Template:Aw mle

MLE Parameter Estimation
The parameters of the 2-parameter Weibull distribution can also be estimated using maximum likelihood estimation (MLE). This log-likelihood function is composed of :


 * $$\begin{align}

& \ln (L)= & \Lambda =\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{\beta }{\eta }{{\left( \frac{\eta } \right)}^{\beta -1}}{{e}^{-{{\left( \tfrac{\eta } \right)}^{\beta }}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }{{\left( \frac{T_{i}^{\prime }}{\eta } \right)}^{\beta }}\overset{FI}{\mathop{+\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }] \end{align}$$
 * where:


 * $$R_{Li}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}}}$$


 * $$R_{Ri}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}}}$$


 * $${{F}_{e}}$$ 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.
 * $$\beta $$ is the Weibull shape parameter (unknown a priori, the first of two parameters to be found).
 * $$\eta $$ is the Weibull scale parameter (unknown a priori, the second of two parameters 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 $${{i}^{th}}$$ 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 is found by solving for a pair of parameters $$\left( \widehat{\beta },\widehat{\eta } \right)$$  so that  $$\tfrac{\partial \Lambda }{\partial \beta }=0$$  and  $$\tfrac{\partial \Lambda }{\partial \eta }=0$$. (Other methods can also be used, such as direct maximization of the likelihood function, without having to compute the derivatives.)
 * $$\begin{align}

&\frac{\partial \Lambda }{\partial \beta }= \frac{1}{\beta }\underset{i=1}{\overset{\mathop{\sum }}}\,{{N}_{i}}+\underset{i=1}{\overset{\mathop{\sum }}}\,{{N}_{i}}\ln \left( \frac{\eta } \right) -\underset{i=1}{\overset{\mathop{\sum }}}\,{{N}_{i}}{{\left( \frac{\eta } \right)}^{\beta }}\ln \left( \frac{{{T}_{i}}}{\eta } \right)-\underset{i=1}{\overset{S}{\mathop{\sum }}}\,N_{i}^{\prime }{{\left( \frac{T_{i}^{\prime }}{\eta } \right)}^{\beta }}\ln \left( \frac{T_{i}^{\prime }}{\eta } \right)+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{-{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}\ln (\tfrac{T_{Li}^{\prime \prime }}{\eta })R_{Li}^{\prime \prime }+{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}\ln (\tfrac{T_{Ri}^{\prime \prime }}{\eta })R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }} \\ &\frac{\partial \Lambda }{\partial \eta }= \frac{-\beta }{\eta }\underset{i=1}{\overset{\mathop{\sum }}}\,{{N}_{i}}+\frac{\beta }{\eta }\underset{i=1}{\overset{\mathop{\sum }}}\,{{N}_{i}}{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta }}+\frac{\beta }{\eta }\underset{i=1}{\overset{S}{\mathop{\sum }}}\,N_{i}^{\prime }{{\left( \frac{T_{i}^{\prime }}{\eta } \right)}^{\beta }}+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{\beta }{\eta }\frac{{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}R_{Li}^{\prime \prime }-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }} \end{align}$$.

Example 4
Using the same data as in the probability plotting example (Example 3), and assuming a 2-parameter Weibull distribution, estimate the parameter using the MLE method.

Solution

In this case we have non-grouped data with no suspensions, therefore the above equations become:


 * $$\frac{\partial \Lambda }{\partial \beta }=\frac{6}{\beta }+\underset{i=1}{\overset{6}{\mathop{\sum }}}\,\ln \left( \frac{\eta } \right)-\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{\left( \frac{\eta } \right)}^{\beta }}\ln \left( \frac{\eta } \right)=0$$


 * and:


 * $$\frac{\partial \Lambda }{\partial \eta }=\frac{-\beta }{\eta }\cdot 6+\frac{\beta }{\eta }\underset{i=1}{\overset{6}{\mathop \sum }}\,{{\left( \frac{\eta } \right)}^{\beta }}=0$$

Solving the above equations simultaneously we get:


 * $$\begin{matrix}

\widehat{\beta }=1.933 \\ \widehat{\eta }=73.526 \\ \end{matrix}$$