Template:Example: Lognormal Distribution MLE: Difference between revisions

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'''Lognormal Distribution MLE Example'''
#REDIRECT [[The Lognormal Distribution]]
 
Using the data of [[Lognormal Example 2 Data|Example 2]] and assuming a lognormal distribution, estimate the parameters using the MLE method.
 
'''Solution'''
In this example we have only complete data. Thus, the partials reduce to:
 
::<math>\begin{align}
  & \frac{\partial \Lambda }{\partial {\mu }'}= & \frac{1}{\sigma'^{2}}\cdot \underset{i=1}{\overset{14}{\mathop \sum }}\,\ln ({{t}_{i}})-{\mu }'=0 \\
& \frac{\partial \Lambda }{\partial {{\sigma'}}}= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{\ln ({{t}_{i}})-{\mu }'}{\sigma'^{3}}-\frac{1}{{{\sigma'}}} \right)=0 
\end{align}</math>
 
Substituting the values of  <math>{{T}_{i}}</math>  and solving the above system simultaneously, we get:
 
::<math>\begin{align}
  & {{{\hat{\sigma' }}}}= & 0.849 \\
& {{{\hat{\mu }}}^{\prime }}= & 3.516 
\end{align}</math>
 
Using the equation for mean and standard deviation in section [[Lognormal Statistical Properties]] we get:
 
::<math>\overline{T}=\hat{\mu }=48.25\text{ hours}</math>
 
and:
 
::<math>{{\hat{\sigma }}}=49.61\text{ hours}.</math>
 
The variance/covariance matrix is given by:
 
::<math>\left[ \begin{matrix}
  \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0515 & {} & \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma'}}}} \right)=0.0000  \\
  {} & {} & {}  \\
  \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma' }}}} \right)=0.0000 & {} & \widehat{Var}\left( {{{\hat{\sigma' }}}} \right)=0.0258  \\
\end{matrix} \right]</math>

Latest revision as of 03:32, 13 August 2012

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