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










share|cite|improve this question











$endgroup$

















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










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










      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 12:41







      rs4rs35

















      asked Jan 10 at 11:57









      rs4rs35rs4rs35

      237




      237






















          1 Answer
          1






          active

          oldest

          votes


















          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











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068547%2fhow-to-prove-alternative-form-of-fisher-information%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068547%2fhow-to-prove-alternative-form-of-fisher-information%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                    SQL update select statement

                    'app-layout' is not a known element: how to share Component with different Modules