chapter 6 functions of random variables:6章对随机变量的函数

chapter 6 functions of random variables:6章对随机变量的函数

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时间:2019-02-21

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1、Chapter6:FunctionsofRandomVariables141Instructor’sSolutionsManualChapter6:FunctionsofRandomVariables6.1ThedistributionfunctionofYis,0≤y≤1.a..Thus,.b..Thus,.c.Thus,.d.e.6.2ThedistributionfunctionofYis,–1≤y≤1.a..Thus,.b..Thus,.c..Thus,.6.3ThedistributionfunctionforY

2、is.a..So,,and.b.E(U)=5.583.c.E(10Y–4)=10(23/24)–4=5.583.6.4ThedistributionfunctionofYis,0≤y.a..Thus,.b.E(U)=13.Chapter6:FunctionsofRandomVariables141Instructor’sSolutionsManual6.1ThedistributionfunctionofYis,1≤y≤5..Differentiating,.6.2RefertoEx.5.10ad5.78.Define.a

3、.Foru≤0,.For0≤u<1,.For1≤u≤2,.Thus,.b.E(U)=1.6.3LetFZ(z)andfZ(z)denotethestandardnormaldistributionanddensityfunctionsrespectively.a.ThedensityfunctionforUisthen.Evaluating,wefind.b.Uhasagammadistributionwithα=1/2andβ=2(recallthatΓ(1/2)=).c.Thisisthechi–squaredistr

4、ibutionwithonedegreeoffreedom.6.4LetFY(y)andfY(y)denotethebetadistributionanddensityfunctionsrespectively.a.ThedensityfunctionforUisthen.b.E(U)=1–E(Y)=.c.V(U)=V(Y).6.5NotethatthisisthesamedensityfromEx.5.12:,0≤y1≤1,0≤y2≤1,0≤y1+y2≤1.a..Thus,,0≤u≤1.b.E(U)=2/3.c.(fou

5、ndinanearlierexerciseinChapter5)E(Y1+Y2)=2/3.Chapter6:FunctionsofRandomVariables141Instructor’sSolutionsManual6.1RefertoEx.5.15andEx.5.108.a.,sothat,u≥0,sothatUhasanexponentialdistributionwithβ=1.b.Frompartaabove,E(U)=1.6.2Itisgiventhatfi(yi)=,yi≥0fori=1,2.LetU=(Y

6、1+Y2)/2.a.sothat,u≥0,agammadensitywithα=2andβ=1/2.b.Frompart(a),E(U)=1,V(U)=1/2.6.3LetFY(y)andfY(y)denotethegammadistributionanddensityfunctionsrespectively.a..ThedensityfunctionforUisthen.Notethatthisisanothergammadistribution.b.Theshapeparameteristhesame(α),butt

7、hescaleparameteriscβ.6.4RefertoEx.5.8;=.Thus,,u≥0.6.5SinceY1andY2areindependent,so,for0≤y1≤1,0≤y2≤1.LetU=Y1Y2.Then,=9u2–8u3+6u3lnu.,0≤u≤1.6.6LetUhaveauniformdistributionon(0,1).ThedistributionfunctionforUis,0≤u≤1.ForafunctionG,werequireG(U)=YwhereYhasdistributionf

8、unctionFY(y)=,y≥0.NotethatFY(y)=P(Y≤y)==u.Soitmustbetruethat=usothatG(u)=[–ln(1–u)]–1/2.Therefore,therandomvariableY=[–ln(U–1)]–1/2hasdistributionfuncti

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