2
$\begingroup$

$X_1, X_2$ 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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.