Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a compound event $E_1E_2$,
Pr($E_1E_2$) = Pr($E_1$)Pr($E_2$|$E_1$)
such that if $E_1$ and $E_2$ are independent events, we can say that:
Pr($E_1E_2$) = Pr($E_1$)Pr($E_2$)

Is there an analogous formula that can be derived for the conditional probability: Pr($E_1E_2$|$E_3$)?

I would like to be able to reduce Pr($E_1E_2$|$E_3$) to terms that do not involve all three of the events $E_1$, $E_2$, $E_3$, given the assumption that only $E_1$ and $E_3$ are independent.

If this cannot be done, then what assumptions would be necessary in order to be able to do it?

share|improve this question

2 Answers 2

Here is one way to write it given the assumption: $$ P(E_1\cap E_2\mid E_3)=P(E_2\mid E_1\cap E_3)P(E_1\mid E_3)=P(E_2\mid E_1\cap E_3)P(E_1). $$

share|improve this answer
    
Is there no way to re-write it as a product of terms, each of which only involves one or two out of the three events? –  user1247 Jan 28 '13 at 10:40
    
I don't think there is, but I am not sure. You should probably edit your question and state you're looking for such an expression. –  Stefan Hansen Jan 28 '13 at 10:45
    
OK I edited my question. I guess this boils down to something like: if A and B are independent, can I write P(A|BC)=P(A|C)? If not, what are the minimal assumptions that are necessary in order for that to be true? –  user1247 Jan 28 '13 at 10:55

Hint: If you write out $$Pr(E2|E1)=\frac{Pr(E_2\cap E_1)}{Pr(E_1)}$$ then you can investigate more complicated formulas involving more events. Don't remember a formula but understand them.

share|improve this answer
    
The problem is that I can't figure out how using that formula would enable me to show, for example, that P(A|BC)=P(A|C), given A and B being independent. –  user1247 Jan 28 '13 at 12:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.