91Ó°ÊÓ

Let \({X_1},{\bf{ }}.{\bf{ }}.{\bf{ }}.{\bf{ }},{\bf{ }}{X_n}\)is drawn from a random sample of a normal distribution where both the mean and the variance are unknown.

Let mean be \(\mu \) and variance\({\sigma ^2}\) .

The probability density function is:

\(f\left( x \right) = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}\)

The 0.95 quantile of\({X_1},{\bf{ }}.{\bf{ }}.{\bf{ }}.{\bf{ }},{\bf{ }}{X_n}\)is when the following parameter’s value \(\theta = \theta \left( {\mu ,\sigma } \right)\) Condition is fulfilled

\(\begin{array}{c}P\left( {X < \theta } \right) = 0.95\\ = P\left( {\frac{{X - \mu }}{\sigma } < \frac{{\theta - \mu }}{\sigma }} \right)\end{array}\)

Let,

\(\begin{array}{l}{\theta _0} = \frac{{\theta - \mu }}{\sigma }\\ \Rightarrow \theta = \sigma {\theta _0} + \mu \ldots \left( 1 \right)\end{array}\)

\(L\left( {\mu ,{\sigma ^2}} \right) = \frac{{ - n}}{2}\ln \left( {2\pi {\sigma ^2}} \right) - {\sum {\frac{{\left( {{X_i} - \mu } \right)}}{{2{\sigma ^2}}}} ^2}\)

One need to find values for parameter in order to maximize the log likelihood functions.

Differentiating with respect to \(\mu \)

\(\begin{array}{c}\frac{{\partial L}}{{\partial \mu }} = \frac{\partial }{{\partial \mu }}\left( {\frac{{ - n}}{2}\ln \left( {2\pi {\sigma ^2}} \right) - \sum {\frac{{\left( {{X_i} - \mu } \right)}}{{2{\sigma ^2}}}} } \right)\\ = 0 - \sum {\frac{{{X_i} - \mu }}{{{\sigma ^2}}}} \\ = - \sum {\frac{{{X_i} - \mu }}{{{\sigma ^2}}}} \end{array}\)

Equating the differentiated equation with zero,

\(\sum {\frac{{{X_i} - \mu }}{{{\sigma ^2}}}} = 0\)

Therefore, for any \({\sigma ^2}\), the likelihood functions is maximized when

\(\begin{array}{c}\mu = \overline X \\ = \frac{1}{n}\sum {{X_i}} \end{array}\)

This is a sample mean.

Differentiating with respect to\({\sigma ^2}\)

\(\begin{array}{c}\frac{{\partial L}}{{\partial {\sigma ^2}}} = \frac{\partial }{{\partial {\sigma ^2}}}\left( {\frac{{ - n}}{2}\ln \left( {2\pi {\sigma ^2}} \right) - \sum {\frac{{\left( {{X_i} - \mu } \right)}}{{2{\sigma ^2}}}} } \right)\\ = \frac{{ - n}}{{2{\sigma ^2}}} + \left( {\sum {\frac{{{{\left( {{X_i} - \mu } \right)}^2}}}{2}} } \right)\frac{1}{{{{\left( {{\sigma ^2}} \right)}^2}}}\end{array}\)

Equating the differentiated equation with 0,

\(\begin{array}{c}\frac{{ - n}}{{2{\sigma ^2}}} + \left( {\sum {\frac{{{{\left( {{X_i} - \mu } \right)}^2}}}{2}} } \right)\frac{1}{{{{\left( {{\sigma ^2}} \right)}^2}}} = 0\\\left( {\sum {\frac{{{{\left( {{X_i} - \mu } \right)}^2}}}{2}} } \right)\frac{1}{{{{\left( {{\sigma ^2}} \right)}^2}}} = \frac{n}{{2{\sigma ^2}}}\\\sum {\frac{{{{\left( {{X_i} - \mu } \right)}^2}}}{2}} = \frac{{n \times \left( {{\sigma ^4}} \right)}}{{2{\sigma ^2}}}\\{\sigma ^2} = \sum {\frac{{{{\left( {{X_i} - \mu } \right)}^2}}}{n}} \end{array}\)

For \(\mu = \overline X \), the likelihood functions is maximized when

\(\begin{array}{c}{\sigma ^2} = \sum {\frac{{{{\left( {{X_i} - \overline X } \right)}^2}}}{n}} \\ = \overline {{{\left( {{X_i} - \overline X } \right)}^2}} \end{array}\)

This is a sample variance.

Substituting the values of\(\mu \,\,and\,\,{\sigma ^2}\) in the quantile expression obtained in (1), the M.L.E. of 0.95 quantile of X is:

\(\begin{array}{c}\theta = \sigma {\theta _0} + \mu \\ = {\theta _0}\sqrt {\overline {{{\left( {{X_i} - \overline X } \right)}^2}} } + \overline X \end{array}\)

The M.L.E. of 0.95 quantile of X is:\({\theta _0}\sqrt {\overline {{{\left( {{X_i} - \overline X } \right)}^2}} } + \overline X \).