91Ó°ÊÓ

\({X_1},...,{X_n}\) form a random sample from a normal distribution for which the mean μ is known, but the variance \({\sigma ^2}\) is unknown. We need to calculate the M.L.E. of \({\sigma ^2}\).

Let \(\theta = {\sigma ^2}\).Then the likelihood function is \(f\left( {x|\theta } \right) = \frac{1}{{{{\left( {2\pi \theta } \right)}^{\frac{n}{2}}}}}\exp \left\{ { - \frac{1}{{2\theta }}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} } \right\}\)

Let \(L\left( \theta \right) = \log f\left( {x|\theta } \right)\)then

\(\frac{{\partial L\left( \theta \right)}}{{\partial \theta }} = - \frac{n}{{2\theta }} + \frac{1}{{2{\theta ^2}}}{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} ^{}}\)

\(L\left( \theta \right) = \log f\left( {x|\theta } \right)\)is maximum when \(\frac{{\partial L\left( \theta \right)}}{{\partial \theta }} = 0\)

This implies

\(\begin{array}{l} - \frac{n}{{2\theta }} + \frac{1}{{2{\theta ^2}}}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} = 0\\\theta = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} \end{array}\)

Hence the M.L.E. of\({\sigma ^2}\)is \(\hat \theta = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} \)