Template:Lognormal distributionrank regression on x - Revision history

Lisa Hacker: Redirected page to The Lognormal Distribution

2012-08-09T09:01:11Z

Redirected page to The Lognormal Distribution

← Older revision		Revision as of 09:01, 9 August 2012
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	~~===Rank Regression on X===~~		#REDIRECT [[The_Lognormal_Distribution]]
	Performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the line is minimized.

	Again, the first task is to bring our <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis and all the equations apply in this case too. The deviation from the previous analysis begins on the least squares fit part, where in this case we treat <math>x</math> as the dependent variable and <math>y</math> as the independent variable. The best-fitting straight line to the data, for regression on X (see Chapter [[~~Parameter Estimation~~]]~~), is the straight line:~~

	~~::<math>x=\widehat{a}+\widehat{b}y</math>~~

	~~The corresponding equations for and <math>\widehat{b}</math> are:~~

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

	~~where:~~

	~~::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(t_{i}^{\prime }) \right]</math>~~

	~~and:~~

	~~::<math>{{x}_{i}}=t_{i}^{\prime }</math>~~

	and the <math>F(t_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, solve the linear equation for the unknown <math>y</math> , which corresponds to:

	~~::<math>y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x</math>~~

	~~Solving for the parameters we get:~~

	~~::<math>a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{{{\mu }'}}{\sigma'}</math>~~

	~~and:~~

	~~::<math>b=\frac{1}{\widehat{b}}=\frac{1}{\sigma'}</math>~~

	~~The correlation coefficient is evaluated as before using equation in the [[The Correlation Coefficient Calculation\|previous section]].~~

	~~'''Example 3:'''~~
	~~{{Example: Lognormal Distribution RRX}}~~

Harry Guo at 21:32, 13 February 2012

2012-02-13T21:32:14Z

← Older revision		Revision as of 21:32, 13 February 2012
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	::<math>b=\frac{1}{\widehat{b}}=\frac{1}{\sigma'}</math>		::<math>b=\frac{1}{\widehat{b}}=\frac{1}{\sigma'}</math>

	The correlation coefficient is evaluated as before using ~~Eqn. (RHOln)~~.		The correlation coefficient is evaluated as before using equation in the [[The Correlation Coefficient Calculation\|previous section]].

	'''Example 3:'''		'''Example 3:'''
	{{Example: Lognormal Distribution RRX}}		{{Example: Lognormal Distribution RRX}}

Harry Guo: /* Rank Regression on X */

2012-02-13T21:28:28Z

Rank Regression on X

← Older revision		Revision as of 21:28, 13 February 2012
Line 2:		Line 2:
	Performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the line is minimized.		Performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the line is minimized.

	Again, the first task is to bring our <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis and ~~Eqns. (lnorm), (yln), (aln) and (bln)~~ apply in this case too. The deviation from the previous analysis begins on the least squares fit part, where in this case we treat <math>x</math> as the dependent variable and <math>y</math> as the independent variable. The best-fitting straight line to the data, for regression on X (see Chapter 3), is the straight line:		Again, the first task is to bring our <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis and all the equations apply in this case too. The deviation from the previous analysis begins on the least squares fit part, where in this case we treat <math>x</math> as the dependent variable and <math>y</math> as the independent variable. The best-fitting straight line to the data, for regression on X (see Chapter [[Parameter Estimation]]), is the straight line:

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

	:where:		where:

	::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(T_{i}^{\prime }) \right]</math>		::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(t_{i}^{\prime }) \right]</math>

	:and:		and:

	::<math>{{x}_{i}}=T_{i}^{\prime }</math>		::<math>{{x}_{i}}=t_{i}^{\prime }</math>

	and the <math>F(T_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, solve ~~Eqn. (xlineln)~~ for the unknown <math>y</math> , which corresponds to:		and the <math>F(t_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, solve the linear equation for the unknown <math>y</math> , which corresponds to:

	::<math>y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x</math>		::<math>y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x</math>

	Solving for the parameters ~~from Eqns. (bln) and (aln)~~ we get:		Solving for the parameters we get:

	::<math>a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{{{\mu }'}}{{{\sigma ~~}_{{{T}~~'~~}}}~~}</math>		::<math>a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{{{\mu }'}}{\sigma'}</math>

	:and:		and:

	::<math>b=\frac{1}{\widehat{b}}=\frac{1}{{{\sigma ~~}_{{{T}~~'~~}}}}\text{~~ }</math>		::<math>b=\frac{1}{\widehat{b}}=\frac{1}{\sigma'}</math>

	The correlation coefficient is evaluated as before using Eqn. (RHOln).		The correlation coefficient is evaluated as before using Eqn. (RHOln).

