单选题
设F
1
(x)，F
2
(x)为两个分布函数，其相应的密度函数f
1
(x)，f
2
(x)是连续函数，则下列函数必为密度函数的是( )．
【正确答案】
D
【答案解析】解析：选项D，由∫
－∞
+∞
[f
1
(x)F
2
(x)+f
2
(x)F
1
(x)]dx =∫
－∞
+∞
d[F
1
(x)F
2
(x)]=F
1
(x)F
2
(x)|
－∞
+∞
=1， 及f
1
(x)F
2
(x)+f
2
(x)F
1
(x)≥0，知D正确． 选项A，由∫
－∞
+∞
f
1
(x)dx=1，∫
－∞
+∞
f
2
(x)dx=1推不出∫
－∞
+∞
f
1
(x)f
2
(x)dx=1，A错误． 选项B，∫
－∞
+∞
[f
1
(x)+f
2
(x)]dx=∫
－∞
+∞
f
1
(x)dx+∫
－∞
+∞
f
2
(x)dx=2≠1，B错误． 选项C，∫
－∞
+∞
f
1
(x)F
2
(x)dx=F
1
(x)F
2
(x)|
－∞
+∞
－∫
－∞
+∞
f
2
(x)F
1
(x)dx≠1，C错误． 故选D．