16.设随机变量(X,Y)具有概率密度-|||-f(x,y)= { (x+y),0leqslant xleqslant 2,0leqslant yleqslant 2 0, . ,-|||-求E(X),E(Y ),Cov(X,Y),ρ XY` D(X+Y)

考查要点：本题主要考查二维连续型随机变量的期望、协方差、相关系数及方差的计算。
解题思路：

1. 期望：利用二重积分计算$E(X)$和$E(Y)$，注意积分区域为$0 \leq x \leq 2$，$0 \leq y \leq 2$。
2. 协方差：先计算$E(XY)$，再通过公式$\text{Cov}(X,Y) = E(XY) - E(X)E(Y)$求解。
3. 相关系数：需先求出方差$D(X)$和$D(Y)$，再结合协方差计算$\rho_{XY} = \frac{\text{Cov}(X,Y)}{\sqrt{D(X)D(Y)}}$。
4. 方差$D(X+Y)$：利用方差性质$D(X+Y) = D(X) + D(Y) + 2\text{Cov}(X,Y)$。
   关键点：正确处理二重积分的计算顺序，注意对称性简化运算。

1. 计算$E(X)$和$E(Y)$

步骤1：计算$E(X)$
$\begin{aligned}E(X) &= \int_{0}^{2} \int_{0}^{2} x \cdot \frac{1}{8}(x+y) \, dy \, dx \\&= \frac{1}{8} \int_{0}^{2} x \left[ \int_{0}^{2} (x+y) \, dy \right] dx \\&= \frac{1}{8} \int_{0}^{2} x \left[ 2x + 2 \right] dx \\&= \frac{1}{8} \int_{0}^{2} (2x^2 + 2x) \, dx \\&= \frac{1}{8} \left[ \frac{2x^3}{3} + x^2 \right]_{0}^{2} = \frac{7}{6}.\end{aligned}$

步骤2：计算$E(Y)$
由对称性，$E(Y) = E(X) = \frac{7}{6}$。

2. 计算$\text{Cov}(X,Y)$

步骤1：计算$E(XY)$
$\begin{aligned}E(XY) &= \int_{0}^{2} \int_{0}^{2} xy \cdot \frac{1}{8}(x+y) \, dy \, dx \\&= \frac{1}{8} \int_{0}^{2} x \left[ \int_{0}^{2} y(x+y) \, dy \right] dx \\&= \frac{1}{8} \int_{0}^{2} x \left[ 2x + \frac{8}{3} \right] dx \\&= \frac{1}{8} \int_{0}^{2} \left( 2x^2 + \frac{8}{3}x \right) dx \\&= \frac{1}{8} \left[ \frac{2x^3}{3} + \frac{4x^2}{3} \right]_{0}^{2} = \frac{4}{3}.\end{aligned}$

步骤2：计算协方差
$\text{Cov}(X,Y) = E(XY) - E(X)E(Y) = \frac{4}{3} - \left( \frac{7}{6} \right)^2 = -\frac{1}{36}.$

3. 计算相关系数$\rho_{XY}$

步骤1：计算方差$D(X)$和$D(Y)$
$\begin{aligned}E(X^2) &= \int_{0}^{2} \int_{0}^{2} x^2 \cdot \frac{1}{8}(x+y) \, dy \, dx = \frac{5}{3}, \\D(X) &= E(X^2) - [E(X)]^2 = \frac{5}{3} - \left( \frac{7}{6} \right)^2 = \frac{11}{36}, \\D(Y) &= D(X) = \frac{11}{36}.\end{aligned}$

步骤2：计算相关系数
$\rho_{XY} = \frac{\text{Cov}(X,Y)}{\sqrt{D(X)D(Y)}} = \frac{-\frac{1}{36}}{\frac{11}{36}} = -\frac{1}{11}.$

4. 计算$D(X+Y)$

$D(X+Y) = D(X) + D(Y) + 2\text{Cov}(X,Y) = \frac{11}{36} + \frac{11}{36} + 2 \cdot \left( -\frac{1}{36} \right) = \frac{5}{9}.$