正态分布的方差怎么求

1）求均 对（*）式两边对u求导： ∫{e^[-(x-u)^2/2(t^2)]*[2(u-x)/2(t^2)]dx= 约去常数，再两边同乘以1/(√2π)t得： ∫[1/(√2π)t]*e^[-(x-u)^2/2(t^2)]*(u-x)dx= 把(u-x)拆开，再移项： ∫x*[1/(√2π)t]*e^[-(x-u)^2/2(t^2)]dx=u*∫[1/(√2π)t]*e^[-(x-u)^2/2(t^2)]d 也就是 ∫x*f(x)dx=u*1= 这样就正好凑出了均值的定义式，证明了均值就是u。

2)方差 过程和求均值是差不多的，我就稍微略写一点了 对(*)式两边对t求导： ∫[(x-u)^2/t^3]*e^[-(x-u)^2/2(t^2)]dx=√2 移项： ∫[(x-u)^2]*[1/(√2π)t]*e^[-(x-u)^2/2(t^2)]dx=t^2 也就是 ∫(x-u)^2*f(x)dx=t^2 正好凑出了方差的定义式，从而结论得证。