题目
5、设Xsim N(mu,sigma^2),Ysim N(mu,sigma^2),X,Y相互独立，求Z_(1)=alpha X+beta Y,Z_(2)=alpha X-beta Y的相关系数.(其中alpha,beta是不为0的常数)
5、设$X\sim N(\mu,\sigma^{2})$,$Y\sim N(\mu,\sigma^{2})$,X,Y相互独立，求$Z_{1}=\alpha X+\beta Y$,$Z_{2}=\alpha X-\beta Y$的相关系数.(其中$\alpha,\beta$是不为0的常数)
题目解答
答案
计算协方差：
\[
\text{cov}(Z_1, Z_2) = \alpha^2 \sigma^2 - \beta^2 \sigma^2 = (\alpha^2 - \beta^2) \sigma^2
\]
计算方差：
\[
D(Z_1) = D(Z_2) = \alpha^2 \sigma^2 + \beta^2 \sigma^2 = (\alpha^2 + \beta^2) \sigma^2
\]
计算相关系数：
\[
\rho_{Z_1 Z_2} = \frac{\text{cov}(Z_1, Z_2)}{\sqrt{D(Z_1)} \sqrt{D(Z_2)}} = \frac{(\alpha^2 - \beta^2) \sigma^2}{(\alpha^2 + \beta^2) \sigma^2} = \frac{\alpha^2 - \beta^2}{\alpha^2 + \beta^2}
\]
**答案：** $\boxed{\frac{\alpha^2 - \beta^2}{\alpha^2 + \beta^2}}$
解析
步骤 1：计算协方差
根据协方差的定义，我们有：
\[ \text{cov}(Z_1, Z_2) = \text{cov}(\alpha X + \beta Y, \alpha X - \beta Y) \]
由于X和Y相互独立，我们有：
\[ \text{cov}(X, Y) = 0 \]
因此：
\[ \text{cov}(Z_1, Z_2) = \alpha^2 \text{cov}(X, X) - \beta^2 \text{cov}(Y, Y) = \alpha^2 \sigma^2 - \beta^2 \sigma^2 = (\alpha^2 - \beta^2) \sigma^2 \]
步骤 2：计算方差
根据方差的性质，我们有：
\[ D(Z_1) = D(\alpha X + \beta Y) = \alpha^2 D(X) + \beta^2 D(Y) = \alpha^2 \sigma^2 + \beta^2 \sigma^2 = (\alpha^2 + \beta^2) \sigma^2 \]
由于$Z_1$和$Z_2$的方差相同，我们有：
\[ D(Z_2) = (\alpha^2 + \beta^2) \sigma^2 \]
步骤 3：计算相关系数
根据相关系数的定义，我们有：
\[ \rho_{Z_1 Z_2} = \frac{\text{cov}(Z_1, Z_2)}{\sqrt{D(Z_1)} \sqrt{D(Z_2)}} = \frac{(\alpha^2 - \beta^2) \sigma^2}{(\alpha^2 + \beta^2) \sigma^2} = \frac{\alpha^2 - \beta^2}{\alpha^2 + \beta^2} \]
根据协方差的定义，我们有：
\[ \text{cov}(Z_1, Z_2) = \text{cov}(\alpha X + \beta Y, \alpha X - \beta Y) \]
由于X和Y相互独立，我们有：
\[ \text{cov}(X, Y) = 0 \]
因此：
\[ \text{cov}(Z_1, Z_2) = \alpha^2 \text{cov}(X, X) - \beta^2 \text{cov}(Y, Y) = \alpha^2 \sigma^2 - \beta^2 \sigma^2 = (\alpha^2 - \beta^2) \sigma^2 \]
步骤 2：计算方差
根据方差的性质，我们有：
\[ D(Z_1) = D(\alpha X + \beta Y) = \alpha^2 D(X) + \beta^2 D(Y) = \alpha^2 \sigma^2 + \beta^2 \sigma^2 = (\alpha^2 + \beta^2) \sigma^2 \]
由于$Z_1$和$Z_2$的方差相同，我们有：
\[ D(Z_2) = (\alpha^2 + \beta^2) \sigma^2 \]
步骤 3：计算相关系数
根据相关系数的定义，我们有：
\[ \rho_{Z_1 Z_2} = \frac{\text{cov}(Z_1, Z_2)}{\sqrt{D(Z_1)} \sqrt{D(Z_2)}} = \frac{(\alpha^2 - \beta^2) \sigma^2}{(\alpha^2 + \beta^2) \sigma^2} = \frac{\alpha^2 - \beta^2}{\alpha^2 + \beta^2} \]