Do I really need to solve 5 nonlinear equations in this Lagrange multiplier problem?












0












$begingroup$


I need to solve the following optimization problem



Let $X=left{ x_{i}right} _{i=1}^{n}$ be an independent sequence of $k$-face die rolls. Where for $jinleft[kright]$ we have $pleft(x_{i}=jright)=theta_{j}$ (So $sum_{i=1}^{k}theta_{i}=1$)



I am requested to find the ML estimator for $theta$ for $k=3$ under the constraint $theta_{1}=theta_{2}+theta_{3}$



My work so far:



The log likelihood function is given by $$ellleft(thetamid Xright)=log p_{theta}left(Xright)=logprod_{i=1}^{n}p_{theta}left(x_{i}right)=sum_{i=1}^{n}log p_{theta}left(x_{i}right)=sum_{i=1}^{n}logtheta_{x_{i}}=sum_{i=1}^{k}n_{i}logtheta_{i}$$
where $n_{i}$ is the number of occurrences of face $i$



So the appropriate Lagrangian function is
$$mathcal{L}left(thetaright)=ellleft(thetamid Xright)-lambda_{0}left(sum_{i=1}^{k}theta_{i}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)=$$
$$sum_{i=1}^{3}n_{i}logtheta_{i}-lambda_{0}left(theta_{1}+theta_{2}+theta_{3}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)$$



and then we need to solve



$$nablamathcal{L}=left(begin{matrix}frac{partialmathcal{L}}{partialtheta_{1}}\
frac{partialmathcal{L}}{partialtheta_{2}}\
frac{partialmathcal{L}}{partialtheta_{3}}\
frac{partialmathcal{L}}{partiallambda_{0}}\
frac{partialmathcal{L}}{partiallambda_{1}}
end{matrix}right)=left(begin{matrix}frac{n_{1}}{theta_{1}}-lambda_{0}-lambda_{1}\
frac{n_{2}}{theta_{2}}-lambda_{0}+lambda_{1}\
frac{n_{3}}{theta_{3}}-lambda_{0}+lambda_{1}\
-theta_{1}-theta_{2}-theta_{3}+1\
-theta_{1}+theta_{2}+theta_{3}
end{matrix}right)=left(begin{matrix}0\
0\
0\
0\
0
end{matrix}right)$$



Is this analysis correct? If so, do I really have to solve 5 (non linear) equations with 5 unknowns to get the answer?
This is my first time working with Lagrange multipliers, and it seems to me that either I am wrong here or else there is some kind of a workaround, especially since the following questions seem like they'l end up with even more equations.






systems-of-equations lagrange-multiplier maximum-likelihood







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I need to solve the following optimization problem



    Let $X=left{ x_{i}right} _{i=1}^{n}$ be an independent sequence of $k$-face die rolls. Where for $jinleft[kright]$ we have $pleft(x_{i}=jright)=theta_{j}$ (So $sum_{i=1}^{k}theta_{i}=1$)



    I am requested to find the ML estimator for $theta$ for $k=3$ under the constraint $theta_{1}=theta_{2}+theta_{3}$



    My work so far:



    The log likelihood function is given by $$ellleft(thetamid Xright)=log p_{theta}left(Xright)=logprod_{i=1}^{n}p_{theta}left(x_{i}right)=sum_{i=1}^{n}log p_{theta}left(x_{i}right)=sum_{i=1}^{n}logtheta_{x_{i}}=sum_{i=1}^{k}n_{i}logtheta_{i}$$
    where $n_{i}$ is the number of occurrences of face $i$



    So the appropriate Lagrangian function is
    $$mathcal{L}left(thetaright)=ellleft(thetamid Xright)-lambda_{0}left(sum_{i=1}^{k}theta_{i}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)=$$
    $$sum_{i=1}^{3}n_{i}logtheta_{i}-lambda_{0}left(theta_{1}+theta_{2}+theta_{3}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)$$



