应用随机过程 概率模型导论 英文版 第九版 (SheldonM.Ross 著) 人民邮电出版社 课后答案 113页

'课后答案网，用心为你服务！大学答案---中学答案---考研答案---考试答案最全最多的课后习题参考答案，尽在课后答案网（www.khdaw.com）！Khdaw团队一直秉承用心为大家服务的宗旨，以关注学生的学习生活为出发点，旨在为广大学生朋友的自主学习提供一个分享和交流的平台。爱校园（www.aixiaoyuan.com）课后答案网（www.khdaw.com）淘答案（www.taodaan.com） 课后答案网www.khdaw.comInstructor’sManualtoAccompanyIntroductiontoProbabilityModelskhdaw.comNinthEditionSheldonM.RossUniversityofCaliforniaBerkeley,Californiakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comContentsChapter1....................................................................1Chapter2....................................................................7Chapter3...................................................................17Chapter4khdaw.com...................................................................33Chapter5...................................................................43Chapter6...................................................................59Chapter7...................................................................71Chapter8...................................................................81Chapter9...................................................................95Chapter10..................................................................99Chapter11.................................................................105khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comChapter11.S={(R,R),(R,G),(R,B),(G,R),(G,G),(G,B),7.If(E∪F)coccurs,thenE∪Fdoesnotoccur,and(B,R),(B,G),(B,B)}.soEdoesnotoccur(andsoEcdoes);Fdoesnotoccur(andsoFcdoes)andthusEcandFcbothTheprobabilityofeachpointinSis1/9.occur.Hence,2.S={(R,G),(R,B),(G,R),(G,B),(B,R),(B,G)}.(E∪F)c⊂EcFc.IfEcFcoccurs,thenEcoccurs(andsoEdoesnot),3.S={(e1,e2,...,en),n≥2}whereei∈(heads,ctails}.Inaddition,en=en−1=headsandforandFoccurs(andsoFdoesnot).Hence,neitherEorFoccurandthus(E∪F)cdoes.Thus,khdaw.comi=1,...,n−2ifei=heads,thenei+1=tails.EcFc⊂(E∪F)cP{4tosses}=P{(t,t,h,h)}+P{(h,t,h,h)}
4andtheresultfollows.11=2=.288.1≥P(E∪F)=P(E)+P(F)−P(EF).4.(a)F(E∪G)c=FEcGc.9.F=E∪FEc,implyingsinceEandFEcaredisjoint(b)EFGc.thatP(F)=P(E)+P(FE)c.(c)E∪F∪G.10.Eitherbyinductionoruse(d)EF∪EG∪FG.nE=E∪EcE∪EcEcE∪···∪Ec···EcE(e)EFG.∪i1121231n−1n,1(f)(E∪F∪G)c=EcFcGc.andaseachofthetermsontherightsidearecccmutuallyexclusive:(g)(EF)(EG)(FG).(h)(EFG)c.P(∪Ei)=P(E1)+P(EcE2)+P(EcEcE3)+···112i+P(Ec···EcEn)1n−135.4.Ifhewins,heonlywins$1,whileifheloses,he≤P(E1)+P(E2)+···+P(En).(why?)loses$3.i−1,i=2,...,76.IfE(F∪G)occurs,thenEoccursandeitherForG11.P{sumisi}=36occur;therefore,eitherEForEGoccursandso13−i,i=8,...,12.36E(F∪G)⊂EF∪EG.12.Eitherusehintorconditiononinitialoutcomeas:Similarly,ifEF∪EGoccurs,theneitherEForEGoccur.Thus,EoccursandeitherForGoccurs;andP{EbeforeF}soE(F∪G)occurs.Hence,=P{EbeforeF|initialoutcomeisE}P(E)EF∪EG⊂E(F∪G),+P{EbeforeF|initialoutcomeisF}P(F)+P{EbeforeF|initialoutcomeneitherEwhichtogetherwiththereverseinequalityprovestheresult.orF}[1−P(E)−P(F)]1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com2AnswersandSolutions=1·P(E)+0·P(F)+P{EbeforeF}17.Prob{end}=1−Prob{continue}=[1−P(E)−P(F)].=1−P({H,H,H}∪{T,T,T})P(E)=1−[Prob(H,H,H)+Prob(T,T,T)].Therefore,P{EbeforeF}=.P(E)+P(F)
11111113.ConditionaninitialtossFaircoin:Prob{end}=1−··+··222222123P{win}=∑P{win|throwi}P{throwi}.=4.i=2
111333Now,Biasedcoin:P{end}=1−··+··444444P{win|throwi}=P{ibefore7}9=.160i=2,12i−118.LetB=eventbotharegirls;E=eventoldestisi=3,...,65+1girl;L=eventatleastoneisagirl.=1i=7,11khdaw.comP(BE)P(B)1/4113−i(a)P(B|E)====.i=8,...,10,P(E)P(E)1/2219−113whereaboveisobtainedbyusingProblems11(b)P(L)=1−P(nogirls)=1−=,44and12.P(BL)P(B)1/41P{win}≈.49.P(B|L)====.P(L)P(L)3/43∞14.P{Awins}=∑P{Awinson(2n+1)sttoss}19.E=eventatleast1sixP(E)n=0∞2nnumberofwaystogetE11=∑(1−P)P==n=0numberofsamplepts36∞=P∑[(1−P)2]nD=eventtwofacesaredifferentP(D)n=01=1−Prob(twofacesthesame)=P1−(1−P)265P(ED)10/361P=1−=P(E|D)===.=366P(D)5/632P−P21=2.20.LetE=eventsamenumberonexactlytwoof−PP{Bwins}=1−P{Awins}thedice;S=eventall3numbersarethesame;D=eventall3numbersaredifferent.These1−P=.3eventsaremutuallyexclusiveanddefinethe2−Pwholesamplespace.Thus,1=P(D)+P(S)+P(E),P(S)=6/216=1/36;forDhave6possible16.P(E∪F)=P(E∪FEc)valuesforfirstdie,5forsecond,and4forthird.=P(E)+P(FEc)∴NumberofwaystogetD=6·5·4=120.sinceEandFEcaredisjoint.Also,P(D)=120/216=20/36P(F)=P(FE∪FEc)∴P(E)=1−P(D)−P(S)c=P(FE)+P(FE)bydisjointness.2015=1−−=.363612Hence,P(E∪F)=P(E)+P(F)−P(EF).21.LetC=eventpersoniscolorblind.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions3P(Male|C)(f)P4,3=P{alwaysahead|a,a}(4/7)(3/6)P(C|Male)P(Male)=(2/7)[1−P{a,a,a,b,b,b|a,a}=P(C|MaleP(Male)+P(C|Female)P(Female)−P{a,a,b,b|a,a}−P{a,a,b,a,b,b|a,a}].05×.5=(2/7)[1−(2/5)(3/4)(2/3)(1/2)=.05×.5+.0025×.5−(3/5)(2/4)−(3/5)(2/4)(2/3)(1/2)]250020==.=1/7.262521(g)P5,1=P{a,a}=(5/6)(4/5)=2/3.22.Lettrial1consistofthefirsttwopoints;trial2the(h)P5,2=P{a,a,a}+P{a,a,b,a}nexttwopoints,andsoon.Theprobabilitythat=(5/7)(4/6)[(3/5)+(2/5)(3/4)]=3/7.eachplayerwinsonepointinatrialis2p(1−p).BythesamereasoningwehavethatNowatotalof2npointsareplayedifthefirst(a−1)trialsallresultineachplayerwinningone(i)P5,3=1/4,ofthepointsinthattrialandthenthtrialresultsin(j)P5,4=1/9.oneoftheplayerswinningbothpoints.Byinde-n−npendence,weobtainthat(k)Inallthecasesabove,Pn,m=.khdaw.comn+nP{2npointsareneeded}25.(a)P{pair}=P{secondcardissame=(2p(1−p))n−1(p2+(1−p)2),n≥1.denominationasfirst}=3/51.TheprobabilitythatAwinsontrialnisn−12(b)P{pair|differentsuits}(2p(1−p))pandsoP{pair,differentsuits}=∞P{differentsuits}P{Awins}=p2(2p(1−p))n−1∑=P{pair}/P{differentsuits}n=123/51p==1/13.=.39/511−2p(1−p)4485239·38·3726.P(E1)==.1121351·50·4923.P(E1)P(E2|E1)P(E3|E1E2)···P(En|E1···En−1)3363926·25P(E2|E1)==.P(E1E2)P(E1E2E3)P(E1···En)1121338·37=P(E1)···P(E1)P(E1E2)P(E1···En−1)P(E|EE)=22426=13/25.31211213=P(E1···En).P(E4|E1E2E3)=1.39·26·1324.LetasignifyavoteforAandboneforB.P(E1E2E3E4)=.51·50·49(a)P2,1=P{a,a,b}=1/3.27.P(E1)=1(b)P3,1=P{a,a}=(3/4)(2/3)=1/2.P(E2|E1)=39/51,since12cardsareintheaceofspadespileand39arenot.(c)P3,2=P{a,a,a}+P{a,a,b,a}P(E3|E1E2)=26/50,since24cardsareinthepiles=(3/5)(2/4)[1/3+(2/3)(1/2)]=1/5.ofthetwoacesand26areintheothertwopiles.(d)P4,1=P{a,a}=(4/5)(3/4)=3/5.P(E4|E1E2E3)=13/49.(e)P4,2=P{a,a,a}+P{a,a,b,a}So=(4/6)(3/5)[2/4+(2/4)(2/3)]=1/3.P{eachpilehasanace}=(39/51)(26/50)(13/49).khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com4AnswersandSolutions28.Yes.P(A|B)>P(A)isequivalenttoP(AB)>Thus,P(A)P(B)whichisequivalenttoP(B|A)>P(B).∑P(Ei1Ei2···Eik)29.(a)P(E|F)=0.i1<···k(1−p)p(2)=3(.3)(.7)2=.441P{X=k−1}↔(n+1)p≥k.p(3)=(.7)3=.343.Theresultfollows.118.p(0)=,p(1)=.16.1−(.95)52−52(.95)51(.05).22111n!9.p(0)=,p(1)=,p(2)=,17.Followssincetherearepermutationsof2105x1!···xr!11nobjectsofwhichx1arealike,x2arealike,...,xrp(3)=,p(3.5)=.arealike.10107khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com8AnswersandSolutions18.Followsimmediately.(b)P{X=3}=p3+(1−p)3.P{X=4}19.P{X1+···+Xk=m}=P{X=4,Ihas2winsinfirst3games}
=n(p+···+p)m(p+···+pr)n−m.+P{X=4,IIhas2winsinfirst3games}m1kk+1=3p2(1−p)p+3p(1−p)2(1−p).
2
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1P{X=5}5!13120.=.054.2!1!2!5102=P{eachplayerhas2winsinthefirst4games}
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23375372221.1−−5−.=6p(1−p).10101021010E[X]=3p3+(1−p)3+12p(1−p)122.32.2222p+(1−p)+30p(1−p).Differentiatingandsettingequalto0shows23.InorderforXtoequaln,thefirstn−1flipsththatthemaximumisattainedwhenp=1/2musthavekhdaw.comr−1heads,andthennflipmustlandheads.Byindependencethedesiredprobabilityis27.P{samenumberofheads}=∑P{A=i,B=i}thus
in−1r−1n−r=∑k(1/2)kn−k(1/2)n−kr−1p(1−p)xp.iiikn−kn=∑(1/2)24.Itisthenumberoftailsbeforeheadsappearsforiiitherthtime.kn−kn=∑(1/2)k−iii25.Atotalof7gameswillbeplayedifthefirst6resultnn=(1/2).in3winsand3losses.Thus,k633Anotherargumentisasfollows.P{7games}=p(1−p).3P{#headsofA=#headsofB}Differentiationyieldsthat=P{#tailsofA=#headsofB}d2332sincecoinisfairP{7}=203p(1−p)−p3(1−p)dp=60p2(1−p)2[1−2p].=P{k−#headsofA=#headsofB}Thus,thederivativeiszerowhenp=1/2.Taking=P{k=total#heads}.thesecondderivativeshowsthatthemaximumisattainedatthisvalue.28.(a)Considerthefirsttimethatthetwocoinsgivedifferentresults.Then26.LetXdenotethenumberofgamesplayed.P{X=0}=P{(t,h)|(t,h)or(h,t)}22=p(1−p)=1(a)P{X=2}=p+(1−p)2p(1−p)2P{X=3}=2p(1−p)22(b)No,withthisprocedureE[X]=2p+(1−p)+6p(1−p)=2+2p(1−p).P{X=0}=P{firstflipisatail}=1−p.Sincep(1−p)ismaximizedwhenp=1/2,weseethatE[X]ismaximizedatthatvalue29.Eachflipafterthefirstwill,independently,resultofp.inachangeoverwithprobability1/2.Therefore,khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions9n−1n−138.c=2.P{kchangeovers}=(1/2).k3139.E[X]=.P{X=i}e−λλi/i!630.==λ/i.P{X=i−1}e−λλi−1/(i−1)!40.LetXdenotethenumberofgamesplayed.Hence,P{X=i}isincreasingforλ≥iand44P{X=4}=p+(1−p)decreasingforλ20}=dx=.20x22nE[Numberofchangeovers]=∑E[Xi]areaofdiskofradiusxi=236.P{D≤x}=areaofdiskofradius1=2(n−1)p(1−p).πx2==x2π42.Supposethecouponcollectorhasidifferenttypes.LetXidenotethenumberofadditionalcoupons37.P{M≤x}=P{max(X1,...,Xn)≤x}collecteduntilthecollectorhasi+1types.ItiseasytoseethattheXiareindependentgeomet-=P{X1≤x,...,Xn≤x}ricrandomvariableswithrespectiveparametersn(n−i)/n,i=0,1,...,n−1.Therefore,=∏P{Xi≤x}
n−1n−1n−1i=1X=n/(n−i)∑∑i∑∑[Xi]=∑=xn.i=0i=0i=0ndn−1=n∑1/j.fM(x)=dxP{M≤x}=nx.j=1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com10AnswersandSolutionsnk43.(a)X=∑Xi.E[X]=(rp−1+(1−p)r)=r−k∑iii=1i=1(b)E[X]=P{X=1}kii+(1−p)r∑i=P{redballiischosenbeforeallni=1blackballs}AnotherwaytosolvethisproblemistoletYdenote=1/(n+1)sinceeachofthesen+1thenumberofboxeshavingatleastonekey,andballsisequallylikelytobethethenusetheidentityX=r−Y,whichistruesinceonlythefirstkeyputineachboxdoesnotresultinonechosenearliestkTherefore,acollision.WritingY=∑I{Ni>0}andtakingni=1E[X]=∑E[Xi]=n/(n+1).expectationsyieldsthati=1kE[X]=r−E[Y]=r−[1−(1−p)r]∑i44.(a)LetYiequal1ifredballiischosenafterthei=1firstbutbeforethesecondblackball,k=r−k+(1−p)ri=1,...,n.Then∑ini=1khdaw.comY=∑Yi.∞i=146.UsingthatX=∑In,weobtain(b)E[Yi]=P{Yi=1}n=1=P{redballiisthesecondchosenfrom∞∞asetofn+1balls}E[X]=∑E[In]=∑P{X≥n}n=1n=1=1/(n+1)sinceeachofthen+1isequallylikelytobethesecondoneMakingthechangeofvariablesm=n−1giveschosen.∞∞E[X]=∑P{X≥m+1}=∑P{X>m}Therefore,m=0m=0E[Y]=n/(n+1).47.LetXibe1iftrialiisasuccessand0otherwise.(c)AnsweristhesameasinProblem41.(a)Thelargestvalueis.6.IfX1=X2=X3,then(d)Wecanlettheoutcomeofthisexperimentbethevector(R1,R2,...,Rn)whereRiisthe1.8=E[X]=3E[X1]=3P{X1=1};numberofredballschosenafterthe(i−1)standsobutbeforetheithblackball.Sinceallorderingsofthen+mballsareequallylikelyitfollowsP{X=3}=P{X1=1}=.6.thatalldifferentorderingsofR1,...,RnwillThatthisisthelargestvalueisseenbyhavethesameprobabilitydistribution.Markov’sinequalitywhichyieldsthatForinstance,P{X≥3}≤E[X]/3=.6.P{R1=a,R2=b}=P{R2=a,R1=b}.(b)Thesmallestvalueis0.Toconstructaproba-FromthisitfollowsthatalltheRihavethebilityscenarioforwhichP{X=3}=0letUsamedistributionandthusthesamemean.beauniformrandomvariableon(0,1),anddefine45.LetNidenotethenumberofkeysinboxi,1ifU≤.6X1=i=1,...,k.Then,withXequaltothenum-0otherwisek1ifU≥.4berofcollisionswehavethatX=∑(Ni−X2=0otherwisei=1k1ifeitherU≤.3orU≥.7+X3=.1)=∑(Ni−1+I{Ni=0})whereI{Ni=0}0otherwisei=1Itiseasytoseethatisequalto1ifNi=0andisequalto0otherwise.Hence,P{X1=X2=X3=1}=0.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions111a∞48..355.1=f(x)dx+f(x)dx0a49.E[X2]−(E[X])2=Var(X)=E(X−E[X])2≥0.aEqualitywhenVar(X)=0,thatis,whenXis≤cdx+P{X>a}constant.0≤ac+P{X>a}.250.Var(cX)=E[(cX−E[cX])]2256.LetXjequal1ifthereisatypeicouponinthe=E[c(X−E(X))]collection,andletitbe0otherwise.Thenumberof2n=cVar(X).distincttypesisX=∑Xi.2i=1Var(c+X)=E[(c+X−E[c+X])]nnn2E[X]=E[X]=P{X=1}=(1−p)k=E[(X−E[X])]∑i∑i∑ii=1i=1i=1=Var(X).TocomputeCov(Xi,Xj)wheni=j,notethatXiXjriseitherequalto1or0ifeitherXiorXjisequalto51.N=∑XjwhereXiisthenumberofflipsbetween0,andthatitwillequal0ifthereiseithernotypeii=1khdaw.comthe(i−1)standithhead.Hence,Xisgeometricortypejcouponinthecollection.Therefore,iwithmean1/p.Thus,P{XiXj=0}=P{Xi=0}+P{Xj=0}rr−P{X=X=0}ijE[N]=∑E[Xi]=.i=1p=(1−p)k+(1−p)kijn−(1−pi−pj)k52.(a).n+1Consequently,fori=j(b)0.(c)1.Cov(Xi,Xj)=P{XiXj=1}−E[Xi]E[Xj]
=1−[(1−pi)k+(1−pj)k211153.n,−.−(1−pi−pj)k]−(1−pi)k+12n+1n+1(1−pj)k54.(a)UsingthefactthatE[X+Y]=0weseethatBecauseVar(X)=(1−p)k[1−(1−p)k]0=2p(1,1)−2p(−1,−1)iiiweobtainwhichgivestheresult.n(b)ThisfollowsinceVar(X)=∑Var(Xi)+2∑∑Cov(Xi,Xj)0=E[X−Y]=2p(1,−1)−2p(−1,1).i=1iF(X3)U3X3X4>X3>X1X2>X4>X1>X3nnX2>X1>X4>X365.Cov(Xi,Xj)=Cov(µi+∑aikZk,µj+∑ajtZt)k=1t=1X4>X2>X3>X1nnX4>X2>X1>X3=∑∑Cov(ajkZk,ajtZt)t=1k=1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions13nnsincetheithballisequallylikelyto=∑∑aikajtCov(Zk,Zt)beeitherofthen+mballs,andsot=1k=1nnE[Xi]=P{Xi=1}==∑aikajkn+mnk=1X=∑Yiwherethelastequalityfollowssincei=1n1ifk=tE[X]=∑E[Yi]Cov(Zk,Zt)=.i=10ifk=tn=P{ithwhiteballisselected}∑i=1X1+···+Xn−nµnknk66.Pn>∈=∑=.n+mn+mi=1=P{|X1+···+Xn−nµ|>n∈}72.Forthematchingproblem,letting≤Var{X1+···+Xn}/n2∈2X=X1+···+XN222where=nσ/n∈1ifithmanselectshisownhatkhdaw.comXi=→0asn→∞.0otherwise,weobtain2N67.P{515}≤2.SinceP{Xi=1}=1/N,wesee1103

11N−15Var(Xi)=1−=2.(ii)P{X1+···+X10>15}≈1−Φ√.NNN10Also
1Cov(Xi,Xj)=E[XiXj]−E[Xi]E[Xj].69.Φ(1)−Φ=.1498.2Now,1iftheithandjthmenbothselect70.LetXibePoissonwithmean1.ThenXiXj=theirownhatsnnnk0otherwise,PX≤n=e−n.∑i∑1k=0k!andthusnE[XiXj]=P{Xi=1,Xj=1}Butfornlarge∑xi−nhasapproximatelyanor-=P{Xi=1}P{Xj=1|Xi=1}1maldistributionwithmean0,andsotheresult=11.follows.NN−1


