Template:Example: Normal Distribution MLE - Revision history

Kate Racaza: Redirected page to The Normal Distribution#MLE Example

2012-08-14T05:59:24Z

Redirected page to The Normal Distribution#MLE Example

← Older revision		Revision as of 05:59, 14 August 2012
Line 1:		Line 1:
	~~'''Normal Distribution MLE Example'''~~		#REDIRECT[[The_Normal_Distribution#MLE_Example]]

	~~Using the data of~~ [[~~Normal distribution example 2\|Example 2]] and assuming a normal distribution, estimate the parameters using the MLE method.~~

	~~'''Solution'''~~

	~~In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, <math>\Lambda ,</math> are given by:~~

	~~::<math>\begin{align}~~
	~~\frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{t}_{i}}-\mu )=0 \\~~
	~~\frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{t}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0~~
	~~\end{align}</math>~~

	(The derivations of these equations are presented in [[Appendix:_Log-Likelihood_Equations\|an appendix]].) Substituting the values of <math>{{t}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>

	~~The Fisher matrix is:~~

	~~::<math>\left[ \begin{matrix}~~
	~~\widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 \\~~
	~~{} & {} & {} \\~~
	~~\widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500 \\~~
	~~\end{matrix} \right]</math>~~

	~~Using Weibull++ , the MLE method can be selected from the Set Analysis page.~~

	~~[[Image:Normal MLE Setting.png\|center\|550px\| ]]~~

	~~The plot of the solution for this example is shown next.~~

	~~[[Image:Normal MLE Plot.png\|center\|550px\|~~ ]]

Richard House at 02:39, 8 August 2012

2012-08-08T02:39:40Z

← Older revision		Revision as of 02:39, 8 August 2012
Line 6:		Line 6:

	In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, <math>\Lambda ,</math> are given by:		In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, <math>\Lambda ,</math> are given by:


	::<math>\begin{align}		::<math>\begin{align}
Line 12:		Line 11:
	\frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{t}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0		\frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{t}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0
	\end{align}</math>		\end{align}</math>


	(The derivations of these equations are presented in [[Appendix:_Log-Likelihood_Equations\|an appendix]].) Substituting the values of <math>{{t}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>		(The derivations of these equations are presented in [[Appendix:_Log-Likelihood_Equations\|an appendix]].) Substituting the values of <math>{{t}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>
Line 23:		Line 21:
	\widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500 \\		\widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500 \\
	\end{matrix} \right]</math>		\end{matrix} \right]</math>


	Using Weibull++ , the MLE method can be selected from the Set Analysis page.		Using Weibull++ , the MLE method can be selected from the Set Analysis page.

	[[Image:Normal MLE Setting.png~~\|thumb~~\|center\|~~250px~~\| ]]		[[Image:Normal MLE Setting.png\|center\|550px\| ]]

	The plot of the solution for this example is shown next.		The plot of the solution for this example is shown next.

	[[Image:Normal MLE Plot.png~~\|thumb~~\|center\|~~250px~~\| ]]		[[Image:Normal MLE Plot.png\|center\|550px\| ]]

Lisa Hacker at 08:40, 3 August 2012

2012-08-03T08:40:38Z

← Older revision		Revision as of 08:40, 3 August 2012
Line 14:		Line 14:


	(The derivations of these equations are presented in [[Appendix: ~~Distribution Log~~-~~Likelihood Equations~~\|~~Appendix~~]].) Substituting the values of <math>{{t}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>		(The derivations of these equations are presented in [[Appendix:_Log-Likelihood_Equations\|an appendix]].) Substituting the values of <math>{{t}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>

	The Fisher matrix is:		The Fisher matrix is:

Nicolette Young at 19:01, 25 April 2012

2012-04-25T19:01:12Z

← Older revision		Revision as of 19:01, 25 April 2012
Line 27:		Line 27:
	Using Weibull++ , the MLE method can be selected from the Set Analysis page.		Using Weibull++ , the MLE method can be selected from the Set Analysis page.

