This post is about the process to calculate the cdf of Cauchy distribution from $X, Y \overset{i.i.d} \sim N(0, 1)$
\begin{aligned} &P(\lbrace X/Y\leqslant t, Y>0 \rbrace\cup\lbrace X/Y\leqslant t, Y<0 \rbrace) \\ =\; &P(X/|Y|\leqslant t, Y>0) + P(-X/|Y|\leqslant t, Y<0) \quad \\ =\; &P(X/|Y|\leqslant t, Y>0) + P(X/|Y|\leqslant t, Y<0)\\ =\; &P(X/|Y|\leqslant t) \end{aligned}
But I still have a problem to understand why $P(\lbrace X/Y\leqslant t, Y>0 \rbrace\cup\lbrace X/Y\leqslant t, Y<0 \rbrace)$ is equivalent to $P(X/|Y|\leqslant t, Y>0) + P(-X/|Y|\leqslant t, Y<0)$
First, why is it valid to covert $Y$ to $\mid Y \mid$? It seems that it is related to the symmetry of the normal distribution, but I failed to make a mathematical proof.
Second, the reason $X$ is converted to $-X$ in second line is both $X$ and $-X$ are same distribution with $N(0, 1)$?
Please give me some detailed and answer. Thanks for your help.
$$ Z:=\frac{X}{Y}1\{Y\ne 0\}=\frac{X}{|Y|}1\{Y>0\}-\frac{X}{|Y|}1\{Y<0\}+0\cdot\{Y=0\}. $$ Since the sets $\{Y>0\}$, $\{Y<0\}$, and $\{Y=0\}$ are disjoint, \begin{align} \mathsf{P}(Z\le t)&=\mathsf{P}(\{Z\le t\}\cap \{Y>0\}) \\ &\quad+\mathsf{P}(\{Z\le t\}\cap \{Y<0\}) \\ &\quad+\mathsf{P}(\{Z\le t\}\cap \{Y=0\}) \\ &=\ldots \end{align} Note that $Z=X/|Y|$ on $\{Y<0\}$, $Z=-X/|Y|$ on $\{Y<0\}$, and $\{Y=0\}$ is a null set.