例 3.4.8

因为

$$\text{Var}(2X - 3Y + 8) = \text{Var}(2X) + \text{Var}(3Y) - 2\text{Cov}(2X, 3Y) = 4\text{Var}(X) + 9\text{Var}(Y) - 12\text{Cov}(X, Y),$$

所以我们先要分别计算 $E(X)$，$E(X^2)$，$E(Y)$，$E(Y^2)$，$E(XY)$。为此先计算两个边际密度函数。

$$p_X(x) = \int_0^2 \frac{1}{3}(x+y) \, \mathrm{d}y = \frac{2}{3}(x+1), \quad 0 < x < 1,$$ $$p_Y(y) = \int_0^1 \frac{1}{3}(x+y) \, \mathrm{d}x = \frac{1}{3}\left(\frac{1}{2} + y\right), \quad 0 < y < 2.$$

然后再计算一、二阶矩，

$$E(X) = \int_0^1 \frac{2}{3} x(x+1) \, \mathrm{d}x = \frac{5}{9},$$ $$E(X^2) = \int_0^1 \frac{2}{3} x^2(x+1) \, \mathrm{d}x = \frac{7}{18},$$ $$E(Y) = \int_0^2 \frac{1}{3} y\left(\frac{1}{2} + y\right) \mathrm{d}y = \frac{11}{9},$$ $$E(Y^2) = \int_0^2 \frac{1}{3} y^2\left(\frac{1}{2} + y\right) \mathrm{d}y = \frac{16}{9}.$$

由此得

$$\text{Var}(X) = \frac{7}{18} - \left(\frac{5}{9}\right)^2 = \frac{13}{162}, \quad \text{Var}(Y) = \frac{16}{9} - \left(\frac{11}{9}\right)^2 = \frac{23}{81}.$$

最后还需要计算 $E(XY)$，它只能从联合密度函数导出。

$$E(XY) = \frac{1}{3} \int_0^1 \int_0^2 xy(x+y) \, \mathrm{d}y \, \mathrm{d}x = \frac{1}{3} \int_0^1 \left(2x^2 + \frac{8}{3}x\right) \mathrm{d}x = \frac{2}{3}.$$

于是得协方差为

$$\text{Cov}(X, Y) = \frac{2}{3} - \frac{5}{9} \times \frac{11}{9} = -\frac{1}{81}.$$

代回原式得

$$\text{Var}(2X - 3Y + 8) = 4 \times \frac{13}{162} + 9 \times \frac{23}{81} - 12 \times \left(-\frac{1}{81}\right) = \frac{245}{81}.$$