	[[Image:Normal MLE Setting.png\|thumb\|center\|~~400px~~\| ]]		[[Image:Normal MLE Setting.png\|thumb\|center\|250px\| ]]

	The plot of the solution for this example is shown next.		The plot of the solution for this example is shown next.

	[[Image:Normal MLE Plot.png\|thumb\|center\|~~400px~~\| ]]		[[Image:Normal MLE Plot.png\|thumb\|center\|250px\| ]]

Harry Guo at 18:24, 13 February 2012

2012-02-13T18:24:27Z

← Older revision		Revision as of 18:24, 13 February 2012
Line 32:		Line 32:

	[[Image:Normal MLE Plot.png\|thumb\|center\|400px\| ]]		[[Image:Normal MLE Plot.png\|thumb\|center\|400px\| ]]

	~~'''Special Note About Bias'''~~

	Estimators (i.e. parameter estimates) have properties such as unbiasedness, minimum variance, sufficiency, consistency, squared error constancy, efficiency and completeness [[Appendix: Weibull References\|[7][5]]]. Numerous books and papers deal with these properties and this coverage is beyond the scope of this reference.

	However, we would like to briefly address one of these properties, unbiasedness. An estimator is said to be unbiased if the estimator <math>\widehat{\theta }=d({{X}_{1,}}{{X}_{2,}}...,{{X}_{n)}}</math> satisfies the condition <math>E\left[ \widehat{\theta } \right]</math> <math>=\theta </math> for all <math>\theta \in \Omega .</math>

	~~Note that <math>E\left[ X \right]</math> denotes the expected value of X and is defined (for continuous distributions) by:~~

	~~::<math>\begin{align}~~
	~~E\left[ X \right]= \int_{\varpi }x\cdot f(x)dx \\~~
	~~X\in & \varpi .~~
	~~\end{align}</math>~~


	It can be shown [[Appendix: Weibull References\|[7][5]]] that the MLE estimator for the mean of the normal (and lognormal) distribution does satisfy the unbiasedness criteria, or <math>E\left[ \widehat{\mu } \right]</math> <math>=\mu .</math> The same is not true for the estimate of the variance <math>\hat{\sigma }^{2}</math> . The maximum likelihood estimate for the variance for the normal distribution is given by:

	~~::<math>\hat{\sigma }^{2}=\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}}</math>~~

	~~with a standard deviation of:~~

	~~::<math>{{\hat{\sigma }}}=\sqrt{\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}}}</math>~~


	These estimates, however, have been shown to be biased. It can be shown [[Appendix: Weibull References\|[7][5]]] that the unbiased estimate of the variance and standard deviation for complete data is given by:


	~~::<math>\begin{align}~~
	\hat{\sigma }^{2}= & \left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}} \right]=\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}} \\
	~~{{{\hat{\sigma }}}}= & \sqrt{\left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}} \right]} \\~~
	~~= & \sqrt{\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}}}~~
	~~\end{align}</math>~~


	~~Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.~~

	~~The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.~~

	When this option is selected, Weibull++ returns the unbiased standard deviation as defined. This is only true for complete data sets. For all other data types, Weibull++ by default returns the biased standard deviation as defined above regardless of the selection status of this option. The next figure shows this setting in Weibull++.


	~~<math></math>~~
	~~[[Image:ldachp8fig3.gif\|thumb\|center\|300px\| ]]~~

Harry Guo at 21:06, 10 February 2012

2012-02-10T21:06:26Z

← Older revision		Revision as of 21:06, 10 February 2012
Line 27:		Line 27:
	Using Weibull++ , the MLE method can be selected from the Set Analysis page.		Using Weibull++ , the MLE method can be selected from the Set Analysis page.

