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An ambiguity in SVM equations about misclassified data



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Moderator Election Q&A - Questionnaire
2019 Community Moderator Election ResultsHow to apply AdaBoost to more “complex” (non-binary) classifications/data fitting?Does importance of SVM parameters vary for subsample of data?Understanding the math behind SVMDeriving backpropagation equations “natively” in tensor formAn formula derivation question about SMO algorithm of SVMSVM on sparse dataOne Class SVM for time series dataSVDD vs once Class SVMDoubt with SVM mathImplementing SVM from scratch?










3












$begingroup$


I have encountered an ambiguity in SVM equations.
As is stated in Chris Bishop's machine learning book, the optimization goal in SVM is to maximize this function:



$$Csumlimits_n = 1^N xi _n + 1 over 2left$$



Subject to this constraints(*):



$$xi _n ge 0$$
$$t_ny(x_n) ge 1 - xi _n$$



where:



$$y(x_n) = w^Tx_n + b$$



so the corresponding Lagrangian function for this problem is:



$$L(w,b,a) = Csumlimits_n = 1^N xi _n + 1 over 2left - sumlimits_n = 1^N a_n t_ny(x_n) - 1 + xi _n - sumlimits_n = 1^N mu _nxi _n $$



and the corresponding KKT conditions are given by (**):



$$a _n ge 0$$



$$t_ny(x_n) - 1 + xi _n ge 0$$



$$a_n(t_ny(x_n) - 1 + xi _n) = 0$$



$$xi _n ge 0$$



$$mu _n ge 0$$



$$mu _nxi _n = 0$$



And if we set



$$partial L over partial xi _n = 0$$



we get (***)



$$a_n = C - mu _n$$



As we know, that subset of data points that have



$$a_n = 0$$



are not support vectors. But for this data points we have (from ***):



$$mu_n = C$$



and therefore (from **)



$$xi _n = 0$$



So here lies the problem. If a data point from this subset is in the wrong side of the decision boundary, then



$$t_ny(x_n) le 0$$



and we will have (from *)
$$xi _n ge 1$$



which is in an obvious conflict with



$$xi _n = 0$$










share|improve this question











$endgroup$
















    3












    $begingroup$


    I have encountered an ambiguity in SVM equations.
    As is stated in Chris Bishop's machine learning book, the optimization goal in SVM is to maximize this function:



    $$Csumlimits_n = 1^N xi _n + 1 over 2left$$



    Subject to this constraints(*):



    $$xi _n ge 0$$
    $$t_ny(x_n) ge 1 - xi _n$$



    where:



    $$y(x_n) = w^Tx_n + b$$



    so the corresponding Lagrangian function for this problem is:



    $$L(w,b,a) = Csumlimits_n = 1^N xi _n + 1 over 2left - sumlimits_n = 1^N a_n t_ny(x_n) - 1 + xi _n - sumlimits_n = 1^N mu _nxi _n $$



    and the corresponding KKT conditions are given by (**):



    $$a _n ge 0$$



    $$t_ny(x_n) - 1 + xi _n ge 0$$



    $$a_n(t_ny(x_n) - 1 + xi _n) = 0$$



    $$xi _n ge 0$$



    $$mu _n ge 0$$



    $$mu _nxi _n = 0$$



    And if we set



    $$partial L over partial xi _n = 0$$



    we get (***)



    $$a_n = C - mu _n$$



    As we know, that subset of data points that have



    $$a_n = 0$$



    are not support vectors. But for this data points we have (from ***):



    $$mu_n = C$$



    and therefore (from **)



    $$xi _n = 0$$



    So here lies the problem. If a data point from this subset is in the wrong side of the decision boundary, then



    $$t_ny(x_n) le 0$$



    and we will have (from *)
    $$xi _n ge 1$$



    which is in an obvious conflict with



    $$xi _n = 0$$










    share|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      I have encountered an ambiguity in SVM equations.
      As is stated in Chris Bishop's machine learning book, the optimization goal in SVM is to maximize this function:



      $$Csumlimits_n = 1^N xi _n + 1 over 2left$$



      Subject to this constraints(*):



      $$xi _n ge 0$$
      $$t_ny(x_n) ge 1 - xi _n$$



      where:



      $$y(x_n) = w^Tx_n + b$$



      so the corresponding Lagrangian function for this problem is:



      $$L(w,b,a) = Csumlimits_n = 1^N xi _n + 1 over 2left - sumlimits_n = 1^N a_n t_ny(x_n) - 1 + xi _n - sumlimits_n = 1^N mu _nxi _n $$



      and the corresponding KKT conditions are given by (**):



      $$a _n ge 0$$



      $$t_ny(x_n) - 1 + xi _n ge 0$$



      $$a_n(t_ny(x_n) - 1 + xi _n) = 0$$



      $$xi _n ge 0$$



      $$mu _n ge 0$$



      $$mu _nxi _n = 0$$



      And if we set



      $$partial L over partial xi _n = 0$$



      we get (***)



      $$a_n = C - mu _n$$



      As we know, that subset of data points that have



      $$a_n = 0$$



      are not support vectors. But for this data points we have (from ***):



      $$mu_n = C$$



      and therefore (from **)



      $$xi _n = 0$$



      So here lies the problem. If a data point from this subset is in the wrong side of the decision boundary, then



      $$t_ny(x_n) le 0$$



      and we will have (from *)
      $$xi _n ge 1$$



      which is in an obvious conflict with



      $$xi _n = 0$$










      share|improve this question











      $endgroup$




      I have encountered an ambiguity in SVM equations.
      As is stated in Chris Bishop's machine learning book, the optimization goal in SVM is to maximize this function:



      $$Csumlimits_n = 1^N xi _n + 1 over 2left$$



      Subject to this constraints(*):



      $$xi _n ge 0$$
      $$t_ny(x_n) ge 1 - xi _n$$



      where:



      $$y(x_n) = w^Tx_n + b$$



      so the corresponding Lagrangian function for this problem is:



      $$L(w,b,a) = Csumlimits_n = 1^N xi _n + 1 over 2left - sumlimits_n = 1^N a_n t_ny(x_n) - 1 + xi _n - sumlimits_n = 1^N mu _nxi _n $$



      and the corresponding KKT conditions are given by (**):



      $$a _n ge 0$$



      $$t_ny(x_n) - 1 + xi _n ge 0$$



      $$a_n(t_ny(x_n) - 1 + xi _n) = 0$$



      $$xi _n ge 0$$



      $$mu _n ge 0$$



      $$mu _nxi _n = 0$$



      And if we set



      $$partial L over partial xi _n = 0$$



      we get (***)



      $$a_n = C - mu _n$$



      As we know, that subset of data points that have



      $$a_n = 0$$



      are not support vectors. But for this data points we have (from ***):



      $$mu_n = C$$



      and therefore (from **)



      $$xi _n = 0$$



      So here lies the problem. If a data point from this subset is in the wrong side of the decision boundary, then



      $$t_ny(x_n) le 0$$



      and we will have (from *)
      $$xi _n ge 1$$



      which is in an obvious conflict with



      $$xi _n = 0$$







      machine-learning svm theory






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Apr 1 at 9:12









      Esmailian

      3,191320




      3,191320










      asked Mar 31 at 20:08









      pythinkerpythinker

      8291213




      8291213




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Good point! Interesting consequence!



          Problem is the $a_n=0$ assumption, i.e. assuming misclassified points are not support vectors.



          Here is the flow. Slack variable $xi_n$ is defined as
          $$xi_n := |t_n - y(boldsymbolx_n)|$$
          where $t_n in +1, -1$ is the true label, and $y(boldsymbolx_n)$ is the prediction. Therefore, for a misclassified point (on the wrong side) we have $$xi_n > 1$$ by definition. Given $mu_n xi_n = 0$, therefore$$mu_n=0$$
          and given $a_n=C - mu_n$, therefore $$a_n = C > 0$$ which means (given $a_n > 0$ only for support vectors)




          Every misclassified point is a support vector.




          This is a nice consequence and should have been stated in the book.



          Although, a remotely! related point has been stated in the book:




          Points with $a_n = C$ can lie inside the margin and can either be
          correctly classified if $xi_n leq 1$ or misclassified if $xi_n > 1$.







          share|improve this answer











          $endgroup$













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            1 Answer
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            active

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            active

            oldest

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            1












            $begingroup$

            Good point! Interesting consequence!



