Conditions for two optimization problems to yield the same solution












4












$begingroup$



Problem: Consider the optimization problems
$$min_beta |y-Xbeta|^2+alpha|beta|^2 tag 1$$
and
$$min_beta |beta|^2 text{ subject to } |y-Xbeta|^2 le c tag 2$$
where $|x|$ is the $2$-norm. Fix $alpha$, and suppose $beta^*$ is the solution to ($1$), and let $c=|y-Xbeta^*|^2$. Is it true that the solution to ($2$) is also $beta^*$?




Attempt: I believe this is true. The argument should be very similar to the one in Why are additional constraint and penalty term equivalent in ridge regression?. However, I was running some numerical experiments and it turns out the two problems have different solutions. Hence my question here: are the two problems really yielding the same solutions? Are there exceptions that I should be careful of?






analysis optimization numerical-methods convex-optimization regression







share|cite|improve this question











$endgroup$

















    4












    $begingroup$



    Problem: Consider the optimization problems
    $$min_beta |y-Xbeta|^2+alpha|beta|^2 tag 1$$
    and
    $$min_beta |beta|^2 text{ subject to } |y-Xbeta|^2 le c tag 2$$
    where $|x|$ is the $2$-norm. Fix $alpha$, and suppose $beta^*$ is the solution to ($1$), and let $c=|y-Xbeta^*|^2$. Is it true that the solution to ($2$) is also $beta^*$?




    Attempt: I believe this is true. The argument should be very similar to the one in Why are additional constraint and penalty term equivalent in ridge regression?. However, I was running some numerical experiments and it turns out the two problems have different solutions. Hence my question here: are the two problems really yielding the same solutions? Are there exceptions that I should be careful of?






    analysis optimization numerical-methods convex-optimization regression







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$



      Problem: Consider the optimization problems
      $$min_beta |y-Xbeta|^2+alpha|beta|^2 tag 1$$
      and
      $$min_beta |beta|^2 text{ subject to } |y-Xbeta|^2 le c tag 2$$
      where $|x|$ is the $2$-norm. Fix $alpha$, and suppose $beta^*$ is the solution to ($1$), and let $c=|y-Xbeta^*|^2$. Is it true that the solution to ($2$) is also $beta^*$?




      Attempt: I believe this is true. The argument should be very similar to the one in Why are additional constraint and penalty term equivalent in ridge regression?. However, I was running some numerical experiments and it turns out the two problems have different solutions. Hence my question here: are the two problems really yielding the same solutions? Are there exceptions that I should be careful of?






      analysis optimization numerical-methods convex-optimization regression







      share|cite|improve this question











      $endgroup$





      Problem: Consider the optimization problems
      $$min_beta |y-Xbeta|^2+alpha|beta|^2 tag 1$$
      and
      $$min_beta |beta|^2 text{ subject to } |y-Xbeta|^2 le c tag 2$$
      where $|x|$ is the $2$-norm. Fix $alpha$, and suppose $beta^*$ is the solution to ($1$), and let $c=|y-Xbeta^*|^2$. Is it true that the solution to ($2$) is also $beta^*$?




      Attempt: I believe this is true. The argument should be very similar to the one in Why are additional constraint and penalty term equivalent in ridge regression?. However, I was running some numerical experiments and it turns out the two problems have different solutions. Hence my question here: are the two problems really yielding the same solutions? Are there exceptions that I should be careful of?






      analysis optimization numerical-methods convex-optimization regression




      analysis optimization numerical-methods convex-optimization regression






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 0:53









      Rócherz

      3,0263823




      3,0263823










      asked Dec 25 '18 at 23:21









      LongtiLongti

      84112




      84112






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          It is true for $alpha>0$. Since $beta^*$ is solution of (1), we have:
          $$|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2.$$
          Reordering:
          $$|beta^*|^2 le frac{1}{alpha}(|y-Xbeta|^2 - |y-Xbeta^*|^2) + |beta|^2.$$
          Now, in (2) we take $beta$ such that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, so we conclude that
          $$|beta^*|^2 le |beta|^2,$$
          which implies that $beta^*$ is a minimum of (2).



