How to prove alternative form of Fisher information












0












$begingroup$


Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
$$
I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
$$
However, an alternative form is given in my studybook
$$
I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
$$
I am not able to proof the alternative form, since I am not able to find why the following equality holds
$$
-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
$$






statistics fisher-information







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    0












    $begingroup$


    Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
    $$
    I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
    $$
    However, an alternative form is given in my studybook
    $$
    I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
    $$
    I am not able to proof the alternative form, since I am not able to find why the following equality holds
    $$
    -frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
    $$






    statistics fisher-information







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
      $$
      I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
      $$
      However, an alternative form is given in my studybook
      $$
      I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
      $$
      I am not able to proof the alternative form, since I am not able to find why the following equality holds
      $$
      -frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
      $$






      statistics fisher-information







      share|cite|improve this question











      $endgroup$




      Assuming the FI regularity conditions hold. The Fisher information matrix $I(theta;X)$ about $theta$ based on $X$ is defined as the matrix with elements
      $$
      I_{i,j}(theta;X)=Cov_{theta}Big(frac{partial}{partialtheta_{i}}log f_{X}(X|theta),frac{partial}{partialtheta_{j}}log f_{X}(X|theta)Big)
      $$
      However, an alternative form is given in my studybook
      $$
      I_{i,j}=-E_{theta}(frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta))
      $$
      I am not able to proof the alternative form, since I am not able to find why the following equality holds
      $$
      -frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=frac{partial}{partialtheta_{i}}log f_{X}(x|theta)frac{partial}{partialtheta_{j}}log f_{X}(x|theta)
      $$






      statistics fisher-information




      statistics fisher-information






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      edited Jan 10 at 12:41







      rs4rs35

















      asked Jan 10 at 11:57









      rs4rs35rs4rs35

      237




      237






















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          $begingroup$

          We have that:



          $-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$



          Now taking the expectation left and right of this relation, we find that:



          $-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$



          This is because of the fact that:



          $-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$



          So the alternative form follows.






          share|cite|improve this answer











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            1 Answer
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            active

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            0












            $begingroup$

            We have that:



            $-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$



            Now taking the expectation left and right of this relation, we find that:



            $-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$



            This is because of the fact that:



            $-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$



            So the alternative form follows.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              We have that:



              $-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$



              Now taking the expectation left and right of this relation, we find that:



              $-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$



              This is because of the fact that:



              $-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$



              So the alternative form follows.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                We have that:



                $-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$



                Now taking the expectation left and right of this relation, we find that:



                $-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$



                This is because of the fact that:



                $-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$



                So the alternative form follows.






                share|cite|improve this answer











                $endgroup$



                We have that:



                $-frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)=-frac{frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}f_{X}(x|theta)}{f_{X}(x|theta)}+frac{frac{partial}{partialtheta_{i}}f_{X}(x|theta)frac{partial}{partialtheta_{j}} f_{X}(x|theta)}{f^{2}_{X}(x|theta)}$



                Now taking the expectation left and right of this relation, we find that:



                $-mathbb{E}_{theta}[frac{partial^{2}}{partialtheta_{i}partialtheta_{j}}log f_{X}(x|theta)]=mathbb{E}[(frac{partial}{partialtheta_{i}}log f_{X}(x|theta)(frac{partial}{partialtheta_{j}}log f_{X}(x|theta))]$



                This is because of the fact that:



                $-mathbb{E}_{theta}[frac{partial}{partialtheta_{i}} log f_X(x;theta)]=frac{partial}{partialtheta_{i}}int f_X(x|theta)text{dx}$



                So the alternative form follows.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 13:03

























                answered Jan 10 at 12:53









                S. CrimS. Crim

                363112




                363112






























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