# 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?

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You're already at the end of the right track. – Hagen von Eitzen Sep 17 '12 at 19:40
Sorry, I forgot to add one part. Question amended. – user41583 Sep 17 '12 at 19:43
For the added question, if $P(Y\gt b)=0$, then (since $a \gt b$), $P(y \gt a)$ is also $0$. How do you define the left side then? – Ross Millikan Sep 17 '12 at 19:56
@RossMillikan, I believe that is where my confusion lies. What is the convention? – user41583 Sep 17 '12 at 19:58
@enduser I think both sides will be zero in this case and one needs to use $\geq$ instead of $>$ – Seyhmus Güngören Sep 17 '12 at 20:06

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 $\frac{1-P(Y\leq a)}{1-P(Y\leq b)}\rightarrow 0$

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Thanks very much @Seyhmus! – user41583 Sep 17 '12 at 22:59
I am happy if it could be of some help. – Seyhmus Güngören Sep 17 '12 at 23:00
I'm just going through some exercises on my own in a probability textbook and trying to learn. This does help me - thanks again. – user41583 Sep 17 '12 at 23:03

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.

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