Question

已知X服从N(1,9)，Y服从N(0,16)，且X与Y的相关系数p(xy)=-1/2，设Z=1/3X+1/2Y，求方差DZ=？

Answer 1

X~N(1,9) Y~N(0,16)
EX=1 DX=9 EY=0 DY=16
DX=E(X^2)-(EX)^2 所以E(X^2)=10
DY=E(Y^2)-(EY)^2 所以E(Y^2)=16

p(XY)=Cov(X,Y)/√DX*√DY 所以Cov(X,Y)=-6
Cov(X,Y)=E(XY)-(EX)(EY)
所以E(XY)=-6

DZ=D(Z^2)-(DZ)^2
=E[(1/3X+1/2Y)^2]-[E(1/3X+1/2Y)}^2
=E[1/9X^2+1/4Y^2+1/3XY]-[1/3EX+1/2EY]^2
=(1/9EX^2+1/4EY^2+1/3EXY)-(1/3EX+1/2EY)^2
=(1/9*10+1/4*16+1/3*(-6))-(1/3*1+1/2*0)^2
=3

上面的又是复制党= =

Answer 2

1）数学期望EZ=E(X/3+Y/2)=EX/3+EY/2=0+1/2=1/2

2）Y与Z的相关系数ρYZ
由ρXY=-1/2=[E(XY)-E(X)E(Y)]/[D(X)D(Y)]^0.5=[E(XY)-0*1]/3*4
所以E(XY)=-6
D(Z)=D(X/3+Y/2)=1/9*D(X)+1/4*D(Y)+2*1/3*1/2*ρXY*[D(X)D(Y)]^0.5
=1/9*3^2+1/4*4^2+2*1/3*1/2*(-1/2)*3*4
=3

ρYZ=[E(XZ)-E(X)E(Z)]/[D(X)D(Z)]^0.5
=[E(1/3X^2+XY/2)-0*1/2]/[3^2*3]^0.5
=[1/3*E(X)*E(X)+1/3*D(X)+1/2E(XY)]/(27)^0.5
0