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#1
09-03-2018, 08:52 PM
 ActuaryHopeful CAS SOA Join Date: Jun 2018 College: Michigan State University alum Posts: 5
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
09-03-2018, 10:34 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,528

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.
#3
09-03-2018, 10:41 PM
 Michael Mastroianni Member SOA Join Date: Jan 2018 Posts: 33

An exponential function does not preserve multiplication. It converts addition to multiplication:

$e^{t(X_{1}+X_{2}+X_{3})}=e^{tX_{1}}\cdot e^{tX_{2}}\cdot e^{tX_{3}}$

$M_{X_{1}X_{2}X_{3}}(t)=E[e^{tX_{1}X_{2}X_{3}}]\ne E[e^{tX_{1}}]E[e^{tX_{2}}]E[e^{tX_{3}}]=M_{X_{1}+X_{2}+X_{3}}(t)$

I hope that clarifies.
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Michael Mastroianni, ASA
Video Course for Exam 1/P: www.ProbabilityExam.com

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