Lagrangian multiplier based on calculus of variation

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The equation (6) and (7) on the chapter 3.2 in http://wnzhang.net/papers/ortb-kdd.pdf , there is Lagrange-multiplier based on calculus of variation.



I'm not sure that (7) is derived from (6).



As i think,



$F(theta, b(theta), b'(theta))$ = $theta w(b(theta))p_theta(theta) - lambda (b(theta)w(b(theta))p_theta(theta))$.



As Euler-Lagrange condition $fracpartial Fpartial b(theta) - fracddthetafracpartial Fpartial b'(theta)=0$.



so,



$theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)] - fracddthetafracpartial Fpartial b'(theta)=0$.



In last term $fracpartial Fpartial b'(theta)=0$.
so,



$theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)]=0$



Is it correct?







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    The equation (6) and (7) on the chapter 3.2 in http://wnzhang.net/papers/ortb-kdd.pdf , there is Lagrange-multiplier based on calculus of variation.



    I'm not sure that (7) is derived from (6).



    As i think,



    $F(theta, b(theta), b'(theta))$ = $theta w(b(theta))p_theta(theta) - lambda (b(theta)w(b(theta))p_theta(theta))$.



    As Euler-Lagrange condition $fracpartial Fpartial b(theta) - fracddthetafracpartial Fpartial b'(theta)=0$.



    so,



    $theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)] - fracddthetafracpartial Fpartial b'(theta)=0$.



    In last term $fracpartial Fpartial b'(theta)=0$.
    so,



    $theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)]=0$



    Is it correct?







    share|cite|improve this question






















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      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      The equation (6) and (7) on the chapter 3.2 in http://wnzhang.net/papers/ortb-kdd.pdf , there is Lagrange-multiplier based on calculus of variation.



      I'm not sure that (7) is derived from (6).



      As i think,



      $F(theta, b(theta), b'(theta))$ = $theta w(b(theta))p_theta(theta) - lambda (b(theta)w(b(theta))p_theta(theta))$.



      As Euler-Lagrange condition $fracpartial Fpartial b(theta) - fracddthetafracpartial Fpartial b'(theta)=0$.



      so,



      $theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)] - fracddthetafracpartial Fpartial b'(theta)=0$.



      In last term $fracpartial Fpartial b'(theta)=0$.
      so,



      $theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)]=0$



      Is it correct?







      share|cite|improve this question












      The equation (6) and (7) on the chapter 3.2 in http://wnzhang.net/papers/ortb-kdd.pdf , there is Lagrange-multiplier based on calculus of variation.



      I'm not sure that (7) is derived from (6).



      As i think,



      $F(theta, b(theta), b'(theta))$ = $theta w(b(theta))p_theta(theta) - lambda (b(theta)w(b(theta))p_theta(theta))$.



      As Euler-Lagrange condition $fracpartial Fpartial b(theta) - fracddthetafracpartial Fpartial b'(theta)=0$.



      so,



      $theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)] - fracddthetafracpartial Fpartial b'(theta)=0$.



      In last term $fracpartial Fpartial b'(theta)=0$.
      so,



      $theta p_theta(theta)fracpartial w(b(theta))partial b(theta) - lambda[w(b(theta))p_theta(theta) + b(theta)fracpartial w(b(theta))partial b(theta)]=0$



      Is it correct?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 26 at 13:53









      watseob

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