          Analogously, for $alpha<0$ you can check that $beta^*$ solution of (1) is also solution of (2), BUT maximizing instead of minimizing:
          $$alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2 le 0 quadLongrightarrowquad |beta^*|^2 ge |beta|^2.$$
          Anyhow, note that you cannot assure equivalence, since the problem becomes non-convex for $alpha<0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I'm not convinced of your final statement. For one thing setting $alpha<0$ may make this problem non-convex. But I see no reason to believe it would give equivalent solutions if I were "maximizing instead of minimizing". After all, you're not negating the entire objective, just the regularizer.
            $endgroup$
            – Michael Grant
            Dec 26 '18 at 22:24










          • $begingroup$
            @MichaelGrant I did not say it is equivalent ;) Anyway, I have added a note to make clear that solution of (1) would be solution of (2) (maximizing instead of minimizing), but not the other way around.
            $endgroup$
            – AugSB
            Dec 27 '18 at 8:39












          • $begingroup$
            I'm afraid I still don't see it for the non-convex case at all. Fortunately I believe the OP is likely concerning himself with the convex case.
            $endgroup$
            – Michael Grant
            Dec 27 '18 at 13:38










          • $begingroup$
            Please, correct me if I'm wrong, because maybe I'm missing something. For any $alpha<0$, $beta^*$ being solution of (1) implies $|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2$ for all $beta$. Therefore, $alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2$ for all $beta$. Assuming that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, we conclude that $|beta^*|^2 ge |beta|^2$ for all $beta$. So $beta^* = max_beta |beta|^2$ subject to $|y-Xbeta|^2 le |y-Xbeta^*|^2$.
            $endgroup$
            – AugSB
            Dec 27 '18 at 18:31












          • $begingroup$
            "Assuming that $|y-Xbeta|^2leq |y-Xbeta^*|$" but that's just it, you can't make that assumption. The point of adding a regularizer like $|beta|^2$ is that you're willing to trade on the optimal value of error in exchange for other criteria.
            $endgroup$
            – Michael Grant
            Jan 7 at 16:51





















          0












          $begingroup$

          Here from (1)



          $$
          f(beta) = y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta+alpha beta'cdotbeta
          $$



          so the minimum condition gives



          $$
          -X'cdot y+X'cdot Xbeta+alphabeta = 0
          $$



          and then



          $$
          beta^* = (Ialpha +X'cdot X)^{-1}X'cdot y
          $$



          and from (2)



          $$
          L(beta,lambda,epsilon)=beta'cdotbeta + lambda(y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta-c+epsilon^2)
          $$



          the stationary points are



          $$
          L_{beta} = 2beta-2lambda X'cdot y + 2lambda X'cdot Xcdotbeta = 0
          $$



          then



          $$
          (I+lambda X'cdot X)cdotbeta^* = lambda X'cdot y
          $$



          or



          $$
          beta^* = (Ifrac{1}{lambda}+X'cdot X)^{-1}cdot X'cdot y
          $$



          but



          $$
          beta'^*cdotbeta^*-lambdabeta'^*cdot X'cdot y+lambda beta'^*cdot X'cdot Xcdotbeta^* = 0
          $$



          and



          $$
          lambda = frac{beta'cdotbeta}{y'cdot y-beta'cdot X'cdot y-c+epsilon^2}
          $$



          so the equivalence between (1) and (2) needs



          $$
          lambda = frac{1}{alpha} = frac{beta'^*cdotbeta^*}{y'cdot y-beta'^*cdot X'cdot y-c+epsilon^2}
          $$



          or



          $$
          alpha = frac{(y-Xcdot beta^*)'cdot y-c+epsilon}{beta'^*cdotbeta^*}
          $$



          which is quite unlikely






          share|cite|improve this answer









          $endgroup$














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            2 Answers
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            2 Answers
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            active

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            3












            $begingroup$

            It is true for $alpha>0$. Since $beta^*$ is solution of (1), we have:
            $$|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2.$$
            Reordering:
            $$|beta^*|^2 le frac{1}{alpha}(|y-Xbeta|^2 - |y-Xbeta^*|^2) + |beta|^2.$$
            Now, in (2) we take $beta$ such that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, so we conclude that
            $$|beta^*|^2 le |beta|^2,$$
            which implies that $beta^*$ is a minimum of (2).



