Proof that two conditional random variables given a condition are independent How to show that the following condition implies random variables X and Y are independent?
$P(X &gt; a | Y &lt; b) = P(X &gt; a)\ \forall\ a,\ b\ \in\ \mathbb{R}$
Let $F_X(a) = P(X \leq a)$, $F_Y(b) = P(Y \leq b)$ and $F_{X,Y}(a,b) = P(X \leq a \mbox{ and } Y \leq b)$ denote the cumulative distribution functions of $X$, $Y$ and $(X, Y)$, respectively. To show independence, we want to prove that $F_X(a)F_Y(b) = F_{X, Y}(a,b)$ for all $a, b \in \mathbb{R}$. Conditional probability tells us that: $$P(X &gt; a | Y &lt; b) = \frac{P(X&gt;a \mbox{ and } Y&lt;b)}{P(Y &lt; b)}$$ and so by the assumptions of the question $$P(X &gt; a) = \frac{P(X&gt;a \mbox{ and } Y&lt;b)}{P(Y &lt; b)}$$ which tells us $$P(X &gt; a)P(Y &lt; b) = P(X&gt;a \mbox{ and } Y&lt;b)$$ for all $a, b \in \mathbb{R}$. Now since $P(X &gt; a) = 1 - F_X(a)$ and $P(Y &lt; b) = F_Y(b)$ we may rewrite the above equation as:
$$F_Y(b) - F_X(a)F_Y(b) = P(X &gt; a \mbox{ and } Y &lt; b)$$
and so $$F_X(a)F_Y(b) = F_Y(b) - P(X &gt; a \mbox{ and } Y &lt; b).$$
Finally, note that $$F_Y(b) = P(X &gt; a \mbox{ and } Y &lt; b) + P(X &lt; a \mbox{ and } Y &lt; b)$$ and we are done.
(&gt;|&lt;)=$\frac{P(X&gt;a,Y&lt;b)}{P(Y&lt;b)}$ = P(X&gt;a)
P(X&gt;a,Y&lt;b)=P(X&gt;a).P(Y&lt;b) so X &amp; Y are independent