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If X,Y are random variables and they follow geometric distribution(p) same for both. Also they are independent r.V's. Define $Z=\frac{X}{Y}$. Find distribution and expectation of $Z$.

I tried solving by assuming $X=k, Y=m$ and $Z=n$ so $X=nY$ but when I did calculations like PMF of $Z=P(Z=n)=\sum_{m=1}^\infty P(X=nm)× P(Y=m)$ But i don't understand how to simplify as Z is not integer or in geometric progression like X and Y ..even if Z was integer then also I got wrong answer I guess. Can you please guide me in right direction where I understand the essence atleast I'm getting confused a lot

Gaurav Mishra
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  • Do $X$ and $Y$ have the same parameter $p$? –  Nov 02 '24 at 19:00
  • Ohh yess i forgot to mention – Gaurav Mishra Nov 02 '24 at 19:01
  • Z will be discrete or continuous? – Gaurav Mishra Nov 02 '24 at 19:14
  • $Z$ is discrete because $\mathbb{P}(Z\in\mathbb{Q}^+)=1$, where $\mathbb{Q}^+$ is countable. –  Nov 02 '24 at 19:44
  • Hint: $Z$ will take on various rational values. Can you thInk of what these might be? Is $\frac 35$ a possible value for $Z$? Of course, $X=3,Y=5$ works, but what about $X=6,Y=10$? $X=9,Y=15$ $X=12,Y=20?$ others? Can you write an explicit expression for $P(X=3,Y=5)$ without any gobbledygook that anyone who has never heard of geometric random variables but has a calculator and is told the value of $p$ can evaluate? Ditto for $P(X=6,Y=10)$? Can you now write a series for $P(Z=\frac 35)$ nd recognize ut as a geometric series? Can you lather, rinse, repeat for other values of $3$ and $5$? – Dilip Sarwate Nov 02 '24 at 19:46

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The range of $X$ and $Y$ is the set of positive integers $\mathbb{N}^+$. So the range of $Z$ is the set of all positive rationals $\mathbb{Q}^+$. By “range” I mean the set of values that can be taken with positive probability.

Let’s start with the expectation of $Z$. Since $X$ and $Y$ are independent, $\mathbb{E}[\frac{X}{Y}]=\mathbb{E}[X]\mathbb{E}[\frac{1}{Y}]$. We know $\mathbb{E}[X]=\frac{1}{p}$. For the other bit, use the Taylor expansion $$\operatorname{ln}(1-t)=-\sum_{n=1}^{+\infty}\frac{t^n}{n}$$ for $|t|<1$. We have $$\mathbb{E}[\frac{1}{Y}]=\sum_{n=1}^{+\infty}\frac{1}{n}(1-p)^{n-1}p=\frac{p}{1-p}\sum_{n=1}^{+\infty}\frac{(1-p)^n}{n}\\=-\frac{p\operatorname{ln}(p)}{1-p}.$$ Ultimately we get $\mathbb{E}[Z]=-\frac{\operatorname{ln}(p)}{1-p}$.

Now, for the distribution of $Z$, take any positive rational $q_0$ and write uniquely $q_0=\frac{a}{b}$ where $a,b$ are positive integers that are prime to each other. Write $$\mathbb{P}(Z=q_0)=\mathbb{P}\Big(\bigcup_{n=1}^{+\infty}\{X=na,Y=nb\}\Big)\\=\sum_{n=1}^{+\infty}\mathbb{P}(X=na)\mathbb{P}(Y=nb)\\=\sum_{n=1}^{+\infty}(1-p)^{na-1}p(1-p)^{nb-1}p\\=\frac{p^2}{(1-p)^2}\sum_{n=1}^{+\infty}\big((1-p)^{a+b}\big)^n\\= \frac{p^2}{(1-p)^2}\Big(\frac{1}{1-(1-p)^{a+b}}-1\Big)\\=\frac{p^2}{(1-p)^2}\frac{(1-p)^{a+b}}{1-(1-p)^{a+b}}.$$