            Analogously, for $alpha<0$ you can check that $beta^*$ solution of (1) is also solution of (2), BUT maximizing instead of minimizing:
            $$alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2 le 0 quadLongrightarrowquad |beta^*|^2 ge |beta|^2.$$
            Anyhow, note that you cannot assure equivalence, since the problem becomes non-convex for $alpha<0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm not convinced of your final statement. For one thing setting $alpha<0$ may make this problem non-convex. But I see no reason to believe it would give equivalent solutions if I were "maximizing instead of minimizing". After all, you're not negating the entire objective, just the regularizer.
              $endgroup$
              – Michael Grant
              Dec 26 '18 at 22:24










            • $begingroup$
              @MichaelGrant I did not say it is equivalent ;) Anyway, I have added a note to make clear that solution of (1) would be solution of (2) (maximizing instead of minimizing), but not the other way around.
              $endgroup$
              – AugSB
              Dec 27 '18 at 8:39












            • $begingroup$
              I'm afraid I still don't see it for the non-convex case at all. Fortunately I believe the OP is likely concerning himself with the convex case.
              $endgroup$
              – Michael Grant
              Dec 27 '18 at 13:38










            • $begingroup$
              Please, correct me if I'm wrong, because maybe I'm missing something. For any $alpha<0$, $beta^*$ being solution of (1) implies $|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2$ for all $beta$. Therefore, $alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2$ for all $beta$. Assuming that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, we conclude that $|beta^*|^2 ge |beta|^2$ for all $beta$. So $beta^* = max_beta |beta|^2$ subject to $|y-Xbeta|^2 le |y-Xbeta^*|^2$.
              $endgroup$
              – AugSB
              Dec 27 '18 at 18:31












            • $begingroup$
              "Assuming that $|y-Xbeta|^2leq |y-Xbeta^*|$" but that's just it, you can't make that assumption. The point of adding a regularizer like $|beta|^2$ is that you're willing to trade on the optimal value of error in exchange for other criteria.
              $endgroup$
              – Michael Grant
              Jan 7 at 16:51


















            3












            $begingroup$

            It is true for $alpha>0$. Since $beta^*$ is solution of (1), we have:
            $$|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2.$$
            Reordering:
            $$|beta^*|^2 le frac{1}{alpha}(|y-Xbeta|^2 - |y-Xbeta^*|^2) + |beta|^2.$$
            Now, in (2) we take $beta$ such that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, so we conclude that
            $$|beta^*|^2 le |beta|^2,$$
            which implies that $beta^*$ is a minimum of (2).



            Analogously, for $alpha<0$ you can check that $beta^*$ solution of (1) is also solution of (2), BUT maximizing instead of minimizing:
            $$alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2 le 0 quadLongrightarrowquad |beta^*|^2 ge |beta|^2.$$
            Anyhow, note that you cannot assure equivalence, since the problem becomes non-convex for $alpha<0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I'm not convinced of your final statement. For one thing setting $alpha<0$ may make this problem non-convex. But I see no reason to believe it would give equivalent solutions if I were "maximizing instead of minimizing". After all, you're not negating the entire objective, just the regularizer.
              $endgroup$
              – Michael Grant
              Dec 26 '18 at 22:24










            • $begingroup$
              @MichaelGrant I did not say it is equivalent ;) Anyway, I have added a note to make clear that solution of (1) would be solution of (2) (maximizing instead of minimizing), but not the other way around.
              $endgroup$
              – AugSB
              Dec 27 '18 at 8:39












            • $begingroup$
              I'm afraid I still don't see it for the non-convex case at all. Fortunately I believe the OP is likely concerning himself with the convex case.
              $endgroup$
              – Michael Grant
              Dec 27 '18 at 13:38










