I have a problem with which I'm struggling for a while.
Suppose that the random variables $X$ and $Y$ are independent and that each has the standard normal distribution. Show that the quotient $X/Y$ has the Cauchy distribution.
I know how to solve it by Jacobian and multivariable transformation but I tried to do this a different way and I can't spot a mistake but my answer gives a contradiction.
My solution: Let $G$ be the cdf of $X/Y $, then
\begin{align*} G(c) &= \mathbb{P}(X/Y \leq c) \\ &= \mathbb{P}(X/Y \leq c \mid Y > 0)\mathbb{P}(Y > 0) + \mathbb{P}(X/Y \leq c \mid Y < 0)\mathbb{P}(Y < 0) \\ &= \frac{1}{2} \big[ \mathbb{P}(X/Y \leq c \mid Y > 0) + \mathbb{P}(X/Y \leq c \mid Y < 0) \big] \\ &= \frac{1}{2} \big[ \mathbb{P}(X \leq cY) + \mathbb{P}(X \geq cY) \big] \\ &= \frac{1}{2}. \end{align*}
Where is the mistake?
Starting from my comment. $$G(c)=\frac{1}{2\pi}(\int_{-\infty}^0\int_{cy}^{\infty}e^{-\frac{x^2+y^2}{2}}dxdy+\int_0^{\infty}\int_{-\infty}^{cy}e^{-\frac{x^2+y^2}{2}}dxdy).$$ Change to polar coordinates and get $$G(c)=\frac{1}{2\pi}\int_0^{\infty}e^-\frac{r^2}{2}rdr\left(\int_{\frac{\pi}{2}}^{\arctan(c)}+\int_{-\frac{3\pi}{2}}^{\arctan(c)}\right)d\theta=\frac{2\arctan(c)+\pi}{2\pi}.$$ When we take the derivative with respect to c we get the density function for the Cauchy distribution $G'(c)=\frac{1}{\pi}\frac{1}{1+c^2}$.