Hence,nmn+m
71.(i)P{X=i}=1121ik−ikCov(Xi,Xj)=N(N−1)−N=N2(N−1)i=0,1,...,min(k,n).and
kNN−11(ii)X=∑XiVar(X)=+222NN(N−1)i=1N−11=+KknNNE[X]=∑E[Xi]=n=1.+mi=1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com14AnswersandSolutions73.AsNiisabinomialrandomvariablewithpara-(iii)Itiseasytoseethattherandomvariablesmeters(n,Pi),wehave(i)E[Ni]=nPji(ii)Var(Xi)I1,I2,...,Inareindependent.Forinstance,for=nPi=(1−Pi);(iii)fori=j,thecovarianceofjn}Cov(Ni,Nj)=∑Cov(Xk,Yk)k1=P{X1isthelargestofX1,...,Xn}=.=∑(E[XkYk]−E[Xk]E[Yk])nkHence,khdaw.com=−∑E[Xk]E[Yk](sinceXkYk=0)∞∞1kE[N]=∑P{N>n}=∑=∞.=−∑PiPjn=1n=1nk=−nPiPj.75.(i)KnowingthevaluesofN1,...,Njisequiva-(iv)Lettinglenttoknowingtherelativeorderingofthe1,ifnotypei’soccurelementsa1,...,a1.Forinstance,ifN1=0,Yi=0,otherwise,N2=1,N3=1thenintherandompermu-tationa2isbeforea3,whichisbeforea1.Thewehavethatthenumberofoutcomesthatneverrindependenceresultfollowsforclearlytheoccurisequalto∑Yiandthus,numberofa1,...,a1thatfollowai+1doesnot1probabilisticallydependontherelativeorder-rringofa1,...,ai.E∑Yi=∑E[Yi]111(ii)P{Ni=k}=,k=0,1,...,i−1ri=∑P{outcomesidoesnotoccur}whichfollowssinceoftheelements1a1,...,ai+1theelementai+1isequallyrst=(1−P)n.likelytobefirstorsecondor...or(i+1).∑i11i−1i−1(iii)E[Ni]=∑k=74.(i)Astherandomvariablesareindependent,ik=02identicallydistributed,andcontinuous,itfol-1i−1(i−1)(2i−1)lowsthat,withprobability1,theywillallE[N2]=k2=i∑havedifferentvalues.Hencethelargestofi6k=0X1,...,XnisequallylikelytobeeitherX1orandsoX2...orXn.Hence,asthereisarecordattime2nwhenXnisthelargestvalue,itfollowsthatVar(i−1)(2i−1)(i−1)(Ni)=−641P{arecordoccursatn}=.i2−1n=.121,ifarecordoccursatj(ii)LetIj=0,otherwise.76.E[XY]=µxµyThenE[(XY)2]=(µ2+σ2)(µ2+σ2)nnn1xxyyE∑Ij=∑E[Ij]=∑.22111jVar(XY)=E[(XY)]−(E[XY])khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions1577.Ifg1(x,y)=x+y,g2(x,y)=x−y,thenEe∑tixi=πetixii∂g1∂g1Ontheotherhand,iftheaboveissatisfied,then∂x∂yJ==2thejointmomentgeneratingfunctionisthatofthe∂g2∂g2sumofnindependentrandomvariablestheithof∂x∂ywhichhasthesamedistributionasxi.AsthejointHence,ifU=X+Y,V=X−Y,thenmomentgeneratingfunctionuniquelydetermines
1u+vu−vthejointdistribution,theresultfollows.fU,V(u,v)=fX,Y,.222∞
79.Eetz2=√1etx2e−x2/2dx221u+v2π−∞=exp−−µ4τσ22σ221∞=√exp{−x2(1−2t)/2}dx
2u−v2π−∞+−µ2=(1−2t)−1/2
e−µ2/σ2uµu2=exp−wherethelastequalityfollowssincewithσ−2=4τσ2σ24σ2khdaw.com1/(1−2t)v2∞exp−1−x2−24σ2√e/2σ=σ¯2π−∞78.(a)φxi(ti)=φ(0,0...0,1,0...0)withthe1inHence,theithplace.EetX=(1−2t)−n/2(b)Ifindependent,thenkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comkhdaw.comkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comChapter3∑xp(x,y)pY(y)(b)Bysamereasoningasin(a),ifY=1,then1.∑p(x|y)===1.X|YpY(y)pY(y)Xhasthesamedistributionasthenumberofxwhiteballschosenwhen5ballsarechosen2.Intuitivelyitwouldseenthatthefirstheadwouldfrom3whiteand6red.Hence,beequallylikelytooccuroneitheroftrials351,...,n−1.Thatis,itisintuitivethatE[X|Y]=5=.93P{X1=i|X1+X2=n}=1/(n−1),khdaw.comi=1,...,n−1.6.pX|Y(1|3)=P{X=1,Y=3}/P{Y=3}=P{1white,3black,2red}Formally,/P{3black}132P{X1=i|X1+X2=n}6!356=1!3!2!141414=P{X1=i,X1+X2=n}33P{X1+X2=n}6!593!3!1414P{X1=i,X2=n−i}=4P{X1+X2=n}=9p(1−p)i−1p(1−p)n−i−1=
8n−1p(1−p)n−2ppX|Y(0|3)=2712=1/(n−1).pX|Y(2|3)=9Intheabove,thenexttolastequalityusestheinde-1pX|Y(3|3)=pendenceofX1andX2toevaluatethenumerator27andthefactthatX1+X2hasanegativebinomial5distributiontoevaluatethedenominator.E[X|Y=1]=.33.E[X|Y=1]=27.GivenY=2,theconditionaldistributionofXand5ZisE[X|Y=2]=3112P{(X,Z)=(1,1)|Y=2}=E[X|Y=3]=.55P{(1,2)|Y=2}=04.No.P{(2,1)|Y=2}=045.(a)P{X=i|Y=3}=P{iwhiteballsselectedP{(2,2)|Y=2}=.5whenchoosing3ballsfrom3whiteand6red}36Soi3−i189=,i=0,1,2,3.E[X|Y=2]=+=95553E[X|Y=2,Z=1]=1.17khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com18AnswersandSolutions11−y1−y8.(a)E[X]=E[X|firstrollis6]expexp6yy515.fX|Y=y(x|y)==y1+E[X|firstrollisnot6]fy(y)exp−ydx60y15=+(1+E[X])166=,01is=Cexp{−(t−nθ)2/2n}f(x)λexp−λx22f(x)==whenx>1=Cexp{(t−nθ)/2n−∑(xi−θ)/2}X|X>1P{X>1}exp−λ=Cexp{t2/2n−θt+nθ2/2−x2/2∑i∞E[X|X>1]=expλxλexp−λxdx=1+1/λ+θt−nθ2/2}1byintegrationbyparts.=Cexp{(x)2/2n−x2/2}∑i∑if(x1,...,xn)f(x)1(b)f(x1,...,xn|T=t)=14.fX|X<1(x)=,xpAthenE[NA]−E[NB]<0,show-1+(1−p)/p−(1−p)ingthatplayingAfirstisbetter.(d)khdaw.comE[X]=(1+answerfrom(a))p+(1+2/p)(1−p)27.Conditionontheoutcomeofthefirstfliptoobtain=(2+p/(1−p)+(1−p)/p)pE[X]=E[X|H]p+E[X|T](1−p)+(1+2/p)(1−p)=(1+E[X])p+E[X|T](1−p)25.LetWdenotethenumberofwins.Conditioningonthenextflipgives(a)E[W]=E[E[W|X]]=E[X+Xp]E[X|T]=E[X|TH]p+E[X|TT](1−p)=(1+p)E[X]=(1+p)np(b)E[W]=E[E[W|Y]]=E[1+Yp]=(2+E[X])p+(2+1/p)(1−p)=1+p/p=2wherethefinalequalityfollowssincegiventhatsinceYisgeometricwithmean1/p.thefirsttwoflipsaretailsthenumberofadditionalflipsisjustthenumberofflipsneededtoobtaina26.LetNAandNBdenotethenumberofgameshead.PuttingtheprecedingtogetheryieldsneededgiventhatyoustartwithAandgiventhatyoustartwithB.ConditioningontheoutcomeofE[X]=(1+E[X])p+(2+E[X])p(1−p)thefirstgamegives+(2+1/p)(1−p)2E[NA]=E[NA|w]pA+E[NA|l](1−pA)orConditioningontheoutcomeofthenextgamegives1E[X]=p(1−p)2E[NA|w]=E[NA|ww]pB+E[NA|wl](1−pB)=2pB+(2+E[NA])(1−pB)28.LetYiequal1ifselectioniisred,andletitequal0=2+(1−pB)E[NA]otherwise.ThenAs,E[NA|l]=1+E[NB],weobtainthatkE[Xk]=∑E[Yi]E[NA]=(2+(1−pB)E[NA])pAi=1+(1+E[NB])(1−pA)E[Y]=r1r+b=1+pA+pA(1−pB)E[NA]r+(1−pA)E[NB]E[X1]=r+bkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions21E[Y2]=E[E[Y2|X1]]h1=E[H|h]p1+E[H|m]q1r+mX1=p1(E[H|h,h]p2+E[H|h,m]q2)+h2q1=Er+b+m=2p1p2+(1+h1)p1q2+h2q1r+mrSimilarly,wehavethatr+b=r+b+mh2=2p1p2+(1+h2)p2q1+h1q2rmr=+andwesolvetheseequationstofindh1r+b+mr+b+mr+b
andh2.rm=1+mmr+b+mr+b130.E[N]=∑E[N|Xo=j]p(j)=∑p(j)=mrj=1j=1p(j)=r+bE[X]=2r31.LetLidenotethelengthofruni.Conditioningon2r+bX,theinitialvaluegivesrToprovebyinductionthatE[Yk]=,assumeE[L1]=E[L1|X=1]p+E[L1|X=0](1−p)r+brkhdaw.comthatforallin1−p+(1+µ2)q1LetSbethenumberoftrialsneededforn=2p1p2+(2+µ1)p1q2+(1+µ2)q1successes,andletFbethenumberneededformfailures.ThenT=max(S,F).TakingexpectationsTheprecedingequationsimplifiestooftheidentityµ1(1−p1q2)=1+p1+µ2q1min(S,F)+max(S,F)=S+FSimilarly,wehavethatyieldstheresultµ2(1−p2q1)=1+p2+µ1q2nmE[min(S,F)]=+−E[T]Solvingtheseequationsgivesthesolution.p1−pkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com22AnswersandSolutions33.LetI(A)equal1iftheeventAoccursandletit35.np1=E[X1]equal0otherwise.n=E[X1|X2=0](1−p2)+E[X|X>0][1−(1−p)n]T∞122E∑Ri=E∑I(T≥i)Rip1ni=1i=1=n(1−p2)1−p2∞+E[X|X>0][1−(1−p)n]=∑E[I(T≥i)Ri]122i=1yieldingtheresult∞=∑E[I(T≥i)]E[Ri]np1(1−(1−p2)n−1)i=1E[X1|X2>0]=n1−(1−p2)∞=∑P{T≥i}E[Ri]i=136.E[X]=E[X|X=0](1−p0)+E[X|X=0]p0∞i−1yieldingthat=∑βE[Ri]i=1E[X]E[X|X=0]=∞1−p0=Eβi−1Rkhdaw.com∑iSimilarly,i=1E[X2]=E[X2|X=0](1−p)+E[X2|X=0]p00yieldingthat34.LetXdenotethenumberofthedicethatlandonsixonthefirstroll.2E[X2]E[X|X=0]=1−p0nnin−iHence,(a)mn=∑E[N|X=i](1/6)(5/6)ii=022nE[X]E[X]nin−iVar(X|X=0)=−=∑(1+mn−i)(1/6)(5/6)1−p0(1−p0)2ii=0n−1µ2+σ2µ2nni=−=1+mn(5/6)+∑mn−i(1/6)1−p0(1−p0)2ii=1(5/6)n−i37.(a)E[X]=(2.6+3+3.4)/3=3(b)E[X2]=[2.6+2.62+3+9+3.4+3.42]/3implyingthat=12.1067,andVar(X)=3.10671+∑n−1mn(1/6)i(5/6)n−imi=1n−ii38.LetXbethenumberofsuccessesinthentrials.n=1−(5/6)nNow,giventhatU=u,Xisbinomialwithpara-meters(n,u).Asaresult,Startingwithm0=0weseethatE[X|U]=nU1m1==6222221−5/6E[X|U]=nU+nU(1−U)=nU+(n−n)U1+m1(2)(1/6)(5/6)m2==96/11Hence,1−(5/6)2andsoon.E[X]=nE[U]=E[X2]=E[nU+(n2−n)U2](b)Sinceeachdierolledwilllandonsixwithprobability1/6,thetotalnumberofdice=n/2+(n2−n)[(1/2)2+1/12]rolledwillequalthenumberoftimesone=n/6+n2/3mustrolladieuntilsixappearsntimes.NHence,Therefore,E∑Xi=6n.Var(X)=n/6+n2/12i=1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions2339.LetNdenotethenumberofcycles,andletXbeTheprocessrestartseachtimetheprisonerthepositionofcard1.returnstohiscell.Therefore,nnE(N|X=1)=2+E(N)11(a)mn=∑E[N|X=i]=∑(1+mn−1)nnE(N|X=2)=3+E(N)i=1i=11n−1E(N|X=3)=0.=1+∑mjnj=1and(b)m1=1E(N)=(.5)(2+E(N))+(.3)(3+E(N))m=1+1=3/2+(.2)(0),22or1m3=1+(1+3/2)=1+1/2+1/3E(N)=9.5days.3=11/6(b)LetNidenotethenumberofadditionaldaystheprisonerspendsafterhavinginitiallycho-1m4=1+(1+3/2+11/6)=25/12sencelli.4(c)mn=1+1/2+1/3+···+1/n111E[N]=(2+E[N1])+(3+E[N2])+(0)khdaw.com(d)Usingtherecursionandtheinductionhypoth-333esisgivesthat51=+(E[N1]+E[N2]).1n−133mn=1+∑(1+···+1/j)Now,nj=1113E[N1]=(3)+(0)=1222=1+(n−1+(n−2)/2+(n−3)/3n11E[N2]=(2)+(0)=1+···+1/(n−1))22andso,15155=1+[n+n/2+···+n/(n−1)E[N]=+=.n3322−(n−1)]41.LetNdenotethenumberofminutesinthemaze.IfListheeventtheratchoosesitsleft,andRthe=1+1/2+···+1/neventitchoosesitsright,wehavebyconditioningn(e)N=∑Xionthefirstdirectionchosen:i=111nnE(N)=E(N|L)+E(N|R)22(f)mn=∑E[Xi]=∑P{iislastof1,...,i}i=1i=11121=(2)+(5+E(N))+[3+E(N)]n2332=∑1/i521i=1=E(N)+66(g)Yes,knowingforinstancethati+1isthelast=21.ofallthecards1,...,i+1tobeseen,tellsusnothingaboutwhetheriisthelastof1,...,i.42.LetXbethenumberofpeoplewhoarrivebeforennyou.Becauseyouareequallylikelytobethefirst,(h)Var(N)=∑Var(Xi)=∑(1/i)(1−1/i)orsecond,orthird,...,oreleventharrivali=1i=11P{X=i}=,i=0,...,1040.LetXdenotethenumberofdoorchosen,andlet10Nbethetotalnumberofdaysspentinjail.Therefore,110(a)ConditioningonX,wegetE[X]=∑i=5.5310i=1E[N]=∑E{N|X=i}P{X=1}.i=1andkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com24AnswersandSolutions110110(11)(21)AsE[X2]=i2==38.5∑10106i=1E[XY]=E[E[XY|X]]=E[XE[Y|X]]=E[X]givingthattheresultfollows.Var(X)=38.5−30.25=8.2548.Usingthehint,weseethat43.UsingExamples4dand4e,mean=µ1µ2,vari-Nance=µσ2+µ2σ2.EX=φ(0)1221∑i144.FromExamples4dand4e,mean=500,variance=E[NE[X]]sinceφ(0)=1,φ(0)=E[X]=E[N]Var(X)+E2(X)Var(N)=E[N]E[X].210(100)2=+(50)(10)212NE∑Xi=φ”(0)=33,333.1=E[N(N−1)E2[X]+NE[X2]]45.Now2=E[N2]E2[X]−E[N]E2[X]E[khdaw.comXn|Xn−1]=0,Var(Xn|Xn−1)=βXn−1.+E[N]E[X2].(a)FromtheaboveweseethatHence,E[Xn]=0.N(b)From(a)wehavethatVar(x2222n)=E[Xn].NowVar∑Xi=E[X](E|N]−E[N]122E[Xn]=E{E[Xn|Xn−1]}+E[N](E[X2]−E2[X])=E[βX2]2n−1=E[X]Var[N]+E[N]Var(X)=βE[X2]n−149.LetAbetheeventthatAistheoverallwinner,and=β2E[X2]letXbethenumberofgamesplayed.LetYequaln−2·tonumberofwinsforAinthefirsttwogames.=βnX2.0P(A)=P(A|Y=0)P(Y=0)46.(a)ThisfollowsfromtheidentityCov(U,V)=+P(A|Y=1)P(Y=1)E[UV]−E[U]E[V]uponnotingthat+P(A|Y=2)P(Y=2)E[XY]=E[E[XY|X]]=E[XE[Y|X]],=0+P(A)2p(1−p)+p2E[Y]=E[E[Y|X]]Thus,(b)Frompart(a)weobtainp2Cov(X,Y)=Cov(a+bX,X)P(A)=1−2p(1−p)=bVar(X).E[X]=E[X|Y=0]P(Y=0)47.E[X2Y2|X]=X2E[Y2|X]+E[X|Y=1]P(Y=1)≥X2(E[Y|X])2=X2+E[X|Y=2]P(Y=2)Theinequalityfollowingsinceforanyrandom=2(1−p)2+(2+E[X])2p(1−p)+2p222variableU.E[U]≥(E[U])andthisremains=2+E[X]2p(1−p)truewhenconditioningonsomeotherrandomvariableX.TakingexpectationsoftheaboveThus,showsthat2E[X]=E[(XY)2]≥E[X2].1−2p(1−p)khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions25110(b)Findtheprobabilitythatyouarethenth50.P{N=n}=(.3)n(.7)10−n3npersontoarrive.10n10−n+(.5)(.5)56.LetY=1ifitrainstomorrow,andletY=0notherwise.+10(.7)n(.3)10−n.E[X]=E[X|Y=1]P{Y=1}n+E[X|Y=0]P{Y=0}Nisnotbinomial.=9(.6)+3(.4)=6.6111E[N]=3+5+7=5.333P{X=0}=P{X=0|Y=1}P{Y=1}+P{X=0|Y=0}P{Y=0}51.Yes.=.6e−9+.4e−352.P{X+Ya)itfollowsthatifAisnotalwaysleadingthen(n−k)!theywillbetiedatsomepoint.(d)n!(b)ConsideranyoutcomeinwhichAreceivesthefirstvoteandtheyareeventuallytied,say72.Forn≥2a,a,b,a,b,a,b,b...WecancorrespondthisP{N>n|U1=y}sequencetoonewhichtakesthepartofthesequenceuntiltheyaretiedinthereverse=P{y≥U2≥U3≥···≥Un}order.Thatis,wecorrespondtheaboveto=P{Ui≤y,i=2,...,n}thesequenceb,b,a,b,a,b,a,a...wheretheremainderofthesequenceisexactlyasintheP{U2≥U3≥···geqUn|original.NotethatthislattersequenceisoneUi≤y,i=2,...,n}inwhichBisinitiallyaheadandthentheyaretied.Asitiseasytoseethatthiscorrespon-=yn−1/(n−1)!denceisonetoone,part(b)follows.(c)Now,∞P{BreceivesfirstvoteandtheyareE[N|U1=y]=∑P{N>n|U1=y}eventuallytied}n=0∞=P{Breceivesfirstvote}=n/(n+m).=2+yn−1/(n−1)!=1+ey∑Therefore,bypart(b)weseethatn=2P{eventuallytied}=2n/(n+m)Also,andtheresultfollowsfrompart(a).P{M>n|U1=1−y}=P{M(y)>n−1}76.Bytheformulagiveninthetextaftertheballot=yn−1/(n−1)!problemwehavethatthedesiredprobabilityis
