Template:Rank Regression on Y for Exponential Distribution - Revision history

Lisa Hacker: Redirected page to The Exponential Distribution

2012-08-09T09:07:02Z

Redirected page to The Exponential Distribution

← Older revision		Revision as of 09:07, 9 August 2012
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	~~===Rank Regression on Y for Exponential Distribution===~~		#REDIRECT [[The_Exponential_Distribution]]
	Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.
	~~The least squares parameter estimation method (regression analysis) was discussed in Chapter~~ [[~~Parameter Estimation~~]]~~, and the following equations for rank regression on Y (RRY) were derived:~~


	~~::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>~~

	~~and:~~

	::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}}</math>

	~~In our case, the equations for <math>{{y}_{i}}</math> and <math>{{x}_{i}}</math> are:~~

	~~::<math>{{y}_{i}}=\ln [1-F({{t}_{i}})]</math>~~

	~~and:~~

	~~::<math>{{x}_{i}}={{t}_{i}}</math>~~


	and the <math>F({{t}_{i}})</math> is estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, then <math>\hat{\lambda }</math> and <math>\hat{\gamma }</math> can easily be obtained from above equations.
	~~For the one-parameter exponential, equations for estimating ''a'' and ''b'' become:~~

	~~::<math>\begin{align}~~
	~~\hat{a}= & 0, \\~~
	~~\hat{b}= & \frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}}~~
	~~\end{align}</math>~~


	~~'''The Correlation Coefficient'''~~

	~~The estimator of <math>\rho </math> is the sample correlation coefficient, <math>\hat{\rho }</math>, given by:~~

	::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>


	~~'''Example 2:'''~~
	~~{{2 parameter exponential distribution RRY example}}~~

Harry Guo: /* Rank Regression on Y for Exponential Distribution */

2012-02-10T23:41:11Z

Rank Regression on Y for Exponential Distribution

← Older revision		Revision as of 23:41, 10 February 2012
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	===Rank Regression on Y for Exponential Distribution===		===Rank Regression on Y for Exponential Distribution===
	Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.		Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.
	The least squares parameter estimation method (regression analysis) was discussed in [[Parameter Estimation~~\|Chapter 4~~]], and the following equations for rank regression on Y (RRY) were derived:		The least squares parameter estimation method (regression analysis) was discussed in Chapter [[Parameter Estimation]], and the following equations for rank regression on Y (RRY) were derived:

Harry Guo at 18:20, 9 February 2012

2012-02-09T18:20:25Z

← Older revision		Revision as of 18:20, 9 February 2012
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	'''Example 2:'''		'''Example 2:'''
	{{2 parameter exponential distribution example}}		{{2 parameter exponential distribution RRY example}}

Harry Guo: /* Rank Regression on Y for Exponential Distribution */

2012-02-09T18:01:51Z

Rank Regression on Y for Exponential Distribution

← Older revision		Revision as of 18:01, 9 February 2012
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	::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>		::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>


			'''Example 2:'''
	{{2 parameter exponential distribution example}}		{{2 parameter exponential distribution example}}

Harry Guo: /* The Correlation Coefficient */

2012-02-08T22:07:37Z

The Correlation Coefficient

← Older revision		Revision as of 22:07, 8 February 2012
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	~~====~~The Correlation Coefficient~~====~~		'''The Correlation Coefficient'''

	The estimator of <math>\rho </math> is the sample correlation coefficient, <math>\hat{\rho }</math>, given by:		The estimator of <math>\rho </math> is the sample correlation coefficient, <math>\hat{\rho }</math>, given by:

Harry Guo at 23:23, 7 February 2012

2012-02-07T23:23:00Z

← Older revision		Revision as of 23:23, 7 February 2012
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	and the <math>F({{t}_{i}})</math> is estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, then <math>\hat{\lambda }</math> and <math>\hat{\gamma }</math> can easily be obtained from above equations.		and the <math>F({{t}_{i}})</math> is estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, then <math>\hat{\lambda }</math> and <math>\hat{\gamma }</math> can easily be obtained from above equations.
	For the one-parameter exponential, equations for estimating 'a' and 'b' become:		For the one-parameter exponential, equations for estimating ''a'' and ''b'' become:

	::<math>\begin{align}		::<math>\begin{align}

Harry Guo: /* Rank Regression on Y for Exponential Distribution */

2012-02-07T23:22:08Z

Rank Regression on Y for Exponential Distribution

← Older revision		Revision as of 23:22, 7 February 2012
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	===Rank Regression on Y for Exponential Distribution===		===Rank Regression on Y for Exponential Distribution===
	Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.		Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.
	The least squares parameter estimation method (regression analysis) was discussed in Chapter 3, and the following equations for rank regression on Y (RRY) were derived:		The least squares parameter estimation method (regression analysis) was discussed in [[Parameter Estimation\|Chapter 4]], and the following equations for rank regression on Y (RRY) were derived:


	::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>		::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>

	:and:		and:

	::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}}</math>		::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}}</math>
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	In our case, the equations for <math>{{y}_{i}}</math> and <math>{{x}_{i}}</math> are:		In our case, the equations for <math>{{y}_{i}}</math> and <math>{{x}_{i}}</math> are:

	::<math>{{y}_{i}}=\ln [1-F({{T}_{i}})]</math>		::<math>{{y}_{i}}=\ln [1-F({{t}_{i}})]</math>

	and:		and:

	::<math>{{x}_{i}}={{T}_{i}}</math>		::<math>{{x}_{i}}={{t}_{i}}</math>


	and the <math>F({{T}_{i}})</math> is estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, then <math>\hat{\lambda }</math> and <math>\hat{\gamma }</math> can easily be obtained from ~~Eqns. (ae) and (be)~~.		and the <math>F({{t}_{i}})</math> is estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, then <math>\hat{\lambda }</math> and <math>\hat{\gamma }</math> can easily be obtained from above equations.
	For the one-parameter exponential, ~~Eqns. (aae)~~ and ~~(bbe)~~ become:		For the one-parameter exponential, equations for estimating 'a' and 'b' become:

	::<math>\begin{align}		::<math>\begin{align}

Harry Guo: moved Template:Rank Regression on Y for One Parameter Exponential to Template:Rank Regression on Y for Exponential Distribution

2012-02-06T23:52:19Z

moved Template:Rank Regression on Y for One Parameter Exponential to Template:Rank Regression on Y for Exponential Distribution

← Older revision	Revision as of 23:52, 6 February 2012
(No difference)

Harry Guo: /* Rank Regression on Y for One Parameter Exponential */

2012-02-06T23:52:02Z

Rank Regression on Y for One Parameter Exponential

← Older revision		Revision as of 23:52, 6 February 2012
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	===Rank Regression on Y for ~~One Parameter~~ Exponential===		===Rank Regression on Y for Exponential Distribution===
	Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.		Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.
	The least squares parameter estimation method (regression analysis) was discussed in Chapter 3, and the following equations for rank regression on Y (RRY) were derived:		The least squares parameter estimation method (regression analysis) was discussed in Chapter 3, and the following equations for rank regression on Y (RRY) were derived:

Harry Guo: /* Rank Regression on Y for One Parameter Exponential */

2012-02-06T23:22:26Z

Rank Regression on Y for One Parameter Exponential

← Older revision		Revision as of 23:22, 6 February 2012
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	::<math>{{y}_{i}}=\ln [1-F({{T}_{i}})]</math>		::<math>{{y}_{i}}=\ln [1-F({{T}_{i}})]</math>

	:and:		and:

	::<math>{{x}_{i}}={{T}_{i}}</math>		::<math>{{x}_{i}}={{T}_{i}}</math>