选择题
设连续型随机变量X
1
,X
2
相互独立,分布函数分别为F
1
(x),F
2
(x),概率密度分别为f
1
(x),f
2
(x),则随机变量min(X
1
,X
2
)的概率密度为______
A、
f
1
(x)f
2
(x)
B、
f
1
(x)+f
2
(x)
C、
f
1
(x)F
2
(x)+f
2
(x)F
1
(x)
D、
f
1
(x)(1-F
2
(x))+f
2
(x)(1-F
1
(x))
【正确答案】
D
【答案解析】
对任意实数x,将min(X
1
,X
2
)的分布函数记为F(x),则
F(x)=P(min(X
1
,X
2
)≤X)=1-P(min(X
1
,X
2
)>X)
=1-P(X
1
>x)P(X
2
>x)
=1-[1-F
1
(x)][1-F
2
(x)],
于是[*]
=f
1
(x)(1-F
2
(x)+f
2
(x)[1-F
1
(x)]. 选D.
提交答案
关闭