            • $begingroup$
              Please, correct me if I'm wrong, because maybe I'm missing something. For any $alpha<0$, $beta^*$ being solution of (1) implies $|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2$ for all $beta$. Therefore, $alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2$ for all $beta$. Assuming that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, we conclude that $|beta^*|^2 ge |beta|^2$ for all $beta$. So $beta^* = max_beta |beta|^2$ subject to $|y-Xbeta|^2 le |y-Xbeta^*|^2$.
              $endgroup$
              – AugSB
              Dec 27 '18 at 18:31












            • $begingroup$
              "Assuming that $|y-Xbeta|^2leq |y-Xbeta^*|$" but that's just it, you can't make that assumption. The point of adding a regularizer like $|beta|^2$ is that you're willing to trade on the optimal value of error in exchange for other criteria.
              $endgroup$
              – Michael Grant
              Jan 7 at 16:51
















            3












            3








            3





            $begingroup$

            It is true for $alpha>0$. Since $beta^*$ is solution of (1), we have:
            $$|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2.$$
            Reordering:
            $$|beta^*|^2 le frac{1}{alpha}(|y-Xbeta|^2 - |y-Xbeta^*|^2) + |beta|^2.$$
            Now, in (2) we take $beta$ such that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, so we conclude that
            $$|beta^*|^2 le |beta|^2,$$
            which implies that $beta^*$ is a minimum of (2).



            Analogously, for $alpha<0$ you can check that $beta^*$ solution of (1) is also solution of (2), BUT maximizing instead of minimizing:
            $$alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2 le 0 quadLongrightarrowquad |beta^*|^2 ge |beta|^2.$$
            Anyhow, note that you cannot assure equivalence, since the problem becomes non-convex for $alpha<0$.






            share|cite|improve this answer











            $endgroup$



            It is true for $alpha>0$. Since $beta^*$ is solution of (1), we have:
            $$|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2.$$
            Reordering:
            $$|beta^*|^2 le frac{1}{alpha}(|y-Xbeta|^2 - |y-Xbeta^*|^2) + |beta|^2.$$
            Now, in (2) we take $beta$ such that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, so we conclude that
            $$|beta^*|^2 le |beta|^2,$$
            which implies that $beta^*$ is a minimum of (2).



            Analogously, for $alpha<0$ you can check that $beta^*$ solution of (1) is also solution of (2), BUT maximizing instead of minimizing:
            $$alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2 le 0 quadLongrightarrowquad |beta^*|^2 ge |beta|^2.$$
            Anyhow, note that you cannot assure equivalence, since the problem becomes non-convex for $alpha<0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 27 '18 at 8:35

























            answered Dec 26 '18 at 0:18









            AugSBAugSB

            3,45421734




            3,45421734












            • $begingroup$
              I'm not convinced of your final statement. For one thing setting $alpha<0$ may make this problem non-convex. But I see no reason to believe it would give equivalent solutions if I were "maximizing instead of minimizing". After all, you're not negating the entire objective, just the regularizer.
              $endgroup$
              – Michael Grant
              Dec 26 '18 at 22:24










            • $begingroup$
              @MichaelGrant I did not say it is equivalent ;) Anyway, I have added a note to make clear that solution of (1) would be solution of (2) (maximizing instead of minimizing), but not the other way around.
              $endgroup$
              – AugSB
              Dec 27 '18 at 8:39












            • $begingroup$
              I'm afraid I still don't see it for the non-convex case at all. Fortunately I believe the OP is likely concerning himself with the convex case.
              $endgroup$
              – Michael Grant
              Dec 27 '18 at 13:38










            • $begingroup$
              Please, correct me if I'm wrong, because maybe I'm missing something. For any $alpha<0$, $beta^*$ being solution of (1) implies $|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2$ for all $beta$. Therefore, $alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2$ for all $beta$. Assuming that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, we conclude that $|beta^*|^2 ge |beta|^2$ for all $beta$. So $beta^* = max_beta |beta|^2$ subject to $|y-Xbeta|^2 le |y-Xbeta^*|^2$.
              $endgroup$
              – AugSB
              Dec 27 '18 at 18:31












            • $begingroup$
              "Assuming that $|y-Xbeta|^2leq |y-Xbeta^*|$" but that's just it, you can't make that assumption. The point of adding a regularizer like $|beta|^2$ is that you're willing to trade on the optimal value of error in exchange for other criteria.
              $endgroup$
              – Michael Grant
              Jan 7 at 16:51




















