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| '''Likelihood Ratio Bound on <math>\lambda </math>'''
| | #REDIRECT [[The Exponential Distribution]] |
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| Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming an exponential distribution, the MLE parameter estimate is calculated to be <math>\hat{\lambda }=0.013514.</math> Calculate the 85% two-sided confidence bounds on these parameters using the likelihood ratio method.
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| '''Solution'''
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| The first step is to calculate the likelihood function for the parameter estimates: | |
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| ::<math>\begin{align}
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| L(\hat{\lambda })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\hat{\lambda })=\underset{i=1}{\overset{N}{\mathop \prod }}\,\hat{\lambda }\cdot {{e}^{-\hat{\lambda }\cdot {{x}_{i}}}} \\
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| L(\hat{\lambda })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,0.013514\cdot {{e}^{-0.013514\cdot {{x}_{i}}}} \\
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| L(\hat{\lambda })= & 3.03647\times {{10}^{-12}}
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| \end{align}</math>
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| where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange the likelihood ratio equation to the form:
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| ::<math>L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>
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| Since our specified confidence level, <math>\delta </math>, is 85%, we can calculate the value of the chi-squared statistic, <math>\chi _{0.85;1}^{2}=2.072251.</math> We can now substitute this information into the equation:
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| ::<math>\begin{align}
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| L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\
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| L(\lambda )-3.03647\times {{10}^{-12}}\cdot {{e}^{\tfrac{-2.072251}{2}}}= & 0, \\
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| L(\lambda )-1.07742\times {{10}^{-12}}= & 0.
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| \end{align}</math>
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| It now remains to find the values of <math>\lambda </math> which satisfy this equation. Since there is only one parameter, there are only two values of <math>\lambda </math> that will satisfy the equation. These values represent the <math>\delta =85%</math> two-sided confidence limits of the parameter estimate <math>\hat{\lambda }</math>. For our problem, the confidence limits are:
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| ::<math>{{\lambda }_{0.85}}=(0.006572,0.024172)</math>
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