Pairwise disjoint?
If Wikipedia says the following:
Does pairwise disjoint mean that the events are stochastically independent of each other or have I misunderstood something?
Lg
(2 rating, 2 votes, rated)You need to be a registered member to rate this.
Loading...
You have a lot of events. Pairwise disjunct means the average of two randomly selected event quantities is empty.
Cube: {1}, {2,5}, {3,4,6}
But why does the system always have to be disjunct in pairs?
So that the sum of the probabilities is 1. If that were not the case, the sum of the probabilities is greater than 1.
Example Dice: {1, 2 }, {2,5},{3,4,6}
Here you would count the event that a 2 is cubed twice. It doesn't make sense.
What is it that if I have all possible events exactly once their sum is exactly 1?
Right, you got it.
Disjunct pairs means that none of the quantities has an element in common with any other of the quantities. That.
for all indices i , j with i ≠ j .
In the Wikipedia article you mentioned, the “in pairs disjunct” is, by the way, also a link to the Wikipedia art that explains what disjunctive quantities are…
https://de.wikipedia.org/wiki/Disjunkt#Definitions
============
No This is not equivalent to stochastic independence.
Stoch's independence is something completely different. Disjunct events are virtually never stochastic independently of one another (except if all up to at most one of the events has probability 0).
For the stochastic independence of two events E _ i , E _ j had
apply. Two disjunctive events E _ i , E _ j is always valid…