33.设随机变量X~N(μ,σ²),Y~N(μ,σ²),且设X,Y相互独立，试求Z_(1)=αX+βY和Z_(2)=αX-βY的相关系数(其中α,β是不为零的常数).

本题考查正态分布随机变量的线性组合的协方差、方差以及相关系数的计算。解题思路是先根据协方差的性质计算$Z_1$和$Z_2$的协方差$\text{Cov}(Z_1, Z_2)$，再分别计算$Z_1$和$Z_2$的方差$D(Z_1)$和$D(Z_2)$，最后根据相关系数的定义计算$\rho_{Z_1Z_2}$。

1. 计算$\text{Cov}(Z_1, Z_2)$

已知$Z_1 = \alpha X + \beta Y$，$Z_2 = \alpha X - \beta Y$，根据协方差的性质$\text{Cov}(aX + bY, cX + dY) = ac\text{Cov}(X, X) + ad\text{Cov}(X, Y) + bc\text{Cov}(Y, X) + bd\text{Cov}(Y, Y)$，可得：
$\begin{align*}\text{Cov}(Z_1, Z_2) &= \text{Cov}(\alpha X + \beta Y, \alpha X - \beta Y)\\&= \alpha^2\text{Cov}(X, X) - \alpha\beta\text{Cov}(X, Y) + \alpha\beta\text{Cov}(Y, X) - \beta^2\text{Cov}(Y, Y)\end{align*}$
因为$\text{Cov}(X, X) = D(X)$，$\text{Cov}(Y, Y) = D(Y)$，且$X$，$Y$相互独立，所以$\text{Cov}(X, Y) = 0$，又已知$X\sim N(\mu, \sigma^2)$，$Y\sim N(\mu, \sigma^2)$，则$D(X) = D(Y) = \sigma^2$，代入上式可得：
$\begin{align*}\text{Cov}(Z_1, Z_2) &= \alpha^2 D(X) - \beta^2 D(Y)\\&= \alpha^2 \sigma^2 - \beta^2 \sigma^2\\&= (\alpha^2 - \beta^2) \sigma^2\end{align*}$

2. 计算$D(Z_1)$和$D(Z_2)$

根据方差的性质$D(aX + bY) = a^2D(X) + b^2D(Y) + 2ab\text{Cov}(X, Y)$，可得：
$\begin{align*}D(Z_1) &= D(\alpha X + \beta Y)\\&= \alpha^2 D(X) + \beta^2 D(Y) + 2\alpha\beta\text{Cov}(X, Y)\end{align*}$
因为$\text{Cov}(X, Y) = 0$，$D(X) = D(Y) = \sigma^2$，代入上式可得：
$\begin{align*}D(Z_1) &= \alpha^2 D(X) + \beta^2 D(Y)\\&= \alpha^2 \sigma^2 + \beta^2 \sigma^2\\&= (\alpha^2 + \beta^2) \sigma^2\end{align*}$
同理可得：
$\begin{align*}D(Z_2) &= D(\alpha X - \beta Y)\\&= \alpha^2 D(X) + (- \beta)^2 D(Y) + 2\alpha(-\beta)\text{Cov}(X, Y)\\&= \alpha^2 D(X) + \beta^2 D(Y) - 2\alpha\beta\text{Cov}(X, Y)\end{align*}$
因为$\text{Cov}(X, Y) = 0$，$D(X) = D(Y) = \sigma^2$，代入上式可得：
$\begin{align*}D(Z_2) &= \alpha^2 D(X) + \beta^2 D(Y)\\&= \alpha^2 \sigma^2 + \beta^2 \sigma^2\\&= (\alpha^2 + \beta^2) \sigma^2\end{align*}$

3. 计算$\rho_{Z_1Z_2}$

根据相关系数的定义$\rho_{Z_1Z_2} = \frac{\text{Cov}(Z_1, Z_2)}{\sqrt{D(Z_1)D(Z_2)}}$，将$\text{Cov}(Z_1, Z_2) = (\alpha^2 - \beta^2) \sigma^2$，$D(Z_1) = D(Z_2) = (\alpha^2 + \beta^2) \sigma^2$代入可得：
$\begin{align*}\rho_{Z_1Z_2} &= \frac{(\alpha^2 - \beta^2) \sigma^2}{\sqrt{(\alpha^2 + \beta^2) \sigma^2 \cdot (\alpha^2 + \beta^2) \sigma^2}}\\&= \frac{(\alpha^2 - \beta^2) \sigma^2}{(\alpha^2 + \beta^2) \sigma^2}\\&= \frac{\alpha^2 - \beta^2}{\alpha^2 + \beta^2}\end{align*}$