Template:Example: Normal Distribution MLE

Normal Distribution MLE Example

Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method. Solution

In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, $$\Lambda ,$$  are given by:


 * $$\begin{align}

\frac{\partial \Lambda }{\partial \mu }= & \frac{1}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{t}_{i}}-\mu )=0 \\ \frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{t}_{i}}-\mu }-\frac{1}{\sigma } \right)=0 \end{align}$$

(The derivations of these equations are presented in an appendix.) Substituting the values of $${{t}_{i}}$$  and solving the above system simultaneously, we get  $$\widehat{\sigma }=29.58$$  hours $$,$$   $$\widehat{\mu }=45$$  hours $$.$$

The Fisher matrix is:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 \\ {} & {} & {} \\   \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500  \\ \end{matrix} \right]$$

Using Weibull++, the MLE method can be selected from the Set Analysis page.



The plot of the solution for this example is shown next.