Question

设连续型随机变量 X 的概率密度为 (1)求常数 A ; (2)求X 的分布函数F(x); (3)求P(0< X<=1/2)？

Answer 1

f(x)
=Ax(1-x)^3 ; 0≤x≤1
=0 ; elsewhere
(1)

利用所有的几率相加=1
∫(0->1) f(x) dx =1
∫(0->1) Ax(1-x)^3 dx =1
A∫(0->1) [-(1-x)+1](1-x)^3 dx =1
A∫(0->1) [(1-x)^3 -(1-x)^4] dx =1
A[ (1-x)^5/5 - (1-x)^4/4]|(0->1) =1
A( 1/4-1/5)=1
A=20
(2)
对于 0≤x≤1
F(x)
=∫(0->x) f(t) dt
=20∫(0->x) t(1-t)^3 dt
=20∫(0->x) [(1-t)^3 -(1-t)^4] dt
=20[ (1-t)^5/5 - (1-t)^4/4]|(0->x)
=20[ (1-x)^5/5 - (1-x)^4/4 - ( 1/5-1/4) ]
=20[ (1-x)^5/5 - (1-x)^4/4 +1/20 ]
=4(1-x)^5 - 5(1-x)^4 +1
得出结果
F(x)
=0 ; x<0
=4(1-x)^5 - 5(1-x)^4 +1 ; 0≤x≤1
=1 ; x>1
(3)
P(0<X≤1/2)
=F(1/2)
=4(1/2)^5 - 5(1/2)^4 +1
=1/8-5/16 +1
=(18-5)/16
=13/16