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Adaboost - Show that adjusting weights brings error of current iteration to 0.5



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
2019 Moderator Election Q&A - Questionnaire
2019 Community Moderator Election ResultsAdjusting weights in an convolutional neural networkMy ADALINE model using Gradient Descent is increasing error on each iterationGeneralization Error DefinitionStep-by-step construction of an RBF neural networkWhich learning algorithms to use in what order - dimensionality reduction, bayesian network structure, regression?boosting an xgboost classifier with another xgboost classifier using different sets of featuresSmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?How to methodologically show that a given 'time-series/sequential' data is not really sequential?Updating weights in AdaboostWhat are some possible reasons that your multiclass classifier is classifying alll the classes in a single class?










1












$begingroup$


I'm trying to solve the following problem but I've gotten sort of stuck.



So for adaboost, $err_t = fracsum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i$



and $alpha_t = frac12ln(frac1-err_terr_t)$



Weights for the next iteration are $w_i' = w_i exp(-alpha_t t^(i) h_t(x^(i)))$
and this assumes $t$ and $h_t$ takes on a value of either $-1$ or $+1$.



I have to show that the error with respect to the new weights $w_i'$ is $frac12$.
i.e., $err_t' = fracsum_i=1^Nw_i' Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i' = frac12$



i.e., we use the weak learner of iteration t and evaluate it according to the new weights, which will be used to learn the $t+1$-st weak learner.



I simplified it so that $w_i'=w_i sqrtfracerr_t1-err_t$ if $w_i$ was correctly classified and $w_i'=w_i sqrtfrac1-err_terr_t$ if $w_i$ was incorrectly classified. I then tried plugging this into the equation for $err_t'=frac12$ and got $fracerr_t1-err_t fracsum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i)) = 1$ but at this point I sort of ran into a dead end and so I'm wondering how one might show the original question.



Thanks for any help!










share|improve this question











$endgroup$
















    1












    $begingroup$


    I'm trying to solve the following problem but I've gotten sort of stuck.



    So for adaboost, $err_t = fracsum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i$



    and $alpha_t = frac12ln(frac1-err_terr_t)$



    Weights for the next iteration are $w_i' = w_i exp(-alpha_t t^(i) h_t(x^(i)))$
    and this assumes $t$ and $h_t$ takes on a value of either $-1$ or $+1$.



    I have to show that the error with respect to the new weights $w_i'$ is $frac12$.
    i.e., $err_t' = fracsum_i=1^Nw_i' Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i' = frac12$



    i.e., we use the weak learner of iteration t and evaluate it according to the new weights, which will be used to learn the $t+1$-st weak learner.



    I simplified it so that $w_i'=w_i sqrtfracerr_t1-err_t$ if $w_i$ was correctly classified and $w_i'=w_i sqrtfrac1-err_terr_t$ if $w_i$ was incorrectly classified. I then tried plugging this into the equation for $err_t'=frac12$ and got $fracerr_t1-err_t fracsum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i)) = 1$ but at this point I sort of ran into a dead end and so I'm wondering how one might show the original question.



    Thanks for any help!










    share|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      I'm trying to solve the following problem but I've gotten sort of stuck.



      So for adaboost, $err_t = fracsum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i$



      and $alpha_t = frac12ln(frac1-err_terr_t)$



      Weights for the next iteration are $w_i' = w_i exp(-alpha_t t^(i) h_t(x^(i)))$
      and this assumes $t$ and $h_t$ takes on a value of either $-1$ or $+1$.



      I have to show that the error with respect to the new weights $w_i'$ is $frac12$.
      i.e., $err_t' = fracsum_i=1^Nw_i' Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i' = frac12$



      i.e., we use the weak learner of iteration t and evaluate it according to the new weights, which will be used to learn the $t+1$-st weak learner.



      I simplified it so that $w_i'=w_i sqrtfracerr_t1-err_t$ if $w_i$ was correctly classified and $w_i'=w_i sqrtfrac1-err_terr_t$ if $w_i$ was incorrectly classified. I then tried plugging this into the equation for $err_t'=frac12$ and got $fracerr_t1-err_t fracsum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i)) = 1$ but at this point I sort of ran into a dead end and so I'm wondering how one might show the original question.