Harry Guo at 18:17, 13 February 2012

2012-02-13T18:17:34Z

@@ Line 36: / Line 36: @@
 The correlation coefficient is evaluated as before using Eqn. (RHOln).
-====Example 3====
+'''Example 3:'''
-Using the data of Example 2 and assuming a lognormal distribution, estimate the parameters and estimate the correlation coefficient,  <math>\rho </math> , using rank regression on X.
+{{Example: Lognormal Distribution RRX}}

Harry Guo: moved Template:Ld rank regression on x to Template:Lognormal distributionrank regression on x

2012-02-10T23:19:35Z

moved Template:Ld rank regression on x to Template:Lognormal distributionrank regression on x

← Older revision	Revision as of 23:19, 10 February 2012
(No difference)

Nicolette Young: Created page with '===Rank Regression on X=== Performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal de…'

2012-01-04T18:32:48Z

Created page with '===Rank Regression on X=== Performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal de…'

New page

===Rank Regression on X===
Performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the line is minimized.

Again, the first task is to bring our <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis and Eqns. (lnorm), (yln), (aln) and (bln) apply in this case too. The deviation from the previous analysis begins on the least squares fit part, where in this case we treat <math>x</math> as the dependent variable and <math>y</math> 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=\widehat{a}+\widehat{b}y</math>

The corresponding equations for and <math>\widehat{b}</math> are:

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

:where:

::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(T_{i}^{\prime }) \right]</math>

:and:

::<math>{{x}_{i}}=T_{i}^{\prime }</math>

and the <math>F(T_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, solve Eqn. (xlineln) for the unknown <math>y</math> , which corresponds to:

::<math>y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x</math>

Solving for the parameters from Eqns. (bln) and (aln) we get:

::<math>a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{{{\mu }'}}{{{\sigma }_{{{T}'}}}}</math>

:and:

::<math>b=\frac{1}{\widehat{b}}=\frac{1}{{{\sigma }_{{{T}'}}}}\text{ }</math>

The correlation coefficient is evaluated as before using Eqn. (RHOln).

====Example 3====
Using the data of Example 2 and assuming a lognormal distribution, estimate the parameters and estimate the correlation coefficient, <math>\rho </math> , using rank regression on X.

=====Solution to Example 3=====
Table 9.2 constructed in Example 2 applies to this example as well. Using the values in this table we get:

::<math>\begin{align}
& \hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime }{{y}_{i}}-\tfrac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime }\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{14}} \\
& & \\
& \widehat{b}= & \frac{10.4473-(49.2220)(0)/14}{11.3646-{{(0)}^{2}}/14}
\end{align}</math>

:or:

::<math>\widehat{b}=0.9193</math>

:and:

::<math>\hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime }}{14}-\widehat{b}\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}</math>

:or:

::<math>\widehat{a}=\frac{49.2220}{14}-(0.9193)\frac{(0)}{14}=3.5159</math>

Therefore, from Eqn. (blnx):

::<math>{{\sigma }_{{{T}'}}}=\widehat{b}=0.9193</math>

and from Eqn. (alnx):

::<math>{\mu }'=\frac{\widehat{a}}{\widehat{b}}{{\sigma }_{{{T}'}}}=\frac{3.5159}{0.9193}\cdot 0.9193=3.5159</math>

Using Eqns. (mean) and (sdv) we get:

::<math>\overline{T}=\mu =51.3393\text{ hours}</math>

:and:

::<math>{{\sigma }_{T}}=59.1682\text{ hours}.</math>

The correlation coefficient is found using Eqn. (RHOln):

::<math>\widehat{\rho }=0.9754.</math>

Note that the regression on Y analysis is not necessarily the same as the regression on X. The only time when the results of the two regression types are the same (i.e. will yield the same equation for a line) is when the data lie perfectly on a line.

Using Weibull++ , with the Rank Regression on X option, the results are:

[[Image:5folio.png|thumb|center|400px| ]]