Template:Example: Lognormal Distribution MLE

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'}}}= & \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]\displaystyle{ {{T}_{i}} }[/math] and solving the above system simultaneously, we get:

[math]\displaystyle{ \begin{align} & {{{\hat{\sigma' }}}}= & 0.849 \\ & {{{\hat{\mu }}}^{\prime }}= & 3.516 \end{align} }[/math]

Using the equation for mean and standard deviation we get:

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

and:

[math]\displaystyle{ {{\hat{\sigma }}}=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'}}}} \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]