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Simplifying the Conditional probability |
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| Sep7-12, 10:37 AM | #1 |
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Simplifying the Conditional probability
P(S1 [itex]\cap[/itex] S2 [itex]\cap[/itex] S3 | r)
How do I simplfy the above equation if S2 and S3 are independent of r ? |
| Sep7-12, 11:55 AM | #2 |
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Recognitions:
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I've don't recall seeing a book that gives the probability law
[itex] P((X|Y)|Z) = P(X | (Y \cap Z) ) [/itex] but I think its true. How to argue it depends on whether you are approaching probability as measure theory or something simpler. You can also apply the usual probability laws with a condition lilke "[itex]| Z[/itex]" tagged onto every term. For example, [itex] P(A \cap B) = P(A|B) P(B) [/itex] so [itex] P(A \cap B | Z) = P( (A|B)|Z) P(B | Z) [/itex] See if you can make progress by applying those ideas. |
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