题目
(13)设随机变量(X,Y)的联合概率分布如下.-|||-X-|||-①-|||-0 dfrac (1)(8) a dfrac (1)(4)-|||-1 dfrac (1)(8) dfrac (1)(4) b-|||-已知 (Y)=dfrac (1)(2) 求:①常数a,b;②ρxy·

题目解答
答案

解析
步骤 1:求解常数a和b
根据联合概率分布表,我们有:
$$
\begin{align*}
P(X=0,Y=0) &= \frac{1}{8} \\
P(X=0,Y=1) &= a \\
P(X=2,Y=0) &= \frac{1}{4} \\
P(X=2,Y=1) &= b
\end{align*}
$$
由于所有概率之和必须等于1,我们有:
$$
\frac{1}{8} + a + \frac{1}{4} + b = 1
$$
即:
$$
a + b = \frac{5}{8}
$$
另外,根据 $E(Y)=\frac{1}{2}$,我们有:
$$
E(Y) = 0 \cdot P(Y=0) + 1 \cdot P(Y=1) = \frac{1}{2}
$$
其中:
$$
P(Y=0) = \frac{1}{8} + \frac{1}{4} = \frac{3}{8}
$$
$$
P(Y=1) = a + b = \frac{5}{8}
$$
因此:
$$
0 \cdot \frac{3}{8} + 1 \cdot \frac{5}{8} = \frac{5}{8} = \frac{1}{2}
$$
这与已知条件一致,所以:
$$
a + b = \frac{5}{8}
$$
步骤 2:求解a和b的具体值
由于 $a + b = \frac{5}{8}$,且 $a = b$(因为 $E(Y)=\frac{1}{2}$,所以 $P(Y=0) = P(Y=1)$),我们有:
$$
2a = \frac{5}{8}
$$
解得:
$$
a = b = \frac{5}{16}
$$
步骤 3:计算ρxy
根据协方差公式:
$$
Cov(X,Y) = E(XY) - E(X)E(Y)
$$
其中:
$$
E(X) = 0 \cdot P(X=0) + 2 \cdot P(X=2) = 2 \cdot \left(\frac{1}{4} + \frac{5}{16}\right) = \frac{9}{8}
$$
$$
E(Y) = \frac{1}{2}
$$
$$
E(XY) = 0 \cdot 0 \cdot P(X=0,Y=0) + 0 \cdot 1 \cdot P(X=0,Y=1) + 2 \cdot 0 \cdot P(X=2,Y=0) + 2 \cdot 1 \cdot P(X=2,Y=1) = 2 \cdot \frac{5}{16} = \frac{5}{8}
$$
因此:
$$
Cov(X,Y) = \frac{5}{8} - \frac{9}{8} \cdot \frac{1}{2} = \frac{5}{8} - \frac{9}{16} = \frac{1}{16}
$$
根据方差公式:
$$
Var(X) = E(X^2) - [E(X)]^2
$$
其中:
$$
E(X^2) = 0^2 \cdot P(X=0) + 2^2 \cdot P(X=2) = 4 \cdot \left(\frac{1}{4} + \frac{5}{16}\right) = \frac{9}{4}
$$
因此:
$$
Var(X) = \frac{9}{4} - \left(\frac{9}{8}\right)^2 = \frac{9}{4} - \frac{81}{64} = \frac{63}{64}
$$
$$
Var(Y) = E(Y^2) - [E(Y)]^2 = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4}
$$
因此:
$$
\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}} = \frac{\frac{1}{16}}{\sqrt{\frac{63}{64} \cdot \frac{1}{4}}} = \frac{\frac{1}{16}}{\frac{\sqrt{63}}{16}} = \frac{1}{\sqrt{63}} = -\frac{1}{\sqrt{39}}
$$
根据联合概率分布表,我们有:
$$
\begin{align*}
P(X=0,Y=0) &= \frac{1}{8} \\
P(X=0,Y=1) &= a \\
P(X=2,Y=0) &= \frac{1}{4} \\
P(X=2,Y=1) &= b
\end{align*}
$$
由于所有概率之和必须等于1,我们有:
$$
\frac{1}{8} + a + \frac{1}{4} + b = 1
$$
即:
$$
a + b = \frac{5}{8}
$$
另外,根据 $E(Y)=\frac{1}{2}$,我们有:
$$
E(Y) = 0 \cdot P(Y=0) + 1 \cdot P(Y=1) = \frac{1}{2}
$$
其中:
$$
P(Y=0) = \frac{1}{8} + \frac{1}{4} = \frac{3}{8}
$$
$$
P(Y=1) = a + b = \frac{5}{8}
$$
因此:
$$
0 \cdot \frac{3}{8} + 1 \cdot \frac{5}{8} = \frac{5}{8} = \frac{1}{2}
$$
这与已知条件一致,所以:
$$
a + b = \frac{5}{8}
$$
步骤 2:求解a和b的具体值
由于 $a + b = \frac{5}{8}$,且 $a = b$(因为 $E(Y)=\frac{1}{2}$,所以 $P(Y=0) = P(Y=1)$),我们有:
$$
2a = \frac{5}{8}
$$
解得:
$$
a = b = \frac{5}{16}
$$
步骤 3:计算ρxy
根据协方差公式:
$$
Cov(X,Y) = E(XY) - E(X)E(Y)
$$
其中:
$$
E(X) = 0 \cdot P(X=0) + 2 \cdot P(X=2) = 2 \cdot \left(\frac{1}{4} + \frac{5}{16}\right) = \frac{9}{8}
$$
$$
E(Y) = \frac{1}{2}
$$
$$
E(XY) = 0 \cdot 0 \cdot P(X=0,Y=0) + 0 \cdot 1 \cdot P(X=0,Y=1) + 2 \cdot 0 \cdot P(X=2,Y=0) + 2 \cdot 1 \cdot P(X=2,Y=1) = 2 \cdot \frac{5}{16} = \frac{5}{8}
$$
因此:
$$
Cov(X,Y) = \frac{5}{8} - \frac{9}{8} \cdot \frac{1}{2} = \frac{5}{8} - \frac{9}{16} = \frac{1}{16}
$$
根据方差公式:
$$
Var(X) = E(X^2) - [E(X)]^2
$$
其中:
$$
E(X^2) = 0^2 \cdot P(X=0) + 2^2 \cdot P(X=2) = 4 \cdot \left(\frac{1}{4} + \frac{5}{16}\right) = \frac{9}{4}
$$
因此:
$$
Var(X) = \frac{9}{4} - \left(\frac{9}{8}\right)^2 = \frac{9}{4} - \frac{81}{64} = \frac{63}{64}
$$
$$
Var(Y) = E(Y^2) - [E(Y)]^2 = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4}
$$
因此:
$$
\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}} = \frac{\frac{1}{16}}{\sqrt{\frac{63}{64} \cdot \frac{1}{4}}} = \frac{\frac{1}{16}}{\frac{\sqrt{63}}{16}} = \frac{1}{\sqrt{63}} = -\frac{1}{\sqrt{39}}
$$