设随机变量X的分布律为
X-101
P1/31/31/3
记Y=X2,
求:
(1)D(X),D(Y);
(2)pXY.
(1)E(X)=-1×1/3+1×1/3=0 E(X2)=(-1)2×1/3+12×1/3=2/3 E(Y)=1×1/3+1×1/3=2/3 E(Y2)=12×1/3+12×1/3=2/3 ∴D (X)=E(X2)-E2(X)=2/3 D(Y)=E(Y2)-E2(Y)=2/3-4/9=2/9 (2)(X,Y)的分布律为 X\Y 0 1 -1 1/9 2/9 0 1/9 2/9 1 1/9 2/9 ∴E(XY)=-1×3/9+1×2/9=0 ∴pXY=Cov(X,Y)/[√D(X)•√D(Y)]=[E(XY)-E(X)•E(Y)]/[√D(X)•√D(Y)] =0