设随机变量X的概率密度函数为f(x)关于x=c对称,即f(x-c)=f(x+c),且EX存在,证明EX=c
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解决时间 2021-04-03 04:55
- 提问者网友：鐵馬踏冰河
- 2021-04-02 22:47
设随机变量X的概率密度函数为f(x)关于x=c对称,即f(x-c)=f(x+c),且EX存在,证明EX=c
最佳答案
- 五星知识达人网友：风格不统一
- 2021-04-03 00:22
EX=在区间(-无穷大, +无穷大)积分xf(x)dx, 换元: u=x-c, du=dx 有
=在区间(-无穷大, +无穷大)积分(u+c)f(u+c)du,
=在区间(-无穷大, +无穷大)积分[uf(u+c) +cf(u+c)]du
容易知道:f(u)=uf(u+c)为奇函数: g(-u)=-uf(-u+c)= -uf(u+c)= -g(u) (由假设得)
故在区间(-无穷大, +无穷大)积分uf(u+c)du=0.
故:EX=0+在区间(-无穷大, +无穷大)积分cf(u+c)du=
=c*{在区间(-无穷大, +无穷大)积分f(u+c)du}=c*1=c,
即EX=c,追问在区间(-无穷大, +无穷大)积分f(u+c)du是等于1的吗，为什么追答因为,在区间(-无穷大, +无穷大)积分f(u)du=1. (概率密度的性质)
故:在区间(-无穷大, +无穷大)积分f(u+c)du = (换元: t=u+c dt=du)
=在区间(-无穷大, +无穷大)积分f(t)dt=1.
=在区间(-无穷大, +无穷大)积分(u+c)f(u+c)du,
=在区间(-无穷大, +无穷大)积分[uf(u+c) +cf(u+c)]du
容易知道:f(u)=uf(u+c)为奇函数: g(-u)=-uf(-u+c)= -uf(u+c)= -g(u) (由假设得)
故在区间(-无穷大, +无穷大)积分uf(u+c)du=0.
故:EX=0+在区间(-无穷大, +无穷大)积分cf(u+c)du=
=c*{在区间(-无穷大, +无穷大)积分f(u+c)du}=c*1=c,
即EX=c,追问在区间(-无穷大, +无穷大)积分f(u+c)du是等于1的吗，为什么追答因为,在区间(-无穷大, +无穷大)积分f(u)du=1. (概率密度的性质)
故:在区间(-无穷大, +无穷大)积分f(u+c)du = (换元: t=u+c dt=du)
=在区间(-无穷大, +无穷大)积分f(t)dt=1.
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