Template:Aae mle

Maximum Likelihood Estimation Method

The log-likelihood function for the exponential distribution is as shown next:

[math]\displaystyle{ \begin{align} & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\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 [\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime })] \end{align} }[/math]

where:

[math]\displaystyle{ R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}} }[/math]

[math]\displaystyle{ R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}} }[/math]

and:
• [math]\displaystyle{ {{F}_{e}} }[/math] is the number of groups of exact times-to-failure data points.

• [math]\displaystyle{ {{N}_{i}} }[/math] is the number of times-to-failure in the [math]\displaystyle{ {{i}^{th}} }[/math] time-to-failure data group.

• [math]\displaystyle{ \lambda }[/math] is the failure rate parameter (unknown).

• [math]\displaystyle{ {{T}_{i}} }[/math] is the exact failure time of the [math]\displaystyle{ {{i}^{th}} }[/math] group.

• [math]\displaystyle{ S }[/math] is the number of groups of suspension data points.

• [math]\displaystyle{ N_{i}^{\prime } }[/math] is the number of suspensions in the [math]\displaystyle{ {{i}^{th}} }[/math] group of suspension data points.

• [math]\displaystyle{ T_{i}^{\prime } }[/math] is the time of the [math]\displaystyle{ {{i}^{th}} }[/math] suspension data group.

• [math]\displaystyle{ FI }[/math] is the number of interval data groups.

• [math]\displaystyle{ N_{i}^{\prime \prime } }[/math] is the number of intervals in the i [math]\displaystyle{ ^{th} }[/math] group of data intervals.

• [math]\displaystyle{ T_{Li}^{\prime \prime } }[/math] is the beginning of the i [math]\displaystyle{ ^{th} }[/math] interval.

• [math]\displaystyle{ T_{Ri}^{\prime \prime } }[/math] is the ending of the i [math]\displaystyle{ ^{th} }[/math] interval.

Substituting the Arrhenius-exponential model into the log-likelihood function yields:

[math]\displaystyle{ \begin{align} & \Lambda = & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{1}{C\cdot {{e}^{\tfrac{B}{{{V}_{i}}}}}}{{e}^{-\tfrac{1}{C\cdot {{e}^{\tfrac{B}{{{V}_{i}}}}}}{{T}_{i}}}} \right] \\ & & -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{1}{C\cdot {{e}^{\tfrac{B}{{{V}_{i}}}}}}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 }] \end{align} }[/math]

where:

[math]\displaystyle{ R_{Li}^{\prime \prime }={{e}^{-\tfrac{T_{Li}^{\prime \prime }}{C{{e}^{\tfrac{B}{{{V}_{i}}}}}}}} }[/math]

[math]\displaystyle{ R_{Ri}^{\prime \prime }={{e}^{-\tfrac{T_{Ri}^{\prime \prime }}{C{{e}^{\tfrac{B}{{{V}_{i}}}}}}}} }[/math]

The solution (parameter estimates) will be found by solving for the parameters [math]\displaystyle{ \widehat{B}, }[/math] [math]\displaystyle{ \widehat{C} }[/math] so that [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial B}=0 }[/math] and [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial C}=0 }[/math] , where:

[math]\displaystyle{ \begin{align} & \frac{\partial \Lambda }{\partial B}= & \frac{1}{C}\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{{{T}_{i}}}{{{V}_{i}}{{e}^{\tfrac{B}{{{V}_{i}}}}}}-\frac{C}{{{V}_{i}}} \right)+\frac{1}{C}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{T_{i}^{\prime }}{{{V}_{i}}{{e}^{\tfrac{B}{{{V}_{i}}}}}} \\ & & \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 })C{{V}_{i}}{{e}^{\tfrac{B}{{{V}_{i}}}}}} \end{align} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial \Lambda }{\partial C}= & \frac{1}{C}\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{{{T}_{i}}}{C{{e}^{\tfrac{B}{{{V}_{i}}}}}}-1 \right)+\frac{1}{{{C}^{2}}}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{T_{i}^{\prime }}{{{e}^{\tfrac{B}{{{V}_{i}}}}}} \\ & & \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 }){{C}^{2}}{{e}^{\tfrac{B}{{{V}_{i}}}}}} \end{align} }[/math]