Template:Example: Lognormal Distribution MLE: Difference between revisions

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::<math>\begin{align}
::<math>\begin{align}
   & \frac{\partial \Lambda }{\partial {\mu }'}= & \frac{1}{\sigma _{{{T}'}}^{2}}\cdot \underset{i=1}{\overset{14}{\mathop \sum }}\,\ln ({{T}_{i}})-{\mu }'=0 \\  
   & \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 }_{{{T}'}}}}= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{\ln ({{T}_{i}})-{\mu }'}{\sigma _{{{T}'}}^{3}}-\frac{1}{{{\sigma }_{{{T}'}}}} \right)=0   
  & \frac{\partial \Lambda }{\partial {{\sigma }_{{{T}'}}}}= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{\ln ({{T}_{i}})-{\mu }'}{\sigma _{{{T}'}}^{3}}-\frac{1}{{{\sigma }_{{{T}'}}}} \right)=0   
\end{align}</math>
\end{align}</math>

Revision as of 22:55, 13 February 2012

Lognormal Distribution MLE Example

Using the data of 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]\displaystyle{ \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 }_{{{T}'}}}}= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{\ln ({{T}_{i}})-{\mu }'}{\sigma _{{{T}'}}^{3}}-\frac{1}{{{\sigma }_{{{T}'}}}} \right)=0 \end{align} }[/math]


Substituting the values of [math]\displaystyle{ {{T}_{i}} }[/math] and solving the above system simultaneously, we get:

[math]\displaystyle{ \begin{align} & {{{\hat{\sigma }}}_{{{T}'}}}= & 0.849 \\ & {{{\hat{\mu }}}^{\prime }}= & 3.516 \end{align} }[/math]


Using Eqns. (mean) and (sdv) we get:

[math]\displaystyle{ \overline{T}=\hat{\mu }=48.25\text{ hours} }[/math]


and:
[math]\displaystyle{ {{\hat{\sigma }}_{{{T}'}}}=49.61\text{ hours}. }[/math]

The variance/covariance matrix is given by:

[math]\displaystyle{ \left[ \begin{matrix} \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0515 & {} & \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma }}}_{{{T}'}}} \right)=0.0000 \\ {} & {} & {} \\ \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma }}}_{{{T}'}}} \right)=0.0000 & {} & \widehat{Var}\left( {{{\hat{\sigma }}}_{{{T}'}}} \right)=0.0258 \\ \end{matrix} \right] }[/math]