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Let $X$ and $Y$ be independent and identically distributed $(i.i.d.)$ r.v.’s, each having the probability distribution, $p(k) = (1 − λ)λ^k$; $k = 0,1,...$ where $λ :(0; 1)$ is a constant. Define $U = min(X; Y )$; $V = max(X; Y )$; $W = V − U$. Determine the joint probability distribution of $U$ and $W$ (taking care with $W = 0$) and verify that $U$ and $W$ are independent r.v.’s.

My work: I set up this: $P(X=W+U, Y=U)$ when $X>Y$ and $P(X=U, Y=W+U)$ when $X<Y$. and the joint distributions are the same in both cases as $W$ is always non-negative. finally I got the following joint pmf: $f(w,u)=(1-λ)^2 λ^(w+2u-2)$ when $X=Y$, $f(w,u)=(1-λ)^2 λ^(2u-2)$ and ${(w,u): w=0,1...; u= 0,1,..}$ is this the correct joint pmf? what will be the final joint pmf?

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  • $\begingroup$ What did you try? If this is some sort of an assignment, consider adding the self-study tag and read the tag wiki. $\endgroup$ – StubbornAtom Oct 18 '18 at 7:16
  • $\begingroup$ Your $W$ is just $|X-Y|$. This might help: math.stackexchange.com/questions/2685256/… $\endgroup$ – StubbornAtom Oct 18 '18 at 7:33
  • $\begingroup$ I saw your mentioned problem. I have solved that problem from Casella and Burger. But in this particular problem, isn't W is always greater than or equal to 0? $\endgroup$ – Dihan Oct 18 '18 at 8:11
  • $\begingroup$ Yes, of course $W$ is non-negative. I was thinking about breaking the problem into cases $X\ge Y$ and $X<Y$. $\endgroup$ – StubbornAtom Oct 18 '18 at 8:17
  • $\begingroup$ I set up this: $P(X=W+U, Y=U)$ when $X>Y$ and $P(X=U, Y=W+U)$ when $X<Y$. and the joint distributions are the same in both cases as $W$ is always positive $\endgroup$ – Dihan Oct 18 '18 at 8:18
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Both values of $U$ and $W$ are non-negative integers, say $u$ and $w$ respectively. We need to find which values of $(x,y)$ are associated with $(u,w).$ This is tantamount to solving the simultaneous equations

$$\begin{array}{rl} \min(x,y) &=u \\ \max(x,y)-\min(x,y)&=w \end{array}$$

for $(x,y).$ There are two possibilities: $\min(x,y)=x$ or $\min(x,y)=y.$ In the first case, $x=u$ whence $y=u+w.$ In the second case $y=u$ whence $x=u+w.$ These cases overlap when $w=0,$ which occurs when $X=Y.$

Therefore, by the probability axioms, when $w\ne 0$

$$\Pr((U,W)=(u,w)) = \Pr((X,Y)=(u,u+w)) + \Pr((X,Y)=(u+w,u))$$

and otherwise when $w=0$

$$\Pr((U,W)=(u,0)) = \Pr((X,Y)=(u,u)).$$

The independence of $X$ and $Y$ means their probabilities multiply, immediately giving

$$\eqalign{ \Pr((U,W)=(u,w)) &= \left\{\begin{array}{rl}(1-\lambda)\lambda^u\,(1-\lambda)\lambda^{u+w} + (1-\lambda)\lambda^{u+w}\,(1-\lambda)\lambda^u & \text{if }w\ne 0 \\ (1-\lambda)\lambda^u\, (1-\lambda)\lambda^u &\text{if } w=0\end{array}\right. \\ &= (1-\lambda)^2\lambda^{2u+w}\left\{\begin{array}{rl}2 & \text{if }w\ne 0 \\ 1 &\text{if } w=0.\end{array}\right. }$$

A convenient way to write that last expression in brackets uses the binary indicator function $\mathcal{I}:$

$$\mathcal{I}(w\ne 0) + 1 = \left\{\begin{array}{rl}2 & \text{if }w\ne 0 \\ 1 &\text{if } w=0.\end{array}\right.$$

Thus

$$\Pr((U,W)=(u,w)) = (1-\lambda)^2\ \left(\color{blue}{\lambda^{2u}}\right)\ \left(\color{red}{\left(\mathcal{I}(w\ne 0) + 1\right)\lambda^w}\right).$$

This is a product of a normalizing constant $(1-\lambda)^2,$ a function of $u$ alone (in blue), and a function of $w$ alone (in red), demonstrating $U$ and $W$ are independent.

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