设 $(X, Y)$ 服从二维正态分布, $X \sim N(1,9), Y \sim N(0,16), \rho_{X Y}=-\frac{1}{2}$, 设 $Z=\frac{X}{3}+\frac{Y}{2}$, 求(1) $E Z, D Z \quad$ (2) $\rho_{X Z} \quad$ (3) $X$ 与 $Z$ 是否相关?
【答案】 $$
\begin{aligned}
&(1) E Z=E\left(\frac{X}{3}+\frac{Y}{2}\right)=\frac{1}{3} E X+\frac{1}{2} E Y=\frac{1}{3} \cdot 1+\frac{1}{2} \cdot 0=\frac{1}{3} \\
& D Z=D\left(\frac{X}{3}+\frac{Y}{2}\right)=D\left(\frac{X}{3}\right)+D\left(\frac{Y}{2}\right)+2 \operatorname{cov}\left(\frac{X}{3}, \frac{Y}{2}\right)=\frac{1}{9} \cdot 9+\frac{1}{4} \cdot 16+2 \operatorname{cov}\left(\frac{X}{3}, \frac{Y}{2}\right) \\
& =5+2 \operatorname{cov}\left(\frac{X}{3}, \frac{Y}{2}\right) \text {, } \\
& \text { 而 } \operatorname{cov}\left(\frac{X}{3}, \frac{Y}{2}\right)=\frac{1}{6} \operatorname{cov}(X, Y)=\frac{1}{6} \rho_{X Y} \cdot \sqrt{D X} \cdot \sqrt{D Y}=\frac{1}{6} \cdot\left(-\frac{1}{2}\right) \cdot 3 \cdot 4=-1 \\
& \therefore D Z=5+2 \cdot(-1)=3
\end{aligned}
$$
(2) $\rho_{X Z}=\frac{\operatorname{cov}(X, Z)}{\sqrt{D X} \sqrt{D Z}}$, 而 $\operatorname{cov}(X, Z)=\operatorname{cov}\left(X, \frac{X}{3}+\frac{Y}{2}\right)$
$$
\begin{aligned}
& =\frac{1}{3} \operatorname{cov}(X, X)+\frac{1}{2} \operatorname{cov}(X, Y) \\
& =\frac{1}{3} D X+\frac{1}{2} \cdot(-6)=\frac{1}{3} \cdot 9-3=0, \therefore \rho_{X Z}=0
\end{aligned}
$$
(3) $Q \rho_{X Z}=0$, 所以 $X$ 与 $Z$ 不相关。


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