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== Estimation of the Weibull Parameters ==
#REDIRECT [[The Weibull Distribution]]
 
The estimates of the parameters of the Weibull distribution can be found graphically via probability plotting paper, or analytically, either using least squares or maximum likelihood.
 
{{weibull parameters probability plotting}}
 
{{weibull parameters Rank Regression on Y}}
 
=== Rank Regression on X ===
 
Performing a rank regression on X is similar to the process for rank regression on Y, with the difference being that the ''horizontal'' deviations from the points to the line are minimized rather than the vertical. Again, the first task is to bring our  function, Eqn. (EQNREF Fw ), into a linear form. This step is exactly the same as in the regression on Y analysis and Eqns. (EQNREF logw ), (EQNREF yw ), (EQNREF aw ) and (EQNREF bw ) apply in this case too. The derivation from the previous analysis begins on the least squares fit part, where in this case we treat  as the dependent variable and  as the independent variable. The best-fitting straight line to the data, for regression on X (see Chapter 3), is the straight line:
 
::<math> x= \hat{a}+\hat{b}y </math> EQNREF xlinew
 
The corresponding equations for <math> \hat{a} </math> and <math> \hat{b} </math> are:
 
::<math> \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\sum\limits_{i=1}^{N}x_{i}}{N} -\hat{b}\frac{\sum\limits_{i=1}^{N}y_{i}}{N} </math>
 
:and
 
::<math> \hat{b}={\frac{\sum\limits_{i=1}^{N}x_{i}y_{i}-\frac{\sum \limits_{i=1}^{N}x_{i}\sum\limits_{i=1}^{N}y_{i}}{N}}{\sum \limits_{i=1}^{N}y_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{N}y_{i}\right) ^{2}}{N}}} </math>
 
:where:
 
::<math> y_{i}=\ln \left\{ -\ln [1-F(T_{i})]\right\} </math> and:
 
<span class="texhtml">''x''<sub>''i''</sub> = ln(''T''<sub>''i''</sub>)</span> and the <span class="texhtml">''F''(''T''<sub>''i''</sub>)</span> values are again obtained from the median ranks.
 
Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, solve Eqn. (EQNREF xlinew ) for <span class="texhtml">''y'',</span> which corresponds to:
 
::<math> y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}}x </math> Solving for the parameters from Eqns. (EQNREF aw ) and (EQNREF bw ) we get
 
::<math> a=-\frac{\hat{a}}{\hat{b}}=-\beta \ln (\eta )</math>  EQNREF awx
 
:and
 
::<math> b=\frac{1}{\hat{b}}=\beta EQNREF bwx </math> The correlation coefficient is evaluated as before using Eqn. (EQNREF RHOw ).
 
==== Example 4 ====
Repeat Example 1 using rank regression on X.
 
===== Solution to Example 4 =====
Solution to Example 4 Table 6.1, constructed in Example 3, can also be applied to this example.
 
Using the values from this table we get:
 
::<math> \hat{b} ={\frac{\sum\limits_{i=1}^{6}(\ln T_{i})y_{i}-\frac{ \sum\limits_{i=1}^{6}\ln T_{i}\sum\limits_{i=1}^{6}y_{i}}{6}}{ \sum\limits_{i=1}^{6}y_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{6}y_{i}\right) ^{2}}{6}}}
</math>
 
::<math>\hat{b} =\frac{-8.0699-(23.9068)(-3.0070)/6}{7.1502-(-3.0070)^{2}/6} </math>
 
:or:
 
::<math> \hat{b}=0.6931 </math>
 
:and:
 
::<math> \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\sum\limits_{i=1}^{6}\ln T_{i} }{6}-\hat{b}\frac{\sum\limits_{i=1}^{6}y_{i}}{6} </math>
 
:or:
 
::<math> \hat{a}=\frac{23.9068}{6}-(0.6931)\frac{(-3.0070)}{6}=4.3318 </math>
 
Therefore, from Eqn. (EQNREF bwx ):
 
::<math> \hat{\beta }=\frac{1}{\hat{b}}=\frac{1}{0.6931}=1.4428 </math>
 
:and from Eqn. (EQNREF awx )
 
::<math> \hat{\eta }=e^{\frac{\hat{a}}{\hat{b}}\cdot \frac{1}{\hat{ \beta }}}=e^{\frac{4.3318}{0.6931}\cdot \frac{1}{1.4428}}=76.0811\text{ hr} </math>
 
The correlation coefficient is found using Eqn. (EQNREF RHOw ):
 
::<math> \hat{\rho }=0.9956 </math>
 
The results and the associated graph using Weibull++ are given next. Note that the slight variation in the results is due to the number of significant figures used in the estimation of the median ranks. Weibull++ by default uses double precision accuracy when computing the median ranks.
 
