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  1. I wonder if it is ture that $P(A\cap B) \geq P(A) \times P(B)$ for any two events?

    My observations so far are:

    When $A \subseteq B$, $P(A\cap B) = P(A) \geq P(A) \times P(B)$.

    When $B = \Omega$, $P(A\cap B) = P(A) = P(A) \times P(B)$.

  2. Can there be an equality relation between $P(A\cap B)$ and $P(A) \times P(B)$, just like the inclusion and exclusion relation between $P(A\cup B)$ and $P(A) + P(B)$?

Thanks!

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If A and B are not independent, think of $P(A \cap B)=P(A|B)P(B)$. Then think of the case when $P(A|B)>P(A)$. –  Alex Oct 21 '12 at 4:15
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What happens when $A$ and $B$ are disjoint? –  Michael Biro Oct 21 '12 at 4:16
    
@Alex: Thanks! Can there be an equality relation between P(A∩B) and P(A)×P(B)? –  Tim Oct 21 '12 at 4:19
    
@Tim: $P(A \cap B)=P(A)P(B)$ iff $A$ and $B$ are statistically independent, because then the outcome of event B does not affect the outcome of event A and $P(A|B)=P(A)$ –  Alex Oct 21 '12 at 4:25
    
@Alex: Thanks! How about in general when A and B are not necessarily independent? –  Tim Oct 21 '12 at 4:27

1 Answer 1

up vote 2 down vote accepted

In general, if $A$ and $B$ are not independent, the probability of the intersect of events is $$ P(A \cap B)=P(A|B)P(B) $$ Iff $A$ and $B$ are statistically independent, then the outcome of event $B$ does not affect the outcome of event $A$, hence $$ P(A|B)=P(A) $$ and the expression above becomes $$ P(A \cap B)=P(A)P(B) $$ Answering the OP, you need to find the case when $P(A \cap B)=P(A|B)P(B) \geq P(A)P(B)$. Clearly the equality holds if events are independent. Regarding the strict inequality, one can think of the (somewhat made-up case of the) probability to observe a 30-year old male who weighs 40 kilos. This probability, $P(A)$, is fairly small. But if you know that this person is 139cm tall, then the probability of him weighing 40 kilos, $P(A|B)$ is much higher than $P(A)$.

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Thanks! (1)there is no equality relation involving both $P(A\cap B)$ and $P(A) P(B)$ (not necessarily equating the two), besides the independence case? (2) Lovasz local lemma provides an inequality relation between $P(A\cap B)$ and something similar to $P(A) P(B)$. I wonder if there are other relations (equality or inequality) between them? –  Tim Oct 21 '12 at 4:52
    
$P(A \cap B)=P(A)P(B)$ holds iff $A$ and $B$ are independent. What you may want to use is called Bayes formula, $P(A|B)P(B)=P(B|A)P(A)$ –  Alex Oct 21 '12 at 6:41

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