Template:Example: MLE for Exponential Distribution

MLE for Exponential Distribution

Using the data of Example 2 and assuming a 2-parameter exponential distribution, estimate the parameters using the MLE method.

Solution

In this example we have complete data only. The partial derivative of the log-likelihood function, $$\Lambda ,$$ is given by:


 * $$\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}}-\gamma \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}}-\gamma  \right) \right]=0$$

Complete descriptions of the partial derivatives can be found in Appendix D. Recall that when using the MLE method for the exponential distribution, the value of $$\gamma $$ is equal to that of the first failure time. The first failure occurred at 5 hours, thus $$\gamma =5$$ hours$$.$$ Substituting the values for $$T$$ and $$\gamma $$ we get:


 * $$\frac{14}{\hat{\lambda }}=560$$

or:


 * $$\hat{\lambda }=0.025\text{ failures/hour}$$

Using Weibull++:



The probability plot is: