# What does $\mathbb{P}(A \cap B) > 0$ tell?

This is related to the conditional chain rule in conditional probability:

If for two sets of a probability space I've been given:

$$\mathbb{P}(A \cap B) > 0$$

Then what does this mean?
Is it related to independence?

• If $A$, $B$ are two events, it means the two events have overlapping outcomes. $A\equiv\{$ the outcome of coin A is H$\}$, $B\equiv\{$the outcome of coin B is T$\}$. we can have the above probability, but the two events are independent. On the other hand, if $A\equiv\{$the out come of flipping A once is H$\}$, $B\equiv\{$the outcome of flipping A twice is HH$\}$, we still have the above probability, but the two events are not independent. – user3813057 Nov 15 '15 at 21:48
• It tells you that the events have a non-negligible intersection and nothing else. It's not really related to independence aside from being half the definition. – dsaxton Dec 28 '15 at 18:50

If we have $P(B) > 0$, then

$$P(A\mid B)=P(A\cap B)/P(B)$$

$P(A\cap B)>0 \Rightarrow P(A\mid B)>0$, which does NOT imply independence.

For $P(B) > 0$, one definition of independence is $P(A\mid B)=P(A)$, which is equivalent to $P(A\cap B)=P(A)P(B)$.

So, we cannot say anything regarding independence from $P(A\cap B)>0$.

• Yeah this is it. The chain rule requires the assumption $P(B)>0$, but I was confused by this way of telling it. – mavavilj Nov 15 '15 at 21:55
• why does $P(A \cap B) > 0$ imply $P(B) > 0$? Also,$P(A|B)$ is meaningless if we do not have $P(B) > 0$ – BCLC Dec 28 '15 at 18:33
• @BCLC $P(A \cap B) \leq P(B)$ – ekvall Dec 28 '15 at 19:25
• @Student001 user3813057 seems to be kind of confusing. I'm going to propose an edit – BCLC Dec 29 '15 at 12:45
• @BCLC I don't see why you think it's confusing, it seems pretty clear to me. (+1) – ekvall Dec 29 '15 at 13:37