x～（u，a^2）证明y=ax+b为正态分布求过程

x～（u，a^2）证明y=ax+b为正态分布求过程

X的期望 = U,
X的方差 = D,
E(X^2) = D + U^2.
问题补充：
原题：X,Y属于相同且独立的的正态分布[N(U,D)],
Z1 = AX + BY,Z2 = AX - BY,
求Z1Z2的相关系数.
答案：
EX = EY = U,
E(X-EX)^2 = E(Y-EY)^2 = D,
E[(X-EX)(Y-EY)] = E[X-EX]E[Y-EY] = 0.
EZ1 = E[AX + BY] = AEX + BEY = AU + BU = (A+B)U,
EZ2 = E[AX - BY] = AEX - BEY = AU - BU = (A-B)U
(U即正态分布期望,到此我还是理解 的）
E(Z1Z2) = E[AX + BY][AX - BY] = E[A^2X^2 - B^2Y^2]
= A^2EX^2 - B^2EY^2
= A^2[D + U^2] - B^2[D + U^2]
= (A^2 - B^2)(D + U^2)
E[Z1 - EZ1]^2 = E[A(X-EX) + B(Y-EY)]^2 = A^2E(X-EX)^2 + B^2E(Y-EY)^2 + 2ABE[(X-EX)(Y-EY)]
= A^2D + B^2D
= (A^2 + B^2)D
E[Z2 - EZ2]^2 = E[A(X-EX) - B(Y-EY)]^2 = A^2E(X-EX)^2 + B^2E(Y-EY)^2 - 2ABE[(X-EX)(Y-EY)]
= A^2D + B^2D
= (A^2 + B^2)D
Z1Z2的相关系数 = E[Z1 - EZ1][Z2 - EZ2]/[E(Z1 - EZ1)^2E(Z2 - EZ2)^2]^(1/2)
= {E[Z1Z2] - EZ1EZ2}/[(A^2+B^2)D*(A^2+B^2)D]^(1/2)
= [(A^2 - B^2)(D + U^2) - (A+B)U*(A-B)U]/[(A^2+B^2)D]
= [(A^2 - B^2)(D + U^2) - (A^2 - B^2)U^2]/[(A^2+B^2)D]
= [(A^2 - B^2)(D + U^2 - U^2)]/[(A^2+B^2)D]
= [A^2 - B^2]/[A^2 + B^2]