Question :

Show that if the event A is completely independent of events B and C, then A is independent of logical sum

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- December 2nd 2009, 05:38 PMzorroShow event is completly independent of another event
Question :

Show that if the event A is completely independent of events B and C, then A is independent of logical sum - December 2nd 2009, 10:56 PMmatheagle
assume that

and

the question is, does

????????

I don't think so.

Try tossing a fair die.

Let

and

I get 1/6 =1/6 for the first two

but then I get 1/6 and 1/4 on the last one. - December 13th 2009, 01:51 AMzorroCould u please explain me
- December 13th 2009, 04:19 AMLaurent
I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that (same with A,C and B,C),

**and**.

Then the property is proved as follows (for instance):

, and the three (main) events on the right-hand side are disjoints, hence

,

and you can compute the latter probabilities using (and similarly for the last term) and the assumptions (recalled above).

You should get , or in fact , which is the same... - December 15th 2009, 04:42 PMzorroI am again stuck

http://www.mathhelpforum.com/math-he...98af3ab8-1.gif,

I am stuck here ???? - December 18th 2009, 02:21 PMLaurent
- December 18th 2009, 02:27 PMzorroI did get u