$X_1,X_2$ are i.i.d with $U=\min(X_1,X_2)$ and $V=X_1-X_2$

Question

$X_1, X_2$ 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.

Answer 1

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.