Proving $a>b$ implies $P(Y>a|Y>b) > P(Y>a)$ Suppose you have a constants $a>b$, and $P(Y>b)<1$, then how can we show that $$P(Y>a|Y>b) > P(Y>a)?$$ 
I used the definition of conditional probability to get that 
$$ P(Y>a|Y>b) = P(Y>a, Y>b)/P(Y>b) = P(Y>a)/P(Y>b) $$
and since $P(Y>b) < 1$, I guess this means $P(Y>a)/P(Y>b) > P(Y>a)$. Am I on the right track?  
What if $P(Y>b) = 0$, though? 
 A: Given $P(Y>b)=0$ we have $P(Y>a)=0$ since $a>b$. In other words, If $Y\leq b$ almost surely so as $Y\leq a$ since $a>b$.
As a result we can write $P(Y\leq a|Y\leq b)P(Y\leq b)=P(Y\leq a)=1$. As $P(Y\leq b)=1$ is also known, we have $P(Y\leq a|Y\leq b)=1$.
From here I think we get $P(Y>a|Y>b)=1-P(Y\leq a|Y\leq b)=0$
EDIT: I think last line seems unclear. Lets try another way. We know
$P(Y>a|Y>b)=\frac{P(Y>a)}{P(Y>b)}$ and similarly $P(Y\leq a|Y\leq b)=\frac{P(Y\leq b)}{P(Y\leq b)}=1$
One can write $P(Y>a|Y>b)=\frac{P(Y>a)}{P(Y>b)}=\frac{1-P(Y\leq a)}{1-P(Y\leq b)}$ and we also know that $P(Y\leq a)\geq P(Y\leq b)$ and $lim_{b\rightarrow a}P(Y\leq a)= P(Y\leq b)$. From here assume that $P(Y\leq a)$ is $\epsilon$-close to $1$. However $P(Y\leq  b)$ is $\kappa\epsilon$-close to $1$ where $\kappa>1$ since $P(Y\leq a)\geq P(Y\leq b)$. Therefore when $P(Y\leq a) \rightarrow 1$,  and as a result $1-P(Y\leq a) \rightarrow 0$ we have $1-P(Y\leq b)>0$ for some $\epsilon$. Therefore $\frac{1-P(Y\leq a)}{1-P(Y\leq b)}\rightarrow 0$
A: By definition, $P(Y>a\mid Y>b) = \frac{P(Y>a \mbox{ and } Y>b)}{P(Y>b)} = 
\frac{P(Y>a)}{P(Y>b)}$.  If $P(Y>b)=0$, the conditional probability is undefined, and what you're trying to prove is not true.  But if $0< P(Y>b) <1$, then $P(Y>a \mid Y>b)>P(Y>a)$, and your conclusion follows.
