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What is the possible range of SVR parameters range?



The Next CEO of Stack Overflow
2019 Community Moderator ElectionSupport vector regression and paremetersChoosing best methods for estimating the unknown parameters in a linear regression modelDoes importance of SVM parameters vary for subsample of data?Difference between rbfnn and svr with gaussian kernelSimulation - identify the input parameters that impact the most the outputGibbs sampling in RUnderstanding Support Verctor Regression (SVR)SVR is giving same prediction for all featuresWhen to use SVR over other regression models










0












$begingroup$


I'm working on a regression problem. While tunning the Parameters of SVR I got the following values c=100, gamma= 10 and epsilon =100. For which I got 95 percent r-square. My question is what is the theoretical range of these parameters values.?










share|improve this question









$endgroup$











  • $begingroup$
    What is $gamma$? Is this the parameter in your basis function?
    $endgroup$
    – MachineLearner
    Mar 22 at 8:44















0












$begingroup$


I'm working on a regression problem. While tunning the Parameters of SVR I got the following values c=100, gamma= 10 and epsilon =100. For which I got 95 percent r-square. My question is what is the theoretical range of these parameters values.?










share|improve this question









$endgroup$











  • $begingroup$
    What is $gamma$? Is this the parameter in your basis function?
    $endgroup$
    – MachineLearner
    Mar 22 at 8:44













0












0








0





$begingroup$


I'm working on a regression problem. While tunning the Parameters of SVR I got the following values c=100, gamma= 10 and epsilon =100. For which I got 95 percent r-square. My question is what is the theoretical range of these parameters values.?










share|improve this question









$endgroup$




I'm working on a regression problem. While tunning the Parameters of SVR I got the following values c=100, gamma= 10 and epsilon =100. For which I got 95 percent r-square. My question is what is the theoretical range of these parameters values.?







python regression svm hyperparameter






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asked Mar 22 at 5:14









imtiaz ul Hassanimtiaz ul Hassan

204




204











  • $begingroup$
    What is $gamma$? Is this the parameter in your basis function?
    $endgroup$
    – MachineLearner
    Mar 22 at 8:44
















  • $begingroup$
    What is $gamma$? Is this the parameter in your basis function?
    $endgroup$
    – MachineLearner
    Mar 22 at 8:44















$begingroup$
What is $gamma$? Is this the parameter in your basis function?
$endgroup$
– MachineLearner
Mar 22 at 8:44




$begingroup$
What is $gamma$? Is this the parameter in your basis function?
$endgroup$
– MachineLearner
Mar 22 at 8:44










1 Answer
1






active

oldest

votes


















1












$begingroup$

I support vector regression the inverse regularization parameter $C$ can be selected from the interval $[0,infty)$. In which $C=0$ means that we are very heavily regularizing and $Cto infty$ no regularization.



The parameter $varepsilon$ is also from the interval $[0,infty)$. In which $varepsilon=0$ forces the regression to penalize every point that is not exactly on the regression line. Whereas $varepsilon > 0$ allows an indifference margin around the regression in which a deviation will not be counted as an error.



Additionally, there are slack variables of $xigeq 0$ and $hatxigeq 0$. These are zero if a point is inside the indifference margin. If a data point lies above and outside the indifference margin we will have $xi>0$ and if a data point lies below and outside the indifference margin we will have $hatxi<0$.



I think you mean the form parameter of your radial basis function when you talk about $gamma$. If we have



$$varphi(boldsymbolx_i,boldsymbolx_j|gamma)=expleft[-gamma||boldsymbolx_i-boldsymbolx_j||^2right]$$



then $gamma in (0,infty)$ (Note the minus sign in front of $gamma$). For $gamma to 0$ we make the kernel flatter as $varphi approx 1$. If $gamma to infty$ we will get a very peaked kernel. Which will be 1 when $boldsymbolx_iapprox boldsymbolx_i$ and almost zero everywhere else.



You should also have a look at the documentation for the implementation of these parameters. It might happen that these parameters are not implemented as you think (see note on $gamma$).






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    I support vector regression the inverse regularization parameter $C$ can be selected from the interval $[0,infty)$. In which $C=0$ means that we are very heavily regularizing and $Cto infty$ no regularization.



    The parameter $varepsilon$ is also from the interval $[0,infty)$. In which $varepsilon=0$ forces the regression to penalize every point that is not exactly on the regression line. Whereas $varepsilon > 0$ allows an indifference margin around the regression in which a deviation will not be counted as an error.



    Additionally, there are slack variables of $xigeq 0$ and $hatxigeq 0$. These are zero if a point is inside the indifference margin. If a data point lies above and outside the indifference margin we will have $xi>0$ and if a data point lies below and outside the indifference margin we will have $hatxi<0$.