[[Image:onevariableplot.png|thumb|center|400px| ]]
 
<br>
 
=== Three-Parameter Weibull Regression ===
 
When the MR versus <span class="texhtml">''T''<sub>''j''</sub></span> points plotted on the Weibull probability paper do not fall on a satisfactory straight line and the points fall on a curve,(Note that other shapes, particularly shapes, might suggest the existence of more than one population. In these cases, the multiple population, mixed Weibull distribution, may be more appropriate. Chapter 10 presents the mixed Weibull distribution.) then a location parameter, <span class="texhtml">γ</span>, might exist which may straighten out these points. The goal in this case is to fit a curve, instead of a line, through the data points using nonlinear regression. The Gauss-Newton method can be used to solve for the parameters, <span class="texhtml">β</span>, <span class="texhtml">η</span> and <span class="texhtml">γ</span>, by performing a Taylor series expansion on <span class="texhtml">''F''(''T''<sub>''i''</sub>;β,η,γ)</span>. Then the nonlinear model is approximated with linear terms and ordinary least squares are employed to estimate the parameters. This procedure is iterated until a satisfactory solution is reached. Weibull++ 7 calculates the value of <span class="texhtml">γ</span> by utilizing an optimized Nelder-Mead algorithm, and adjusts the points by this value of <span class="texhtml">γ</span> such that they fall on a straight line, and then plots both the adjusted and the original unadjusted points. To draw a curve through the original unadjusted points, if so desired, select Weibull 3P Line Unadjusted for Gamma from the ''Show Plot Line'' submenu under the ''Plot Options'' menu.  The returned estimations of the parameters are the same when selecting RRX or RRY. To display the unadjusted data points and line along with the adjusted data points and line, select ''Show/Hide Items'' under the ''Plot Options ''menu and include the unadjusted data points and line as follows:
 
[[Image:showhideplotitems.png|thumb|center|300px]]
 
[[Image:showhideplotwindow.png|thumb|center|300px]]
 
The results and the associated graph for the previous example using the three-parameter Weibull case are shown next:
 
[[Image:3parameterweibullplot.png|thumb|center|400px| ]]
<br>
 
=== Maximum Likelihood Estimation ===
 
As outlined in Chapter 3, maximum likelihood estimation works by developing a likelihood function based on the available data and finding the values of the parameter estimates that maximize the likelihood function.  This can be achieved by using iterative methods to determine the parameter estimate values that maximize the likelihood function, but this can be rather difficult and time-consuming, particularly when dealing with the three-parameter distribution.  Another method of finding the parameter estimates involves taking the partial derivatives of the likelihood function with respect to the parameters, setting the resulting equations equal to zero and solving simultaneously to determine the values of the parameter estimates. ( Note that MLE asymptotic properties do not hold when estimating <span class="texhtml">γ</span> using MLE [27].) The log-likelihood functions and associated partial derivatives used to determine maximum likelihood estimates for the Weibull distribution are covered in Appendix C.
 
 
====Example 5====
Repeat Example 1 using maximum likelihood estimation.
 
 
=====Solution to Example 5=====
In this case, we have non-grouped data with no suspensions or intervals, i.e. complete data. The equations for the partial derivatives of the log-likelihood function are derived in Appendix C and given next:
::<math> \frac{\partial \Lambda }{\partial \beta }=\frac{6}{\beta } +\sum_{i=1}^{6}\ln \left( \frac{T_{i}}{\eta }\right) -\sum_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }\ln \left( \frac{T_{i}}{\eta }\right) =0
</math>
 
:and:
 
::<math> \frac{\partial \Lambda }{\partial \eta }=\frac{-\beta }{\eta }\cdot 6+\frac{ \beta }{\eta }\sum\limits_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }=0 </math>
 
Solving the above equations simultaneously we get:
 
::<math> \hat{\beta }=1.933,</math> <math>\hat{\eta }=73.526 </math>
 
<br>
The variance/covariance matrix is found to be,
 
::<math> \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.4211 & \hat{Cov}( \hat{\beta },\hat{\eta })=3.272  \\
 
\hat{Cov}(\hat{\beta },\hat{\eta })=3.272 & \hat{Var} \left( \hat{\eta }\right) =266.646 \end{array} \right] </math>
 
The results and the associated graph using Weibull++ (MLE) are shown next.
 
[[Image:weibullfolio16plot.png|thumb|center|400px| ]]
 
You can view the variance/covariance matrix directly by clicking the ''Quick Calculation Pad ''(QCP) icon
 
[[Image:qcpicon.gif|thumb|center|400px| ]]
 
 
[[Image:qcpfolio16.png|thumb|center|400px| ]]
 
 
<br> Note that the decimal accuracy displayed and used is based on your individual User Setup.

Latest revision as of 09:07, 3 August 2012

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