    and then we need to solve



    $$nablamathcal{L}=left(begin{matrix}frac{partialmathcal{L}}{partialtheta_{1}}\
    frac{partialmathcal{L}}{partialtheta_{2}}\
    frac{partialmathcal{L}}{partialtheta_{3}}\
    frac{partialmathcal{L}}{partiallambda_{0}}\
    frac{partialmathcal{L}}{partiallambda_{1}}
    end{matrix}right)=left(begin{matrix}frac{n_{1}}{theta_{1}}-lambda_{0}-lambda_{1}\
    frac{n_{2}}{theta_{2}}-lambda_{0}+lambda_{1}\
    frac{n_{3}}{theta_{3}}-lambda_{0}+lambda_{1}\
    -theta_{1}-theta_{2}-theta_{3}+1\
    -theta_{1}+theta_{2}+theta_{3}
    end{matrix}right)=left(begin{matrix}0\
    0\
    0\
    0\
    0
    end{matrix}right)$$



    Is this analysis correct? If so, do I really have to solve 5 (non linear) equations with 5 unknowns to get the answer?
    This is my first time working with Lagrange multipliers, and it seems to me that either I am wrong here or else there is some kind of a workaround, especially since the following questions seem like they'l end up with even more equations.






    systems-of-equations lagrange-multiplier maximum-likelihood







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I need to solve the following optimization problem



      Let $X=left{ x_{i}right} _{i=1}^{n}$ be an independent sequence of $k$-face die rolls. Where for $jinleft[kright]$ we have $pleft(x_{i}=jright)=theta_{j}$ (So $sum_{i=1}^{k}theta_{i}=1$)



      I am requested to find the ML estimator for $theta$ for $k=3$ under the constraint $theta_{1}=theta_{2}+theta_{3}$



      My work so far:



      The log likelihood function is given by $$ellleft(thetamid Xright)=log p_{theta}left(Xright)=logprod_{i=1}^{n}p_{theta}left(x_{i}right)=sum_{i=1}^{n}log p_{theta}left(x_{i}right)=sum_{i=1}^{n}logtheta_{x_{i}}=sum_{i=1}^{k}n_{i}logtheta_{i}$$
      where $n_{i}$ is the number of occurrences of face $i$



      So the appropriate Lagrangian function is
      $$mathcal{L}left(thetaright)=ellleft(thetamid Xright)-lambda_{0}left(sum_{i=1}^{k}theta_{i}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)=$$
      $$sum_{i=1}^{3}n_{i}logtheta_{i}-lambda_{0}left(theta_{1}+theta_{2}+theta_{3}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)$$



      and then we need to solve



      $$nablamathcal{L}=left(begin{matrix}frac{partialmathcal{L}}{partialtheta_{1}}\
      frac{partialmathcal{L}}{partialtheta_{2}}\
      frac{partialmathcal{L}}{partialtheta_{3}}\
      frac{partialmathcal{L}}{partiallambda_{0}}\
      frac{partialmathcal{L}}{partiallambda_{1}}
      end{matrix}right)=left(begin{matrix}frac{n_{1}}{theta_{1}}-lambda_{0}-lambda_{1}\
      frac{n_{2}}{theta_{2}}-lambda_{0}+lambda_{1}\
      frac{n_{3}}{theta_{3}}-lambda_{0}+lambda_{1}\
      -theta_{1}-theta_{2}-theta_{3}+1\
      -theta_{1}+theta_{2}+theta_{3}
      end{matrix}right)=left(begin{matrix}0\
      0\
      0\
      0\
      0
      end{matrix}right)$$



      Is this analysis correct? If so, do I really have to solve 5 (non linear) equations with 5 unknowns to get the answer?
      This is my first time working with Lagrange multipliers, and it seems to me that either I am wrong here or else there is some kind of a workaround, especially since the following questions seem like they'l end up with even more equations.






      systems-of-equations lagrange-multiplier maximum-likelihood







      share|cite|improve this question









      $endgroup$




      I need to solve the following optimization problem



      Let $X=left{ x_{i}right} _{i=1}^{n}$ be an independent sequence of $k$-face die rolls. Where for $jinleft[kright]$ we have $pleft(x_{i}=jright)=theta_{j}$ (So $sum_{i=1}^{k}theta_{i}=1$)



      I am requested to find the ML estimator for $theta$ for $k=3$ under the constraint $theta_{1}=theta_{2}+theta_{3}$



      My work so far:



