已知X、Y分别服从正态分布N（0,9）和N（1,16）,且X与Y的相关系数ρXY=－1/2,设Z=X/3+Y/2,求

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