let $X,Y\sim Geo(p)$, $X,Y$ are i.i.d
i know that $X+Y \sim NB(2,p)$
im trying to find how $Z= X-Y, W= \frac{X}{Y}$ are distributed
my attemp: $P(Z\le k) = \sum_{i=-\infty}^k\sum_{j=-\infty}^\infty P(Y=j)P(X=i+j)=\sum_{i=-\infty}^k\sum_{j=-\infty}^\infty p(1-p)^{j-1}(1-p)^{i+j-1}p= \sum_{i=-\infty}^k p^2(1-p)^{i-2}\sum_{j=-\infty}^\infty(1-p)^{2j} = \sum_{i=-\infty}^k \frac{p^2(1-p)^{i-2}}{1-(1-p)^2}+\sum_{i=-\infty}^k\frac{p^2(1-p)^{i-2}}{(1-(1-p)^{-2})^2} = \frac{p^2}{(1-p)^2} \left (\frac{1}{1-(1-p)^2} + \frac{1}{(1-(1-p)^{-2})^2} \right ) \sum_{i=-\infty}^k (1-p)^i = \frac{p^2}{(1-p)^2} \left (\frac{1}{1-(1-p)^2} + \frac{1}{(1-(1-p)^{-2})^2} \right ) \left( \frac{1}{1-(1-p)^{-1}} + \frac{1-(1-p)^k}{1-(1-p)}\right)$
from here i really struggled with the fractions and didnt get any known distirbution or something that i think is correct, the same goes for $W$ as well, did i get something wrong along the way? how do i continue from here and also how to i calculate $W$ distirbution?
For $\frac{X}{Y}$ to be finite this would need $Y\not=0$ and so suggests the "trials until success" version of the geometric distribution with the support $\{1,2,3,\ldots\}$.
So $\mathbb P(X=n) =\mathbb P(Y=n) = p(1-p)^{n-1}$ for $n\ \in \{1,2,\ldots\}$ and $0$ otherwise.
For $z$ an integer: $$\mathbb P(X-Y=z) = \sum\limits_{n=|z|+1}^\infty p^2(1-p)^{2n-2+|z|} = \frac{p}{2-p} (1-p)^{|z|}$$
For $w=\frac{a}{b}$ a positive rational in lowest terms: $$\mathbb P\left(\frac XY=\frac{a}{b}\right) = \sum\limits_{n=1}^\infty p^2(1-p)^{an+bn-2} = p^2 \frac{(1-p)^{a+b-2}}{1-(1-p)^{a+b}}$$
As a check, the peak occurs when $X=Y$, with $\mathbb P(X-Y=0) =\mathbb P\left(\frac XY=1\right)=\frac{p}{2-p}$