最尤推定
0 = ∂ ∂ μ log  ( ( 1 2 π σ 2 ) n 2 e − ∑ i = 1 n ( x i − x ̄ ) 2 + n ( x ̄ − μ ) 2 2 σ 2 ) = ∂ ∂ μ ( log  ( 1 2 π σ 2 ) n 2 − ∑ i = 1 n ( x i − x ̄ ) 2 + n ( x ̄ − μ ) 2 2 σ 2 ) = 0 − − 2 n ( x ̄ − μ ) 2 σ 2 {\displaystyle {\begin{matrix}0&=&{\frac {\partial }{\partial \mu }}\log \left(\left({\frac {1}{2\pi \sigma ^{2}}}\right)^{\frac {n}{2}}e^{-{\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}+n({\bar {x}}-\mu )^{2}}{2\sigma ^{2}}}}\right)\\&=&{\frac {\partial }{\partial \mu }}\left(\log \left({\frac {1}{2\pi \sigma ^{2}}}\right)^{\frac {n}{2}}-{\frac {\sum _{i=1}^{n}(x_{i}-{\bar {x}})^{2}+n({\bar {x}}-\mu )^{2}}{2\sigma ^{2}}}\right)\\&=&0-{\frac {-2n({\bar {x}}-\mu )}{2\sigma ^{2}}}\\\end{matrix}}}