            Problem is the $a_n=0$ assumption, i.e. assuming misclassified points are not support vectors.



            Here is the flow. Slack variable $xi_n$ is defined as
            $$xi_n := |t_n - y(boldsymbolx_n)|$$
            where $t_n in +1, -1$ is the true label, and $y(boldsymbolx_n)$ is the prediction. Therefore, for a misclassified point (on the wrong side) we have $$xi_n > 1$$ by definition. Given $mu_n xi_n = 0$, therefore$$mu_n=0$$
            and given $a_n=C - mu_n$, therefore $$a_n = C > 0$$ which means (given $a_n > 0$ only for support vectors)




            Every misclassified point is a support vector.




            This is a nice consequence and should have been stated in the book.



            Although, a remotely! related point has been stated in the book:




            Points with $a_n = C$ can lie inside the margin and can either be
            correctly classified if $xi_n leq 1$ or misclassified if $xi_n > 1$.







            share|improve this answer











            $endgroup$

















              1












              $begingroup$

              Good point! Interesting consequence!



              Problem is the $a_n=0$ assumption, i.e. assuming misclassified points are not support vectors.



              Here is the flow. Slack variable $xi_n$ is defined as
              $$xi_n := |t_n - y(boldsymbolx_n)|$$
              where $t_n in +1, -1$ is the true label, and $y(boldsymbolx_n)$ is the prediction. Therefore, for a misclassified point (on the wrong side) we have $$xi_n > 1$$ by definition. Given $mu_n xi_n = 0$, therefore$$mu_n=0$$
              and given $a_n=C - mu_n$, therefore $$a_n = C > 0$$ which means (given $a_n > 0$ only for support vectors)




              Every misclassified point is a support vector.




              This is a nice consequence and should have been stated in the book.



              Although, a remotely! related point has been stated in the book:




              Points with $a_n = C$ can lie inside the margin and can either be
              correctly classified if $xi_n leq 1$ or misclassified if $xi_n > 1$.







              share|improve this answer











              $endgroup$















                1












                1








                1





                $begingroup$

                Good point! Interesting consequence!



                Problem is the $a_n=0$ assumption, i.e. assuming misclassified points are not support vectors.



                Here is the flow. Slack variable $xi_n$ is defined as
                $$xi_n := |t_n - y(boldsymbolx_n)|$$
                where $t_n in +1, -1$ is the true label, and $y(boldsymbolx_n)$ is the prediction. Therefore, for a misclassified point (on the wrong side) we have $$xi_n > 1$$ by definition. Given $mu_n xi_n = 0$, therefore$$mu_n=0$$
                and given $a_n=C - mu_n$, therefore $$a_n = C > 0$$ which means (given $a_n > 0$ only for support vectors)




                Every misclassified point is a support vector.




                This is a nice consequence and should have been stated in the book.



                Although, a remotely! related point has been stated in the book:




                Points with $a_n = C$ can lie inside the margin and can either be
                correctly classified if $xi_n leq 1$ or misclassified if $xi_n > 1$.







                share|improve this answer











                $endgroup$



                Good point! Interesting consequence!



                Problem is the $a_n=0$ assumption, i.e. assuming misclassified points are not support vectors.



                Here is the flow. Slack variable $xi_n$ is defined as
                $$xi_n := |t_n - y(boldsymbolx_n)|$$
                where $t_n in +1, -1$ is the true label, and $y(boldsymbolx_n)$ is the prediction. Therefore, for a misclassified point (on the wrong side) we have $$xi_n > 1$$ by definition. Given $mu_n xi_n = 0$, therefore$$mu_n=0$$
                and given $a_n=C - mu_n$, therefore $$a_n = C > 0$$ which means (given $a_n > 0$ only for support vectors)




                Every misclassified point is a support vector.




                This is a nice consequence and should have been stated in the book.



                Although, a remotely! related point has been stated in the book:




                Points with $a_n = C$ can lie inside the margin and can either be
                correctly classified if $xi_n leq 1$ or misclassified if $xi_n > 1$.








                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Mar 31 at 22:33

























                answered Mar 31 at 21:38









                EsmailianEsmailian

                3,191320




                3,191320



























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