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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?

My answer was:

$A$ 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?

Filippo
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aaazalea
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  • Does the formula $P(A\cap B)=\frac{P(A|B)}{P(B)}$ help you? – Sigur Jul 17 '13 at 18:34
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    @JakobWeisblat I see nothing wrong with your proof. – Nick Peterson Jul 17 '13 at 18:35
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    @Sigur I don't see why I'd need it here. – aaazalea Jul 17 '13 at 18:36
  • @NicholasR.Peterson That's good to hear. – aaazalea Jul 17 '13 at 18:36
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    Two events are independent iff the equality in your third line holds, so your proof is correct. – Alex Jul 17 '13 at 18:37
  • @JakobWeisblat, I was trying to decide if $P(A|B)$ or $P(B|A)$ are zero or not. – Sigur Jul 17 '13 at 18:38
  • Print out this page and show it to your instructor) – Alex Jul 17 '13 at 18:40
  • To chime in, if this was the precise question then the answer is correct, – André Nicolas Jul 17 '13 at 18:45
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    I'd like to hear the "explanation" (an incorrect one, of course) you were given. – David Mitra Jul 17 '13 at 19:21
  • @DavidMitra "mathematically it may seem that way, but probabilistically it it isn't." – aaazalea Jul 17 '13 at 23:12
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    @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. – SBF Jul 18 '13 at 07:39
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    Btw, before we have been given a formal definition of the independence we were asked whether intuitively disjoint events are independent or not - and around a half of the class said "yes". In the beginning the lack of independence in such case may be non-intuitive, but then it becomes obvious: if $A$ happened, you know for sure that $B\subseteq \Omega\setminus A$ did not. – SBF Jul 18 '13 at 07:43
  • If it depends on conditions, then how did they declared that events are independent? See, the case is that, when any condition will make them related somehow, then they won't be disjointed anymore. – Vicrobot Aug 06 '19 at 13:46

2 Answers2

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Your argument is valid, as was already stated in the comments. I am posting this as CW answer so that this question no longer shows as unanswered.

Gustav Delius
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As so many people have already told you in the comments, your argument is perfectly correct. Perhaps your instructor was thinking of a slightly different question

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.

Dilip Sarwate
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