11510573.Conditiononthevalueofthesumpriortogoing(18/38)(20/38).35over100.Inallcasesthemostlikelyvalueis101.(Forinstance,ifthissumis98thenthefinalsum77.WewillproveitwhenXandYarediscrete.isequallylikelytobeeither101,102,103,or104.Ifthesumpriortogoingoveris95thenthefinal(a)Thispartfollowsfrom(b)bytakingsumis101withcertainty.)g(x,y)=xy.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions29(b)E[g(X,Y)|Y=y]=∑∑g(x,y)80.Conditiononthetotalnumberofheadsandthenyxusetheresultoftheballotproblem.LetpdenoteP{X=x,Y=y|Y=y}.thedesiredprobability,andletjbethesmallestNow,integerthatisatleastn/2.P{X=x,Y=y|Y=y}nnp=∑pi(1−p)n−i2i−n0,ify=yini=j=P{X=x,Y=y},ify=y.1So,81.(a)f(x)=E[N]=E[N|X1=y]dy,0E[g(X,Y)|Y=y]=∑g(x,y)P{X=x|Y=y}.1ifyx.(c)E[XY]=E[E[XY|Y]]Hence,1=E[YE[X|Y]]by(a).f(x)=1+f(y)dy.x78.LetQn,mdenotetheprobabilitythatAisnever(b)f(x)=−f(x).behind,andPn,mtheprobabilitythatAisalwayskhdaw.com(c)f(x)=ce−x.Sincef(1)=1,weobtainthatahead.ComputingPn,mbyconditioningonthec=e,andsof(x)=e1−x.firstvotereceivedyieldsn(d)P{N>n}=P{xn}andsothedesiredprobabilityisn=0=(1−x)n/n!=e1−x.n+1−m∑Qn,m=.nn+1Thisalsocanbesolvedbyconditioningonwho82.(a)LetAdenotetheeventthatXisthekthiiobtainsthelastvote.ThisresultsintherecursionlargestofX1,...,Xi.ItiseasytoseethatthesenmareindependenteventsandP(Ai)=1/i.Qn,m=Qn−1,m+Qn,m−1,cccn+mn+mP{Nk=n}=P(AkAk+1···An−1An)whichcanbesolvedtoyieldk−1kn−21=···n+1−mkk+1n−1nQn,m=.n+1k−1=n(n−1)79.Letussupposewetakeapictureoftheurnbefore(b)Sinceknowledgeofthesetofvalueseachremovalofaball.Ifattheendoftheexper-{X1,...,Xn}givesusnoinformationaboutimentwelookatthesepicturesinreverseordertheorderoftheserandomvariablesitfollows(i.e.,lookatthelasttakenpicturefirst),wewillthatgivenNk=n,theconditionaldistribu-seeasetofballsincreasingateachpicture.ThetionofXNisthesameasthedistributionofksetofballsseeninthisfashionalwayswillhaveththeklargestofnrandomvariableshavingmorewhiteballsthanblackballsifandonlyifindistributionF.Hence,theoriginalexperimenttherewerealwaysmore∞whitethanblackballsleftintheurn.Therefore,fXN(x)=∑k−1n!kn=kn(n−1)(n−k)!(k−1)!thesetwoeventsmusthavesameprobability,i.e.,n−m/n+mbytheballotproblem.×(F(x))n−k(F(x))k−1f(x)khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com30AnswersandSolutionsNowmakethechangeofvariablei=n−k.(c)Asfollowthehint(d)Itfollowsfrom(b)and(c)thatf(x)=f(x).P{Neithereiorejwaseverrequestedbytimet}XNk=(1−P−P)t−1.ij83.LetIjequal1ifballjisdrawnbeforeballiandletitequal0otherwise.Thentherandomvariablewehaveofinterestis∑Ij.Now,byconsideringthefirstE[Positionofeiattimet]j=itimethateitheriorjiswithdrawnweseethat=1+∑1(1−Pi−Pj)t−1P{jbeforei}=wj/(wi+wj).Hence,j=i2Pjt−1wj+P+P(1−(1−Pi−Pj))E∑Ij=∑jij=ij=iwi+wjand84.WehaveE[Positionofelementrequestedatt]=∑PjE[Positionofeiattimet].E[Positionofelementrequestedattimet]n85.Considerthefollowingordering:khdaw.com=∑E[Positionattimet|eiselected]Pii=ie1,e2,...,el−1,i,j,el+1,...,enwherePi0,andsothesecondorderingisbetter.Thisshows=1+∑E(Ij)thateveryorderingforwhichtheprobabilitiesarej=inotindecreasingorderisnotoptimalinthesense=1+∑P{ejprecedeseiattimet}.thatwecandobetter.Sincethereareonlyafinitej=inumberofpossibleorderings,theorderingforGiventhatarequesthasbeenmadeforeitherwhichp1≥p2≥p3≥···≥pnisoptimum.eiorej,theprobabilitythatthemostrecentone87.(i)Thiscanbeprovedbyinductiononm.ItiswasforejisPj/(Pi+Pj).Therefore,obviouswhenm=1andthenbyfixingtheP{ejprecedeseiattimet|eiorejwasrequested}valueofx1andusingtheinductionhypoth-nPn−i+m−2jesis,weseethatthereare∑=.m−2Pi+Pji=0n−i+m−2suchsolutions.AsequalstheOntheotherhand,m−2numberofwaysofchoosingm−1itemsfromP{ejprecedeseiattimet|neitherwaseverasetofsizen+m−1undertheconstraintthatrequested}thelowestnumbereditemselectedisnum-1beri+1(thatis,noneof1,...,iareselected=.wherei+1is),weseethat2khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions31nn−i+m−2n+m−1n−1∑=.=P{∑jIj+n≤K}1/2i=0m−2m−1j=1Italsocanbeprovenbynotingthateachsolu-n−1tioncorrespondsinaone-to-onefashionwith+P{∑jIj≤K}1/2j=1apermutationofnonesand(m−1)zeros.Thecorrespondencebeingthatx1equalsthe=[Pn−1(k−n)+Pn−1(K)]/2.numberofonestotheleftofthefirstzero,x2thenumberofonesbetweenthefirstandsec-1ondzeros,andsoon.Asthereare(n+m−90.(a)e−552/2!·5e−5·e−51)!/n!(m−1)!suchpermutations,theresult11follows.(b)+e−552/2!·5e−5·e−5·e−552/2!e−552/2!(ii)Thenumberofpositivesolutionsofx1+···+xm=nisequaltothenumberofnonnegative111solutionsofy1+···+ym=n−m,andthus91.p5(1−p)3+p2(1−p)+pn−1therearesuchsolutions.m−192.LetXdenotetheamountofmoneyJoshpicksup(iii)Ifwefixasetofkofthexiandrequirethemwhenhespotsacoin.Thenkhdaw.comtobetheonlyzeros,thenthereareby(ii)n−1E[X]=(5+10+25)/4=10,(withmreplacedbym−k)suchm−k−1E[X2]=(25+100+625)/4=750/4mn−1Therefore,theamounthepicksuponhiswaysolutions.Hence,therearetoworkisacompoundPoissonrandomvariablekm−k−1withmean10·6=60andvariance6·750/4=outcomessuchthatexactlykoftheXiare1125.Becausethenumberofpickupcoinsthatequaltozero,andsothedesiredprobabilityJoshspotsisPoissonwithmean6(3/4)=4.5,wecanalsoviewtheamountpickedupasacom-mn−1n+m−1.Niskm−k−1m−1poundPoissonrandomvariableS=∑Xiwherei=1NisPoissonwithmean4.5,and(with5centsas88.(a)SincetherandomvariablesU,X1,...,Xnaretheunitofmeasurement)theXiareequallylikelyallindependentandidenticallydistributedittobe1,2,3.EitherusetherecursiondevelopedfollowsthatUisequallylikelytobetheithinthetextorconditiononthenumberofpickupssmallestforeachi+1,...,n+1.Therefore,todetermineP(S=5).Usingthelatterapproach,withP(N=i)=e−4.5(4.5)i/i!,givesP{X=i}=P{Uisthe(i+1)stsmallest}P(S=5)=(1/3)P(N=1)+3(1/3)3P(N=3)=1/(n+1).+4(1/3)4P(N=4)+5(1/3)5P(N=5)(b)GivenU,eachXiislessthanUwithprobabil-ityU,andsoXisbinomialwithparameters94.UsingthatE[N]=rw/(w+b)yieldsn,U.Thatis,giventhatU0(b)P4=.44440ij∑ikkj1,2k19.Thelimitingprobabilitiesareobtainedfrom14.(i){0,1,2}recurrent.r0=.7r0+.5r1(ii){0,1,2,3}recurrent.r1=.4r2+.2r3(iii){0,2}recurrent,{1}transient,{3,4}recur-r2=.3r0+.5r1rent.r0+r1+r2+r3=1,(iv){0,1}recurrent,{2}recurrent,{3}transient,{4}transient.andthesolutionis1339r0=,r1=,r2=,r3=.15.Consideranypathofstateskhdaw.comi0=i,i1,i2,...,in=j4202020suchthatPikik+1>0.Callthisapathfromitoj.ThedesiredresultisthusIfjcanbereachedfromi,thentheremustbea2pathfromitoj.Leti0,...,inbesuchapath.Ifallr0+r1=.5ofthevaluesi0,...,inarenotdistinct,thenthereisasubpathfromitojhavingfewerelementsm(forinstance,ifi,1,2,4,1,3,jisapath,thensois20.If∑Pij=1forallj,thenrj=1/(M+1)i,1,3,j).Hence,ifapathexists,theremustbeonei=0withalldistinctstates.satisfiesmm16.IfPwere(strictly)positive,thenPnwouldbe0rj=∑riPij,∑rj=1.ijjii=00foralln(otherwise,iandjwouldcommunicate).Butthentheprocess,startingini,hasapositiveHence,byuniquenessthesearethelimitingprob-probabilityofatleastPijofneverreturningtoi.abilities.Thiscontradictstherecurrenceofi.HencePij=0.21.Thetransitionprobabilitiesaren1−3α,ifj=i17.∑Yj/n→E[Y]bytheStronglawoflargenum-Pi,j=α,ifj=ii=1bers.NowE[Y]=2p−1.Hence,ifp>1/2,thenBysymmetry,E[Y]>0,andsotheaverageoftheYisconverges1inthiscasetoapositivenumber,whichimpliesPn=(1−Pn),j=iij3iinthat∑Yi→∞asn→∞.Hence,state0canbeSo,letusprovebyinductionthat1visitedonlyafinitenumberoftimesandsomust13n+(1−4α)ifj=ibetransient.Similarly,ifp<1/2,thenE[Y]<0,n44nPi,j=andsolim∑Yi=−∞,andtheargumentis1−1(1−4α)nifj=i144similar.Astheprecedingistrueforn=1,assumeitforn.Tocompletetheinductionproof,weneedtoshow18.Ifthestateattimenisthenthcointobeflip-thatpedthensequenceofconsecutivestatesconsti-tuteatwostateMarkovchainwithtransition13n+1+(1−4α)ifj=iprobabilitiesn+144P=i,j11n+1P1,1=.6=1−P1,2,P2,1=.5=P2,2−(1−4α)ifj=i44khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions35Now,Pn+1=PnP+PnP1/504/5i,ii,ii,i∑i,jj,ij=iP=2/73/72/713n3/94/92/9=+(1−4α)(1−3α)4411Solveforthestationaryprobabilitiestoobtainthe+3−(1−4α)nα44solution.13n=+(1−4α)(1−3α−α)25.LettingXndenotethenumberofpairsofshoes44atthedoortherunnerdepartsfromatthebegin-13n+1=+(1−4α)ningofdayn,then{Xn}isaMarkovchainwith44transitionprobabilitiesBysymmetry,forj=iPi,i=1/4,00.BecauseπPnisthelongrunpro-i,jii,j34.(a)πi,i=1,2,3,whicharetheuniquesolutionsportionoftimethechainiscurrentlyinstatejandofthefollowingequations.hadbeeninstateiexactlyntimeperiodsago,theinequalityfollows.π1=q2π2+p3π3π2=p1π1+q3π339.Becauserecurrenceisaclasspropertyitfollowsπ1+π2+π3=1thatstatej,whichcommunicateswiththerecur-rentstatei,isrecurrent.Butifjwerepositive(b)Theproportionoftimethatthereisarecurrent,thenbythepreviousexercisesowouldcounterclockwisemovefromiwhichbei.Becauseiisnot,wecanconcludethatjisnullisfollowedby5clockwisemovesisrecurrent.πiqipi−1pipi+1pi+2pi+3,andsotheanswer340.(a)Followsbysymmetry.to(b)is∑πiqipi−1pipi+1pi+2pi+3.Ini=1(b)Ifπi=a>0then,foranyn,theproportionofthepreceding,p0=p3,p4=p1,p5=p2,timethechainisinanyofthestates1,...,nisp6=p3.na.Butthisisimpossiblewhenn>1/a.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com38AnswersandSolutions41.(a)Thenumberoftransitionsintostateibytimen,44.GivenXn,Xn=1isbinomialwithparametersmandthenumberoftransitionsoriginatingfromstatep=Xn/m.Hence,E[Xn+1|Xn]=m(Xn/m)=Xn,ibytimen,andthenumberoftimeperiodsandsoE[Xn+1]=E[Xn].SoE[Xn]=iforalln.thechainisinstateibytimenalldifferbyatTosolve(b)notethatasallstatesbut0andmaremost1.Thus,theirlongrunproportionsmusttransient,itfollowsthatXnwillconvergetoeitherbeequal.0orm.Hence,fornlarge(b)riPijisthelongrunproportionoftransitionsE[Xn]=mP{hitsm}+0P{hits0}thatgofromstateitostatej.=mP{hitsm}.(c)∑riPijisthelongrunproportionoftransi-jButE[Xn]=iandthusP{hitsm}=i/m.tionsthatareintostatej.(d)Sincerjisalsothelongrunproportionoftran-45.(a)1,sinceallstatescommunicateandthusallaresitionsthatareintostatej,itfollowsthatrecurrentsincestatespaceisfinite.rj=∑riPij.(b)Conditiononthefirststatevisitedfromi.N−1jxi=∑Pijxj+PiN,i=1,...,N−1j=142.(a)Thisisthelongrunproportionoftransitionsx0=0,xN=1.khdaw.comthatgofromastateinAtooneinAc.(c)Mustshow(b)ThisisthelongrunproportionoftransitionsN−1cijthatgofromastateinAtooneinA.=∑Pij+PiNNN(c)BetweenanytwotransitionsfromAtoAcj=1theremustbeonefromActoA.SimilarlyNjbetweenanytwotransitionsfromActoAthere=∑PijNmustbeonefromAtoAc.Therefore,thelongj=0runproportionoftransitionsthatarefromAandfollowsbyhypothesis.toAcmustbeequaltothelongrunproportionoftransitionsthatarefromActoA.46.(i)Letthestatebethenumberofumbrellashehasathispresentlocation.Thetransition43.Consideratypicalstatesay,123.Wemustshowprobabilitiesare∏123=∏123P123,123+∏213P213,123P0,x=1,Pi,r−i=1−p,Pi,r−i+1=p,i=1,...,r.+∏231P231,123.NowP123,123=P213,123=P231,123=P1andthus,(ii)Wemustshowthat∏j=∑1riPijissatisfiedbythegivensolution.Theseequationsreduceto∏123=P1∏123+∏213+∏231.rr=r0+r1pWemustshowthatrj=rr−j(1−p)+rr−j+1p,j=1,...,r−1P1P2P2P1P2P3∏123=1−P,∏213=1−P,∏231=1−Pr0=rr(1−p),122satisfiestheabovewhichisequivalenttoanditiseasilyverifiedthattheyaresatisfied.pqP2P1P2P3(iii)pr0=.P1P2=P1+r+q1−P21−P2dp(1−p)(4−p)(1−2p)+p(1−p)P1(iv)=2=P2(P1+P3)dp4−p(4−p)1−P2p2−8p+4=P1P2(sinceP1+P3=1−P2).=2(4−p)√Bysymmetryalloftheotherstationaryequations8−48p2−8p+4=0⇒p==.55.alsofollow.2khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions3947.{Yn,n≥1}isaMarkovchainwithstates(i,j).53.Withπi(1/4)equaltotheproportionoftimeapolicyholderwhoseyearlynumberofacci-0,ifj=kP(i,j),(k,
)=P,ifj=k,dentsisPoissondistributedwithmean1/4isinj
Bonus-Malusstatei,wehavethattheaveragepre-miumiswherePj
isthetransitionprobabilityfor{Xn}.21(326.375)+[200π1(1/4)+250π2(1/4)limP{Yn=(i,j)}=limP{Xn=i,Xn+1=j}33n→∞n+400π3(1/4)+600π4(1/4)]=lim[P{Xn=i}Pij]n54.E[Xn+1]=E[E[Xn+1|Xn]].=riPij.NowgivenXn,M−XnXn+1,withprobability49.(i)No.MXn+1=limP{X=i}=pr1(i)+(1−p)r2(i).XnnXn−1,withprobability.M(ii)Yes.Hence,khdaw.comP=pP(1)+(1−p)P(2).ijM−XnXnijijE[Xn+1|Xn]=Xn+−MM50.UsingtheMarkovchainofExercise9,µh,t=1/.3,2Xn=Xn+1−,µt,h=1/.6.Also,thestationaryprobabilitiesofMthischainareπh=2/3,πt=1/3.Therefore,21andsoE[Xn+1]=1−ME[Xn]+1.E[A(t,t)]==578.7(1/3)(.4)(.6)(.3)(.6)(.3)(.4)Itisnoweasytoverifybyinductionthatthegivingformulapresentedin(ii)iscorrect.E[N(tththtt)|X0=h]=E[N(t,t)|X0=h]55.S11=P{offspringisaa|bothparentsdominant}+E(A(t,t)]P{aa,bothdominant}Also,=P{bothdominant}1E[N(t,t)|X0=h]=E[N(t)|X0=h]+21(1/3)(.4)rr213=4=.==10.8(1−q)24(1−q)21.2Therefore,E[N(tththtt)|X0=h]=589.5P{aa,1dominantand1recessiveparent}S10=P{1dominantand1recessiveparent}52.Letthestatebethesuccessivezonalpickuploca-P{aa,1parentaAand1parentaa}tions.ThenPA,A=.6,PB,A=.3.Thelongrun=proportionsofpickupsthatarefromeachzoneare2q(1−q)1πA=.6πA+.3πB=.6πA+.3(1−πA)2qr2=Therefore,πA=3/7,πB=4/7.LetXdenotethe2q(1−q)profitinatrip.Conditioningonthelocationoftherpickupgives=.2(1−q)34E[X]=E[X|A]+E[X|B]56.Thisisjusttheprobabilitythatagamblerstart-77ingwithmreacheshergoalofn+mbeforegoing341−(q/p)m=[.6(6)+.4(12)]+[.3(12)+.7(8)]broke,andisthusequalto,77n+m1−(q/p)=62/7whereq=1−p.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com40AnswersandSolutions57.LetAbetheeventthatallstateshavebeenvisited(b)Conditiononthefirsttransition.SupposeitbytimeT.Then,conditioningonthedirectionofistotheright.Inthiscasetheprobabilityisthefirststepgivesjusttheprobabilitythatagamblerwhoalwaysbets1andwinseachbetwithprobabilitypP(A)=P(A|clockwise)pwill,whenstartingwith1,reachγbefore+P(A|counterclockwise)qgoingbroke.Bythegamblersruinproblemthisprobabilityisequalto1−q/p1−p/q=p1−(q/p)n+q1−(p/q)n1−q/pγ.1−(q/p)Theconditionalprobabilitiesintheprecedingfollowbynotingthattheyareequaltotheproba-Similarly,ifthefirstmoveistotheleftthenbilityinthegamblersruinproblemthatagamblertheproblemisagainthesamegamblersruinthatstartswith1willreachnbeforegoingbrokeproblembutwithpandqreversed.Thewhenthegamblerswinprobabilitiesarepandq.desiredprobabilityisthusp−qq−p58.Usingthehint,weseethatthedesiredproba-γ=γ.1−(q/p)1−(p/q)bilityis∞∞Pkhdaw.com{Xn+1=i+1|Xn=i}64.(a)E∑Xk|X0=1=∑E[Xk|X0=1]P{limXm=N|Xn=i,Xn+1=i+1}k=0k=0P{limXm=N|Xn=1}∞1=µk=.∑pPi+1k=01−µ=Pi∞n(b)E∑Xk|X0=n=.andtheresultfollowsfromEquation(4.74).k=01−µ59.Conditionontheoutcomeoftheinitialplay.65.r≥0=P{X0=0}.Assumethatr≥P{Xn−1=0}.61.WithP0=0,PN=1P{Xn=0=∑P{Xn=0|X1=j}PjPi=αiPi+1+(1−αi)Pi−1,i=1,...,N−1j=P{X=}jPTheselatterequationscanberewrittenas∑n−1jj≤rjPPi+1−Pi=βi(Pi−Pi−1)∑jjwhereβi=(1−αi)/αi.Theseequationscannow=r.besolvedexactlyasintheoriginalgamblersruinproblem.Theygivethesolution166.(a)r0=1+∑i−1C3j=1jPi=N−1,i=1,...,N−1(b)r0=1.1+∑j=1Cj√(c)r0=3−12.wherej67.(a)Yes,thenextstatedependsonlyonthepresentCj=∏βiandnotonthepast.i=1(b)Oneclass,periodis1,recurrent.(c)PN−i,whereαi=(N−i)/N(c)P=PN−i,i=0,1,...,N−1i,i+1N62.