We know, by definition, that two random variables are independent if $$P(X\leq x, Y\leq y)= P(X\leq x)P(Y\leq y).$$ If, insted, I have that $$P(X>x, Y>y)= P(X>x)P(Y>y),$$ does this also imply independence between $X$ and $Y$?
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$\begingroup$ I assume you mean $P(X \le x, Y \le y)$? $\endgroup$– Cliff ABJul 8, 2019 at 15:03
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1$\begingroup$ You currently have $P(X \le x, Y \le x)$ and same for the next expression with the inequality reversed $\endgroup$– Cliff ABJul 8, 2019 at 15:17
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$\begingroup$ @CliffAB Ah! You are right. Thanks for pointing that out. $\endgroup$– July4Jul 8, 2019 at 15:30
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2 Answers
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Yes!
Claim: $$P(X>x,Y>y) = P(X>x)P(Y>y) \implies P(X\le x,Y\le y) = P(X\le x)P(Y\le y)$$
Proof:
Let's start by assuming $P(X>x,Y>y) = P(X>x)P(Y>y)$. Now let's expand the right side.
$$ P(X>x)P(Y>y) = (1-P(X\le x)) (1-P(Y\le y)) $$
$$= 1- P(X\le x)-P(Y\le y)+P(X\le x)P(Y\le y)$$
Now let's expand the left side:
$$P(X>x,Y>y) =P(X>x\bigcap Y>y)$$
$$=P((X\le x \bigcup Y\le y)^C) $$
$$= 1- P(X\le x \bigcup Y\le y)$$
$$= 1-\bigg[P(X\le x) + P(Y\le y) - P(X\le x,Y\le y)\bigg] $$
$$= 1- P(X\le x) - P(Y\le y) + P(X\le x,Y\le y)$$
Let's set the sides equal and cancel terms.
$$1- P(X\le x) - P(Y\le y) + P(X\le x,Y\le y) = 1- P(X\le x)-P(Y\le y)+P(X\le x)P(Y\le y)$$
$$ \implies P(X\le x,Y\le y) = P(X\le x)P(Y\le y) $$
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Note that those two statements are equivalent. Dave's proof is great, but I would go for a simpler approach:
- $-X$ and $-Y$ are independent if, and only if, $X$ and $Y$ are indepedent.
- Apply the definition for $-X$ and $-Y$
WARNING: I am not sure about how tough to proof the first statement really is (probably not too much). However, from an intuitive perpective, it is reasonable to assume as true (otherwise independence sucks!). The second part is quite straight-forward.
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$\begingroup$ This was my first thought, but it seems like it assumes the proof I gave to be true. $\endgroup$– DaveJul 8, 2019 at 15:17
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$\begingroup$ @Dave You are probably right that this is not a rigorous proof. However, from an intuitive perpective, the first statement is easy to assume as true (otherwise independence sucks!) and the second part is quite easy $\endgroup$– DavidJul 8, 2019 at 15:29