数理统计问题2

f(x;μ)=1/[根号(2π)]e^[-(x-u)^2/2]
L(μ)=∏(i=1,...n)1/[根号(2π)]e^[-(xi-u)^2/2]
=(2π)^(-n/2)e^(-1/2∑(xi-u)^2)
lnL=-n/2ln2π-1/2∑(xi-u)^2
lnL对μ求导数=∑xi-nμ=0
μ=∑xi/n=x-
所以μ的极大似然估计量为X-

x- X- 代表均值

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