Question:
Let $X$ and $Y$ be two independent identically distributed random variables with probability mass function $$f(k) = (1-p)p^{k-1}\;\;\;k=1,2,...$$ where $0<p<1$. Let $U = $ min$\{X,Y\},\;$ $V = $ max$\{X,Y\},\;$ and $W = V - U$. Determine the joint probability mass function for $U$ and $W$ and show that $U$ and $W$ are independent.
My attempt:
I have managed to find PMFs for $U$, $V$, and $W$, and I have verified that these PMFs all sum to 1. I found: $$P(U = k) = (1-p^2)(p^2)^{k-1}\;\;\;k=1,2,...$$ $$P(V = k) = 2(1-p)p^{k-1} - (1-p^2)(p^2)^{k-1}\;\;\;k=1,2,... $$ $$P(W = k) = \left\{\begin{aligned}
&p^k\frac{1-p}{1+p}\; ,&& k=0\\
&2(p^k)\frac{1-p}{1+p}\; ,&& k=1,2,...
\end{aligned}
\right.$$
But when finding the joint PMF of $U$ and $W$ I determined that $P(W=k \;|\; U=j) = P(V=k+j \;|\; V\geq j)$. So in order to show that $U$ and $W$ are independent, I would have to show that $P(W=k) = P(V=k+j | V \geq j)$. However, I cannot seem to show this.
In addition, the PMF I found for $W$ is piecewise (since if $W = k > 0$ then $V = U + k$ which happens if either $X = Y+k$ or $X = Y-k$, whereas if $W = 0$ then $V = U$ which happens only if $X=Y$, hence why the PMF is multiplied by $2$ for $k>0$), whereas the PMF for $V$ is not piecewise, so it seems unlikely to me that
$P(V=k+j | V \geq j)$ would end up being piecewise.
So I'm hoping someone can guide me in the right direction. Are the PMFs I found for $U$, $V$, and $W$ correct? If so, what am I missing to find the joint PMF of $U$ and $W$ and show independence? And if not, where have I gone wrong?
I have linked to scanned photos of the work I have done so far to find these PMFs for reference. I am a third year undergraduate student studying math and statistics. Any help would be very much appreciated.
