Probability and Random Processes for Electrical and Computer Engineers

《Probability and Random Processes for Electrical and Computer Engineers》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、SolutionsManualforProbabilityandRandomProcessesforElectricalandComputerEngineersJohnA.GubnerUniversityofWisconsin–MadisonFileGeneratedJuly13,2007CHAPTER1ProblemSolutions1.Ω={1,2,3,4,5,6}.2.Ω={0,1,2,...,24,25}.3.Ω=[0,∞).RTT>10msisgivenbytheevent(10,∞).4.(a)Ω={(x,y)∈IR2:x2+y2≤100}.(b){(x,y)∈IR

2、2:4≤x2+y2≤25}.5.(a)[2,3]c=(−∞,2)∪(3,∞).(b)(1,3)∪(2,4)=(1,4).(c)(1,3)∩[2,4)=[2,3).(d)(3,6](5,7)=(3,5].6.Sketches:yyy11−1xxx−1BBB01−1yyy3xxx3C1H3J312Chapter1ProblemSolutionsyy33xx33UH3J3=M3H3UJ3=N3yy423xx234UUM2N3=M2M4N37.(a)[1,4]∩[0,2]∪[3,5]=[1,4]∩[0,2]∪[1,4]∩[3,5]=[1,2]∪[3,4].(b)c[0

3、,1]∪[2,3]=[0,1]c∩[2,3]chihi=(−∞,0)∪(1,∞)∩(−∞,2)∪(3,∞)hi=(−∞,0)∩(−∞,2)∪(3,∞)hi∪(1,∞)∩(−∞,2)∪(3,∞)=(−∞,0)∪(1,2)∪(3,∞).∞(c)(−1,1)={0}.nnn=1∞(d)[0,3+1)=[0,3].2nn=1[∞(e)[5,7−1]=[5,7).3nn=1[∞(f)[0,n]=[0,∞).n=1Chapter1ProblemSolutions38.WefirstletC⊂AandshowthatforallB,(A∩B)∪C=A∩(B∪C).WriteA∩(B

4、∪C)=(A∩B)∪(A∩C),bythedistributivelaw,=(A∩B)∪C,sinceC⊂A⇒A∩C=C.Forthesecondpartoftheproblem,suppose(A∩B)∪C=A∩(B∪C).WemustshowthatC⊂A.Letω∈C.Thenω∈(A∩B)∪C.Butthenω∈A∩(B∪C),whichimpliesω∈A.9.LetI:={ω∈Ω:ω∈A⇒ω∈B}.WemustshowthatA∩I=A∩B.⊂:Letω∈A∩I.Thenω∈Aandω∈I.Therefore,ω∈B,andthenω∈A∩B.⊃:Letω∈A∩B.

5、Thenω∈Aandω∈B.Wemustshowthatω∈Itoo.Inotherwords,wemustshowthatω∈A⇒ω∈B.Butwealreadyhaveω∈B.10.Thefunctionf:(−∞,∞)→[0,∞)withf(x)=x3isnotwelldefinedbecausenotallvaluesoff(x)lieintheclaimedco-domain[0,∞).11.(a)ThefunctionwillbeinvertibleifY=[−1,1].(b){x:f(x)≤1/2}=[−π/2,π/6].(c){x:f(x)<0}=[−π/2,0)

6、.12.(a)Sincefisnotone-to-one,nochoiceofco-domainYcanmakef:[0,π]→Yinvertible.(b){x:f(x)≤1/2}=[0,π/6]∪[5π/6,π].(c){x:f(x)<0}=∅.13.ForB⊂IR,X,0∈Band1∈B,−1A,1∈Bbut0∈/B,f(B)=cA,0∈Bbut1∈/B,∅,0∈/Band1∈/B.14.Letf:X→Ybeafunctionsuchthatftakesonlyndistinctvalues,sayy1,...,yn.LetB⊂Ybesuchthatf−1(

7、B)isnonempty.Bydefinition,eachx∈f−1(B)hasthepropertythatf(x)∈B.Butf(x)mustbeoneofthevaluesy1,...,yn,sayyi.Nowf(x)=yiifandonlyifx∈Ai:=f−1({yi}).Hence,[f−1(B)=A.ii:yi∈B15.(a)f(x)∈Bc⇔f(x)∈/B⇔x∈/f−1(B)⇔x∈f−1(B)c.[∞(b)f(x)∈Bnifandonlyiff(x)∈Bnforsomen;i.