Distribution of Maximum Likelihood EstimatorMaximum likelihood of function of the mean on a restricted parameter spaceMaximum Likelihood Estimator (MLE)Maximum Likelihood Estimator of the exponential function parameter based on Order StatisticsFind maximum likelihood estimateMaximum Likelihood Estimation in case of some specific uniform distributionsMaximum likelihood estimate for a univariate gaussianFind the Maximum Likelihood Estimator given two pdfsMaximum Likelihood Estimate for a likelihood defined by partsVariance of distribution for maximum likelihood estimatorHow is $P(D;theta) = P(D|theta)$?

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Distribution of Maximum Likelihood Estimator


Maximum likelihood of function of the mean on a restricted parameter spaceMaximum Likelihood Estimator (MLE)Maximum Likelihood Estimator of the exponential function parameter based on Order StatisticsFind maximum likelihood estimateMaximum Likelihood Estimation in case of some specific uniform distributionsMaximum likelihood estimate for a univariate gaussianFind the Maximum Likelihood Estimator given two pdfsMaximum Likelihood Estimate for a likelihood defined by partsVariance of distribution for maximum likelihood estimatorHow is $P(D;theta) = P(D|theta)$?













1












$begingroup$


Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)



Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$



Then after taking sample of size n



$$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$



And we want to find $theta_max$ such that $L(theta)$ is maximized and $theta_max$ is our estimate (once a sample has actually been selected)



Since $theta_max$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$



where



$$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$



Taking the derivative with respect to $theta$



$$fracf'(x_1;theta)f(x_1;theta)+fracf'(x_2;theta)f(x_2;theta)...+fracf'(x_n;theta)f(x_n;theta)$$



$theta_max$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?










share|cite|improve this question







New contributor




Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
















    1












    $begingroup$


    Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)



    Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$



    Then after taking sample of size n



    $$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$



    And we want to find $theta_max$ such that $L(theta)$ is maximized and $theta_max$ is our estimate (once a sample has actually been selected)



    Since $theta_max$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$



    where



    $$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$



    Taking the derivative with respect to $theta$



    $$fracf'(x_1;theta)f(x_1;theta)+fracf'(x_2;theta)f(x_2;theta)...+fracf'(x_n;theta)f(x_n;theta)$$



    $theta_max$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?










    share|cite|improve this question







    New contributor




    Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      1












      1








      1





      $begingroup$


      Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)



      Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$



      Then after taking sample of size n



      $$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$



      And we want to find $theta_max$ such that $L(theta)$ is maximized and $theta_max$ is our estimate (once a sample has actually been selected)



      Since $theta_max$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$



      where



      $$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$



      Taking the derivative with respect to $theta$



      $$fracf'(x_1;theta)f(x_1;theta)+fracf'(x_2;theta)f(x_2;theta)...+fracf'(x_n;theta)f(x_n;theta)$$



      $theta_max$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?










      share|cite|improve this question







      New contributor




      Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Why is the Maximum Likelihood Estimator Normally distributed? I can't figure out why it is true for large n in general. My attempt (for single parameter)



      Let $L(theta)$ be the maximum likelihood function for the distribution $f(x;theta)$



      Then after taking sample of size n



      $$L(theta)=f(x_1;theta)cdot f(x_2theta)...f(x_n;theta)$$



      And we want to find $theta_max$ such that $L(theta)$ is maximized and $theta_max$ is our estimate (once a sample has actually been selected)



      Since $theta_max$ maximizes $L(theta)$ it also maximizes $ln(L(theta))$



      where



      $$ln(L(theta))=ln(f(x_1;theta))+ln(f(x_2;theta))...+ln(f(x_n;theta))$$



      Taking the derivative with respect to $theta$



      $$fracf'(x_1;theta)f(x_1;theta)+fracf'(x_2;theta)f(x_2;theta)...+fracf'(x_n;theta)f(x_n;theta)$$



      $theta_max$ would be the solution of the above when set to 0 (after selecting values for all $x_1,x_2...x_n$) but why is it normally distributed and how do I show that it's true for large n?







      probability distributions normal-distribution estimation sampling






      share|cite|improve this question







      New contributor




      Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 3 hours ago









      Colin HicksColin Hicks

      1353




      1353




      New contributor




      Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Colin Hicks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          MLE requires $$fracpartial ln L(theta)partial theta = sum_i=1^n frac f'(x_i;theta)f(x_i;theta),$$
          where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac f'(x;theta)f(x;theta).$ Then $g(x_i;theta)_i=1^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
          $$sqrtn(barg_n(theta)-Eg(x_1;theta))=sqrtnbarg_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $barg_n(theta)=frac1n sum_i=1^n g(x_i;theta).$ The ML estimator solves the equation
          $$barg_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hattheta=barg_n^-1(0).$$
          So long as the set of discontinuity points of $barg_n^-1(z)$, i.e. the set of all values of $z$ such that $barg_n^-1(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.






          share|cite|improve this answer










          New contributor




          dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            $fracf'(x)f(x)$ is just another function of x so central limit theorem applies thank you for that
            $endgroup$
            – Colin Hicks
            3 hours ago











          • $begingroup$
            You're welcome :)
            $endgroup$
            – dlnB
            3 hours ago