	[[Image:~~weibullfolio1~~.png\|thumb\|center\|400px\| ]]		[[Image:Normal MLE Setting.png\|thumb\|center\|400px\| ]]

	The plot of the solution for this example is shown next.		The plot of the solution for this example is shown next.

	[[Image:~~weibullfolioplot1~~.png\|thumb\|center\|400px\| ]]		[[Image:Normal MLE Plot.png\|thumb\|center\|400px\| ]]

	'''Special Note About Bias'''		'''Special Note About Bias'''

Harry Guo at 20:41, 10 February 2012

2012-02-10T20:41:17Z

← Older revision		Revision as of 20:41, 10 February 2012
Line 67:		Line 67:

	Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.		Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.

	~~Weibull++ by default returns the unbiased standard deviation given above.~~

	The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.		The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.

	When this option is selected, Weibull++ returns the unbiased standard deviation as defined. This is only true for complete data sets. For all other data types, Weibull++ by default returns the standard deviation as defined ~~by Eqn. (normbias2)~~ regardless of the selection status of this option. The next figure shows this setting in Weibull++.		When this option is selected, Weibull++ returns the unbiased standard deviation as defined. This is only true for complete data sets. For all other data types, Weibull++ by default returns the biased standard deviation as defined above regardless of the selection status of this option. The next figure shows this setting in Weibull++.


	<math></math>		<math></math>
	[[Image:ldachp8fig3.gif\|thumb\|center\|300px\| ]]		[[Image:ldachp8fig3.gif\|thumb\|center\|300px\| ]]

Harry Guo at 19:15, 10 February 2012

2012-02-10T19:15:33Z

← Older revision		Revision as of 19:15, 10 February 2012
Line 9:		Line 9:

	::<math>\begin{align}		::<math>\begin{align}
	\frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{T}_{i}}-\mu )=0 \\		\frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{t}_{i}}-\mu )=0 \\
	\frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{T}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0		\frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{t}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0
	\end{align}</math>		\end{align}</math>


	(The derivations of these equations are presented in [[Appendix: Distribution Log-Likelihood Equations\|Appendix]].) Substituting the values of <math>{{T}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>		(The derivations of these equations are presented in [[Appendix: Distribution Log-Likelihood Equations\|Appendix]].) Substituting the values of <math>{{t}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>

	The Fisher matrix is:		The Fisher matrix is:
Line 24:		Line 24:
	\end{matrix} \right]</math>		\end{matrix} \right]</math>


	Using Weibull++ , the MLE method can be selected from the Set Analysis page.		Using Weibull++ , the MLE method can be selected from the Set Analysis page.

Line 35:		Line 35:
	'''Special Note About Bias'''		'''Special Note About Bias'''

	Estimators (i.e. parameter estimates) have properties such as unbiasedness, minimum variance, sufficiency, consistency, squared error constancy, efficiency and completeness [7][5]. Numerous books and papers deal with these properties and this coverage is beyond the scope of this reference.		Estimators (i.e. parameter estimates) have properties such as unbiasedness, minimum variance, sufficiency, consistency, squared error constancy, efficiency and completeness [[Appendix: Weibull References\|[7][5]]]. Numerous books and papers deal with these properties and this coverage is beyond the scope of this reference.

	However, we would like to briefly address one of these properties, unbiasedness. An estimator is said to be unbiased if the estimator <math>\widehat{\theta }=d({{X}_{1,}}{{X}_{2,}}...,{{X}_{n)}}</math> satisfies the condition <math>E\left[ \widehat{\theta } \right]</math> <math>=\theta </math> for all <math>\theta \in \Omega .</math>		However, we would like to briefly address one of these properties, unbiasedness. An estimator is said to be unbiased if the estimator <math>\widehat{\theta }=d({{X}_{1,}}{{X}_{2,}}...,{{X}_{n)}}</math> satisfies the condition <math>E\left[ \widehat{\theta } \right]</math> <math>=\theta </math> for all <math>\theta \in \Omega .</math>

