Distribution of Maximum Likelihood Estimator












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 $theta$



$$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$



$theta_{max}$ 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







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    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 $theta$



    $$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$



    $theta_{max}$ 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.







    $endgroup$















      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 $theta$



      $$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$



      $theta_{max}$ 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.







      $endgroup$




      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 $theta$



      $$frac{f'(x_1;theta)}{f(x_1;theta)}+frac{f'(x_2;theta)}{f(x_2;theta)}...+frac{f'(x_n;theta)}{f(x_n;theta)}$$



      $theta_{max}$ 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




      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.
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      asked 4 hours ago









      Colin HicksColin Hicks

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          $begingroup$

          MLE requires $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
          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,
          $$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
          $$bar{g}_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hat{theta}=bar{g}_n^{-1}(0).$$
          So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ 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.






          $endgroup$













          • $begingroup$
            $frac{f'(x)}{f(x)}$ 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











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          $begingroup$

          MLE requires $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
          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,
          $$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
          $$bar{g}_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hat{theta}=bar{g}_n^{-1}(0).$$
          So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ 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.






          $endgroup$













          • $begingroup$
            $frac{f'(x)}{f(x)}$ 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 $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
          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,
          $$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
          $$bar{g}_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hat{theta}=bar{g}_n^{-1}(0).$$
          So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ 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.






          $endgroup$













          • $begingroup$
            $frac{f'(x)}{f(x)}$ 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 $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
          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,
          $$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
          $$bar{g}_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hat{theta}=bar{g}_n^{-1}(0).$$
          So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ 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 $$frac{partial ln L(theta)}{partial theta} = sum_{i=1}^n frac{ f'(x_i;theta)}{f(x_i;theta)},$$
          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,
          $$sqrt{n}(bar{g}_n(theta)-Eg(x_1;theta))=sqrt{n}bar{g}_n(theta) rightarrow_D N(0,E(g(x;theta)g(x;theta)'),$$
          where $bar{g}_n(theta)=frac{1}{n} sum_{i=1}^n g(x_i;theta).$ The ML estimator solves the equation
          $$bar{g}_n(theta)=0.$$
          It follows that the ML estimator is given by
          $$hat{theta}=bar{g}_n^{-1}(0).$$
          So long as the set of discontinuity points of $bar{g}_n^{-1}(z)$, i.e. the set of all values of $z$ such that $bar{g}_n^{-1}(z)$ 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





















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          answered 3 hours ago









          dlnBdlnB

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          Check out our Code of Conduct.












          • $begingroup$
            $frac{f'(x)}{f(x)}$ 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


















          • $begingroup$
            $frac{f'(x)}{f(x)}$ 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
















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






          $begingroup$
          $frac{f'(x)}{f(x)}$ 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.










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
















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