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 Short-Term Actuarial Math Old Exam C Forum

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#11
02-01-2019, 01:40 PM
 gauchodelpaso Member SOA Join Date: Feb 2012 College: Eastern Michigan U Posts: 136

Quote:
 Originally Posted by Academic Actuary What I am saying is assume you have an urn with different coins with the probability of heads on the coins varying according to a beta distribution. You draw 20 different coins with replacement and flip them where Xi is one if you have a head on the ith flip. The xi's are mutually independent so the variance of the sum will be of the variance of an individual flip: ( E[p(1-p)] + V(p) ) x number of flips

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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Last edited by gauchodelpaso; 02-01-2019 at 03:33 PM..
#12
02-01-2019, 03:18 PM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,719

Quote:
 Originally Posted by Academic Actuary What I am saying is assume you have an urn with different coins with the probability of heads on the coins varying according to a beta distribution. You draw 20 different coins with replacement and flip them where Xi is one if you have a head on the ith flip. The xi's are mutually independent so the variance of the sum will be of the variance of an individual flip: ( E[p(1-p)] + V(p) ) x number of flips
Aha. I see that we were talking about two different things.

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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