Template:MLE Example (EXP1-2)

MLE Parameter Estimation
Let's assume six identical units are reliability tested at the same application and operation stress levels. All of these units fail during the test after operating for the following times (in hours), $${{T}_{i}}$$ : 96, 257, 498, 763, 1051 and 1744.

The parameter of the exponential distribution can also be estimated using the maximum likelihood estimation (MLE) method.

This log-likelihood function is:


 * $$\ln (L)=\Lambda =\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]$$

where:
 * $$R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}$$


 * $$R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}$$

and
 * •	 $${{F}_{e}}$$ is the number of groups of times-to-failure data points.
 * •	 $${{N}_{i}}$$ is the number of times-to-failure in the  $${{i}^{th}}$$  time-to-failure data group.
 * •	 $$\lambda $$ is the failure rate parameter (unknown a priori, the only parameter to be found).
 * •	 $${{T}_{i}}$$ is the time of the  $${{i}^{th}}$$  group of time-to-failure data.
 * •	 $$S$$ is the number of groups of suspension data points.
 * •	 $$N_{i}^{\prime }$$ is the number of suspensions in the  $${{i}^{th}}$$  group of suspension data points.
 * •	 $$T_{i}^{\prime }$$ is the time of the  $${{i}^{th}}$$  suspension data group.
 * •	 $$FI$$ is the number of interval data groups.
 * •	 $$N_{i}^{\prime \prime }$$ is the number of intervals in the i $$^{th}$$  group of data intervals.
 * •	 $$T_{Li}^{\prime \prime }$$ is the beginning of the i $$^{th}$$  interval.
 * •	 $$T_{Ri}^{\prime \prime }$$ is the ending of the i $$^{th}$$  interval.

The solution will be found by solving for a parameter $$\widehat{\lambda }$$  so that  $$\tfrac{\partial \Lambda }{\partial \lambda }=0$$  where:
 * $$\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\left( \frac{1}{\lambda }-{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }T_{i}^{\prime }-\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }}$$

For the given data then the equation is
 * $$\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}} \right) \right]=0$$

Substituting the values for $$T$$  we get:


 * $$\begin{align}

\frac{6}{\lambda }= & 4409,\text{ or:} \\ \lambda = & 0.00136\text{ failure/hr} \end{align}$$