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  • The event of getting a 6 the first time a die is rolled and the event of getting a 6 the second time are ''independent''.

  • By contrast, the event of getting a 6 the first time a die is rolled and the event that the sum of the numbers seen on the first and second trials is 8 are ''dependent''.

  • If two cards are drawn ''with'' replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are ''independent''.

  • By contrast, if two cards are drawn ''without'' replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are ''dependent''.


Similarly, two Random Variable s are independent if the conditional probability distribution of either given the observed value of the other is the same as if the other's value had not been observed.


INDEPENDENT EVENTS


The standard definition says:

:Two events ''A'' and ''B'' are independent If And Only If Pr(''A'' ∩ ''B'') = Pr(''A'')Pr(''B'').

Here ''A'' ∩ ''B'' is the Intersection of ''A'' and ''B'', that is, it is the event that both events ''A'' and ''B'' occur.

More generally, any collection of events -- possibly more than just two of them -- are mutually independent if and only if for any finite subset ''A''1, ..., ''A''''n'' of the collection we have

:\Pr(A_1 \cap \cdots \cap A_n)=\Pr(A_1)\,\cdots\,\Pr(A_n). \!\,

This is called the ''multiplication rule'' for independent events.

If two events ''A'' and ''B'' are independent, then the Conditional Probability of ''A'' given ''B'' is the same as the unconditional (or marginal) probability of ''A'', that is,

:\Pr(A\mid B)=\Pr(A). \!\,

There are at least two reasons why this statement is not taken to be the definition of independence: (1) the two events ''A'' and ''B'' do not play symmetrical roles in this statement, and (2) problems arise with this statement when events of probability 0 are involved.

  : P(''X'' &le''x'', ''Y'' &le''y'' ''Z'' ''z'') = P(''X'' &le''x'' ''Z'' = ''z'') &middot P(''Y'' &le ''y'' ''Z'' = ''z'')
  : ''p''<sub>''XY''''Z''</sub>(''x'', ''y'' ''z'') ''p''<sub>''X''''Z''</sub>(''x'' ''z'') &middot ''p''<sub>''Y''''Z''</sub>(''y'' ''z'')
  : P(''X'' ''x'' ''Y'' = ''y'', ''Z'' = ''z'') = P(''X'' = ''x'' ''Z'' = ''z'')