为自己加油

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p_XY

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．