《概率论与数理统计》习题参考答案第四章

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1、15VÇ؆ênÚO6SK‰YSK4(A)1.w,,EX=1;EY=1:1.P5P5C60P53002.d1=PfX=kg=´•C=,ÏdEX=kPfX=kg=5C=.k=1k=1k137k=11373.w,,XŠþ•1;0;1VÇ©O•1;1;1,ÏdEX=1.63R2R3R1bc4.EX6=1.‡y{:eEX=1,K1=EX=xf(x)dx=x
cx=.q1=f(x)dx=RR0b+2R1bccxdx=,gñ.0b+15.Øu.Ï•E(X2)=0:8,(EX)2=0:64.6.dK¿Œ•ZZ1Z2n+1nnn

2、n2(21)E(X)=xf(x)dx=xxdx+x(2x)dx=R01(n+1)(n+2)7.dK¿Œ•(1)PfX=1g=0:4;PfX=2g=0:2;PfX=3g=0:4;PfY=1g=0:3;PfY=0g=0:4;PfY=1g=0:3EX=2;EY=0YXXyj1(2)E(Z)=E=pij=Xxi15ijXX(3)E(Z)=E(XY)2=(xy)2p=5ijijij8.duX;YƒpÕá,ÏdkZZ2E(X
Y)=EX
EY=fX(x)dx
fY(y)dy=
6=4xy39.DX=EX2(

3、EX)2=1:7R10.duf(x)3Rþ´ó¼ê,ÏdEX=xf(x)dx=0,RZZ12Z12222x2xDX=E(X)(EX)=xf(x)dx=pdx=pdxR11x201x2Z2Z22sin221cos21=
cosdx=d=(Cþ“†x=sin)0cos022R11.duf(x)3Rþ´ó¼ê,ÏdEX=xf(x)dx=0,RZZ+1Z+122221x2xDX=E(X)(EX)=xf(x)dx=x
edx=xedx=2R02012.¯‡3˜gÁ¥u)gê•X,u

4、)VÇ•p,KXÑlëê•p01©Ù(0

5、98115.duX;YƒpÕá,ÏdX2;Y2•ƒpÕá.Ïd†>=D(XY)=E(X2Y2)[E(XY)]2=E(X2)E(Y2)(EX
EY)2m>=[E(X2)(EX)2][E(Y2)(EY)2]+(EX)2[E(Y2)(EY)2]+(EY)2[E(X2)(XY)2]=E(X2)E(Y2)(EX
EY)216.dK¿Œ•ppD(X+Y)=DX+DY+2cov(X;Y)=DX+DY+2XYDXDY=85ppD(XY)=DX+DY2cov(X;Y)=DX+DY2XYDXDY=37Cov(X+Y;

6、XY)=E[(X+Y)(XY)]E(X+Y)
E(XY)=E(X2)(EX)2[E(Y2)(EY)2]=DXDY=1117.dK¿Œ•Cov(X;Y)D(X+Y)DXDYXY=pp=pp=0:6DXDYDXDY18.duf(x;y)'u:¥%é¡,Ïdkxyf(x;y);xf(x;y);yf(x;y)©O•'ux;x;yÛ¼ê,ÏdkCov(X;Y)=E(XY)EX
EY=00
0=019.dK¿Œ•Cov(;)Cov(aX+b;cY+d)acCov(X;Y)ac=pp=pp=p

7、p=XYDDD(aX+b)D(cY+d)a2DXc2DYjacj20.w,,F(x;y)=F(x)
F(y);8(x;y)2R2¤á,X;YƒpÕá.XY(B)p1.(1);(2)(D(X+Y)=D(XY)()Cov(X;Y)=0);(3)(dD(X+Y)=D(XY)Œ•Cov(X;Y)=0,lX;YÃ');pp(4);(5)(dX+Y=nŒ•Cov(X;Y)=Cov(X;nX)=Cov(X;X)=DX=ppppppXYDXDY=XYDXD(nX)=XYDXDX,lXY=1);p(6)(C

8、ov(X+Y;XY)=DXDY=0).2.(1)1;6Rw,,f(x)•Rþó¼ê,ÏdEX=0,E(X2)=21x2(1x)dx=14,Ïd,DX=06E(X2)(EX)2=1.6(2)1;5;2qp
2(w,,=100p(1p)100p+1p=5,…=p=1p=p=1ž•ŒŠ);2