hw5sol - 36-225 - Homework 5 SOLUTIONS1.16 points: (a) 8...

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Unformatted text preview: 36-225 - Homework 5 SOLUTIONS1.16 points: (a) 8 points (b) 8 points (3 forE(Y), 3 forE(Y2)and 2 forV(X)(a)MY(t) =E(etY) =XyetyPY(y) =Xy=0etye-yy!=e-Xy=0(et)yy!=e-eet=e(et-1).(b)E(Y) =ddtMY(t)|t=0=e(et-1)(et)|t=0=E(Y2) =d2dt2MY(t)|t=0=e(et-1)(et)2+e(et-1)(et)|t=0=2+.Thus,V(Y) =E(Y2)-E2(Y) =2+-()2=.2.12 points: 5 forE(Y), 5 forE(Y2), 2 forV(Y)E(Y) =ddtMX(t)t=0=pet(1-qet)-pet(-qet)(1-qet)2t=0=p(1-q) +pq(1-q)2(q=1-p)z}|{=p2+p(1-p)p2=p+ (1-p)p=1pE(Y2) =d2dt2MX(t)t=0=ddtpet-pqe2t+pqe2t(1-qet)2t=0=pet(1-qet)2-2pet(1-qet)(-qet)(1-qet)4t=0=p(1-q)2+ 2pq(1-q)(1-q)4(q=1-p)z}|{==p3+ 2p2(1-p)p4=p+ 2(1-p)p2=2-pp2This implies thatV(Y) =E(Y2)-E2(Y) =2-pp2-1p2=1-pp2.3.6 pointsFrom lecture #10 (6/1), we know that them.g.f.of the binomial distribution isMY(t) =(pet+ 1-p)nTherefore the distribution in question isBin(n= 3, p=.6).We know that the geometricm.g.f.isMY(t) =pet1-qet. Therefore the distribution in questionisG(p=.3)....
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hw5sol - 36-225 - Homework 5 SOLUTIONS1.16 points: (a) 8...

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