information:filtering, stylistic:tone
Example: Quotient of two standard normals

Given two standard normal variables U and V, the quotient can be computed as follows. First, the variables have the following density functions: p (u) = 1 2 π e − u 2 / 2 p (v) = 1 2 π e − v 2 / 2 {\displaystyle {\begin{aligned}p(u)&={\frac {1}{\sqrt {2\pi }}}e^{-{u^{2}}/{2}}\\p(v)&={\frac {1}{\sqrt {2\pi }}}e^{-{v^{2}}/{2}}\end{aligned}}}

We transform as described above: Y = U / V Z = V {\displaystyle {\begin{aligned}Y&=U/V\\Z&=V\end{aligned}}}

This leads to: p (y) = ∫ − ∞ ∞ p U (y z) p V (z) | z | d z = ∫ − ∞ ∞ 1 2 π e − 1 2 y 2 z 2 1 2 π e − 1 2 z 2 | z | d z = ∫ − ∞ ∞ 1 2 π e − 1 2 (y 2 + 1) z 2 | z | d z = 2 ∫ 0 ∞ 1 2 π e − 1 2 (y 2 + 1) z 2 z d z = ∫ 0 ∞ 1 π e − (y 2 + 1) u d u u = 1 2 z 2 = − 1 π (y 2 + 1) e − (y 2 + 1) u | u = 0 ∞ = 1 π (y 2 + 1) {\displaystyle {\begin{aligned}p(y)&=\int _{-\infty }^{\infty }p_{U}(yz)\,p_{V}(z)\,|z|\,dz\\&=\int _{-\infty }^{\infty }{\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}y^{2}z^{2}}{\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}z^{2}}|z|\,dz\\&=\int _{-\infty }^{\infty }{\frac {1}{2\pi }}e^{-{\frac {1}{2}}\left(y^{2}+1\right)z^{2}}|z|\,dz\\&=2\int _{0}^{\infty }{\frac {1}{2\pi }}e^{-{\frac {1}{2}}\left(y^{2}+1\right)z^{2}}z\,dz\\&=\int _{0}^{\infty }{\frac {1}{\pi }}e^{-\left(y^{2}+1\right)u}\,du&&u={\tfrac {1}{2}}z^{2}\\&=\left.-{\frac {1}{\pi \left(y^{2}+1\right)}}e^{-\left(y^{2}+1\right)u}\right|_{u=0}^{\infty }\\&={\frac {1}{\pi \left(y^{2}+1\right)}}\end{aligned}}}

This is the density of a standard Cauchy distribution.