            • $begingroup$
              I'm not convinced of your final statement. For one thing setting $alpha<0$ may make this problem non-convex. But I see no reason to believe it would give equivalent solutions if I were "maximizing instead of minimizing". After all, you're not negating the entire objective, just the regularizer.
              $endgroup$
              – Michael Grant
              Dec 26 '18 at 22:24










            • $begingroup$
              @MichaelGrant I did not say it is equivalent ;) Anyway, I have added a note to make clear that solution of (1) would be solution of (2) (maximizing instead of minimizing), but not the other way around.
              $endgroup$
              – AugSB
              Dec 27 '18 at 8:39












            • $begingroup$
              I'm afraid I still don't see it for the non-convex case at all. Fortunately I believe the OP is likely concerning himself with the convex case.
              $endgroup$
              – Michael Grant
              Dec 27 '18 at 13:38










            • $begingroup$
              Please, correct me if I'm wrong, because maybe I'm missing something. For any $alpha<0$, $beta^*$ being solution of (1) implies $|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2$ for all $beta$. Therefore, $alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2$ for all $beta$. Assuming that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, we conclude that $|beta^*|^2 ge |beta|^2$ for all $beta$. So $beta^* = max_beta |beta|^2$ subject to $|y-Xbeta|^2 le |y-Xbeta^*|^2$.
              $endgroup$
              – AugSB
              Dec 27 '18 at 18:31












            • $begingroup$
              "Assuming that $|y-Xbeta|^2leq |y-Xbeta^*|$" but that's just it, you can't make that assumption. The point of adding a regularizer like $|beta|^2$ is that you're willing to trade on the optimal value of error in exchange for other criteria.
              $endgroup$
              – Michael Grant
              Jan 7 at 16:51


















            $begingroup$
            I'm not convinced of your final statement. For one thing setting $alpha<0$ may make this problem non-convex. But I see no reason to believe it would give equivalent solutions if I were "maximizing instead of minimizing". After all, you're not negating the entire objective, just the regularizer.
            $endgroup$
            – Michael Grant
            Dec 26 '18 at 22:24




            $begingroup$
            I'm not convinced of your final statement. For one thing setting $alpha<0$ may make this problem non-convex. But I see no reason to believe it would give equivalent solutions if I were "maximizing instead of minimizing". After all, you're not negating the entire objective, just the regularizer.
            $endgroup$
            – Michael Grant
            Dec 26 '18 at 22:24












            $begingroup$
            @MichaelGrant I did not say it is equivalent ;) Anyway, I have added a note to make clear that solution of (1) would be solution of (2) (maximizing instead of minimizing), but not the other way around.
            $endgroup$
            – AugSB
            Dec 27 '18 at 8:39






            $begingroup$
            @MichaelGrant I did not say it is equivalent ;) Anyway, I have added a note to make clear that solution of (1) would be solution of (2) (maximizing instead of minimizing), but not the other way around.
            $endgroup$
            – AugSB
            Dec 27 '18 at 8:39














            $begingroup$
            I'm afraid I still don't see it for the non-convex case at all. Fortunately I believe the OP is likely concerning himself with the convex case.
            $endgroup$
            – Michael Grant
            Dec 27 '18 at 13:38




            $begingroup$
            I'm afraid I still don't see it for the non-convex case at all. Fortunately I believe the OP is likely concerning himself with the convex case.
            $endgroup$
            – Michael Grant
            Dec 27 '18 at 13:38












            $begingroup$
            Please, correct me if I'm wrong, because maybe I'm missing something. For any $alpha<0$, $beta^*$ being solution of (1) implies $|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2$ for all $beta$. Therefore, $alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2$ for all $beta$. Assuming that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, we conclude that $|beta^*|^2 ge |beta|^2$ for all $beta$. So $beta^* = max_beta |beta|^2$ subject to $|y-Xbeta|^2 le |y-Xbeta^*|^2$.
            $endgroup$
            – AugSB
            Dec 27 '18 at 18:31






