Distribution of a division of two absolutely continuous random variables Assume $Z,Y $ are independent standard normal random variables.
Find the distribution of $ Z/Y $.
The answer is that $ Z/Y $ is absolutely continuous random variable with density
$$ f_{\frac{Z}{Y}}(x)=\frac{1}{\pi(1+x^2)} $$
Here's what I don't understand:
I tried to write:
\begin{align}
F_{\frac{Z}{Y}}(t) & =\mathbb{P}\left(\frac{Z}{Y}\leq t\right)=\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y<0\right)+\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y>0\right) \\[8pt]
& =\mathbb{P} (Z\geq tY\cap Y<0)+\mathbb{P}\left(Z\leq tY\cap Y>0\right) \\[8pt]
& =\mathbb{P}(Z\geq tY) \mathbb{P}(Y<0)+\mathbb{P}(Z\leq tY) \mathbb{P}(Y>0) \\[8pt]
& =\frac{1}{2}\mathbb{P}(Z\geq tY)+\frac{1}{2}\mathbb{P} (Z\leq tY)
\end{align}
And then to calculate each term. for example :
$$ \mathbb{P}(Z\leq tY)=\mathbb{P} ((Z,Y)\in \{ (z,y):z\leq ty \} ) = \intop_{-\infty}^\infty \intop_{-\infty}^{ty} \frac{1}{2\pi} e^{-\frac{z^2+y^2}{2}} \, dz\, dy $$
But this leads me to a wrong answer, and I'm pretty sure that the problem is somewhere in what I wrote before.
Any help would be appreciated, thanks in advance.
 A: Draw a picture of a $(y,z)$-plane and sketch the region consisting of points $(y,z)$ satisfying $z/y \le t$. This should look like two cones (or pizza slices) with points at the origin and extending infinitely away from the origin. Because the distribution of $(Y,Z)$ is rotationally symmetric, the probability that $(Y,Z)$ lies in this region is proportional to the total angle that these cones sweep out. The angle in question will involve arctangent and $t$, and differentiating to get the PDF will lead to the $\frac{1}{1+t^2}$ expression.
A: Let $W:=(Y,Z)$, then as $Y$ and $Z$ are independent $f_W=f_Y\cdot f_Z$, therefore for $A_c:=\{(s,t)\in \mathbb{R}^2 :s/t\leqslant c\}$ we have that
$$
\begin{align*}
\Pr [Z/Y\leqslant c]&=\Pr [W\in A_c]\\&=\int_{\mathbb{R}^2}\mathbf{1}_{A_c}(s,t)f_Y(s)f_Z(t)\mathop{}\!d(s,t)\\
&=\frac1{2\pi }\left(\int_{(0,\infty )}e^{-t^2/2}\int_{(-\infty ,tc]}e^{-s^2/2}\mathop{}\!d s \mathop{}\!d t+\int_{(-\infty ,0)}e^{-t^2/2}\int_{[tc,\infty )}e^{-s^2/2}\mathop{}\!d s \mathop{}\!d t\right)
\end{align*}
$$
Differentiating respect to $c$ gives
$$
f_{Z/Y}(c)=\frac1{2\pi}\left(\int_{(0,\infty )}te^{-\frac1{2}(1+c)^2t^2}\mathop{}\!d t-\int_{(-\infty ,0)}te^{-\frac1{2}(1+c)^2t^2}\mathop{}\!d t\right)\\
=\frac1{\pi }\int_{(0,\infty )}te^{-\frac1{2}(1+c)^2t^2}\mathop{}\!d t=\frac1{\pi(1+c^2)}
$$
∎

EDIT: alternatively, using the change to polar coordinates $(t,s)=(r\cos \alpha ,r\sin \alpha )$ we have that the condition $s/t\leqslant c$ is equivalent to $\tan \alpha \leqslant c \Rightarrow  -\pi/2<\alpha \leqslant \arctan c$, and as the tangent have two full periods on $(-\pi,\pi)$ then
$$
\int_{\mathbb{R}^2}\mathbf{1}_{A_c}(s,t)f_Y(s)f_Z(t)\mathop{}\!d(s,t)=\frac1{\pi }\int_{[0,\infty )\times (-\pi/2,\arctan c]}re^{-r^2/2}\mathop{}\!d (r,\alpha )
=\frac1{\pi}(\arctan c+\frac{\pi}{2})
$$
where the result follows again by differentiation.
A: When you are separating two conditions (here: ${Y>0}$ and ${Y<0}$) you have to take it into account. For instance:
$ \Pr\{Z\leq tY \cap Y>0\} = \Pr\{Z\leq tY | Y>0\} \Pr\{Y>0\}$.
The condition effects the limits of integral that follows, i.e., where you have conditioned $Y$ to be positive, $y$ starts at $0$, not $-\infty$.
A: You seem to have concluded that $\Pr (Z\geq tY\cap Y>0) = \Pr(Z\geq tY) \Pr(Y>0).$
But these events are not independent, as may be seen by drawing a graph. Try it, for example, with $t=3.$ Look at the region of the $y,z$-plane where $y>0$ and $z>3y$ and the region where $y<0$ and $z>3y.$ These will get two very different probabilities.
You have $\Pr(Y>0) = 1/2,$ so $\Pr(Z\ge tY\mid  Y>0) = \dfrac{\Pr(Y\ge tY \cap Y>0)}{1/2}.$
This is $\displaystyle 2\int_0^{+\infty} \int_{-\infty}^{ty} f(y,z)\,dz\,dy,$ where $f$ is the joint density.
PS: If you want to use polar coordinates here, notice that $\big[y>0 \text{ and } z>ty\big]$ amounts to $\big[ \arctan t < \theta < \pi/2 \big],$ and from there to the bottom line is a very short route.
