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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?

StubbornAtom
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Dong Le
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1 Answers1

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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.

NCh
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