题目
14.(单选题,4.0分) 设随机变量X_(1),X_(2)同分布,其分布律为: PX_{i)=-1}=1/2,PX_{i)=0}=1/2,i=1,2.若PX_{1)X_(2)=0}=1,则PX_{1)=X_(2)}=( ).A. 1B. (1)/(4)C. 0D. (1)/(2)
14.(单选题,4.0分) 设随机变量$X_{1}$,$X_{2}$同分布,其分布律为: $P\{X_{i}=-1\}=1/2$,$P\{X_{i}=0\}=1/2$,$i=1,2$.若$P\{X_{1}X_{2}=0\}=1$,则$P\{X_{1}=X_{2}\}=( )$.
A. 1
B. $\frac{1}{4}$
C. 0
D. $\frac{1}{2}$
题目解答
答案
C. 0
解析
步骤 1:理解条件
由题意,$X_1$ 和 $X_2$ 的可能取值为 $-1$ 和 $0$,且 $P\{X_i = -1\} = P\{X_i = 0\} = \frac{1}{2}$。条件 $P\{X_1 X_2 = 0\} = 1$ 表明 $X_1$ 和 $X_2$ 中至少有一个为 $0$,即 $P\{X_1 = -1, X_2 = -1\} = 0$。
步骤 2:利用边缘分布
- $P\{X_1 = -1\} = P\{X_1 = -1, X_2 = 0\} = \frac{1}{2}$
- $P\{X_1 = 0\} = P\{X_1 = 0, X_2 = -1\} + P\{X_1 = 0, X_2 = 0\} = \frac{1}{2}$
- $P\{X_2 = -1\} = P\{X_1 = 0, X_2 = -1\} = \frac{1}{2}$
- $P\{X_2 = 0\} = P\{X_1 = -1, X_2 = 0\} + P\{X_1 = 0, X_2 = 0\} = \frac{1}{2}$
步骤 3:求联合概率
解得:
- $P\{X_1 = -1, X_2 = 0\} = \frac{1}{2}$
- $P\{X_1 = 0, X_2 = -1\} = \frac{1}{2}$
- $P\{X_1 = 0, X_2 = 0\} = 0$
步骤 4:计算$P\{X_1 = X_2\}$
因此,$P\{X_1 = X_2\} = P\{X_1 = -1, X_2 = -1\} + P\{X_1 = 0, X_2 = 0\} = 0 + 0 = 0$。
由题意,$X_1$ 和 $X_2$ 的可能取值为 $-1$ 和 $0$,且 $P\{X_i = -1\} = P\{X_i = 0\} = \frac{1}{2}$。条件 $P\{X_1 X_2 = 0\} = 1$ 表明 $X_1$ 和 $X_2$ 中至少有一个为 $0$,即 $P\{X_1 = -1, X_2 = -1\} = 0$。
步骤 2:利用边缘分布
- $P\{X_1 = -1\} = P\{X_1 = -1, X_2 = 0\} = \frac{1}{2}$
- $P\{X_1 = 0\} = P\{X_1 = 0, X_2 = -1\} + P\{X_1 = 0, X_2 = 0\} = \frac{1}{2}$
- $P\{X_2 = -1\} = P\{X_1 = 0, X_2 = -1\} = \frac{1}{2}$
- $P\{X_2 = 0\} = P\{X_1 = -1, X_2 = 0\} + P\{X_1 = 0, X_2 = 0\} = \frac{1}{2}$
步骤 3:求联合概率
解得:
- $P\{X_1 = -1, X_2 = 0\} = \frac{1}{2}$
- $P\{X_1 = 0, X_2 = -1\} = \frac{1}{2}$
- $P\{X_1 = 0, X_2 = 0\} = 0$
步骤 4:计算$P\{X_1 = X_2\}$
因此,$P\{X_1 = X_2\} = P\{X_1 = -1, X_2 = -1\} + P\{X_1 = 0, X_2 = 0\} = 0 + 0 = 0$。