	Note that <math>E\left[ X \right]</math> denotes the expected value of X and is defined (for continuous distributions) by:		Note that <math>E\left[ X \right]</math> denotes the expected value of X and is defined (for continuous distributions) by:

Line 46:		Line 47:


	It can be shown [7][5] that the MLE estimator for the mean of the normal (and lognormal) distribution does satisfy the unbiasedness criteria, or <math>E\left[ \widehat{\mu } \right]</math> <math>=\mu .</math> The same is not true for the estimate of the variance <math>\hat{\sigma ~~}_{T~~}^{2}</math> . The maximum likelihood estimate for the variance for the normal distribution is given by:		It can be shown [[Appendix: Weibull References\|[7][5]]] that the MLE estimator for the mean of the normal (and lognormal) distribution does satisfy the unbiasedness criteria, or <math>E\left[ \widehat{\mu } \right]</math> <math>=\mu .</math> The same is not true for the estimate of the variance <math>\hat{\sigma }^{2}</math> . The maximum likelihood estimate for the variance for the normal distribution is given by:

	::<math>\hat{\sigma ~~}_{T~~}^{2}=\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}</math>		::<math>\hat{\sigma }^{2}=\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}}</math>

	with a standard deviation of:		with a standard deviation of:

	::<math>{{\hat{\sigma }~~}_{T~~}}=\sqrt{\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}</math>		::<math>{{\hat{\sigma }}}=\sqrt{\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}}}</math>


	These estimates, however, have been shown to be biased. It can be shown [7][5] that the unbiased estimate of the variance and standard deviation for complete data is given by:		These estimates, however, have been shown to be biased. It can be shown [[Appendix: Weibull References\|[7][5]]] that the unbiased estimate of the variance and standard deviation for complete data is given by:


	::<math>\begin{align}		::<math>\begin{align}
	\hat{\sigma ~~}_{T~~}^{2}= & \left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]=\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \\		\hat{\sigma }^{2}= & \left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}} \right]=\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}} \\
	{{{\hat{\sigma }}~~}_{T~~}}= & \sqrt{\left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]} \\		{{{\hat{\sigma }}}}= & \sqrt{\left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}} \right]} \\
	= & \sqrt{\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}		= & \sqrt{\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{t}_{i}}-\bar{T})}^{2}}}
	\end{align}</math>		\end{align}</math>

Line 67:		Line 68:
	Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.		Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.

	Weibull++ by default returns the standard deviation ~~as defined by Eqn. (NormSt2)~~.		Weibull++ by default returns the unbiased standard deviation given above.

	The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.		The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.

	When this option is selected, Weibull++ returns the standard deviation as defined ~~by Eqn. (NormSt2)~~. This is only true for complete data sets. For all other data types, Weibull++ by default returns the standard deviation as defined by Eqn. (normbias2) regardless of the selection status of this option. The next figure shows this setting in Weibull++.		When this option is selected, Weibull++ returns the unbiased standard deviation as defined. This is only true for complete data sets. For all other data types, Weibull++ by default returns the standard deviation as defined by Eqn. (normbias2) regardless of the selection status of this option. The next figure shows this setting in Weibull++.


	<math></math>		<math></math>
	[[Image:ldachp8fig3.gif\|thumb\|center\|300px\| ]]		[[Image:ldachp8fig3.gif\|thumb\|center\|300px\| ]]

Harry Guo at 19:05, 10 February 2012

2012-02-10T19:05:17Z

← Older revision		Revision as of 19:05, 10 February 2012
Line 1:		Line 1:

	'''Normal Distribution MLE Example'''		'''Normal Distribution MLE Example'''

	Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method.		Using the data of [[Normal distribution example 2\|Example 2]] and assuming a normal distribution, estimate the parameters using the MLE method.

