设随机变量（X,Y)具有概率密度f(x)=1/8(x+y),0

Question

设随机变量（X,Y)具有概率密度f(x)=1/8(x+y),0

Answer 1

设随机变量（X,Y)具有概率密度f(x,y)=1/8(x+y),0f(x)=1/4*(x+1),0f(y)=1/4*(y+1),0EX=∫xf(x)dx=7/6
EY=∫yf(y)dy=7/6
EX^2=∫x^2f(x)dx=5/3
EY^2=∫y^2f(y)dy=5/3
DX=EX^2-(EX)^2=11/36
DY=EY^2-(EY)^2=11/36
EXY=∫∫xyf(x,y)dxdy=4/3
cov(X,Y)=EXY-EXEY=4/3-7/6*7/6=-1/36
ρXY=cov(X,Y)/√DXDY=-1/11
解毕