      Thanks for any help!










      share|improve this question











      $endgroup$




      I'm trying to solve the following problem but I've gotten sort of stuck.



      So for adaboost, $err_t = fracsum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i$



      and $alpha_t = frac12ln(frac1-err_terr_t)$



      Weights for the next iteration are $w_i' = w_i exp(-alpha_t t^(i) h_t(x^(i)))$
      and this assumes $t$ and $h_t$ takes on a value of either $-1$ or $+1$.



      I have to show that the error with respect to the new weights $w_i'$ is $frac12$.
      i.e., $err_t' = fracsum_i=1^Nw_i' Pi (h_t(x^(i)) neq t^(i))sum_i=1^Nw_i' = frac12$



      i.e., we use the weak learner of iteration t and evaluate it according to the new weights, which will be used to learn the $t+1$-st weak learner.



      I simplified it so that $w_i'=w_i sqrtfracerr_t1-err_t$ if $w_i$ was correctly classified and $w_i'=w_i sqrtfrac1-err_terr_t$ if $w_i$ was incorrectly classified. I then tried plugging this into the equation for $err_t'=frac12$ and got $fracerr_t1-err_t fracsum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i)) = 1$ but at this point I sort of ran into a dead end and so I'm wondering how one might show the original question.



      Thanks for any help!







      machine-learning algorithms boosting






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Feb 1 at 23:04







      Saad Hussain

















      asked Feb 1 at 21:54









      Saad HussainSaad Hussain

      62




      62




















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          You're nearly there. The quantity $err_t/(1-err_t)$ is exactly what you need it to be. It might be easier to see if you think about
          $$sum_i=1^N w_i Pi(h_t(x^(i))=t^(i))$$
          as
          $$sum_i: x^(i)text is correctly classified w_i$$
          (just using the indicator function to reduce the summation range).






          share|improve this answer









          $endgroup$




















            0












            $begingroup$

            For simplicity, lets define some variables as follows:



            $W_C := sum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))$



            $W_I := sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))$



            Therefore, $err_t = W_I/(W_C+W_I)$



            $a := sqrtfracerr_t1-err_t = sqrtfracW_IW_C$



            Now, for new weights we have



            $W'_C := sum_i=1^Nw'_i Pi (h_t(x^(i)) = t^(i)) = aW_C$



            $W'_I := sum_i=1^Nw'_i Pi (h_t(x^(i)) neq t^(i)) = (1/a)W_I$



            Now, as the final step:



            $err'_t = fracW'_IW'_I + W'_C = frac(1/a)W_I(1/a)W_I + aW_C oversettimes a= fracW_IW_I + a^2W_C = fracW_IW_I + fracW_IW_CW_C=frac12$






            share|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

              votes









              0












              $begingroup$

              You're nearly there. The quantity $err_t/(1-err_t)$ is exactly what you need it to be. It might be easier to see if you think about
              $$sum_i=1^N w_i Pi(h_t(x^(i))=t^(i))$$
              as
              $$sum_i: x^(i)text is correctly classified w_i$$
              (just using the indicator function to reduce the summation range).






              share|improve this answer









              $endgroup$

















                0












                $begingroup$

                You're nearly there. The quantity $err_t/(1-err_t)$ is exactly what you need it to be. It might be easier to see if you think about
                $$sum_i=1^N w_i Pi(h_t(x^(i))=t^(i))$$
                as
                $$sum_i: x^(i)text is correctly classified w_i$$
                (just using the indicator function to reduce the summation range).






                share|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  You're nearly there. The quantity $err_t/(1-err_t)$ is exactly what you need it to be. It might be easier to see if you think about
                  $$sum_i=1^N w_i Pi(h_t(x^(i))=t^(i))$$
                  as
                  $$sum_i: x^(i)text is correctly classified w_i$$
                  (just using the indicator function to reduce the summation range).






                  share|improve this answer









                  $endgroup$



                  You're nearly there. The quantity $err_t/(1-err_t)$ is exactly what you need it to be. It might be easier to see if you think about
                  $$sum_i=1^N w_i Pi(h_t(x^(i))=t^(i))$$
                  as
                  $$sum_i: x^(i)text is correctly classified w_i$$
                  (just using the indicator function to reduce the summation range).







