Is this a consequence of independence?

Suppose the real random variables $X$ and $Y$ are independent; i.e. $$\mathbb{P}[\{X \leq x\} \cap \{Y \leq y\}] = \mathbb{P}[\{X \leq x\}] \cdot \mathbb{P}[\{Y \leq y\}].$$

Does it follow that $$\mathbb{P}[\{X > x\} \cap \{Y > y\}] = \mathbb{P}[\{X > x\}] \cdot \mathbb{P}[\{Y > y\}]?$$

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Yes. The definition (essentially) of independence of random variables is that for any two Borel sets $A$ and $B$, $\mathbb P(X \in A, Y \in B) = \mathbb P(X \in A) \mathbb P(Y \in B)$ . – cardinal Sep 24 '11 at 19:13

Previous comments and answers are extremely helpful and right since they are pointing you and the most useful definition of the independence and the way how to show that these two definitions are equivalent.

On the other hand, you can think of just direct proof from the formulation that you have. Let $A = \{X\leq x\}$ and $B = \{Y\leq y\}$. We know that $$P(A \cap B) = P(A)P(B)$$ and we should prove that $P(A^c \cap B^c) = P(A^c)P(B^c)$. Let us start from the right-hand side: $$P(A^c)P(B^c) = (1-P(A))(1-P(B)) = 1-P(A)-P(B)+P(A)P(B)$$ but $P(A)P(B) = P(A\cap B)$ so $$P(A^c)P(B^c) = (1-P(A))(1-P(B)) = 1-P(A)-P(B)+P(A\cap B).$$ For the left-hand side: $$P(A^c\cap B^c) = 1- P(A\cup B)$$ by De Morgan law how to change intersection and the union. But: $$P(A\cup B) = P(A)+P(B) - P(A\cap B)$$ so $$P(A^c\cap B^c) = 1- P(A\cup B) = 1-P(A)-P(B)+P(A\cap B)$$ which conicides with the expression for the right-hand side.

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I can't think of a demonstration of that right now. But a more general definition would be that: $X_1, X_2, \dots, X_n$ are independent random variables if:

$$P(X_1\in A_1, X_2\in A_2, \dots, X_n\in A_n)=\prod_{i=1}^nP(X_i\in A_i)$$

Which is obvious that is followed by any of the two statements above and a little less obvious that there's an equivalence between the three of them.

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The equivalence follows from an application of the $\pi-\lambda$ theorem and the recognition that either class of intervals is a $\pi$ system. – cardinal Sep 24 '11 at 19:21

If $$P(A \cap B) = P(A) P(B)$$ then $$P(A^c \cap B) = P(B - A \cap B) = P(B)-P(A \cap B)=\left(1-P(A)\right)P(B) = P(A^c) P(B).$$ Applying this argument twice - to the first event as just above, and then to the second - gives the implication you seek.

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