题目
4.设随机变量 -N(1,9) . -N(0,16), 相关系数 (rho )_(x)=-dfrac (1)(2), 设-|||-=dfrac (X)(3)+dfrac (Y)(2) 求:(1)随机变量Z的期望E(Z)与方差D Z);(2)随-|||-机变量X与Z的相关系数ρxz·

题目解答
答案

解析
步骤 1:计算期望E(Z)
根据期望的线性性质,我们有:
$$E(Z) = E\left(\frac{X}{3} + \frac{Y}{2}\right) = \frac{1}{3}E(X) + \frac{1}{2}E(Y)$$
由于$E(X) = 1$,$E(Y) = 0$,代入上式得:
$$E(Z) = \frac{1}{3} \times 1 + \frac{1}{2} \times 0 = \frac{1}{3}$$
步骤 2:计算方差D(Z)
根据方差的性质,我们有:
$$D(Z) = D\left(\frac{X}{3} + \frac{Y}{2}\right) = \left(\frac{1}{3}\right)^2D(X) + \left(\frac{1}{2}\right)^2D(Y) + 2 \times \frac{1}{3} \times \frac{1}{2} \times \text{Cov}(X,Y)$$
由于$D(X) = 9$,$D(Y) = 16$,$\text{Cov}(X,Y) = \rho_{xy} \times \sqrt{D(X)D(Y)} = -\frac{1}{2} \times \sqrt{9 \times 16} = -6$,代入上式得:
$$D(Z) = \left(\frac{1}{3}\right)^2 \times 9 + \left(\frac{1}{2}\right)^2 \times 16 + 2 \times \frac{1}{3} \times \frac{1}{2} \times (-6) = 1 + 4 - 2 = 3$$
步骤 3:计算相关系数ρxz
根据相关系数的定义,我们有:
$$\rho_{xz} = \frac{\text{Cov}(X,Z)}{\sqrt{D(X)D(Z)}}$$
其中,$\text{Cov}(X,Z) = \text{Cov}(X, \frac{X}{3} + \frac{Y}{2}) = \frac{1}{3}D(X) + \frac{1}{2}\text{Cov}(X,Y) = \frac{1}{3} \times 9 + \frac{1}{2} \times (-6) = 3 - 3 = 0$,代入上式得:
$$\rho_{xz} = \frac{0}{\sqrt{9 \times 3}} = 0$$
根据期望的线性性质,我们有:
$$E(Z) = E\left(\frac{X}{3} + \frac{Y}{2}\right) = \frac{1}{3}E(X) + \frac{1}{2}E(Y)$$
由于$E(X) = 1$,$E(Y) = 0$,代入上式得:
$$E(Z) = \frac{1}{3} \times 1 + \frac{1}{2} \times 0 = \frac{1}{3}$$
步骤 2:计算方差D(Z)
根据方差的性质,我们有:
$$D(Z) = D\left(\frac{X}{3} + \frac{Y}{2}\right) = \left(\frac{1}{3}\right)^2D(X) + \left(\frac{1}{2}\right)^2D(Y) + 2 \times \frac{1}{3} \times \frac{1}{2} \times \text{Cov}(X,Y)$$
由于$D(X) = 9$,$D(Y) = 16$,$\text{Cov}(X,Y) = \rho_{xy} \times \sqrt{D(X)D(Y)} = -\frac{1}{2} \times \sqrt{9 \times 16} = -6$,代入上式得:
$$D(Z) = \left(\frac{1}{3}\right)^2 \times 9 + \left(\frac{1}{2}\right)^2 \times 16 + 2 \times \frac{1}{3} \times \frac{1}{2} \times (-6) = 1 + 4 - 2 = 3$$
步骤 3:计算相关系数ρxz
根据相关系数的定义,我们有:
$$\rho_{xz} = \frac{\text{Cov}(X,Z)}{\sqrt{D(X)D(Z)}}$$
其中,$\text{Cov}(X,Z) = \text{Cov}(X, \frac{X}{3} + \frac{Y}{2}) = \frac{1}{3}D(X) + \frac{1}{2}\text{Cov}(X,Y) = \frac{1}{3} \times 9 + \frac{1}{2} \times (-6) = 3 - 3 = 0$,代入上式得:
$$\rho_{xz} = \frac{0}{\sqrt{9 \times 3}} = 0$$