例 4.2.2 (求随机向量的方差)

设 $(X, Y)$ 的联合分布密度函数为

f_{X, Y} (x, y) = {8 x y, 0, 0 \leq y \leq x, 0 \leq x \leq 1. 其他 .

试求 $E [X] ， E [Y] ， Var [X] ， Var [Y] ， Cov (X, Y)$ 和 $r (X, Y)$ .

解

$E [X] = \int_{0}^{1} d x \int_{0}^{x} x \cdot 8 x y d y = \frac{4}{5} .$

E [Y] = \int_{0}^{1} d y \int_{y}^{1} y \cdot 8 x y d x = \frac{8}{15} .

Var (X) = E [X^{2}] - (E [X])^{2} = \int_{0}^{1} d x \int_{0}^{x} x^{2} \cdot 8 x y d y - (\frac{4}{5})^{2} = \frac{2}{3} - (\frac{4}{5})^{2} = \frac{2}{75} .

Var (Y) = E [Y^{2}] - (E [Y])^{2} = \int_{0}^{1} d y \int_{y}^{1} y^{2} \cdot 8 x y d x - (\frac{8}{15})^{2} = \frac{1}{3} - (\frac{8}{15})^{2} = \frac{11}{225} .

Cov (X, Y) = E [X Y] - E [X] \cdot E [Y] = \int_{0}^{1} d x \int_{0}^{x} x y \cdot 8 x y d y - \frac{4}{5} \cdot \frac{8}{15} = \frac{4}{9} - \frac{4}{5} \cdot \frac{8}{15} = \frac{4}{225} .

r (X, Y) = \frac{Cov ( X , Y )}{Var ( X ) \cdot Var ( Y )} = \frac{\frac{4}{225}}{\frac{2}{75} \cdot \frac{11}{225}} = \frac{2 66}{33} \approx 0.4924.