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I know that if $A$ and $B$ are disjoint events, then $P(A \cup \ B) = P(A) + P(B)$. However, is the converse true as well? Thanks.

Cm7F7Bb
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user136503
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    Disjoint events in the sense of probability simply means $P(A \cap B) = 0$, not necessarily that as sets $A\cap B = \emptyset$. For this reason (avoiding confusion) it is sometimes stated instead that $A,B$ are mutually exclusive events. – hardmath Jan 29 '16 at 15:13
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    @hardmath, I don't agree with your definition of disjoint. Can you cite any references? – TonyK Jan 29 '16 at 15:18
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    @hardmath I don't agree as well. If they are disjoint, then $P(A \cap B) = 0$, but the converse doesn't necessarily hold. – Clarinetist Jan 29 '16 at 15:20
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    @TonyK: I'm drawing a distinction between the use of "disjoint" in probabilities and in sets, and suggesting to avoid confusion by using "mutually exclusive" instead of "disjoint". The terminology that events $A,B$ are disjoint (while not ideal) is certainly tolerable where the probabilistic context is clear (esp. when there is no ascription of sets to the events). – hardmath Jan 29 '16 at 15:23
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    @hardmath: I am asking whether you have any evidence for your claim that this is the default meaning of "disjoint" in probability theory. – TonyK Jan 29 '16 at 15:26
  • @Clarinetist: I understand the basis for your objection, but I'm trying to clarify the terminology in a sympathetic way. Many elementary treatments will "abuse" the term disjoint in a purely probabilistic way as zero joint probability. Google's first hit on "disjoint probability definition" is This: Disjoint: P(A and B) = 0 – hardmath Jan 29 '16 at 15:28
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    @hardmath As you can see in that page, that isn't taken as a definition. Right before that line, "If two events are disjoint, then the probability of them both occurring at the same time is 0." – Clarinetist Jan 29 '16 at 15:31
  • @hardmath: I looked at the first eight Google hits for "disjoint events". Of these, three gave a definition of "disjoint'; all three disagreed with your definition. I think you are outvoted :-/ – TonyK Jan 29 '16 at 15:32
  • Note that Comments are intended to clarify the Question. Here the choice of terminology is open to interpretation and I was calling attention to the meaning of the term as crucial to giving a useful Answer. @Clarinetist: I don't think your line proves that the line I quoted is not offered as definition. – hardmath Jan 29 '16 at 15:39
  • @TonyK: The process of clarification should be a sympathetic one. I suggested a "better" term in my Comment. Possibly you disagree with the use of "mutually exclusive" as well. However you should then suggest what term could be used to better ask this Question. – hardmath Jan 29 '16 at 15:43
  • As discussed, your equation implies $P(A \cap B) = 0$. In probability, we don't necessarily say that these two events are disjoint, but we would say that they are disjoint almost surely. – Marcus M Jan 29 '16 at 19:55

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No, all you can deduce is that $P(A \cap B) = 0$, because $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ It doesn't mean that $A \cap B$ is empty.

Kamil Jarosz
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TonyK
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No for example suppose you have a uniform distribution on $[0,1]$. Let $A=[0,1/2]$ and $B=[1/2,1]$. Then $P(A\cap B)=0$ but $A\cap B=\{1/2\}\not=\emptyset$.

Gregory Grant
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  • Thank you for coming up with an excellent counterexample! (+1) – Clarinetist Jan 29 '16 at 15:25
  • Why is $P(A\cap B)=0$? – Inuyaki Jan 29 '16 at 15:29
  • Because the uniform distribution on $[0,1]$ is continuous: the probability of any single particular point is $0$. – Clement C. Jan 29 '16 at 15:31
  • @Inuyaki If you have any continuous random variable X, $$\mathbb{P}(X = c) = \int_{c}^{c}f_{X}(x)\text{ d}x = 0$$ – Clarinetist Jan 29 '16 at 15:32
  • @Inuyaki Another way to see that $P(A \cap B) = 0$ is to note that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ for any subsets $A$ and $B$ of $[0,1]$ . Now $P(A \cup B) = 1$, $P(A) = 1/2$, and $P(B) = 1/2$. Plugging in these values reveals $P(A \cap B) = 0$. – Joel Jan 29 '16 at 15:42
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EDIT: This is incorrect, disregard. The comments have valuable discussion and counterexamples.

Yes. Assuming $A$ and $B$ are independent, by definition, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If we have $P(A \cup B) = P(A) + P(B)$, then $P(A \cap B) = 0$, which is the definition of disjoint events.

DylanSp
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    Why do you need independence? – Clement C. Jan 29 '16 at 15:10
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    Hmm, is that true? Doesn't it just mean that $A\cap B$ has probability zero, not that it is empty. (Unless only empty sets have probability zero.) – MPW Jan 29 '16 at 15:10
  • Why do we need to assume that $A$ and $B$ are independent? What do you mean "by definition"? – Ben Grossmann Jan 29 '16 at 15:11
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    Is that really the definition of disjoint events? What if $X$ is uniformly distributed on $[0,1]$, $A$ is $(X \le \frac12)$, and $B$ is $(X \ge \frac12)$? – TonyK Jan 29 '16 at 15:11
  • A counterexample can even be given on a finite probability space, e.g. if $X$ is a random variable with values in ${1,2}$, with $\mathbb{P}(X=1)=1$ and $\mathbb{P}(X=2) = 0$. Letting $A = { 1, 2 }$ and $B = { 2 }$, we see $A$ and $B$ are independent, non-disjoint events such that $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B)$. – Clive Newstead Jan 29 '16 at 15:18