I know that event with zero probability is independent of any other (because $P(A\cap B)=P(A)\cdot P(B)$). I've got mixed feelings about this. Consider disjoint events $A$ and $B$, where $P(A)=0$. Isn't it weird to say that $A$ and $B$ are independent? It seems to me that it's just as natural to call them "dependent" (since they are disjoint) as to call them "independent" (since $P(A|B)=P(A)$, assuming that $P(B)>0$). I guess my questions are:

  1. Is there any intuition/motivation on this?
  2. In other words why do we do it that way? Why not just stick to a definition by conditional probability instead of the product one and leave zero probability events out of the dependence business?
  • 1
    $\begingroup$ Think of something with $0$ probability, something that will never happen. For example consider the probability that $3=5$. If anything anywhere happens $5=3$ is still false no matter what. So, the outcome of any event doesn't change the chance of the event with $0$ probability happening. That's why they're "independent" to each other. $\endgroup$ – Anthony Ter May 11 '19 at 6:46
  • $\begingroup$ @AnthonyTer Thank you. That seems fair but actually i have problems the other way around. Consider geometric probability on $[0,1]$ and let $A=\{\frac{1}{2}\}$, $B=[\frac{2}{3},1]$. Suppose that $A$ happens. It changes probability of $B$ happening drastically. $\endgroup$ – snak May 11 '19 at 9:55
  • $\begingroup$ @snak Going from $0$ to $0$ is not "changing drastically." We have to accept that "probability 0" does not coincide with "intuitively impossible" in order to work with infinite sample spaces at all. $\endgroup$ – Ned May 11 '19 at 10:23
  • $\begingroup$ @Ned Thank you; i'd say it changes from $\frac{1}{3}$ to $0$ though. Am i missing something? $\endgroup$ – snak May 14 '19 at 10:03
  • $\begingroup$ But A "can not happen" in the sense of probability measure. You can't always apply finite intuitions to infinite settings without intuitive conundrums, just like with infinite cardinalities. I had missed what you meant by the interval [2/3,1], but in any case, to make sense of all this you must accept that you can't condition on an event of probability 0 -- probability measures can't distinguish probability 0 from the empty event. $\endgroup$ – Ned May 14 '19 at 10:43

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.