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

Lisa Hacker: Redirected page to The Exponential Distribution

2012-08-10T08:19:34Z

Redirected page to The Exponential Distribution

← Older revision		Revision as of 08:19, 10 August 2012
Line 1:		Line 1:
	~~===Rank Regression on X for Exponential Distribution===~~		#REDIRECT [[The Exponential Distribution]]
	Similar to rank regression on Y, 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 exponential <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis. The deviation from the previous analysis begins on the least squares fit step, since 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=\hat{a}+\hat{b}y</math>~~

	~~The corresponding equations for <math>\hat{a}</math> and <math>\hat{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}}=\ln [1-F({{t}_{i}})]</math>~~

	~~and:~~

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


	The values of <math>F({{t}_{i}})</math> are estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, solve for the unknown <math>y</math> value, which corresponds to:

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

	~~Solving for the parameters from above equations we get:~~

	~~::<math>a=-\frac{\hat{a}}{\hat{b}}=\lambda \gamma \Rightarrow \gamma =\hat{a}</math>~~

	~~and:~~

	~~::<math>b=\frac{1}{\hat{b}}=-\lambda \Rightarrow \lambda =-\frac{1}{\hat{b}}</math>~~

	~~For the one-parameter exponential case, 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 }}}\,y_{i}^{2}}~~
	~~\end{align}</math>~~


	The ~~correlation coefficient is evaluated as before.~~


	~~'''Example 3:'''~~
	~~{{Example: 2 Parameter~~ Exponential Distribution ~~RRX}}~~

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

2012-02-10T23:01:43Z

Rank Regression on X for Exponential Distribution

← Older revision		Revision as of 23:01, 10 February 2012
Line 2:		Line 2:
	Similar to rank regression on Y, 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.		Similar to rank regression on Y, 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 exponential <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis. The deviation from the previous analysis begins on the least squares fit step, since 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 ([[Parameter Estimation~~\|see Chapter 4)~~]], is the straight line:		Again the first task is to bring our exponential <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis. The deviation from the previous analysis begins on the least squares fit step, since 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=\hat{a}+\hat{b}y</math>		::<math>x=\hat{a}+\hat{b}y</math>

Harry Guo at 18:08, 9 February 2012

2012-02-09T18:08:31Z

← Older revision		Revision as of 18:08, 9 February 2012
Line 47:		Line 47:

	'''Example 3:'''		'''Example 3:'''
	{{~~Rank Regression on X for Exponential Distribution}}~~		{{Example: 2 Parameter Exponential Distribution RRX}}

	~~'''~~2 Parameter Exponential Distribution RRX~~'''~~

	Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate, <math>\hat{\rho }</math>, using rank regression on X.

	~~''' Solution to Example 3'''~~

	~~Table 7.2 constructed in Example 2 applies to this example also. Using the values from this table, we get:~~

	~~::<math>\begin{align}~~
	\hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}\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}} \\
	\\
	~~\hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{22.1148-{{(-13.2315)}^{2}}/14}~~
	~~\end{align}</math>~~

	~~or:~~

	~~::<math>\hat{b}=-34.5563</math>~~

	~~and:~~

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

	~~or:~~

	~~::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>~~

	~~Therefore:~~

	~~::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>~~

	~~and:~~

	~~::<math>\hat{\gamma }=\hat{a}=12.3406</math>~~

	~~The correlation coefficient is found to be:~~

	~~::<math>\hat{\rho }=-0.9679</math>~~

	Note that the equation for regression on Y is not necessarily the same as that for the regression on X. The only time when the two regression methods yield identical results is when the data lie perfectly on a line. If this were the case, the correlation coefficient would be <math>-1</math>. The negative value of the correlation coefficient is due to the fact that the slope of the exponential probability plot is negative.

	~~This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on X (RRX) methods for analysis, as shown below.~~
	~~The estimated parameters and the correlation coefficient using Weibull++ were found to be:~~

	~~::<math>\begin{array~~}~~{*{35~~}~~{l}}~~
	~~\hat{\lambda }= &0.0289 \text{failures/hour} \\~~
	~~\hat{\gamma}= & 12.3395 \text{hours} \\~~
	~~\hat{\rho} = &-0.9679 \\~~
	~~\end{array}</math>~~


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


	~~The probability plot can be obtained simply by clicking the '''Plot''' icon.~~


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

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

2012-02-09T18:07:05Z

Rank Regression on X for Exponential Distribution

← Older revision		Revision as of 18:07, 9 February 2012
Line 44:		Line 44:

	The correlation coefficient is evaluated as before.		The correlation coefficient is evaluated as before.


			'''Example 3:'''
			{{Rank Regression on X for Exponential Distribution}}

	'''2 Parameter Exponential Distribution RRX'''		'''2 Parameter Exponential Distribution RRX'''

Harry Guo: /* Example 3: 2 Parameter Exponential Distribution RRX */

2012-02-09T18:02:42Z

Example 3: 2 Parameter Exponential Distribution RRX

← Older revision		Revision as of 18:02, 9 February 2012
Line 45:		Line 45:
	The correlation coefficient is evaluated as before.		The correlation coefficient is evaluated as before.

	~~====Example 3:~~ 2 Parameter Exponential Distribution RRX~~====~~		'''2 Parameter Exponential Distribution RRX'''

	Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate, <math>\hat{\rho }</math>, using rank regression on X.		Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate, <math>\hat{\rho }</math>, using rank regression on X.

