Let $X$ and $Y$ be independent random variables with the same geometric distribution $P(X=k)=P(Y=k)=(1-p)p^k$ for $k=0,1,\dots, $. Let $U=\min\{X, Y\}$ and $V=\max\{X, Y\}$ and $W=V-U$.
(a) Find $P(U=i)$ and $P(W=j)$ for $i,j=0,1,\dots$.
(b) Show that $U$ and $W$ are independent.
My solution:
(a) For the first distribution, note that for $i=1,2,\dots$ $$ P(U=i)=P(U\le i)-P(U\le i-1). $$
Since $$ P(U\le i)=1-P(U>i)=1-P(X>i, Y>i)=1-(1-F(i))^2=1-p^{2i+2} $$ where $F(i)=\sum_{k=0}^i (1-p)p^k=1-p^{i+1}$.
Hence, $P(U=i)=p^{2i}-p^{2i+2}$.
For the second distribution, note that for $j=1,2,\dots$, $$ P(W=j)=P(W\le j)-P(W\le j-1). $$
So it is enough to get $$ P(W\le j)=P(V-U\le j)=\sum_{i=0}^{\infty} P(V\le U+j)=\sum_{i=0}^{\infty} P(V\le U+j|U=i)P(U=i) $$ $$ =\sum_{i=0}^{\infty}P(V\le i+j)P(U=i)=\sum_{i=0}^{\infty}(F(i+j))^2P(U=i)=\sum_{i=0}^{\infty} (1-p^{i+j+1})^2(p^{2i}-p^{2i+2}). $$
Is there a problem with my (a) asking where it was done? Also, I can't further simplify...
(b) To prove this one, we need to get $$ P(U=i, W=j)=P(V=i+j)=P(V\le i+j)-P(V\le i+j-1). $$
Note that $$ P(V\le i+j)=P(X\le i+j, Y\le i+j)=[F(i+j)]^2=(1-p^{i+j+1})^2 $$
