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#11




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I believe this case is the one where q is different for each Xi. It's important to know if it's the same for each or not. Like if some coins are weighted, then the q is not probably the same. For the binomial case you need to have the same q, I understand.
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German ______________ Prelims: VEE: Last edited by gauchodelpaso; 02012019 at 03:33 PM.. 
#12




Quote:
I was saying that if, given q, X1 and X2 were mutually independent (both with the same q), then the unconditional X1 and unconditional X2 need not be mutually independent. It seems to me that you are assuming the each Xj  qj is Bernoulli(qj), with the set of qj's being mutually independent and identically distributed. Then the unconditional Xj's are indeed mutually independent and are in fact Bernoulli(E[qj]), so the sum of m unconditionally is Binomial(m, E[qj]), whose variance is m E[qj] {1  E[qj]} which turns out to be the same as what you wrote.
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