Assume $X$ and $Y$ are two continuous random variables. Assuming the given conditions to be true does any of them concretely tell us that X and Y are independent. $$1.\ P(X > a | Y > a) = P(X > a)\ \forall\ a\ \in\ \mathbb{R}$$
$$2.\ P(X > a| Y < b) = P(X > a)\ \forall\ a,\ b\ \in\ \mathbb{R}\\$$
Why would $(2)$ always imply random variables $X$ and $Y$ to be independent but not $(1)$?
Please help me understand.
Let's say we had some random variables $X$ and $Y$ such that $P(X > a \cap Y > b) = \frac{a}{b}P(X > a)P(Y > b)$ for $b > a > 0$, else $P(X > a)P(Y > b).$
Clearly, letting $a = b$ would satisfy the first criterion:
$$P(X > a | Y > a) = \frac{P(X > a \cap Y > a)}{P(Y > a)} = \frac{\frac{a}{a}P(X>a)P(Y>a)}{P(Y>a)} = P(X>a)$$
However, for $a = 2, b = 1$ we have $P(X > a | Y > b) = \frac{1}{2}P(X > a) \neq P(X > a),$ so the variables are not independent.
Now, suppose we have our second criterion, $P(X > a|Y < b) = P(X > a).$
Let's rewrite the events $X > a$ and $Y < b$ as $A$ and $B$ for simplicity.
By Bayes' rule:
$$P(A|\bar{B}) = \frac{P(\bar{B}|A)P(A)}{P(\bar{B})} = (1 - P(B|A))\frac{P(A)}{1 - P(B)}$$ $$ = \frac{P(A) - P(A|B)P(B)}{P(A)}\cdot \frac{P(A)}{1 - P(B)} = \frac{P(A)-P(A|B)P(B)}{1-P(B)}$$
Now, notice that from our second criterion we have $P(A|B) = P(A).$ So:
$$\frac{P(A)-P(A|B)P(B)}{1-P(B)} = P(A)\frac{1 - P(B)}{1 - P(B)} = P(A)$$
Which is to say that $P(X>a|Y>b)=P(X>a).$