Final_Fall12_sol_v2.pdf

《Final_Fall12_sol_v2.pdf》由会员上传分享，免费在线阅读，更多相关内容在学术论文-天天文库。

1、MassachusettsInstituteofTechnologyDepartmentofElectricalEngineering&ComputerScience6.041/6.431:ProbabilisticSystemsAnalysis(Fall2012)Fall2012FinalExam:SolutionsProblem1:(24points)LetXnbethestateattimenoftheMarkovchainshowninthediagram.1113213123113211344561223211343(a)(4points)Iden

2、tifythetransientstatesandtherecurrentclasses.Solution:States2and3aretransient.State1formsitsownrecurrentclass.States4,5,and6formanotherrecurrentclass.(b)(4points)SupposethatX0=2.Findlimn!1P(Xn=3).Solution:Sincestate3istransient,thesteady-stateprobabilitythatthechainisinstate3is0.(c

3、)(4points)SupposethatX0=4.DoesXnconvergeinprobability?Solution:Thef4;5;6grecurrentclassisaperiodic(asevidentthroughtheself-transitions),andthereforesteady-stateprobabilitiesexistandareallnonzero(afactthatcanbeconrmedinpart(d)below)forthese3states.Sincethechainbeginswithinthisrecur

4、rentclass,itwillnotsettleonanysinglestate,andhenceXnwillnotconvergeinprobability.(d)(6points)SupposethatX0=3.Findlimn!1P(Xn=4).Solution:Startingatstate3,thechainwilleventuallytransitiontostate5andtherebyenterthef4;5;6grecurrentclass.Hence,1=2=3=0.Tond4;5;6,wesolvethefollowin

5、gsetofsteady-stateprobabilityequations:214=4+5;321125=4+5+6;343116=5+6;431=4+5+6:1Page1of7MassachusettsInstituteofTechnologyDepartmentofElectricalEngineering&ComputerScience6.041/6.431:ProbabilisticSystemsAnalysis(Fall2012)Thesolutionis12834=;5=;6=:232323Hence,limP(

6、X=4)==12:n!1n423(e)(6points)SupposethatX0=4andthatnislargeenoughsothattheMarkovchaincanbeassumedtobeinsteady-state.Giventhatthelasttransitionwasanupward"transition(thatis,giventhatXn+1=Xn+1),ndtheconditionalprobabilitythatXn=4.(Toavoiddoublejeopardy,incaseyoucannotsolvefortheste

7、ady-stateprobabilitiesinpart(d),orifyouareunsureaboutyouranswer,youcanexpressyouranswertopart(e)intermsofthesymbols1;:::;6.)Solution:ThedesiredprobabilityisP(Xn=4;Xn+1=5)P(Xn=4jXn+1=Xn+1)=:P(Xn+1=Xn+1)Sinceweareassumingthechainisinsteady-state,wecanusetheapproximationP(Xn=k)=kto

8、obtain:1214P(Xn=4;Xn+1=5)=