(a)Sinceri=1/5isequaltotheinverseoftheiexpectednumberoftransitionstoreturntoPi,i−1=(1−P),i=1,2,...,NNstatei,itfollowsthattheexpectednumberofi(N−i)stepstoreturntotheoriginalpositionis5.Pi,i=P+(1−p),i=0,1,...,N.NNkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions41(d)See(e).Pij71.Ifrj=c,thenPji(e)r=Npi(1−p)N−i,i=0,1,...,N.PijPjkiirjPjk=cPji(f)DirectsubstitutionoruseExample7a.PjkPkjrkPkj=cN−1Pki(g)Time=∑Tj,whereTjisthenumberofj=iandarethusequalbyhypothesis.flipstogofromjtoj+1heads.Tjisgeo-72.Rateatwhichtransitionsfromitojtokoccur=metricwithE[Tj]=N/j.Thus,E[time]=riPijPjk,whereastherateinthereverseorderisN−1∑N/j.rkPkjPji.So,wemustshowj=iriPijPjk=rkPkjPji.68.(a)∑riQij=∑rjPji=rj∑Pji=rj.Now,riPijPjk=rjPjiPjkbyreversibilityiii=rjPjkPji(b)Whetherperusingthesequenceofstatesinkhdaw.comtheforwarddirectionoftimeorinthereverse=rkPkjPjibyreversibility.directiontheproportionoftimethestateisiwillbethesame.73.ItisstraightforwardtocheckthatriPij=rjPji.Forinstance,considerstates0and1.ThenMM!169.r(n1,...,nm)=n,...,n!m.r0p01=(1/5)(1/2)=1/101mWemustnowshowthatwhereasrnj+11r1p10=(2/5)(1/4)=1/10.(n1,...,ni−1,...,nj+1,...)MM−1i1=r(n1,...,ni,...,nj,...)74.(a)ThestatewouldbethepresentorderingoftheMM−1nprocessors.Thus,therearen!states.nj+1nior=,whichfollows.(b)Considerstatesx=(x1,...,xi−1,xi,xi+1,...,(ni−1)!(nj+1)!ni!nj!xn)andx1=(x1,...,xi−1,xi+1,xi,...,xn).Withqtequalto1−ptthetimereversible(m−i)2i2equationsare70.(a)Pi,i+1=m2,Pi,i−1=m2,i−1i−12i(m−i)1qPi,i=r(x)qxipxi+1∏qxk=r(x)qxi+1pxi∏xkm2k=1k=1(b)Since,inthelimit,thesetofmballsinurn1isorequallylikelytobeanysubsetofmballs,itis−11intuitivelyclearthatr(x)=(qxi+1/pxi+1)(qxi/pxi)r(x).2mmmim−iiSupposenowthatwesuccessivelyutilizetheπi=2m=2maboveidentityuntilwereachthestate(1,2,...,n).Notethateachtimejismovedtotheleftwemul-mm(c)Wemustverifythat,withtheπgivenin(b),tiplybyqj/pjandeachtimeitmovestotherightiwemultiplyby(q/p)−1.Sincex,whichisini-πP=πPjjjii,i+1i+1i+1,itiallyinpositionj,istohaveanetmoveofj−xjThatis,wemustverifythatpositionstotheleft(soitwillendupinpositionmmj−(j−xj)=xj)itfollowsfromtheabovethat(m−i)=(i+1)ii+1j−xr(x)=C∏(qxj/pxj)j.whichisimmediate.jkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com42AnswersandSolutions∞ThevalueofC,whichisequaltor(1,2,...,n),n+1canbeobtainedbysummingoverallstatesx=bj+∑a∑EβI{Xn=i,an=a}Pij(a)n=0i,aandequatingto1.Sincethesolutiongivenbytheabovevalueofr(x)satisfiesthetimereversibil-=b+aanEIP(a)ityequationsitfollowsthatthechainistimej∑∑β(Xn=i,an=a}iji,anreversibleandthesearethelimitingprobabilities.=bj+a∑yiaPij(a).75.Thenumberoftransitionsfromitojinanyinter-i,avalmustequal(towithin1)thenumberfromjtoi(c)Letdj,adenotetheexpecteddiscountedtimesinceeachtimetheprocessgoesfromitojinordertheprocessisinj,andaischosenwhenpolicytogetbacktoi,itmustenterfromj.βisemployed.Thenbythesameargumentasin(b):76.Wecanviewthisproblemasagraphwith64nodeswherethereisanarcbetween2nodesifaknight∑djacangofromonenodetoanotherinasinglemove.aTheweightsoneachareequalto1.Itiseasy=b+aanE[I{X=a}]P(a)tocheckthat∑∑wij=336,andforacornerj∑∑βn=i,aniji,anijnodei,∑wij=2.Hence,fromExample7b,fornyiakhdaw.comj=bj+a∑∑aEβI{Xn=i}∑yPij(a)i,aniaoneofthe4cornernodesi,∏=2/336,andthusaiyiathemeantimetoreturn,whichequals1/ri,is=bj+a∑∑dia,Pij(a),336/2=168.i,aa∑yiaa77.(a)y=EanIandweseefromEquation(9.1)thattheabove∑ja∑β∑{Xn=j,an=a}aanissatisfieduponsubstitutionofdia=yia.As1=EanIitiseasytoseethat∑i,adia=1−a,theresultβ∑∑{Xn=j,an=a}nafollowssinceitcanbeshownthattheselinearequationshaveauniquesolution.=EanI.β∑{Xn=j}n(d)Followsimmediatelyfrompreviousparts.Itisawell-knowresultinanalysis(andeasily(b)y=EanIn∑∑jaβ∑∑{Xn=j}proven)thatiflimn→∞∑iai/nalsoequalsjanja.Theresultfollowsfromthissincen1=Eβ∑a=.1−αE[R(Xn)]=∑R(j)P{Xn=j}∑yjaja∞=∑R(j)rj.=b+E=anIijβ∑{Xn=j}n=178.Letπj,j≥0,bethestationaryprobabilitiesofthe∞n+1underlyingchain.=bj+Eβ∑aI{Xn+1=j}n=0∞(a)∑jπjp(s|j)=b+E=an+1Ijβ∑∑{Xn=i,an=a}n=0i,aπjp(s|j)(b)p(j|s)=∑jπjp(s|j)I(Xn+1=j}khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comChapter51.(a)e−1,(b)e−1.7.P{Xt}=letSbeyourservicetime.Then,P{X1=t,X2>t}+P{X2=t,X1>t}f1(t)F¯2(t)E[T]=E[R+S1+S2+S3+S4+S]=khdaw.com4f1(t)F¯2(t)+f2(t)F¯1(t)=E[R]+∑E[Si]+E[S]=6/µi=1DividingthoughbyF¯1(t)F¯2(t)yieldstheresult.(Foramorerigorousargument,replace

=t”Wherewehaveusedthelackofmemorypropertyby”∈(t,t+)”throughout,andthenlet→0.)toconcludethatRisalsoexponentialwithrateµ.8.LetXihavedensityfiandtaildistributionF¯i.3.TheconditionaldistributionofX,giventhatnX>1,isthesameastheunconditionaldistribu-∑P{T=i}fi(t)i=1tionof1+X.Hence,(a)iscorrect.r(t)=n∑P{T=j}F¯j(t)11j=14.(a)0,(b),(c).n274∑P{T=i}ri(t)F¯i(t)i=1=n5.e−1bylackofmemory.∑P{T=j}F¯j(t)j=16.Conditiononwhichserverinitiallyfinishesfirst.TheresultnowfollowsfromNow,P{T=i}F¯i(t)P{T=i|X>t}=nP{Smithislast|server1finishesfirst}∑P{T=j}F¯j(t)=P{server1finishesbeforeserver2}j=1bylackofmemory9.Conditiononwhethermachine1isstillworkingλ1=.attimet,toobtaintheanswer,λ1+λ21−e−λ1t+e−λ1tλ1Similarly,λ1+λ2λ211.(a)UsingEquation(5.5),thelackofmemoryP{Smithislast|server2finishedfirst}=λ1+λ2propertyoftheexponential,aswellasthefactthattheminimumofindependentexponen-andthustialsisexponentialwitharateequaltothe
2
2sumoftheirindividualrates,itfollowsthatλ1λ2P{Smithislast}=+.nµλ1+λ2λ1+λ2P(A1)=λ+nµ43khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com44AnswersandSolutions
and,forj>1,λiλj1(d)∑i=j=kλ1+λ2+λ3λj+λkλ1+λ2+λ3(n−j+1)µP(Aj|A1···Aj−1)=11λ+(n−j+1)µ++λj+λkλkHence,wherethesumisoverall6permutationsof1,2,3.n(n−j+1)µp=∏λ+(n−j+1)µ13.LetTndenotethetimeuntilthenthpersoninlinej=1departstheline.Also,letDbethetimeuntilthe(b)Whenn=2.firstdeparturefromtheline,andletXbetheaddi-P{maxYi1.=1/(µ+µ)2+2[p/µ2+(1−p)/µ2]1221(d)Tisthesumofn−1independentexponen-−(p/µ+(1−p)/µ)2.tialswithrate2µ(sinceeachtimeafailure21occursthetimeuntilthenextfailureisexpo-µ1nentialwithrate2µ).20.(a)PA=µ1+µ2(e)Gammawithparametersn−1and2µ.2µ2(b)PB=1−24.LetTdenotethetimebetweenthe(i−1)thandtheµ1+µ2iithjobcompletion.ThentheTareindependent,(c)E[T]=1/µ1+1/µ2+PA/µ2+PB/µ2iwithTi,i=1,...,n−1beingexponentialwithrateµ121.E[time]=E[timewaitingat1]+1/µ1µ1+µ2.Withprobability,Tnisexponen-µ1+µ2+E[timewaitingat2]+1/µ2.µ2tialwithrateµ2,andwithprobabilityitisNowµ1+µ2khdaw.comexponentialwithrateµ1.Therefore,E[timewaitingat1]=1/µ1,n−1E[timewaitingat2]=(1/µ)µ1.E[T]=∑E[Ti]+E[Tn]2µ1+µ2i=11µ11µ21Thelastequationfollowsbyconditioningon=(n−1)++whetherornotthecustomerwaitsforserver2.µ1+µ2µ1+µ2µ2µ1+µ2µ1Therefore,n−1Var(T)=∑Var(Ti)+Var(Tn)E[time]=2/µ1+(1/µ2)[1+µ1/(µ1+µ2)].i=11=(n−1)+Var(Tn)(µ+µ)222.E[time]=E[timewaitingforserver1]+1/µ112Nowuse+E[timewaitingforserver2]+1/µ2.Var(Tn)=E[Tn2]−(E[Tn])2Now,thetimespentwaitingforserver1istheremainingservicetimeofthecustomerwithserverµ12µ221plusanyadditionaltimeduetothatcustomer=µ+µ2+µ1+µ212µ22µ1blockingyourentrance.Ifserver1finishesbeforeserver2thisadditionaltimewillequaltheaddi-µ11µ212−(+)tionalservicetimeofthecustomerwithserver2.µ1+µ2µ2µ1+µ2µ1Therefore,25.Parts(a)and(b)followuponintegration.ForpartE[timewaitingforserver1](c),conditiononwhichofXorYislargerandusethelackofmemorypropertytoconcludethatthe=1/µ1+E[Additional]amountbywhichitislargerisexponentialrateλ.=1/µ1+(1/µ2[µ1/(µ1+µ2)].Forinstance,forx<0,Sincewhenyouenterservicewithserver1thecus-fx−y(x)dxtomerprecedingyouwillbecateringservicewith=P{XX}server2,itfollowsthatyouwillhavetowaitforserver2ifyoufinishservicefirst.Therefore,con-1λeλxdx=ditioningonwhetherornotyoufinishfirst2For(d)and(e),conditiononI.E[timewaitingforserver2]13µ1=(1/µ2)[µ1/(µ1+µ2)].26.(a)+∑iµ1+µ2+µ3i=1µ1+µ2+µ3µiThus,4E[time]=2/µ1+(2/µ2)[µ1/(µ1+µ2)]+1/µ2.=µ1+mu2+µ3khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions471530.Conditiononwhichanimaldiedtoobtain(b)+(a)=µ1+µ2+µ3µ1+µ2+µ3E[additionallife]27.(a)µ1=E[additionallife|dogdied]µ1+µ3µ1µ2λdλc(b)+E[additionallife|catdied]µ1+µ3µ2+µ3λc+λdλc+λd1µ1µ21=1λd+1λc(c)∑µ+µ+µµ+µµλcλc+λdλdλc+λdii13233
1µ11µ2131.Conditiononwhetherthe1PMappointmentis(d)∑++iµiµ1+µ2µ2µ2+µ3µ3stillwiththedoctorat1:30,andusethefactthatifsheorheisthentheremainingtimespentisexpo-µ2µ1µ21+nentialwithmean30.Thisgivesµ1+µ2µ1+µ3µ2+µ3µ3E[timespentinoffice]28.Forbothparts,conditiononwhichitemfailsfirst.=30(1−e−30/30)+(30+30)e−30/30=30+30e−1khdaw.com(a)λiλ1∑ni=1λ∑λj32.LetTidenotethetimespentonjobi.Then∑jj=ij=1X=T1+T1+T2+T1+T2+T3=3T1+2T2+T3nandthus1λi1(b)n+∑nE[X]=6,Var(X)=9+4+1=14λi=1λ∑λj∑j∑jj=ii=1j=133.(a)Bythelackofmemoryproperty,nomatterwhenYfailstheremaininglifeofXisexpo-nentialwithrateλ.29.(a)fX|X+Y(x|c)=CfX.X+Y(x,c)(b)E[min(X,Y)|X>Y+c]=C1fXY(x,c−x)=E[min(X,Y)|X>Y,X−Y>c]=fX(x)fY(c−x)=E[min(X,Y)|X>Y]=Ce−λxe−µ(c−x),090}≈PZ>05.6=P{Z>2.07}=.019245.E[N(T)]=E[E[N(T)|T]]=E[λT]=λE[T]100−78.4E[TN(T)]=E[E[TN(T)|T]]=E[TλT]=λE[T2](e)P{L>100}≈PZ>khdaw.com5.6E[N2(T)]=EE[N2(T)|T]=E[λT+(λT)2]=P{Z>3.857}=.00006=λE[T]+λ2E[T2]40.Theeasiestwayistousedefinition5.1.ItiseasyHence,toseethat{N(t),t≥0}willalsopossessstation-Cov(T,N(T))=λE[T2]−E[T]λE[T]=λσ2aryandindependentincrements.SincethesumoftwoindependentPoissonrandomvariablesisalsoandPoisson,itfollowsthatN(t)isaPoissonrandomvariablewithmean(λ1+λ2)t.Var(N(T))=λE[T]+λ2E[T2]−(λE[T])2=λµ+λ2σ241.λ1/(λ1+λ2).N(t)N(t)42.(a)E[S4]=4/λ.46.E[∑Xi]=EE[∑Xi|N(t)]i=1i=1(b)E[S4|N(1)=2]=E[µN(t)]=µλt=1+E[timefor2moreevents]=1+2/λ.(c)E[N(4)−N(2)|N(1)=3]=E[N(4)−N(2)]N(t)N(t)E[N(t)∑Xi]=EE[N(t)∑Xi|N(t)]=2λ.i=1i=1Thefirstequalityusedtheindependentincre-=E[µN2(t)]=µ(λt+λ2t2)mentsproperty.Therefore,N(t)43.LetSidenotetheservicetimeatserveri,i=1,2Cov(N(t),X)=µ(λt+λ2t2)−λt(µλt)=µλt∑iandletXdenotethetimeuntilthenextarrival.i=1Then,withpdenotingtheproportionofcustomersthatareservedbybothservers,wehave47.(a)1(2µ)+1/λp=P{X>S1+S2}(b)LetTidenotethetimeuntilbothserversarebusywhenyoustartwithibusyserversi==P{X>S1}PX>S1+S2|X>S1}0.1.Then,µ1µ2=µ+λµ+λE[T0]=1/λ+E[T1]12Now,startingwith1serverbusy,letTbethe44.(a)e−λTtimeuntilthefirstevent(arrivalordeparture);khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions49n(n−j+1)µletX=1ifthefirsteventisanarrivalandP{N=0}=∏letitbe0ifitisadeparture;letYbetheλ+(n−j+1)µj=1additionaltimeafterthefirsteventuntilbothserversarebusy.P{N=n−i}λi(n−j+1)µE[T1]=E[T]+E[Y]=∏λ+(n−i)µλ+(n−j+1)µj=11λ=+E[Y|X=1]λ+µλ+µ−λ(T−s)49.(a)P{N(T)−N(s)=1}=λ(T−s)eµ+E[Y|X=0]λ+µ(b)Differentiatingtheexpressioninpart(a)andthensettingitequalto0,gives1µ=+E[T0]λ+µλ+µe−λ(T−s)=λ(T−s)e−λ(T−s)Thus,implyingthatthemaximizingvalueis11µE[T0]−=+E[T0]s=T−1/λλλ+µλ+µor(c)Fors=T−1/λ,wehavethatλ(T−s)=1khdaw.comandthus,2λ+µE[T0]=−1λ2P{N(T)−N(s)=1}=eAlso,50.LetTdenotethetimeuntilthenexttrainarrives;λ+µandsoTisuniformon(0,1).Notethat,conditionalE[T1]=λ2onT,XisPoissonwithmean7T.(c)LetLidenotethetimeuntilacustomerislost(a)E[X]=E[E[X|T]]=E[7T]=7/2.whenyoustartwithibusyservers.Then,(b)E[X|T]=7T,Var(X|T)=7T.Bythecondi-reasoningasinpart(b)givesthattionalvarianceformula1µVar(X)=7E[T]+49Var[T]=7/2+49/12=E[L2]=+E[L1]λ+µλ+µ91/12.1µ=λ+µ+(E[T1]+E[L2])λ+µ51.ConditiononX,thetimeofthefirstaccidenttoobtain1µµ=++E[L2]∞λ+µλ2λ+µE[N(t]=E[N(t)|X=s]βe−βsdsThus,0t1µ(λ+µ)=(1+α(t−s))βe−βsdsE[L2]=+30λλ52.Thisisthegambler’sruinprobabilitythat,start-48.GivenT,thetimeuntilthenextarrival.N,theingwithk,thegambler’sfortunereaches2knumberofbusyserversfoundbythenextarrival,before0whenherprobabilityofwinningeachisabinomialrandomvariablewithparametersnbetisp=λ1/(λ1+λ2).Thedesiredprobabilityisandp=e−µT.1−(λ/λ)k21.1−(λ/λ)2k21(a)E[N]=E[N|T=t]λe−λtdt53.(a)e−1−µt−λtnλ=neλedt=−1−1−1λ+µ(b)e+e(.8)eFor(b)and(c),youcaneitherconditiononT,−λm54.(a)P{L1=0}=eorusetheapproachofpart(a)ofExercise11toobtain(b)P{Lx}=e−λ(x−m)1(b)2p.m.1(e)E[R]=P{R>x}dx058.LetL=P{iisthelasttypecollected}.i1=m+P{R>x}dxLi=P{Xi=maxXj}mj=1,...,n1−nλ(x−m)∞−px−px=m+edx=piei∏(1−ej)dxm0j=i1−e−nλ(1−m)1=m+=∏(1−ypj/pi)dy(y=e−pix)nλ0j=iNow,usingthatP{L>x}=1−P{L≤x}=1−e−nλ(m−x),=E∏(1−Upj/pi)0C;iftheinequalityisrateλ,ithasagrazedistributionwithparam-reversedthent=0isbest.etersnandλ.(c)Sincethenumberofitemsnotfoundbyany(b)UsetheresultthatgivenSn=tthesetoftimestimetisindependentofthenumberfoundatwhichthefirstn−1ridersdepartedare(sinceeachofthePoissonnumberofitemsindependentuniform(0,t)randomvariables.willindependentlyeitherbecountedwithprobability1−e−µtoruncountedwithprob-Therefore,eachoftheseriderswillstillbeabilitye−µt)thereisnoaddedgaininlet-walkingattimetwithprobabilitytingthedecisiononwhethertostopattimet1−e−µt−µ(t−s)tdependonthenumberalreadyfound.p=eds/t=.0µt75.(a){Yn}isaMarkovchainwithtransitionproba-Hence,theprobabilitythatnoneofthearebilitiesgivenbywalkingattimetis(1−p)n−1.P0j=aj,Pi,i−1+j=aj,j≥0,73.(a)Itisthegammadistributionwithparameterswherenandλ.−λtje(λt)(b)Forn≥1,aj=dG(t).j!P{N=n|T=t}n−1(b){Xn}isaMarkovchainwithtransitionprob-P{T=t|N=n}p(1−p)=abilitiesfT(t)∞(λt)n−1=C(1−p)n−1Pi,i+1−j=βj,j=0,1,...,i,Pi,0=∑βj,(n−1)!k=i+1(λ(1−p)t)n−1where=C(n−1)!−µtje(µt)n−1βj=dF(t).=e−λ(1−p)t(λ(1−p)t)j!(n−1)!wherethelastequalityfollowssincethe76.LetYdenotethenumberofcustomersservedprobabilitiesmustsumto1.inabusyperiod.NotethatgivenS,theservice(c)ThePoissoneventsarebrokenintotwotimeoftheinitialcustomerinthebusyperiod,itclasses,thosethatcausefailureandthosethatfollowsbytheargumentpresentedinthetextthatdonot.ByProposition5.2,thisresultsin2theconditionaldistributionofY−1isthatofthekhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com54AnswersandSolutionsN(S)(iii)P{T>t}=P{N(t)=0}=e−m(t)where1compoundPoissonrandomvariable∑Yi,ti=1m(t)=λ(s)ds.wheretheYihavethesamedistributionasdoes0Y.Hence,81.(i)LetSidenotethetimeoftheithevent,i≥1.E[Y|S]=1+λSE[Y]Letti+his,X2>t}−max(min(X1,X2),...,min(X1,Xn)).=P{T1>s,T3>s,T2>t,T3>t}Nowusetheinductionhypothesis.Asecondmethodisasfollows:=P{T1>s,T2>t,T3>max(s,t)}SupposeX1≤X2≤···≤Xn).Thenthecoef-=e−λ1se−λ2te−λ3max(s,t)ficientofXontheright-sideisi


