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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!

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  • $\begingroup$ Yes, that is the definition of independence. $\endgroup$ – Gregory Grant Feb 12 '16 at 0:34
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    $\begingroup$ Hint: $\{X\leqslant a\}=\bigcup\limits_{j=0}^{\lfloor a\rfloor} \{X=j\}$. $\endgroup$ – Stefan Hansen Feb 12 '16 at 7:56
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We want to show that

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

  1. If a or b is a negative number, then

$$LHS = 0 = RHS$$

  1. If a and b are zero, then

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

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

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

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

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

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

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

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


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


  1. If a and b are nonnegative numbers not both zero, then same as above but replace

  2. $a$ with $\lfloor a \rfloor$

  3. $b$ with $\lfloor b \rfloor$

except

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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.

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