Prove quotient of two $N(0,1)$ is $\text{Cauchy}(0,1)$ Problem: Show that if $X$ and $Y$ are independent $N(0,1)$-distributed random variables, then $X/Y ∈ C(0,1)$.
Question: I don't know how to proceed below. I want to prove that the PDF of $X/Y$ is Cauchy. PS. I looked on wiki and they defined $C$ as $X/Y$
Attempt: 
The expression looked hairy and wolfram could not integrate it. 
Note to self: problem 8.
 A: Let $Z=X/Y$.
Then
$$
Ef(Z) = \int f(x/y)n(x)n(y) dxdy
= \int f(z) n(yz)n(y)|y|dydz
$$
With $n(x) = \frac1{\sqrt{2\pi}} \exp(-x^2/2)$
and $\sigma(z)^2 = 1/(1+z^2)$,
$$
n(y)n(yz) = \frac1{2\pi} \exp((1+z^2)(-y^2/2))
= \frac 1{2\pi} \exp(-y^2/2\sigma(z)^2)
\\
\int n(yz)n(y)|y|dy
=  2\int_0^\infty \frac 1{2\pi} \exp(-y^2/2\sigma(z)^2)ydy\\
= \frac{\sigma(z)^2}{\pi} \int_0^\infty \exp(-y^2/2\sigma(z)^2)\frac{y}{\sigma(z)}
\frac{dy}{\sigma(z)} = \frac{\sigma(z)^2}{\pi}
\\
Ef(Z) = \int f(z) n(yz)n(y)|y|dydz
= \int f(z)  \frac{\sigma(z)^2}{\pi} dz
= \int f(z)  \frac{dz}{\pi(1+z^2)} 
$$
A: An alternative proof taken from here - I find it helpful as most steps are explained intuitively. Please correct me if something is incorrect. 
Proof
Start with the CDF which is $P(\frac{X}{Y} \leq t)$ which is an event, and we want to find the probability of this event.
$$ P(\frac{X}{Y} \leq t)  = P(\frac{X}{|Y|} \leq t) = P(X \leq |Y|t) \text{  by symmetry of } N(0,1) $$
To get a probability, integrate the joint pdf over the region of interest. For $Y$ this is between $- \infty$ to $\infty$ but for $X$ this is from $- \infty$ to $t |y|$ !
$$P(X \leq |Y|t)  = \int_{- \infty}^{\infty} \int_{- \infty}^{t |y|} f(x,y) \ dx \ dy $$ 
As $X$ and $Y$ are i.i.d. the joint PDF can be found by multiplying the pdf : $f_{X,Y}(x,y) = f_X(x) f_Y(y)$ 
\begin{aligned}
 P(X \leq |Y|t)  &= \int_{- \infty}^{\infty} \int_{- \infty}^{t |y|} \frac{1}{\sqrt{2 \pi}} \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2}(x^2 + y^2)}\ dx \ dy 
 \\
 & = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty}  e^{- \frac{1}{2} y^2} \int_{- \infty}^{t |y|}  \frac{1}{\sqrt{2 \pi}} e^{- \frac{1}{2} x^2}\ dx \ dy 
\end{aligned}
Recognize that the inner integral cannot be computed analytically, however this is the normal CDF $\Phi(t |y|)$ 
\begin{aligned}
 P(X \leq |Y|t)   & = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty}  e^{- \frac{1}{2} y^2} \Phi(t |y|) \ dy 
\end{aligned}
This is an even function so we can remove the absolute value and multiply by 2. 
\begin{aligned}
 P(X \leq |Y|t)   & = \frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty}  e^{- \frac{1}{2} y^2} \Phi(t y) \ dy 
\end{aligned}
The PDF is the derivative of the CDF, so we can take the derivative of the above equation. Under mild technical conditions we can swap the derivative and the integral. Note that $y$ is a dummy variable so here we need to take the derivative w.r.t. $t$
\begin{aligned}
 PDF = F'(t) & = \frac{d}{dt}  \frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- \frac{1}{2} y^2} \Phi(t y) \ dy 
 \\
  & = \frac{2}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{- \frac{1}{2} y^2} \ \frac{d}{dt} \Phi(t y) \ dy 
\end{aligned}
To compute the derivative use the chain rule:  $\frac{d}{dt} \Phi(t y) = \frac{d \Phi(t y) }{d(ty)} \frac{d (t y) }{d(t)} = y \frac{d \Phi(t y) }{d(ty)} = y \phi(ty)$ where $\phi(ty)$ is the normal pdf of $(ty)$
\begin{aligned}
 F'(t) & = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty}  y \ e^{- \frac{1}{2} y^2} \  \phi(t y) \ dy 
 \\
 & = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} y \  e^{- \frac{1}{2} y^2} \frac{1}{\sqrt{2 \pi}} e^{- \frac{t^2y^2 }{2}} \ dy 
 \\
 & = \frac{1}{\pi} \int_{0}^{\infty} y e^{ - \frac{1}{2}(1 + t^2)y^2 } dy
\end{aligned}
Which can be solved using substitution $u = \frac{1}{2}(1 + t^2)y^2$ so $du = y (1 + t^2) dy$
\begin{aligned}
 F'(t) & =  \frac{1}{\pi} \int_{0}^{\infty} y e^{ - \frac{1}{2}(1 + t^2)y^2 } dy
 \\
 & =  \frac{1}{\pi(1 + t^2)} \int_{0}^{\infty} e^{-u}  \ du
 \\
 & =  \frac{1}{\pi(1 + t^2)}  \ \ \forall t
\end{aligned}
Additional detail about the symmetry of $N(0,1)$ - Taken from here
The probability $P( X/Y\leqslant t)$ may be viewed as
\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)\\
=\; &P(X/|Y|\leqslant t, Y>0) + P(X/|Y|\leqslant t, Y<0)\\
=\; &P(X/|Y|\leqslant t)
\end{aligned}
by symmetry of the standard normal distribution, independence and the fact that $P(Y=0)=0$.