      The log likelihood function is given by $$ellleft(thetamid Xright)=log p_{theta}left(Xright)=logprod_{i=1}^{n}p_{theta}left(x_{i}right)=sum_{i=1}^{n}log p_{theta}left(x_{i}right)=sum_{i=1}^{n}logtheta_{x_{i}}=sum_{i=1}^{k}n_{i}logtheta_{i}$$
      where $n_{i}$ is the number of occurrences of face $i$



      So the appropriate Lagrangian function is
      $$mathcal{L}left(thetaright)=ellleft(thetamid Xright)-lambda_{0}left(sum_{i=1}^{k}theta_{i}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)=$$
      $$sum_{i=1}^{3}n_{i}logtheta_{i}-lambda_{0}left(theta_{1}+theta_{2}+theta_{3}-1right)-lambda_{1}left(theta_{1}-theta_{2}-theta_{3}right)$$



      and then we need to solve



      $$nablamathcal{L}=left(begin{matrix}frac{partialmathcal{L}}{partialtheta_{1}}\
      frac{partialmathcal{L}}{partialtheta_{2}}\
      frac{partialmathcal{L}}{partialtheta_{3}}\
      frac{partialmathcal{L}}{partiallambda_{0}}\
      frac{partialmathcal{L}}{partiallambda_{1}}
      end{matrix}right)=left(begin{matrix}frac{n_{1}}{theta_{1}}-lambda_{0}-lambda_{1}\
      frac{n_{2}}{theta_{2}}-lambda_{0}+lambda_{1}\
      frac{n_{3}}{theta_{3}}-lambda_{0}+lambda_{1}\
      -theta_{1}-theta_{2}-theta_{3}+1\
      -theta_{1}+theta_{2}+theta_{3}
      end{matrix}right)=left(begin{matrix}0\
      0\
      0\
      0\
      0
      end{matrix}right)$$



      Is this analysis correct? If so, do I really have to solve 5 (non linear) equations with 5 unknowns to get the answer?
      This is my first time working with Lagrange multipliers, and it seems to me that either I am wrong here or else there is some kind of a workaround, especially since the following questions seem like they'l end up with even more equations.






      systems-of-equations lagrange-multiplier maximum-likelihood




      systems-of-equations lagrange-multiplier maximum-likelihood






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 11 '18 at 20:59









      D.M. D.M.

      494




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          $begingroup$

          These aren't that hard to solve, despite being nonlinear. Referring to the equations by row number, $0=(2)-(3)$ gives $theta_3 = n_3theta_2/n_2$.



          Next, $0=(4)-(5)=-2theta_2+1-2theta_3$. Plug in your $theta_3$ to solve for $theta_2$. Keep going...






          share|cite|improve this answer









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            $begingroup$

            These aren't that hard to solve, despite being nonlinear. Referring to the equations by row number, $0=(2)-(3)$ gives $theta_3 = n_3theta_2/n_2$.



            Next, $0=(4)-(5)=-2theta_2+1-2theta_3$. Plug in your $theta_3$ to solve for $theta_2$. Keep going...






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              These aren't that hard to solve, despite being nonlinear. Referring to the equations by row number, $0=(2)-(3)$ gives $theta_3 = n_3theta_2/n_2$.



              Next, $0=(4)-(5)=-2theta_2+1-2theta_3$. Plug in your $theta_3$ to solve for $theta_2$. Keep going...






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                These aren't that hard to solve, despite being nonlinear. Referring to the equations by row number, $0=(2)-(3)$ gives $theta_3 = n_3theta_2/n_2$.



                Next, $0=(4)-(5)=-2theta_2+1-2theta_3$. Plug in your $theta_3$ to solve for $theta_2$. Keep going...






                share|cite|improve this answer









                $endgroup$



                These aren't that hard to solve, despite being nonlinear. Referring to the equations by row number, $0=(2)-(3)$ gives $theta_3 = n_3theta_2/n_2$.



                Next, $0=(4)-(5)=-2theta_2+1-2theta_3$. Plug in your $theta_3$ to solve for $theta_2$. Keep going...







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 11 '18 at 21:18









                Alex R.Alex R.

                24.9k12452




                24.9k12452






























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