	'''Solution'''		'''Solution'''
Line 15:		Line 14:


	(The derivations of these equations are presented in Appendix C.) Substituting the values of <math>{{T}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>		(The derivations of these equations are presented in [[Appendix: Distribution Log-Likelihood Equations\|Appendix]].) Substituting the values of <math>{{T}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>

	The Fisher matrix is:		The Fisher matrix is:

Harry Guo: Created page with ' '''Normal Distribution MLE Example''' Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method. '''Solution''' In this e…'

2012-02-10T19:03:10Z

Created page with ' '''Normal Distribution MLE Example''' Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method. '''Solution''' In this e…'

New page

'''Normal Distribution MLE Example'''

Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method.

'''Solution'''

In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function, <math>\Lambda ,</math> are given by:

::<math>\begin{align}
\frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{T}_{i}}-\mu )=0 \\
\frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{T}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0
\end{align}</math>

(The derivations of these equations are presented in Appendix C.) Substituting the values of <math>{{T}_{i}}</math> and solving the above system simultaneously, we get <math>\widehat{\sigma }=29.58</math> hours <math>,</math> <math>\widehat{\mu }=45</math> hours <math>.</math>

The Fisher matrix is:

::<math>\left[ \begin{matrix}
\widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 \\
{} & {} & {} \\
\widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500 \\
\end{matrix} \right]</math>

Using Weibull++ , the MLE method can be selected from the Set Analysis page.

[[Image:weibullfolio1.png|thumb|center|400px| ]]

The plot of the solution for this example is shown next.

[[Image:weibullfolioplot1.png|thumb|center|400px| ]]

'''Special Note About Bias'''

Estimators (i.e. parameter estimates) have properties such as unbiasedness, minimum variance, sufficiency, consistency, squared error constancy, efficiency and completeness [7][5]. Numerous books and papers deal with these properties and this coverage is beyond the scope of this reference.

However, we would like to briefly address one of these properties, unbiasedness. An estimator is said to be unbiased if the estimator <math>\widehat{\theta }=d({{X}_{1,}}{{X}_{2,}}...,{{X}_{n)}}</math> satisfies the condition <math>E\left[ \widehat{\theta } \right]</math> <math>=\theta </math> for all <math>\theta \in \Omega .</math>
Note that <math>E\left[ X \right]</math> denotes the expected value of X and is defined (for continuous distributions) by:

::<math>\begin{align}
E\left[ X \right]= \int_{\varpi }x\cdot f(x)dx \\
X\in & \varpi .
\end{align}</math>

It can be shown [7][5] that the MLE estimator for the mean of the normal (and lognormal) distribution does satisfy the unbiasedness criteria, or <math>E\left[ \widehat{\mu } \right]</math> <math>=\mu .</math> The same is not true for the estimate of the variance <math>\hat{\sigma }_{T}^{2}</math> . The maximum likelihood estimate for the variance for the normal distribution is given by:

::<math>\hat{\sigma }_{T}^{2}=\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}</math>

with a standard deviation of:

::<math>{{\hat{\sigma }}_{T}}=\sqrt{\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}</math>

These estimates, however, have been shown to be biased. It can be shown [7][5] that the unbiased estimate of the variance and standard deviation for complete data is given by:

::<math>\begin{align}
\hat{\sigma }_{T}^{2}= & \left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]=\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \\
{{{\hat{\sigma }}}_{T}}= & \sqrt{\left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]} \\
= & \sqrt{\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}
\end{align}</math>

Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.

Weibull++ by default returns the standard deviation as defined by Eqn. (NormSt2).
The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.

When this option is selected, Weibull++ returns the standard deviation as defined by Eqn. (NormSt2). This is only true for complete data sets. For all other data types, Weibull++ by default returns the standard deviation as defined by Eqn. (normbias2) regardless of the selection status of this option. The next figure shows this setting in Weibull++.

<math></math>
[[Image:ldachp8fig3.gif|thumb|center|300px| ]]