已知随机变量X~N(0,1),则随机变量Y=2X+1的概率密度fY(y)=____.
【正确答案】:(1/√2π)e-(y-12/8),x∈R
解析:X~N(0,1),则fX(X)=(1/√2π)e-(x2/2)
fY(y)=F′Y (y)=F′x [(Y-1)/2]•1/2=fx[(y-1)/2]•1/2
=(1/√2π)e-(y-12/8),x∈R
已知随机变量X~N(0,1),则随机变量Y=2X+1的概率密度fY(y)=____.
- 2024-11-07 16:21:12
- 概率论与数理统计(工)(13174)