Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to show that

$P(E\mid E\bigcup F) \geq P(E \mid F)$.

This is intuitively clear. But when expanding I get $P(E)\ P(F)\geq P(E\bigcup F)\ P(E \bigcap F)$. How to continue?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Let $a=P(E\cap F^c)$, $b=P(E\cap F)$ and $c=P(F\cap E^c)$. You have that $P(E)=a+b$, $P(F)=b+c$. Since $E\cup F=((E\cap F^c)\cup(E\cap F)\cup (F\cap E^c))$ and since the the union is disjoint you have that $P(E\cup F)=a+b+c$. Therefore, the problem you stated reduces to showing $(a+b)(b+c)\geq b(a+b+c)$ which follows trivially since $ac=P(E\cap F^c)P(F\cap E^c)\geq 0$.

share|improve this answer

Hint: the difference between these two terms is $P(E\cap F^c)P(F\cap E^c)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.