            $begingroup$
            Please, correct me if I'm wrong, because maybe I'm missing something. For any $alpha<0$, $beta^*$ being solution of (1) implies $|y-Xbeta^*|^2 + alpha|beta^*|^2 le |y-Xbeta|^2 + alpha|beta|^2$ for all $beta$. Therefore, $alpha(|beta^*|^2 - |beta|^2) le |y-Xbeta|^2 - |y-Xbeta^*|^2$ for all $beta$. Assuming that $|y-Xbeta|^2 le |y-Xbeta^*|^2$, we conclude that $|beta^*|^2 ge |beta|^2$ for all $beta$. So $beta^* = max_beta |beta|^2$ subject to $|y-Xbeta|^2 le |y-Xbeta^*|^2$.
            $endgroup$
            – AugSB
            Dec 27 '18 at 18:31














            $begingroup$
            "Assuming that $|y-Xbeta|^2leq |y-Xbeta^*|$" but that's just it, you can't make that assumption. The point of adding a regularizer like $|beta|^2$ is that you're willing to trade on the optimal value of error in exchange for other criteria.
            $endgroup$
            – Michael Grant
            Jan 7 at 16:51






            $begingroup$
            "Assuming that $|y-Xbeta|^2leq |y-Xbeta^*|$" but that's just it, you can't make that assumption. The point of adding a regularizer like $|beta|^2$ is that you're willing to trade on the optimal value of error in exchange for other criteria.
            $endgroup$
            – Michael Grant
            Jan 7 at 16:51













            0












            $begingroup$

            Here from (1)



            $$
            f(beta) = y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta+alpha beta'cdotbeta
            $$



            so the minimum condition gives



            $$
            -X'cdot y+X'cdot Xbeta+alphabeta = 0
            $$



            and then



            $$
            beta^* = (Ialpha +X'cdot X)^{-1}X'cdot y
            $$



            and from (2)



            $$
            L(beta,lambda,epsilon)=beta'cdotbeta + lambda(y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta-c+epsilon^2)
            $$



            the stationary points are



            $$
            L_{beta} = 2beta-2lambda X'cdot y + 2lambda X'cdot Xcdotbeta = 0
            $$



            then



            $$
            (I+lambda X'cdot X)cdotbeta^* = lambda X'cdot y
            $$



            or



            $$
            beta^* = (Ifrac{1}{lambda}+X'cdot X)^{-1}cdot X'cdot y
            $$



            but



            $$
            beta'^*cdotbeta^*-lambdabeta'^*cdot X'cdot y+lambda beta'^*cdot X'cdot Xcdotbeta^* = 0
            $$



            and



            $$
            lambda = frac{beta'cdotbeta}{y'cdot y-beta'cdot X'cdot y-c+epsilon^2}
            $$



            so the equivalence between (1) and (2) needs



            $$
            lambda = frac{1}{alpha} = frac{beta'^*cdotbeta^*}{y'cdot y-beta'^*cdot X'cdot y-c+epsilon^2}
            $$



            or



            $$
            alpha = frac{(y-Xcdot beta^*)'cdot y-c+epsilon}{beta'^*cdotbeta^*}
            $$



            which is quite unlikely






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Here from (1)



              $$
              f(beta) = y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta+alpha beta'cdotbeta
              $$



              so the minimum condition gives



              $$
              -X'cdot y+X'cdot Xbeta+alphabeta = 0
              $$



              and then



              $$
              beta^* = (Ialpha +X'cdot X)^{-1}X'cdot y
              $$



              and from (2)



              $$
              L(beta,lambda,epsilon)=beta'cdotbeta + lambda(y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta-c+epsilon^2)
              $$



              the stationary points are



              $$
              L_{beta} = 2beta-2lambda X'cdot y + 2lambda X'cdot Xcdotbeta = 0
              $$



              then



              $$
              (I+lambda X'cdot X)cdotbeta^* = lambda X'cdot y
              $$



              or



              $$
              beta^* = (Ifrac{1}{lambda}+X'cdot X)^{-1}cdot X'cdot y
              $$



              but



              $$
              beta'^*cdotbeta^*-lambdabeta'^*cdot X'cdot y+lambda beta'^*cdot X'cdot Xcdotbeta^* = 0
              $$



