Density of a quotient. Let $A,B\overset{\tt iid}{\sim} \mathcal{N}(0,1)$, $X=ae^A$ and $Y=be^B$. What is the density of $Z=X/Y$?
Assume that $A$ and $B$ are independent random variables.
Note that $Z=X/Y = \frac{ae^{A}}{be^{B}} = \frac{a}{b}e^{A-B}$.
Since $A$ and $B$ are independent $\mathcal{N}(0,1)$, we have that $A-B\sim\mathcal{N}(0,2)$. So $e^{A-B}\sim \textrm{LogNormal}(0,2)$.
Hence $Z$ is $\frac{a}{b}$ times a $\textrm{LogNormal}(0,2)$ random variable. See if you can write out the density of this now.
Thank you to Minus One-Twelfth https://math.stackexchange.com/users/643882/minus-one-twelfth for the hints and help.
There is a $\tt Theorem$ that states:
$\textbf{IF}$ ${X}_{{j}={1}:{n}}\overset{\tt \perp}{\sim}\mathcal{N}({\mu}_{j},{\sigma}^2_{j})$, $\bf THEN$
$$
\sum_{{j}={1}}^{n} {\alpha}_{j}{X}_{j}\sim\mathcal{N}\bigg( \sum_{{j}={1}}^{n}{\alpha}_{j}{\mu}_{j}, \sum_{{j}={1}}^{n}{\alpha}^2_{j}{\sigma}^2_{j} \bigg)
$$
So $\bf IF$ ${\alpha}_{1}={1},{X}_{1}={A},{\mu}_{1}={0},{\sigma}_{1}={1},{\alpha}_{2}=-{1},{X}_{2}={B},{\mu}_{2}={0}$, and ${\sigma}_{2}={1}$, $\bf THEN$ $$ \sum_{{j}={1}}^{2}{\alpha}_{j}{X}_{j}={A}-{B}={V}\sim \mathcal{N}({0},{2}) \tag*{$(1)^2(1)^2+(-1)^2(1)^2=2=\sqrt{2}^2$} $$
$\bf IF$ ${V}\sim\mathcal{N}({\mu}_{v},{\sigma}^2_{v})$, $\bf THEN$ ${f}_{V}({v})=\frac{1}{\sqrt{{2}{\pi}}{\sigma}} {e}^{-\frac{1}{2}\left(\frac{{v}-{\mu}_{v}}{{\sigma}_{v}}\right)^2}$
$\huge \bf{solution}$
$\displaystyle {Z}=\frac{X}{Y}=\frac{{a}{e}^{A}}{{b}{e}^{B}}=\frac{a}{b}{e}^{{A}-{B}}$
Letting ${V}={A}-{B}$ and ${k}=\frac{a}{b}$
\begin{align*} {F}_{Z} ({z} )
&
=
\mathbb{P} ({Z}<{z})
\\&
=
\mathbb{P} ({k}{e}^{V}<{z} )
\\& =
\mathbb{P} \bigg({V}<\ln{\frac{z}{k}} \bigg)
\\&
=
\int_{-\infty}^{{v}=\ln{\frac{z}{k}}} \left[ {f}_{V} ( {v} )=\displaystyle \frac{1}{\sqrt{{2}{\pi}}{\sigma}}{e}^{-\frac{1}{2}\left(\frac{{v}-{\mu}}{\sigma}\right)^2}\right] d{v}
\\&
=
\frac{1}{\sqrt{{2}{\pi}}{\sigma}}\int_{-\infty}^{{v}=\ln{\frac{z}{k}}} {e}^{-\frac{1}{2}\left(\frac{{v}-{\mu}}{\sigma}\right)^2} d{v}
\end{align*}
\begin{align*}
{f}_{Z} ({z} )
&
=
{F}_{Z}' ({z} )
\\&
\\&
= \frac{1}{\sqrt{{2}{\pi}}{\sigma}}\left( {e}^{-\frac{1}{2}\left(\frac{ \ln{\frac{z}{k}}-{\mu} } {\sigma}\right)^2} \frac{d }{d{z}}\ln{\frac{z}{k}}-{e}^{-\frac{1}{2}\left(\frac{{-\infty}-{\mu}}{\sigma}\right)^2}\frac{d }{d{z}}\ln{\frac{z}{k}}\right) \\& = \frac{1}{\sqrt{{2}{\pi}}{\sigma}}\left( {e}^{-\frac{1}{2}\left(\frac{ \ln{\frac{z}{k}}-{\mu} } {\sigma}\right)^2} \frac{1}{\frac{z}{k}}(\frac{1}{k})-{0}\right) \\& = \frac{1}{\sqrt{{2}{\pi}}{\sigma}{z}}{e}^{-\frac{1}{2}\left(\frac{ \ln{z}-({\mu}+\ln{k}) } {\sigma}\right)^2} \tag*{$\color{red}{\blacksquare}$} \end{align*} In this example ${\mu}={0}$, ${\sigma}=\sqrt{2}$ and ${k}=\frac{a}{b}$ $$\frac{1}{\sqrt{{2}{\pi}}\sqrt{2}{z}}{e}^{-\frac{1}{2}\left(\frac{ \ln{z}-({0}+\ln{\frac{a}{b}}) }{\sqrt{2}}\right)^2} = \frac{1}{{2}\sqrt{{\pi}}{z}}{e}^{-\frac{1}{4}\left( \ln{\left( \frac{b}{a}{z}\right) }\right)^2} \tag*{$\color{red}{\blacksquare}$}$$
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