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outlier

9 messages · kan Liu, Spencer Graves, Duncan Murdoch +1 more

Brian Ripley
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On Tue, 17 Jun 2003, kan Liu wrote:

            

      
      
      
    

        
Easy: you don't.  It make no sense to consider R^2 after arbitrary outlier 
removal: if I remove all but two points I get R^2 = 1!

R^2 is normally used to measure the success of a multiple regression, but 
as you mention two variables, did you just mean the Pearson 
product-moment correlation?  It makes more sense to use a robust measure 
of correlation, as in cov.rob (package lqs) or even Spearman or Kendall 
measures (cov.test in package ctest).

If you intended to do this for a multiple regression, you need to do some 
sort of robust regression and a use a robust measure of fit.

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Spencer Graves
    

    

      
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It is also wise to make scatterplots, as shown by the famous examples 
produced of 4 scatterplots with the same R^2, where the first shows the 
standard ellipsoid pattern implied by the assumptions while the other 
three indicate very clearly that the assumptions are incorrect.  See 
Anscombe (1973) "Graphs in Statistical Analysis", The American 
Statistician, 27: 17-22, reproduced in, e.g., du Toit, Steyn and Stumpf 
(1986) Graphical Exploratory Data Analysis (Springer).

hth.  spencer graves

        
Prof Brian Ripley wrote:

            

      
      
      
    

  
  


  


      

    

    

      
      

        


  

        

      kan Liu
    

    

      
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Hi, many thanks for your advice. I appreciate very
much. Maybe I can make the question more clear: I want
to evaluate the correlation between two variables: one
is the actual outputs of a system, another is the
predicted values of the outputs of the system using
neural networks. When I made scatterplots in excel, I
can get the linear equation and the corresponding
R-squared. In the bottom of the page
http://www.statsoftinc.com/textbook/stathome.html, it
mentioned that sometimes outliers will affect
correlation coefficient biasly. So I thought it might
be worth to remove outlier before  calculating
R-squared in R. It seems to be a bad idea according to
your comments. Now can you make comments on how to
evaluate the performance of the neural network model
in predicting the actual outputs?

Kan

        
--- Spencer Graves <spencer.graves at PDF.COM> wrote:

            

      
      
      
    

        
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      Brian Ripley
    

    

      
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On Tue, 17 Jun 2003, kan Liu wrote:

        

            

      
      
      
    

        
Yes. That's the whole point of robust methods: compensate rather than 
reject.

            

      
      
      
    

        
If you are interested in correlation coefficients, use cov.rob. However,
this is predicted vs actual, and you probably do want to penalize bad
predictions, not reject them.  It's up to you to choose a suitable loss
function for your application.  In particular, if the predicted values
were always 1e-45 times the actual values minus 1e310, the correlation
would be one and the predictions would be derisory.

            

      
      
      
    

        

      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      kan Liu
    

    

      
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which R environmen variable can be used to point to
x1.R so that it can be sourced in any directory?

Kan

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      Duncan Murdoch
    

    

      
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On Wed, 18 Jun 2003 11:09:41 +0100 (BST), you wrote:

        

            

      
      
      
    

        
I don't think there is such a thing.  source() looks in the current
working directory, it doesn't use environment variables to do a wider
search.

Of course, you can always make up your own variable, and then do
something like

source(paste(getenv('MYDIR'),'/x1.R',sep=''))

and that will work from anywhere if MYDIR is defined properly.

Duncan Murdoch

  
  


  


      

    

    

      
      

        


  

        

      kan Liu
    

    

      
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I wrote a .R file (see below)to calculate robust
measure of correlation using cov.rob. I got different
correlation coefficients (0.70, 0.79, 0.63, ...) when
I run the file different times. Can you tell me what
this means or what is wrong in using cov.rob?
-------------

library(lqs)
a <-
c(5.41,4.67,5.88,2.38,4.79,5.30,1.94,3.40,5.05,3.31,5.88,4.92,5.08,4.58,4.59,4.77,5.25,3.77,2.88,5.30,5.32,2.56,4.29,5.54,4.53,3.51,4.93,2.49,2.85,5.04,2.51,2.60,3.58,2.11,1.70,5.20,5.08,4.48,3.96,4.87,4.98,2.56,1.69,4.28,1.70,2.91,5.37,2.16,3.04,1.69,1.88,5.36,1.70,3.81,1.70,5.88,3.52)
p <-
c(5.30,4.78,4.79,0.62,4.32,2.33,0.64,3.14,3.06,4.73,5.72,2.21,4.81,1.74,4.93,4.74,5.81,3.88,3.03,4.72,5.79,3.43,4.07,5.93,2.26,3.70,5.32,4.56,1.52,2.54,0.26,2.79,3.67,4.44,1.46,4.26,4.49,5.29,3.26,3.87,3.12,3.97,3.49,0.45,0.76,4.49,5.29,1.94,4.69,2.80,2.75,5.16,0.74,5.81,1.46,5.24,4.00)
ap <- cbind(a,p)
cov.rob(ap, cor=TRUE)

  
  


  


      

    

    

      
      

        


  

        

      Brian Ripley
    

    

      
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Please do read the help page. which explains this is a random algorithm.
In your example you can try cov.rob(ap, cor=TRUE, nsamp="exact")

        
On Wed, 18 Jun 2003, kan Liu wrote: