# Joint probability distribution of geometric distribution

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 $$X. and the joint distributions are the same in both cases as $$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?

• What did you try? If this is some sort of an assignment, consider adding the self-study tag and read the tag wiki. – StubbornAtom Oct 18 '18 at 7:16
• Your $W$ is just $|X-Y|$. This might help: math.stackexchange.com/questions/2685256/… – StubbornAtom Oct 18 '18 at 7:33
• 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? – Dihan Oct 18 '18 at 8:11
• Yes, of course $W$ is non-negative. I was thinking about breaking the problem into cases $X\ge Y$ and $X<Y$. – StubbornAtom Oct 18 '18 at 8:17
• 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 – Dihan Oct 18 '18 at 8:18

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.