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Old 09-03-2018, 08:52 PM
ActuaryHopeful ActuaryHopeful is offline
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Default Number 98 on SOA 328 sample questions

I get how the answer is A, but I donít 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, shouldnít the moment generating function of Y equal the cube of the moment generating function of X, which is the identical distribution of the X_iís (I believe itís implied that the X_iís 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 itís wrong. Can anyone offer an explanation?
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Old 09-03-2018, 10:34 PM
Academic Actuary Academic Actuary is offline
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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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Old 09-03-2018, 10:41 PM
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Michael Mastroianni Michael Mastroianni is offline
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An exponential function does not preserve multiplication. It converts addition to multiplication:

I hope that clarifies.
Michael Mastroianni, ASA
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