90.P{X1>s}=P{X1>s,X2>0}1−n−i+n−i−n−i+···123=e−λ1se−λ3sn−i=e−(λ1+λ3)s.=(1−1)0,i=n91.Tobegin,notethat=1,i=n,nPX1>∑XiandsobothsidesequalXn.Bysymmetrythe2resultfollowsforallotherpossibleorderingsoftheX
s.=P{X1>X2}P{X1−X2>X3|X1>X2}(iii)Takingexpectationsof(ii)whereXiisthe=P{X1−X2−X3>X4|X1>X2+X3}...timeofthefirsteventoftheithprocessyields=P{X1−X2···−Xn−1>Xn|X1>X2λ−1−(λ+λ)−1∑i∑∑ij+···+Xn−1}ii∑Xi−M=∑PX1>∑Xi1i=1i−1j=i=n/2n−1.94.(i)P{X>t}=P{noeventsinacircleofareart2}92.M2(t)=∑Ji−λrt2i=e.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions57∞E[N(s)|N(t)=n](ii)E[X]=P{X>t}dt0=E[E[N(s)|N(t)=n,L]|N(t)=n]∞e−λrt2=dt=E[n+L(s−t)|N(t)=n]0=n+(s−t)E[L|N(t)=n]∞√=√1e−x2/2dxbyx=t2λr2rλ0For(c),usethatforanyvalueofL,giventhattherehavebeenneventsbytimet,thesetofnevent1=√timesaredistributedasthesetofnindependent2λuniform(0,t)randomvariables.Thus,fors1.Letuschecktheforwardandbackwardequationsforthestate{(0,0);(0,0)}.Nowsubstituteintothebackwardequations.Backwardequation9.Sincethedeathrateisconstant,itfollowsthatasWeshouldhavelongasthesystemisnonempty,thenumberofdeathsinanyintervaloflengthtwillbeaPoissonP(t)=(λ1+λ2)λ2P(0,1)(0,0)(t)randomvariablewithmeanµt.Hence,(0,0),(0,0)λ1+λ2λ1P(t)=e−µt(µt)i−j/(i−j)!,02,µλ1λ012i×log−−.µ−λµ2µµ1=µ2=12.Hence,13.Withthenumberofcustomersintheshopasthe2state,wegetabirthanddeathprocesswith555P1=P0,P2=P1=P0,3333λ0=λ1=3µ1=µ2=4.55P3=P2=P033Therefore33332andas∑Pi=1,wehaveP1=P0P2=P1=P0.044422−1255527P0=1+++=.Andsince∑Pi=1,weget3332720khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions63µ(a)Thefractionoftheattendant’stimespentser-P12=P0vicingcarsisequaltothefractionoftimeλtherearecarsinthesystemandisthereforeλ(1−α)(1−α)µ2P2=P1=P0.1−P0=245/272.µ1αµ12(b)ThefractionofpotentialcustomersthatareSince∑Pi=1,wegetlostisequaltothefractionofcustomersthat0arrivewhentherearethreecarsinthestation−1µ21−αµ2andisthereforeP0=1++λααµ13λαµ15=.P3=P0=125/272.λαµ1+µ1µ2+λ(1−α)µ23P0isthepercentageoftimethesiteisoccupiedbyanacceptablemolecule.15.Withthenumberofcustomersinthesystemasthestate,wegetabirthanddeathprocesswithThepercentageoftimethesiteisoccupiedbyanunacceptablemoleculeisλ0=λ1=λ2=3λi=0,i≥4,µ1=2µ2=µ3=4.P=1−αµ2P=λ(1−α)µ2.20αµ1λαµ1+µ1+λ(1−α)µ2khdaw.comTherefore,thebalanceequationsreduceto33932717.Saythestateis0ifthemachineisup,sayitisiP1=2P0P2=4P1=8P0P3=4P2=32P0.whenitisdownduetoatypeifailure,i=1,2.Thebalanceequationsforthelimitingprobabili-Andtherefore,tiesareasfollows.−1392732λP0=µ1P1+µ2P2P0=1+++=.2832143µ1P1=λpP0(a)Thefractionofpotentialcustomersthatenterµ2P2=λ(1−p)P0thesystemisP0+P1+P2=1.λ(1−P3)2732116Theseequationsareeasilysolvedtogivethe=1−P3=1−×=.λ32143143results(b)WithaserverworkingtwiceasfastwewouldP0=(1+λp/µ1+λ(1−p)/µ2)−1get23P1=λpP0/µ1,P2=λ(1−p)P0/µ2.3333P1=P0P2=P1=P0P3=P0,4444−118.Therearek+1states;state0meansthemachine23isworking,stateimeansthatitisinrepairphase33364andP0=1+4+4+4=175.i,i=1,...,k.Thebalanceequationsforthelimit-ingprobabilitiesareSothatnow2764148λP0=µkPk1−P3=1−=1−=.64175175µ1P1=λP016.LetthestatebeµiPi=µi−1Pi−1,i=2,...,kP0+···+Pk=1.0:anacceptablemoleculeisattachedTosolve,notethat1:nomoleculeattached2:anunacceptablemoleculeisattached.µiPi=µi−1Pi−1=µi−2Pi−2=···=λP0.Hence,ThenthisisabirthanddeathprocesswithbalanceequationsPi=(λ/µi)P0,khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com64AnswersandSolutionsand,uponsumming,21.Howwehaveabirthanddeathprocesswithparametersk1=P01+∑(λ/µi).λi=λ,i=1,2i=1µi=iµ,i=1,2.Therefore,Therefore,−1k1+λ/µP0=1+∑(λµi),Pi=(λ/µi)P0,P0+P1=1+λ/µ+(λ/µ)2/2,i=1i=1,...,k.andsotheprobabilitythatatleastonemachineisupishigherinthiscase.Theanswertopart(a)isPiandtopart(b)isP0.22.Thenumberinthesystemisabirthanddeathpro-19.Thereare4states.Letstate0meanthatnocesswithparametersmachinesaredown,state1thatmachineoneisλn=λ/(n+1),n≥0downandtwoisup,state2thatmachineoneisµn=µ,n≥1.upandtwoisdown,and3thatbothmachinesaredown.Thebalanceequationsareasfollows.khdaw.comFromEquation(5.3),(λ1+λ2)P0=µ1P1+µ2P2∞1/P=1+(λ/µ)n/n!=eλ/µ0∑(µ1+λ2)P1=λ1P0+µ1P3n=1(λ1+µ2)P2=λ2P0andµ1P3=µ2P1+µ1P2Pn=P0(λ/µ)n/n!=e−λ/µ(λ/µ)n/n!,n≥0.P0+P1+P2+P3=1.23.LetthestatedenotethenumberofmachinesthatTheseequationsareeasilysolvedandthearedown.Thisyieldsabirthanddeathprocessproportionoftimemachine2isdownisP2+P3.with32120.Lettingthestatebethenumberofdownmachines,λ0=,λ1=,λ2=,λi=0,i≥3101010thisisabirthanddeathprocesswithparameters122µ1=,µ2=,µ3=.888λi=λ,i=0,1µi=µ,i=1,2.ThebalanceequationsreducetoBytheresultsofExample3g,wehavethat3/1012P1=P0=P01/85E[timetogofrom0to2]=2/λ+µ/λ2.2/10448P2=P1=P1=P0UsingtheformulaattheendofSection3,wehave2/8525that1/104192P3=P2=P3=P0.2/810250Var(timetogofrom0to2)3=Var(T0)+Var(T1)Hence,using∑Pi=1,yields011µµ22=λ2+λ(λ+µ)+λ3+µ+λ(2/λ+µ/λ).−11248192250P0=1+++=.5252501522UsingEquation(5.3)forthelimitingprobabilitiesofabirthanddeathprocess,wehavethat(a)Averagenumbernotinuse1+λ/µ21361068P0+P1=1+λ/µ+(λ/µ)2.=P1+2P2+3P3==.1522761khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions65(b)ProportionoftimebothrepairmenarebusyThereforeasolutionoftheformCαnβnmustbegivenby672336=P2+P3=1522=761.nmλλλλPn,m=1−1−.24.Wewillletthestatebethenumberoftaxiswait-µ1µ1µ2µ2ing.Then,wegetabirthanddeathprocesswithItiseasytoverifythatthisalsosatisfies(c)andλn=1µn=2.ThisisaM/M/1,andtherefore,(d)andisthereforethesolutionofthebalanceequations.1(a)Averagenumberoftaxiswaiting=µ−λ26.SincethearrivalprocessisPoisson,itfollowsthat1==1.thesequenceoffuturearrivalsisindependentof2−1thenumberpresentlyinthesystem.Hence,by(b)Theproportionofarrivingcustomersthatgettimereversibilitythenumberpresentlyinthesys-taxisistheproportionofarrivingcustomerstemmustalsobeindependentofthesequenceofthatfindatleastonetaxiwaiting.Theratepastdepartures(sincelookingbackwardsintimeofarrivalofsuchcustomersis2(1−P0).Thedeparturesareseenasarrivals).proportionofsucharrivalsistherefore2(1−P0)λλ127.ItisaPoissonprocessbytimereversibility.Ifkhdaw.com=1−P0=1−1−==.λ>δµ,thedepartureprocesswill(inthelimit)2µµ2beaPoissonprocesswithrateδµsincetheservers25.IfNi(t)isthenumberofcustomersintheithwillalwaysbebusyandthusthetimebetweensystem(i=1,2),thenletustake{N1(t),N2(t)}departureswillbeindependentrandomvariablesasthestate.Thebalanceequationarewitheachwithrateδµ.n≥1,m≥1.28.LetPx,VxdenotetheparametersoftheX(t)andiji(a)λP=µPyy0,020,1Pij,VioftheY(t)process;andletthelimiting(b)Pn,0(λ+µ1)=λPn−1,0+µ2Pn,1probabilitiesbyPx,Py,respectively.Byindepen-ii(c)P0,m(λ+µ2)=µ1P1,m−1+µ2P0,m+1dencewehavethatfortheMarkovchain(d)Pn,m(λ+µ1+µ2)=λPn−1,m+µ1Pn+1,m−1{X(t),Y(t)}itsparametersare+µ2Pn,m+1.V=Vx+Vy(i,
)i
WewilltryasolutionoftheformCαnβm=Pn,m.VxP=iPxFrom(a),weget(i,
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nn−1nand(λ+µ1)Cα=λCα+µ2Cαβ,limPxyor{(X(t),Y(t))=(i,j)}=PiPj.t→∞λ(λ+µ1)α=λ+µ2αβ=λ+µ2α=λ+λα,Hence,weneedshowthatµλPxPyVxPx=PxPyVxPx.andµ1α=λ⇒α=.i
iijj
jjiµ1[Thatis,ratefrom(i,
)to(j,
)equalstherateTogetC,weobservethat∑Pn,m=1,from(j,
)to(i,
)].Butthisfollowsfromthefactn,mthattheratefromitojinX(t)equalstheratefrombutjtoi;thatis,Pnm11xxxxxx∑n,m=C∑α∑β=C1−α1−β,PiViPij=PjVjPji.n,mnmλλTheanalysisissimilarinlookingatpairs(i,
)andandC=1−1−.µ1µ2(i,k).khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com66AnswersandSolutions29.(a)LetthestatebeS,thesetoffailedmachines.(c)TheprocessistimereversibleifwecanfindcprobabilitiesP(n)thatsatisfytheequations(b)Fori∈S,j∈S,P(n)µ/(r−1)=P(n)µ/(r−1)ijqS,S−i=µi/|S|,qS,S+j=λj,wherenandnareasgiveninpart(b).ThewhereS−iisthesetSwithideletedandS+jaboveequationsareequivalenttoissimilarlySwithjadded.Inaddition,|S|µP(n)=µ/P(n).denotesthenumberofelementsinS.ij(c)PSqS,S−i=PS−iqS−i,S.Sincen=n+1andn=n+1(whereiijjnreferstothekthcomponentofthevectorn),(d)Theequation(c)areequivalenttoktheaboveequationsuggeststhesolutionPSµi/|S|=PS−iλirP(n)=C(1/µ)nkor∏kk=1PS=PS−i|S|λi/µi.whereCischosentomaketheprobabili-tiessumto1.AsP(n)satisfiesallthetimeIteratingthisrecursiongivesreversibilityequationsitfollowsthatthechainistimereversibleandP(n)givenabovearethekhdaw.comPS=P0(|S|)!∏(λi/µi),limitingprobabilities.i∈Swhere0istheemptyset.SummingoverallS32.Thestatesare0,1,1,n,n≥2.State0meansthegivessystemisempty,state1(1)meansthatthereisoneinthesystemandthatoneiswithserver1(2);1=P0∑(|S|)!∏(λi/µi),staten,n≥2,meansthattherearencustomersinSi∈Sthesystem.Thetimereversibilityequationsareasandsofollows.(|S|)!∏(λi/µi)(λ/2)P0=µ1P1P=i∈S.(λ/2)P0=µ2P1S∑(|S|)!∏(λi/µi)Si∈SλP1=µ2P2AsthissolutionsatisfiesthetimereversibilityλP1=µ1P2equations,itfollowsthat,inthesteadystate,λPn=µPn+1,n≥2thechainistimereversiblewiththeselimitingprobabilities.whereµ=µ1+µ2.Solvingthelastsetofequa-tions(withn≥2)intermsofP2gives30.Sinceλijistherateitentersjwheninstatei,allPn+1=(λ/µ)PnweneeddotoprovebothtimereversibilityandthatPjisasgivenistoverifythat=(λ/µ)2P=···=(λ/µ)n−1P.n−12nThatis,λkjPk=λjkPj∑Pj=1.n1Pn+2=(λ/µ)P2,n≥0.Sinceλkj=λjk,weseethatPj≡1/nsatisfiestheEquationsthreeandfouryieldthatabove.P1=(µ2/λ)P231.(a)ThisfollowsbecauseofthefactthatallofP1=(µ1/λ)P2theservicetimesareexponentiallydistributedThesecondequationyieldsthatandthusmemoryless.P=(2µ/λ)P2)P.,...,n,...,n,...,n021=(2µ1µ2/λ2(b)Letn=(n1ijr),wheren>0andletn=(n,...,n−1,...,Thusalltheotherprobabilitiesaredeterminedini1inj−1,...,nr).Thenqn,n=µi/(r−1).termsofP0.However,wemustnowverifythatthekhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions67topequationholdsforthissolution.Thisisshownincludesall16statesexcept(0,0,0,0).Hence,asfollows.forthetruncatedmodelP=(2µ/λ)P=(2µµ/λ2)P.P{allworking/truncated}011122Thusallthetimereversibleequationsholdwhen=P{allworking}/(1−P(0,0,0,0)theprobabilitiesaregiven(intermsofP2)as4shownabove.ThevalueofP2isnowobtained∏(µi/(µi+λi)byrequiringalltheprobabilitiestosumto1.Thei=1=.factthatthissumwillbefinitefollowsfromthe4assumptionthatλ/µ<1.1−∏(λi/(λi+µi)i=133.Supposefirstthatthewaitingroomisofinfinitesize.LetX(t)denotethenumberofcus-35.WemustfindprobabilitiesPnsuchthatiitomersatserveri,i=1,2.ThensinceeachofPnqn=PnqntheM/M/1processes{Xi(t)}istimereversible,iijjjiitfollowsbyProblem28thatthevectorprocess{(X1(t),X2(t)),t≥0}isatimereversibleMarkovorchain.Nowtheprocessofinterestisjustthetrun-cPnq=Pnq,ifi∈A,j∈/Aiijjjikhdaw.comcationofthisvectorprocesstothesetofstatesAPq=cPnq,ifi∈/A,j∈AwhereiijjjiPiqij=Pjqji,otherwise.A={(0,m):m≤4}∪{(n,0):n≤4}Now,Piqij=Pjqjiandsoifwelet∪{(n,m):nm>0,n+m≤5}.Hence,theprobabilitythattherearenwithserver1nkPi/cifi∈APi=kPiifi∈/Aandnwithserver2isPn,m=k(λ1/µ1)n(1−λ1/µ1)(λ2/µ2)m(1−λ2/µ2),thenwehaveasolutiontotheaboveequations.BychoosingktomakethesumofthePnequalto1,wenmj=C(λ1/µ1)(λ2/µ2),(n,m)∈A.havethedesiredresult.Thatis,TheconstantCisdeterminedfrom−1∑Pn,n=1,k=∑Pi/c−∑Pi.i∈Ai∈/Awherethesumisoverall(n,m)inA.36.InProblem3,withthestatebeingthenumberof34.Theprocess{Xi(t)}isatwostatecontinuoustimemachinesdown,wehaveMarkovchainanditslimitingprobabilityislimP{Xi(t)=1}=µi/(µi+λi),i=1,...,4v0=2λP0,1=1t→∞µv1=λ+µP1,0=P1,2=1(λ+µ)(λ+µ)(a)Byindependence,v2=µP2,1=1.proportionoftimeallworking4Wewillchoosev=2λ=2µ,thentheuniformized=∏µi/(µi+λi).versionisgivenbyi=1(b)ItisacontinuoustimeMarkovchainsincethevn=2(λ+µ)fori=0,1,2processes{Xi(t)}areindependentwitheachibeingacontinuoustimeMarkovchain.n2λλP00=1−=(c)Yes,byProblem28sinceeachoftheprocesses2(λ+µ)(λ+µ){Xi(t)}istimereversible.n2λλP01=·1=(d)Themodelwhichsupposesthatoneofthe2(λ+µ)(λ+µ)phonesisdownisjustatruncationofthepro-nλ+µµµcess{X(t)}tothesetofstatesA,whereAP10=2(λ+µ)·(λ+µ)=2(λ+µ)khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com68AnswersandSolutionsnλ+µ1IntegratinggivesP11=1−=2(λ+µ)2λµtλ−(λ+µ)tn+µλλn(t)=−2e+CP12==λ+µ(λ+µ)2(λ+µ)(λ+µ)2(λ+µ)µSincem(0)=0itfollowsthatC=λ/(λ+µ)2.Pn=212(λ+µ)39.E[0(t)|x(0)=1]=t−E[timein1|X(0)=1]nµ2λ+µP22=1−=.2(λ+µ)2(λ+µ)λtµ−(λ+µ)t=t−−[1−e].λ+µ(λ+µ)237.Thestateofanytimeisthesetofdowncomponentsatthattime.ForS⊂{1,2,...,n},ThefinalequalityisobtainedfromExample7b(ori∈/S,j∈SProblem38)byinterchangingλandµ.q(S,S+i)=λi40.Cov[X(s),X(t)]=E[X(s)X(t)]−E[X(s)]EX(t)]q(S,S−j)=µjα|S|Now,whereS+i=S∪{i},S−j=S∩{j}c,|S|=num-1ifX(s)=X(t)=1berofelementsinS.X(s)X(t)=.khdaw.com0otherwiseThetimereversibleequationsareTherefore,fors≤tP(S)µα|S|=P(S−i)λ,i∈SiiE[X(s)X(t)]Theaboveissatisfiedwhen,forS={i1,i2,...,ik}=P{X(s)=X(t)=1|X(0)=0}λi1λi2···λik=P00(s)P00(t−s)bytheMarkovianpropertyP(S)=P(φ)µiµi···µiαk(k+1)/21−(λ+µ)s−(λ+µ)(t−s)12k=[µ+λe][µ+λe].(λ+µ)2whereP(φ)isdeterminedsothatAlso,∑P(S)=1nE[X(s)]E[X(t)]wherethesumisoverallthe2subsetsof{1,2,...,n}.=1[µ+λe−(λ+µ)s][µ+λe−(λ+µ)t].2(λ+µ)38.Saythattheprocessis“on”wheninstate0.Hence,(a)E[0(t+h)]=E[0(t)+ontimein(t,t+h)]Cov[X(s),X(t)]=n(t)+E[ontimein(t,t+h)]=1[µ+λe−(λ+µ)s]λe−(λ+µ)t[e(λ+µ)s−1].2Now(λ+µ)E[ontimein(t,t+h)|X(t)=0]=h+o(h)41.(a)LettingTidenotethetimeuntilatransitionE[ontimein(t,t+h)|X(t)=1]=o(h).outofioccurs,wehaveSo,bytheabovePij=P{X(Y)=j}=P{X(Y)=j|TiT.71khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com72AnswersandSolutionsHence,10.Yes,ρ/µ
TE[F]=xf(x)dx+(T+E[F])[1−F(T)]N(t)1numberofrenewalsin(X1,t)011.=+orttt
Txf(x)dx+T[1−F(T)]SinceX1<∞,Proposition3.1impliesthatE[F]=0,F(T)numberofrenewalsin(X1,t)1−ast−∞.andtheresultfollowsfromProposition3.1.tµ9.Ajobcompletionconstitutesareneval.LetTdenote12.LetXbethetimebetweensuccessived-events.thetimebetweenrenewals.TocomputeE[T]startConditioningonT,thetimeuntilthenexteventbyconditioningonW,thetimeittakestofinishthefollowingad-event,givesnextjob.
