设连续随机向量x,y的密度函数F(xy)={a(x^2+y^2),(0≤x≤1,0≤y≤1)；0,其

设连续随机向量x,y的密度函数F(xy)={a(x^2+y^2),(0≤x≤1,0≤y≤1)；0,其

设连续随机向量x,y的密度函数F(xy)={a(x^2+y^2),(0≤x≤1,0≤y≤1)；0,其他.求设连续随机向量x,y的密度函数F(xy)={a(x^2+y^2),(0≤x≤1,0≤y≤1)；0,其他.①求x,y,xy的期望；②判断x和y是否相关；③求Cov（ax+B,y）.(图1)答案网 www.Zqnf.com 答案网 www.Zqnf.com ======以下答案可供参考======供参考答案1:注意：F通常用于分布函数，f通常用于密度函数。 ∫(0,1)dx∫(0,1)a(x^2+y^2)dy=a∫(0,1)dx∫(0,1)(x^2+y^2)dy=a∫(0,1)[x^2y+(1/3)y^3](0,1)dx=a∫(0,1)[x^2+(1/3)]dx=a[(1/3)x^3+(1/3)x](0,1)=2a/3=1 a=3/2 f(x)=∫(0,1)(3/2)(x^2+y^2)dy=(3/2)[x^2y+(1/3)y^3](0,1)=(3/2)x^2+(1/2) 0≤x≤1 f(y)=∫(0,1)(3/2)(x^2+y^2)dx=(3/2)[y^2x+(1/3)x^3](0,1)=(3/2)y^2+(1/2) 0≤y≤11.E(X)=∫(0,1)x[(3/2)x^2+(1/2)]dx=[(3/2)(1/4)x^4+(1/2)(1/2)x^2](0,1)=3/8+1/4=5/8 E(Y)=∫(0,1)y[(3/2)y^2+(1/2)]dy=[(3/2)(1/4)y^4+(1/2)(1/2)y^2](0,1)=3/8+1/4=5/8 E(XY)=∫(0,1)dx∫(0,1)(3/2)xy(x^2+y^2)dy=(3/2)∫(0,1)[x^3(1/2)y^2+x(1/4)y^4](0,1)dx =(3/2)∫(0,1)[(1/2)x^3+(1/4)x]dx=(3/2)2(1/2)(1/4)=3/82.cov(X,Y)=E(XY)-E(X)E(Y)=(3/8)-(5/8)(5/8)=-1/64 ∴X和Y相关。3.cov(aX+b,Y)=acov(X,Y)=-a/64

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