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Feb 4 at 2:33









                  Ben ReinigerBen Reiniger

                  458212




                  458212





















                      0












                      $begingroup$

                      For simplicity, lets define some variables as follows:



                      $W_C := sum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))$



                      $W_I := sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))$



                      Therefore, $err_t = W_I/(W_C+W_I)$



                      $a := sqrtfracerr_t1-err_t = sqrtfracW_IW_C$



                      Now, for new weights we have



                      $W'_C := sum_i=1^Nw'_i Pi (h_t(x^(i)) = t^(i)) = aW_C$



                      $W'_I := sum_i=1^Nw'_i Pi (h_t(x^(i)) neq t^(i)) = (1/a)W_I$



                      Now, as the final step:



                      $err'_t = fracW'_IW'_I + W'_C = frac(1/a)W_I(1/a)W_I + aW_C oversettimes a= fracW_IW_I + a^2W_C = fracW_IW_I + fracW_IW_CW_C=frac12$






                      share|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        For simplicity, lets define some variables as follows:



                        $W_C := sum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))$



                        $W_I := sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))$



                        Therefore, $err_t = W_I/(W_C+W_I)$



                        $a := sqrtfracerr_t1-err_t = sqrtfracW_IW_C$



                        Now, for new weights we have



                        $W'_C := sum_i=1^Nw'_i Pi (h_t(x^(i)) = t^(i)) = aW_C$



                        $W'_I := sum_i=1^Nw'_i Pi (h_t(x^(i)) neq t^(i)) = (1/a)W_I$



                        Now, as the final step:



                        $err'_t = fracW'_IW'_I + W'_C = frac(1/a)W_I(1/a)W_I + aW_C oversettimes a= fracW_IW_I + a^2W_C = fracW_IW_I + fracW_IW_CW_C=frac12$






                        share|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          For simplicity, lets define some variables as follows:



                          $W_C := sum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))$



                          $W_I := sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))$



                          Therefore, $err_t = W_I/(W_C+W_I)$



                          $a := sqrtfracerr_t1-err_t = sqrtfracW_IW_C$



                          Now, for new weights we have



                          $W'_C := sum_i=1^Nw'_i Pi (h_t(x^(i)) = t^(i)) = aW_C$



                          $W'_I := sum_i=1^Nw'_i Pi (h_t(x^(i)) neq t^(i)) = (1/a)W_I$



                          Now, as the final step:



                          $err'_t = fracW'_IW'_I + W'_C = frac(1/a)W_I(1/a)W_I + aW_C oversettimes a= fracW_IW_I + a^2W_C = fracW_IW_I + fracW_IW_CW_C=frac12$






                          share|improve this answer









                          $endgroup$



                          For simplicity, lets define some variables as follows:



                          $W_C := sum_i=1^Nw_i Pi (h_t(x^(i)) = t^(i))$



                          $W_I := sum_i=1^Nw_i Pi (h_t(x^(i)) neq t^(i))$



                          Therefore, $err_t = W_I/(W_C+W_I)$



                          $a := sqrtfracerr_t1-err_t = sqrtfracW_IW_C$



                          Now, for new weights we have



                          $W'_C := sum_i=1^Nw'_i Pi (h_t(x^(i)) = t^(i)) = aW_C$



                          $W'_I := sum_i=1^Nw'_i Pi (h_t(x^(i)) neq t^(i)) = (1/a)W_I$



                          Now, as the final step:



                          $err'_t = fracW'_IW'_I + W'_C = frac(1/a)W_I(1/a)W_I + aW_C oversettimes a= fracW_IW_I + a^2W_C = fracW_IW_I + fracW_IW_CW_C=frac12$







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Mar 6 at 14:57









                          EsmailianEsmailian

                          3,771420




                          3,771420



























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