Harry Guo: /* Solution to Example 3 */

2012-02-08T22:13:50Z

Solution to Example 3

← Older revision		Revision as of 22:13, 8 February 2012
Line 48:		Line 48:
	Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate, <math>\hat{\rho }</math>, using rank regression on X.		Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate, <math>\hat{\rho }</math>, using rank regression on X.

	~~=====~~ Solution to Example 3~~=====~~		''' Solution to Example 3'''

	Table 7.2 constructed in Example 2 applies to this example also. Using the values from this table, we get:		Table 7.2 constructed in Example 2 applies to this example also. Using the values from this table, we get:

Harry Guo at 01:10, 8 February 2012

2012-02-08T01:10:44Z

← Older revision		Revision as of 01:10, 8 February 2012
Line 63:		Line 63:
	and:		and:

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

	or:		or:

Harry Guo: /* Solution to Example 3 */

2012-02-08T01:09:51Z

Solution to Example 3

← Older revision		Revision as of 01:09, 8 February 2012
Line 52:		Line 52:

	::<math>\begin{align}		::<math>\begin{align}
	\hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}}\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}} \\		\hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}\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}} \\
	\\		\\
	\hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{22.1148-{{(-13.2315)}^{2}}/14}		\hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{22.1148-{{(-13.2315)}^{2}}/14}

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

2012-02-07T23:33:08Z

Rank Regression on X for Exponential Distribution

← Older revision		Revision as of 23:33, 7 February 2012
Line 2:		Line 2:
	Similar to rank regression on Y, 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.		Similar to rank regression on Y, 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 exponential <math>cdf</math> function~~, Eqn. (Fe),~~ into a linear form. This step is exactly the same as in regression on Y analysis ~~and Eqns. (loge), (ye), (ae), and (be) again apply in this case~~. The deviation from the previous analysis begins on the least squares fit step, since 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 exponential <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis. The deviation from the previous analysis begins on the least squares fit step, since 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 ([[Parameter Estimation\|see Chapter 4)]], is the straight line:

	::<math>x=\hat{a}+\hat{b}y</math>		::<math>x=\hat{a}+\hat{b}y</math>
Line 16:		Line 16:
	where:		where:

	::<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>


	The values of <math>F({{T}_{i}})</math> are estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, solve ~~Eqn. (xline)~~ for the unknown <math>y</math> value, which corresponds to:		The values of <math>F({{t}_{i}})</math> are estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, solve for the unknown <math>y</math> value, which corresponds to:

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

	Solving for the parameters from ~~Eqns. (ae) and (be)~~ we get:		Solving for the parameters from above equations we get:

	::<math>a=-\frac{\hat{a}}{\hat{b}}=\lambda \gamma \Rightarrow \gamma =\hat{a}</math>		::<math>a=-\frac{\hat{a}}{\hat{b}}=\lambda \gamma \Rightarrow \gamma =\hat{a}</math>
Line 35:		Line 35:
	::<math>b=\frac{1}{\hat{b}}=-\lambda \Rightarrow \lambda =-\frac{1}{\hat{b}}</math>		::<math>b=\frac{1}{\hat{b}}=-\lambda \Rightarrow \lambda =-\frac{1}{\hat{b}}</math>

	For the one-parameter exponential case, ~~Eqns. (aaexz)~~ and ~~(bbexz)~~ become,		For the one-parameter exponential case, equations for estimating a and b become:

	::<math>\begin{align}		::<math>\begin{align}
Line 43:		Line 43:


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

	====Example 3: 2 Parameter Exponential Distribution RRX====		====Example 3: 2 Parameter Exponential Distribution RRX====
Line 69:		Line 69:
	::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>		::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>

	Therefore~~, from Eqn. (bex)~~:		Therefore:

	::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>		::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>

	and ~~from Eqn. (aex)~~:		and:

	::<math>\hat{\gamma }=\hat{a}=12.3406</math>		::<math>\hat{\gamma }=\hat{a}=12.3406</math>

	:The correlation coefficient is found ~~using Eqn. (RHOe)~~:		The correlation coefficient is found to be:

	::<math>\hat{\rho }=-0.9679</math>		::<math>\hat{\rho }=-0.9679</math>

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

2012-02-07T23:04:07Z

Rank Regression on X for Exponential Distribution

← Older revision		Revision as of 23:04, 7 February 2012
Line 10:		Line 10:
	::<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}}=\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>
Line 31:		Line 31:
	::<math>a=-\frac{\hat{a}}{\hat{b}}=\lambda \gamma \Rightarrow \gamma =\hat{a}</math>		::<math>a=-\frac{\hat{a}}{\hat{b}}=\lambda \gamma \Rightarrow \gamma =\hat{a}</math>

	:and:		and:

	::<math>b=\frac{1}{\hat{b}}=-\lambda \Rightarrow \lambda =-\frac{1}{\hat{b}}</math>		::<math>b=\frac{1}{\hat{b}}=-\lambda \Rightarrow \lambda =-\frac{1}{\hat{b}}</math>
Line 57:		Line 57:
	\end{align}</math>		\end{align}</math>

	:or:		or:

	::<math>\hat{b}=-34.5563</math>		::<math>\hat{b}=-34.5563</math>

	:and:		and:

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

	:or:		or:

	::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>		::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>

	:Therefore, from Eqn. (bex):		Therefore, from Eqn. (bex):

	::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>		::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>

	:and from Eqn. (aex):		and from Eqn. (aex):

	::<math>\hat{\gamma }=\hat{a}=12.3406</math>		::<math>\hat{\gamma }=\hat{a}=12.3406</math>