Re: [R] Robustness of Segmented Regression Contributed by Muggeo

From: vito muggeo <vmuggeo_at_dssm.unipa.it>
Date: Fri 10 Jun 2005 - 18:10:02 EST

Hi,
sorry for my delay..

In addition to valuable Achim's comments.

As Achim said, you can try different starting values to assess how the final solution depends on them. Then select one having the best logLik (or the minimum RSS).

Everybody dealing with nonlinear models knows that the logLik may be not concave. This is particulary true for broken-line model, so different starting values (psi0) sometimes can lead to different solutions (segmented performs "just" an iterative estimating algorithm..). This sensitivity depends on your data: the more clear-cut the relationship, the stabler the algorithm, i.e. more indipendent of psi0 the estimates are. Of course here, a grid-search is able to fix the problem.

Furthermore comparing the results with a visual inspection of a (possibly smoothed) scatterplot can lead to different location of psi. Usually visual location of changepoints through plots is based on local fitting, while segmented, in its standard usage, performs global fitting (on each side of the range of the explanatory variable Z). Influential points on extreme limits of the Z-range may influence the slopes and then the breakpoint location.

Hope this helps.

vito

Achim Zeileis wrote:

> On Wed, 8 Jun 2005 08:25:16 -0400  Park, Kyong H Mr. RDECOM wrote:
> 
> 

>>Hello, R users,
>>I applied segmented regression method contributed by Muggeo and got
>>different slope estimates depending on the initial break points. The
>>results are listed below and I'd like to know what is a reasonable
>>approach handling this kinds of problem. I think applying various
>>initial break points is certainly not a efficient approach. Is there
>>any other methods to deal with segmented regression? From a graph, v
>>shapes are more clear at 1.2 and 1.5 break points than 1.5 and 1.7.
>>Appreciate your help.
> 
> 
> When you keep the number of break points fixed, then there is a unique
> solution to the problem of fitting a segmented regression: the solution
> which maximizes the likelihood (or for linear models equivalently
> minimizes the RSS). Vito's segmented package gives an iterative method
> which can be shown to converge to this unique solution. If empirically
> you find different solutions with different starting values, you can
> always compare them using the RSS or log-likelihood and choose the one
> which fits better (because the other one can't be the optimal solution).
> 
> The function breakpoints() in package strucchange computes (as
> opposed to approximates) the unique solution for a fully segmented model
> instead of a broken line trend.
> 
> Another nonparametric solution using quantreg was already pointed out by
> Roger.
> 
> hth,
> Z
>  
> 

>>Result1:
>>Initial break points are 1.2 and 1.5. The estimated break points and
>>slopes:
>>
>> Estimated Break-Point(s):
>> Est. St.Err
>>Mean.Vel 1.285 0.05258
>> 1.652 0.01247
>>
>> Est. St.Err. t value
>> CI(95%).l
>>CI(95%).u
>>slope1 0.4248705 0.3027957 1.403159 -0.1685982
>>1.018339 slope2 2.3281445 0.3079903 7.559149 1.7244946
>> 2.931794
>>slope3 9.5425516 0.7554035 12.632390 8.0619879
>>11.023115 Adjusted R-squared: 0.9924.
>>
>>Result2:
>>Initial break points are 1.5 and 1.7. The estimated break points and
>>slopes:
>>
>>Estimated Break-Point(s):
>> Est. St.Err
>>Mean.Vel 1.412 0.02195
>> 1.699 0.01001
>>
>> Est. St.Err. t value
>> CI(95%).l
>>CI(95%).u
>>slope1 0.7300483 0.1381587 5.284129 0.4592623
>>1.000834 slope2 3.4479466 0.2442530 14.116289 2.9692194
>> 3.926674
>>slope3 12.5000000 1.7783840 7.028853 9.0144314
>>15.985569
>>
>>Adjusted R-squared: 0.995.
>>
>>
>>
>>
>> [[alternative HTML version deleted]]
>>
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> 
> 
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-- 
====================================
Vito M.R. Muggeo
Dip.to Sc Statist e Matem `Vianelli'
UniversitÓ di Palermo
viale delle Scienze, edificio 13
90121 Palermo - ITALY
tel: 091 6626240
fax: 091 485726/485612

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Received on Fri Jun 10 18:42:16 2005

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