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 $Z=X/Y$?

Question

This a question from the Stat 134 midterm that ended last week(2021.11.21) at the University of California, Berkeley. I'm not in the class but I'm helping fellow classmates review this midterm (I'm very bad at statistics). Over two days, I could do every question but the very last one. I have changed the variables but left the concept.

Let ${A},{B}\overset{\tt iid}{\sim} \mathcal{N}(0,1)$, ${X}={a}{e}^{A}>{0}$ and ${Y}={b}{e}^{B}>{0}$. What is the density of ${Z}={X}/{Y}$?

I've tried $\texttt{change of variable}$, $\texttt{convolution}$, and $\texttt{CDF}$ I'm just lost.

My $\tt CDF$ approach, which ended up being the correct approach. This was as far as I got before great hints from Minus One-Twelfth \begin{align*} {Z}=\frac{X}{Y}=\frac{{a}{e}^{A}}{{b}{e}^{B}}=\frac{a}{b}{e}^{{A}-{B}} \end{align*}

\begin{align*} {F}_{Z}({z})&= \mathbb{P}({Z}<{z}) \\ &= \mathbb{P}\bigg(\frac{a}{b}{e}^{{A}-{B}}<{z}\bigg) \\ &= \mathbb{P}\bigg({{A}-{B}}<\ln{\frac{b}{a}{z}}\bigg) \end{align*} and this is where I got stuck. Many hours of $\tt convolution$ attempts wore me down

I made an edit to an answer so that the one (Minus One-Twelfth) who gave the hints would get the credit but it got rejected. It took forever to type. I retyped it as an answer...hopefully it doesn't get rejected !! -.-

Thanks, Jason
Math undergrad
University of California, Berkeley

Answer 1

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.

Answer 2

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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