已知随机向量X服从标准正态分布N(0，1)，令Y=X3，求X与y的相关系数pXY
【正确答案】：E(XY)=E(X4) =(1/√2π)∫+∞-∞x4e-(x2/2)dx=-(1/2π) ∫+∞-∞x3de-(x2/2) =-(1/√2π)x3e-(x2/2)∣+∞-∞+(3/√2π) ∫+∞-∞x2e-(x2/2)dx =(3/√2π)∫+∞-∞x2e-(x2/2)dx=3, D(X)=1， D(Y)=E(Y2)-(E(Y))2=E(X6)-(E(X3))2． 而E(X6)=(1/√2π)∫+∞-∞x6•e-(x2/2)dx=15， E(X3)=(3/√2π)∫+∞-∞x2e-(x2/2)dx=0. 则D(Y)=15， ρxy= Cov(X,Y)/[√D(X)√D(y) ] = [ E(XY)-E(X)E(Y)]/[√D(X)√D(y)] =(3-0)/(1•√15)=√15/5．
