as $T$ Let T be the ratio X/Y of two i.i.d. N (0, 1) r.v.s. X, Y . This is the Cauchy distribution and, as shown  it has PDF $f_T(t)= \frac{1}{\pi (1+t^2)}$
(a) Show that $1/T$ has the same distribution as $T$ using calculus, after first finding the CDF of $1/T$.
(b) Show that $1/T$ has the same distribution as $T$ without using calculus, in 140 characters or fewer.

a) $Y= \frac 1T$, $P(Y \le y)=P(\frac1T \le y)=P(T \le 1/y)= \int_{-\infty}^{1/y} \frac{1}{\pi} \frac {1}{(1+t^2)}dt=\frac {tan^{-1}(1/y)}{\pi}$; $y \in (-\infty;\infty)$
$f_Y(y)= \frac {-1}{\pi x^2+\pi}$
b) Can you help me with $b$ and check $a$?
If $y&gt;0$ then $1/T\le y$ if $T&lt;0$ or $T\ge1/y,$ so $$F_Y(y) = 1/2 + P(T&gt;1/y)= 1- \frac{1}{\pi}\tan^{-1}(1/y).$$ If $y&lt;0$ then $1/T \le y$ if $1/y \le T &lt; 0.$ So $$F_Y(y) = P(1/y &lt;T&lt;0) = -\frac{1}{\pi}\tan^{-1}(1/y).$$ This can be condensed into $$ F_Y(y) = 1/2+\frac{1}{\pi}\cot^{-1}(1/y) $$ for all $y\ne 0$ and $1/2$ for $y=0.$
$$P(\frac{1}{T} \le y) = P(T \ge \frac{1}{y}) =  \int_{\frac{1}{y}}^{\infty} \frac{1}{\pi(1+t^2)}dt$$

$$=\left[\frac{1}{2}-\frac{cot^{-1}(y)}{\pi} \right]$$

$$ F_{Y}(y)= \left[\frac{1}{2}-\frac{cot^{-1}(y)}{\pi} \right]$$

$$f_{Y}(y) = \frac{d}{dy}F_{Y}(y)$$

$$f_{Y}(y) = \frac{1}{\pi(1+y^2)}$$

For Part B)

As @Hw Chu said, 1/T is the ratio Y/X, but X and Y are i.i.d. so this is really the same as X/Y and hence the same density.