d
∞E[X]=xλe−λxdx+(x+E[X]λe−λxdxE[T]=E[E[T|W]].0dNow,todetermineE[T|W=w]conditiononS,=1/λ+E[X]e−λdthetimeofthenextshock.Thisgives1
∞Therefore,E[X]=λ(1−e−λd)E[khdaw.comT|W=w]=E[T|W=w,S=x]λe−λxdx01−λd(a)=λ(1−e)Now,ifthetimetofinishislessthanthetimeoftheE[X]shockthenthejobiscompletedatthefinishtime;(b)1−e−λdotherwiseeverythingstartsoverwhentheshockoccurs.Thisgives13.(a)N1andN2arestoppingtimes.N3isnot.x+E[T]ifxt,E[e−λW]=e−λwf(w)dw.andsoN(t)+1=ndependsonlyon0X1,...,Xn.ThusN(t)+1isastoppingtime.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions73(b)FollowsuponapplicationofWald’s17.(i)Yes.(ii)No—Yes,ifFexponential.equation—usingN(t)+1asthestoppingtime.18.WecanimaginethatarenewalcorrespondstoaN(t)+1machinefailure,andeachtimeanewmachineis(c)∑Xiisthetimeofthefirstrenewalputinuseitslifedistributionwillbeexponentiali=1aftert.Theinequalityfollowsdirectlyfromwithrateµ1withprobabilityp,andexponentialthisinterpretationsincetheremustbeatleastwithrateµ2otherwise.Hence,ifourstateistheonerenewalintheintervalbetweentandindexoftheexponentiallifedistributionofthet+m.machinepresentlyinuse,thenthisisa2-statecon-N(t)+1tinuoustimeMarkovchainwithintensityrates(e)t<∑Xit]e−λttherewardduringacycleequaltoW1+···+WN.ThusE[W],theaveragerewardperunitkhdaw.comtimeisE[W1+···+WN]/E[N].=E[X|Xt]e−λt+t+1e−λtλ+µbothincreaseatthesamerate.Hence,theyarealwaysequal.=1−t+1e−λt+t+1e−λt(c)Thisfollowsfrom(b)bylookingatthevalueµµλ+µofthetwototalsattheendofthefirstbusyperiod.1λ=1−e−λt(d)ItiseasytoseethatNisastoppingtimeµλ+µfortheLi,i≥1,andso,byWald’sEquation,NMoreintuitively,writingX=B+(X−B),andE[∑Li]=E[L]E[N].Thus,from(a)and(c),notingthatX−Bistheadditionalamountofser-i=1vicetimeremainingwhenthecycleends,givesweobtainthatE[W]=E[L].E[B]=E[X]−E[X−B]A(t)t−SN(t)30.=tt11=−P(X>B)µµSN(t)=1−11λt=−e−λtµµλ+µSN(t)N(t)=1−.ThelongrunproportionoftimethattheserverisN(t)tE[B]TheresultfollowssinceSN(t)/N(t)—µ(bythebusyist+1/λ.Stronglawoflargenumbers)andN(t)/t—1/µ.34.Acyclebeginsimmediatelyafteracleaningstarts.31.P{E(t)>x|A(t)=s}LetCbethecostofacycle.=P{0renewalsin(t,t+x]|A(t)=s}
3T/4E[C]=λC2T/4+C1λG¯(y)dy=P{interarrival>x+s|A(t)=s}0=P{interarrival>x+s|interarrival>s}wheretheprecedingusesthatthenumberofcus-tomersinanM/G/∞systemattimetisPoisson1−F(x+s)
=.t1−F(s)distributedwithmeanλG¯(y)dy.Thelongrun0khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com76AnswersandSolutionsaveragecostisE[C]/T.Thelongrunproportionof(iii)Inthiscasethelocationofskieri,whetherT/4goingupordown,isa2-statecontinuoustimetimeofthesystemisbeingcleanedis=1/4.TMarkovchain.Lettingstate0correspondtogoingup,thensinceeachskieractsindepen-35.(i)WecanviewthisasanM/G/∞systemwheredentlyaccordingtothesameprobability,weasatellitelaunchingcorrespondstoanarrivalhaveandFistheservicedistribution.Hence,nkn−kP{U(t)=k}=[P00(t)][1−P00(t)]P{X(t)=k}=e−λ(t)[λ(t)]k/k!,k
whereP(t)=(λe−(λ+µ)t+µ)/(λ+µ).t00whereλ(t)=λ(1−F(s))ds.037.(a)Thisisanalternatingrenewalprocess,with(ii)Byviewingthesystemasanalternatingthemeanofftimeobtainedbyconditioningonrenewalprocessthatisonwhenthereisatwhichmachinefailstocausetheoffperiod.leastonesatelliteorbiting,weobtain3E[off]=∑E[off|ifails]P{ifails}1/λi=1limP{X(t)=0}=,1/λ+E[T]λ1λ2=(1/5)+(2)khdaw.comwhereT,theontimeinacycle,isthequantityλ1+λ2+λ3λ1+λ2+λ3ofinterest.Frompart(i)λ3+(3/2)limP−λµ,λ1+λ2+λ3{X(t)=0}=e
∞Astheontimeinacycleisexponentialwithwhereµ=(1−F(s))dsisthemeantimerateequaltoλ1+λ2+λ3,weobtainthat0p,theproportionoftimethatthesystemisthatasatelliteorbits.Hence,workingise−λµ=1/λ,1/(λ1+λ2+λ3)p=1/λ+E[T]E[C]whereandsoE[C]=E[cycletime]1−e−λµE[T]=.λe−λµ=1/(λ1+λ2+λ3)+E[off](b)Thinkofthesystemasarenewalrewardpro-36.(i)IfweletNi(t)denotethenumberoftimescessbysupposingthatweearn1perunittimepersonihasskieddownbytimet,thenthatmachine1isbeingrepaired.Then,r1,the{Ni(t)}isa(delayed)renewalprocess.Asproportionoftimethatmachine1isbeingN(t)=∑Ni(t),wehaverepairedisN(t)Ni(t)1λ1lim=∑lim=∑,(1/5)titiµi+θir=λ1+λ2+λ31E[C]whereµiandθiarerespectivelythemeanof(c)Byassumingthatweearn1perunittimethedistributionsFiandGi.whenmachine2isinastateofsuspendedani-(ii)Foreachskier,whethertheyareclimbingupormation,showsthat,withs2beingthepropor-skiingdownconstitutesanalternatingrenewaltionoftimethat2isinastateofsuspendedprocess,andsothelimitingprobabilitythatanimation,skieriisclimbingupispi=µi/(µi+θi).Fromλ1λ3thisweobtain(1/5)+(3/2)λ1+λ2+λ3λ1+λ2+λ3s2=limP{U(t)=k}=∑{∏pi∏(1−pi)},E[C]Si∈Si∈Scn38.LetTe,fdenotethetimeittakestogofrometowheretheabovesumisoverallofthekf,andletdbethedistancebetweenAtoB.Then,subsetsSofsizek.withSbeingthedriver’sspeedkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions77
601Hence,E[TA,B]=E[TA,B|S=s]ds20401/(2λ)E[idle]=
1/(2λ)+E[B]160d=ds2040s1/(1−P1)40.Proportionoftime1shoots=by3=dlog(3/2)∑1/(1−Pj)20j=1alternatingrenewalprocess(orbysemi-MarkovAlso,process)since1/(1−Pj)isthemeantimemarks-E[TB,A]=E[TB,A|S=40](1/2)+E[TB,A|Smanjshoots.Similarly,proportionoftimeishoots1/(1−Pi)=.1∑1/(1−Pj)=60](1/2)=(d/40+d/60)2
1=d/48(1−F(x)dx41.0µ1
1E[T]log(3/2)2−x3A,B20dx=inpart(i)(a)=024khdaw.comE[TA,B]+E[TB,A]1log(3/2)+1/48=20
1−x−1edx=1−einpart(ii).(b)Byassumingthatarewardisearnedatarate0of1perunittimewheneverheisdrivingata
xspeedof40milesperhour,weseethatp,the1−y/µ−x/µ42.(a)Fe(x)=edy=1−e.proportionoftimethisisthecase,isµ01
x1(1/2)d/4080(b)Fe(x)=dy=x/c,0≤x≤c.p==cE[TA,B]+E[TB,A]1log(3/2)+1/48020(c)Youwillreceiveaticketif,startingwhenyoupark,anofficialappearswithin1hour.39.LetBbethelengthofabusyperiod.WithSFromExample5.1cthetimeuntiltheoffi-equaltotheservicetimeofthemachinewhosecialappearshasthedistributionFe,which,byfailureinitiatedthebusyperiod,andTequaltoparta,istheuniformdistributionon(0,2).theremaininglifeoftheothermachineatthatThus,theprobabilityisequalto1/2.moment,weobtain
43.Sincehalftheinterarrivaltimeswillbeexponen-E[B]=E[B|S=s]g(s)dstialwithmean1andhalfwillbeexponentialwithmean2,itwouldseemthatbecausetheexponen-Now,tialswithmean2willlast,onaverage,twiceaslong,thatE[B|S=s]=E[B|S=s,T≤s](1−e−λs)+E[B|SF¯2e−x/2+1e−x=s,T>s]e−λse(x)=33Withµ=(1)1/2+(2)1/2=3/2equaltothemean=(s+E[B])(1−e−λs)+se−λsinterarrivaltime−λs
∞F¯(y)=s+E[B](1−e)F¯e(x)=dyxµSubstitutingbackgivesthatandtheearlierformulaisseentobevalid.E[B]=E[S]+E[B]E[1−e−λs]44.LetTbethetimeittakestheshuttletoreturn.orNow,givenT,XisPoissonwithmeanλT.Thus,E[X|T]=λT,Var(X|T)=λTE[S]E[B]=E[e−λs]Consequently,khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com78AnswersandSolutions(a)E[X]=E[E[X|T]]=λE[T](a)r=2.1(b)Var(X)=E[Var(X|T)]+Var(E[X|T])5riµi(b)Pi=andso,=λE[T]+λ2Var(T)∑iriµi(c)Assumethatarewardof1isearnedeachtime243P1=,P2=,P3=.theshuttlereturnsempty.Then,fromrenewal999rewardtheory,r,therateatwhichtheshuttle46.Continuous-timeMarkovchain.returnsempty,isP{empty}47.(a)Byconditioningonthenextstate,weobtainr=E[T]thefollowing:P{empty|T=t}f(t)dtµj=E[timeini]=E[T]=∑E[timeini|nextstateisj]Pije−λtf(t)dt=∑tP.=ijijE[T]iE[e−λT](b)Usethehint.Then,=khdaw.comE[T]E[rewardpercycle](d)Assumethatarewardof1isearnedeachtime=E[rewardpercycle|nextstateisj]Pijthatacustomerwritesanangryletter.Then,withNaequaltothenumberofangryletters=tijPij.writteninacycle,itfollowsthatra,therateatAlso,whichangrylettersarewritten,isE[timeofcycle=E[timebetweenvisitstoi].ra=E[Na]/E[T]Now,ifwehadsupposedarewardof1per
unittimewhenevertheprocesswasinstate=E[Na|T=t]f(t)dt/E[T]iand0otherwisethenusingthesamecycletimesasabovewehavethat
∞=λ(t−c)f(t)dt/E[T]E[rewardiscycle]µicPi==.E[timeofcycle]E[timeofcycle]=λE[(T−c)+]/E[T]Hence,Sincepassengersarriveatrateλ,thisimpliesthattheproportionofpassengersthatwriteE[timeofcycle]=µi/Pi,angrylettersisra/λ.andso(e)Becausepassengersarriveataconstantrate,theproportionofthemthathavetowaitmoreaveragerewardperunittime=tijPijPi/µi.thancwillequaltheproportionoftimethatTheaboveestablishestheresultsincetheaver-theageoftherenewalprocess(whoseeventagerewardperunittimeisequaltothepro-timesarethereturntimesoftheshuttle)isportionoftimetheprocessisiniandwillnextgreaterthanc.ItisthusequaltoF¯e(c).enterj.45.ThelimitingprobabilitiesfortheMarkovchainare48.Letthestatebethepresentlocationifthetaxiisgivenasthesolutionofwaitingorletitbethemostrecentlocationifit1isontheroad.Thelimitingprobabilitiesofther1=r2+r32embeddedMarkovchainsatisfyr2=r12r1=r3r1+r2+r3=1.31orr2=r1+r3321r1=r2=5,r3=5.r1+r2+r3=1.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions79Solving,yieldswhich,assuchvisitscanbethoughtofasbeingrenewals,convergesto13r1=,r2=r3=.(E[numberoftransitionsbetweenvisits])−148ThemeantimespentinstateibeforeenteringbyProposition3.1.But,byMarkov-chainthe-anotherstateisory,thismustequalxi.Asthequantityin(d)isclearlyunaffectedbytheactualtimesbetweenµ1=1+10=11,µ2=2+20=22,transition,theresultfollows.2167Equation(6.2)nowfollowsbydividingnumer-µ3=4+15+25=,333atoranddenominatorof(b)bym;bywritingandsothelimitingprobabilitiesareXjXjiiNi(m)=66198201mNi(m)(m)P1=,P2=,P3=.465465465andbyusing(c)and(d).Thetimethestateisiisbrokeninto2parts—the51.Itisanexampleoftheinspectionparadox.Becausetimetiwaitingati,andthetimetraveling.Hence,everytouristspendsthesametimeindepartingtheproportionoftimethetaxiiswaitingatstateithecountry,thosequestionedatdepartureconsti-khdaw.comisPiti/(ti/µi).Theproportionoftimeitistravel-tutearandomsampleofallvisitingtourists.OningfromitojisPimij/(ti+µi).theotherhand,ifthequestioningisofrandomlychosenhotelgueststhen,becauselongerstaying49.Thinkofeachinterarrivaltimeasconsistingofnguestsaremorelikelytobeselected,itfollowsthatindependentphases—eachofwhichisexponen-theaveragetimeoftheonesselectedwillbelargertiallydistributedwithrateλ—andconsiderthethantheaverageofalltourists.Thedatathatthesemi–Markovprocesswhosestateatanytimeistheaverageofthoseselectedfromhotelswasapprox-phaseofthepresentinterarrivaltime.Hence,thisimatelytwiceaslargeasfromthoseselectedatsemi-Markovprocessgoesfromstate1to2to3...todepartureareconsistentwiththepossibilitythatnto1,andsoon.Alsothetimespentineachstatethetimespentinthecountrybyatouristisexpo-hasthesamedistribution.Thus,clearlythelimit-nentialwithameanapproximatelyequalto9.ingprobabilitiesofthissemi-MarkovchainisPi=1/n,i=1,...,n.TocomputelimP{Y(t)Xntheamountbywhichitisgreateris,Ni(m)bythelackofmemoryproperty,alsoexpo-j50.(a)∑Xi.nentialwithrateλ.Repeatingthisargumentj=1yieldstheresult.Ni(m)∞j∑Xi(b)E[N(Y)]=∑P{N(Y)≥n}j=1(b).n=1Ni(m)∞j∑∑Xi=∑P{X1+···+Xn≤Y}ij=1n=1(c)FollowsfromtheStronglawoflargenumbers∞Pnj=∑P{Xx}=P{∑Xi>x|T=0}(1−ρ)andsoi=1i=1Var(T)=Var(T)+Var(T∞)≈6.96×107T1,2+P{∑Xi>x|T>0}ρi=155.E[T(1)]=(.24)−2+(.4)−1=19.8611,TE[T(2)]=24.375,E[T]=21.875,=P{∑Xi>x|T>0}ρ12i=1E[T2,1]=17.3611.Thesolutionoftheequations
∞TF¯(y)=ρP{∑Xi>x|T>0,X1=y}dy19.861=E[M]+17.361P(2)0i=1µ24.375=E[M]+21.875P(1)ρ
xT=P{∑Xi>x|T>0,X1=y}F¯(y)dyµ0i=11=P(1)+P(2)
∞ρgivetheresults+F¯(y)dyµxP(2)≈.4425.E[M]≈12.18

ρxρ∞=h(x−y)F¯(y)dy+F¯(y)dyµ0µx(10)10956.(a)i!/(10)i10!∑i=0ρ
xρ
x=h(0)+h(x−y)F¯(y)dy−F¯(y)dy(b)Definearenewalprocessbysayingthataµ0µ0renewaloccursthefirsttimethatarunofwherethefinalequalityusedthat5consecutivedistinctvaluesoccur.Also,letarewardof1beearnedwhenevertheprevi-
∞ρh(0)=ρ=F¯(y)dyous5datavaluesaredistinct.Then,lettingµ0khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comChapter81.(a)E[numberofarrivals]µM=20µA=30.=E[E{numberofarrivals|serviceSetLM=averagenumberofcustomersinperiodisS}]queuewhenMaryworksand=E[λS]LA=averagenumberofcustomersinqueuewhenAliceworks.=λ/µ.10(b)P{0arrivals}ThenusingEquation(3.2),LM==1(20−10)khdaw.com=E[P{0arrivals|serviceperiodisS}]101LA==.=E[P{N(S)=0}](20−10)2−λS=E[e]SoCM=$3+$1/customer×LMcustomers
x−λs−µs=$3+$1=eµeds0=$4/hour.µ=.λ+µAlso,CA=$C+$1/customer×LAcustomers12.ThisproblemcanbemodeledbyanM/M/1=$C+$1×queueinwhichλ=6,µ=8.Theaveragecostrate12willbe=$C+/hour.2$10perhourpermachine×averagenumberofbrokenmachines.(b)Wecanrestatetheproblemthisway:IfCA=CM,solveforC.TheaveragenumberofbrokenmachinesisjustL,whichcanbecomputedfromEquation(3.2):14=C+⇒C=$3.50/hour,2L=λ/(µ−λ)i.e.,$3.50/houristhemosttheemployer6shouldbewillingtopayAlicetowork.At==3.2ahigherwagehisaveragecostislowerwithHence,theaveragecostrate=$30/hour.Maryworking.3.LetCM=Marysaveragecost/hourandCA=4.LetNbethenumberofothercustomersthatwereAlicesaveragecost/hour.inthesystemwhenthecustomerarrived,andletC=1/fW∗(x).ThenThen,CM=$3+$1×(Averagenumberofcus-QtomersinqueuewhenMaryworks),fN|W∗(n|x)=CfW∗|N(x|n)P{N=n}QQandCA=$C+$1×(Averagenumberofcus-n−1tomersinqueuewhenAliceworks).=Cµe−µx(µx)(λ/µ)n(1−λ/µ)(n−1)!Thearrivalstreamhasparameterλ=10,and(λx)n−1therearetwoserviceparametersoneforMary=KandoneforAlice:(n−1)!81khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com82AnswersandSolutionswhere7.TocomputeWfortheM/M/2,setupbalance1−µxequationsasK=µe(λ/µ)(1−λ/µ)fW∗(x)Qλp0=µp1(eachserverhasrateµ)Using∞∞n−1(λ+µ)p1=λp0+2µp2(λx)λx1=∑fN|W∗(n|x)=K∑=KeQ(n−1)!(λ+2µ)pn=λpn−1+2µpn+1n≥2.n=1n=1showsthat(λx)n−1ThesehavesolutionsPn/2n−1pwhere−λxn=ρ0fN|W∗(n|x)=e,n>0ρ=λ/µ.Q(n−1)!∞Thus,N−1isPoissonwithmeanλx.Theboundarycondition∑Pn=1impliesn=0Theprecedingalsoyieldsthatforx>0fW∗(x)=eλxµe−µx(λ/µ)(1−λ/µ)1−ρ/2(2−ρ)QP0==.1+ρ/2(2+ρ)=µλ(µ−λ)e−(µ−λ)xNowwehavePn,sowecancomputeL,andhenceHence,forkhdaw.comx>0WfromL=λW:
xP{W∗≤x}=P{W∗=0}+f∗(y)dy∞∞ρn−1QQWQL=∑np=ρp∑n0n02n=0n=0=1−µλ+µλ(1−e−(µ−λ)x)∞nρ=2p0∑n2∗n=05.LetIequal0ifWQ=0andletitequal1otherwise.Then,(2−ρ)(ρ/2)=2∗(2+ρ)(1−ρ/2)2E[W|I=0]=0Q4ρ∗−1=(2+ρ)(2−ρ)E[W|I=1]=(µ−λ)Q∗4µλVar(W|I=0)=0=.Q(2µ+λ)(2µ−λ)Var(W∗|I=1)=(µ−λ)−2QFromL=λWwehaveHence,4µE[Var(W∗|I]=(µ−λ)−2λ/µW=Wm/m/2=(2µ+λ)(2µ−λ).QVar(E[W∗|I])=(µ−λ)−2λ/µ(1−λ/µ)TheM/M/1queuewithservicerate2µhasQConsequently,bytheconditionalvariance1formula,Wm/m/1=2µ−λ∗λλVar(WQ)=2+2fromEquation(3.3).Weassumethatintheµ(µ−λ)µ(µ−λ)M/M/1queue,2µ>λsothatthequeueissta-6.Letthestatebetheidleserver.Thebalanceequa-4µble.Butthen4µ>2µ+λ,or>1,whichtionsare2µ+λimpliesWm/m/2>Wm/m/1.RateLeave=RateEnter,(µ+µ)P=µ1P+µ1P,Theintuitiveexplanationisthatifonefindsthe231µ1+µ23µ1+µ32queueemptyintheM/M/2case,itwoulddono(µ1+µ3)P2=µ2P1+µ2P3,goodtohavetwoservers.Onewouldbebetteroffµ2+µ3µ2+µ1withonefasterserver.µ1+µ2+µ3=1.1NowletWQ=WQ(M/M/1)ThesearetobesolvedandthequantityPirepre-sentstheproportionoftimethatserveriisidle.W2=WQ(M/M/2).Qkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions83Then,(c)TheaveragenumberofitemsinstockisL.W1=Wm/m/1−1/2µ(d)Thestatespaceisj,−n≤j≤k.ThestateQisj,j≥0,whentherearejitemsinstockW2=Wm/m/2−1/µ.andnowaitingcustomers;itisj,j<0,whenQtherearenoitemsinstockandnowaitingcus-So,tomers.ThebalanceequationsareW1=λ(3.3)µPk=λPk−1Q2µ(2µ−λ)(λ+µ)Pj=λPj−1+µPj+1,−nWQ2>(2µ+λ)∑jm−njλ<2µ.10.Thestateisthenumberofcustomersinthesystem,Sinceweassumeλ<2µforstabilityintheandthebalanceequationsarekhdaw.comM/M/1,W20pcl+∑pn=pcl+.µ(2µ+λ)(d)Therateatwhichanarrivalfindsnisequalton=2therateatwhichadepartureleavesbehindn.λ14.Thesystemhasgivenstateswhosetransitiondia-n−1(e)Pn=Pn−1gramisµnλn−1λn−240404040=Pn−2µnµn−101234λ0λ1···λn−160606060=P0µ1µ2···µnHence,thebalanceequationsare∞Usingthat∑n=0Pngives40p0=30p1P=170p1=40p0+30p20λλ···λ∞01n−11+∑n=1µµ···µn70p2=40p1+60p312Thenecessaryconditionforasolutionisthat100p3=40p2+60p4∞λλ···λ∑01n−1<∞60p4=40p3.n=1µ1µ2···µnSolutionofthesegivesp=4/3p,p=16/9p,1020∞(f)λa=∑n=0λnPnp3=32/27p0,p4=64/81p0.Thecondition∑pi=∞nPn1impliesp0is81/493,whichisabout1/6.L∑n=0(g)W==∞λa∑n=0λnPnClearly,p0=Proportionoftimebothserversarefree.13.(a)λp0=µp1(b)Theoriginalattendantworksaproportion(λ+µ)p1=λp0+2µp2(1−p0)ofthetime,andthesecondatten-(λ+2µ)pn=λpn−1+2µpn+1n≥2.dantworksaproportionp3+p4ofthetime.SoifthefirstattendantreceivesSz,thesecondThesearethesamebalanceequationsasfortheM/M/2queueandhavesolutionshouldreceiveonlyS(p3+p4)(1−p0)z.Computationshowsthisfactortobeabout2µ−λλnp0=,pn=n−1np0.0.414,so0.414x+x=100,orx=70.72,and2µ+λ2µthesecondattendantreceives$29.28.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions8515.Therearefourstates=0,1A,1B,2.Balance(a)W=L/λa=[1(P01+P10)+2P11]/[λ(1−equationsareP11)]=56/119.Anotherwayistoconditiononthestateas2P0=2P1Bseenbythearrival.LettingTdenotethetime4P1A=2P0+2P2spent,thisgives4P1B=4P1A+4P2W=E[T|00]128/338+E[T|01]100/3386P2=2P1B+E[T|10]110/3383P0+P1A+P1B+P2=1⇒P0=9,=(1/4)(228/338)+(1/2)(110/338)231P1A=,P1B=,P2=.=56/119.9992(b)P01+P11=275/513.(a)P0+P1B=.317.Thestatespacecanbetakentoconsistofstates(b)Byconditioninguponwhetherthestatewas0(0,0),(0,1),(1,0),(1,1),wheretheithcomponentor1Bwhenheenteredwegetthatthedesiredofthestatereferstothenumberofcustomersatprobabilityisgivenbykhdaw.comserveri,i=1,2.Thebalanceequationsare1124+=.22662P0,0=6P0,178P0,1=4P1,0+4P1,1(c)P1A+P1B+2P2=9.6P1,0=2P0,0+6P1,1(d)Again,conditiononthestatewhenheenterstoobtain10P1,1=2P0,1+2P1,0111112171=P0,0+P0,1+P1,0+P1,1+++=.242246212SolvingtheseequationsgivesP0,0=1/2,Thiscouldalsohavebeenobtainedfrom(a)P0,1=1/6,P1,0=1/4,P1,1=1/12.Land(c)bytheformulaW=.λa7(a)P1,1=1/12.9=7LP0,1+P1,0+2P1,17Thatis,W=.