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I am reading Wasserman's book and his class notes. In the notes, I found the following statement

If A, B are disjoint, both having positive probability, then A and B cannot be independent.

Now, if A and B are disjoint, $\mathbb{P}(A\cap B) = \mathbb{P}(\phi) =0$. So mathematically $\mathbb{P}(A)$ or $\mathbb{P}(B)$ cannot both be positive. But I am getting utterly confused when I am thinking abuot 2 events that are dependent and disjoint. Can someone please clarify the relation between independence and being disjoint of 2 events?

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As you correctly point out, if $A$ and $B$ are disjoint then you have $A \cap B = \varnothing$ so that:

$$\mathbb{P}(A \cap B) = \mathbb{P}(\varnothing) = 0.$$

Independence of $A$ and $B$ would therefore require that $\mathbb{P}(A) \mathbb{P}(B) = \mathbb{P}(A \cap B) = 0$, which can only occur if at least one of the events has zero probability. Thus, Wasserman is correct to assert that if the events are disjoint, with positive probability, they cannot be independent.

Your confusion appears to come from your assertion that $\mathbb{P}(A)$ and $\mathbb{P}(B)$ cannot both be positive. That is wrong; they can both be positive so long as the events are not independent. An example would be a sporting competition between two teams (where there are no draws) where the event $A$ is that Team A wins and the event $B$ is that Team B wins. In this case it may well be the case that each team has some positive chance of winning, and the events are disjoint (they can't both win), so those events are not independent.

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Somewhat edited from an earlier answer of mine:

If $0 < P(A), P(B) < 1$, then mutual independence and mutual exclusion (or disjointedness) are mutually exclusive properties. If one property holds, the other cannot. Of course, the most commonly encountered case is that neither property holds. Said out loud and clear

  • If $A$ and $B$ are mutually independent events, then they cannot be mutually exclusive (i.e. disjoint) events.

  • If $A$ and $B$ are mutually exclusive (i.e. disjoint) events, then they cannot be mutually independent events.

In the first case, note that mutual independence implies that $P(A\cap B) = P(A)P(B) > 0$ (since both $P(A)$ and $P(B)$ are greater than $0$), and so the intersection of $A$ and $B$ has positive probability, that is, $A$ and $B$ cannot possibly be disjoint. In the second case, $P(A\cap B) = 0$ cannot equal $P(A)P(B)$ since neither $P(A)$ nor $P(B)$ is $0$ by assumption and so their product is a positive number.

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  • $\begingroup$ Missing ) after P(B in the last paragraph of your answer. :-) $\endgroup$ – MarianD Feb 14 at 23:06
  • $\begingroup$ @MarianD. Thanks! I was missing a ) after "greater than $0$" too.... $\endgroup$ – Dilip Sarwate Feb 14 at 23:09
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The law of total probability gives us: P(A) = P(A|B)P(B) + P(A|not B)P(not B). Since the probabilities are positive, P(A|B) = 0 and P(A|not B) > P(A) which is one definition of dependence.

If I never carry my umbrella when it's sunny, I'm much more likely to carry my umbrella if it's not sunny. Sunny weather and carrying my umbrella are disjoint, dependent events.

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To "clarify the relation between independence and being disjoint" consider the definitions of independent and disjoint events:

A, B independent <--> $P(A\cap B)= P(A)P(B)$.

Since $P(A) > 0$ and $P(B)>0$, obviously, $P(A\cap B) >0$ .

A, B disjoint <--> $P(A\cap B)= 0$.

Hence, A and B can't at the same time be disjoint and independent if their probabilities are positive.

Furthermore, disjoint events (with positive probabilities) are very (extremely!) dependent since occurrence of one mandates that the other does not occur.

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