Is it possible to prove this if the function is not invertible?











up vote
0
down vote

favorite












Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



I have successfully shown:




  1. $T_n to T_infty$ in probability


  2. $G(T_n) to g(T_infty)$ in probability


  3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


  4. Derived the expression of $sigma (t)$.



However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



Does anyone see a way forward?



Edit:



Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






probability convergence self-learning







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



    Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



    My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



    I have successfully shown:




    1. $T_n to T_infty$ in probability


    2. $G(T_n) to g(T_infty)$ in probability


    3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


    4. Derived the expression of $sigma (t)$.



    However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



    Does anyone see a way forward?



    Edit:



    Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






    probability convergence self-learning







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



      Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



      My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



      I have successfully shown:




      1. $T_n to T_infty$ in probability


      2. $G(T_n) to g(T_infty)$ in probability


      3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


      4. Derived the expression of $sigma (t)$.



      However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



      Does anyone see a way forward?



      Edit:



      Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






      probability convergence self-learning







      share|cite|improve this question















      Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



      Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



      My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



      I have successfully shown:




      1. $T_n to T_infty$ in probability


      2. $G(T_n) to g(T_infty)$ in probability


      3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


      4. Derived the expression of $sigma (t)$.



      However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



      Does anyone see a way forward?



      Edit:



      Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






      probability convergence self-learning




      probability convergence self-learning






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 at 7:44









      daw

      24k1544




      24k1544










      asked Nov 27 at 7:19









      Xiaomi

      972115




      972115






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
          $$
          G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
          $$          ($f_1$ is partial derivative with respect to $t$.)
          What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
          $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
          $$           almost surely. Moreover, we know from CLT the limiting distribution of $S$:
          $$sqrt{n}S_n to N(0,sigma^2),$$
          where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
          $$          From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
          $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
          $$          (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
          $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
          $$






          share|cite|improve this answer





















            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',
            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%2f3015449%2fis-it-possible-to-prove-this-if-the-function-is-not-invertible%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








            up vote
            1
            down vote



            accepted










            I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
            $$
            G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
            $$            ($f_1$ is partial derivative with respect to $t$.)
            What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
            $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
            $$             almost surely. Moreover, we know from CLT the limiting distribution of $S$:
            $$sqrt{n}S_n to N(0,sigma^2),$$
            where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
            $$            From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
            $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
            $$            (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
            $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
            $$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
              $$
              G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
              $$              ($f_1$ is partial derivative with respect to $t$.)
              What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
              $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
              $$               almost surely. Moreover, we know from CLT the limiting distribution of $S$:
              $$sqrt{n}S_n to N(0,sigma^2),$$
              where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
              $$              From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
              $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
              $$              (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
              $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
              $$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
                $$
                G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
                $$                ($f_1$ is partial derivative with respect to $t$.)
                What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
                $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
                $$                 almost surely. Moreover, we know from CLT the limiting distribution of $S$:
                $$sqrt{n}S_n to N(0,sigma^2),$$
                where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
                $$                From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
                $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
                $$                (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
                $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
                $$






                share|cite|improve this answer












                I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
                $$
                G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
                $$                ($f_1$ is partial derivative with respect to $t$.)
                What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
                $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
                $$                 almost surely. Moreover, we know from CLT the limiting distribution of $S$:
                $$sqrt{n}S_n to N(0,sigma^2),$$
                where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
                $$                From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
                $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
                $$                (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
                $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 at 8:02









                Song

                3,435315




                3,435315






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f3015449%2fis-it-possible-to-prove-this-if-the-function-is-not-invertible%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

                    Wiesbaden

                    Image
                    Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                    Read more

                    Marschland

                    Image
                    Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                    Read more

                    Dieringhausen

                    Image
                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                    Read more