Joint distribution of $X$ and $Y-X$ (discrete case)

Question

The joint distribution of the random variable $(X,Y)$ is $\mathbb{P}(X=x,Y=y)=\frac{e^{-2}}{x!(y-x)!},x=0,1,...,y=x=x+1,...$

1. Find the marginal distribution of $X$ and $Y$.

$\rightarrow \mathbb{P}(X=x)=\sum_{y=x}^{+\infty}\mathbb{P}(X=x,Y=y)=\frac{e^{-2}}{x!}\sum_{s=0}^{+\infty}\frac{1}{s!}\Rightarrow X\sim Poi(1)$;

$\rightarrow \mathbb{P}(Y=y)=\sum_{x=0}^{y}\mathbb{P}(X=x,Y=y)=\frac{e^{-2}}{y!}\sum_{x=0}^{y}\frac{y!}{x!(y-x)!}\cdot 1^x \Rightarrow Y\sim Poi(2);$

$\rightarrow \mathbb{P}(X=x)\cdot \mathbb{P}(Y=y)\neq \mathbb{P}(X=x,Y=y)\Rightarrow X$ and $Y$ are not independent.

2. Find the joint distribution of $X$ and $Y-X$, and say if are independent or not.

---

If we had been in the continuous case, for $X$ and $Y$ not independent, we'd have $f_{VZ}(v,z)=f_{XY}(v,z)|J|$ with $\left\{\begin{matrix} v=x\\ z=y-x \end{matrix}\right. \Rightarrow \left\{\begin{matrix} x=v\\ y=z+v \end{matrix}\right.$. How does it works in the discrete case? What is the formule to apply?

Thanks in advance.

Answer 1

$P(X=x,Y-X=z)=P(X=x,Y=x+z)=\frac {e^{-2}} {x!z!}$ for $x,z =0,1,2,...$. Note that $P(X=x,Y-X=z)=P(X=x)P(Y-X=z)$ so $X$ and$Y-X$ are independent.