Let $X$ and $Y$ be independent random variables with geometric probability function $p(k) = (1-\pi)\pi^{k}$ for $k = 0,1,…$ and $0 < \pi < 1$. Let $U = \max(X,Y)$ and $V = \min (X,Y)$. Define $W = V – U$.
I want to find the joint probability mass function of $U$ and $W$.
My attempt:
I found the distribution of $U$ by cumulative function of geometric distribution and independness:
$$
P(X \leq k) = 1 – (1 – \pi)^{k}
$$
$$P(U \leq u) = P (X \leq u, Y \leq v) = P(X\leq u) P(Y\leq u) $$
How do I find the distribution of $W$ or any attempt to solve this problem?
Best Answer
For $k=1,2,\ldots$ $$ \mathbb P(W=-k)=\mathbb P(\max(X,Y)-\min(X,Y)=k)=\mathbb P(X<Y, Y=X+k)+\mathbb P(Y<X, X=Y+k) = 2\mathbb P(X<Y, Y=X+k)=2\sum_{i=0}^\infty \mathbb P(X=i, Y=i+k) = \ldots $$ You need to calculate it.
For $k=0$ $$ \mathbb P(W=0)=\mathbb P(\max(X,Y)=\min(X,Y))=\mathbb P(X=Y) = \sum_{i=0}^\infty \mathbb P(X=i, Y=i) = \ldots $$
And the distribution of $W$ is found.
All the rest is to find $$ \mathbb P(W=-k, U=m)=\mathbb P(\max(X,Y)=\min(X,Y)+k=m). $$ For $k=1,2,\ldots$, $m\geq k$ $$ \mathbb P(W=-k, U=m)=\mathbb P(\max(X,Y)=m, \min(X,Y)=m-k) = 2\mathbb P(Y=m, X= m-k). $$ For $k=0$, $m\geq 0$ $$ \mathbb P(W=0, U=m)=\mathbb P(Y=m, X= m). $$ After all you need in $\mathbb P(U=m)$, find it from cdf of $U$ which you already found.