Question about the ratio of two random variables

Question

We have a couple $(X,Y)$ of random variables and its joint density function is: $$p_{X,Y}(x,y)=\begin{cases} \dfrac{e^{-1}}{3(x-1)!} & \mbox{if }x \in \mathbb{N}\setminus \{0\},\ y \in \{-x,0,x\} \\ 0 & \mbox{otherwise}\end{cases}$$ Find the distribution of $W=Y/X$.

The first thing I noticed is that $W \in \{-1,0,1\}$. Then I calculated the marginal distributions: $$p_X(x)=\sum _{y \in \{-x,0,x\}} p_{X,Y}(x,y)=3\dfrac{e^{-1}}{3(x-1)!}=\dfrac{1}{e(x-1)!} \\ p_Y(y)=\sum _{x=1}^{\infty} \dfrac{e^{-1}}{3(x-1)!}=\dfrac{1}{3e}\sum_{x=1}^{\infty} \dfrac{1}{(x-1)!}=\dfrac{1}{3e}\sum_{x=0}^{\infty}\dfrac{1}{x!}=\dfrac{1}{3e}e=\dfrac{1}{3}$$ And then I proceeded to calculate $p_W$ in its three values using the disintegration formula as it follows: $$P(W=-1)=P(Y/X=-1)=P(Y=-X)=\sum_{x=1}^{\infty}P(Y=-x,X=x)P(X=x)\\ =\sum_{x=1}^{\infty} \dfrac{e^{-1}}{3(x-1)!}\dfrac{1}{e(x-1)!}=\dfrac{1}{3e^2}\sum_{x=0}^{\infty}\left(\dfrac{1}{x!}\right)^2$$ At this point I don't know what to do, it seems reasonable to me that $W$ is uniform (Also because of the fact that $P(W=0)=P(Y=0)=1/3$), and so my thought was that I was able to obtain an $e^2$ from the sum. I'm not able to understand if I did something wrong or not.

Answer 1

I think you meant $P(Y = -X) = \sum_{x = 1}^{\infty} P(Y = -x \mid X = x) P(X = x)$ rather than $\sum_{x=1}^{\infty} P(Y = -x, X = x) P(X = x)$. This leads to $$ P(Y = -X) = \sum_{x = 1}^{\infty} P(Y = -x, X = x) $$ which then gives $P(W = -1) = 1/3$ as expected.