solutions_6.pdf

Solutions 6Guanghua School of ManagementPeking UniversityNovember 9,20121/10Question 1(X,Y)=Cov(X,Y)pVar(X)pVar(Y),Cov(X,Y)=E(XY)E(X)E(Y)From the calculation in Homework 5,we haveE(XY)=1/2.The marginal distribution of X and Y,for x (0,1),f1(x)=Zx012y2dy=4x3.For y (0,1),f2(y)=Z1y12y2dx=12y2(1 y).2/10Question 1 continuedCalculate E(X),Var(X),E(Y),Var(Y),E(X)=Z10 xf1(x)dx=Z104x4dx=45E(X2)=Z10 x2f1(x)dx=Z104x5dx=23.Var(X)=E(X2)E(X)2=275.E(Y)=Z10yf2(y)dy=Z1012y3(1 y)=35.E(Y2)=Z10y2f2(y)dy=Z1012y4(1 y)=25.Var(Y)=E(Y2)E(Y)2=125(1)Calculte(X,Y).3/10Question 2Use the formula:Var(X+Y+Z)=Var(X)+Var(Y)+Var(Z)+2Cov(X,Y)+2Cov(X,Z)+2Cov(Y,Z)Var(3X Y 2Z+1)=Var(3X Y 2Z)=9Var(X)+Var(Y)+4Var(Z)6Cov(X,Y)12Cov(X,Z)+4Cov(Y,Z).4/10Question 3.We have =6.5,E(Xn)=6.5,Var(Xn)=4/n.From the Chebyshev inequalityP(6 Xn 7)=P(0.5 Xn 6.5 0.5)=P(|Xn 6.5|0.5)1 Var(Xn)0.52=1 16n.1 16n=0.8,thenn=80.5/10Question 4.Proof:E(Zn)=n2P(Zn=n2)+0P(Zn=0)=n .For any?0,P(|Zn|?)=P(Zn=n2)=1n 0.Then Znp 0.6/10Question 5.Suppose X N(,2),thenP(X 1 2/n2/16=1 16n.Let1 16n=0.99,then n=1600.8/10Question 6.Method 2:Use the factXnis also a normal distributionN(,2/n),then we haven(Xn)N(0,1).P(n|(Xn)|n4)=(n4)(n4)=2(n4)1.Where()is the cumulative distribution function ofstandard normal distribution.Then2(n4)=1.999/10Question 6.From the fact(2.575829)=0.995,we have n=106.15.Let n=107,the sample size is much smaller because we havemore information about the distribution.10/10