    I think you mean the form parameter of your radial basis function when you talk about $gamma$. If we have



    $$varphi(boldsymbolx_i,boldsymbolx_j|gamma)=expleft[-gamma||boldsymbolx_i-boldsymbolx_j||^2right]$$



    then $gamma in (0,infty)$ (Note the minus sign in front of $gamma$). For $gamma to 0$ we make the kernel flatter as $varphi approx 1$. If $gamma to infty$ we will get a very peaked kernel. Which will be 1 when $boldsymbolx_iapprox boldsymbolx_i$ and almost zero everywhere else.



    You should also have a look at the documentation for the implementation of these parameters. It might happen that these parameters are not implemented as you think (see note on $gamma$).






    share|improve this answer









    $endgroup$

















      1












      $begingroup$

      I support vector regression the inverse regularization parameter $C$ can be selected from the interval $[0,infty)$. In which $C=0$ means that we are very heavily regularizing and $Cto infty$ no regularization.



      The parameter $varepsilon$ is also from the interval $[0,infty)$. In which $varepsilon=0$ forces the regression to penalize every point that is not exactly on the regression line. Whereas $varepsilon > 0$ allows an indifference margin around the regression in which a deviation will not be counted as an error.



      Additionally, there are slack variables of $xigeq 0$ and $hatxigeq 0$. These are zero if a point is inside the indifference margin. If a data point lies above and outside the indifference margin we will have $xi>0$ and if a data point lies below and outside the indifference margin we will have $hatxi<0$.



      I think you mean the form parameter of your radial basis function when you talk about $gamma$. If we have



      $$varphi(boldsymbolx_i,boldsymbolx_j|gamma)=expleft[-gamma||boldsymbolx_i-boldsymbolx_j||^2right]$$



      then $gamma in (0,infty)$ (Note the minus sign in front of $gamma$). For $gamma to 0$ we make the kernel flatter as $varphi approx 1$. If $gamma to infty$ we will get a very peaked kernel. Which will be 1 when $boldsymbolx_iapprox boldsymbolx_i$ and almost zero everywhere else.



      You should also have a look at the documentation for the implementation of these parameters. It might happen that these parameters are not implemented as you think (see note on $gamma$).






      share|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        I support vector regression the inverse regularization parameter $C$ can be selected from the interval $[0,infty)$. In which $C=0$ means that we are very heavily regularizing and $Cto infty$ no regularization.



        The parameter $varepsilon$ is also from the interval $[0,infty)$. In which $varepsilon=0$ forces the regression to penalize every point that is not exactly on the regression line. Whereas $varepsilon > 0$ allows an indifference margin around the regression in which a deviation will not be counted as an error.



        Additionally, there are slack variables of $xigeq 0$ and $hatxigeq 0$. These are zero if a point is inside the indifference margin. If a data point lies above and outside the indifference margin we will have $xi>0$ and if a data point lies below and outside the indifference margin we will have $hatxi<0$.



        I think you mean the form parameter of your radial basis function when you talk about $gamma$. If we have



        $$varphi(boldsymbolx_i,boldsymbolx_j|gamma)=expleft[-gamma||boldsymbolx_i-boldsymbolx_j||^2right]$$



        then $gamma in (0,infty)$ (Note the minus sign in front of $gamma$). For $gamma to 0$ we make the kernel flatter as $varphi approx 1$. If $gamma to infty$ we will get a very peaked kernel. Which will be 1 when $boldsymbolx_iapprox boldsymbolx_i$ and almost zero everywhere else.



        You should also have a look at the documentation for the implementation of these parameters. It might happen that these parameters are not implemented as you think (see note on $gamma$).






        share|improve this answer









        $endgroup$



        I support vector regression the inverse regularization parameter $C$ can be selected from the interval $[0,infty)$. In which $C=0$ means that we are very heavily regularizing and $Cto infty$ no regularization.



        The parameter $varepsilon$ is also from the interval $[0,infty)$. In which $varepsilon=0$ forces the regression to penalize every point that is not exactly on the regression line. Whereas $varepsilon > 0$ allows an indifference margin around the regression in which a deviation will not be counted as an error.



        Additionally, there are slack variables of $xigeq 0$ and $hatxigeq 0$. These are zero if a point is inside the indifference margin. If a data point lies above and outside the indifference margin we will have $xi>0$ and if a data point lies below and outside the indifference margin we will have $hatxi<0$.



        I think you mean the form parameter of your radial basis function when you talk about $gamma$. If we have



        $$varphi(boldsymbolx_i,boldsymbolx_j|gamma)=expleft[-gamma||boldsymbolx_i-boldsymbolx_j||^2right]$$



        then $gamma in (0,infty)$ (Note the minus sign in front of $gamma$). For $gamma to 0$ we make the kernel flatter as $varphi approx 1$. If $gamma to infty$ we will get a very peaked kernel. Which will be 1 when $boldsymbolx_iapprox boldsymbolx_i$ and almost zero everywhere else.



        You should also have a look at the documentation for the implementation of these parameters. It might happen that these parameters are not implemented as you think (see note on $gamma$).







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 22 at 9:04









        MachineLearnerMachineLearner

        36910




        36910



























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