题目
9.设二维随机变量(X,Y)的联合密度函数为 f(x,y)= ) Axy,0lt xlt 1.0lt ylt x 0. .

题目解答
答案

解析
步骤 1:确定常数A
根据联合密度函数的性质,联合密度函数在整个定义域上的积分应等于1。因此,我们有:
$$
\int_{0}^{1}\int_{0}^{x} Axy \, dy \, dx = 1
$$
计算内层积分:
$$
\int_{0}^{x} Axy \, dy = A \int_{0}^{x} xy \, dy = A \left[ \frac{xy^2}{2} \right]_{0}^{x} = A \frac{x^3}{2}
$$
计算外层积分:
$$
\int_{0}^{1} A \frac{x^3}{2} \, dx = A \frac{1}{2} \int_{0}^{1} x^3 \, dx = A \frac{1}{2} \left[ \frac{x^4}{4} \right]_{0}^{1} = A \frac{1}{2} \cdot \frac{1}{4} = A \frac{1}{8}
$$
因此,我们有:
$$
A \frac{1}{8} = 1 \Rightarrow A = 8
$$
步骤 2:计算 $P\{ 0\lt X\lt 0.5,0\lt Y\lt 0.25\}$
根据联合密度函数,我们有:
$$
P\{ 0\lt X\lt 0.5,0\lt Y\lt 0.25\} = \int_{0}^{0.5}\int_{0}^{0.25} 8xy \, dy \, dx
$$
计算内层积分:
$$
\int_{0}^{0.25} 8xy \, dy = 8x \int_{0}^{0.25} y \, dy = 8x \left[ \frac{y^2}{2} \right]_{0}^{0.25} = 8x \cdot \frac{0.25^2}{2} = 8x \cdot \frac{1}{32} = \frac{x}{4}
$$
计算外层积分:
$$
\int_{0}^{0.5} \frac{x}{4} \, dx = \frac{1}{4} \int_{0}^{0.5} x \, dx = \frac{1}{4} \left[ \frac{x^2}{2} \right]_{0}^{0.5} = \frac{1}{4} \cdot \frac{0.5^2}{2} = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32}
$$
步骤 3:计算 $P\{ X=Y\}$
由于 $X$ 和 $Y$ 的联合密度函数在 $X=Y$ 的直线上是零,因此:
$$
P\{ X=Y\} = 0
$$
根据联合密度函数的性质,联合密度函数在整个定义域上的积分应等于1。因此,我们有:
$$
\int_{0}^{1}\int_{0}^{x} Axy \, dy \, dx = 1
$$
计算内层积分:
$$
\int_{0}^{x} Axy \, dy = A \int_{0}^{x} xy \, dy = A \left[ \frac{xy^2}{2} \right]_{0}^{x} = A \frac{x^3}{2}
$$
计算外层积分:
$$
\int_{0}^{1} A \frac{x^3}{2} \, dx = A \frac{1}{2} \int_{0}^{1} x^3 \, dx = A \frac{1}{2} \left[ \frac{x^4}{4} \right]_{0}^{1} = A \frac{1}{2} \cdot \frac{1}{4} = A \frac{1}{8}
$$
因此,我们有:
$$
A \frac{1}{8} = 1 \Rightarrow A = 8
$$
步骤 2:计算 $P\{ 0\lt X\lt 0.5,0\lt Y\lt 0.25\}$
根据联合密度函数,我们有:
$$
P\{ 0\lt X\lt 0.5,0\lt Y\lt 0.25\} = \int_{0}^{0.5}\int_{0}^{0.25} 8xy \, dy \, dx
$$
计算内层积分:
$$
\int_{0}^{0.25} 8xy \, dy = 8x \int_{0}^{0.25} y \, dy = 8x \left[ \frac{y^2}{2} \right]_{0}^{0.25} = 8x \cdot \frac{0.25^2}{2} = 8x \cdot \frac{1}{32} = \frac{x}{4}
$$
计算外层积分:
$$
\int_{0}^{0.5} \frac{x}{4} \, dx = \frac{1}{4} \int_{0}^{0.5} x \, dx = \frac{1}{4} \left[ \frac{x^2}{2} \right]_{0}^{0.5} = \frac{1}{4} \cdot \frac{0.5^2}{2} = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32}
$$
步骤 3:计算 $P\{ X=Y\}$
由于 $X$ 和 $Y$ 的联合密度函数在 $X=Y$ 的直线上是零,因此:
$$
P\{ X=Y\} = 0
$$