(b)W===212λa2(1−P1,1)2223P0,0+P0,18(c)=1−P1,11116.Letthestatesbe(0,0),(1,0),(0,1),and(1,1),wherestate(i,j)meansthatthereisicustomers18.(a)Thestatesare0,1,2,3wherethestateisiwithserver1andjwithserver2.Thebalancewhenthereareiinthesystem.equationsareasfollows.(b)ThebalanceequationsareλP00=µ1P10+µ2P01(λ+µ)P=λP+µPλP0=µP111000211(λ+µ)P=µP(λ+µ)P1=λP0+2µP2201111(µ+µ)P=λP+λP(λ+2µ)P2=λP1+2µP3121101102µP3=λP2P00+P01+P10+P11=1P0+P1+P2+P3=1Substitutingthevaluesλ=5,µ1=4,µ2=2andThesolutionoftheseequationsissolvingyieldsthesolutionP=128/513,P=110/513,P=100/513,P1=(λ/µ)P0,P2=(λ2/2µ2)P0,P3=(λ3/4µ3)P0001001P11=175/513P0=[1+λ/µ+λ2/(2µ2)+λ3/(4µ3)]−1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com86AnswersandSolutions(c)E[Time]=E[Timeinqueue]22.Thestateisthepair(i,j),i=0,1,0≤j≤nwhereisignifiesthenumberofcustomersinserviceand+E[timeinservice]jthenumberinorbit.Thebalanceequationsare=1/(2µ)+1/µ.(d)1−P.(λ+jθ)P0,j=µP1,j,j=0,...,N3(e)Conditioningonthestateasseenbythe(λ+µ)P1,j=λP0,j+(j+1)θP0,j+1,arrivalj=0,...,N−1W=[(1/µ)(P0+P1)+(2/µ)P2]/(1−P3).CouldalsouseW=L/λa.µP1,N=λP0,N19.(a)Saythatthestateis(n,1)wheneveritisa(c)1−P1,Ngoodperiodandthereareninthesystem,and(d)Theaveragenumberofcustomersinthesys-saythatitis(n,2)wheneveritisabadperiodtemisandthereareninthesystem,n=0,1(b)(λ1+α1)P0,1=µP1,1+α2P0,2L=∑(i+j)Pi,ji,j(λ2+α2)P0,2=µP1,2+α1P0,1khdaw.comHence,theaveragetimethatanenteringcus-(µ+α1)P1,1=λ1P0,1+α2P1,2tomerspendsinthesystemisW=L/λ(1−P1,N),andtheaveragetimethatanentering(µ+α2)P1,2=λ2P0,2+α1P1,1customerspendsinorbitisW−1/µ.P0,1+P0,2+P1,1+P1,2=123.(a)Thestatesaren,n≥0,andb.Statenmeans(c)P0,1+P0,2thereareninthesystemandstatebmeans(d)λ1P0,1+λ2P0,2thatabreakdownisinprogress.20.(a)Thestatesare0,(1,0),(0,1)and(1,1),(b)βPb=a(1−P0)where0meansthatthesystemisempty,(1,0)λP0=µP1+βPbthatthereisonecustomerwithserver1andnonewithserver2,andsoon.(λ+µ+a)Pn=λPn−1+µPn+1,n≥1(b)(λ1+λ2)P0=µ1P10+µ2P01∞(c)W=L/λn=∑nPa/[λ(1−Pb)].(λ1+λ2+µ1)P10=λ1P0+µ2P11n=1(d)Sincerateatwhichservicesarecompleted=(λ1+µ2)P01=λ2P0+µ1P11µ(1−P0−Pb)itfollowsthattheproportion(µ1+µ2)P11=λ1P01+(λ1+λ2)P10ofcustomersthatcompleteserviceisµ(1−P0−Pb)/λaP0+P10+P01+P11=1=µ(1−P0−Pb)/[λ(1−Pb)].(c)L=P01+P10+2P11Anequivalentanswerisobtainedbycondi-(d)W=L/λa=L/[λ1(1−P11)+λ2(P0+P10)]tioningonthestateasseenbyanarrival.Thisgivesthesolution21.(a)λ1P10∞Pn+1(b)λ2(P0+P10)∑n[µ/(µ+a)],(c)λP/[λP+λ(P+P)]n=01101102010(d)Thisisequaltothefractionofserver2scus-wheretheaboveusesthattheprobabilitytomersthataretype1multipliedbythepro-thatn+1servicesofpresentcustomersoccurportionoftimeserver2isbusy.(Thisistruebeforeabreakdownis[µ/(µ+a)]n+1.sincetheamountoftimeserver2spendswith(e)Pb.acustomerdoesnotdependonwhichtypeofcustomeritis.)By(c)theansweristhus24.Thestatesarenown,n≥0,andn,n≥1where(P01+P11)λ1P10/[λ1P10+λ2(P0+P10)].thestateisnwhenthereareninthesystemandnokhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions87breakdown,anditisnwhenthereareninthesys-(c)λ<µ.temandabreakdownisinprogress.Thebalanceequationsare27.(a)Thespecialcustomersarrivalrateisactθbecausewemusttakeintoaccounthisser-λP0=µP1vicetime.Infact,themeantimebetweenhis(λ+µ+α)Pn=λPn−1+µPn+1+βPn,n≥1arrivalswillbe1/θ+1/µ1.Hence,thearrivalrateis(1/θ+1/µ)−1.1(β+λ)P1=αP1(b)Clearlyweneedtokeeptrackwhetherthe(β+λ)Pn=αPn+λP(n−1),n≥2specialcustomerisinservice.Forn≥1,set∞∞Pn=Pr{ncustomersinsystemregularcus-∑Pn+∑Pn=1.tomerinservice},n=0n=1PS=Pr{ncustomersinsystem,specialcus-nIntermsofthesolutiontotheabove,tomerinservice},and∞P0=Pr{0customersinsystem}.L=∑n(Pn+Pn)Sn=1(λ+θ)P0=µP1+µ1P1andsoS(λ+θ+µ)Pn=λPn−1+µPn+1+µ1Pn+1W=L/λα=L/λ.khdaw.com(λ+µ)PnS=θPn−1+λPS,n−125.(a)λP0=µAPA+µBPBS.n≥1P0=P0(λ+µA)PA=aλP0+µBP2(c)Sinceserviceismemoryless,onceacustomer(λ+µB)PB=(1−a)λP0+µAP2resumesserviceitisasifhisservicehasstartedanew.Oncehebeginsaparticularser-(λ+µA+µB)Pn=λPn−1+(µA+µB)Pn+1vice,hewillcompleteitifandonlyifthen≥2whereP1=PA+PB.nextarrivalofthespecialcustomerisafterhis∞service.TheprobabilityofthisisPr{Service(b)L=PA+PB+∑nPnk+1−λE{eδD|systemleftempty}µkk−1∞(b)−1P+k−1−n+nP+Pnn.λµλδ(X+Y)λ0∑λµn∑µ=+1−E{e},n=1n−1µµ−δµkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com88AnswersandSolutionswhereXhasthedistributionofinterarrivaltimes,L49(ii)W==.Yhasthedistributionofservicetimes,andXand∑Tj180Yareindependent.32.LettingthestatebethenumberofcustomersatThenserver1,thebalanceequationsareE{eδ(X+Y)}=E{eδXeδY)}(µ2/2)P0=(µ1/2)P1δXδY)=E[e]E[e]byindependence(µ1/2+µ2/2)P1=(µ2/2)P0+(µ1/2)P2λµ=.(µ1/2)P2=(µ2/2)P1λ−δµ−δSo,P0+P1+P2=1δDλµλλµE{e}=µµ−δ+1−µλ−δµ−δSolvingyieldsthat=λ.P1=(1+µ1/µ2+µ2/µ1)−1,P0=µ1/µ2P1,(λ−δ)P2=µ2/µ1P1Bytheuniquenessofgeneratingfunctions,itfol-lowsthatkhdaw.comDhasanexponentialdistributionwithHence,lettingLibetheaveragenumberofcus-parameterλ.tomersatserveri,then29.(a)Letstate0meanthattheserverisfree;letstateL1=P1+2P2,L2=2−L11meanthatatype1customerishavingawash;letstate2meanthattheserveriscut-Theservicecompletionrateforserver1istinghair;andletstate3meanthatatype3isµ1(1−P0),andforserver2itisµ2(1−P2).gettingawash.(b)λP0=µ1P1+µ2P233.(a)UsetheGibbssamplertosimulateaMarkovchainwhosestationarydistributionisthatofµ1P1=λp1P0thequeuingnetworksystemwithm−1cus-tomers.Usethissimulatedchaintoestimateµ2P2=λp2P0+µ1P3Pi,m−1,thesteadystateprobabilitythatthereareicustomersatserverjforthissystem.µ1P3=λp3P0Since,bythearrivaltheorem,thedistributionP0+P1+P2+P3=1functionofthetimespentatserverjinthem−1mcustomersystemis∑i=0Pi,m−1Gi+1(x),(c)P2whereGk(x)istheprobabilitythatagamma(d)λP0(k,µ)randomvariableislessthanorequaltoDirectsubstitutionnowverifiestheequation.x,thisenablesustoestimatethedistributionfunction.31.Solvingforthetotalarrivalrateswehave(b)Thisquantityisequaltotheaveragenumberofcustomersatserverjdividedbym.λ1=5λ2=10+1λ1+1λ3λ2j22∑1jµj(µj−λj)λ3=15+2λ1+λ234.WQ=LQ/λα=∑jrjimplyingthatλ1=5,λ2=85/2,andλ3=60.35.LetSandUdenote,respectively,theservicetimeλi585/260andvalueofacustomer.ThenUisuniformon(i)L=∑=++(0,1)andµi−λi550−85/2100−6049=.E[S|U]=3+4U,Var(S|U)=5.6khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions89Hence,(b)TakingexpectationswegetE[S]=E{E[S|U]}=3+4E[U]=5EXn+1=EXn−1+EYn+Eδn.Var(S)=E[Var(S|U)]+Var(E[S|U])Lettingn→∞,EXn+1andEXncancel,andEY∞=EY1.Therefore,=5+16Var(U)=19/3.Eδ∞=1−EY1.Therefore,TocomputeEY1,conditiononthelengthof2serviceS;E[Y1|S=t]=λtbyPoissonarrivals.E[S]=19/3+25=94/3.ButE[λS]isjustλES.Hence,94λ/3Eδ∞=1−λES.(a)W=WQ+E[S]=+5.1−δλ(c)SquaringEquation(8.1)weget94λ/3(b)WQ+E[S|U=x]=+3+4x.(∗)X2=Xn2+1+Yn2+2(XnYn−Xn)−2Yn1−δλn+1+δn(2Yn+2Xn−1).36.Thedistributionsofthequeuesizeandbusyperiodarethesameforallthreedisciplines;thatofButtakingexpectations,thereareafewfactskhdaw.comthewaitingtimeisdifferent.However,themeanstonotice:areidentical.ThiscanbeseenbyusingW=EδnSn=0sinceδnSn≡0.L/λ,sinceListhesameforall.Thesmallestvari-anceinthewaitingtimeoccursunderfirst-come,YnandXnareindependentrandomvariablesfirst-servedandthelargestunderlast-come,first-becauseYn=numberofarrivalsduringthe(n+1)stservice.Hence,served.EXnYn=EXnEYn.37.(a)Theproportionofdeparturesleavingbehind0workForthesamereason,Ynandδnareindepen-dentrandomvariables,soEδnYn=EδnEYn.=proportionofdeparturesleavinganEY2=λES+λ2ES2bythesameconditioningemptysystemnargumentofpart(b).=proportionofarrivalsfindinganempty2≡δsystemFinallyalsonoteδnn.=proportionoftimethesystemisemptyTakingexpectationsof(*)gives(byPoissonarrivals)2222EXn+1=EXn+1+λE(S)+λE(S)=P0.+2EXn(λE(S)−1)(b)Theaverageamountofworkasseenbyadepartureisequaltotheaveragenumber−2λE(S)+2λE(S)Eδn−Eδn.itseesmultipliedbythemeanservicetime(sincenocustomersseenbyadeparturehasLettingn→∞cancelsEX2andEX2,andnn+1yetstartedservice).Hence,Eδn→Eδ∞=1−λE(S).ThisleavesAverageworkasseenbyadeparture220=λE(S)+2EX∞(λE(S)−1)+2λE(S)=averagenumberitsees×E[S][1−λE(S)],=averagenumberanarrivalsees×E[S]=LE[S]byPoissonarrivalswhichgivestheresultuponsolvingforEX∞.(d)IfcustomernspendstimeWninsystem,=λ(WQ+E[S])E[S]thenbyPoissonarrivalsE[Xn|Wn]=λWn.22Hence,EXn=λEWnandlettingn→∞yieldsλE[S]E[S]2=+λ(E[S]).EX∞=λW=L.Italsofollowssincetheaver-λ−λE[S]agenumberasseenbyadepartureisalwaysequaltotheaveragenumberasseenbyan38.(a)Yn=numberofarrivalsduringthe(n+1)starrival,whichinthiscaseequalsLbyPoissonservice.arrivals.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com90AnswersandSolutions39.(a)a0=P0duetoPoissonarrivals.AssumingSincethepriorityruledoesnotaffecttheamountthateachcustomerpays1perunittimewhileofworkinsystemcomparedtoFIFOandinservicethecostidentity(2.1)statesthatWQ=V,wecanuseEquation(6.5)forWQ.FIFOFIFOAveragenumberinservice=λE[S],NowW=αW1+(1−α)W2byaveragingoverQQQorbothclassesofcustomers.ItiseasytocheckthatWQthenbecomes1−P0=λE[S].λES2+λES2[α(1−ρ−ρ)+(1−α)](b)Sincea0istheproportionofarrivalsthathave112212servicedistributionG1and1−a0thepropor-WQ=2(1−ρ−ρ)(1−ρ),121tionhavingservicedistributionG2,theresultfollows.whichwewishtocompareto(c)WehaveλES2+λES2Q1122(1−ρ1)E[I]WFIFO=2(1−ρ−ρ)·(1−ρ).P0=121E[I]+E[B]QandE[I]=1/λandthus,ThenWQ(1−α)ρ1λP0=E[S].⇔λ1·λ2E(S2)1−λE[S]λ1+λ2Nowfrom(a)and(b)wehave>λ2·λ1ES1λ1+λ2E[S]=(1−λE[S])E[S1]+λE[S]E[S2],⇔E(S2)>E(S1).orE[S1]43.Problem42showsthatifµ1>µ2,thenserving1sE[S]=.1+λE[S1]+λE[S2]firstminimizesaveragewait.Butthesameargu-mentworksifc1µ1>c2µ2,i.e.,SubstitutionintoE[B]=E[S]/(1−λE[S])nowyieldstheresult.E(S1)E(S2)<.c1µ140.(a)(i)Alittlethoughtrevealsthattimetogofromnton−1isindependentofn.44.(a)Aslongastheserverisbusy,workdecreasesnE[S](ii)nE[B]=.by1perunittimeandjumpsbytheservice1−λE[S]ofanarrivaleventhoughthearrivalmaygo(b)(i)E[T|N]=A+NE[B].directlyintoservice.Sincethebumpedcus-(ii)E[T]=A+E[N]E[B]tomersremainingservicedoesnotchangebybeingbumped,thetotalworkinsys-λAE[S]A=A+=.temremainsthesameasfornonpreemptive,1−λE[S]1−λE[S]whichisthesameasFIFO.41.E[N]=2,E[N2]=9/2,E[S2]=2E2[S]=1/200(b)AsfarastypeIcustomersareconcerned,thetypeIIcustomersdonotexist.AtypeIcus-15tomersdelayonlydependsonothertypeI/4+4·2/40041W=202=customersinsystemwhenhearrives.There-1−8/2048011fore,WQ=V=amountoftypeIworkin41117system.WQ=−=.48020480Bypart(a),thisisthesameV1asforthenonpreemptivecase(6.6).Therefore,42.Fornotationalease,setα=λ1/(λ1+λ2)=pro-portionofcustomersthataretypeI.2λ1ES1W1=λE(S)W1+,ρ1=λ1E(S1),ρ2E(S2).Q11Q2khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions91orfirsttwomomentsofaservicetime.NowthetimeofaserviceisthetimeTuntilsomethinghappensλES2111(eitheraservicecompletionorabreakdown)plusWQ=.2(1−λ1E(S1)]anyadditionaltimeA.Thus,NotethatthisisthesameasforanM/G/1E[S]=E[T+A]queuewhichhasonlytypeIcustomers.(c)Thisdoesnotaccountforthefactthatsome=E[T]+E[A]typeIIworkinqueuemayresultfromcus-tomersthathavebeenbumpedfromservice,TocomputeE[A]weconditionuponwhethertheandsotheiraverageworkwouldnotbeE[S].happeningisaserviceorabreakdown.Thisgives(d)IfatypeIIarrivalfindsabumpedtypeIIµinqueue,thenatypeIisinservice.ButinE[A]=E[A|service]thenonpreemptivecase,theonlydifferenceisµ+αthatthetypeIIbumpedcustomerisservedα+E[A|breakdown]aheadofthetypeI,bothofwhomstillgoµ+αbeforethearrival.Sothetotalamountofworkαfoundfacingthearrivalisthesameinboth=E[A|breakdown]µ+αcases.Hence,khdaw.comα=(1/β+E[S]).2µ+αVQ(nonpreemptive)+E(extratime)2W5QSince,E[T]=1/(α+µ)weobtainthattotalworkfoundextratimeduebytypeIItobeingbumped1αE[S]=+(1/β+E[S])(e)AssoonasatypeIIisbumped,hewillnotα+µµ+αreturntoserviceuntilalltypeIsarrivingdur-oringthefirsttypeIsservicehavedeparted,allfurthertypeIswhoarrivedduringtheaddi-tionaltypeIserviceshavedeparted,andsoE[S]=1/µ+α/(µβ)on.Thatis,eachtimeatypeIIcustomeris2bumped,hewaitsbackinqueueforonetypeIWealsoneedE[S],whichisobtainedasfollows.busyperiod.BecausethetypeIcustomersdonotseethetypeIIsatall,theirbusyperiodisE[S2]=E[(T+A)2]justanM/G1/1busyperiodwithmean22=E[T]+2E[AT]+E[A]E(S1).1−λ1E(S1)=E[T2]+2E[A]E[T]+E[A2]SogiventhatacustomerisbumpedNtimes,wehaveTheindependenceofAandTfollowingbecausethetimeofthefirsthappeningNE(S1)E{extratime|N}=.isindependentofwhetherthehappen-1−λ1E(S1)ingwasaserviceorabreakdown.Now,22(f)SincearrivalsarePoisson,E[N|S2]=λ1S2,E[A]=E[A|breakdown]αµ+αandsoEN=λ1ES2.αα2=E[(downtime+S)](g)From(e)and(f),µ+αλ1E(S2)E(S1)αE(extratime)=.Combining=E[down2]+2E[down]E[S]+E[S2]1−λ1E(S1)µ+αthiswith(e)givestheresult.α221α2=+++E[S].45.Byregardinganybreakdownsthatoccurduringaµ+αβ2βµµβserviceasbeingpartofthatservice,weseethatthisisanM/G/1model.WeneedtocalculatetheHence,khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com92AnswersandSolutions22αTheequationchecksout.E[S]=+2(µ+β)2β(µ+α) α1α++47.Fork=1,Equation(8.1)givesµ+αµµβα221α2P=1=(λ)P=λ(ES)++++E[S].01µ+αβ2βµµβ1+λE(S)(λ)+E(S)1+λE(S)2E(S)NowsolveforE[S].Thedesiredansweris=.λ+E(S)2λE[S]WQ=2(1−λE[S])Intheabove,SαistheadditionalserviceneededOnecanthinkoftheprocessasanalteractingafterthebreakdownisover.Sαhasthesamedis-renewalprocess.SincearrivalsarePoisson,thetimetributionasS.Theabovealsousesthefactthatuntilthenextarrivalisstillexponentialwiththeexpectedsquareofanexponentialistwicetheparameterλ.squareofitsmean.AnotherwayofcalculatingthemomentsofSistoendofendofservicearrivalserviceusetherepresentationkhdaw.comNS=∑(Ti+Bi)+TN+1ASstatesi=1ASwhereNisthenumberofbreakdownswhileacus-tomerisinservice,Tiisthetimestartingwhenser-vicecommencesfortheithtimeuntilahappeningThebasicresultofalternatingrenewalprocessesisoccurs,andBisthelengthoftheithbreakdown.thatthelimitingprobabilitiesaregivenbyiWenowusethefactthat,givenN,alloftheran-domvariablesintherepresentationareindepen-E(S)Pr{beinginstateS}=anddentexponentialswiththeTihavingrateµ+αE(A)+E(S)andtheBihavingrateβ.ThisyieldsE[S|N]=(N+1)/(µ+α)+N/βPR{beinginstateA}=E(A)E(A)+E(S)Var(S|N)=(N+1)/(µ+α)2+N/β2.Therefore,since1+NisgeometricwithmeanTheseareexactlytheErlangprobabilitiesgiven(µ+α)/µ[andvarianceα(α+µ)/µ2]weobtainabovesinceEA=1/λ.NotethisusesPoissonarrivalsinanessentialway,viz.,toknowthedis-E[S]=1/µ+α/(µβ);tributionoftimeuntilthenextarrivalafteraser-and,usingtheconditionalvarianceformula,viceisstillexponentialwithparameterλ.Var(S)=[1/(µ+α)+1/β]2α(α+µ)/µ2+1/[µ(µ+α)]+α/µβ2).48.TheeasiestwaytocheckthatthePiarecorrectissimplytocheckthattheysatisfythebalanceequa-46.βistobethesolutionofEquation(7.3):tions:
∞β=e−µt(1−β)dG(t)λp0=µp10IfG(t)=1−e−λt(λ<µ)andβ=λ/µ(λ+µ)p1=λp0+2µp2
∞
∞(λ+2µ)p=λp+3µpe−µt(1−λ/µ)dG(t)=e−µt(1−λ/µ)λe−λtdt21300
∞(λ+iµ)pi=λpi−1+(i+1)µpi+10=r2222Withqi=1−Pi,thestructurefunctionisNowr(p)=p1p2p3+p3p4p5−p1p2p3p4p5,r(p)=P{eitherof1,2,or3works}andsoP{eitherof4or5works}1111Psystemlife<=1−−+28832=(1−q1q2q3)(1−q4q5).25=.3213.Takingexpectationsoftheidentity19.X(i)isthesystemlifeofann−i+1ofnsys-φ(X)=Xiφ(1i,X)+(1−Xi)φ(0i,X),temeachhavingthelifedistributionF.Hence,theresultfollowsfromExample5e.notingtheindependenceofXiandφ(1i,X)andofφ(0i,X).20.Thedensitiesarerelatedasfollows.14.r(p)=p3P{max(X1,X2)=1=max(X4,X5)}a−1khdaw.comg(t)=a[F¯(t)]f(t).+(1−p3)P{max(X1X4,X2X5)=1}Therefore,=p3(p1+p2−p1p2)(p4+p5−p4p5)λC(t)=a[F¯(t)]a−1f(t)/[F¯(t)]a+(1−p3)(p1p4+p2p5−p1p4p2p5).=af(t)/F¯(t)7173169=aλF(t).15.(a)≤r≤1−=.3228512Theexactvalueisr(1/2)=7/32,which21.(a)(i),(ii),(iv)−(iv)becauseitistwo-of-three.agreeswiththeminimalcutlowerboundsincetheminimalcutsets{1},{5},{2,3,4}(b)(i)becauseitisseries,(ii)becauseitcanbedonotoverlap.thoughtofasbeingaseriesarrangementof1andtheparallelsystemof2and3,whichas17.E[N2]=E[N2|N>0]P{N>0}F2=F3isIFR.2(c)(i)becauseitisseries.≥(E[N|N>0])P{N>0},sinceE[X2]≥(E[X])2.22.(a)Ft(a)=P{X>t+a|X>t}Thus,P{X>t+a}F(t+a)==.P{X>t}F(t)E[N2]P{N>0}≥(E[N|N>0]P{N>0})2(b)Supposeλ(t)isincreasing.Recallthat2=(E[N]).−tλ(s)dsF(t)=e0.LetNdenotethenumberofminimalpathsetshavingallofitscomponentsfunctioning.ThenHence,F(t+a)t+ar(p)=P{N>0}.=e−0λ(s)ds,whichdecreasesintF(t)Similarly,ifwedefineNasthenumberofminimalsinceλ(t)isincreasing.Togotheotherway,cutsetshavingallofitscomponentsfailed,thensupposeF(t+a)/F(t)decreasesint.Nowfor1−r(p)=P{N>0}.asmallInbothcaseswecancomputeexpressionsforE[N]F(t+a)/F(t)=e−aλ(t).andE[N2]bywritingNasthesumofindicator(i.e.,Bernoulli)randomvariables.ThenwecanuseHence,e−aλ(t)mustdecreaseintandthusλ(t)theinequalitytoderiveboundsonr(p).increases.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions97n23.(a)F(t)=∏Fi(t)sinceIFRA.i=1Hence,nd∑Fj(t)∏Fj(t)1−F(x)≥(1−p)x/ξ=e−θx.F(t)dtj=1i=jλF(t)==nF(t)26.Eitherusethehintinthetextorthefollowing,∏Fi(t)whichdoesnotassumeaknowledgeofconcavei=1functions.n∑Fj(t)Toshow:h(y)≡λαxα+(1−λα)yαj=1=F(t)−(λx+(1−λ)y)α≥0,jn0≤y≤x,=∑λj(t).where0≤λ≤1,0≤α≤1.j=1(b)Ft(a)=P{additionallifeoft-year-old>a}Note:h(0)=0,assumey>0,andletg(y)=h(y)/yanF(t+a)αα∏iλxαλx1g(y)=+1−λ−+1−λ.=,yykhdaw.comFi(t)Letz=x/y.Nowg(y)≥0∀01.Thisfollowssince24.Itiseasytoshowthatλ(t)increasingimpliesthatα−1α−1tf(z)=αλ(λz)−αλ(λz+1−λ)0λ(s)ds/talsoincreases.Forinstance,ifwedif-t2f(z)≥0⇔(λz)α−1≥(λz+1−λ)α−1ferentiate,wegettλ(t)−0λ(s)ds/t,whichisttnonnegativesince0λ(s)ds≤0λ(t)dt=tλ(t).⇔(λz)1−α≤(λz+1−λ)1−αAcounterexampleis⇔λz≤λz+1−λ⇔λ≤1.
