[R] Replies of the question about robustness of segmented regression

From: Park, Kyong H Mr. RDECOM <kyong.ho.park_at_us.army.mil>
Date: Fri 10 Jun 2005 - 21:20:37 EST


I appreciate to Roger Koenker, Achim Zeileis and Vito Muggeo for their informative answers. Listed below is unedited replies I got followed by the question I posted.

Kyong

  1. Roger Koenker: You might try rqss() in the quantreg package. It gives piecewise linear fits for a nonparametric form of median regression using total variation of the derivative of the fitted function as a penalty term. A tuning parameter
    (lambda) controls the number of distinct segments. More details are
    available in the vignette for the quantreg package.
  2. Achim Zeileis:

hen 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.

3. Vito Muggeo:

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.

the question:

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.

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.

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