I've seen examples (more appropriately, counter-examples) which show that:

Mutual independence does not imply pairwise independence and vice versa.

I've tried looking around for examples where events are neither pairwise or mutually independent. I couldn't find any so I came up with some examples. To keep things simple, I restricted them to only 3 events. I began to think that at least one pair had to be disjoint (it's sufficient though) in order for the 3 events to be neither mutually or pairwise independent but this wasn't the case.

Question: Is it common for events to be neither mutually or pairwise independent?

  • $\begingroup$ Silly example: an event is $A$ always not independent with itself, so $A, A, A, A$ are neither pairwise or mutually independent. $\endgroup$ – angryavian Feb 27 at 23:24
  • $\begingroup$ Most examples are neither. Though I would be interested in an example which had mutually independent events but not pairwise independent events $\endgroup$ – Henry Feb 27 at 23:24
  • $\begingroup$ (1) Mutual independence does imply pairwise independence, but not vice versa. see e.g. en.wikipedia.org/wiki/… esp. the last sentence in that section. (2) It is really easy to come up with dependent events! e.g. if $3$ events $X=Y=Z$ then they are all dependent (pairwise and mutually). a less degenerate example: let $R$ be result of a 6-sided die roll and consider the events $R \ge 3, R \ge 4, R \ge 5$. $\endgroup$ – antkam Feb 27 at 23:26
  • $\begingroup$ @Henry Check example 4 here faculty.washington.edu/fm1/394/Materials/2-3indep.pdf $\endgroup$ – Tomás Palamás Feb 27 at 23:28
  • 3
    $\begingroup$ @TomásPalamás Mutual independence of three events $A,B,C$ is not just $P(A \cap B \cap C) = P(A) P(B) P(C)$, but also requires pairwise independence for each pair of events. See Wikipedia: "Note that it is not a single condition involving only the product of all the probabilities of all single events; it must hold true for all subsets of events." $\endgroup$ – angryavian Feb 27 at 23:32

Your Answer

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

Browse other questions tagged or ask your own question.