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\[f(x) = \frac 1 {\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2} {2\sigma^2}}\] with \(\mu\) the mean of the distribution and \(\sigma\) the standard deviation
\[F(x) =\int_{-\infty}^{x}\frac 1 {\sigma\sqrt{2\pi}} e^{-\frac{(y-\mu)^2} {2\sigma^2}}dy =\int_{-\infty}^{\frac {x-\mu}{\sigma}}\frac 1 {\sqrt{2\pi}} e^{-\frac{z^2} {2}}dz =\frac 1 2 \left[1+\text{erf}\left(\frac {x-\mu} {\sigma\sqrt{2}} \right)\right]\] with \(\text{erf}\) being the error function.
\[L(\mu,\sigma;X)=\sum_i\left[-\frac 1 2 \ln(2\pi)-\ln(\sigma)-\frac{1}{2\sigma^2}(X_i-\mu)^2\right]\]
\[V(\mu,\sigma;X) =\left( \begin{array}{c} \frac{\partial L}{\partial \mu} \\ \frac{\partial L}{\partial \sigma} \end{array} \right) =\sum_i\left( \begin{array}{c} \frac {X_i-\mu}{\sigma^2} \\ \frac {(X_i-\mu)^2-\sigma^2}{\sigma^3} \end{array} \right) \]
\[\mathcal J (\mu,\sigma;X)= -\left( \begin{array}{cc} \frac{\partial^2 L}{\partial \mu^2} & \frac{\partial^2 L}{\partial \mu \partial \sigma} \\ \frac{\partial^2 L}{\partial \sigma \partial \mu} & \frac{\partial^2 L}{\partial \sigma^2} \end{array} \right) =\sum_i \left( \begin{array}{cc} \frac{1}{\sigma^2} & \frac{2(X_i-\mu)}{\sigma^3} \\ \frac{2(X_i-\mu)}{\sigma^3} & \frac{3(X_i-\mu)^2-\sigma^2}{\sigma^4} \end{array} \right) \]
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