# Independence of disjoint events

I'm taking a class in Probability Theory, and I was asked this question in class today:

Given disjoint events $A$ and $B$ for which $$P(A)>0\\ P(B)>0$$ Can $A$ and $B$ be independent?

$A$ and $B$ are disjoint, so $P(A\cap B)=0$.
$P(A)>0$ and $P(B)>0$, so $P(A)P(B)>0$.
$P(A\cap B)\not =P(A)P(B)$, so $A$ and $B$ are not independent.

However, I was told that I am wrong and we cannot know whether or not $A$ and $B$ are independent from the given information, but I did not receive a satisfactory explanation. Is my argument valid? If not, where do I go wrong?

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@JakobWeisblat I see nothing wrong with your proof. – Nick Peterson Jul 17 '13 at 18:35
@Sigur I don't see why I'd need it here. – Jakob Weisblat Jul 17 '13 at 18:36
Two events are independent iff the equality in your third line holds, so your proof is correct. – Alex Jul 17 '13 at 18:37
I'd like to hear the "explanation" (an incorrect one, of course) you were given. – David Mitra Jul 17 '13 at 19:21
@JakobWeisblat: +1: you proof is perfectly correct - and this is one of the basic exercises to show that two disjoint event of positive probability are not independent. This also helps better understanding, what can be independent and what cannot. Regarding the explanation you received: after we got the axiomatization of probability (and, in particular, of independence) - this is a bit awkward to claim that mathematical and probabilistic reasoning may lead to different results. One may have a serious concern which kind of knowledge you'll get in the end from such class. – Ilya Jul 18 '13 at 7:39

Given disjoint events $A$ and $B$ for which $P(A\cup B)>0$, can $A$ and $B$ be independent?
to which the answer is Yes, and it happens when one of $A$ and $B$ is an event of probability $1$ that is a proper subset of the sample space $\Omega$ and the other is a subset of the complement, and hence has probability $0$.
For example, if $X$ is a continuous random variable, then $A = \{X \neq a\}$ and $B =\{X = a\}$ are disjoint events satisfying the condition $P(A \cup B) > 0$, and of course $$P(A\cap B) = P(\emptyset) = 0 = 1\times 0 = P(A)P(B)$$ showing that that $A$ and $B$ are independent.