          Your Answer





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          1 Answer
          1






          active

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          active

          oldest

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          active

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          3












          $begingroup$

          MLE requires $$fracpartial ln L(theta)partial theta = sum_i=1^n frac f'(x_i;theta)f(x_i;theta),$$
          where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac f'(x;theta)f(x;theta).$ Then $g(x_i;theta)_i=1^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
          $$sqrtn(barg_n(theta)-Eg(x_1;theta))=sqrtnbarg_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $barg_n(theta)=frac1n sum_i=1^n g(x_i;theta).$ The ML estimator solves the equation
          $$barg_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hattheta=barg_n^-1(0).$$
          So long as the set of discontinuity points of $barg_n^-1(z)$, i.e. the set of all values of $z$ such that $barg_n^-1(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.






          share|cite|improve this answer










          New contributor




          dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            $fracf'(x)f(x)$ is just another function of x so central limit theorem applies thank you for that
            $endgroup$
            – Colin Hicks
            3 hours ago











          • $begingroup$
            You're welcome :)
            $endgroup$
            – dlnB
            3 hours ago















          3












          $begingroup$

          MLE requires $$fracpartial ln L(theta)partial theta = sum_i=1^n frac f'(x_i;theta)f(x_i;theta),$$
          where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac f'(x;theta)f(x;theta).$ Then $g(x_i;theta)_i=1^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
          $$sqrtn(barg_n(theta)-Eg(x_1;theta))=sqrtnbarg_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $barg_n(theta)=frac1n sum_i=1^n g(x_i;theta).$ The ML estimator solves the equation
          $$barg_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hattheta=barg_n^-1(0).$$
          So long as the set of discontinuity points of $barg_n^-1(z)$, i.e. the set of all values of $z$ such that $barg_n^-1(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.






          share|cite|improve this answer










          New contributor




          dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            $fracf'(x)f(x)$ is just another function of x so central limit theorem applies thank you for that
            $endgroup$
            – Colin Hicks
            3 hours ago











          • $begingroup$
            You're welcome :)
            $endgroup$
            – dlnB
            3 hours ago













          3












          3








          3





          $begingroup$

          MLE requires $$fracpartial ln L(theta)partial theta = sum_i=1^n frac f'(x_i;theta)f(x_i;theta),$$
          where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac f'(x;theta)f(x;theta).$ Then $g(x_i;theta)_i=1^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
          $$sqrtn(barg_n(theta)-Eg(x_1;theta))=sqrtnbarg_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $barg_n(theta)=frac1n sum_i=1^n g(x_i;theta).$ The ML estimator solves the equation
          $$barg_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hattheta=barg_n^-1(0).$$
          So long as the set of discontinuity points of $barg_n^-1(z)$, i.e. the set of all values of $z$ such that $barg_n^-1(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.






          share|cite|improve this answer










          New contributor




          dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          MLE requires $$fracpartial ln L(theta)partial theta = sum_i=1^n frac f'(x_i;theta)f(x_i;theta),$$
          where $f'(x_i;theta)$ could denote a gradient (allowing for the multivariate case, but still sticking to your notation). Define a new function $g(x;theta)=frac f'(x;theta)f(x;theta).$ Then $g(x_i;theta)_i=1^n$ is a new iid sequence of random variables, with $Eg(x_1;theta)=0$. If $Eg(x_1;theta)g(x_1;theta)'<infty$, CLT implies,
          $$sqrtn(barg_n(theta)-Eg(x_1;theta))=sqrtnbarg_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $barg_n(theta)=frac1n sum_i=1^n g(x_i;theta).$ The ML estimator solves the equation
          $$barg_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hattheta=barg_n^-1(0).$$
          So long as the set of discontinuity points of $barg_n^-1(z)$, i.e. the set of all values of $z$ such that $barg_n^-1(z)$ is not continuous, occur with probability zero, the continuous mapping theorem gives us asymptotic normality of $theta$.







          share|cite|improve this answer










          New contributor




          dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago





















          New contributor




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          answered 3 hours ago









          dlnBdlnB

          3917




          3917




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          dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          dlnB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.











          • $begingroup$
            $fracf'(x)f(x)$ is just another function of x so central limit theorem applies thank you for that
            $endgroup$
            – Colin Hicks
            3 hours ago











          • $begingroup$
            You're welcome :)
            $endgroup$
            – dlnB
            3 hours ago
















          • $begingroup$
            $fracf'(x)f(x)$ is just another function of x so central limit theorem applies thank you for that
            $endgroup$
            – Colin Hicks
            3 hours ago











          • $begingroup$
            You're welcome :)
            $endgroup$
            – dlnB
            3 hours ago















          $begingroup$
          $fracf'(x)f(x)$ is just another function of x so central limit theorem applies thank you for that
          $endgroup$
          – Colin Hicks
          3 hours ago





          $begingroup$
          $fracf'(x)f(x)$ is just another function of x so central limit theorem applies thank you for that
          $endgroup$
          – Colin Hicks
          3 hours ago













          $begingroup$
          You're welcome :)
          $endgroup$
          – dlnB
          3 hours ago




          $begingroup$
          You're welcome :)
          $endgroup$
          – dlnB
          3 hours ago










          Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.









          draft saved

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          Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.












          Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.











          Colin Hicks is a new contributor. Be nice, and check out our Code of Conduct.














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