设随机变量X的分布律为X-101P1/31/31/3记Y=X2，

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)ρ_XY．
【正确答案】：(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 所以 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 所以 E(XY)=1×(2/9)+1×(2/9)=0 所以 [Cov(X，Y)]/√D(X)•√D(Y)=[E(XY)-E(X)•E(Y)]/√D(X)•√D(Y)=0