$X_1,X_2$ are i.i.d with $U=\min(X_1,X_2)$ and $V=X_1-X_2$ $X_1, X_2$ are independent,identically distributed non-negative integer-valued Random Variables, $U=\min(X_1,X_2)$ and $V=X_1-X_2$
If $P[U=u,V=0]=P[U=u]P[V=0]$ holds for $u=0,1,2,\ldots$, prove that $X_i$s follow geometric distribution.
I found the df of $U$ as $F_U(u)=1-(1-F_{X_i}(u))^2$
But cannot proceed with the given condition.
 A: Let $p(x)$ be the common pmf of $X_1, X_2$. Consider the LHS
$$ \Pr\{U = u, V = 0\} = \Pr\{\min\{X_1, X_2\} = u, X_1 - X_2 = 0\}
= \Pr\{X_1 = u, X_2 = u\} = p(u)^2$$
and
$$ \begin{align} 
\Pr\{U = u\} &= \Pr\{\min\{X_1, X_2\} = u\} \\
&= \Pr\{X_1 = u, X_2 > u\} + \Pr\{X_2 = u, X_1 > u\} + \Pr\{X_1 = u, X_2 = u\}\\
&= 2p(u)\sum_{x=u+1}^{\infty}p(x) + p(u)^2
\end{align}$$
for $u = 0, 1, 2, \ldots$. In particular consider the condition when $u = 0, 1$, we have
$$ \frac {\Pr\{U = 0\}} {\Pr\{U = 0, V = 0\}} 
= \frac {1} {\Pr\{V = 0\}} = \frac {\Pr\{U = 1\}} {\Pr\{U = 1, V = 0\}} $$
And thus put back all the result,
$$ \begin{align} 
\frac {1} {p(0)^2} \left(2p(0)\sum_{x=1}^{\infty}p(x) + p(0)^2\right)
&= \frac {1} {p(1)^2} \left(2p(1)\sum_{x=2}^{\infty}p(x) + p(1)^2\right)\\
\Rightarrow \frac {1 - p(0)} {p(0)} &= \frac {1 - p(0) - p(1)} {p(1)} \\
\Rightarrow \frac {p(1)} {p(0)} &= 1 - p(0)
\end{align}$$
Similarly, for $u = 0, 2$, we have
$$ \frac {p(2)} {p(0)} = 1 - p(0) - p(1) = 1 - p(0) - p(0)(1 - p(0)) = (1 - p(0))^2 $$
Assume
$$ \frac {p(x)} {p(0)} = (1 - p(0))^x $$ 
for all non-negative integers $x \leq m$, for some integers $m$. When $x = m+1$,
$$ \frac {p(m+1)} {p(0)} = 1 - \sum_{x=0}^m p(x) 
= 1 - p(0)\sum_{x=0}^m (1 - p(0))^x 
= 1 - p(0)\frac {1 - (1 - p(0))^{m+1}} {1 - (1 - p(0))} = (1 - p(0))^{m+1}$$
So by induction, we conclude
$$ p(x) = p(0)(1 - p(0))^x, x = 0, 1, 2, \ldots $$ 
which is the pmf of a geometric distribution.