              and



              $$
              lambda = frac{beta'cdotbeta}{y'cdot y-beta'cdot X'cdot y-c+epsilon^2}
              $$



              so the equivalence between (1) and (2) needs



              $$
              lambda = frac{1}{alpha} = frac{beta'^*cdotbeta^*}{y'cdot y-beta'^*cdot X'cdot y-c+epsilon^2}
              $$



              or



              $$
              alpha = frac{(y-Xcdot beta^*)'cdot y-c+epsilon}{beta'^*cdotbeta^*}
              $$



              which is quite unlikely






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Here from (1)



                $$
                f(beta) = y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta+alpha beta'cdotbeta
                $$



                so the minimum condition gives



                $$
                -X'cdot y+X'cdot Xbeta+alphabeta = 0
                $$



                and then



                $$
                beta^* = (Ialpha +X'cdot X)^{-1}X'cdot y
                $$



                and from (2)



                $$
                L(beta,lambda,epsilon)=beta'cdotbeta + lambda(y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta-c+epsilon^2)
                $$



                the stationary points are



                $$
                L_{beta} = 2beta-2lambda X'cdot y + 2lambda X'cdot Xcdotbeta = 0
                $$



                then



                $$
                (I+lambda X'cdot X)cdotbeta^* = lambda X'cdot y
                $$



                or



                $$
                beta^* = (Ifrac{1}{lambda}+X'cdot X)^{-1}cdot X'cdot y
                $$



                but



                $$
                beta'^*cdotbeta^*-lambdabeta'^*cdot X'cdot y+lambda beta'^*cdot X'cdot Xcdotbeta^* = 0
                $$



                and



                $$
                lambda = frac{beta'cdotbeta}{y'cdot y-beta'cdot X'cdot y-c+epsilon^2}
                $$



                so the equivalence between (1) and (2) needs



                $$
                lambda = frac{1}{alpha} = frac{beta'^*cdotbeta^*}{y'cdot y-beta'^*cdot X'cdot y-c+epsilon^2}
                $$



                or



                $$
                alpha = frac{(y-Xcdot beta^*)'cdot y-c+epsilon}{beta'^*cdotbeta^*}
                $$



                which is quite unlikely






                share|cite|improve this answer









                $endgroup$



                Here from (1)



                $$
                f(beta) = y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta+alpha beta'cdotbeta
                $$



                so the minimum condition gives



                $$
                -X'cdot y+X'cdot Xbeta+alphabeta = 0
                $$



                and then



                $$
                beta^* = (Ialpha +X'cdot X)^{-1}X'cdot y
                $$



                and from (2)



                $$
                L(beta,lambda,epsilon)=beta'cdotbeta + lambda(y'cdot y-2beta'cdot X'y+beta'cdot X'cdot Xcdotbeta-c+epsilon^2)
                $$



                the stationary points are



                $$
                L_{beta} = 2beta-2lambda X'cdot y + 2lambda X'cdot Xcdotbeta = 0
                $$



                then



                $$
                (I+lambda X'cdot X)cdotbeta^* = lambda X'cdot y
                $$



                or



                $$
                beta^* = (Ifrac{1}{lambda}+X'cdot X)^{-1}cdot X'cdot y
                $$



                but



                $$
                beta'^*cdotbeta^*-lambdabeta'^*cdot X'cdot y+lambda beta'^*cdot X'cdot Xcdotbeta^* = 0
                $$



                and



                $$
                lambda = frac{beta'cdotbeta}{y'cdot y-beta'cdot X'cdot y-c+epsilon^2}
                $$



                so the equivalence between (1) and (2) needs



                $$
                lambda = frac{1}{alpha} = frac{beta'^*cdotbeta^*}{y'cdot y-beta'^*cdot X'cdot y-c+epsilon^2}
                $$



                or



                $$
                alpha = frac{(y-Xcdot beta^*)'cdot y-c+epsilon}{beta'^*cdotbeta^*}
                $$



                which is quite unlikely







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '18 at 11:55









                CesareoCesareo

                9,7863517




                9,7863517






























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