2
$\begingroup$

When I study Statistical Theory, I find that these two concepts confuse me a lot.

By definition, if we say two events are PAIRWISE DISJOINT, that means the intersection of these two event is empty set. If we say that two events are MUTUALLY EXCLUSIVE, that means if one of these two events happens, the other will not. But doesn't it means that these two events are PAIRWISE DISJOINT?

If we say two events are MUTUALLY EXCLUSIVE, then they are not INDEPENDENT. Can we say that two PAIRWISE DISJOINT events are not INDEPENDENT as well?

If these two concepts are different (actually my teacher told me they are), could you please give me an example that two events are MUTUALLY EXCLUSIVE but not PAIRWISE DISJOINT, or they are PAIRWISE DISJOINT but not MUTUALLY EXCLUSIVE.

Thank you for your help.

$\endgroup$
3
$\begingroup$

"Disjoint" is a property of sets. Two sets are disjoint if there is no element in both of them, that is if $A \cap B = \emptyset$.

In some (but not all!) texts, "mutually exclusive" is a slightly different property of events (sets in a probability space). Two events are mutually exclusive if the probability of them both occurring is zero, that is if $\operatorname{Pr}(A \cap B) = 0$. With that definition, disjoint sets are necessarily mutually exclusive, but mutually exclusive events aren't necessarily disjoint.

Consider points in the square with each coordinate uniformly distributed from 0 to 1. Let $A$ be the event where the $x$-coordinate is 0, and $B$ be the event that the $y$-coordinate is 0. $A \cap B = \{(0,0)\}$ so $A$ and $B$ are not disjoint, but $\operatorname{Pr}(A \cap B) = 0$ so they are mutually exclusive.

As a second (silly, but finite) example, let the sample space be $S = \{x, y, z\}$ with probabilities $\operatorname{Pr}(\{x\}) = 0$, $\operatorname{Pr}(\{y\}) = \frac{1}{2}$, and $\operatorname{Pr}(\{z\}) = \frac{1}{2}$. If $A = \{x, y\}$ and $B = \{x, z\}$, then $A \cap B = \{x\}$, but $\operatorname{Pr}(A \cap B) = \operatorname{Pr}(\{x\}) = 0$. They are mutually exclusive but not disjoint.

$\endgroup$
2
$\begingroup$

Good discussion and I agree. This is an interesting result that seems to be ignored most of the time and disjoint and mutually exclusive are taken as equivalent. If we have finite sample spaces without 0 probability outcomes then the counter-examples above suggest they are then equivalent, and these are the properties of the common examples used when first teaching students these concepts. The current Wikipedia entry for Mutual Exclusive conflates the two concepts. The exceptions to equivalence seem to occur only when 0 probability outcomes exist in the sample space, but this is a requirement of continuous sample spaces and is not precluded in the definition of discrete sample spaces, both definitions usually state the outcomes must be mutually exclusive. However this seems to be a redundant condition since the different simple events of any sample space are disjoint by the definition of a set (i.e. multiplicities of elements are reduced) in any case, which thus implies mutual exclusivity by default.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.