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:


 * $$\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 }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{\ln ({{t}_{i}})-{\mu }'}{\sigma'^{3}}-\frac{1} \right)=0 \end{align}$$

Substituting the values of $${{T}_{i}}$$  and solving the above system simultaneously, we get:


 * $$\begin{align}

& = & 0.849 \\ & {{{\hat{\mu }}}^{\prime }}= & 3.516 \end{align}$$

Using the equation for mean and standard deviation in section Lognormal Statistical Properties we get:


 * $$\overline{T}=\hat{\mu }=48.25\text{ hours}$$

and:


 * $$=49.61\text{ hours}.$$

The variance/covariance matrix is given by:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0515 & {} & \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }}, \right)=0.0000 \\ {} & {} & {} \\   \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }}, \right)=0.0000 & {} & \widehat{Var}\left(  \right)=0.0258  \\ \end{matrix} \right]$$