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Probability Old Exam P Forum 

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#1




Number 98 on SOA 328 sample questions
I get how the answer is A, but I don’t see how C is wrong. Since X_1, X_2, and X_3 is a random sample, they are mutually independent. If any random variables U and V are independent, then E[g(U)h(V)] = E[g(U)]E[h(V)]. Since the moment generating function is an expected value of a function of a random variable, shouldn’t the moment generating function of Y equal the cube of the moment generating function of X, which is the identical distribution of the X_i’s (I believe it’s implied that the X_i’s follow an identical distribution)? And for any X_i, the moment generating function is 1/3 + (2/3)e^t, so I thought the answer would just be this cubed, but it’s wrong. Can anyone offer an explanation?

#2




If you were dealing with the sum the MGF would be cubed. You have to find the probability distribution of Y which is actually very simple in this case and Y can only be 0 or 1.

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exam p, mgf 
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