Independence of two non-negative integer valued random variables Let $X,Y$ be two non-negative integer valued random variables defined on a probability space $(\Omega,\cal F, \Bbb P)$. The question is,


  If $\Bbb P\{X=i,Y=j\}=\Bbb P\{X=i\}P\{Y=j\}$ for every $i,j\ge0$, then are $X,Y$ independent random variables? i.e. the events $\{X\le a\}$ and $\{Y\le b\}$ are independent for any $a,b\in\Bbb R$?


Anyone can help with a proof of this? Thank you!
We want to show that

$$P(X \le a, Y \le b) = P(X \le a) P(Y \le b)$$


If a or b is a negative number, then


$$LHS = 0 = RHS$$


If a and b are zero, then


$$LHS = P(X = 0, Y = 0)$$

$$RHS = P(X = 0) P(Y = 0)$$


If a and b are nonnegative integers not both zero, then




$$LHS = P\left([\bigcup_{x=0}^{a} X = x] \bigcap [\bigcup_{y=0}^{b} Y = y]\right)$$

$$ = P(X=0, Y=0 \ or ... \ or X=0, Y=b \ or X=1, Y=0 \ or ... \ or X=1, Y=b \ or ... \ or X=a, Y=0 \ or ... \ or X=a, Y=b)$$

$$ = P(X=0, Y=0)+ ... +P(X=0, Y=b) + P(X=1, Y=0) + ... + P(X=1, Y=b)$$

$$+ ... + P(X=a, Y=0) + ... + P(X=a, Y=b) \tag{*}$$

$$ = P(X=0) P(Y=0)+ ... +P(X=0) P(Y=b) + P(X=1) P(Y=0) + ... + P(X=1) P(Y=b) + ... + P(X=a) P(Y=0) + ... + P(X=a) P(Y=b)$$



$$RHS = P(\bigcup_{x=0}^{a} X = x) P(\bigcup_{y=0}^{b} Y = y) = [P(X=0) + ... + P(X=a)][P(Y=0) + ... + P(Y=b)] = (*)$$




If a and b are nonnegative numbers not both zero, then same as above but replace
$a$ with $\lfloor a \rfloor$
$b$ with $\lfloor b \rfloor$


except
Yes, that is the definition of independence of two discrete random variables.  And since $X$ and $Y$ are independent then if $A$ is any event for $X$ and $B$ is any event for $Y$ then $A$ and $B$ are independent.  In particular the two events you specify.