(t)27.Ifp>p0,thenp=p0αforsomea∈(0,1).Hence,r(p)=r(pα)≥[r(p)]α=pα=p.000Ifpcαandlettingitbe0reducestootherwise.Then,n−1nn−1=−nnn∞i−2i−1i−1E∑Ii=∑E[Ii]=∑P{Xi>c}i=1i=1i=1whichistrue.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comChapter101.X(s)+X(t)=2X(s)+X(t)−X(s).Part(b)canbeprovenbyusing
∞Now2X(s)isnormalwithmean0andvariance4sE[Ta]=P{Ta>t}dtandX(t)−X(s)isnormalwithmean0andvari-0ancet−s.AsX(s)andX(t)−X(s)areindepen-inconjunctionwith(2.3).dent,itfollowsthatX(s)+X(t)isnormalwithmean0andvariance4s+t−s=3s+t.5.P{T1x}.Conditionon=E[X(t1)E[X2(t2)|X(t1)]](∗)X(t1)toobtain2
∞=E[X(t1){(t2−t1)+X(t1)}]P(M)=P(M|X(t)=y)√1e−y2/2t1dy1=E[X3(t1)]+(t2−t1)E[X(t1)]−∞2πt1Now,usethat=0wheretheequality(∗)followssincegivenX(t1),P(M|X(t1)=y)=1y≥xX(t2)isnormalwithmeanX(t1)andvarianceand,foryx−y}0x−y}t→∞
∞=√2e−y2/2dyby(10.6)8.(a)LetX(t)denotethepositionattimet.Then2r0√[t/∆t]X(t)=∆t∑Xi=2P{N(0,1)>0}=1.i=199khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com100AnswersandSolutionswhereNow√th√1/∆t1/h+1ifistepisup1−µ∆t1−µhXi=lim√=lim−1ifithstepisdown.∆t→01+µ∆th→01+µhAsµn1−=limnE[X1]=p−1(1−p)n→∞1+µn=2p−1byn=1/h√=µ∆te−µ==e−2µandeµwherethelastequalityfollowsfromVar(X)=EX2−(E[X])2iiixnlim1+=ex.=1−µ2∆tsinceX2=1n→∞niHencethelimitingvalueof(∗)as∆t→0isweobtainthat√√1−e−2µBkhdaw.comt.E[X(t)]=∆t∆tµ∆t1−e−2µ(A+B)→µtas∆t→011.LetX(t)denotethevalueoftheprocessattimet=nh.LetX=1iftheithchangeresultsintheiVar(X(t))=∆tt(1−µ2∆t)statevaluebecominglarger,andletXi=0other-√√∆tσh−σhwise.Then,withu=e,d=e→tas∆t→0.∑nXin−∑nXiX(t)=X(0)ui=1di=1∑nX(b)Bythegambler’sruinproblemtheprobabilitynui=1i=X(0)dofgoingupAbeforegoingdownBisdTherefore,1−(q/p)BXn(t)1−(q/p)A+BlogX(0)=nlog(d)+∑Xilog(u/d)i=1wheneachstepiseitherup1ordown1t√√t/h=−σh+2σh∑Xiwithprobabilitiespandq=1−p.(Thisishi=1theprobabilitythatagamblerstartingwithBythecentrallimittheorem,theprecedingBwillreachhisgoalofA+Bbeforegoing1√becomesanormalrandomvariableash→0.broke.)Now,whenp=(1+µ∆t),q=Moreover,becausetheXiareindependent,itis21√easytoseethattheprocesshasindependentincre-1−p=(1−µ∆t)andsoq/p=√2ments.Also,1−µ√∆t.Hence,inthiscasetheprobabilityX(t)Elog1+µ∆t√X(0)ofgoingup√A/∆t√beforegoingdownt√√t1µ√B/∆t(wedivideby∆tsinceeachstepis=−σh+2σh(1+h)hh2σnowofthissize)is=µt√√B/∆tand1−µ∆t1−√X(t)t1+µ∆tVarlog=4σ2hp(1−p)(∗)√√.X(0)h(A+B/∆t)→σ2t1−µ∆t1−√1+µ∆twheretheprecedingusedthatp→1/2ash→0.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions10112.Ifwepurchasexunitsofthestockanyyofthe14.Purchasingthestockwillbeafairbetunderoptionthenthevalueofourholdingsattime1isprobabilities(p1,p2,1−p1−p2)on(50,100,200),thesetofpossiblepricesattime1,if150x+25yifpriceis150value=.25xifpriceis25100=50p1+100p2+200(1−p1−p2),Soiforequivalently,if150x+25y=25x,ory=−5x3p1+2p2=2.thenthevalueofourholdingsis25xnomat-(a)Theoptionbetisalsofairiftheprobabilitiesterwhatthepriceisattime1.Sincethecostofalsosatisfypurchasingxunitsofthestockand−5xunitsofoptionsis50x−5xcitfollowsthatourprofitfromc=80(1−p1−p2).suchapurchaseisSolvingthisandtheequation3p1+2p2=225x−50x+5xc=x(5c−25).forp1andp2givesthesolution(a)Ifcp1=c/40,p2=(80−3c)/80,=5thenthereisnosurewin.khdaw.com(b)Sell|x|unitsofthestockandbuy−5|x|units1−p1−p2=c/80.ofoptionswillrealizeaprofitof5|x|nomatterHence,noarbitrageispossibleaslongasthesewhatthepriceofthestockisattime1.(Thatpiallliebetween0and1.However,thiswillis,buyxunitsofthestockand−5xunitsofbethecaseifandonlyiftheoptionsforx<0.)80≥3c.(c)Buyingxunitsofthestockand−5xunitsof(b)Inthiscase,theoptionbetisalsofairifoptionswillrealizeapositiveprofitof25xwhenx>0.c=20p2+120(1−p1−p2).(d)Anyprobabilityvector(p,1−p)on(150,25),Solvinginconjunctionwiththeequationthepossiblepricesattime1,underwhichbuy-ingthestockisafairbetsatisfiesthefollowing3p1+2p2=2givesthesolution50=p(150)+(1−p)(25),p1=(c−20)/30,p2=(40−c)/20,or1−p1−p2=(c−20)/60.Thesewillallbebetween0and1ifandonlyifp=1/5.20≤c≤40.Thatis,(1/5,4/5)istheonlyprobabilityvec-torthatmakesbuyingthestockafairbet.15.TheparametersofthisproblemareThus,inorderfortheretobenoarbitragepos-sibility,thepriceofanoptionmustbeafairσ=.05,σ=1,xo=100,t=10.betunderthisprobabilityvector.Thismeansthatthecostcmustsatisfy(a)IfK=100thenfromEquation(4.4)c=25(1/5)=5.√b=[.5−5−log(100/100)]/1013.Iftheoutcomeisithenourtotalwinningsare√=−4.510=−1.423oi(1+oi)−1−∑(1+oj)−1andj=ixioi−∑xj=1−∑(1+o)−1√−.5j=ikc=100φ(10−1.423)−100eφ(−1.423)k(1+oi)(1+oi)−1−∑(1+oj)−1=100φ(1.739)−100e−.5[1−φ(1.423)]j=1−∑(1+ok)−1=91.2.k=1.Theotherpartsfollowsimilarly.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com102AnswersandSolutions16.Takingexpectationsofthedefiningequationofa19.SinceknowingthevalueofY(t)isequivalenttoMartingaleyieldsthatknowingB(t)wehavethatE[Y(s)]=E[E[Y(t)/Y(u),0≤u≤s]]=E[Y(t)].E[Y(t)|Y(u),0≤u≤s]Thatis,E[Y(t)]isconstantandsoisequalto=e−c2t/2E[ecB(t)|B(u),0≤u≤s]E[Y(0)].=e−c2t/2E[ecB(t)|B(s)]17.E[B(t)|B(u),0≤u≤s]=E[B(s)+B(t)−B(s)|B(u),0≤u≤s]Now,givenB(s),theconditionaldistributionofB(t)isnormalwithmeanB(s)andvariancet−s.=E[B(s)|B(u),0≤u≤s]Usingtheformulaforthemomentgeneratingfunctionofanormalrandomvariableweseethat+E[B(t)−B(s)|B(u),0≤u≤s]e−c2t/2E[ecB(t)|B(s)]=B(s)+E[B(t)−B(s)]byindependent=e−c2t/2ecB(s)+(t−s)c2/2increments=e−c2s/2ecB(s)khdaw.com=B(s).=Y(s).18.E[B2(t)|B(u),0≤u≤s]=E[B2(t)|B(s)],Thus,{Y(t)}isaMartingale.wheretheabovefollowsbyusingindependentincrementsaswasdoneinProblem17.SincetheE[Y(t)]=E[Y(0)]=1.conditionaldistributionofB(t)givenB(s)isnor-malwithmeanB(s)andvariancet−sitfollows20.BytheMartingalestoppingtheoremthatE[B(T)]=E[B(0)]=0.E[B2(t)|B(s)]=B2(s)+t−s.However,B(T)=2−4TandsoHence,2−4E[T]=0E[B2(t)−t|B(u),0≤u≤s]=B2(s)−s.or,E[T]=1/2.Therefore,theconditionalexpectedvalueofB2(t)−t,givenallthevaluesofB(u),0≤u≤s,21.BytheMartingalestoppingtheoremdependsonlyonthevalueofB2(s).FromthisitE[B(T)]=E[B(0)]=0.intuitivelyfollowsthattheconditionalexpectationgiventhesquaresofthevaluesuptotimesisalsoBut,B(T)=(x−µT)/σandsoB2(s)−s.Aformalargumentisobtainedbycon-E[(x−µT)/σ]=0,ditioningonthevaluesB(u),0≤u≤sandusingtheabove.ThisgivesorE[B2(t)−t|B2(u),0≤u≤s]E[T]=x/µ.=EE[B2(t)−t|B(u),0≤u≤s]|B2(u),22.(a)ItfollowsfromtheresultsofProblem19andtheMartingalestoppingtheoremthat0≤u≤s]2E[exp{cB(T)−cT/2}]=E[B2(s)−s|B2(u),0≤u≤s]=E[exp{cB(0)}]=1.2SinceB(T)=[X(T)−µT]/σpart(a)follows.=B(s)−s(b)Thisfollowsfrompart(a)sincewhichprovesthat{B2(t)−t,t≥0}isaMartin-gale.Bylettingt=0,weseethat−2µ[X(T)−µT]/σ2−(2µ/σ)2T/2E[B2(t)−t]=E[B2(0)]=0.=−2µX(T)/σ2.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions103(c)SinceTisthefirsttimetheprocesshitsAor25.Themeansequal0.−Bitfollowsthat

111VartdX(t)=t2dt=A,withprobabilitypX(T)=.003−B,withprobability1−p

111Vart2dX(t)=t4dt=.Hence,weseethat0051=E[e−2µX(T)/σ2]=pe−2µA/σ2+(1−p)e2µB/σ226.(a)Normalwithmeanandvariancegivenby:andsoE[Y(t)]=tE[X(1/t)]=01−e2µB/σ2p=−2µA/σ22µB/σ2.Var(Y(t))=t2Var[X(1/t)]=t2/t=t.e−e(b)Cov(Y(s),Y(t))=Cov(sX(1/s),tX(1/t))23.BytheMartingalestoppingtheoremwehavethat=stCov(X(1/s),X(1/t))E[B(T)]=E[B(0)]=0.Since,B(T)=[X(T)−µT]/σthisgivesthe1=stwhens≤tkhdaw.comequalitytE[X(T)−µT]=0=swhens≤t.(c)Clearly{Y(t)}isGaussian.AsithasthesameormeanandcovariancefunctionastheBrown-E[X(T)]=µE[T].ianmotionprocess(whichisalsoGaussian)itfollowsthatitisalsoBrownianmotion.Now212E[X(T)]=pA−(1−p)B27.E[X(at)/a]=E[X(at)]=0.awhere,frompart(c)ofProblem22,Forss−Cov(N(t),N(t+s+1)−N(t+s))Hence,Cov(Y(t),Y(t+s))=Cov(N(t+1),N(t+s+1)−N(t+s))(∗)
s
∞1wheretheequality(∗)followssinceN(t)isinde-=ye−yλdy+y(y−s)λe−λydy−.pendentofkhdaw.comN(t+s+1)−N(t+s).Now,fors≤t,0sλ2Cov(N(s),N(t))=Cov(N(s),N(s)+N(t)−N(s))32.(a)Var(X(t+s)−X(t))=Cov(N(s),N(s))=Cov(X(t+s)−X(t),X(t+s)−X(t))=λs.=R(0)−R(s)−R(s)+R(0)Hence,from(∗)weobtainthat,whens<1,=2R(0)−2R(s).Cov(X(t),X(t+s))=Cov(N(t+1),N(t+s+1))(b)Cov(Y(t),Y(t+s))−Cov(N(t+1),N(t+s))=Cov(X(t+1)−X(t),X(t+s+1)=λ(t+1)−λ(t+s)−X(t+s))=Rx(s)−Rx(s−1)−Rx(s+1)+Rx(s)=λ(1−s)=2Rx(s)−Rx(s−1)−Rx(s+1),s≥1.Whens≥1,N(t+1)−N(t)andN(t+s+1)−N(t+s)are,bytheindependentincrementsprop-erty,independentandsotheircovarianceis0.33.Cov(X(t),X(t+s))31.(a)Startingatanytimetthecontinuationofthe=Cov(Y1coswt+Y2sinwt,PoissonprocessremainsaPoissonprocessY1cosw(t+s)+Y2sinw(t+s))withrateλ.(b)E[Y(t)Y(t+s)]=coswtcosw(t+s)+sinwtsinw(t+s)
∞=cos(w(t+s)−wt)=E[Y(t)Y(t+s)|Y(t)=y]λe−λydy0=cosws.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comChapter11i−1j−11.(a)Letubearandomnumber.If∑Pj1/3Xj=1ifUf(1)=0,weorseethatc=30/16,andsothealgorithmis−logU≥(n−1)λY/n+1−nStep1:GeneraterandomnumbersU1andU2.−(n−1)log(λY/n).Step2:IfU≤16(U2−2U3+U4),setX=U.Now,Y1=−logUisexponentialwithrate1,21111OtherwisereturntoStep1.andY2=λY/nisalsoexponentialwithrate1.Hence,thealgorithmcanbewrittenasgivenλe−λx(λx)n−1inpart(c).8.(a)Withf(x)=(n−1)!(d)Uponacceptance,theamountbywhichY1λe−λx/nexceeds(n−1){Y2−log(Y2)−1}isexpo-andg(x)=,nentialwithrate1.nn(λx)n−1e−λx(1−1/n)10.Wheneveriisthechosenvaluethatsatisfiesf(x)/g(x)=.(i)(n−1)!Lemma4.1nametheresultantQasQ.khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.comAnswersandSolutions10712.Let(b)P{X=n}1ifXi=jforsomei=P{U1>λ(1),U2>λ(2),...,Un−1Ij=0otherwise>λ(n−1),Un≤λ(n)}then=(1−λ(1))(1−λ(2))···nD=∑Ij(1−λ(n−1))λ(n).j=1(c)Sinceλ(n)≡pitsetsandsoX=min{n:U≤p}.nn−1kE[D]==1nE[I]=1−Thatisifeachtrialisasuccesswithprobabil-∑j∑njj=1itypthenitstopsatthefirstsuccess.kn−1(d)GiventhatX≥n,then=n1−nλ(n)P{X=n|X>n}=P=λ(n)kk(k−1)p≈n1−1+−.n2n2khdaw.com15.Use2µ=X.13.P{X=i}=P{Y=i|U≤PY/CQY}th16.(b)LetIjdenotetheindexofthejsmallestXi.P{Y=i,U≤PY/CQY}=17.(a)GeneratetheX(i)sequentiallyusingthatKgivenX(1),...,X(i−1)theconditionaldistri-QiP{U≤PY/CQY|Y=i}butionofX(i)willhavefailureratefunction=Kλi(t)givenbyQiPi/CQi0tX(i−1)CK(b)ThisfollowssinceFisanincreasingfunctionwhereK=P{U≤PY/CQY}.SincetheaboveisathedensityofU(i)isprobabilitymassfunctionitfollowsthatKC=1.f(i)(t)=n!(F(t))i−1(i−1)!(n−i)14.(a)Byinductionweshowthat(∗)P{X>k}=(1−λ(1))···(1−λ(k)).×(F(t))n−if(t)Theaboveisobviousfork=1andsoassume=n!ti−1(1−t)n−i,ittrue.Now(i−1)!(n−i)P{X>k+1}0k+1|X>k}P{X>k}whichshowsthatU(i)isbeta.(c)InterpretYastheithinterarrivaltimeofi=(1−λ(k+1))P{X>k}aPoissonprocess.NowgivenY1+···+Y=t,thetimeofthe(n+1)stevent,itwhichproves(*).Nown+1followsthatthefirstneventtimesaredistri-P{X=n}butedastheorderedvaluesofnuniform(0,t)randomvariables.Hence,=P{X=n|X>n−1}P{X>n−1}Y1+···+Yi,i=1,...,n=λ(n)P{X>n−1}Y1+···+Yn+iandtheresultfollowsfrom(*).willhavethesamedistributionasU(1),...,U(n).khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 课后答案网www.khdaw.com108AnswersandSolutions(d)fU(1),...|U(n)(y1,...,yn−1|yn)Nowsupposetheresultistrueforn−1.Thennn−1=f(y1,...,yn)Var∑λiXi=Var∑λiXi+Var(λnXn)fU(yn)i=1i=1(n)n−1λ=n!=(1−λn)2Var∑iXin−11−λnnyi=1(n−1)!+λn2VarXn.=n−1,0